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and identify the coeﬃcients a1 and a2, e.g., by using the results of part (a). Show that the corresponding dve−ikvp(v, t) is given Fokker–Planck equation for the generating function g(k, t) = by ∂ ∂t + γk ∂ ∂k + γT m k2 g(k, t) = 0. (10.118) (c) The ﬁrst order partial diﬀerential equation Eq. (10.118) can be solved for...
gator (left) and interaction vertex (right), respectively. ﬁeld renormalization exponent χ so as to make the diﬀusion constant D and the coeﬃcient of frequency invariant under renormalization. Show that this condition generates the equations . The “mass term” τ and the interaction constant κ ﬂow according to Eq. 10.115...
oeﬃcients, . . . f . We repeat that these latter ﬂuctuations may eﬀectively be generated by a quantum average (i.e. a trace operation) over microscopic degrees of freedom of the system. The discussion above suggests the following transcription of the probabilistic elements of classical nonequilibrium mechanics into qua...
emission process wherein bath bosons are resonantly created from oscillator bosons. Emission, too, enters the master equation as a sum of an in and an out process. A few more comments can be made on the structure of the master equation: The concerted appearance of in- and out- terms implies the conservation of the tra...
density operators ˆU ˆρ ˆU †, ˆU ≡ ˆU (t0, −t0). The trace over these operators, tr( ˆU ˆρ ˆU †) = tr( ˆU † ˆU ˆρ) =Z can be interpreted as a variant of the above partition function, where the curve [0, S] is parameterized in terms of a time argument running from −t0 to t0 and back. Pictorially, (cf. Fig. 11.1), the tr...
alternative representation of the block Green function Gtt →G tt ≡ U Gtt U † = = −ie−iω(t−t) 1 2 1 1 1 −1 G++ G+− G−+ G−− 1 + 2n(ω) Θ(t − t) −Θ(t − t) 0 ≡ 1 1 1 −1 tt GK G+ G− 0 , tt (11.19) where the three block Green functions are deﬁned by10 tt ≡ −iΘ(t − t)e−iω(t−t), G+ tt ≡ +iΘ(t − t)e−iω(t−t), G− tt ≡ −i(1 + 2n(ω)...
raction vertex at this stage in generality. Notice, however, that the interaction vanishes for ψq = 0, which is a manifestation of the vanishing of the action of purely classical ﬁelds.12 In later perturbative applications, it will 12 The particular two body interaction considered here also vanishes on a pure quantum c...
) ei C sC dt (pC ∂tqC −H(qC ,pC )) C sC Dq ei dt ( m 2 (∂tqC )2−V (qC )) Dq ei dt m∂tqq∂tqc−V 1√ 2 (qc+qq) )+V 1√ 2 (qc−qq) . (11.42) To explore the classical limit of this expression, we re-introduce as iS → i−1S. Alluding to the discussion of the quantum-nature of conﬁguration diﬀerences q+ − q− ∼ qq above, we also r...
1.9), and of the equilibrium density operator (11.10) remain unchanged, only that the normalizing factor is now given by Z0 = 1 + e−β(ω−μ). Again, we will want to represent the partition function in terms of (now fermionic) coherent states. It is straightforward to verify that the auxiliary identity Eq. (11.11) general...
ner transformed representation, ˆG(t1, t2) → G(, t), where t = (t1 + t2)/2 and is conjugate to Δt = t1 − t2. Technically, the Wigner representation of an arbitrary operator ˆg ≡ {g(t, t)} is deﬁned by the transform g(, t) ≡ dΔt eiΔtg(t + Δt/2, t − Δt/2), g(t1, t2) = d 2π e−i(t1−t2)g(, t). (11.63) In applications, we of...
ype are attracting a lot of current attention: advances in experimentation make it possible to explore the physics of nanoscopic systems that are strongly inﬂuenced by quantum interference phenomena – from ultracold atom physics, quantum information devices, and spintronics, to single molecule electronics, and beyond. ...
ger steps, i.e. the classical charge nc(t) is quantized. T EXERCISE If you ﬁnd the argument above too vague, you may derive charge quantization in rigorous terms. Do not ignore the W -dependence of the tunneling term. Rather, expand exp(iS) schematically to nth order in Stun. The winding numbers then enter as a pure ph...
o the dot are limited to short duration t ∼ E−1 . Tunneling into a metallic system leads to more intricate behavior. The tunneling process generates an initial charge distribution which is energetically unfavourable (sub-barrier). In the course of time, that initial distribution will relax in what typically will be a d...
e charge of the dot (left–hand side) to current ﬂow (right–hand side). Speciﬁcally, Cη ≡ δIL + δIR is to be interpreted as the sum of current ﬂuctuations through the left and right tunnel resistor. Each of the two contributions Ix, x = L, R is expected to express Poissonian shot noise statistics, Ix(t)Ix(t) = ¯Iδ(t − t...
l of currents (the Fano factor) signaled the presence of two tunneling barriers instead of just one. Statistical analysis of current ﬂuctuations – both experimental and theoretical – has become a prominent tool of diagnostics to explore the microscopic nature of condensed matter systems through observable transport coe...
