category
stringclasses 107
values | title
stringlengths 15
179
| question_link
stringlengths 59
147
| question_body
stringlengths 53
33.8k
| answer_html
stringlengths 0
28.8k
| __index_level_0__
int64 0
1.58k
|
|---|---|---|---|---|---|
wave-particle duality
|
Wave-particle duality
|
https://physics.stackexchange.com/questions/53959/wave-particle-duality
|
<p>I have been trying to understand "wave-particle duality" and other cases related to it. I am currently a college level student. I have few question which I am not getting answers clearly. </p>
<p>In double slit experiment, A particle behave like a wave, then how is "wave-particle duality" explained? I mean, If the particle behave like wave, then is it generating a wave or behaving as a wave? Is that wave going horizontally through slits (Double Slit Experiment) or vertically up and down or in which direction/axis the particle is vibrating to have a specific frequency? How does light behave as waves in it? and how does observer modify the experiment?</p>
<p>I may be thankful to you, If you clear my problems. I have read the theories on Wikipedia and other informational sites and tried to understand it.</p>
|
<p>This <a href="http://en.wikipedia.org/wiki/Two_slit_experiment#With_particle_detectors_at_the_slits" rel="nofollow noreferrer">wikipedia article</a> has a double slit experiment with individual electrons which really shows the particle/wave duality.</p>
<p>What one should keep clear in one's head is that</p>
<p>a) the particle nature is given by the ability to locate a "particle" at a specific (x,y,z,t) maybe with some delta errors.</p>
<p>b) the wave nature is found in the probability distributions of these particles, the probability of finding them at a specific (x,y,z,t)</p>
<p>The following experiment is accumulating electrons one by one .</p>
<p><img src="https://i.sstatic.net/CIeS9.jpg" alt="electrons through slits"></p>
<p>*successive accumulation of electrons *</p>
<p>In the first exposures one does not see much of a pattern. When enough statistics is accumulated the interference pattern is seen in the probability distribution for these electrons. </p>
<p>An experiment <a href="http://adsabs.harvard.edu/abs/1987FoPh...17..891M" rel="nofollow noreferrer">in 1987</a> first determined the "particle" position and the wave nature simultaneously.</p>
<p>In <a href="http://arstechnica.com/science/2012/05/disentangling-the-wave-particle-duality-in-the-double-slit-experiment/" rel="nofollow noreferrer">this recent</a> experiment the slit the photon passed through is ingeniously known, and it shows that the photon passes from a unique slit.</p>
<p>That the interference pattern appears to be destroyed when the slit the "particle" passed through is known, is because the older measurements destroyed the information by disturbing the path when trying to detect it. </p>
<p>"Particle" in quotes because in the microcosm world the only concept of a particle is a disturbance in the experimental apparatus, that can be described as a billiard ball is described macroscopically, by its position . What we have is a quantum mechanical entity that has an (x,y,z) location within the uncertainty principle, which (x,y,z) is given by the quantum mechanical probability function describing the dynamics of its path. It is a probability wave , not a "mass" wave, i.e. it is a bad visualization that the "particle" is distributed in the whole possible space . Just the probability of finding the "particle" depends on a wave function distribution which creates an interference wave pattern at detection.</p>
| 600
|
wave-particle duality
|
Wave particle duality
|
https://physics.stackexchange.com/questions/500353/wave-particle-duality
|
<p>Can someone please explain wave particle duality for large bodies? Why don't large bodies exhibit wave like nature for example if I am walking with some momentum, the wavelength associated with me is <span class="math-container">$h/p$</span> what does this mean?</p>
|
<p>de Broglie wave length is defined as : <span class="math-container">$$\lambda _{B}={\frac {h}{p}}$$</span>
So if for example your mass is 60 kg and you are moving with a sports car with a speed of 200 km/h, then <em>your</em> de Broglie wave-length <a href="https://calistry.org/calculate/deBroglieEquation" rel="nofollow noreferrer">would be</a> on the order of <span class="math-container">$10^{-37}$</span>m. No physical length can be smaller than <a href="https://en.wikipedia.org/wiki/Planck_length" rel="nofollow noreferrer">Plank length</a> which is <span class="math-container">$1.616\;229\times 10^{-35}\ \mathrm {m}$</span>. Thus de Broglie wave lengths for macroscopic big objects means nothing. </p>
| 601
|
wave-particle duality
|
What is wave particle duality?
|
https://physics.stackexchange.com/questions/101049/what-is-wave-particle-duality
|
<p>I am sort of confused about this. Wave particle duality says that sub atomic particles are waves. There is something more though. What is the actual meaning of <a href="http://en.wikipedia.org/wiki/Wave%E2%80%93particle_duality" rel="nofollow">wave particle duality</a>?</p>
|
<p><a href="https://physics.stackexchange.com/a/100458/38111">I answered a question related to this a few days ago</a>, so I suppose I'll try to summarize it here.</p>
<p>Wave particle duality doesn't really say that waves <em>are</em> particles. It says that "particles" aren't really particles, nor are they really waves, they're just little objects that have some properties of waves and some properties of particles, and there are certain situations where one is more visible than the other. I've heard it said (in a <em>very</em> rough sense) that subatomic objects travel like waves, and interact like particles. Again, this is a huge simplification, but there's an important intuition, which is that these objects are always a little like waves and a little like particles. We can describe their position by a function that tells you the <em>probability</em> that the object will be at a particle point in space at a particular time; this function takes the mathematical form of a wave, so we call it a <em>wavefunction</em>, and this is the sense in which particles are like waves. When these objects interact, however, we tend to see them more as particles, like little classical marbles.</p>
<p>The <a href="http://en.wikipedia.org/wiki/Double_slit_experiment" rel="nofollow noreferrer">double-slit experiment</a> is a good example of this. Once more, I emphasize that this is a very big simplification, but just for the purposes of giving you a bit of context, we can imagine that as the electron travels through the slits, its wavelike character is more obvious, and so there are noticeable behaviors we normally attribute to classical waves, like interference. When it collides with the backboard, however, its particle-like character is more obvious, and so we see a single point where the electron collided with the wall. <em>But at all times, the electron had both wave and particle characteristics</em>, and that's the essence of wave-particle duality.</p>
| 602
|
wave-particle duality
|
Matter's wave-particle duality - true?
|
https://physics.stackexchange.com/questions/268360/matters-wave-particle-duality-true
|
<p>I had an interesting conversation with CuriousOne the other day about the question <a href="https://physics.stackexchange.com/questions/268250/experiment-that-demonstrates-the-wave-particle-duality-of-electrons/268252#268252">Experiment that demonstrates the wave-particle duality of electrons</a>. I thought that wave-particle duality existed, CuriousOne thought it didn't (whether CuriousOne thought this wasn't true for both light and matter, or just matter I'm not really sure). Some of the comments were really interesting, and I wanted to know what some of the physicists here thought. </p>
<p>The comments on the question are just below the main question, and are just below my answer to the question (now moved to chat). Please don't close this with "not enough research" as the reason, I did look into it, but some of CuriousOne's comments left me kind of confused. </p>
<p>In terms of people thinking this is a duplicate: I am asking whether there is experimental evidence for electrons having a wave-particle duality (this duality being like the one light has) and whether the double-slit experiment is an example of experimental evidence for this. </p>
<p><strong>Question:</strong></p>
<p>Was de Broglie's hypothesis that electrons (and other matter) have wave-particle duality correct? Does the double-slit experiment prove this, or have anything to do with this? What is some experimental evidence for de Broglie's hypothesis?</p>
<p>Thanks!</p>
|
<p>Yes, there is a duality and in the framework of quantum field theory (QFT) it is not even a contradiction at all. It seems pretty natural.</p>
<p>All fields and particles are treated very similarly in the QFT language. Both are fields in space-time, so “waves”. There is a suble difference in the <em>spin statistics</em>, namely that fields corresponding to ordinary matter (fermions) have canonical anticommutation relations whereas “ordinary fields” like photons (bosons) have commutation relations. One consequence is that no two fermions can be in the same state (momentum, position, …).</p>
<p>So far this only describes waves. The commutation relations then limit the excitation of the fields to integer numbers. Those things are called particles.</p>
<p>Therefore QFT describes particles as the smallest quantized excitation of the field.</p>
| 603
|
wave-particle duality
|
Photons-Wave/particle duality
|
https://physics.stackexchange.com/questions/69954/photons-wave-particle-duality
|
<p>I know that photons and electrons and such are said to have a wave particle duality, but what does that mean for a photon? When light strikes an object, are many photons emitted, enough to draw infinitely many rays, is only one emitted, or something in between?</p>
<p>In particular, I'm having trouble with thin film interference:
<img src="https://i.sstatic.net/E53nS.gif" alt="enter image description here"></p>
<p>The two resulting rays are said to constructively interfere, which is confusing to me. The two rays are clearly parallel, but not coinciding, so how do the two interfere at all? I think my problem is that I imagine light to be a single ray, with a linear oscillating magnetic field- what is the proper way to address these rays?
Are they photons? Or are they small instances of a wave front? I've heard Huygens' Principle, but in this case we present single rays at the end, so I'm led to believe they really ARE rays, in which case they would be photons, and the interference problem would be a result of the wave/particle duality.
The only other thought I've had with regards to the interference is that, as opposed to looking at the rays as one dimensional rays, they could be some kind of representation of a wave 'centered' around that vector, but that doesn't make sense either.
I know it's a heavy question, but it's really confusing me.</p>
|
<p>Anna v's answer is entirely correct, but a simple explanation to merely the thin-film problem is that rays are really waves that have width, so they overlap.
<img src="https://www.ualberta.ca/~pogosyan/teaching/PHYS_130/images/1000px-Refraction_-_Huygens-Fresnel_principle.svg.png" alt="plane wave refracting"></p>
<p>So yes, the ray represents kind of a wave "centered" around that vector. Within a small region of space you can approximate a more realistic wave as propagating only in the forward direction, exactly like a ray.</p>
<p><img src="https://www.uaudio.com/webzine/2005/december/graphics/02_askthedoctors/fig2.gif" alt="plane wave approximation"></p>
| 604
|
wave-particle duality
|
Wave Particle Duality - Weight?
|
https://physics.stackexchange.com/questions/205852/wave-particle-duality-weight
|
<p>Regards the issue of wave particle duality and the double slit experiment. </p>
<p>If the experiment was run with the ’screen’ and detector as a 'box', with electrons being sent into the box from an external source and this box was placed on a set of scales and measurements taken before and after the experiment. </p>
<p>If the electron is a particle, then over time would many of the ‘particles' remain within the box and as they have mass would they add to the weight of the box? </p>
<p>If the electron is a wave then would the wave eventually dissipate and the weight of the box remain unchanged? </p>
|
<p>If you distribute the boxes at various positions and placed them on scales, then over time boxes would one by one get heavier. </p>
<p>And eventually you would notice that some boxes got much more heavy than others. There would be regions where the boxes got lots heaver and region where the boxes only got a little bit heavier and regions in between where the boxes got an in between amount heavier.</p>
<p>The function of how much heavier the box got would look wavey.</p>
| 605
|
wave-particle duality
|
wave-particle duality and entanglement
|
https://physics.stackexchange.com/questions/109853/wave-particle-duality-and-entanglement
|
<p>By fundamental definition of a entangled system we can say that if we know the quantum state of one subsystem then we can describe the state of another subsystem.
A particle possess wave-particle duality. If one experiment verify the wave nature of particle then we can not see its particle behaviour in same exp and vice-versa.
Can we say that wave and particle behaviour are in some sort of entanglement state.</p>
|
<p>No they are different things.</p>
<p><strong>Entanglement:</strong></p>
<p>Imagine that you are in lab, you have two electron spin measuring devices and you have a particle which has spin 0.
after few seconds it decays into two electrons. After you will measure first electron's spin and for example device says that it has spin $\uparrow$ then according to spin conservation: sum of spins should equal $0$ and because of that second electron will have spin $\downarrow$ ($S=\uparrow+\downarrow=0$), no matter how many times you will do experiment you always will get same result: if one electron's spin is down second's will be up and vice versa (unless you are measuring spin of electrons in different axes in other words EPR paradox). Here's Sample image of that experiment about entanglement:
<img src="https://i.sstatic.net/2nuIw.gif" alt="enter image description here"></p>
<p><strong>Wave-particle duality:</strong></p>
<p>Wave-particle duality in simple words means that any matter can behave like a wave and a particle. Wavelength of that wave can be calculated using De-Broglie wavelength:
$$
\lambda=\frac{h}{mv}
$$
Where $\lambda$ is wavelength, $h$ is Plank's Constant, $m$ is mass and $v$ is velocity. So for example imagine youngs double slit experiment, when you aren't observing light it behaves like a wave and you will get interference pattern but when you will observe it (in other words you will know that in which slit each photon passed through) interference pattern will dissapear. Here's sample image of young's double slit experiment.</p>
<p><img src="https://i.sstatic.net/0d0UP.jpg" alt="Young's Double slit experiment"></p>
<p><strong>Conclusion:</strong></p>
<p>So in short, Quantum Entanglement means that when you have two entangled particles, and you will measure one particle's spin you will "collapse" wavefunction and second particle will definitely have opposite spin or in other words in you know state of one of two entangled particles you will definitely know state of the other entangled particle (so it means that state of one of two(or $n$) entangled particles depends on the state of the other and vice versa), and Wave-Particle duality means that matter can sometimes behave like a wave and sometimes behave like a particle, in aforementioned example about wave-particle duality you can see that when you don't observe particle (in example case you don't know in which slit photon passed though) light will behave like a wave but when you will start observing it (Using polarizers (It's called <a href="http://en.wikipedia.org/wiki/Quantum_eraser_experiment" rel="nofollow noreferrer">Quantum Eraser</a> or <a href="http://en.wikipedia.org/wiki/Delayed_choice_quantum_eraser" rel="nofollow noreferrer">Delayed choice Quantum Eraser</a>)) it will behave like a particle. So as you can see entanglement and wave-particle duality are different things. So we can't say that wave and particle behaviour are in some sort of entanglement state.</p>
| 606
|
wave-particle duality
|
Questions on wave-particle duality
|
https://physics.stackexchange.com/questions/22685/questions-on-wave-particle-duality
|
<p>Wave-particle duality states that a particle has both wave properties and particle properties when one is <em>not</em> observing it.</p>
<p>1) What is an observer? Need it be anything living or can other particles also act as observers?</p>
<p>2) When doing the electron double slit experiment--shooting just one electron at a time, the electron goes through both slits at the same time (if one is not observing it). Does that say that the electron is on every single geographical point at the same time?</p>
|
<p>The wave particle duality is a man-made simple solution to understand the properties of electron/photon. The wave-particle duality does NOT say that a particle has both a wave property and a particle property when NOT observing it.. it in fact says that to understand the properties of a particle, one must consider it to be a wave sometimes and a particle at some other times. </p>
<ol>
<li><p>An observer is simply defined as something that increases the entropy of a system.</p></li>
<li><p>Here, you must not consider the "geographical location" of an electron. Treat the double slit experiment as a transfer of wavepacket from point A to point B. </p></li>
</ol>
| 607
|
wave-particle duality
|
Does sound show wave-particle duality?
|
https://physics.stackexchange.com/questions/202058/does-sound-show-wave-particle-duality
|
<p>We know that light and electrons both show wave-particle duality. Or in other words we can say that they can be both seen as a wave and a particle. Can a similar theory be applicable for sound? Can sound also be explained as a particle as well as a wave?</p>
|
<p>The notion you should look up and learn about is the <a href="https://en.wikipedia.org/wiki/Phonon" rel="nofollow">phonon</a>. It is a quasiparticle that arises in the quantum description of acoustics in condensed matter. The description is simplest and clearest in regular lattices of atoms / quantum particles, so it doesn't work so well for sound in a gas. But phonons can be thought of as quantums of sound in solid lattices.</p>
<p>Basically, a lattice is modelled as a system of coupled quantum harmonic oscillators, whose Schrödinger equation is very like a classical model of point masses linked by ideal massless springs. The system has eigenmodes with natural frequencies $\omega_j$, and the energy level of $j^{th}$ eigenmode can change only by integer multiples of $\hbar\,\omega_j$, whilst its ground state has energy $\frac{1}{2}\,\hbar\,\omega_j$. The quantum of this energy change $\hbar\,\omega_j$ corresponds to the phonons of the acoustic eigenmode in question.</p>
| 608
|
wave-particle duality
|
Interpreting wave-particle duality due to wave crests
|
https://physics.stackexchange.com/questions/471025/interpreting-wave-particle-duality-due-to-wave-crests
|
<p>I have been thinking a lot about the double slit experiment and am wondering whether any theorist has ever considered the following interpretation for wave-particle duality:</p>
<p>Could the reason we sometimes detect particles actually be because we are detecting the crest of the wave? This interpretation would suggest there is only waves, no duality, and that certain observations measure crests which then appear to us as particles. Imagine, for example, a sine wave on an oscilloscope. If you cover everything but the very top line it looks like a stream of particles.</p>
<p>In the research I have done, I have not come across this idea, though it seems consistent with various theories and the underlying mathematics. Some ideas, for example that the electron is a standing wave, do imply we interpret the crest as a particle. </p>
<p>Can anyone tell me if any theories consider the observation of wave crests as the reason for the apparent wave-particle duality? </p>
|
<p>No. No interpretations of quantum mechanics consider wave crests to be particles. And for good reason: this interpretation doesn’t work.</p>
<p>Consider the ground state of a particle in a 3D box. The wavefunction is a standing wave with its crest at the center of the box. But if you measure where the particle is in the box, you find it all around, with various probabilities, not just at the center.</p>
<p>The waves in nonrelativistic quantum mechanics are waves of <em>complex probability amplitude</em>. Thinking of them as similar to water waves, sound waves, or light waves is a conceptual error. There are some mathematical similarities, which is why NRQM is sometimes called wave mechanics, but the conceptual differences are more important.</p>
| 609
|
wave-particle duality
|
Does String Theory explain wave-particle duality?
|
https://physics.stackexchange.com/questions/89383/does-string-theory-explain-wave-particle-duality
|
<p>Does string theory explain the weird things that happens at the quantum level, especially wave-particle duality? </p>
|
<p>I do not think so. As a matter of fact strings must be "quantized" to produce "quantum" physics. So the quantum structure (including all apparently weird things) is assumed <em>a priori</em> even dealing with strings. It is deeper than string theory.</p>
| 610
|
wave-particle duality
|
Is the wave-particle duality a real duality?
|
https://physics.stackexchange.com/questions/46237/is-the-wave-particle-duality-a-real-duality
|
<p>I often hear about the wave-particle duality, and how particles exhibit properties of both particles and waves. However, I wonder, is this actually a duality? At the most fundamental level, we 'know' that everything is made up out of particles, whether those are photons, electrons, or maybe even strings. That light for example, also shows wave-like properties, why does that even matter? Don't we know that everything is made up of particles? In other words, wasn't Young wrong and Newton right, instead of them both being right?</p>
|
<p><a href="http://en.wikipedia.org/wiki/Duality">Duality</a> is the relationship between two entities that are claimed to be fundamentally equally important or legitimate as features of the underlying object.</p>
<p>The precise definition of a "duality" depends on the context. For example, in string theory, a duality relates two seemingly inequivalent descriptions of a physical system whose physical consequences, when studied absolutely exactly, are absolutely identical.</p>
<p>The wave-particle duality (or dualism) isn't far from this "extreme" form of duality. It indeed says that the objects such as photons (and electromagnetic waves composed of them) and electrons exhibit both wave and particle properties and they are equally natural, possible, and important.</p>
<p>In fact, we may say that there are two equivalent descriptions of particles – in the position basis and the momentum basis. The former corresponds to the particle paradigm, the latter corresponds to the wave paradigm because waves with well-defined wavelengths are represented by simple objects.</p>
<p>It's certainly not true that Young was wrong and Newton was right. Up to the 20th century, it seemed obvious that Young was more right than Newton because light indisputably exhibits wave properties, as seen in Young's experiments and interference and diffraction phenomena in general. The same wave phenomena apply to electrons that are also behaving as waves in many contexts.</p>
<p>In fact, the state-of-the-art "theory of almost everything" is called quantum field theory and it's based on fields as fundamental objects while particles are just their quantized excitations. A field may have waves on it and quantum mechanics just says that for a fixed frequency $f$, the energy carried in the wave must be a multiple of $E=hf$. The integer counting the multiple is interpreted as the number of particles but the objects are more fundamentally waves.</p>
<p>One may also adopt a perspective or description in which particles look more elementary and the wave phenomena are just a secondary property of them.</p>
<p>None of these two approaches is wrong; none of them is "qualitatively more accurate" than the other. They're really equally valid and equally legitimate – and mathematically equivalent, when described correctly – which is why the word "duality" or "complementarity" is so appropriate.</p>
| 611
|
wave-particle duality
|
Does wave-particle duality exist for gravitational waves?
|
https://physics.stackexchange.com/questions/235587/does-wave-particle-duality-exist-for-gravitational-waves
|
<p>For electromagnetic waves there exists a wave/particle duality: light sometimes behaves as a wave, and other times as a particle (photons).</p>
<p>Does such a duality exist for gravitational waves? In other words would we expect gravitational waves to sometimes behave has particles (gravitons)?</p>
|
<p>Yes. Gravitational waves <a href="https://www.ligo.caltech.edu/news/ligo20160211" rel="nofollow">have been observed</a>, and assuming that quantum mechanics is the right way to think about the universe, then weak gravitational waves of the sort that can be observed at LIGO can be thought of as coherent ensembles of 'graviton' particles. </p>
<p>Now, theoretically this picture is 'OK' because although any quantum field theory that describes gravity is nonrenormalizable, the energy scale at which we expect new physics associated with gravity to be detectable is extremely large ($\sim 10^{19} \text{GeV}$). Hence, we can use an effective theory valid at low energies to describe gravity in terms of particles, even if the 'true' theory of gravity valid at arbitrary energies or length scales is somewhat different conceptually.</p>
<p>Unfortunately, it's impossible with current technology to observe or manipulate single gravitons in a lab, so gravitons are still theoretical.</p>
| 612
|
wave-particle duality
|
Particle associated with material waves according to Wave particle duality
|
https://physics.stackexchange.com/questions/676039/particle-associated-with-material-waves-according-to-wave-particle-duality
|
<p>What would be the particle associated with the material waves (like water waves, sound waves) according to <strong>Wave particle duality</strong> and <strong>de Broglie hypothesis</strong>? Are those the medium particles (or molecules) themselves?