Now, the transformations Λ → exp(iχσ1/2)Λ exp(−iχσ1/2) render the matrices Λx(χx) non-triangular. This means that stationary conﬁgurations Λd will in general no longer be of the form Eq. (11.76). However, the transformation by the counting ﬁeld does not alter the “normalization condition,” Λ2 = σ0 remains valid. It is...
phenomena. Practically all other theoretical nonequilibrium tools (at least those familiar to the authors of this text) can be recovered as restrictions of the Keldysh approach. Speciﬁcally, we have discussed the derivation of quantum master and quantum kinetic equations, the connection to equilibrium ﬁeld theory, the ...
cause a single quantum oscillator mode does not behave as a bath. Indeed, the mode–atom coupling is strongest at resonance, Δ = 0, when “system” and “bath” ﬂuctuate at comparable time scales. Technically this means that the (Markovian) assumption of constancy of the time evolution operator under the integral Eq. (11.10...
± before doing the integration, and mind the quantization condition nc(0) ∈ Z). to obtain the representation of the Keldysh partition function Z = ∞ m=0 − g 2 1 m! m {σ} Dt e−iEc dt (n2 +(t)−n2 −(t)) m' k=1 Lσ2k−1,σ2k (t2k−1 − t2k), (11.114) 11.9 Problems 761 where Dt = {σ} is a sum over all sign conﬁgurations σ ∈ {−1,...
aw, 549 Anderson Impurity Hamiltonian, 91 Anderson localization, 320 Anderson transition, 488 Anderson–Higgs mechanism, 294 angle resolved photoemission spectroscopy (ARPES), 366 annihilation operators, 45 anomalous dimension, 429, 439 anomaly, 175 antiferromagnetic exchange, 64 anyons, 42, 578 Arrhenius factor, 647 as...
ll eﬀect, 290 quantum Hall transition, 531 quantum harmonic oscillator, 21 quantum optics, 754 quantum phase transition, 478 quantum vacuum, 28 quantum wire, 67 quasi long–range order, 465 quasiclassical theory, 724 Rabi oscillations, 757 Raman spectroscopy, 365 random phase approximation, 217 random variable, 604 rand...
s when they try to solve a loop-the-loop problem using only an expression for centripetal force, perhaps with Newton’s laws, but ignore the fact that energy is conserved too. In electrodynamics it more often comes from e.g. starting with the wave equation (correctly) but failing to re-insert individual Maxwell equation...
hile the divergence theorem is: ∇ · ZV A d3x = A d2x ˆn · IS/V there is a “gradient theorem”: ∇f d3x = ˆnf d2x IS/V ZV and so on. (3.6) (3.7) To prove Jackson’s expression we might therefore try to ﬁnd a suitable product whose divergence contains J as one term. This isn’t too easy, however. The problem is ﬁnding the ri...
tude z that is the polar distance of the complex number from the \| complex origin. The usual polar angle θ can then be swept out from the positive real axis to identify the complex number on the circle of radius . \| This representation can then be expressed in trigonometric forms as + iy = cos(θ) + i \| (cos(θ) + i sin(...
ers associated with geometric algebras of higher grade. It’s characteristic deﬁning feature is that is is invariant under transformations of the underlying coordinate system. All of the following are algebraic examples of scalar quantities: x, 1.182, π, Ax, A B... · 1st rank tensor or vector. This is a set of scalar nu...
excellent desk reference for both undergraduate and graduate level classical physics in general! Section 27 in this book covers simple cartesian tensors, section 31 tensors deﬁned in terms of transformations. Schaum’s Outline series has a volume on vectors and tensors. Again an excellent desk reference, it has very co...
rms and then repeating the reasoning for B): n · E = 0 and n B = 0. · (9.24) Which basically means for a real unit vector n that E and B are perpendicular to n, the direction of propagation! A plane electromagnetic wave is therefore a transverse wave. This seems like it is an important thing to know, and is not at all ...