Edit:
for eg, photon is associated with electromagnetic wave. Just like this which particle is associated with material waves?</p>
|
<p>Assume we have a travelling particle, e.g. an electron.</p>
<p>With it is an associated <a href="https://en.wikipedia.org/wiki/Wave_function" rel="nofollow noreferrer">wave function</a> which is the solution to the <a href="https://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation" rel="nofollow noreferrer">Schrödinger equation</a>.</p>
<p>The wave function is a mathematical object which contains all the information there is to know about the particle.</p>
<p>But the wave function is not a material wave as you state. For one, often wave functions are <a href="https://en.wikipedia.org/wiki/Complex_number" rel="nofollow noreferrer">Complex valued</a> (<span class="math-container">$\in \mathbb{C}$</span>), so they don't 'live in the Real world'.</p>
<p>Wave functions (often noted as <span class="math-container">$\psi$</span>) are related to the probability of finding the particle in a part of space.</p>
<blockquote>
<p>Are those the medium particles (or molecules) themselves?</p>
</blockquote>
<p>In the case of the electron, clearly 'yes', but not as a 'material wave'. The material waves idea is really an old and now obsolete interpretation of Quantum Mechanics.</p>
| 613
|
wave-particle duality
|
Gravity: Wave/Particle Duality of the Graviton
|
https://physics.stackexchange.com/questions/599225/gravity-wave-particle-duality-of-the-graviton
|
<p>It just occurred to me that the graviton is still a <em>hypothetical</em> quantum of gravity yet gravitational waves are proven and measurable. Should we not expect that gravity follows the same wave/particle duality of other occurring quanta (radiation) given its ability to behave like a wave?</p>
<p>What limitations exist for establishing a "double-slit" graviton experiment?</p>
|
<p>"Should we not expect that gravity follows the same wave/particle duality of other occurring quanta (radiation) given its ability to behave like a wave?"</p>
<p>We expect it and we do it. we can quantize the gravitational wave using regular QFT and call it graviton. The problem is when we calculate the cross section of graviton, we get infinities. One way to get rid of these infinities is string theory.</p>
<p>"What limitations exist for establishing a "double-slit" graviton experiment?"</p>
<p>I do not think that it is a possibility. In order to do the double split experiment, you put a sheet which BLOCKS the light and put two holes in it. There is nothing to block gravitational wave. The gravitational wave passes through everything.</p>
| 614
|
wave-particle duality
|
How does wave-particle duality describe Photoelectric effect?
|
https://physics.stackexchange.com/questions/246250/how-does-wave-particle-duality-describe-photoelectric-effect
|
<p>I don't know if electrons work as particles or waves or maybe both in photoelectric effect.</p>
<p>How is Photoelectric Effect actually described by Wave-Particle Duality?</p>
|
<p>Depending on the experiment electrons can behave either like a wave or a particle. </p>
<p>In the photoelectric effect is the electron is mainly exhibiting particle behavior. The kinetic energy of the electron is equal to the energy of the photon minus the binding energy of the electron. </p>
| 615
|
wave-particle duality
|
Does anything exist only as ‘‘pure wave’’ without wave-particle duality?
|
https://physics.stackexchange.com/questions/389279/does-anything-exist-only-as-pure-wave-without-wave-particle-duality
|
<p>Are sound waves “pure waves”? Or does sound also have a particle nature according to wave-particle duality?</p>
|
<blockquote>
<p>Are sound waves “pure waves”? Or sound also has a particle nature according to wave-particle duality?</p>
</blockquote>
<p>Waves are described by solutions of wave equations, mainly sinusoidal ones.</p>
<p>There are classical waves where the amplitude of the functions is related to a propagation of energy in a medium, as sound and water waves, and electromagnetic waves which do not need a medium to propagate the energy in space. The mathematics allows to build up <a href="http://hyperphysics.phy-astr.gsu.edu/hbase/Waves/wpack.html" rel="nofollow noreferrer">wavepackets,</a> from waves.</p>
<p><a href="https://i.sstatic.net/iXYCF.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/iXYCF.gif" alt="wavepacket"></a></p>
<p>Have a look at this <a href="https://www.youtube.com/watch?v=MADng1fqECY" rel="nofollow noreferrer">soliton in a tank of water.</a>.Wave packets in classical waves, like this soliton in the water, have some characteristics of particles, i.e. are <em>localized</em> and carry momentum and energy. </p>
<p>Now wave particle duality is a quantum mechanical issue, not classical. The waves in quantum mechanics are not energy amplitude waves. The amplitude of the quantum mechanical wave is related to the <em>probability</em> of detection/scattering/decaying... of "particles" as described in quantum mechanics. It is called a duality because there is a probability distribution which is wave like, shows the sinusoidal interferences, as seen clearly in <a href="https://en.wikipedia.org/wiki/Double-slit_experiment#Interference_of_individual_particles" rel="nofollow noreferrer">the double slit experiment of single electrons at a time,</a> where both aspects of the electron are visible.</p>
<p><a href="https://i.sstatic.net/NEjga.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/NEjga.png" alt="enter image description here"></a></p>
<p>The accumulation of individual electron footprints, a quantum mechanical probability distribution for the experiment "electron scattering on two slits of specific width and distance apart", shows the interference pattern expected from a wave solution. The single points in the detecting screen have a specific (x,y,z) denoting the impact of a particle. This is the wave particle duality.</p>
<p>It is evident that this probability interpretation does not exist in the classical waves, solitons or not.</p>
<p>On the other hand, the wavepacket solutions are the mathematical tool which allows the description of particles in quantum field theory, which uses the probabilistic wave theory, as wavepackets in space and time. In this formulation the <a href="https://en.wikipedia.org/wiki/File:Standard_Model_of_Elementary_Particles.svg" rel="nofollow noreferrer">point particles of the standard model</a>, as the electron etc, are a theoretical base represented by sinusoidal solutions, and the real interacting particles are wavepackets in space time built up by these elementary particle probability waves. </p>
<p>So your question can be reversed, wave-particle duality can be described by the use of soliton type solutions, the mathematics similar to classical wave equations.</p>
<p>One should always have in mind that it is the probability that is waving in quantum mechanics, the amplitude corresponds to a probability and not to the energy/mass of a particle.</p>
| 616
|
wave-particle duality
|
Modern interpretation of wave-particle duality
|
https://physics.stackexchange.com/questions/257060/modern-interpretation-of-wave-particle-duality
|
<p>As far as I understand, in the early days of quantum theory there was quite a lot of debate over how to interpret what it meant for a quantum mechanical object to exhibit both wave-like and particle-like properties. </p>
<p>Is it correct to say that the modern interpretation (as in the one arrived at at the end of the original construction of quantum mechanics) is that there is an intrinsic uncertainty in the measurement of properties of a quantum mechanical object, such as its position, momentum, etc. As such one can at best describe the state that it is in by a wave function that encodes all the statistical information about the possible values of its observables. Hence there is no wave-particle duality - the wave-like properties arise due to the intrinsic uncertainty arising in the measurements of a particles physical observables.</p>
<p>Is something like this a correct understanding at all?</p>
|
<p>It probably depends what interpretation of quantum physics you subscribe to. That sounds approximately right for the Copenhagen interpretation, in which you aren't allowed to analyze where the wave function comes from.</p>
<p>For those who appreciate more what de Broglie, Einstein, Bell and others have put into quantum physics, there's always the interpretation that a real, physical system underlies the probabilistic description, a system with a localized component (particle) interacting with a non-localized one (waves). In this case the wave-like properties really do come from the wave part and the particle-like properties from the particle part. See bouncing droplets, for example: <a href="https://www.youtube.com/watch?v=nmC0ygr08tE" rel="nofollow">https://www.youtube.com/watch?v=nmC0ygr08tE</a>, <a href="http://math.mit.edu/~bush/wordpress/wp-content/uploads/2015/08/Bush-PHYSICS-TODAY2015.pdf" rel="nofollow">http://math.mit.edu/~bush/wordpress/wp-content/uploads/2015/08/Bush-PHYSICS-TODAY2015.pdf</a></p>
| 617
|
wave-particle duality
|
Are there theories that explain wave-particle duality?
|
https://physics.stackexchange.com/questions/59448/are-there-theories-that-explain-wave-particle-duality
|
<p>I'm confused by the famous wave-particle duality mystery:</p>
<p>When a particle is left unobserved, it acts like a wave and can explore all classically available particle trajectories simultaneously. By looking at it, you force it to decide on a single trajectory, like going through the left or right slit, or like Schrödinger's cat that ends up being either dead or alive; the wave-like characteristics are lost.</p>
<p>Are there theories that actually <em>explain</em> this behaviour?</p>
|
<p>The answer by JKL is sufficient but I want to address particularly the <em>Why?</em>.</p>
<blockquote>
<p>I am so confused. Why does it act the way it does?</p>
</blockquote>
<p>If one reads a bit about the history of science and physics in particular, it becomes clear that physics at the ultimate end does not answer the ultimate <em>Why?</em>. Physics posits laws and uses sophisticated mathematical tools to theorize from "axioms", get equations, and check against the data experimentally. It finds <em>How</em>, from "axioms" one ends with predictions for measurements that validate them.</p>
<p>The <em>Why?</em> questions addressed to the "axioms" has the only answer: <em>Because</em> .</p>
<p>When one is validating a theory, as for example Newton's gravitational theory, and one hits a disagreement with the data, new "axioms" and new theoretical tools are developed to explain the <em>Why</em> of the disagreement and the new theory validated for the regime of disagreement. Special and General Relativity are an example. The history of physics has other examples : Thermodynamics, developed theoretically to explain bulk behavior, statistical mechanics, out of classical mechanics emerge out of asking <em>How</em> and assuming "axioms" to contain the <em>Why</em>. Each theory with its own regime of validity.</p>
<p>Within a theory a question with a <em>Why</em> is answered by proofs of <em>How</em> finally hitting on the "axioms". Quantum mechanics is the last in the series of exploring the microcosm. The <em>Why</em> you are asking hits against the <em>Because</em> of its "axioms".</p>
<p>There are people who continue the exploration, trying new axiomatic theories of how a quantum mechanical theory can emerge from a smaller regime where we are back to classical concepts, and contain the <em>Why</em> in "axioms" for their new theory. They are not successful except with small models. The bulk of theoretical physicists either ignores their efforts or proves that their new theories would violate a basic validated law as, for example, Lorentz invariance. And that is the story of <em>Why</em> in Physics. Ultimately the answer is <em>Because</em>.</p>
<p>Edit: "axioms" in quotation marks , to include mathematial axioms and physics postulates. Physical theories start from strict axiomatic mathematical theories where everything is contained and rigorously self consistent and add postulates in order to connect physical observables to the mathematical variables/functions. <a href="http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/qm.html" rel="nofollow">Postulates</a> are a necessary choice to turn a mathematical theory into a physics theory and in classical theories are referred as "laws". I call them "axioms" because they are the foundation stones of the theories, if one goes the whole theory goes.</p>
| 618
|
wave-particle duality
|
Explanation of electric current and wave-particle duality of electron
|
https://physics.stackexchange.com/questions/285388/explanation-of-electric-current-and-wave-particle-duality-of-electron
|
<p>Edit note: As I've got a downvote and some negative comments, I will try to make myself very clear.</p>
<p>Electrons are thought to be particles, classically. Electric current is defined to be the movement of electrons.
But, electrons have wave-particle duality.</p>
<p>Q.No:1: How is electric current(defined as movement of electrons,classically) explained with keeping in mind the wave-particle duality of electrons, as electrons are not really classical particles ?</p>
<p>Q:2: Has QM to do anything with this explanation?</p>
|
<blockquote>
<p>How is electric current(defined as movement of electrons,classically)</p>
</blockquote>
<p>Incorrect.</p>
<p>Electric current is defined as the motion of electric charge. In the natural world, most currents are electrolytic, and involve the motion of fairly massive ions both positive and negative. For example, in acid solutions the electric current is proton-flow. In saline, the current is Na+ ions moving one way and Cl- ions moving opposite. In <em>solid</em> metals the electric current is the <em>difference between</em> the motion of conduction-band electrons versus the solid lattice composed of positive metal ions. (That's why, when we physically move a wire, we're not producing lots of amperes.) Also, in liquid metals we'll have motion of positive metal ions, not only a pure electron current.</p>
<p>IIRC, I saw the answer to your #2 question some years back. For electric current in metal wires, the QM effects of a large group of electrons at low average velocity are insignificant. So, while single/few electrons are quantum objects, an entire "electron sea" or large population of interacting electrons are not.</p>
<p>Similar concept: charge up a metal ball, then move it along, and you've created electric current on a microamperes scale, but without QM phenomena becoming significant. Both the crystal lattice and the electron sea of the entire metal ball have well defined position and velocity, and behave like macroscopic materials. But zoom in on a single metal ion or a single conduction-band electron, and such is not the case.</p>
<p>I recall an interesting/silly experiment from The Physics Teacher magazine. If we perform a Hall-effect experiment on a large current in a metal strip, performed in order to identify the polarity of the moving charges, and if we then slide this strip along at exactly the opposite of electron-drift velocity, then the results (correctly) show that the moving charge-carriers are positive, not negative. After all, by moving the wire backwards, we've made the electrons stop drifting wrt the lab frame! The grid of metal atoms with their positive charges becomes the only "electric current" in the metal.</p>
<p>Knowing the above, it's obvious that a macroscopic copper wire, when moving wrt the <em>stationary</em> electron-sea within the wire, demonstrates an electric current lacking micro-scale QM phenomena. The same is true if we hold the copper wire still, and only move the electron-sea.</p>
| 619
|
wave-particle duality
|
Higgs Mechanism and Wave-Particle Duality
|
https://physics.stackexchange.com/questions/415791/higgs-mechanism-and-wave-particle-duality
|
<p>According to Higgs Mechanism a particle acquires mass when it couples to the higgs field. Now consider Wave-Particle Duality. Suppose I am doing a Young's Double Slit Experiment using electrons. When they pass through the slits they behave like waves and quantum waves have no mass, there is just momentum given by de-Broglie relation. Now when they "impinge" upon a screen their behaviour is like that of a particle. So they must have mass. When did the electrons interact with the Higgs field to acquire mass when they did not have any mass while "passing" through the slits? Does this imply that Higgs field is decoupling and coupling with the electron field? Also, I could make the distance between the slits and the screen arbitrarily small, doesn't that imply that the coupling "speed" (for lack of a better word on my part) of Higgs field is infinite?</p>
|
<blockquote>
<p>quantum waves have no mass, there is just momentum ...</p>
</blockquote>
<p>is a fundamental misconception about QM. The probability amplitude wavepacket desribing the motion of the electron fully accounts for the movement of its mass (and charge, and...) because this wavepacket, moving freely is restricted by the dispersive equation of Schroedinger, here in one dimension w.l.o.g.,
$$
\left (i\hbar\partial_t + \frac{\hbar^2}{2m} \partial_x^2\right ) \psi =0,
$$
which is manifestly "aware" of its mass.</p>
<p>The relevant solution is essentially a spreading wavepacket $\psi (x,t)$, yielding a probability
$$
\Large \rho(x,t)=|\psi|^2=\frac{1}{\sqrt{1+4(t\hbar/m)^2}} e^{\frac{-2(x-vt)^2}{1+4(t\hbar/m)^2}} ,
$$
in natural wavepacket length units, where <em>v</em> is the group velocity of the wavepacket.</p>
<p>The relevant probability flow current to the right can be seen to be ~ $v \rho$, and so the rate of transport of mass to the right is $mv \rho$, as you should confirm. </p>
<p>The takeaway is that the "wave" of your mental picture <em>does</em> very much transport mass, charge, spin, etc... to the screen. Calling it "massless" is meaningless. The Higgs mechanism has done its job of giving the electron a mass and completely decoupled from the problem, conceptually. The wave went through the slits and interfered with no less mass than a corpuscle.</p>
| 620
|
wave-particle duality
|
Reviewing the wave-particle duality
|
https://physics.stackexchange.com/questions/843444/reviewing-the-wave-particle-duality
|
<p>This could sound an elementary question, but the more I think about it, more convinced I am that there could be a different perspective.</p>
<p>I seriously doubt about what is called "wave particle duality" as a "fact". Let me explain with detail what I want to say.</p>
<p>For terms of this question, let us to confine to non-relativistic quantum mechanics of point particles. Let us leave aside quantum field theory at the moment.</p>
<p>For terms of this question, let us start with an important definition:</p>
<p><strong>Let us define an electron as a physical entity with charge <span class="math-container">$-1.602\times 10^{19} C$</span> and mass <span class="math-container">$9.109\times10^{-31} kg$</span> both, up to experimental errors, well defined.</strong></p>
<p>With this definition at hand, let us analyze the double slit experiment:</p>
<p>In this case, if a beam of electrons hits the screen, a wave pattern is produced. However, it is possible to make the same experiment, but letting each electron hit the screen one at a time. The result is given by the next figure (which is taken from Brian Hall's book, "Quantum theory for mathematicians")
<a href="https://i.sstatic.net/IYih9IIW.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/IYih9IIW.png" alt="Four images showing the
impact of individual electrons gradually building up to form an interference pattern." /></a></p>
<p>Quoting this same reference:</p>
<blockquote>
<p>...each electron passes through the slits and
then strikes a screen, causing a small spot of light to appear. The location of
this spot is then recorded for each electron, one at a time. The key point is
that each individual electron strikes the screen at a single point. That is to
say, individual electrons are not smeared out across the screen in a wavelike
pattern, but rather behave like point particles, in that the observed location
of the electron is indeed a point. Each electron, however, strikes the screen
at a different point, and once a large number of the electrons have struck
and their locations have been recorded, an interference pattern emerges.</p>
</blockquote>
<p>This is, essentially, the same explanation given in The Feynmann Lectures on physics</p>
<p>Here, <a href="https://physics.stackexchange.com/questions/15854/why-does-davisson-germer-experiment-confirm-electrons-wave-particle-duality">Why does Davisson-Germer experiment confirm electron's wave-particle duality?</a>, the answer given by Vineet Menon is a very valid perspective: electron behaves as particle as well as wave, under appropriate conditions. However, if we use the definition that I stated above, then an electron is never a wave. I haven't found any experiment where the wave associated to a single electron is detected, or the charge or mass of an electron (a single electron) could be associated to a wave.</p>
<p>I think that, what is really happening is that the electron, as defined above, IS a particle. It's his behavior what is associated to a wave-like pattern. I am not trying to say, for example, that the wave function only describes a bunch or ensemble of similarily prepared systems, as in the statistical interpretation of quantum mechanics (like here <a href="https://journals.aps.org/rmp/abstract/10.1103/RevModPhys.42.358" rel="nofollow noreferrer">https://journals.aps.org/rmp/abstract/10.1103/RevModPhys.42.358</a>). No. What I am trying to say is that the wave associated to an electron is associated, directly, to his behavior, or evolution, according to the conditions over its located (external potentials, barriers, etc), and is not the physical entity called ``electron'' (as defined above) itself a wave.</p>
<p>This interpretation is reforced by the statistical interpretation of the wave function. <span class="math-container">$\psi(x,t)$</span> represents a <strong>probability amplitude</strong>. It does not represent a charge or a mass itself but the behavior of such mass-charge configuration.</p>
<p>In this way, I seriously doubt if a single electron (again as defined above) could be a wave. By "be a wave" I mean that we could be able to detect the wave pattern over a, for example, a screen or detector, (again using a single electron), and we could be able to associate the charge or mass of the electron to such a wave pattern.</p>
<p>An argument against this could be that the wave is not a "classical wave", but a "quantum wave". However, it is important, then, to define unambiguously what you call a "quantum wave" and a "classical wave". Again, remember that, according to all books and articles, <span class="math-container">$\psi$</span> represents a probability amplitude, and that the probability given by such a function is measured only when we make a series of copies of exactly the same experiment, one after another, and we record the results.</p>
<p>Again, we are able to detect the "wave-like" behavior of quantum particles only when we use a bunch of such a particles. Never when we use only one. And again, I am not telling you that <span class="math-container">$\psi$</span> only describes an ensemble of paticles. <span class="math-container">$\psi$</span> could describe the behavior of a single particle. What <span class="math-container">$\psi$</span> does is to tell you the "chance" you have of "seeing" an electron "here or there". It is still a particle (remember my definition of electron given above), but with a probabilistic and randomly behavior.</p>
<p>So, where is the wave associated to the electron? Answer: the wave is in his bevahior (a wave-like behavior) but the physical entity itself(again, as defined above) is still a particle.</p>
<p>Another argument againts this could be: Ok, what if I don't make any measurement over the electron? If anybody measures the electron itself, but we know it exists in some part of the Universe, is a wave or a particle? and, if its is a particle, it should have a definited position even after the measurement, right? My answer would be the same; The electron is a particle. But is a quantum particle. And, as Feynmann says in his Lectures, if you don't make any measurement over the electron, you cannot say where it is. You cannot speak about it's position before a measurement. And you don't need it. Nothing in the Universe depends on the nature or position of the electron before a measurement. And no, the electron does not appear because you made a measurement, and the charge and mass are not "spread over the space" before a measurement. The electron is a particle with a quantum behavior, with a well defined charge and mass. And that's it.</p>
<p>After this long explanation, my question is:</p>
<p><strong>What could be the problems you see in this understanding?</strong></p>
<p><strong>Is there any experiment or physical fact that contradicts all this?</strong></p>
<p>A common claim is that "the electron sometimes behave like a particle, but in other situations behave like a wave"</p>
<p><strong>Is there any experiment showing the wave pattern associated with a single electron (single electron as defined above)?</strong></p>
<p>If the electron were a wave in some situations, then we would be able to detect the wave itself when we make a single measurement over a single electron, I think.</p>
<p>In my understanding, I don't see any problem in the explanation I give you above, just a tremendously counterintuitive new image of the physical world that, in some way, also contradicts the popular claim "the wave-particle duality". In any case, any help or kind and well structured argument will be welcome.</p>
|
<blockquote>
<p>I think that, what is really happening is that the electron, as defined above, IS a particle. It's his behavior what is associated to a wave-like pattern.</p>
</blockquote>
<p>Have you come across the Bohm-de Broglie pilot wave theory? This is an interpretation of quantum mechanics where the electron is always a particle and its behaviour is influenced by an associated pilot wave. The theory is deterministic and nonlocal. Moreover, unlike the usual quantum mechanics each electron has a well defined position and velocity and hence also a well defined trajectory. It does not suffer from a measurement problem.</p>
| 621
|
wave-particle duality
|
Photoelectric effect and wave particle duality
|
https://physics.stackexchange.com/questions/248132/photoelectric-effect-and-wave-particle-duality
|
<p>In a vacuum, if electrons are accelerated by a certain voltage, giving the electrons a specific de Broglie wavelength and were incident on a piece of metal, providing the wavelength was roughly the diameter/distance between two of the atoms in the metal, would the electrons interact with the metal as a wave due to diffraction of the elections taking place?</p>
<p>If so, would this mean that the delocalised electrons further in the metal would receive the energy carried by the accelerated electrons in the form of a wave and as a result electrons in the metal would gain energy over a period of time as appose to instantaneous electron emission due to the transfer of energy from electron-electron collisions, as particles? </p>
<p>This is all theoretical, assuming perfect 'conditions'. </p>
<p>I apologise if this makes no sense what so ever. I was just looking at the photoelectric effect and wave particle duality and this question just arose out of curiosity. </p>
<p>Thanks. </p>
|
<p>The diffraction pattern is due to elastic scattering from the "ion core", which is the stationary net charge of the atomic nucleus and it's bound electrons; these elastically scattered electrons don't lose any energy. The electrons which interact with the free electrons are inelastically scattered, and contribute a foggy background to the diffraction pattern.</p>
<p>An electron which scatters more than once contributes to the background, so very thin crystals are used in transmission. For my research I found 100 nm worked well for gold and platinum, and several hundred nm for graphite.</p>
<p>Because of the energies involved this latter process may eject some of the free electrons, or generate x-rays. For an electron microscope the energies may vary from 20 to 200 keV or more; for low energy electron probes the electrons may have only a few eV.</p>
<p>There are many valuable applications for electron probes, including imaging, diffraction, and analysis. Electrons, x-rays, and neutrons each has it's own unique advantages. Electrons require ultrahigh vacuums, while x-rays can be done in air; electrons have much shorter wavelengths than x-rays of the same energy, and interact more strongly. In addition, it is easy to focus electrons, which allows you to image the surface.</p>
| 622
|
wave-particle duality
|
Wave/particle duality
|
https://physics.stackexchange.com/questions/43592/wave-particle-duality
|
<p>Apologies if this has been asked before (I did check and I believe it wasn't). I have a question about the <a href="http://en.wikipedia.org/wiki/Wave%E2%80%93particle_duality" rel="nofollow">particle/wave duality</a> of photons (or other particles). Depending on what and how we measure the photon turns out to be either a wave or a particle. Recently I saw some web page (and I can't remember where) that maybe neither is true. Both the wave perception and the particle perception are just that: perceptions, strengthened by our perhaps limited ability to observe. What if the reality of the photon is something else, something "on top" of our two perceptions??
Could someone direct me to a site that could tell me more about it (or maybe debunk the whole idea?)?</p>
|
<p>This is the idea: when you see the high school equations that describe a parabolic motion, you consequently visualize a flying stone in your mind. Physics does not care about what you imagine, but rather it deals with the equations and their ability to tell you the distance the stone will reach. Because the equations were invented and tuned for that purpose, the numeric results of that mathematical tricks are equal to what happens in nature.</p>
<p>Around the 1930s, some men eventually managed to build a consistent sets of mathematical tricks that correctly give the same results as nature, when dealing with a wide range of problems for which the known theories had failed. It is known as Quantum Mechanics. But, unlike the high school parabolic shot, the equations of QM are there but nobody can imagine a graphical representation.</p>
<p>Whatever it may be happening down there, there is no possible way of depicting it in your human brain. It is just so. You may spend your life thinking about a wave as a transparent diffuse entity that suddenly feels observed and transforms into a tiny hard ball... But the result will be nothing but a headache. With Quantum Mechanics you see the equations and check the result, but there is never a parallel mental image of a flying thing or whatever.</p>
<p>There is however good news, because the most basic of the theory is not difficult to grasp. But bear in mind that it consists on math (by the way, Physics may be well defined as the science that invents mathematical models that resemble the behavior of nature). Google for a small collections of video lectures called "Quantum Entanglements 1" by Susskind. After that, the third volume of the Feynman Lectures is a good choice (but that requires more work).</p>
<p>Of course, there is room for much more than pure math alone, but that room is only accessible in a mentally healthy way after you know the math. With no knowledge of the math, words about Quantum Mechanics are meaningless, and intuitive ideas are unavoidable wrong. Whatever meaning you give to the words "wave" and "particle", there is no way to combine them in a satisfactory explanation without at least some elementary knowledge of Quantum Mechanics.</p>
| 623
|
wave-particle duality
|
Is a conservation/symmetry law behind wave-particle duality?
|
https://physics.stackexchange.com/questions/673102/is-a-conservation-symmetry-law-behind-wave-particle-duality
|
<p>Please help me in solving a question stuck in my head about symmetry and conservation laws?