\| ⊥ Similarly for B: ˆn (9.60) (B0 + B′′0) · ˆn = B′0 · ˆn = 1 µ′ 1 µ ˆn × (B0 + B′′0) B′0 × (where, recall, B = (k E)/(vk) etc.) Again, one usually would not use this cross product algebraically, but would simply formulate the problem in a convenient coordinate system and take advantage of the fact that: E0\| v = √µǫ (...
media. 9.3.1 Static Case Recall, (from sections 4.5 and 4.6) that when the electric ﬁeld penetrates a medium made of bound charges, it polarizes those charges. The charges themselves then produce a ﬁeld that opposes, and hence by superposition reduces, the applied ﬁeld. The key assumption in these sections was that th...
h frequency behavior which is “universal”. 9.3.6 Low Frequency Behavior Near ω = 0 the qualitative behavior depends upon whether or not there is a “resonance” there. If there is, then ǫ(ω 0) can begin with a complex component that attenuates the propagation of EM energy in a (nearly static) applied electric ﬁeld. This ...
t z = ω + iδ where δ 0+ (or deform the integral a bit below the → singular point on the Re(ω) axis). From the Plemlj Relation: 1 ω = P 1 ω′ − (see e.g. Wyld, Arfkin). If we substitute this into the integral above along the real axis only, do the delta-function part and subtract it out, cancel a factor of 1/2 that thus ...
e used the ﬁrst result and substituted δ2 = 2/(µcωσ). This is a well-known diﬀerential equation that can be written any of several ways. Let κ2 = 2i δ2 . It is equivalent to all of: ( ∂2 ∂ξ2 + κ2)( ˆn × H c) = 0 ( H c) ˆn = 0 ∂2 ∂ξ2 + κ2)( ˆn × × ∂2 ∂ξ2 + κ2)H ∂2 ∂ξ2 + κ2)H c = 0 = 0 ( ( \|\| Where: as noted above. The s...
z ⊥ + ˆz ∇ Ez ⊥ × (10.55) (10.56) and ikB iωµǫ(ˆz ⊥ − ikB E ) = ∇ Bz ⊥ Bz = iωµǫ(ˆz ⊥ × ∇ i k2 ωµǫ B ⊥ − ⊥ − k ωµǫ ∇ ⊥ ⊥ Bz = ik(ˆz × E ) ⊥ × E ) ⊥ i k2 ωµǫ B ⊥ − k ωµǫ ∇ ⊥ Bz = iωB + ˆz ⊥ ∇ Ez ⊥ × or B E ⊥ ⊥ = = i µǫω2 − i µǫω2 − k2 (k∇ k2 (k∇ Bz + µǫω(ˆz ⊥ Ez − ⊥ ω(ˆz × × ∇ ∇ ⊥ Ez)) Bz)) ⊥ (10.57) (10.58) (10.59) (wh...
rier transform at a particular frequency of a general time dependent distribution; however, this is a very involved issue that we will examine in detail later in the semester. The form of the charge distribution we will study for the next few weeks is: ρ(x, t) = ρ(x)e− iωt (11.1) 107 J (x, t) = J (x)e− iωt. (11.2) The ...
the stationary wave Green’s function is just perfect. [As a parenthetical aside, you will often see people get carried away in the literature and connect the outgoing wave Green’s function for the IHE to the retarded Green’s function for the Wave Equation (fairly done – they are related by a contour integral as we shal...
cal frequencies. If the atoms are in a liquid or solid, there is a near ﬁeld interaction (implicitly alluded to in chapter 4 and 7) that may be important in determining optical dispersion and other observable phenomena. Only for microwave frequencies and less does the near zone become relevant on a macroscopic scale. F...
±L (r) For convenience, we deﬁne the following: JL(r) = jℓ(kr)YL(ˆr) NL(r) = nℓ(kr)YL(ˆr) H ±L (r) = h±ℓ (kr)YL(ˆr) (11.102) (11.103) (11.104) These are the basic solutions to the HHE that are also eigenfunctions of L2 and Lz. Clearly there is an implicit label of k (or k2) for these solutions. A general solution (on ...
ntire solid angle (as part of your assignment) you obtain the total power radiated: P = c2k4 12π s µ0 ǫ0 \| p 2 \| (11.132) The most important feature of this is the k4 dependence which is, after all, why the sky is blue (as we shall see, never fear). Example: A centerfed, linear antenna In this antenna, d << λ and I(z, ...
(or do the sum in the expression above) and show that: P = c2k6 960π s µ0 ǫ0 Q2 0. (11.162) At this point it should be clear that we are oﬀ on the wrong track. To quote Jackson: The labor involved in manipulating higher terms in (the multipolar expansion of A(x)) becomes increasingly prohibitive as the expansion is ext...