Can wave-particle duality be considered an atomic-particle symmetry? And if so, what is the conservation law behind this symmetry?</p>
|
<p>Wave-particle duality is not a symmetry, although based on other places the word "duality" is used in physics, I could see why you might think it is.</p>
<p><em>If</em> wave-particle duality were a symmetry, you would expect something like the following to be true. You can do a calculation with "wave" variables and get an answer. Then, in the original problem, you transform the "wave" variables to "particle" variables and repeat the calculation. At the end, you find the answer is exactly the same, regardless of whether the original problem was phrased in "wave" or "particle" variables. <strong>This is not a good description of wave-particle duality.</strong></p>
<p>Instead, a "quantum thing" is <em>neither</em> a wave <em>nor</em> a particle. It is simply a different, quantum thing, that obeys its own rules. <em>In certain limits</em>, a "quantum thing" behaves like a wave. <em>In other limits</em>, a "quantum thing" behaves like a particle. <em>There is no limit</em> where a "quantum thing" <em>simultaneously</em> behaves like a wave <em>and</em> a particle. The fact that the "quantum thing" can sometimes behave like a wave, and sometimes like a particle, but never like both things, is the origin of the name "wave-particle duality."</p>
<p>A more advanced way of expressing this, is to talk about the operators used to represent a quantum field. You can express quantum fields in terms of the <em>number operator</em>, which counts how many particles (or "quanta") are present in a given state. States which are eigenstates of the number operator, have a clear particle interpretation. You can <em>also</em> express quantum fields in terms of the <em>phase operator</em><span class="math-container">$^\star$</span>, which tells you the phase of the field -- where the peaks and troughs are. States with a definite <em>phase</em> have a clear wave interpretation. However, the number and phase operators <em>do not commute.</em> Therefore, it is typically not possible for a system to be simultaneously in a state that both has a definite particle number, and a definite phase. Here we see explicitly that the quantum field is the "quantum thing", and if we use the quantum field to describe the physics, we always get the right answer (even though the right answer may be very counterintuitive). Meanwhile, we see in some special cases (if the quantum state is as a particle number eigenstate of a phase eigenstate) that we can give a particle <em>or</em> wave interpretation to the state, but there are no states where both interpretations are simultaneously possible. (With the exception of the vacuum state, where there are no particles and also no wave)</p>
<hr />
<p><span class="math-container">$^\star$</span> I'm simplifying a bit to keep this answer short. Really the "phase operator" is not Hermitian, and in a rigorous analysis one should break the field into operators associated sine and cosine components (or <em>quadratures</em>), as described for instance in <a href="https://journals.aps.org/ppf/pdf/10.1103/PhysicsPhysiqueFizika.1.49" rel="noreferrer">https://journals.aps.org/ppf/pdf/10.1103/PhysicsPhysiqueFizika.1.49</a>. However, the net result is that there is an uncertainty relationship between phase and number, which is the main point of this paragraph.</p>
| 624
|
wave-particle duality
|
Wave particle duality or complementarity?
|
https://physics.stackexchange.com/questions/152760/wave-particle-duality-or-complementarity
|
<p>First off I have found several different definitions of <em>duality</em> and <em>complementarity</em>, so if anyone has a clear idea on what it meant with these terms please do share.</p>
<p>Now, what I mean is the following: in the wave-particle picture for light and for massive particles, can <strong>all</strong> phenomena be interpreted in both pictures, or do certian problems rely exclusively on one picture?</p>
|
<p>In physics, complementarity is a fundamental principle of quantum mechanics, closely associated with the Copenhagen interpretation. It holds that objects have complementary properties which cannot be measured accurately at the same time. The more accurately one property is measured, the less accurately the complementary property is measured, according to the Heisenberg uncertainty principle</p>
<p>On the other hand, the wave–particle duality is the concept that every elementary particle or quantic entity exhibits the properties of not only particles, but also waves. It addresses the inability of the classical concepts "particle" or "wave" to fully describe the behavior of quantum-scale objects.</p>
<p>To answer your question more specifically, any quatum system exibits both phenomena simultaneously, so they are not alternative interepretations of the same phenomenon, but rather two different characteristics shared by any quantum system. </p>
| 625
|
wave-particle duality
|
Wavelength of a baseball in the context of wave-particle duality
|
https://physics.stackexchange.com/questions/511710/wavelength-of-a-baseball-in-the-context-of-wave-particle-duality
|
<p>This MIT professor teaches wave-particle duality of matter <a href="https://youtu.be/Qg7pQ_CYaIQ?t=1452" rel="nofollow noreferrer">here on YouTube</a>. The formula is:</p>
<pre><code>wavelength = h/mv
</code></pre>
<p>Her conclusion is that the wavelength is too small to be detected. Well, I can always make the speed <code>v</code> small enough so that the wavelength is 1 meter.</p>
<p>In the context of wave-particle duality, what does 1 meter wavelength of a baseball mean? What can you observe?</p>
|
<p>I have not done the maths, but I suspect that were you to calculate the speed you would find it to be infinitely smaller than the jiggling of the atoms in the ball due to their temperature. In any case, the wave particle characteristics of the ball would in fact be the composite of the characteristics of its constituent particles, and since they would be randomly out of phase with each other you might consider it unlikely that the ball itself has an identifiable wavelike characteristic.</p>
| 626
|
wave-particle duality
|
Does the magnetic field lie in the Wave-Particle duality?
|
https://physics.stackexchange.com/questions/93620/does-the-magnetic-field-lie-in-the-wave-particle-duality
|
<p>There is <a href="https://en.wikipedia.org/wiki/Wave%E2%80%93particle_duality" rel="nofollow noreferrer">Wave–particle duality</a>. According to this theory, light is a wave and a particle at once.</p>
<p>What about magnetic field? Can it be so, that it is also a wave and particle, but this particle has not yet been discovered?</p>
<p>Is magnetic field discrete?</p>
<p><img src="https://i.sstatic.net/SMLlc.png" alt="enter image description here"></p>
|
<p>Current physics formulation has two frameworks. One is the macroscopic one, in dimensions comensurate with our physical faculties of observation, larger than micrometer sizes. This is the classical framework which was studied and formulated mathematically until the beginning of the twentieth century. Since then we have found out that the classical framework emerges from a more fundamental one, <a href="http://en.wikipedia.org/wiki/Quantum_mechanics" rel="nofollow">the quantum mechanical</a> framework that describes the microscopic world, molecules, atoms, nuclei etc.</p>
<blockquote>
<p>There is Wave–particle duality. According to this theory, light is a wave and a particle at once.</p>
</blockquote>
<p>The statement on wave particle duality is a statement on the quantum mechanical framework. In that framework yes light can behave like a billiard ball ( particle) or like a wave ( with a sine/cosine wave like behavior in the possibility of detecting it).</p>
<blockquote>
<p>What about magnetic field? Can it be so,</p>
</blockquote>
<p>The magnetic field is a macroscopic concept described by the <a href="http://en.wikipedia.org/wiki/Classical_electromagnetism" rel="nofollow">classical electromagnetic theory.</a></p>
<blockquote>
<p>that it is also a wave and particle, but this particle has not yet been discovered? </p>
</blockquote>
<p>so no, it is not a wave and a particle in the quantum mechanical sense. The magnetic field emerges from the underlying quantum mechanical framework in a strict mathematical way, that you might learn if you continue your physics studies, but has <strong>no</strong> description as a quantum mechanical/wave-particle. </p>
<p>As a handwaving explanation, the magnetic field emerges from the virtual coherent overlapping of innumerable photons, which photons are the particle/wave form of the electromagnetic force in the quantum mechanical framework. Virtual because of the mathematics involved which do not allow them to light up as a light source :) but remain tied up to the atoms and molecules creating the classical field. Non virtual photons do obey the wave particle duality, depending on the way they are detected.</p>
<blockquote>
<p>Is magnetic field discrete?</p>
</blockquote>
<p>It is discrete only in the sense that it is generated by individual atoms that carry a magnetic field, not in the sense of the particle/wave duality.</p>
| 627
|
wave-particle duality
|
Does wave-particle duality rely on accepting the Copenhagen interpretation?
|
https://physics.stackexchange.com/questions/478985/does-wave-particle-duality-rely-on-accepting-the-copenhagen-interpretation
|
<p>If you're a scientist that subscribes to the many worlds theorem, does that mean you do not accept wave particle duality? Seeing as MW postulates that the wave or particle form has always existed that way in your world (If I understood it correctly)</p>
|
<p>No, there is no logical dependence on the two. Recall that particle wave duality existed long before the Copenhagen interpretation of QM. Relative to that interpretation the wave and the particle acquire specific meaning but they existence of duality is not dependent on, nor does it require, the Copenhagen view of QM. Particles and Waves were two competing paradigms for describing phenomena that go way back. I say competing since we usually say, based on experience, that something is either a particle or a wave. In fact the "things" to which we attribute these behaviors have proven us wrong. Each is an abstraction based on our experiences. Duality means that the "things" we play with can behave in both ways and there was not reason to choose one paradigm over the other. Now the two views are seen as complementary rather than competing. </p>
| 628
|
wave-particle duality
|
Does the wave/particle duality exist across the entire electromagnetic spectrum?
|
https://physics.stackexchange.com/questions/169547/does-the-wave-particle-duality-exist-across-the-entire-electromagnetic-spectrum
|
<p>Does the wave/particle duality exist across the entire electromagnetic spectrum?</p>
<p>If theory says so, then to what extent have physicists confirmed by experimental means?</p>
|
<p>Wave/particle duality is present across all particles, an equation to show this is:
$$
p=h/\lambda
$$
where p is the momentum of the "particle", lambda is the wavelength and h is Planck's constant. From this it can be seen that anything can be considered a wave, but they must have a very small mass to have a wavelength that isn't negligible.</p>
| 629
|
wave-particle duality
|
Which types of particles are affected by the wave-particle duality?
|
https://physics.stackexchange.com/questions/103074/which-types-of-particles-are-affected-by-the-wave-particle-duality
|
<p>If we take the double slit experiment as a way of demonstrating the wave-particle duality, which types of particles would show an interference pattern?</p>
<p>For example, I know that electrons show such a pattern. But do protons, too? What about atoms? Where is the boundary between "wavey particles" and "classical particles"?</p>
|
<p>All of them. Even molecules show their wave-like nature, as does, in principle, every object. Speaking of these topics an interesting read about diffraction of C60 molecules is: <a href="http://www.univie.ac.at/qfp/research/matterwave/c60/">http://www.univie.ac.at/qfp/research/matterwave/c60/</a></p>
<p>The point is that the wave-like nature of objects can only be observed at lengths comparable the object's De Broglie wavelength, defined by:</p>
<p>$$ \lambda = \frac{h}{p} $$</p>
<p>where $h$ is the Planck constant and $p$ is the object momentum. For an object moving at non-relativistic speeds you may remember that the momentum is defined as $\vec{p}=m\vec{v}$; the De Broglie wavelength is then inversely proportional to the object mass. The bigger an object is, the less relevant its wave-like nature is, and that's why in everyday experience we are not used to observing the wave-like nature of massive objects. </p>
| 630
|
wave-particle duality
|
Is wave–particle duality considered a valid interpretation of the behavior of photons?
|
https://physics.stackexchange.com/questions/43992/is-wave-particle-duality-considered-a-valid-interpretation-of-the-behavior-of-ph
|
<p>There are a number of questions on this site that explain the many wave-like behaviors of photons by making reference to <a href="http://en.wikipedia.org/wiki/Wave%E2%80%93particle_duality" rel="noreferrer">wave-particle duality</a>. </p>
<p>However, I have just finished reading Feynman's book <em>QED</em>, where he seems to go to great effort to explain things without resorting to this wave-particle duality. He explains the double-slit interference pattern entirely as a matter of the interaction of probability amplitudes. If anything, he seems to consider wave-particle duality to be an obsolete concept that physicists used to rely on to explain photon's behavior, before we "knew better".</p>
<p>I know that Feynman's book is > 20 years old by now, but even the recent new introduction doesn't try to explain how things have evolved since then. Am I interpreting Feynman's explanation correctly, and if so, is this still considered the "best" way of explaining the wave-like behavior of particles?</p>
|
<p>Physics is all about constructing approximate mathematical models to describe the real world. For example Newton's laws are a mathematical model, and they do a pretty good job of describing the motion of the planets round the Sun. Pretty good, but not perfect, because Newton's laws fail to fully describe the <a href="http://en.wikipedia.org/wiki/Anomalous_perihelion_precession#Perihelion_precession_of_Mercury" rel="noreferrer">motion of Mercury</a>. To fix this we use a more accurate, but also far more complicated, mathematical model called General Relativity. But for most purposes, e.g. sending spaceships to Mars, we don't need the accuracy of General Relativity and Newton's laws will do, which is just as well as GR is extremely hard to do calculations with.</p>
<p>The point of this rambling is that physicists continually face decisions as to what mathematical models to use, and they'll normally choose the simplest one that works. Quantum Electrodynamics is the most accurate theory we have for describing the interactions of light, but it's also very hard and in many cases we don't need to go that far. If you're asked to calculate the diffraction pattern from a Young's slits experiment you can do it to better than experimental accuracy using the wave model: you don't need to use QED.</p>
<p>Feynmann is quite correct that the wave/particle duality is a naive idea, and one now superceded by quantum field theory. But just because an idea is naive doesn't mean it isn't useful. There isn't a <em>best</em> way of calculating the behaviour of physical systems just like there isn't a <em>best</em> car. You could use QED to calculate what happens in the Young's slits experiment, but you wouldn't, any more than you'd use a Ferrari to go to the supermarket.</p>
| 631
|
wave-particle duality
|
Why does Davisson-Germer experiment confirm electron's wave-particle duality?
|
https://physics.stackexchange.com/questions/15854/why-does-davisson-germer-experiment-confirm-electrons-wave-particle-duality
|
<p>First I apologize if my question is trivial and for my poor English.
I was wondering why my teacher states that "electron's wave-particle duality is verified if we observe diffraction of the electron flux when fired at a crystal"
I mean, if the diffraction was considered to be only shown by waves, then why did the Davisson-Germer experiment lead to think that electrons behave like wave at specific scales?</p>
<p>My point is: what is the reason that makes the duality (applied to all particles) the more correct approach over considering that diffraction could be 'performed' by specific particles (in this case the electron) under specific conditions? </p>
|
<p>Divisson Germer experiment proved that electron behaves as wave, since diffraction was caused by waves; its a property associated with wave.</p>
<p>But then, long ago when electron was discovered by J.J. Thompson. He found it as a particle in cathode ray tube.</p>
<p>So, Because of validity of both of these experiments, we led to the conclusion that electron behaves as particle as well as wave, under appropriate conditions.</p>
<p>See the image below, well its for photons but applies to electron as well,
<img src="https://i.sstatic.net/ccTpf.jpg" alt="enter image description here"></p>
| 632
|
wave-particle duality
|
Wave/particle duality of an alpha-particle that is emitted by alpha decay
|
https://physics.stackexchange.com/questions/639762/wave-particle-duality-of-an-alpha-particle-that-is-emitted-by-alpha-decay
|
<p>I am wondering about the wave/particle duality of an alpha particle in vacuum. Suppose a U-238 nucleus emits an alpha-particle in vacuum. Is the alpha particle initially a spherical wave propagating in all directions? Does the wave function of the alpha particle collapse only later, when it interacts with a remote object, like an atom or a detector? Does the recoil of the mother/daughter nucleus occur when the alpha particle is emitted, or later, when the wave function collapses?</p>
|
<p>If you measure the recoil immediately after the emission, then you will know immediatly know which direction the particle is going in. If you wait until the first atom has been ionized then you will know which direction the nucleus is recoiling in without needing to measure it. The recoil and the ionization are <em>correlated events</em>. One implies and is implied by the other.</p>
<p>Actually this situation is one of the clearest example that shows that the notion of wavefunction collapse is <em>unneccesary</em>, but is a useful <em>approximation</em>. The best account I know of this is in Schiff's <em>Quantum Mechanics</em> and is based on earlier work by Nevill Mott. The idea is compute the amplitude for the ionization of <em>two</em> atoms by the emitted s-wave electron. He shows that the amplitude is zero unless the two atoms and the nucleus emitting the s-wave alpha particle all lie on a straight line, and that the amplitude is same for all such linear arrangements. Thus, once one atom is ionized you know that only other atoms on the nucleus-first-atom line will can be ionized i.e the ionization trail is a straight line. Showing that the two atoms and nucleus need to be a straight line requires <em>second order</em> perturbation theory in the Hilbert space of the alpha-particle's motion and internal states of the <em>two</em> atoms.</p>
<p>If you assume the alpha particle s-wave wavefunction "collapses" to a narrow beam at the moment that the first atom is ionized you can get away with <em>first order</em> perturbation in just the alpha particle wavefunction and the internal state of the second atom. The full "non-collapse" calculation shows that the first event is inevitably correlated with with the second and so on.
The bottom line is that "wavefunction collapse" is unnecessary to get the straight line track as long as you use a big enough Hilbert space, but if you are lazy you can use the "collapse" notion to save your labour.</p>
| 633
|
wave-particle duality
|
Why complex functions for explaining wave particle duality?
|
https://physics.stackexchange.com/questions/91147/why-complex-functions-for-explaining-wave-particle-duality
|
<p>I have this very bad habit of going to the scratch, discarding all the developments of a theory and worldly knowledge, and ask some fundamental (mostly stupid and naive, as some may say) questions as to why why we needed so and so assumption, why we had to consider this way, could we assume $X$ instead of $Y$ and get a different theory and so on and so forth. As a part of this, is the following question :</p>
<p>Way back in the early twentieth century, physicists struggled to explain certain phenomenon like photo electric effect which needed light waves to behave a particles (photons), and the interference effect of electrons (diffraction) which needed them to behave as waves.</p>
<p>So they said, hey, consider a wave (don't ask me know what it is, but just consider it)... ok
$$\Psi(x,t) = e^{i(kx-\omega t)},$$ now without asking what is $\Psi$, we can explain the interference of electrons and also the photo electric effect, basically the wave/particle duality, if we make an analogy between wave and particle nature as $p = \hbar k$, $E = \hbar \omega$.</p>
<p>My seemingly blunt question is, if you want to explain wave nature and interference by considering a wave function $\Psi$, why the heck do we need complex numbers, why not just real function? Cant we consider something like $\Psi(x,t) : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ or something like that. Hold on, Water waves are successfully explained by real wave function for example $\Psi(x,t) = \cos(kx-\omega t)$, so why the heck we need complex waves for making an anology for wave-particle duality? What if we just unlearn all the QM and start with a real function for waves, what is going to happen? </p>
<p>Sorry for being a bit presumptuous, but some gentlemen will start talking about probability amplitude, uncertainty principle, and so on and so forth, but Gentlemen, wait, I haven't gone that far! I don't have Born interpretation and I don't have uncertainty principle or for that matter the Schrodinger equation yet, So your logic will lead to circular arguments!</p>
<p>After all our goal is to explain physical phenomenon...what if we venture into this jungle of real functions and come up with a totally different theory which explains physical phenomenon.</p>
<p>((If you ask me to do research in theoretical physics, I'll throw all the QM books in garbage (no disregard though) and start thinking from this point of view...Thats my style of working!)</p>
<p>My expected answer is in this spirit, "Hey if you go in that direction, you are bound to end up in a quick sand, for so and so reason"</p>
|
<blockquote>
<p>After all our goal is to explain physical phenomenon...what if we venture into this jungle of real functions and come up with a totally different theory which explains physical phenomenon.</p>
</blockquote>
<p>Good luck to you.
First of all you are wrong that <em>classical</em> physics did not use imaginary functions. The solutions of Maxwell's equations expressed as imaginary functions are more general and universal than sines and cosines. </p>
<blockquote>
<p><a href="http://en.wikipedia.org/wiki/Electromagnetic_wave_equation" rel="nofollow noreferrer">The simplest set of solutions</a> to the wave equation result from assuming sinusoidal waveforms of a single frequency in separable form</p>
</blockquote>
<p>$$\mathbf{E}(\mathbf{r}, t)=\mathrm{Re}\{\mathbf{E}(\mathbf{r} e^{i\omega t}\}$$</p>
<p>Imaginary functions are a useful tool in integrations and descriptions of real data.</p>
<blockquote>
<p>((If you ask me to do research in theoretical physics, I'll throw all the QM books in garbage (no disregard though) and start thinking from this point of view...Thats my style of working!)</p>
</blockquote>
<p>With such blind spots I am sure nobody will ask you to do research in theoretical physics.</p>
<p><a href="http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/qm.html" rel="nofollow noreferrer">The difference between the classical</a> use of imaginary functions from the solutions of the wave equations and the quantum mechanical one is the postulate the posits that the square of the wavefunction is real and gives the probability of an elementary particle (or nuclear) interaction to be observed. When in the microcosm quantum mechanics reigns. There one cannot take a ruler, mark it and measure, it was found that the theories and data agreed when the probability postulate was imposed. One has to make many measurements and get the probability distribution for a particularly desired value.</p>
<p>The above link discusses the postulates of quantum mechanics which were not imposed out of a freak imagination, but were necessary to be able to calculate and fit known observations, like the hydrogen atom, and predict the outcome of experiments and observations.</p>
<p><strong>EDIT</strong> to address the last part of the question:</p>
<blockquote>
<p>((If you ask me to do research in theoretical physics, I'll throw all the QM books in garbage (no disregard though) and start thinking from this point of view...Thats my style of working!)</p>
</blockquote>
<p>That works for art, art is much less dependent on data bases of observations and the tools that can be used.</p>
<p>The fact that for two thousand years people have been creating models of physical observations, and particularly the last 300 a data base of mathematical tools too, constrains creativity in science. The mathematical tools have been used to model all observations up to now. These models are in a way a shorthand description of nature that could be used in many ways instead of going back to the data itself. There exists a frontier of experimental research where the models have not been validated , and that is where new thinking can come in. </p>
<blockquote>
<p>My expected answer is in this spirit, "Hey if you go in that direction, you are bound to end up in a quick sand, for so and so reason"</p>
</blockquote>
<p>If you go into the direction of throwing everything away you will end up with vague models like the Democritus atomic model, or the phlogiston theory, in your own words. The mathematical models used now are validated, some of them to great accuracy. New mathematical tools to model the already modeled data would only be worth the attention if something new and unexpected is predicted and <em>found</em> in the experiments. </p>
<p>There are people working off the beaten track theories, trying to explain quantum mechanics by underlying deterministic theories. <em>These people have a thorough knowledge of existing mathematical tools and the physics models that have been validated.</em> They just want to work at the frontier by ignoring that mainstream physics considers their effort contradictory or impossible/prohibited by the postulates of quantum mechanics and special relativity. An example is the current research interests of G.'t Hooft who has also <a href="https://physics.stackexchange.com/users/11205/g-t-hooft">participated here</a> a while ago . </p>
<p>So if you go in that direction you will end in quick sand surely if you do not have a thorough knowledge of the data and mathematical tools used by physics up to now. If you make the effort to acquire them, then of course you are free to prove mainstream physics "wrong" , as long as your new theory can accommodate the data shorthand of the models up to now . All new theories as they appeared in physics joined smoothly with the old ones, as limiting cases.</p>
| 634
|
wave-particle duality
|
Wave-particle duality or particle-wave duality
|
https://physics.stackexchange.com/questions/751663/wave-particle-duality-or-particle-wave-duality
|
<p>Is there a difference between { a particle that acts as a wave} & { a wave that acts as a particle} ??
Ex: when u consider electrons, they have a specific mass and an inconstant velocity, but when we consider photons, they neither have a specific mass nor a changing velocity.</p>
|
<p>Wave-particle duality is best seen with the help of the de Broglie relation between matter and wave:
<span class="math-container">$$ h=p \lambda \tag 1 \label{eq:1}$$</span></p>
<p>(or between wave and matter, if you switch terms <span class="math-container">$\lambda p$</span>, it doesn't matter how you say it). What is a most profound statement here is that if something is material,- it has to have characteristic wave-length; otherwise, <span class="math-container">$\eqref{eq:1}$</span> would not hold. And in reverse, if there are some oscillations (not dependent on roots,- mechanical, electrical, gravitational, etc.),- they must impart some momentum <span class="math-container">$p$</span>, which basically is attributed to particles; likewise, <span class="math-container">$\eqref{eq:1}$</span> would not hold either.</p>
<p>Unless something doesn't have momentum at all, then as per <span class="math-container">$\eqref{eq:1}$</span> definition <span class="math-container">$\lambda \to \infty$</span>, which is indeterminate form <span class="math-container">$0 \cdot \infty$</span>, and then we can't be sure that happens there,- wave-particle principle is then destroyed.
Keep in mind that this situation is pretty much unphysical because neither there exist waves with infinite wavelengths, nor can you have zero momentum because, at the microscopic level, everything moves more or less because it's bounded by Heisenberg uncertainty principle.</p>
<p>As this question is with a bit of philosophical taste, I'll finish my answer with the Latin roots of the word "<a href="https://www.etymonline.com/search?q=duality" rel="nofollow noreferrer">duality</a>",</p>
<blockquote>
<p>"twofold nature, state of being two or divided in two," late 14c., from Late Latin dualitas, from Latin dualis "that contains two; the dual number, duality," from duo "two" (from PIE root *dwo- "two").</p>
</blockquote>
<p>So matter and wave properties are different sides of the same "thing", like two sides of the same cent.</p>
| 635
|
wave-particle duality
|
Is wavelenth of a particle relative according to Wave Particle Duality?
|
https://physics.stackexchange.com/questions/716128/is-wavelenth-of-a-particle-relative-according-to-wave-particle-duality
|
<p>De Broglie's equation regarding the wave particle duality of matter <span class="math-container">$\lambda =\frac{h}{mv}$</span> depends on velocity. <strong>Now, velocity is relative and depends on the frame of reference of the observer. But does this mean that even the wavelength is relative and will be different for different observers at different frames of reference?</strong> At first I almost readily agreed to this conclusion but then I realized that this means that two people at different speeds will perceive wave"length" at different amounts. This is what struck me as odd, as the only place where I've heard of length disparities among different observers is in special relativity</p>
<p>Can someone explain what does it really mean to have <span class="math-container">$\lambda = \frac{h}{mv}$</span> when velocity is relative?</p>
|
<p>The de <a href="https://en.wikipedia.org/wiki/Wave%E2%80%93particle_duality" rel="nofollow noreferrer">Broglie</a> relation is a forerunner of the quantum mechanical wave equations. We now know that the wavelength in the wave particle duality is not a wavelength in space. It is the wavelength of the <em>wavefunction</em> describing the quantum mechanical probability distribution of finding the particle at (x,y,z) at time t. This means that to measure the wavelength you need an accumulation of measurements with the same boundary conditions.</p>
<blockquote>
<p>wavelength is relative and will be different for different observers at different frames of reference</p>
</blockquote>
<p>Yes, <em>distributions</em> can differ if the measurement happens in different inertial frames. But an accumulation of measurements is always necessary to see the wave-particle duality.</p>
<p>To get an intuition of wave particle duality , see how the distribution of electrons from a <a href="https://en.wikipedia.org/wiki/Double-slit_experiment#Interference_of_individual_particles" rel="nofollow noreferrer">double slit experiment</a> show the wavelength for those particular boundary condition</p>
<p><a href="https://i.sstatic.net/poLvs.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/poLvs.gif" alt="dblel" /></a></p>
<p>The wavelength referred for the probability distribution should be referred to the rest frame of the experiment.</p>
<p><a href="https://iopscience.iop.org/article/10.1088/1367-2630/15/3/033018" rel="nofollow noreferrer">New link</a> might interest people:</p>
<blockquote>
<p>In 1965, Richard Feynman presented a thought experiment to show these features. Here we demonstrate the full realization of his famous thought experiment</p>
</blockquote>
| 636
|
wave-particle duality
|
Wave-particle duality?
|
https://physics.stackexchange.com/questions/830257/wave-particle-duality
|
<p>What does it mean for an electron to behave as a wave? I can visualize electrons or other subatomic things as particles. But what do we mean when say it's all a wave. What is waving actually? Waves are just a way by which energy can get transferred right with the oscillation of particles. But in this case the particle is an electron and is it oscillating to produce waves? And then we say that normal day to day objects also have wave properties but not really significant. How can I visualize it?</p>
|
<p>What behaves like a wave is the quantum wave function
<span class="math-container">$$\Psi(\vec x_1, \dots, \vec x_n)$$</span>
defined on configuration space. It evolves according to the Schrödinger equation
<span class="math-container">\begin{align}
\label{eq:S_eq}
i \hbar \frac{\partial}{\partial t} \Psi(t, \vec x_1, \dots, \vec x_n) = - \sum_{i= 1}^n \frac{\hbar ^2}{2 m_i} \Delta_i \Psi(t,\vec x_1, \dots, \vec x_n) + V(\vec x_1, \dots, \vec x_n) \Psi(t, \vec x_1, \dots, \vec x_n)\,.
\end{align}</span></p>
<p>It must be stressed that this is not a classical wave like and EM-waves, as it is not defined on physical space. This creates a lot of confusion.</p>
<p>According to quantum mechanics, the electrons' positions are random with probability density</p>
<p><span class="math-container">$$
\rho(\vec x_1, \dots, \vec x_1) = |\Psi(\vec x_1, \dots, \vec x_n)|^2.
$$</span>
The waves are not produced by the particle.</p>
<p>As an example, consider the double-slit experiment with a single particle. In this case, it follows from the Schrödinger equation that <span class="math-container">$\rho(x)$</span> has the same interference pattern as a classical wave passing through the slit.
By doing many double-slit experiments in a row, we can sample the probability density <span class="math-container">$\rho$</span>, i.e., the interference pattern builds up on the screen.</p>
<p>The similarity between this statistical pattern and the classical wave interference can be regarded as the wavelike behaviour of the particle.
On the other hand, the flashes on the screen in each individual run are sharply localized. This part of the phenomenology is rather similar to the behaviour of classical particles. Thus, we have also particle-like behaviour.</p>
<p>However, both the classical wave picture and the classical particle picture are wrong. Instead, we have a new theory called quantum mechanics that describes what is going on.</p>
<p>While the classical wave picture and the classical particle picture are incompatible, there is no such logical inconsistency with the quantum mechanics.</p>
| 637
|
wave-particle duality
|
Questions about the Double Slit Experiment and wave-particle duality
|
https://physics.stackexchange.com/questions/396787/questions-about-the-double-slit-experiment-and-wave-particle-duality
|
<p><strong>Re: The Double Slit Experiment and wave-particle duality.</strong></p>
<p><strong>Q1.1: Is this interpretation correct:</strong> When just one single particle (and no more) is sent through the two slits but is not measured, it shows interference as if a wave of particles, equal to every possible path that could be followed by that single particle, has also been sent through the slits.
Once it is measured and we discover the actual path that the single particle took, then we can see that the implied wave amplitudes <em>did</em> somehow interfere with the single particle in its path of travel, and it changed direction to account for those supposed impacts.</p>
<p><strong>Q1.2: How do we know the particle behaves as a wave if we cannot measure it without that measurement collapsing the wave information?</strong> </p>
<p><em>(original post included additional questions regards many world interpretation, and virtual reality theory but have been removed to adher to posting guidelines)</em></p>
|
<blockquote>
<p>When just one single particle (and no more) is sent through the two slits but is not measured, it shows interference...</p>
</blockquote>
<p>Quote of the comment from Bill Alsept:</p>
<blockquote>
<blockquote>
<p>One particle will not make an interference pattern. It can make one mark on the detector and that’s it. It takes many individual impacts to form a pattern.</p>
</blockquote>
</blockquote>
<p>Interestingly, the detector can be placed as closed to the slit as you want and the impact always will be a dot from the single photon. Being close enough <strong>you are able to observe through which slit the photon was slipping through</strong>.</p>
<p>Furthermore behind <strong>single sharp edges</strong> and even with single photons after a while one will observe an intensity distribution on a detector screen. So slits with its difference in the path length from the right edge and from the left edge are not necessary to form an intensity pattern.</p>
<p>There is another point one can state. Th intensity distribution is a stationary pattern on the screen. That is an astonishing fact because <a href="https://en.wikipedia.org/wiki/Young%27s_interference_experiment" rel="nofollow noreferrer">Young</a> derived the wave characteristics of light from the interference pattern of two water waves:</p>
<p><a href="https://i.sstatic.net/UJScdm.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/UJScdm.jpg" alt="enter image description here"></a></p>
<p>This sketch shows one moment of the “living” pattern. In reality the points C and D are moving to the top and the points E and F are moving downwards. This behavior is not observed for the intensity pattern from light.</p>
<p>What you can conclude from these observations?</p>
<ol>
<li>As you state right ...</li>
</ol>
<blockquote>
<p>How do we know the particle behaves as a wave if we cannot measure it without that measurement collapsing the wave information?</p>
</blockquote>
<p>... the wave character is an interpretation and could not be observed directly.</p>
<ol start="2">
<li><p>The interaction between sharp edges and photons is not an object of consideration. All the more a quantized force field between the surface electrons from the sharp edge and the photons are not discussed.</p></li>
<li><p>The above phenomena are proved. The conclusion in point 2 is not recognized.</p></li>
</ol>
| 638
|
wave-particle duality
|
How does the Wave Particle Duality fit with Quantum Field Theory?
|
https://physics.stackexchange.com/questions/343237/how-does-the-wave-particle-duality-fit-with-quantum-field-theory
|
<p>It's heard quite often that fundamental particles (photons, quarks, etc) act as both particles and waves. </p>
<p>Now, I'm looking at it from a Quantum Field perspective. Is this localized energy <em>ripple</em> what the wave is? And is the fact that it is <em>localized</em> make it a particle?</p>
<p><a href="https://i.sstatic.net/ldqch.gif" rel="noreferrer"><img src="https://i.sstatic.net/ldqch.gif" alt="Quantum Field"></a></p>
<p>As shown in the GIF above, there is a red lattice, which is a quantum field. This is an oscillating ripple. Is this the wave nature of the particle? Is that what causes the result of Thomas Young's double slit experiment? Also, there is a green layer representing the particle's position, representing the disturbance in the quantum field as a particle. Is this the explanation for the photoelectric effect?</p>
<p><strong>Ultimately, is this the explanation for the wave-particle duality?</strong></p>
|
<p>No, this is not at all how quantum field theory works.</p>
<ol>
<li><p>A "quantum field" does not have a definite value at any time, it is an <em>operator</em> in the quantum theory, not something that has a fixed numerical value, therefore representing it as a lattice as you have done does not reflect the quantum nature of the field. This is the <em>classical</em> picture of the field, just like a point particle is the classical picture of the electron, not its quantum picture.</p></li>
<li><p>The quantum field and the particle states are <em>different things</em> - the field is an operator and the particle is a <em>state</em> in the quantum theory. You can use (parts of) the quantum field operator to create particles, but the notion of particle is much more elusive than it being a simple ripple in a classical field. For more on this see <a href="https://physics.stackexchange.com/a/163705/50583">this answer of mine</a> on real particles and <a href="https://physics.stackexchange.com/q/230113/50583">this question and its answers</a> on virtual particles.</p></li>
<li><p>The "wave-particle duality" is, in any case, a somewhat vague notion that has no real formal counterpart in modern quantum mechanics. Quantum objects are just that, quantum objects. They have aspects of waves (e.g. they can "interfere", they can obey wave-like equations, they "spread") and they have aspects of particles (e.g. they can (but not must be) localized at "points", they have mass) but <strong>they are neither</strong>. And I'm sure you can find quantum behaviour that you'll not be able to attribute to either a wavy or a particle nature, such as Bell experiments about entanglement (which cannot be explained classically, and hence any attempt to explain them with a particle or wave picture must necessarily fail).</p></li>
</ol>
| 639
|
wave-particle duality
|
What is the mechanism behind the wave - particle duality
|
https://physics.stackexchange.com/questions/172510/what-is-the-mechanism-behind-the-wave-particle-duality
|
<p>I'm not exactly sure how to phrase this question; I've been reading about wave -particle duality, its history and how it works. But it's really bothering me, whenever I watch YouTube videos about it or read about it, physicists seems to be careful to say that a wave such as light can <strong>BEHAVE</strong> like a particle but never at the same time.</p>
<p>What I am wondering is, whether light is a wave that only behaves like a particle at times or an actual particle. Or is an electron a particle that behaves like a wave at times or can it be an actual wave?</p>
<p>And also what is the actual mechanism behind it because I can't seem to find the answer or is it simply that no one knows?</p>
<p>There does seem to be one explanation, but I'm not sure if it's correct or that I understood it right. It states that a wave is continuous and infinite but when several waves are joined, they can form a pulse and that is a particle.</p>
<p>I studied maths at college level, computer science at university level and physics only in high school (UK education, for US college=high school and high school=secondary education). But I keep up with physics as a hobby, I'm quite familiar with classical Newtonian physics and know some amount of general and special relativity. I'm really only just getting into the quantum world but this is one of the first things that struck me.</p>
|
<h1>The Trouble with Models</h1>
<p>An honest answer is that we use models to simulate how the universe behaves, and sometimes our models just do not accurately display what something is. This is why there have been, are, and will be so many models in physics. Our models fail every so often. We try to keep the best models by updating and replacing as needed.</p>
<p>Light can behave as if it were a wave or as a particle, but it does not display behavior that only one model can totally explain. This is why the idea of the duality has been taken up. Both models are viewed as "equally correct," so both are applied. The most accurate way to describe light is that it acts like light, but that is not helpful.</p>
<p>In quantum mechanics, particles' behavior can be described by their wave functions. As the name implies, these functions often look more like waves rather than anything else. It turns out that these wave functions allow for particles to act as if they were a wave. On top of that, the situations where it acts like a wave is inexplicable if we think of electrons as hard particles. You can see this with phenomena like <a href="http://en.wikipedia.org/wiki/Electron_diffraction" rel="noreferrer">electron diffraction</a>. In such cases, it is better to think of these electrons as waves.</p>
<p>Due to the fact that they exhibit so many wave-like and particle-like properties, we do not call them one or the other because the individual models do not account for the behavior we see well enough. Once again, our models are the sources of the confusion, and we are stuck with something that we apply two models to. We compromise and call it a duality; the electron is an electron, and we model it like a wave or a particle.</p>
<h1>The Simple Answer</h1>
<p>The simplest answer is that we do not really know what these things are.<span class="math-container">$^1$</span> We have nothing on our scale or within our experience that we can compare them to. Therefore, we use models of things we do know and <em>carefully</em> apply them. The particle-wave duality of these things is a by-product of the models we use to understand them.</p>
<hr />
<p>Footnote</p>
<ol>
<li>When I say "we don't know what they are" I really mean "we don't know what one thing to compare them to as to understand everything about that." We <em>do</em> know many of the properties of light/electrons/etc, and we <em>can</em> predict their behavior with amazing accuracy.</li>
</ol>
| 640
|
wave-particle duality
|
How does the Pauli exclusion principle apply to wave-particle duality?
|
https://physics.stackexchange.com/questions/273646/how-does-the-pauli-exclusion-principle-apply-to-wave-particle-duality
|
<p>Based on the Pauli exclusion principle , no two particles can have the same quantum state. However, in the double slit experiment with electrons (in which we observe wave-particle duality), at some points the wave functions add up to each other. In those specific spatial spots, two electrons have exactly the same quantum states, but not only do they not exclude each other, they are adding to the probability of each other's presence. How that is explained? Is this because only the two electrons with different spins are adding up to each other in this experiment? If so, I think the number of electrons in those specific points should be statistically half of the expected. Is that true?</p>
|
<p>Electrons only interfere with themselves. The dual slit experiment is basically a single electron case. Therefore Pauli exclusion dies not come into play. As to the question in the title, also particle wave duality is associated with single particles. </p>
| 641
|
wave-particle duality
|
Does Wave-Particle Duality Mean "Particles" are Just Waves With Short Wavelengths?
|
https://physics.stackexchange.com/questions/666557/does-wave-particle-duality-mean-particles-are-just-waves-with-short-wavelength
|
<p>I have the following question about wave-particle duality:</p>
<blockquote>
<p>Are particles really just waves with short wavelengths?</p>
</blockquote>
<p>If this is correct, would it then be accurate to say:</p>
<blockquote>
<p>"<em>everything in the universe is a wave, but when a wavelength is short, it acts like our macroscopic conception of a particle. However, on a quantum level, everything is really just a wave</em>"</p>
</blockquote>
<p>For years, I have thought about it like I stated above and it makes perfect sense to me. Indeed, the de-Broglie relation
<span class="math-container">$$\text{wavelength} = \frac{h}{mv}$$</span>
shows that all matter exhibits wave like properties seems to confirm my understanding that they are "really" just waves with short wavelengths.</p>
<p>But I ask the question because I hear quotes like "we don't know if things are particles or waves" and "our brains can't comprehend it", etc. I want to make sure I am not missing something.</p>
<p>The following quote also seems to justify the interpretation I have given above:</p>
<blockquote>
<p>"If the distance between wave peaks is much smaller than the size of an object, the object will block the waves. But if the distance between wave peaks is much larger than the size an object, the waves will go around the object."</p>
</blockquote>
<p>Thus anytime we use the word "particle" really it would be a wave with a very short wavelength given by the DeBroglie formula.</p>
<p>Any input would be appreciated.</p>
|
<p>Well, one can argue that there is no duality what so ever but that all particles are simply excitations of some quantum fields. For example, I do not think that it makes sense to say things like "an electron is both a wave and a particle at once" since after all it is neither classical wave nor classical particle but an excitation of the electron field and that's it.</p>
<p>The idea of wave particle duality is still useful though. It simply has to be used with more care. For example it makes sense to say:</p>
<p>"Under certain conditions, electrons may create interference patterns that are similar to the patterns observable for classical waves."</p>
<p>or</p>
<p>"Under other conditions, electrons can be scattered by a target and behave just like classical point particles."</p>
<p>The argument about the wave length etc. that you are making is a statement about the energy scales of the specific situation at hand. This energy scale determines which - if any - classical analogon (i.e. wave or particle) comes closest to the behaviour of your quantum system.</p>
| 642
|
wave-particle duality
|
What is the role of the magnetic moment on wave particle duality
|
https://physics.stackexchange.com/questions/645939/what-is-the-role-of-the-magnetic-moment-on-wave-particle-duality
|
<p>An electron, travelling at high speed relative to an observer, does not radiate unless it undergoes some form of acceleration. Yet we can observe wave like properties under certain measurement conditions. If we accept that the electron has a magnetic moment, then there is a high probability that the electron gained some angular momentum on leaving the source. This angular momentum of the magnetic moment must exponentially dissipate its energy via radiation. <em><strong>Is this wave particle duality?</strong></em> Could this explain interference in the double slit experiment with very well spaced electrons?</p>
| 643
|
|
wave-particle duality
|
Light wave particle duality
|
https://physics.stackexchange.com/questions/217646/light-wave-particle-duality
|
<p>I have studied about the dual nature of light and all the experiments that proved light was a wave and sometimes a particle, and I am comfortable with the concept that it can be both. However, I have a few questions I am confused about.</p>
<p>1) If I have two different colours of light, I know that they are waves with different frequencies if I look at their wave nature, but what causes colour difference according to particle nature? Is it number of photons or energy that the photon contains?</p>
<p>2) according to wave nature, energy is proportional to amplitude of a wave. Does it mean a red beam of light can have more energy than another one if I increase its amplitude? And how do I increase its amplitude? (By shining more light?)</p>
<p>3) what exactly does amplitude and frequency correspond to in particle nature and wave nature and how do I physically see it (color, brightness, etc)
If some one could explain using analogies, that would be great.</p>
|
<p>In optics, you rather speak of intensity of light which is the energy per time and per surface. The brighter the light the more intensity it has. The energy transported by a light beam per minute is proportional to the squared amplitude of a wave or the number of photons times their energy.</p>
<p>$E\propto |E_0|^2$ for waves and</p>
<p>$E=n \cdot E_{photon}=n \cdot h \cdot f$ (h is the Planck constant)</p>
<p>To increase the intensity of a light beam, you can either increase the number of photons emitted per second or you can increase the frequency of light. In both cases it will increase the amplitude of the electromagnetic wave.</p>
<p>To make confusion perfect, sometimes the energy of single photon does matter and only increasing their number wouldn't help (photons don't "arrive" at exactly the same time so if a certain threshold energy is needed for a specific process, you can't just add up their energies). A classic example is the <a href="https://en.wikipedia.org/wiki/Photoelectric_effect" rel="nofollow">photoelectric effect</a> which got Albert Einstein his Nobel Prize</p>
| 644
|
wave-particle duality
|
Trust in wave-particle duality
|
https://physics.stackexchange.com/questions/609407/trust-in-wave-particle-duality
|
<p>I am doing an essay in school to consider how much trust is needed to accept knowledge within the natural sciences. For one of my points I decided to look into W-P duality. I know the photoelectric effect shows that light is made up of photons, diffraction patterns suggest EM wave, and the Davisson-Germer exp suggest both (with debroglie wavelength). However, my teacher says that in classical mechanics light cannot be both and for quantum mechanics we can. Does this mean we need more trust in the validity of quantum mechanics because there is a sense of ambiguity?</p>
|
<blockquote>
<p>However, my teacher says that in classical mechanics light cannot be both and for quantum mechanics we can. Does this mean we need more trust in the validity of quantum mechanics because there is a sense of ambiguity?</p>
</blockquote>
<p>We only trust that we know more about nature than a hundred and fifty years ago, not because of ambiguity. We have gone to much smaller dimensions and much larger energies, and have discovered quantum mechanics which predicts probabilities of observing states, whereas classical physics is deterministic,</p>
<p>The wave particle duality of light in particular is clear in <a href="https://www.sps.ch/artikel/progresses/wave-particle-duality-of-light-for-the-classroom-13/" rel="nofollow noreferrer">this one photon</a> at a time experiment, where single photons compose the light , as they accumulate from left to right the classical interference patterns is seen. That is the duality, the wave is a probability of finding the photon
wave.</p>
<p><a href="https://i.sstatic.net/8obhg.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/8obhg.jpg" alt="enter image description here" /></a></p>
<blockquote>
<p>Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.</p>
</blockquote>
<p>The same is seen with the electrons, <a href="https://en.wikipedia.org/wiki/Double-slit_experiment#Interference_of_individual_particles" rel="nofollow noreferrer">here</a></p>
<p><a href="https://i.sstatic.net/M2F4U.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/M2F4U.gif" alt="enter image description here" /></a></p>
<p>So we trust in the data measured the last hundred years, and use/trust the theory that describes them well, knowing that any new theory in the future will have to include these observations, because they have small measurement errors.</p>
| 645
|
wave-particle duality
|
Wave-Particle Duality in the Confinement of an Electron in a Box
|
https://physics.stackexchange.com/questions/151592/wave-particle-duality-in-the-confinement-of-an-electron-in-a-box
|
<p><a href="https://i.sstatic.net/wycgI.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/HBOeX.png" title="source: imgur.com" /></a>
According to the wave particle duality, one can say that an electron is both a wave and a particle. If we confine it in a box, it can only form standing waves at particular wavelengths, which leads to the quantization of energy. <strong>When one draws the second excited state, do we mean to say that we want three antinodes and four nodes in the box?</strong></p>
<p>Are we talking about the same type of standing waves? Are the following formulas applicable:</p>
<p>$$\lambda=\frac{2L}{m}~?$$</p>
<p>Using it, I found for m=3, the second excited state, a wavelenth of $6*10^{-11}$ m. However, this does not work with the answer in my book. Apparently, the answer is 0.515 nm. <strong>Are standing waves in this confinement the same type as the standing waves of strings for example?</strong></p>
| 646
|
|
wave-particle duality
|
What exactly is meant by "observed" when talking about the wave-particle duality?
|
https://physics.stackexchange.com/questions/129522/what-exactly-is-meant-by-observed-when-talking-about-the-wave-particle-duality
|
<p>When talking about the wave-particle duality, teachers and books say that when you send a single photon through a slit, it makes a wave pattern. But if you send that particle through the slit and "you observe it directly", then it appears as a single point (a particle).</p>
<p>What is meant by "observe"? Is that like with your eyeballs? Or is that with some measuring device? It's unclear what that word means, it could mean anything haha.</p>
<p>One reason I am wondering is because, it seems like the act of trying to measure it directly (if "measure with a device" is what is meant by observe) would mean sending out some sort of radiation or particle itself, so yes, that would mess with the experiment. But maybe I just don't quite understand yet, so looking forward to a bit of clarification.</p>
|
<p>I would generally support <a href="https://physics.stackexchange.com/a/129524/670">Ayesha's answer</a>, in that it explains that decoherence is instigated by the microscopic interacting with the macroscopic. </p>
<p>As with many things in Quantum Physics, this is evidently true at the extremes (e.g. putting a detector in the path of a photon in the dual slit experiment), but it is not clear when something is considered macroscopic. For example, somewhere between coupling an atomic spin to a single other atom and coupling to a million other atoms, we (and presumably the universe) say that decoherence has occurred, and the wavefunction has collapsed.</p>
<p>One answer is to look at the <a href="http://en.wikipedia.org/wiki/Density_matrix" rel="nofollow noreferrer">Density matrix</a>. Consider that a pair of photons have a 2x2 density matrix. If we entangle the photon polarisations like-for-like, then we have either HH or VV, which would be a density matrix thus:</p>
<pre><code> H V
H 0.5 0
V 0 0.5
</code></pre>
<p>Von Neumann tells us that measurement always increases entropy inside the system, thus decreasing the quantum information contained in it. So even if we perform a measurement that is unrelated, like checking for HV states, we start to corrupt the matrix. So perhaps it becomes:</p>
<pre><code> First
H V
Second H 0.44 0
V 0.1 0.46
</code></pre>
<p>Suppose we further entangle this pair with another pair (still in its clean state), such that if the second photon of the first pair is V, then so is the first photon of the second pair (and the same for H):</p>
<pre><code> First pair
HH HV VH VV
HH 0.44
Second HV
Pair VH
VV 0.1 0.46
</code></pre>
<p>If that second pair was a bit sullied, and looked more like: </p>
<pre><code>HH 0.48, HV 0.03, VH 0.07, VV 0.42
</code></pre>
<p>Then we have:</p>
<pre><code> HH HV VH VV
HH 0.48x0.44 0.48x0
HV 0.03x0.44 0.03x0
VH 0.07x0.1 0.07x0.46
VV 0.42x0.1 0.42x0.46
</code></pre>
<p>Renormalised (/0.4988):</p>
<pre><code> HH HV VH VV
HH .4234
HV .0265
VH .0140 .0646
VV .0842 .3873
</code></pre>
<p>We can see how the original purity of the superposition is being polluted by the entropy, and as the space increases with the addition of more particles, this will further erode the original state. Note in particular how the HHHH state has P=42.34%, while the VVVV state has P=38.73%. Clearly this erosion will erode one of the original states more than the other, so after a few interactions we might expect one to disappear below the noise floor, leaving a single preferred state. In the experiment, this would come out as a 'decoherence' in which the increasing interaction with the environment dissipates the original superposition through an ever larger space until random noise takes over.</p>
<p>The signal vs noise metaphor is apt, in that it appears to be the quantum information stored in the states that it eroded. This is the origin of the Quantum Error Correction methods designed to use encodings to sustain a coherent state for longer.</p>
<p>Thus, 'observation' really means the coupling of a quantum state to a much larger system, and in doing so polluting the quantum information stored within with entropy and thus undergoing decoherence.</p>
| 647
|
wave-particle duality
|
Is the probabilistic behaviour of quantum mechanics something "new" (not directly connected) to wave-particle duality?
|
https://physics.stackexchange.com/questions/630001/is-the-probabilistic-behaviour-of-quantum-mechanics-something-new-not-directl
|
<p>Is the probabilistic behaviour of quantum mechanics theory a direct consequence of the particle-wave duality ? Or is probabilistic behaviour an additive, <em>independent</em> feature (with respect to particle-wave duality) , of quantum mechanics.</p>
<p>To reformulate in another way : if we assume that a particle is actually both a particle or a wave, would the theory <em>necessary</em> have for consequence to be a probabilistic theory ?</p>
<p>If the answer is that probabilistic behaviour is a direct consequence of wave-particle duality, then I could not understand why Einstein was puzzled and against the idea of probabilistic behaviour ("god does not play with dice"). Indeed, Einstein did show that photon was <em>both</em> a particle and a wave, so he showed already the duality. If the consequence of duality is the probabilistic behaviour, how could he be shocked of probabilistic behaviour ?</p>
|
<p>It's something new.</p>
<p>Generally, probability was seen as an outcome of not knowing everything about a system. This is how thermodynamiics explain the bulk properties of many particle systems.</p>
<p>In QM, on the other hand, probability was seen as ontologically basic. It's not due to the lack of information we have about the system.</p>
<p>Having said it is something new, Aristotle said that 'some philosophers claim that chance is a cause'. This means at the very beginnings of physics a claim was made that chance should be taken as ontologically basic. This was forgotten in the enthusiasm that greeted Newtonian physics which was wholly deterministic.</p>
<p>Quantum reality is value indefinite in some sense. And it's because of this that chance is ontologically basic. Though in this view, it is also derived, since by being indefinite and so indeterminiate we cannot say that something can be determined and so cause here must be by chance.</p>
| 648
|
wave-particle duality
|
Dark Matter Wave Particle Duality
|
https://physics.stackexchange.com/questions/482654/dark-matter-wave-particle-duality
|
<p>CERN is looking for dark photons, and they are seeking to look for interactions with the Higgs boson with the dark photon.<a href="https://home.cern/news/news/physics/cms-hunts-dark-photons-coming-higgs-boson" rel="nofollow noreferrer">See link here</a>. So in other words a dark photon - a mediator for dark matter is not unthinkable. Like wise, as photons acts as a mediator for light, and an interference pattern just underlines the same. So, what interference pattern, could a dark photon exhibit, (if such a photon exists) dark bands with dark bands? As a dark photon is still a photon, duality is an existential property for the same(unless dark photons do not have duality!) . </p>
<p>What if any interference pattern could one expect for "dark photons"(or any way of detecting dark interference patterns- interaction with the Higgs field?)? </p>
<p><strong>EDIT</strong> Or a dark matter radio such as the one described here <a href="https://www.symmetrymagazine.org/article/a-radio-for-dark-matter" rel="nofollow noreferrer">help</a> tune to interference in the cosmos, or in labs? </p>
|
<p>All particles exhibit wave-particle duality and interference effects, not just photons. However, it would be difficult to observe this aspect of dark matter because of its extremely limited interactions with our own kind of matter.</p>
<p>For example, in the CERN experiment you mention, they are simply looking for "missing energy" - less energy in the products of a collision, than went into it. In some of the theories being tested, there are messenger particles that interact with our Higgs and with dark matter, and if the Higgs produces a messenger that decays into dark matter rather than ordinary matter, then the energy of the messenger is lost to detection, thus "missing energy".</p>
<p>On the other hand, to observe interference effects in such a theory, you might need e.g. a collision between two Higgses, each of which emits a messenger, and then the messengers exchange a dark photon. This would be a very rare event, so hard to detect and analyze... Since dark photons by definition interact with dark matter, your best chance of seeing them interfere, is if you yourself, and your sense organs or experimental apparatus, are made of dark matter.</p>
<p>There are a few ways in which the quantum nature of dark matter might show up. If there are "dark atoms", they may emit dark photons at specific wavelengths, and this might eventually, indirectly, have a manifestation in the world of ordinary matter, like the distribution of cosmic rays from the galactic center, an environment in which extreme interactions would occur in great quantity.</p>
<p>Alternatively, there are theories of dark matter in which the halo of dark matter surrounding a galaxy, rather than being like a big cloud of dust, exhibits some quantum properties on astronomical scales, like Pauli exclusion or superfluidity. So there might be traces of this on a vast scale, in the distribution of ordinary matter (stars, gas).</p>
<p><strong>edit</strong>: There's a paper today on <a href="https://arxiv.org/abs/2009.14201" rel="nofollow noreferrer">"Dark Matter Interferometry"</a>.</p>
| 649
|
wave-particle duality
|
Wave particle duality and gravity
|
https://physics.stackexchange.com/questions/450263/wave-particle-duality-and-gravity
|
<p>Is a particle's center of gravity at the center of its wave function or is it where we would measure the particle to be? When we measure a particle does its center of gravity shift to where the particle is measured?</p>
|
<p>The wave function only predicts the probability distribution of finding a particle or a system of particles at an (x,y,z). The center of a wavefunction <span class="math-container">$Ψ$</span> is not a well defined concept, as it is different for different boundary conditions. It obeys the symmetries imposed by the interactions that define the <span class="math-container">$Ψ$</span> .</p>
<p>In any case at present the <a href="https://en.wikipedia.org/wiki/Standard_Model" rel="noreferrer">standard model of particle physics</a> has elementary particles defined as point particles,i.e. no dimensions, and the center of mass/gravity of course is at that point, and is the point whose measurement will verify the probability distribution, <span class="math-container">$Ψ*Ψ$</span>, when enough events/measurements with the same boundary conditions are accumulated.</p>
<p>Such elementary paticles are described by plane wave wavefunctions which are spread all over the place. A free electron has to be represented by a <a href="http://hyperphysics.phy-astr.gsu.edu/hbase/Waves/wpack.html" rel="noreferrer">wavepacket wave function</a> ,</p>
<p><a href="https://i.sstatic.net/dpLEx.gif" rel="noreferrer"><img src="https://i.sstatic.net/dpLEx.gif" alt="wavepacket"></a></p>
<p>but as the center of gravity/mass is where the electron is, the wave function squared will give the <em>probability</em> of finding it away from the maximum probability, which is the center for this wavepacket.</p>
<p>Complex systems, as <a href="https://profmattstrassler.com/articles-and-posts/largehadroncolliderfaq/whats-a-proton-anyway/" rel="noreferrer">the proton</a> and the neutron, will have an extension in space, and the instantaneous center of mass would be part of the model calculating the quantity of interest to measure. The <a href="https://epub.uni-regensburg.de/10760/1/NDA.pdf" rel="noreferrer">models for hadrons</a> are complicated .</p>
<p>In general, in quantum mechanical problems, it is the symmetries of the interaction that are embedded in the wavefunction, <a href="https://web.phys.ksu.edu/vqm/tutorials/interpretingwavefunctions/exploring.html" rel="noreferrer">the wave function</a> itself may be all over a large space, depending on the problem under study, and its center has no usable meaning with respect to the center of mass of the particle-system modeled. </p>
| 650
|
wave-particle duality
|
Wave-particle duality and temperature
|
https://physics.stackexchange.com/questions/182669/wave-particle-duality-and-temperature
|
<p>Today it was told me that wave properties of a particle increase if the temperature decreases. I'm surprised because I have never listened a similar thing, but I think that it's very interesting. </p>
<p>Could you explain me why it happens? </p>
|
<p>I think that your teacher (?) asked you about <a href="http://en.wikipedia.org/wiki/Thermal_de_Broglie_wavelength" rel="nofollow">thermal de Broglie wavelength</a>, where $$\lambda_T \propto\frac{1}{\sqrt{T}}.$$ You get this expression when you express the momentum in $\lambda=h/p$ in in terms of kinetic energy and the kinetic energy itself in terms of the energy due to temperature. (The derivation is also in the wikipedia article...) Indeed, when you drop the temperature the thermal de Broglie wavelength increases.</p>
<p>As far as I understand it, the thermal de Broglie wavelength indicitates <em>roughly</em> when you have to treat a gas clasically or quantum mechanically. In the sense that when $\lambda_T$ is much smaller than the particle separation then the particles "have to do with each other" and if the wavelength is approximately the same as the particle separation then the particles "have to do with each other". Depending on how they "have to do with each other" you then have to treat the gas as a Bose- or a Fermi gas.</p>
<p>Some time ago I read about the interference of Bose-Einstein-condensates. I thought that in this context this concept could be applicable and in fact this is true: when you, for example, quickly read through this <a href="http://www.kip.uni-heidelberg.de/matterwaveoptics/teaching/archive/ss07/Ketterle.pdf" rel="nofollow">paper</a>, which describes work which was awarded the <a href="http://www.nobelprize.org/nobel_prizes/physics/laureates/2001/popular.html" rel="nofollow">Nobel prize in 2001</a>, then you will find the sentence (page three)</p>
<blockquote>
<p>The critical number of atoms ... is determined by the condition that the number of atoms per cubic thermal de broglie wavelength exceeds... .</p>
</blockquote>
<p>So even if it's just for estimation purposes, this concept is surely used in research and definitely not "nonsense". </p>
| 651
|
wave-particle duality
|
Are all elementary particles exhibit wave-particle duality?
|
https://physics.stackexchange.com/questions/346287/are-all-elementary-particles-exhibit-wave-particle-duality
|
<p>I have watched a lot of videos about quantum theory/mechanics and other and when they speak, say, about double slit experiment they say that 'if you shoot an electron...'. Then I read somewhere that photon exhibits w-p duality. And on some other video the narrator said that all elementary particles can behave as a wave and a particle. So if it is true, it means that you can actually shoot a neutron or a proton or a photon in the double slit experiment and it will be the same?</p>
| 652
|
|
wave-particle duality
|
Doubt on "wave-particle duality" in quantum mechanics
|
https://physics.stackexchange.com/questions/594303/doubt-on-wave-particle-duality-in-quantum-mechanics
|
<p>I'm reading the book <span class="math-container">$[1]$</span> (which is not a scientific communication book, rather a student-friendly introduction to Quantum Mechanics).</p>
<p>Jakob <span class="math-container">$[1]$</span> then writes:</p>
<blockquote>
<p><em>Many people unfamiliar with quantum mechanics may wonder how an electron could be a partile and a wave at the same time. Please ignore this kind of idle speculation. The situation is not as crazy as some would lead you believe. Electrons, photons and all other elementary particles are particle. Period. This is what every experiment tell us. Our detector make "click, click, click"<span class="math-container">$^{(*)}$</span>. Waves are merely one convenient mathematical tool for describing the behavior of these particles.</em></p>
<p><span class="math-container">$^{(*)}$</span>Here the author is talking about the double slit experiment using electrons.</p>
</blockquote>
<p>Considering the realization of the author, I can conclude that, when the books (modern physics mostly and some introductory texts on quantum mechanics also) said the famous idea "the nature of particles in quantum mechanics have a dual behavior: a electron can be a wave and a particle at the same time! This is called particle-wave duality" they acctualy want to mean: <em>Electrons, photons and all other elementary particles are particle. Period. This is what every experiment tell us (...) Waves are merely one convenient mathematical tool for describing the behavior of these particles.</em></p>
<p><strong>So, can I say that particle-wave duality is mostly a mathematical formalism rather than a huge physical fact?</strong></p>
<p><span class="math-container">$$ --\circ --$$</span></p>
<p><span class="math-container">$[1]$</span> Jakob Schwichtenberg. <em>No-Nonsense Quantum Mechanics</em>. No-nonsense Books. 2ed. 2020.</p>
|
<p>The definition of particles in QFT is a bit technical than our usual notion of particles. A particle is an excitation of a field. For example, the Higgs boson is an excitation of the Higgs field. With this notion, we can say electrons are particles. However, the wave notion is also built-in in the excitation part of the definition.</p>
<p>In the usual sense, we cannot say that electron is only a particle and the wave nature is just a mathematical tool. This is not a correct statement. In some experiments, it behaves as a particle and it some other experiments it behaves as a wave. This is because neither description is the full fledged QFT description of electrons. The price we pay is we have to choose the electron either as a particle or as a wave according to the needs, while in truth they are not two different things.</p>
<p>For example, if you consider that the electron is a particle, you cannot have double slit experiment (just put a detector on one of the slits and the pattern will be destroyed) , and if you consider electron as waves in the usual sense, photoelectric effect cannot be explained.</p>
<p>While the author is correct in saying that electrons are particles, his emphasis on the wave nature being just a mathematical convenience is a bit oversimplification to make the book readable to beginners, a trait that is often found in these books but can be harmful sometimes.</p>
| 653
|
wave-particle duality
|
Why does the wave-particle duality become unnoticeable with more mass?
|
https://physics.stackexchange.com/questions/348602/why-does-the-wave-particle-duality-become-unnoticeable-with-more-mass
|
<p>According to the formula $λ=h/mv$ for the De Broglie wavelength, as the mass increases, it becomes a greater coefficient to multiply the velocity by, and the larger number in the denominator makes the wavelength so small that it can't be detected for high mass objects. My physics teacher used this reasoning to explain how even large macroscopic objects may have their own wave functions, but are so small we can't really detect them.</p>
<p>However, aren't all larger objects made of smaller ones? Like how molecules are made of atoms, and atoms are made of the elementary particles, and etc. Since this is the case, why can't we apply the λ=h/mv formula to all the particles that make up the atoms and molecules that are inside of say, a baseball, and average find the average of those waves? Since on average, half of those waves should be constructive, and the other half destructive, shouldn't the average of all the waves of the particles that make up the larger object be the true wavelength of the object? This obviously isn't the case though, since a baseball doesn't exhibit wave-like behavior.</p>
<p>What's the flaw in my reasoning? Why don't high-mass objects behave like I expect them to? I tried to read the Wikipedia page on the wave-particle duality, but I still haven't been able to come to a conclusion.</p>
|
<p>There exist wave equations , second degree differential equations which have as solutions sinusoidal functions. These functions can model perfectly water waves, sound waves, and even light waves. What is modeled as waving is the amplitude of the transfer of energy in water waves, acoustic waves, and light waves. </p>
<p>The energy "waves", i.e. if one sits at an (x,y,z) and an acoustic wave passes, the appropriate detector will go up and down at the frequency of the passing wave. Appropriate means that its dimensions are commensurate with the wavelength of the passing energy transferring wave.</p>
<p>In quantum mechanics similar differential wave equations give solutions that model the particle experiment results, wave solutions, sinusoidal ones, that is why they are called wavefunctions. <strong>BUT</strong>, and it is a huge "but" what is "waving" is not the energy of the particles. It is the probability of finding the particle at (x,y,z) at time t, that changes in in wave pattern, a sinusoidal pattern.</p>
<p>This becomes clear watching single electrons going through the double slit.</p>
<p><a href="https://i.sstatic.net/05esD.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/05esD.png" alt="dblslit"></a></p>
<p>Starting at the top frame, each electron manisfests into a dot, that looks random, what you might expect if you threw balls at double slits, the balls scattering randomly at the edges. But going down in frames and accumulating many electrons, an interference pattern appears. The accumulation of measurements gives a probability distribution for the experiment " electron hitting double slits" , a quantum mechanical experiment, with a given energy of electrons and distances and widths of slits connected with the quantum mechanical wavelength. BUT it has nothing to do with energy , <em>it is how probable it is to see an electron hitting the screen at (x,y).</em></p>
<p>At the particle level, ensembles of particles interact with each other and have a wavefunction which is a solutions of the quantum mechanical problem, but it is again on probability distributions. The solutions are coherent, i.e. the phases are known and what is called a <a href="https://en.wikipedia.org/wiki/Density_matrix" rel="nofollow noreferrer">density matrix</a> "sums" all the individual wavefunctions.As the distances between atoms and molecules grows
going into bulk matter the coherence of the solutions is lost, and one goes into the classical physics regime. </p>
<p>Coherence is necessary for any interference effects, wave effects to show up, both classically and quantum mechanically. A baseball can be described by a quantum mechanical density matrix , but the coherence is lost, the matrix has only diagonal elements. In quantum mechanical terms it means that the probability of a baseball to appear in a spot different than the classical calculation is zero. <a href="https://www.livescience.com/19268-quantum-double-slit-experiment-largest-molecules.html" rel="nofollow noreferrer">Large molecules</a> have been seen to interfere.</p>
| 654
|
wave-particle duality
|
Wave-particle duality seems like an obvious example of a contradiction that people refuse to accept. What am I missing?
|
https://physics.stackexchange.com/questions/636192/wave-particle-duality-seems-like-an-obvious-example-of-a-contradiction-that-peop
|
<p>We state that electrons are subatomic particles with no known subcomponents.</p>
<p>We discover that these electrons behave as waves.</p>
<p>We also discover that sometimes, these electrons behave as point-particles.</p>
<p>We conclude that they're the both at the same time.</p>
<p>This, to me, sounds nonsensical. If this happened in math, and one discovered that 1 = 2, one would not go around saying that we have one-two duality, would we? No, we would question the axioms that we used to perform our calculations. If one-two duality is an outcome of those axioms, we should <strong>not</strong> accept one-two duality, rather we should reject the axioms!</p>
<p>Yet, in physics, this does not seem to be the case. Why? Why is wave-particle duality accepted, rather than the axioms rejected?</p>
<p>And what are the axioms in this case? Well, it's the first line in this question. We assumed electrons have no known subcomponents. Well, clearly our experimens show we're wrong. They <strong>do</strong> have subcomponents, and when the wave-like behavior of electrons occurs, it occurs precisely because of interactions between those subcomponents that we are not familiar with and therefore cannot understand.</p>
<p>So what am I missing? It seems like physicists have accepted this bizarre notion of wave-particle duality that even Einstein thought was an embarrassment of physics, rather than consider that some premises may be wrong and it is those wrong premises that lead to the wave-particle duality.</p>
|
<p>Maybe, just maybe, we don't have a good word for it and we should call it a "warticle" or a "pave"? Why would you expect that a phenomenon at such a small scale can be described by words we use for the macroscopic world? Sometimes it behaves like how we know a wave behaves, sometimes it behaves like how
we know a particle behaves. There is nothing more to it. Learn to live with it!</p>
| 655
|
wave-particle duality
|
Clarification about Wave-particle duality
|
https://physics.stackexchange.com/questions/208818/clarification-about-wave-particle-duality
|
<p>Okay,so I am learning about the double slit experiment done with electrons. I saw this picture, which shows the interference pattern being built up slowly with increasing number of electrons:
<a href="https://i.sstatic.net/ACeod.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ACeod.gif" alt="enter image description here"></a></p>
<p>I just wanted to confirm whether I have the correct understanding. The fact that the first image has a random distribution, shows that each electron interferes with itself and strikes a point on on the screen which would be dictated by the probability function.
The interference pattern is the result of the same interference of many electrons and is a statistical property of many electrons.</p>
<p>Also, does this mean the electron travels as a wave, but then it obviously must strike as a particle since it hits a well defined spot on the screen?</p>
|
<blockquote>
<p>The fact that the first image has a random distribution, shows that each electron interferes with itself and strikes a point on on the screen which would be dictated by the probability function. </p>
</blockquote>
<p>Yes.</p>
<blockquote>
<p>The interference pattern is the result of the same interference of many electrons and is a statistical property of many electrons.</p>
</blockquote>
<p>Sort of. Each electron impact obeys (technically, samples) the probability distribution, which contains the interference. You need many hits for the probability distribution to become evident, but saying that the interference is <em>exclusively</em> a statistical phenomenon is slightly contentious.</p>
<blockquote>
<p>Also, does this mean the electron travels as a wave, but then it obviously must strike as a particle since it hits a well defined spot on the screen?</p>
</blockquote>
<p>Yes. There is a disparity in the evolution of quantum systems: wavelike, continuous, and linear ("unitary") when they're left 'by themselves' and discrete, particle-like, discontinuous, nonlinear, when they're 'measured'. The current state of affairs is not really satisfactory, as there isn't an ironclad rule to say which situations are 'systems by themselves' and which situations are 'measurements', so there's still much to understand here. The overall problem is known as the <a href="https://en.wikipedia.org/wiki/Measurement_problem" rel="nofollow"><strong>measurement problem</strong></a>, and while there's been some impressive progress recently, we're still far from anything like a satisfactory understanding of these matters.</p>
| 656
|
wave-particle duality
|
Wave-particle duality for Higgs boson
|
https://physics.stackexchange.com/questions/600069/wave-particle-duality-for-higgs-boson
|
<p>I know that all matter particles have a dual nature, particle and wave. And apart from matter, photons also have dual nature. But what about bosons, specifically Higgs bosons? Do they show both wave and particle characters?</p>
|
<p><em>All</em> particles have a wave nature. Therefore, the Higgs boson must has a wave nature.</p>
<p>This is true for all particles whether they are massive or not, or whether they are bosons or fermions.</p>
| 657
|
wave-particle duality
|
particle wave duality
|
https://physics.stackexchange.com/questions/376333/particle-wave-duality
|
<p>A single photon travelling within a single wavelength contains a dual nature, in that it can behave as a particle or a wave depending upon the chosen experiment or measure.</p>
<p>When the duality behaves as a particle, is it true to say that at the point of measurement or detection that the wave-form of the duality has collapsed, that the energy within the wave is now confined to a particle?</p>
<p>Or is it the case that the wave and particle form of the duality remain in tact and that the measurement is simply confined to either form?</p>
<p>Two examples I am thinking of are the photoelectric effect and the detection of photons within the double slit experiment.</p>
|
<p>A photon does not "collapse" into a particle or a wave depending on which measurement you make. <em>That</em> is specifically a property of wavefunctions, which probabilistically describe the <strong><em>location</em></strong> of the photon. When people talk about wave-particle duality, they are instead talking about the fact that the photon acts like an EM wave, which deals with the distribution of the photon's <strong><em>energy</strong>.</em></p>
<p>As such, I have found it best to think about the wave and particle properties of a photon as two separate, but ever-present attributes of the object. (As an analogy, perhaps think of cars. All cars have a color and a speed. When you measure the speed, you don't change the color, and vise versa. Sometimes one property is more useful to measure than the other). So, in the photoelectric effect, we are most interested in investigating the particle-like properties of the photons generated, so we sometimes ignore the wave-like properties, but they are still there!</p>
| 658
|
wave-particle duality
|
Wave particle duality/double slit experiment
|
https://physics.stackexchange.com/questions/275671/wave-particle-duality-double-slit-experiment
|
<p>In double slit experiment, if flux of electrons are very low, then..</p>
<ol>
<li><p>if one "observes" the electrons before it has reached the slit, it "behaves" as a particle and this is due to the interaction (reflection?) of photons with the electrons.</p></li>
<li><p>The screen depicts an interference pattern but the individual spots are still due to particle nature. therefore we can say that <em>whenever waves are interacting it interacts through particle/quanta (in this context).</em></p></li>
</ol>
<p>Here are the questions:</p>
<ol>
<li><p>If no one is observing, then the electron is suppose to behave as a wave.</p></li>
<li><p>Even when no one is observing, there should be interactions (collisions/reflection) between photons and electrons. If so, then how does it behave as a wave?</p></li>
</ol>
<p>Is it something to do with the reference system? i.e. "observers" who observe the reflected photons observe that the electron is a particle and those who don't observe it as a wave?..
if so, why the information of the interaction is carried only by the photons and not retained in the electrons? </p>
|
<p>Electrons are quantum mechanical entities. They are successfully modeled by wavefunctions which are solutions of a wave equation. These wave functions are probabilistic, the complex conjugate square of the function gives the probability of finding the electron at (x,y,z,t) in a dV, where V is the four vector volume element. A probability density function.</p>
<blockquote>
<p>If no one is observing, then the electron is suppose to behave as a wave.</p>
</blockquote>
<p>Whether an observer exists or not, the probability density is given by the boundary conditions of the experiment at hand. It will have sinusoidal forms which can show interference according to the boundary conditions. </p>
<blockquote>
<p>Even when no one is observing, there should be interactions (collisions/reflection) between photons and electrons. If so, then how does it behave as a wave?</p>
</blockquote>
<p>The small interactions are very weak with respect to the boundary conditions of "electron impinging on double slit of distance x between slits" and do not destroy the main solution. If one is interested in the solutions of "free electron interacting with ambient photons" it is another story and different boundary conditions apply. </p>
<p><a href="http://www.physicstogo.org/features/featureSummary.cfm?FID=302" rel="nofollow">An electron in a bubble chamber</a> leaves an ionization track from the small energy interactions with the hydrogen atoms in the chamber. It is always described by a probability density. Its footprint is that of a particle loosing energy with small interactions with matter. Any sinusoidal behavior is not within the boundary conditions of the small collisions and thus no "wave like" behavior is observed.</p>
| 659
|
wave-particle duality
|
Are there new concepts for the explanation of the wave-particle duality?
|
https://physics.stackexchange.com/questions/122536/are-there-new-concepts-for-the-explanation-of-the-wave-particle-duality
|
<p>Whenever we can observe photons immediate, they are particles. That includes that photons have a inner structure with periodically varying electric and magnetic fields. The EM field of a radio antenna exists because this field consists a lot of photons with their periodically changing EM components. Whenever we observe statistical manifestation of interaction between photons and certain physical states (most of them based on diffraction) we interpret the fringes on a screen as waves manifestation of particles. And at the same moment we always emphasize that these states are not observable. It's an interpretation of what we see.
To interpret the fringes as a result of the interaction between photons (or electrons, ...) and the EM field of certain physical states is not common but has some charme. No more need in interference of an electron (or photon) with itself in the single particle experiments. No more sentences like "we can mathematically write down but not describe what happens".</p>
<p>Yes we always describe that photons interfer EM fields (static between condenser plates or in interaction with other particles or especially with other photons (photon bunching)) but we don't articulate that. We repeat what physicists understood 90 years ago. We came to QED and work with the quantization of fields but our verbal expression is from 1920.</p>
<p>Are there new concepts for the explanation of the wave-particle duality?</p>
|
<p>@anna v: It's not an answer but it's to long for comments.
I could not stick together some facts.</p>
<p>1st: "The electromagnetic wave is composed by a huge number of photons." with what I fully agree. But the following sentence "They (photons) do not carry explicitly electric or magnetic fields." I could not stick to the first. Why? How a radio wave could be a EM field without the photons are having a EM field? There is a huge difference between radio waves and photons EM wave. Electrons will be accelerated in the antenna rod and they emit photons. Electrons are doing this more or less synchronous with the AC from the antenna generator. That is the reason of the radio wave with its amplitude (depends only from the rod length) and its frequency (depends from the generator and more or less from the rod length). Some evidence for this is the fact that the receiver not at all needs an antenna with the amplitudes length of the emitter. It's enough to use some tiny piece of conductor that will be hit by a small stream of photons.</p>
<p>2nd: Electrons in the antenna in total are accelerated parallel to the antennas rod. The resulting magnetic field is perpendicular to the rod and the resulting electric field (except the rods end) is again parallel to the rod. This situation changes rapidly. There is some divergence of the radio wave. Does it come it from the interaction of the photons or better to say from the EM fields of the photons?</p>
<p>3rd: What makes the change between the nearfield radio wave with it shift difference of 90° between electric and magnetic component and the farfield without this difference? Are the single photons are responsible for this state?</p>
<p>4th: What about the bouncing (I couldn't find at the moment the right word, it sounds like what I wrote) of photons from far away stars? Two photons are traveling together. How they can do this without interaction of their EM fields?</p>
| 660
|
wave-particle duality
|
Is the Uncertainty Principle a logical consequence from the Wave-Particle duality?
|
https://physics.stackexchange.com/questions/197821/is-the-uncertainty-principle-a-logical-consequence-from-the-wave-particle-dualit
|
<p>I always thought of the Uncertainty Principle as a logical consequence that follows from the Wave-Particle duality, or more precise, from the fact that all particles behave as waves as long as they do not interact with other particles.</p>
<p>From basic wave properties, it's clear that you cannot know both the position and the speed/impulse of a wave exactly. A strongly localized wave dissipates fast, so it has a "blurry" impulse. Moreover, a wave with an exact impulse has no precise location. So, if a fundamental "particle" behaves like a wave, then the Uncertainty Principle (for the position/impulse variable pair) is a logical consequence.</p>
<p>Is that correct so far?</p>
<p>So I wonder, why do many physicists describe the Uncertainty Principle like it was some axiom that is built into this world, and that has no reason or explanation for <strong>why</strong> it exists? There are even physicists who describe it like Nature playing masquerade. <em>"The more we look at the position, the more Nature conceals the impulse."</em></p>
<p>No, that's not what the Uncertainty Principle is about.</p>
<p>The fundamental fact is that everything behaves like waves. This is the axiom, the fact without a known reason behind it. The Uncertainty Principle is just a consequence of that fact, one of the first consequences discovered by mankind.</p>
<p><strong>EDIT</strong> @ACuriousMind: Well that makes sense, but I have the impression that this just shifts the discussion over to the fact that there are non-commutating operators. It's clear that <strong>if</strong> two operators don't commutate, then there's the uncertainty that is quantitatively described by the formula you wrote. But I'm still convinced that the fact that there are non-commutating operators has a qualitative reason. After all, if classical mechanics is formulated using the Hilbert Spaces tools, your inequality becomes trivial because $[A, B] = 0$ for all operators in classical mechanics.</p>
|
<p>The uncertainty principle is much more general than anything you might say about the wave-particle duality. In particular, wave-particle duality is a vague and imprecise statement about how certain types of quantum systems qualitatively behave, while the uncertainty principle is a very general and <em>quantitative</em> statement about the standard deviations of operators.</p>
<p>While, in settings like the double-slit, it is true that you <em>may</em> think about the quantum objects as being represented by a probability wave, this breaks down whenever one considers finite-dimensional Hilbert spaces, as they occur e.g. in the setting of quantum information and its qubits. There's no continuous set of generalized position operators - not ever a position operator at all - and hence no "wavefunction". Nevertheless, the relation
$$ \sigma_A(\psi)\sigma_B(\psi) \geq \frac{1}{2}\lvert\langle \psi \vert [A,B] \vert \psi \rangle\rvert$$
holds for all operators $A,B$ and all states $\psi$.</p>
<p>And even in the infinite-dimensional setting where you might claim that we have a "wave nature" and a "particle nature", this relation holds for <em>all</em> operators, not just position and momentum, and the proof just relies on basic properties of Hilbert spaces like the Cauchy-Schwarz inequality.</p>
<p>To stress this crucial fact: The uncertainty relation is a general consequence of the axioms that states are rays in a Hilbert space and the rule how these states give expectation values. No conception of "particle" or "wave" <em>ever</em> enters into the derivation, and the fact that waves also exhibit a type of uncertainty relation in their widths is a simple consequence of the properties of the Fourier transform. Since the Fourier transform is also intimately related to the position and momentum operators by the <a href="https://en.wikipedia.org/wiki/Stone%E2%80%93von_Neumann_theorem">Stone-von Neumann theorem</a> about their essentially unique representation as multiplication and differentiation, this explains the similarity without any reference to "wave-particle duality".</p>
| 661
|
wave-particle duality
|
Could wave-particle duality not be duality, but a constant switching between particulate and waveform?
|
https://physics.stackexchange.com/questions/297497/could-wave-particle-duality-not-be-duality-but-a-constant-switching-between-par
|
<p>SO, I'm just an enthusiast who's been doing some reading - and I don't have the level of math training to get my answer from the equations - so I apologize in advance if this is a stupid question. </p>
<p>I have been reading about the photon interference patterns in the double-slit and interferometer experiments...and how photons, even sent individually, will create an 'interference pattern with themselves' due to the duality of their wave-particle nature. I've also been reading about the theoretical (almost philosophical) explanations for decoherence such as 'multi-world' and the Copenhagen Wavefunction Collapse theories. </p>
<p>The question: </p>
<p>It seems that all of these experiments could be explained not by the photon having a constant dual wave-particle nature, but if it swapped completely in and out of one or the other. </p>
<p>I.e. when traveling 'uninterfered'* through empty spacetime (vacuum) the photon is entirely waveform with no particle characteristics. Upon interacting with something that can absorb and reflect (or eject new energy from a metastable state) the entire energy of the wave (not wavefunction but actual energy of the wave) collapses back into a point-area, before metastasizing (correct term?) back into wave energy (reflected). </p>
<p>Using a classical analogy, a (particulate) droplet of water falls onto the surface of a pond. At the point of contact, the droplet - for all intents and purposes - ceases to exist and becomes entirely wave energy in the form of a ripple. </p>
<p>The analogy breaks down when the wave, instead of expending it's energy evenly over the length of the shore; at the first point of contact with the shore, the ENTIRETY of the ripple's energy collapses - physically - back into a 'point', after which some of the energy is absorbed, but the remaining energy is again expressed - from that point - back out into the pond entirely as wave energy. </p>
<p>*I use the term 'interfered with' instead of observed, to avoid the idea that actual measurement must occur. Based on the above, anytime the wave energy interacts with something that can absorb/reflect any ratio of it - it collapses and then re-expresses - regardless of whether observed or not. </p>
<p>Thus, if the photon was reflected off two mirrors before being collected and 'observed', the process would still occur. </p>
<p>Has this already been addressed and discarded?</p>
|
<p>Ignoring the very difficult idea of all of the energy somehow magically collapsing back to a point, you'll find that the delayed quantum eraser experiments thoroughly challenge your idea. In those experiments, the light acts like a particle or a wave, but the decision as to which it behaves like occurs <em>after</em> the detector has sensed the light. With mainstream quantum mechanics, this is easily explained without causality violation, but trying to explain the behavior of this experiment with your system would require the future to interact with the past.</p>
<p>(Which is something some theories do explore, but with great mathematical rigor)</p>
| 662
|
wave-particle duality
|
Wave Particle duality because of discrete time?
|
https://physics.stackexchange.com/questions/73254/wave-particle-duality-because-of-discrete-time
|
<p>If time is discrete, such as the Planck's length, would the transition from one frame of time to the next explain why it appears matter changes from a particle to a wave? During that infinitely small space between each frame we can not measure the particle and it appears as a wave?</p>
| 663
|
|
wave-particle duality
|
What is the upper limit of objects behaving as a wave-particle?
|
https://physics.stackexchange.com/questions/254367/what-is-the-upper-limit-of-objects-behaving-as-a-wave-particle
|
<p>Photons, electrons are subatomic particles to which the wave-particle duality applies. Protons are heavier quantum objects and posses quantum tunneling which is a wave character. </p>
<p>What is the upper limit in size or mass after which quantum effects (wave-particle duality) no longer hold?</p>
|
<p>Quantum interference experiments have been done even by using large molecules, such as the <a href="http://www.nature.com/nature/journal/v401/n6754/abs/401680a0.html" rel="nofollow">C-60 fullerene</a>; although there is no such thing as a 'limit', as quantum effects hold good for all objects, as it is already pointed out in the comments.</p>
| 664
|
wave-particle duality
|
What exactly is wave particle duality?
|
https://physics.stackexchange.com/questions/689143/what-exactly-is-wave-particle-duality
|
<p>I am a 12th grade student I want to clarify some fundamentals about waves.It's clear to me when some one says sound propagates
As a wave(the variations of air densities as a function of distance from the source at an instant of time represent a sinusoidal function.).But what does it mean when someone says electron behaves as a wave?Does it mean it's trajectory represents a wave?</p>
|
<p>The “wave” of the electron is a probability wave used to describe quantum effects. For quantum-scale particles that have not had any effects measured (e.g. spin, momentum, position), the “particle” cannot be described as a discrete object in one spot/state, but as a wave existing in and moving through multiple states at once. This wave is described by the wavefunction, and by taking the absolute value and then squaring it, you get the probability distribution. This <span class="math-container">$|\psi|^2$</span> tells you where you are most likely to find the electron upon measurement.</p>
<p>It’s a very weird and intriguing phenomenon that has no classical analogue.</p>
| 665
|
wave-particle duality
|
Slit screen and wave-particle duality
|
https://physics.stackexchange.com/questions/22923/slit-screen-and-wave-particle-duality
|
<p>In a double-slit experiment, interference patterns are shown when light passes through the slits and illuminate the screen. So the question is, if one shoots a single photon, does the screen show interference pattern? Or does the screen show only one location that the single photon particle is at?</p>
|
<p>The answer is <strong>yes</strong> to both questions: yes, the screen does show one location for one particle and yes, the accumulated picture after repeating the experiment many, many times does show the interference pattern.</p>
<p>There is a set of beautiful pictures and a video of the double slit experiment in one-particle-per-time mode that can be <a href="http://www.hitachi.com/rd/research/em/doubleslit.html">found here</a> (the experiment is with electron but conceptually there is no difference).</p>
| 666
|
wave-particle duality
|
Is wave-particle duality not clear from the single-slit experiment?
|
https://physics.stackexchange.com/questions/197350/is-wave-particle-duality-not-clear-from-the-single-slit-experiment
|
<p>In experiments it is easy to discern between 2 and more-than-2 fringes on a screen, making the double-slit experiment the default one for wave-particle tests.</p>
<p>Let's say we shoot massive particles (e.g. electrons) towards a slit. Would the image behind it be the same no matter if we consider the electrons to be classical particles or wave-packets?</p>
<p>My interpretation, using an ideal (infinitely-narrow) slit, is that the (interpretation of the particles as) classical particles would produce an image with sharp boundaries, while a wave would imprint a gaussian-like distribution on the screen.</p>
|
<p>There is still interference at a single slit resulting in a Fraunhofer pattern. Just consider both edges of the split as starting point of a new wave.</p>
<p>Generally you're right. But, in a single slit, the electrons could still be deflected by the atoms that make up the slit. This - I think - leaves more room for discussion than the double-slit. It is propably just a matter of what resonated with people first and would gives students the least amount of headaches..</p>
| 667
|
wave-particle duality
|
Evaluating double-slit experiment for wave-particle duality
|
https://physics.stackexchange.com/questions/250545/evaluating-double-slit-experiment-for-wave-particle-duality
|
<p>Is it possible that the wave-like behavior of particles in double slit experiments is just an outcome of particle distribution? Can we regard or treat a normal or Gaussian distribution as wave-like? </p>
<p>Supposing the pegs on a <a href="http://en.wikipedia.org/wiki/Bean_machine" rel="nofollow">Galton board</a> could be arranged to produce distributions of marbles analogous to particles on the screen in the double-slit experiment. Would the marbles be considered to be behaving as a "wave"? </p>
|
<p>It is possible that the wavelike behavior in a double slit experiment is just the outcome of particle distribution. One example is on my link at the top of my page.</p>
<p>I do not believe that propagating photons, in slit experiments are being influenced by a Galton board type medium. I do agree with Einstein, that somethings happening but we don’t have all the answers.</p>
<p>On the other hand, something like a Galton board can offer a good analogy. For instance, say we have a box sitting on four short legs with a small hole in the center on top. You can’t see, hear or feel what’s happening inside but when you drop balls into the hole, they always form a wave/interference pattern at the bottom where they fall out.</p>
<p>No one can explain this strange behavior, but a few geniuses come up with an extremely accurate formula that perfectly predicts the observed results. They call it QM, and some even say the mystery inside this box has in it the heart of QM.</p>
<p>Einstein agrees the formula works but still believes there’s some unknown hidden process happening inside. With all his genius even, he fails to come up with a physical model to describe this strange behavior. He never gives up trying until the day he dies.</p>
<p>Most everyone was happy with the formula and as time went by less and less people cared about what caused it.</p>
<p>Every now and then someone takes the time to re-visit this mystery and wonders what’s really going on inside the box. They think about it and right or wrong they offer solutions.</p>
<p>Even though some of these physical solutions match the QM prediction, the consensus is to just except the formula (because its perfect) even though it has no physical explanation.</p>
<p>So, in my opinion, it is possible that the wave/interference behavior happening at the bottom of the box is the outcome of some deterministic, hidden, and physical process that we cannot see.</p>
<p>If someone placed this box on your table and demonstrated this great mystery, would you wonder (like people use to) what's going on inside the box, or would you just except the formula as the explanation?</p>
| 668
|
wave-particle duality
|
If photons have a wave particle duality would gravitons have a wave particle duality?
|
https://physics.stackexchange.com/questions/217043/if-photons-have-a-wave-particle-duality-would-gravitons-have-a-wave-particle-dua
|
<p>If electromagnetic waves are the same as photons would gravitational waves be the same as gravitons?</p>
|
<p>Yes, all particles, atoms, and molecules smaller than about a C60 molecule exhibit both wave and particle properties. It's not until the object is larger than its wave function that it begins to lose its wave nature.</p>
| 669
|
wave-particle duality
|
Isn't wave particle duality of light actually cheating?
|
https://physics.stackexchange.com/questions/8715/isnt-wave-particle-duality-of-light-actually-cheating
|
<p>When answering questions about light, I see that we conveniently shift between wave and particle nature of light to match the answer-- isn't this really cheating? </p>
<p>Or, is it the principle that the observation changes the nature of the object?</p>
|
<p>You wouldn't think it's cheating if you understood the "wave" involved to be NOT a physical wave in some ether-like medium, but a measure of the information we have about the system (technically, a wavefunction -- the complete description of the system).</p>
<p>When there is no detector to access that information, the wave is spread out in space, more wave-like because of its spatial extent and displays all of the typical wavy behaviour of interference and diffraction. If you now have a detector that measures position, it will localize the wave in space to such a small region that it will <em>appear</em> to behave like a particle, but really still is a wavepacket -- albeit very sharply peaked around the measured position.</p>
<p>The details of how detectors and measurement devices are able to do this is partially described by the theory of quantum decoherence (some would argue, fully described). But I hope it's clear to you that the wave-particle duality is not some kind of ad hoc prescription; it's a natural consequence of asserting that the wavefunction is a complete description of the system.</p>
<p>More here:</p>
<p><a href="http://en.wikipedia.org/wiki/Wave%E2%80%93particle_duality#Treatment_in_modern_quantum_mechanics">Modern treatment of the wave-particle duality</a></p>
<p><a href="http://en.wikipedia.org/wiki/Quantum_decoherence">Quantum Decoherence</a></p>
| 670
|
wave-particle duality
|
Does Wave-Particle duality exist at high speed?
|
https://physics.stackexchange.com/questions/421341/does-wave-particle-duality-exist-at-high-speed
|
<p>I know that relativistic DeBroglie wavelength is given by
$λ = h/γmv$.
And $γ ≥ 1$, so at higher speed $λ$ will get shorter and shorter, does this mean it will start behaving like a particle and wave picture would be destroyed?</p>
|
<p>Wave Particle Duality principle is valid for particles of any speeds. In a double slit experiment with electron(s), interference fringes are formed for any speed of electron(s).</p>
<p>Of course, the details of the interference fringes seen on the screen will depend on the momentum, and hence on the speed, of the electron(s).</p>
| 671
|
wave-particle duality
|
Wave/particle-duality as result of taking different limits of a QFT
|
https://physics.stackexchange.com/questions/154510/wave-particle-duality-as-result-of-taking-different-limits-of-a-qft
|
<p>There is an account on dualities in quantum field theories and string theories by Polchinski from last week </p>
<p><a href="http://arxiv.org/abs/1412.5704">http://arxiv.org/abs/1412.5704</a></p>
<p>At the end of page 4, he writes the wave/particle dichotomy arises from <em>different limits</em> you can take in a quantum field theory.</p>
<p>Which limits are meant here exactly, and can one give a proper example? I assume it might relate to many/few quanta states.</p>
|
<p>There are probably various answers to this question and I will try to provide one that I consider quite interesting. It is a specific realization/example of the fact that the path integral is dominated by estrema of the action.</p>
<p>The wave aspect of a QFT is probably trivial as QFT is dealing with wave equations. This is particularly apparent for massless particles and I will not discuss it any further.<br>
Let me thus focus instead on the opposite limit when the particles are very heavy. I will use the Schwinger proper time and heavily follow Matt Schwartz textbook.</p>
<p>For simplicity, consider the propagator of a scalar particle in an external field source $A_\mu$ that in the Schwinger proper time takes a path-integral form over the particle trajectory
$$
G_A(x,y)=\langle A|T\phi(x)\phi(y)|A\rangle=\int_0^\infty ds e^{-is m^2}\langle y| e^{-i\hat{H}s}|x\rangle
$$
where
$$
\langle y| e^{-i\hat{H}s}|x\rangle =\int_{z(0)=x}^{z(s)=y} [dz(\tau)] e^{i\mathcal{L}(z,\dot{z})}
$$
with
$$
\mathcal{L}=-\int_0^s d\tau \left(\frac{dz^\mu(\tau)}{2d\tau}\right)^2+e \int A_\mu(z) dz^\mu\,.
$$
It is convenient to rescale the variables with the mass, $s\rightarrow s/m^2$ and $\tau\rightarrow/m^2$ so that the path-integral is clearly dominated by the free kinetic energy when the mass is large
$$
G_A(x,y)=\frac{1}{m^2}\int_0^\infty ds e^{-is}\int_{z(0)=x}^{z(s/m^2)=x} [dz(\tau)]e^{-i\int_0^s d\tau m^2(\frac{dz^\mu}{d\tau})^2+i\int eA_\mu dz^\mu}
$$
This is the limit of particle that takes a well definite trajectory since the path-integral is dominated by the point of stationary phase that corresponds to the free particle solution
$$
z^\mu(\tau)=x^\mu+\tau v^\mu\qquad v^\mu=(y-x)^\mu/s\,.
$$
Moreover, on this solution the propagator becomes (after rescaling back to the original variables)
$$
G_A(x,y)=\int_0^\infty ds e^{-i\left[s m^2+\frac{(y-x)^2}{4s}-ev^\mu\int_0^s d\tau A_\mu z(\tau)\right]}
$$
where the last term is the same that one get by adding the source current
$$
J_\mu=v_\mu \delta(x-v\tau)
$$
so that the heavy particle creates the field $A_\mu$ as if moving in a classical trajectory at constant speed.
As Schwartz says, when a particle is heavy the QFT can be approximated by treating the particle as a classical source (but treating everything else as quantum, e.g. the particle can possibly generates quantum radiation $A_\mu$ upon which we haven't integrated over yet).</p>
| 672
|
wave-particle duality
|
Bohm's view of double-slit experiment, wave-particle duality
|
https://physics.stackexchange.com/questions/435630/bohms-view-of-double-slit-experiment-wave-particle-duality
|
<p>I gather that Bohm denies the notion that the act of measurement decides whether a photon will be a wave or a particle. Bohm's idea seems to be that the photon is always a particle with a real trajectory that always passes through one OR the other slit, NOT bizarrely through both slits simultaneously, in the double-slit experiment. The photon, Bohm says, also has a "pilot wave" that causes the photon to land on a photographic detecting plate according to an interference pattern (which pattern shows up on the plate after the physicists fire off, one at a time, a sufficient number of photons). </p>
<p>My question: if the photon always has Bohm's pilot wave, and if both the photon itself and its pilot wave objectively exist prior to and independent of measurement and detection, then how does Bohm explain the lack of any interference pattern when the double-slit experiment is set up to detect not waves but particles? I gather that when the physicists set up the double-slit experiment to detect a particle, and the physicists fire off one photon at a time toward the double slit barrier, then after enough photons have been fired off, the result on the detecting photographic plate is NOT an interference pattern, but only two bands of light corresponding to the two slits, just as one would expect if light were corpuscular. In other words, no indication of wave behavior, which means no indication of Bohm's pilot wave. If, as Bohm says, the photon is always a real particle that has a pilot wave that really exists objectively, prior to and independent of measurement or detection, shouldn't an interference pattern ALWAYS show up with the two slit experiment (after enough photons have been fired off), even when physicists set up the experiment to detect particles, not waves? </p>
<p>If possible, explain your answer without assuming I have any knowledge of physics. Thanks very much.</p>
|
<p>If you have a detector at one of the slits to detect which slit the particle goes through, the interference pattern disappears. In the de Broglie-Bohm pilot wave theory, there is still a wavefunction which is defined at both slits. However, each particle's trajectory is well-defined, and carries it through only one slit. </p>
<p>The full answer to this question is rather complicated due to the fancy "conditional wavefunction" used in de Broglie-Bohm theory. In dBB theory, there is a universal wavefunction whose evolution is governed by the Schroedinger equation. However, we can describe smaller systems using this so-called conditional wavefunction. An interesting aspect of these conditional wavefunctions is that they do not necessarily evolve based on the Schroedinger equation, although they often do.</p>
<p>In pilot wave theory, the future positions of a particle are fully knowable given the initial conditions. The double slit experiment in this theory takes as a presumption that the initial states of the particles shot at the slit are distributed randomly, and that these positions are individually unknowable. Because of this, we observe an interference pattern. If we measure the slit through which a particle travels, though, then we necessarily know its state and so can know with certainty its future trajectory. Essentially, when we register a detection, i.e., make a measurement, we necessarily have to interact with the pilot wave (that is what it means to take a measurement, after all), which causes the collapse of the wavepacket traveling through the other slit, removing anything with which to interfere.</p>
<p>Any more "why" than this requires getting into the math, which without a physics background is relatively complicated.</p>
| 673
|
wave-particle duality
|
Wave-particle duality: interactions of like / different quantum fields
|
https://physics.stackexchange.com/questions/826859/wave-particle-duality-interactions-of-like-different-quantum-fields
|
<p>With my pop-sci level of understanding, it seems to me that quantum fields exhibit particle-like properties only when interacting with a different quantum field - i.e. electromagnetic field interacts with electron field to produce an excitation of an electron and this interaction is localized so both fields appear as particles to us (photon / electron respectively). A field seems to interact only in a wave-like fashion with itself, i.e. constructive or destructive interference.</p>
<p>Is this a valid impression? Are there examples of particle-like interactions of the same field or wave-like interactions of different types of fields?</p>
|
<p>On the question: <em>"Is this a valid impression".</em> The answer is "meh". It's not great, it's not terrible. A photon interacting with an electron has a quantum field description (e.g. Feynman diagrams), or a Fock state description in quantum optics. Both of those are technical.</p>
<p>Any general statement about "it's a particle" or "it's a wave" is, well, too general, to be of any pedagogical value.</p>
<p>An example of particle behavior that is wavelike is the Pomeron in diffractive (wave-like) electro-rho (particle like) production. Ofc, that example is useless at a pop-sci level of understanding.</p>
<p>A general statement you can make is: classical analogies of quantum phenomenon with out actual quantum mechanics will lead you astray if you take them too seriously. The reason is: while classical physics is the <span class="math-container">$\hbar \rightarrow 0$</span> limit of quantum mechanics, you can't go the other way. A classical analogy will break.</p>
<p>When you replace quantum mechanics with quantum field theory--it only gets worse. We don't even solve real problems exactly; rather, we compute corrections to approximate states, or transition probabilities between approximate initial and final states.</p>
| 674
|
wave-particle duality
|
Understanding the interpretation of wave-particle duality by W.L.Bragg
|
https://physics.stackexchange.com/questions/167195/understanding-the-interpretation-of-wave-particle-duality-by-w-l-bragg
|
<blockquote>
<p>W.L.Bragg, the pioneer in x-ray diffraction, gave this lucid but vivid interpretation:"The dividing line between the wave & particle nature of matter & radiation is the moment <strong>now</strong>. As this moment steadily advances through time, it coagulates <strong>a wavy future into a particle past.</strong>. . . . Everything in the future is a wave, everything in the past is a particle." If "the moment 'now' " is understood to be the time a measurement is performed, this is a reasonable way to think about the situation.( The philosopher Søren Kierkegaard may have been anticipating this aspect of modern physics when he wrote, "Life can only be understood backwards, but it must be lived forwards.") $^\text{1}$</p>
</blockquote>
<p>What is meant by <strong>now</strong>? What does the statement actually mean? </p>
<p>$^\text{1}$ <strong>Concepts of Modern Physics</strong> by Arthur Beiser, Shobhit Mahajan , S Rai Choudhury.</p>
|
<p>Bragg <em>appears</em> to to try to explain wave-particle duality by waving his hands rapidly.</p>
<p>The moment of observation (the present) is <strong>now</strong>. Extrapolating backwards in time from that moment (when I observe a particle) I can deduce that this thing I observed was a particle in the past. But I cannot deduce much about its future - the further out you try to predict, the more uncertain that future looks. Knowing about the past is not very helpful - that's the point of the Kierkegaard quote.</p>
<p>This is how some people thought about uncertainty and wave-particle duality <em>a long time ago</em> when this whole science was in its infancy. I wouldn't get too hung up about it.</p>
| 675
|
wave-particle duality
|
Should particle-wave duality be understood as a description of light's dual nature or as a description of two observable states of light?
|
https://physics.stackexchange.com/questions/503720/should-particle-wave-duality-be-understood-as-a-description-of-lights-dual-natu
|
<p>In double slit experiments, light is observed in two distinct conditions (no measurement of trajectory / measurement of trajectory) that bring two different results (no interferences / interferences).</p>
<p>In such a context, light is given two types of representation (wave / particle) which, while matching observed behaviour, are of incompatible nature : the wave is probabilistic, infinite, while the particle is finite and localized.
Because of that, one talks about light's "particle-wave duality."</p>
<p>What if those two observations (wave, particle) were not to be taken as two representations of a unique phenomenon (light), but rather as two intrinsically different states of that phenomenon ?</p>
<p>In such a perspective, saying that "there is a paradox in the wave / particle duality" would as be inadequate as, for instance, saying that "there is a paradox in the liquid/ice duality of water."</p>
<p>Rather than seeing irreconcilable and opposite sides of a quite abstract concept (light as a quantum field), can one consider "wave" and "particle" just as two distinct states of light, that are brought by the varying experimental conditions ?</p>
<p>In other words, as we know that wave and particle properties will never be simultaneously observable, can we consider decoherence in the double slit experiment as a state change of light? </p>
<p>In that perspective, can we think of lightwaves as an excited state of vacuum, and of photons as an event more excited state of the same thing?</p>
|
<p>The whole "wave-particle duality" mess is just that: a misleading mess, and it better if one banishes this term from one's repertoire altogether.</p>
<p>First off, what we <em>observe</em> that leads to positing this notion is this:</p>
<ul>
<li>At very low intensities, when light hits a detector, it hits it like little speckles. This is perhaps the most direct evidence suggesting that light can be thought of as being composed of particles. There are actually some more pieces we need, some of which are more complex, to make it airtight, but it's a good start.</li>
<li>Moreover, we find that the energy of light at a given frequency, likewise, accumulates in anything that is put to do so, in regular steps or jumps, the size of each jump being proportional to the light's frequency, again, as though it were composed of little bits with a set amount of energy in each one. This was what led Einstein to postulate the photon.</li>
<li>Bulk light - i.e. where the particles are too numerous to count, when we shine it through an opening that is suitably small, or we try to view things that are suitably small under a light microscope - shows effects that look like wave propagation: we get interference patterns and diffraction blurring of tiny objects (which is why that electron microscopes were developed, and that further shows and was actually based on, that similarly, bulk electron beams also act as a wave).</li>
</ul>
<p>The first two points favor the "particle" concept, while the last one favors the "wave" concept. Historically, the last point of evidence came before the first two, as the first two require considerably more technological sophistication. We may ask, then, as to how we can reconcile these notions. One way to do that would be to take such a bulk wave setup and then diminish the intensity of the light to the point where the effects in the first bullet point become noticeable. When we do that, we observe this:</p>
<ul>
<li>The light hits the detector behind the opening like little speckles, as before. But if we record their positions and let them build up over time, the speckles begin to trace out the wave pattern. Moreover, this happens no matter how few we let through - even one at a time.</li>
</ul>
<p>So what does this mean? Well, we also have one more important piece to take into account:</p>
<ul>
<li>Under <em>no</em> circumstance can we <em>ever</em> register a single "particle" of light as though it were some kind of physically-extended object, like a wave.</li>
</ul>
<p>To me, it is these last two points that put the kosh on this notion. We can, of course, posit that perhaps maybe it's a wave when we're not looking, and always is when we are, but that is about as valid or not valid as positing that there is always a silent gnome behind your back when you aren't looking, and never there when you turn around to look. It's possible, but then again you can also insert an infinite number of other things.</p>
<p>Instead, the most cogent idea is that (we should think) <strong>it is a particle</strong>, but the <em>propagation</em> of such particle from emitter to receiver is (highly) nonclassical, and in "just so" a way that each such particle ends up having the capability to single-handedly contribute its own part to the wave pattern that would be formed if one sent a "true" wave through the two slits. Getting into the details of how our best descriptions of this work is a different matter, saved for a different post. To answer the question as posed though more head-on, "wave-particle duality", if you want to call it that, means that particles have the ability to create seemingly non-local and/or holistic patterns even without all the pieces of the whole being present at the same time, so long as they all come together eventually.</p>
<p>Ultimately, science does not give "absolute truths": rather it gives <em>useful stories</em> we can tell about the world. It doesn't tell us "what things are", but rather provides ideas we can use to reliably answer questions about <em>what will happen</em> if we do/don't do something, or <em>what we will see</em> given a set of circumstances, and it is up to us to figure out which ones maximize their utility while sacrificing as little understandability as possible. Scientific "truths" are really about actions and consequences, not about things. Scientific models, or theories, are social constructs. What aren't, are the guidance they provide.</p>
| 676
|
wave-particle duality
|
Misunderstood of wave-particle dualism?
|
https://physics.stackexchange.com/questions/408599/misunderstood-of-wave-particle-dualism
|
<p>Reading about dual nature of light, and atomic transitions, it seems to me, maybe wrongly, that the dual nature depends on the way we look at the phenomena. </p>
<p>Suppose a water wave travels and reach another water source. Why can't we talk about a condition like:</p>
<p><span class="math-container">$$\Delta E=h\nu$$</span></p>
<p>where <span class="math-container">$\nu$</span> is the frequency of water molecules impacts? Another point of view should be the interference of waves.</p>
<p>Isn't talking about light dualism quite the same? When a photon interacts with an electron, is must have an exact frequency, but also it might be understood as a wave interference (if I'm not mistaken).</p>
|
<p>A classical wave emerges in a medium of a huge number of molecules , we cannot identify individual molecules with the frequency of the wave, the molecules are the "coordinates" on which the energy and momentum of the wave motion can be mathematically modeled, by their small motions up or down (transverse wave), left or right (longitudinal wave) as the wave passes. It is like a <a href="https://www.youtube.com/watch?v=H0K2dvB-7WY" rel="nofollow noreferrer">wave in a stadium</a>, the frequency has a very wide spectrum, depending on how fast the people can respond.</p>
<p>A photon and an electron are point particles, i e. have a definite position $(x,y,z,t)$ in the model that describes their interactions.
When a photon interacts with an electron, its wave nature is not in energy, but in probability of interaction. It is the <strong>probability of interaction</strong> that is modeled as a wave, for the point particles which are the electron and the photon.</p>
<p>The classical wave in water, and <strong>does not</strong> behave as a particle, i.e. it is not defined by an $(x,y,z,t)$ point isolated in space, which is the definition of a point particle. One could extend the definition by allowing a $Δ(χ)$ ... In this extended case yes, there can be <a href="http://hyperphysics.phy-astr.gsu.edu/hbase/Waves/wpack.html" rel="nofollow noreferrer">wave packets</a> called <a href="https://en.wikipedia.org/wiki/Soliton" rel="nofollow noreferrer">solitons</a> in water, which do behave as a particle <a href="https://www.youtube.com/watch?v=MADng1fqECY" rel="nofollow noreferrer">carrying energy and momentum</a> , in this case a one dimensional one. <a href="https://www.youtube.com/watch?v=909o_kbCdFg" rel="nofollow noreferrer">This is also interesting</a>.</p>
| 677
|
wave-particle duality
|
Light has a wave particle duality, how do we know?
|
https://physics.stackexchange.com/questions/129892/light-has-a-wave-particle-duality-how-do-we-know
|
<p>I've been told my whole life that light is either a wave or a particle. When it's traveling through space, it's a wave. When it hits a wall, or a photo-sensitive chemical strip or something similar, it's a particle. </p>
<p>However, upon looking back all of the examples I've seen I can only recall instances in which we observe light as a particle. Are there in fact ways we can measure it as a wave?</p>
|
<blockquote>
<p>When it's traveling through space, it's a wave. When it hits a wall, or a photo-sensitive chemical strip or something similar, it's a particle.</p>
</blockquote>
<p>No, this is wrong. It's not sometimes a particle and sometimes a wave. It's always a particle and always a wave. <a href="http://www.lightandmatter.com/html_books/lm/ch34/ch34.html#Section34.3">Here</a> is an example of an experiment whose results can't be explained by a pure wave model or a pure particle model.</p>
<blockquote>
<p>Are there in fact ways we can measure it as a wave?</p>
</blockquote>
<p>The <a href="http://en.wikipedia.org/wiki/Young%27s_interference_experiment">classic evidence</a> is the existence of diffraction.</p>
| 678
|
wave-particle duality
|
A Simple Question on wave-particle duality
|
https://physics.stackexchange.com/questions/515298/a-simple-question-on-wave-particle-duality
|
<p>This was an mcq question in which only <strong>1 option</strong> is correct . The question stated -</p>
<p>Two Photons having -</p>
<ol>
<li><p>equal wavelength have equal linear momenta</p></li>
<li><p>equal energy have equal linear momenta </p></li>
<li><p>equal frequency have equal linear momenta </p></li>
<li><p>equal linear momenta have equal wavelengths</p></li>
</ol>
<p>Now i was able to rule to 2 and 3 option easily. </p>
<p>But I am not able to decide btw 1 and 4. Aren't they the same options ,if one is true other has to be true.</p>
<p>The book gave answer as option D but no reason is mentioned.</p>
<p>Can anybody provide me reason that why option (1) is not true</p>
|
<p>Option 1 is that equal wavelength implies equal momenta.
Option 4 is that equal momenta implies equal wavelength.</p>
<p>They are not the same statement. The implication goes in opposite directions. (Similar to how "human" implies "mammal", but "mammal" does not imply "human".)</p>
<p>From the overall momentum vector, you can infer the magnitude of the momentum, and from that, the wavelength. But from the wavelength you can infer the magnitude of the momentum, but not the direction.</p>
| 679
|
wave-particle duality
|
Could particle wave duality be caused by gravity?
|
https://physics.stackexchange.com/questions/246243/could-particle-wave-duality-be-caused-by-gravity
|
<p>We know that light (and other particles) displays particle wave duality, or the ability to be a particle and a wave at the same time. After that it becomes confusing. We also know that gravity is a sort of indentation in the space time continuum. Finally, we know that light has a gravitational field. So could particle wave duality be caused by a particle, let's say light, traveling through space-time, and creating a wave with its gravitational field? I have tried to research this question but I haven't gotten any kind of consistent results. I understand that this might be a dumb question but I'm no physicist.</p>
|
<p>The short answer is no.</p>
<p>The longer answer requires a background in physics, to understand that measurements have told us that there are four interactions, (strong, weak, electromagnetic, gravitational) and in order of strength gravity is the weakest of all and will be the last to play any measurable role and is undetectable for elementary particles .</p>
<p>The wave property of particles appears in <a href="https://en.wikipedia.org/wiki/Double-slit_experiment#Interference_of_individual_particles" rel="nofollow">cumulative probability distributions</a>, and it is not a mass or energy wave. Understanding requires studying physics and <a href="http://hyperphysics.phy-astr.gsu.edu/hbase/quacon.html" rel="nofollow">quantum mechanics</a> in particular.</p>
| 680
|
wave-particle duality
|
Why the wave-particle duality cannot be explained as a traveling-standing wave duality?
|
https://physics.stackexchange.com/questions/78074/why-the-wave-particle-duality-cannot-be-explained-as-a-traveling-standing-wave-d
|
<p>This would explain why speed and position cannot be measured at the same time, since either the wave would be traveling (speed) or enclosed and standing (position). The act of enclosing it (to be observed) could explain the effects of observation as well.</p>
<p>I asked a similar question some time ago and apparently it was deleted.</p>
<p>In the previous question I asked for the name of the theory that holds this.</p>
<p>Apparently there was something I misunderstood, there is no such a theory and this is impossible for some reasons that were not specified (because that was not the question).</p>
<p>So that is the question now.</p>
<p>I understand the reasons may be long to explain. If that is the case then a reference should be enough and the pointer would be greatly appreciated. If it is possible to provide a short explanation then the explanation will be even more appreciated.</p>
<p>In any case thank you very much for your answers and sorry for my ignorance.</p>
|
<p>You idea of "standing enclosed waves" with moving boundaries sounds exactly like the wave packet description of particles in QM.</p>
| 681
|
wave-particle duality
|
Does this experiment on wave-particle duality and complementarity disprove quantum mechanics and prove the EPR viewpoint on it?
|
https://physics.stackexchange.com/questions/54385/does-this-experiment-on-wave-particle-duality-and-complementarity-disprove-quant
|
<p>I recently read</p>
<blockquote>
<p><a href="http://arstechnica.com/science/2012/05/disentangling-the-wave-particle-duality-in-the-double-slit-experiment/" rel="nofollow"><em>Photons act like they go through two paths, even when we know which they took</em></a>, at Ars Technica,</p>
</blockquote>
<p>which reports on the paper</p>
<blockquote>
<p>Wave-particle dualism and complementarity unraveled by a different mode. R. Menzel et al. <a href="http://dx.doi.org/10.1073/pnas.1201271109" rel="nofollow"><em>PNAS</em> <strong>109</strong> (2012), 9314</a>.</p>
</blockquote>
<p>Please refer the Ars Technica link and its conclusion.</p>
<p>I am an Engineer. What I infer from this is :-</p>
<ol>
<li>This proves ERP.</li>
<li>Einstein Wins.</li>
<li>This basically proves that quantum mechanics is incomplete/incorrect.</li>
<li>There is a requirement for an extension for QM.</li>
</ol>
<p>What this does is :</p>
<ol>
<li>"Declaration of completeness of quantum mechanics" by Heisenberg needs to be pulled down officially</li>
<li>Other theories needs to be thought about, like Bohm's.</li>
</ol>
<blockquote>
<p>Can somebody confirm my understanding ? Any help is appreciated.</p>
</blockquote>
|
<p>In my answer to <a href="https://physics.stackexchange.com/questions/53959/wave-particle-duality/53970#53970">wave-particle duality</a> I expain the key to this misunderstanding in popular expositions on quantum mechanics.</p>
<p>The basic problem is that people think that the wave nature of the elementary particle entity, is exhibited in its mass distribution. They think that the entity is spread out in space according to the wave function of the quantum mechanical solution. <strong>It is the probability of finding the particle at (x,y,z,t) that is predicted by the quantum mechanical solutions.</strong></p>
<p>When one looks at a probability curve, for example the life expectancy as a function of current age, one does not see oneself spreading out in fractions or continually until 100 years. One knows that it is just a probability distribution derived by the census bureau. It is the same with the quantum mechanical probability distributions ( the square of the wave function). The experiment you link to is just a smarter experiment than the traditional "check the slit" experiments which introduced large distortion to the measurements and created the artifact of a spread particle magically gathering itself up.</p>
| 682
|
wave-particle duality
|
waves particle duality of human body
|
https://physics.stackexchange.com/questions/185681/waves-particle-duality-of-human-body
|
<p>De Broglie stated electrons can travel as particles and waves. Electrons show its waves properties when they can diffract through a carbon layer. So, I am not sure about his statement which human also can diffract through a gap. </p>
| 683
|
|
wave-particle duality
|
How does wave-particle duality explain the interference pattern?
|
https://physics.stackexchange.com/questions/668399/how-does-wave-particle-duality-explain-the-interference-pattern
|
<p>The screen is the Y axis and the line perpendicular to it is the X axis. We fire an electron with a well-defined momentum <span class="math-container">$p_x$</span> in the X direction. Shouldn't the x-co-ordinate of the particle follow a probabilistic distribution of wavelength <span class="math-container">$\frac{p_x}{h}$</span>? Then how come we observe a probabilistic distribution along the Y axis instead (the screen)?</p>
|
<p>As soon as the electron meets the screen, you know its <span class="math-container">$x$</span> coordinate, so its <span class="math-container">$p$</span> is uncertain and so is its <span class="math-container">$y$</span> coordinate.</p>
| 684
|
wave-particle duality
|
Light-Particle Wave Duality
|
https://physics.stackexchange.com/questions/64406/light-particle-wave-duality
|
<p>There is a lot of reading to do on this to fully understand it, but without doing that reading is there a short explanation as to why and how light behaves as a wave and a particle?</p>
|
<p>The <a href="http://en.wikipedia.org/wiki/Interference_%28wave_propagation%29" rel="nofollow noreferrer">interference patterns</a> of light show the wave nature of light.</p>
<p><img src="https://i.sstatic.net/eF34O.png" alt="two point sources"></p>
<blockquote>
<p>Optical interference between two point sources for different wavelengths and source separations</p>
</blockquote>
<p>The <a href="http://en.wikipedia.org/wiki/Photoelectric_effect" rel="nofollow noreferrer">photoelectric effect</a> shows the particle nature of light, because light hits an electron and transfers its energy to the electron.</p>
<p><a href="http://hyperphysics.phy-astr.gsu.edu/hbase/mod2.html" rel="nofollow noreferrer">This link</a> gives an clear explanation and can be a start in understanding the reason light is also a particle, the photon:</p>
<p><img src="https://i.sstatic.net/oTxy1.gif" alt="photoelectric"></p>
| 685
|
wave-particle duality
|
Wave-particle duality - particles as a special case of waves?
|
https://physics.stackexchange.com/questions/194397/wave-particle-duality-particles-as-a-special-case-of-waves
|
<p>This may be an incredibly dumb question, but I'm asking it anyway.</p>
<p><strong>What is wrong in thinking that particles are just waves with amplitude zero?</strong></p>
|
<p>By the Born interpretation, the probability of finding a particle near a point $x$ is $|\psi(x)|^2$. If $\psi(x) = 0$, then the probability of finding the particle is zero for all locations, which means that the particle doesn't exist.</p>
<p><strong>EDIT:</strong> The "amplitude" of a general wave-function isn't really well-defined, but roughly speaking it tells us "how big" the wavefunction is. In other words, we could take a wavefunction over all of space and multiply it by a constant $A$ to get the same wavefunction, just with higher/lower "peaks" and deeper/shallower "valleys". For example, compare the functions $\psi(x) = \cos x$ and $\psi(x) = 2 \cos x$. This is, I think, the notion of "amplitude" that most people are familiar with.</p>
<p>The thing about wavefunctions, though, is that their amplitude is fixed by the Born interpretation mentioned above. In particular, the probability that we find the particle <em>somewhere</em>—anywhere—must be equal to 1. This means that we must have
$$
\int_\text{all space} |\psi(x)|^2 dx = 1.
$$
If we take this wavefunction and change its amplitude, we would get
$$
\int_\text{all space} |A \psi(x)|^2 dx = |A|^2 \int_\text{all space} |\psi(x)|^2 dx = |A|^2
$$
and so the probability of the particle's existence would be something other than 1. In particular, if $A = 0$, then the probability of the particle being found would be zero; this is the point I was trying to make above.</p>
<p>Now, what I think you're trying to get at is something like the following: in classical mechanics, we know the particle's position exactly. This means that there a probability of 1 to find the particle at one specific location $x = x_0$, and <em>zero</em> probability to find it everywhere else. In particular, this means that $\psi(x) = 0$ for $x \neq x_0$; but since we must have the square-integral of $\psi$ equal to one (as above), this means that $\psi(x_0)$ must be infinite. Physicists use the <a href="https://en.wikipedia.org/wiki/Dirac_delta_function" rel="nofollow">Dirac delta function</a> to model such situations; in this case, we would have $|\psi(x)|^2 = \delta(x - x_0)$. But you can't really say that this function has "zero amplitude", since its maximum value is, well, infinite.</p>
<p>(I'm running roughshod over a lot of precise mathematics in my description of the $\delta$-function, by the way. Click through to the Wikipedia article to see how it's defined more rigorously.)</p>
| 686
|
wave-particle duality
|
Particle- and wave-like properties
|
https://physics.stackexchange.com/questions/668148/particle-and-wave-like-properties
|
<p>Is wave-particle duality a real concept or a pedagogical tool? In a less opinion-based way: what are <em>particle properties</em> and <em>wave properties</em> of a particle (that we speak of particle properties of particles nicely encompasses the problem).</p>
<p>Judging by the recurrent themes in the questions about the basics of quantum mechanics in this community, many difficulties arise from not being able to distinguish the <em>particle properties</em> from the <em>classical properties</em> (there is less confusion about waves, perhaps because people new to QM rarely have much intuition from wave theory).</p>
<p>Let me give two examples:</p>
<p><strong>Two-slit experiment</strong><br />
We say that, if the size of slits is comparable to the de Broglie wave length, the electron behaves as a wave. In the other limit, if the slits are very wide, it will behave as a particle. This is the expression of the duality.</p>
<p>However, if we were to approach this situation rigorously, we would treat the electron as a wave: solve the Schrödinger equation and demonstrate that for wide slits the coarse-grained probability distribution is particle like. Particle-like behavior here refers to a classical trajectory, and is an approximation to "true" wave-like behavior.</p>
<p><strong>Photons</strong><br />
Many questions arise from naively associating photons with particles, having momentum and trajectory. In reality, what makes the photons particle-like is their countability.</p>
<p><strong>Reprise</strong><br />
Thus, I would say that the main particle-like property is the countability, while the wave-like properties are due to the mode structure described by a partial differential equation (Schrödinger, Dirac, Maxwell, etc.) On the other hand, having simultaneously measured velocity and position is not a particle-like, but a classical particle property. Am I still missing something?</p>
<hr />
<p><strong>Related</strong></p>
<ul>
<li><a href="https://physics.stackexchange.com/q/46237/247642">Is the wave-particle duality a real duality?</a></li>
<li><a href="https://physics.stackexchange.com/q/473566/247642">How does light 'choose' between wave and particle behaviour?</a></li>
</ul>
|
<p>I would not say the main particle like property is countability. I would say that it is the all or nothing outcome of interactions. Though perhaps that isn't all that different.</p>
<p>Suppose you have a very dim plane wave approaching a phosphor screen with a pinhole. Behind it is another screen. You will see individual flashes of light where photons hit. Each photon hits one spot and misses all others.</p>
<p>If a photon goes through the pinhole, there is no interaction with any spot on the screen. Never the less, the photons diffract. From the diffraction pattern from many photons, you can infer the state of the photons that pass through. The wave function fills the pinhole.</p>
| 687
|
wave-particle duality
|
Why doesn’t simultaneous wave particle observation collapse the wave function?
|
https://physics.stackexchange.com/questions/422889/why-doesn-t-simultaneous-wave-particle-observation-collapse-the-wave-function
|
<p>My question is pretty much as the subject suggests. Recently Fabrizio Carbonne and a team from EPFL have managed to image the wave particle duality of light. I thought however that this was a technical impossibility given the measurement effect and what Bell’s inequalities told us about the reality of superposition, the absence of local hidden variables (and that Schrodingers cat is really both alive and dead). My understanding was that the wave exists as a probability function only and that to observe light as a wave is really to take many measurements of many different photons each obeying this probability function. Is that essentially what the team has imaged and in which case why is this success so special?</p>
|
<p>The experiment <a href="https://actu.epfl.ch/news/the-first-ever-photograph-of-light-as-both-a-parti/" rel="nofollow noreferrer">described here</a> is not with one particle at a time, so the wave particle duality is healthy and not attacked by these observations. It is working with ensembles of particles , and though the results depend on the wave particle duality, all they show are probability distributions for complicated scatterings.</p>
<p>Look at this <a href="https://www.sps.ch/artikel/progresses/wave-particle-duality-of-light-for-the-classroom-13/" rel="nofollow noreferrer">single photon at a time experiment</a> that does demonstrate the duality: the accumulated crossection of single photons shows a wave nature.</p>
<p><a href="https://i.sstatic.net/JI9Xr.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/JI9Xr.jpg" alt="enter image description here" /></a></p>
<p>The hit of the individual photon shows the footprint of a particle, while the accumulation shows the wave nature.</p>
<p>The EPFL experiment, may be special, but imo depends on the "eye of the beholder effect", (actually the "interpretation of the observer").</p>
<p><a href="https://i.sstatic.net/0USiP.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/0USiP.jpg" alt="epfl" /></a></p>
<blockquote>
<p>As the electrons pass close to the standing wave of light, they “hit” the light’s particles, the photons. As mentioned above, this affects their speed, making them move faster or slower. This change in speed appears as an exchange of energy “packets” (quanta) between electrons and photons. The very occurrence of these energy packets shows that the light on the nanowire behaves as a particle.</p>
</blockquote>
<p>So, in my view, to interpret the probability/crossection seen at the picture, a cumulative ensemble of photons, the quantum mechanical distinction of particles and probability waves has to be mathematically invoked, but it is not a simple demonstration of the particle/wave duality. The single photon double slit is.</p>
| 688
|
wave-particle duality
|
Experiment that demonstrates the wave-particle duality of electrons
|
https://physics.stackexchange.com/questions/268250/experiment-that-demonstrates-the-wave-particle-duality-of-electrons
|
<p>EDIT : You're about to read the first iteration of my question which is flawed. Please go to the end to see an illustration of what I meant to say. The phenomenon I was talking about is called emission spectrum.</p>
<hr>
<p>I remember watching an experiment in a documentary, <em>The Elegant Universe</em> by Brian Greene I believe. We put an atom, charge it with energy and then filter its energy through a prism.</p>
<p>The result was a distinct pattern of colour, darkness, colour, darkness, colour, darkness.</p>
<p>I believe that the problem was that each colour was emitted by an electron that were stimulated, yet we ended with more colours than electrons in this atom.</p>
<p>Can anyone point me to the scientific title of this experiment? Other experiments and reading material (non-scientist oriented) would be much appreciated.</p>
<p>EDIT :
<a href="https://www.youtube.com/watch?v=eCFTVdExxPA" rel="nofollow">Here</a>'s a link to the episode of the documentary in question. Experiment begins at 08:52. </p>
<hr>
<p>It was actually a heated gas that emitted a spectrum following this pattern.
The distinction is due to electrons changing orbits when heated.
<a href="https://upload.wikimedia.org/wikipedia/commons/thumb/5/52/Emission_spectrum-Fe.svg/757px-Emission_spectrum-Fe.svg.png" rel="nofollow"><img src="https://upload.wikimedia.org/wikipedia/commons/thumb/5/52/Emission_spectrum-Fe.svg/757px-Emission_spectrum-Fe.svg.png" alt="Emission spectrum"></a></p>
<p>The result is a distinct pattern of colours with interruption of darkness in between.</p>
|
<p><strong>Double-Slit Experiment</strong></p>
<p>I believe you are describing the double slit experiment with electrons (as opposed to with light). The pattern you are describing is called an interference pattern (much like two pebbles producing ripples in a pond and there are parts where the ripples cancel out). Below is a diagram of the double slit experiment.</p>
<p><a href="https://i.sstatic.net/1Kkrf.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/1Kkrf.jpg" alt="double slit experiment"></a></p>
<p>One way of showing the interference pattern is shown below:</p>
<p><a href="https://i.sstatic.net/YI0W1.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/YI0W1.jpg" alt="interference pattern"></a></p>
<p>To summarize how this experiment works, an electron gun is used to fire electrons one by one at a screen with two slits in it. Behind the double slit screen is another screen where the electrons build up. The importance of this experiment is that the electrons build up in an interference pattern (as shown below, where a is the beginning and e is the end) which proves that electrons have a wave-particle duality like light, as de Broglie predicted. This phenomenon has been shown to occur with photons, electrons, atoms, and some molecules (such as buckyballs, though, referencing the comments, I'm not sure about water).</p>
<p><a href="https://i.sstatic.net/TdDvZ.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/TdDvZ.jpg" alt="electron pattern"></a></p>
<p>There are two other important things to note about the double slit experiment. The first is the idea behind quantum mechanics - the probability wave. In quantum mechanics (unlike in Newtonian mechanics) nothing can be predicted exactly. A baseball will probably end up at home plate, but there's always a chance, however minuscule, that it'll end up over in the Andromeda Galaxy. The probabilities of where things will be are represented by what is called a probability wave. You can think of a probability wave like the one in the picture below (which isn't the probability wave of the baseball). </p>
<p><a href="https://i.sstatic.net/FdkGk.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/FdkGk.jpg" alt="probability wave"></a></p>
<p>The basic idea is that where the probability wave is high, the particle represented is likely to end up; where it is low, it is not likely to end up. The reason I'm bringing up probability waves is because they are in the experiment. The electron's probability wave goes through both slits at once (interestingly, if you close one slit, the results are not at all what you'd expect with Newtonian mechanics, but it makes sense in the context of the electron going through both slits at once) and produces the interference pattern. </p>
<p>The second is an interesting thought experiment. The idea of this thought experiment is that if particle detectors are placed behind each slit to show how many photons/electrons/particles come through, the interference pattern will disappear, because of the fact that particles cannot be observed as both wave and particle at once, resulting in the wavefunction collapsing and the particle settling on a single slit. This can be somewhat represented by the diagram below.</p>
<p><a href="https://i.sstatic.net/dzQpP.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/dzQpP.jpg" alt="wave collapsing detectors"></a></p>
<p>The one difference is that the screen would have two slits with detectors A and B in front of those two slits; otherwise, the basic idea is the same.</p>
<p><a href="https://en.m.wikipedia.org/wiki/Double-slit_experiment" rel="nofollow noreferrer">Here</a> is a website that the double slit experiment further. </p>
<p><strong>Emission Spectrum</strong></p>
<p>The emission spectrum of a chemical element or chemical compound is the spectrum of frequencies of electromagnetic radiation (e.g., visible light, microwaves, radio waves, gamma waves, etc) emitted due to an atom or molecule making a transition from a higher energy state to a lower energy state. The photon energy of the emitted photon (the photon is the carrier of the electromagnetic force) is equal to the energy difference between the two energy levels. Below shows the electromagnetic spectrum:</p>
<p><a href="https://i.sstatic.net/FXuGH.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/FXuGH.png" alt="EM spectrum"></a></p>
<p>In the context of the video, when one of the electrons makes a quantum leap to one of the lower energy levels closer to the nucleus, it emits electromagnetic radiation at certain wavelengths. This emitted electromagnetic radiation is its emission spectrum.</p>
<p>The experiment is done by heating gases up until they glow - agitating the electrons in the process - and then observing the light they gave off through a prism, which produces very defined lines - the emission spectrum. This could not be explained with the Newtonian picture of the atom (which, honestly, wasn't really defined - definitive proof of atoms was only provided by Einstein during his annus mirabilis in 1905, with his paper on Brownian Motion) but <em>could</em> be explained with the Bohr picture of the atom. Though Bohr's picture turned out to be partially incorrect (never, ever picture an atom as a miniature solar system, or physicists will pull their hair out) one idea remained the same, and was further emphasized by the Pauli Exclusion Principle. This idea was that electrons couldn't be in just any spot - they had to be in certain shells, like layers of an onion. They could never, ever be in the spots between the shells. </p>
<p>This was kind of important because one, electrons couldn't spiral into the nucleus (always good to keep the fundamental structure of matter from collapsing) and two, it explained this experiment. When electrons do a little quantum leap between the energy levels, they either have to gain energy (to move away from the nucleus, toward the outer energy levels) or lose energy (to move towards the nucleus and the inner energy levels). Now, because of the Law of Conservation of Energy, the energy lost cannot just vanish - it has to go somewhere. Enter the emission spectrum. The electrons emit electromagnetic radiation, the energy of that radiation depending on how much energy the electrons are losing, and the wavelength of that radiation depending on the atom - which is where the science of spectroscopy comes from. Below is a diagram of an atom's shells and how electrons emit or absorb energy when moving between these.</p>
<p><a href="https://i.sstatic.net/F7mIy.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/F7mIy.jpg" alt="atom shells"></a></p>
<p>Emission spectrums vary from atom to atom. Below shows the emission spectrums of a few elements.</p>
<p><a href="https://i.sstatic.net/VOmwC.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/VOmwC.jpg" alt="emission spectrum elements"></a></p>
<p>Hope this helps!</p>
| 689
|
wave-particle duality
|
Is there a 5th fundamental force which may be responsible for the behavior of matter to be wave or a particle?
|
https://physics.stackexchange.com/questions/482902/is-there-a-5th-fundamental-force-which-may-be-responsible-for-the-behavior-of-ma
|
<p>We already know about the 4 fundamental forces in nature which are gravitational,electromagnetic,weak nuclear and strong nuclear forces.</p>
<p>There are several other questions on this website which are somewhat related to this question but I want to know if there is any other fundamental force which is responsible for wave particle duality and is internal for the matter as a system and decides when matter would chose between particle and wave behaviour?</p>
<p>If I had not addressed the problem properly then I ask it in a short version below(I take matter as a system):</p>
<p><strong>Is there any other internal fundamental force which is responsible for the wave particle duality of matter?</strong></p>
|
<p>Quantum mechanics is a physics theory , and physics theories impose extra axioms on mathematical solutions of equations in order to fit observations and predict future behaviors. These axioms are called: postulates, laws, principles and depend directly on experimental observations.</p>
<p><a href="http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/qm.html" rel="nofollow noreferrer">The postulate</a>s of quantum mechanics make it a probabilistic theory, not a deterministic one. The wave function is postulated as defining the probability distribution by evaluation of <span class="math-container">$Ψ^*Ψ$</span> , a real number between 0. and 1.</p>
<p>The wave particle duality is the observation that when interacting, quantum mechanically described particles act like a classical particle. An accumulation of measurements though is needed for the QM solutions to predict the probable location of a particle, and that distribution has a wave behavior, because in general the quantum mechanical equations are wave equations. This can be seen clearly in the accumulation of data in a <a href="https://en.wikipedia.org/wiki/Double-slit_experiment#Interference_of_individual_particles" rel="nofollow noreferrer">single electron at a time double slit experiment.</a></p>
<p>The <a href="https://en.wikipedia.org/wiki/Standard_Model" rel="nofollow noreferrer">standard model,</a> and any extensions of it are a meta level on the wave function solutions of the quantum mechanical boundary conditions problem. So within the present framework any new forces from <a href="https://en.wikipedia.org/wiki/Grand_Unified_Theory" rel="nofollow noreferrer">GUT theories</a> or supersymmetry etc will not change the axiomatic status of the probabilistic nature of quantum mechanics.</p>
<p>There does exist serious research which tries to make the quantum mechanical level a meta-level of a deterministic system, which will reproduce all the successes of quantum mechanics in describing the data as emergent from an underlying theory. For example the <a href="https://en.wikipedia.org/wiki/De_Broglie%E2%80%93Bohm_theory" rel="nofollow noreferrer">Bohmian mechanics.</a>, or the research of <a href="https://physics.stackexchange.com/users/11205/g-t-hooft">G. 't Hooft,</a> who has honored us by discussing it a few years ago, but these are not main stream research directions. </p>
<p>In deterministic theories, the probabilistic nature used in the postulates of quantum mechanics will be seen as emergent, but not by any particular forces at the standard model level and extensions. These will be emergent meta levels out of the hypothetical deterministic forces.</p>
| 690
|
wave-particle duality
|
Quantization and wave-particle dualism of light
|
https://physics.stackexchange.com/questions/414084/quantization-and-wave-particle-dualism-of-light
|
<p>I'm studying atomic spectras and got puzzled about <em>light-quantization</em>. I'll expose my effort to understand it so far.</p>
<p><strong>Blackbody radiation</strong></p>
<p>Around the year $1900$ Planck explained <a href="https://en.wikipedia.org/wiki/Black-body_radiation" rel="nofollow noreferrer">blackbody radiation</a> including a term of the form $A=k\,\nu$ which is reasonable because of the experimental results: radiating energy had been shown to decrease with frequency. If I'm not mistaken $k$ was a constant obtained by fitting the experimental data (then called Planck's constant $h$).</p>
<p>A couple of years later it comes the <a href="https://en.wikipedia.org/wiki/Photoelectric_effect" rel="nofollow noreferrer">photoelectric effect</a>, which was explained just using Planck's idea and little more. Everyone knows the formula so I won't paste it here. </p>
<p><strong>Question</strong></p>
<p>Does <em>photoelectric effect</em> suggest that light can be thought as particle? How?</p>
<p>I understand perfectly $E=h\,\nu$ and that atoms have quantized energy levels, but is light energy quantized? </p>
|
<p>The formula $E=h\nu$ implies that light is quantised. Planck introduced the quantum in 1900 as a heuristic trick Then Einstein's 1905 explanation of the photoelectric effect proved that the light quantum was physical. </p>
<p>The reason is that the number of electrons emitted is proportional to the energy of the irradiating light divided by $h\nu$. The electron kinetic energy however increases with frequency. This means that each electron is emitted by a discrete process in which a single energy quantum $h\nu$ is absorbed. </p>
| 691
|
wave-particle duality
|
What is the basis of description of matter and energy in our universe in the form of wave and particle?
|
https://physics.stackexchange.com/questions/643335/what-is-the-basis-of-description-of-matter-and-energy-in-our-universe-in-the-for
|
<ol>
<li><p>What exactly does wave particle duality mean does it mean that an electron is a particle which is moving like a wave or does it mean that an electron and a photon is neither a wave nor a particle and something completely different or is it as if it’s sometimes a wave and the other times a particle or if it’s both simultaneously? Similarly for photons.
(Is it like mass is nothing but compactly packed energy and similarly for energy, as in <span class="math-container">$E=mc^2$</span>, which is described by wave particle duality as in the answer to the first question?)</p>
</li>
<li><p>Based on what have we got the two ideas of wave and particle? That is, how have we-on what basis, classified the behaviour of matter and energy as wave and particle? (Like we have classified matter into solid, liquid, gas, Bose-Einstein condensate and plasma based broadly on the inter particle forces of attraction, how is matter packed, the relative energies etc.)</p>
</li>
</ol>
|
<p>The basis is observation. The way to understand observation proved to be quantum mechanics. See <a href="https://en.wikipedia.org/wiki/Quantum_mechanics" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Quantum_mechanics</a>.</p>
| 692
|
wave-particle duality
|
What is a wave function?
|
https://physics.stackexchange.com/questions/712920/what-is-a-wave-function
|
<p>I read about Erwin Schrödinger describing wave particle duality with something called a <a href="https://en.wikipedia.org/wiki/Wave_function" rel="nofollow noreferrer">wave function</a>. What does a wave function mean?</p>
|
<p>So before we get into this there is a few levels of explanation to this and as this seems to be one of your first encounters with the subject I will skimp on mathematical details and rigor and keep the discussion rather shallow. If you are interested in a more detailed account let me know.</p>
<p>So first of all let us quickly recap how we normally describe an object, say a particle in classical physics:</p>
<p>Assuming the particle itself has no internal structure ( or spin for that matter) we can describe how the particle behaves by looking at its position <span class="math-container">$x(t)$</span> as a function of time <span class="math-container">$t$</span>. So at each time, we can look at this function and it tells us where we will find our particle. Of course the hard bit if figuring out what this function actually looks like which we normally do by using the equation of motion of the particle if we know the forces acting on it. (Thing along the lines of <span class="math-container">$F = ma$</span> which gives a differential equation for <span class="math-container">$x(t)$</span>.)</p>
<p>Now when we move to quantum mechanics however things are not as simple anymore. You might have heard of the idea that in Quantum mechanics the notion of a point particle as we used before is not really sensible anymore as a particle is no longer at any single point in space but rather "smeared" out across a region of space. This behavior is exactly what is captured by the wavefunction. A wavefunction <span class="math-container">$\Psi(x,t)$</span> of a particle now no longer only depends on time but also depends on space. Roughly speaking we can look at this function plug in a point in space and a time and the value of the function will tell us what the probability is to find our particle there if we were to e.g. place a detector there. (Actually, the probability is given by <span class="math-container">$|\Psi(x,t)|^2$</span> but the intuition still holds).</p>
<p>So that is what the Wavefunction actually is. Now one question you might ask is why can't we write down something like this for a classical particle? The answer is that we can however the function will be trivial in the sense that it is <span class="math-container">$1$</span> along the path <span class="math-container">$x(t)$</span> and <span class="math-container">$0$</span> everywhere else, because out trajectory <span class="math-container">$x(t)$</span> is unique if we know all the forces and the initial conditions in our problem.
Now you might wonder why the converse does not work. Why can't we just write down something like <span class="math-container">$x(t)$</span> for our quantum particle if we know all the forces etc.? The answer is connected to a different concept in Quantum mechanics called Heisenbergs uncertainty principle, which roughly tells us that there is no way to both restrict the initial position and momentum of the particle to a single value at the same time. Thus the path the particle will take is no longer unique but rather a smeared out function.</p>
<p>So to recap:
A wavefunction is a function which tells us (approximately) what the probability is to find a particle at a certain point <span class="math-container">$x$</span> and time <span class="math-container">$t$</span>.</p>
| 693
|
wave-particle duality
|
Why do we need a double slit experiment, when a single slit experiment shows that an electron/photon behaves like a wave?
|
https://physics.stackexchange.com/questions/720559/why-do-we-need-a-double-slit-experiment-when-a-single-slit-experiment-shows-tha
|
<p>To demonstrate wave-particle duality, it is often stated that we need to perform a double slit experiment. However, it seems that in a single slit experiment, individual photons or electrons behave like a wave. This is a reference for diffraction with a single slit: <a href="https://opentextbc.ca/universityphysicsv3openstax/chapter/single-slit-diffraction/" rel="nofollow noreferrer">https://opentextbc.ca/universityphysicsv3openstax/chapter/single-slit-diffraction/</a></p>
<p>Is it then necessary to perform a double slit experiment on top of a single slit experiment to show wave-particle duality?</p>
|
<p>As explained in the reference you cite, the single slit is not behaving as a <em>single source</em>; in that example it is wide enough to behave like many sources and thereby interfere <em>with itself</em>.</p>
<p>In the classic double-slit arrangement, the two slits are purposely made as narrow as possible so as to better approximate two single sources and are spaced apart by such a distance as to furnish the most dramatic phase cancellation pattern for demonstration purposes.</p>
| 694
|
wave-particle duality
|
Is electron/photon wave or particle in Feynman sum over histories formulation?
|
https://physics.stackexchange.com/questions/412294/is-electron-photon-wave-or-particle-in-feynman-sum-over-histories-formulation
|
<p>In the famous double slit experiment, a photon (say) can behave as wave or particle depending on whether there is (or how) an outside observer measuring the experiment.</p>
<p>Copenhagen interpretation interpret this wave-particle duality by saying that the photon behaves as wave when unobserved, and the act of interpretation "collapses" the wave function makes it behaves as particle.</p>
<p>How would Feynman sum over histories formulation interpret this duality? Or to put it the other way, how does Feynman sum over histories interpret Quantum decoherence? Or it is <em><strong>just</strong></em> a mathematical tool, nothing more, so it doesn't really care about whether a photon is a wave or particle and <em>there is no underlying QM interpretation</em>?</p>
|
<p>The sum over histories formulation is the path <a href="https://en.wikipedia.org/wiki/Path_integral_formulation" rel="nofollow noreferrer">integral</a> method:</p>
<blockquote>
<p>The path integral formulation of quantum mechanics is a description of quantum theory that generalizes the action principle of classical mechanics. It replaces the classical notion of a single, unique classical trajectory for a system with a sum, or functional integral, over an infinity of quantum-mechanically possible trajectories to compute a quantum amplitude.</p>
</blockquote>
<p>Thus it is a way of calculating that is mathematically equivalent to the others:</p>
<blockquote>
<p>The path-integral approach has been proved to be equivalent to the other formalisms of quantum mechanics and quantum field theory. Thus, by deriving either approach from the other, problems associated with one or the other approach (as exemplified by Lorentz covariance or unitarity) go away.</p>
</blockquote>
<p>The particles in the <a href="https://en.wikipedia.org/wiki/Standard_Model" rel="nofollow noreferrer">standard model of particle physics</a> are particles in the sense that when measured they have a unique mass and the quantum numbers identifying them in the table. The wave nature appears in the <strong>probability distributions</strong> for the given interactions, as with the double slit experiment.</p>
<p>So yes, it is a mathematical tool equivalent to the others used in quantum mechanical calculations.</p>
| 695
|
wave-particle duality
|
Is $E=hf$ applicable for all types of particle?
|
https://physics.stackexchange.com/questions/591998/is-e-hf-applicable-for-all-types-of-particle
|
<p>According to Planck's law, <span class="math-container">$E=hf$</span> is applicable for photon and photons show wave-particle duality. But De Broglie proved that electrons and other substances also show wave-particle duality and he showed that
<span class="math-container">$\lambda=h/p$</span>. But I have some problem about this topic: <span class="math-container">$mc^2=hf$</span> or, <span class="math-container">$mc^2=hc/\lambda$</span> or, <span class="math-container">$mc=h/\lambda$</span> or, <span class="math-container">$\lambda=h/mc$</span>. But the term <span class="math-container">$mc$</span> is the momentum of photon but not the momentum of other dynamic substance. So how does <span class="math-container">$\lambda=h/mc=h/p$</span> show the wavelength of any substance? Isn't Planck's law <span class="math-container">$E=hf$</span> applicable for other substances except photons?</p>
|
<p>The momentum of a photon is not given by</p>
<p><span class="math-container">$$p = \frac{h}{mc}$$</span></p>
<p>since a photon has no mass. Instead momentum of a photon is given by</p>
<p><span class="math-container">$$p = \frac{E}{c}$$</span></p>
<p>where <span class="math-container">$E$</span> is the energy of the photon (and this equation applies to other massless particles). The de Broglie wavelength of a particle with mass <span class="math-container">$m$</span> and velocity <span class="math-container">$v$</span> is given by</p>
<p><span class="math-container">$$\lambda = \frac{h}{p} = \frac{h}{mv}$$</span></p>
<p>And the equation <span class="math-container">$E = hf$</span> applies to photons and all fundamental particles (and certain composite particles).</p>
| 696
|
wave-particle duality
|
Hamilton's characteristic function, wave-particle duality and constant-action surfaces
|
https://physics.stackexchange.com/questions/823685/hamiltons-characteristic-function-wave-particle-duality-and-constant-action-su
|
<p>So, I'm currently doing some research about the way in which classical physics connects to quantum physics, and I came across the Hamilton-Jacobi equation, and the implications of Hamilton's characteristic function.</p>
<p>I found out that, since:</p>
<p><span class="math-container">$$ \frac{\partial{S(q_i,t)}}{\partial{q_i}} = p_i$$</span></p>
<p>and, consequently:</p>
<p><span class="math-container">$$ \frac{\partial{S(\textbf{q} ,t)}}{\partial\textbf{q}} = \textbf{p} \Rightarrow \nabla_q S(\textbf{q},t) = \textbf{p}$$</span></p>
<p>the vector composed of <a href="https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Variational_Principles_in_Classical_Mechanics_(Cline)/07%3A_Symmetries_Invariance_and_the_Hamiltonian/7.02%3A_Generalized_Momentum" rel="nofollow noreferrer">generalized momenta</a> is normal to each iso-action surface.</p>
<p>However, I found in multiple articles, a claim that I don't quite understand: this particular relation between iso-action surfaces and the momentum vector, appears to suggest that the trajectory taken by the system in the configuration space is perpendicular to such surfaces.</p>
<p>The thing about such a conclusion that bothers me, is the apparent assumption that the momentum vector is always tangent to the trajectory.</p>
<p>Can somebody explain under which conditions does such an assumption hold or, alternatively, what am I getting wrong about those suggestions?</p>
|
<p>You are right, trajectories of the solutions of EL equations are not normal to the surfaces at constant <span class="math-container">$S$</span> in general, if using the Euclidean metric in coordinates which, however, has no physical meaning in general.</p>
<p>However, <strong>for Lagrangians whose kinetic energy is quadratic in <span class="math-container">$\dot{q}^k$</span> with a non degenerate quadratic form</strong>
<span class="math-container">$$L(q,\dot{q})= \sum_{h,k=1}^n \frac{1}{2} g_{hk}(q) \dot{q}^k\dot{q}^h - U(q)$$</span>
we can use this quadratic form as the physically meaningful metric in the configuration space. There, <span class="math-container">$p$</span> and <span class="math-container">$\dot{q}$</span> are nothing but the contravariant and the covariant form of the normal vector to the <span class="math-container">$S$</span> constant surfaces:
<span class="math-container">$$dS_k = p_k\:, \quad p_k = \sum_h g_{kh}\dot{q}^h$$</span>
from HJ equations and the definition of <span class="math-container">$p_k$</span> with respect to <span class="math-container">$L$</span> respectively.</p>
<p>Referring to that metric the solutions of the EL equations define normal (contravariant) vectors to the <span class="math-container">$S$</span>-constant surfaces.</p>
| 697
|
wave-particle duality
|
Is a large system just a set of smaller systems?
|
https://physics.stackexchange.com/questions/52101/is-a-large-system-just-a-set-of-smaller-systems
|
<p>All particles exhibit wave-particle duality. And I have a strange question. </p>
<p>Why does a larger system, liken an atom that is just a set of smaller systems, itself exhibit wave-particle duality?</p>
<p>In principle all large systems can be defined as a set of smaller systems. An atom is a set of nucleus (a smaller set of up and down quarks held together by the strong interaction) and some electron(s) bound to the set of quarks by the electromagnetic force.</p>
<p>How is this set of smaller systems being able exhibit wave-particle duality as a whole as if it is a particle itself?</p>
<p>Does this imply that electron and all other elementary particles are indeed just another set of smaller sets and any sets can exhibit wave-particle duality?</p>
<p>And here comes the question: how do you define a set? We naturally define composition such as atom, molecule or "elementary particle" like electron as a set. Can the composition of electrons and up quarks be defined as a set (a system with wavefunction) and the down quarks that present in the atom be defined as a separated set (another system)?</p>
|
<p>According to Quantum mechanics, every particle has a wavefunction which completely describes it. The behavior of the particle, including its time evolution and the distribution of outcomes to any measurement performed on the particle, is determined by its wavefunction (<strong>Edit:</strong> Michael Brown correctly notes in the comments that if the particle is part of a bigger composite system, we often cannot use a wavefunction to describe it, and we need to use a <em>density matrix</em> instead).<p></p>
<p>Some properties of these quantum particles are similar to properties of classical particles. Some properties are similar to classical waves. The coexistence of properties of both kinds is what some people describe as "wave-particle duality" (I have to say that personally I dislike this term).<p></p>
<p>A composite system, like an atom which is composed of a nucleus and electrons, also has a wavefunction (or density matrix), and therefore also has both kinds of properties (particle-like and
wave-like), and therefore can also be said to have "wave-particle duality".<p></p>
<p>The question of whether a certain elementary particle is indeed elementary or is composed of smaller particles is interesting, but unrelated. As of now mainstream physics considers electrons elementary particles, but of course we have no way of knowing whether or not this will change in the future.<p></p>
<p><strong>Edit:</strong> To answer your last question - yes, in principle you can take any set of particles and consider them as one system. The only question is whether it is useful to do so. For example the nucleus and the electrons of an atom are often easy to treat as one system - an atom. You can in principle consider, like you suggest in your question, only part of the quarks in the nucleus together with the electrons as one system, but you are not likely to achieve any useful insight by doing that. The rest of the quarks in the atom interact with the quarks in your system very strongly, and any meaningful description will need to include them.</p>
| 698
|
wave-particle duality
|
Dual nature of Matter at gross level
|
https://physics.stackexchange.com/questions/86270/dual-nature-of-matter-at-gross-level
|
<p>Is the Dual nature (wave - particle duality) of Matter completely proved or just a theory and are the objects (water,rubber ball, car, apple etc.) that we see all around us in day to day life exhibit dual nature (wave - particle duality) in their natural state of existence (as perceived by the human eye) or only when they are broken down to the molecular level?</p>
|
<p>There are two approaches to this.</p>
<p>Firstly, we can steadily increase the size of objects and try to measure quantum properties. At the moment the record is for a <a href="http://www.scientificamerican.com/article.cfm?id=quantum-microphone" rel="nofollow">quantum tuning fork</a> that contains around 10 trillion atoms. It gets harder and harder to measure quantum behaviour as objects get bigger (for reasons discussed below) but there is no sharp cutoff so there is no indication that baseballs or cars can't (in principle) display quantum behaviour.</p>
<p>Secondly we have a theory called <a href="http://en.wikipedia.org/wiki/Quantum_decoherence" rel="nofollow">decoherence</a> to explain why it gets harder to measure quantum behaviour as objects get bigger. I don't fully understand decoherence so you'll need to ask an expert for the details. However the basic idea is that any object interacts with it's environment, and this interaction destroys quantum superpositions. The bigger an object, or more precisely the more degrees of freedom it has, the faster the interaction occurs. So we can easily measure quantum behaviour for electrons, but no experiment we can (currently) do can put a baseball in a superposition of states for long enough for us to measure it.</p>
<p>The upshot is that we believe the same equations that describe elementary particles also describe baseballs and cars, but these theories also tell us we'll never see quantum behaviour for such large objects.</p>
| 699
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.