(12.25) If we know a solution to this equation for X = E we can obtain B from its curl from the equation above: ∇ B = i ω and vice versa. However, this is annoying to treat directly, because of the vector charactor of E and B which complicate the description (as we have seen – transverse electric ﬁelds are related to ...
reader) to verify that JzY m jj = mY m jj. (12.74) One simply plugs in the explicit form of Jz and commutes the resultant Lz with L to cancel the “spin” part. Solutions with j = ℓ If j = ℓ, we see from the equation after (12.62) that L Y = 0. To go further we have to go back to (12.62) and follow a diﬀerent line. If we...
ch is connected (as we shall see) to the continuity equation for charge and current. E and B are now (as usual) determined from the vector potential by the full relations, i. e. – we make no assumption that we are outside the region of sources: E = − B = ∇ ∇Φ − A, × ∂A ∂t (13.23) (13.24) Using the methods discussed bef...
we have developed above to obtain the radiation pattern of a dipole antenna, this time without assuming that it’s length is small w.r.t. a wavelength. Jackson solves more or less the same problem in his section 9.12, so this will permit the direct comparison of the coeﬃcients and constants in the ﬁnal expressions for ...
d(rh−ℓ′ (kr)) dr 1 kr q iˆr × ℓ(ℓ + 1) q ℓ(ℓ + 1)h−ℓ′ (kr)(ˆr Y m′ ∗ℓ′ℓ′ (ˆr) × +nLm∗L′ nLm∗L′ − nLn∗L′ − ℓ′(ℓ′ + 1) h−ℓ′ (kr) kr (Y m ℓℓ(ˆr) × ˆr)Y ∗L′(ˆr) ˆr) ℓ′(ℓ′ + 1) h−ℓ′ (kr) kr Y ∗L′(ˆr) (ˆr × Y m′ ∗ℓ′ℓ′ (ˆr)) (13.105) The lowest order term in the asymptotic form for the spherical bessel functions makes a contr...
if you prefer, short wavelengths (still long with respect to the size of the scatterer and only if the dipole term in the scattering dominates) are scattered more strongly than long wavelengths. This is the original “blue sky” theory and probably the origin of the phrase! To go further in our understanding, and to gai...
tion), we get: dσ dΩ = k4a6 ˆǫ0 − · 1 2 ( ˆn ˆǫ∗) ( ˆn0 × · × . (14.47) 2 ˆǫ0) ˆǫ∗ After much tedious but straightforward work, we can show (or rather you can show for homework) that: dσ k dΩ dσ ⊥ dΩ = = 2 1 2 cos θ − 1 2 cos θ 1 − k4a6 2 k4a6 2 2 so that the total diﬀerential cross section is: dσ dΩ = k4a6 5 8 (1 + co...
ntz Transformation To motivate the Lorentz transformation, recall the Galilean transformation between moving coordinate systems: vt x′1 = x1 − x′2 = x2 x′3 = x3 t′ = t (15.1) (15.2) (15.3) (15.4) (where K is ﬁxed and K ′ is moving in the 1–direction at speed v). Then Fj = m¨xj = m¨x′j = F ′j (15.5) or Newton’s Laws are...
er. Note that this does not mean that earlier (and later) events on each world line to not connect. The events are disconnected, not the world lines themselves. In this case, a suitable Lorentz transformation can make t′1 = t′2, but x′1 6 = x′2 always. lightlike separation S2 ⇒ 12 = 0 t2)2 = c2(t1 − Both events are on ...
telling us that selecting clever initial directions with an eye to simplifying the algebra “lacks motivation” he derives a result by selecting particular initial directions. The guy loves algebra, what can I say. Feel free to study his approach. It works. I, on the other hand, am too lazy to spend most of a period der...
k to completely determine a theory. In the part of this book devoted to mathematical physics is an entire chapter that discusses tensors, in particular the deﬁnitions of covariant and contravariant tensors, how to contract (Einstein sum) pairs of tensors to form tensors of lower rank, and the role of the metric tensor ...
educes the rank of the expression by one, the indices being summed must occur as a co/contra pair (in either order). If both are covariant, or both are contravariant, one or the other must be raised or lowered by contracting it with the metric tensor before contracting the overall pair! We use this repeatedly in the al...
he matrices L are called the generators of the linear transformation. Thus, whenever we write A = eL (16.64) where the L’s are (to be) the generators of the Lorentz group transformations we should remember what it stands for. Let’s ﬁnd the distinct L. Each one 4 real, traceless matrix that is (as we shall see) antisymm...
h, 2 too large. That is, the ﬁne–structure intervals observed in nature were only half the theoretically predicted values. If g = 1 was chosen instead, the splittings were correct but the Zeeman eﬀect was then normal (instead of anomalous, as observed). I don’t have the time to go into more detail on, for example, what...
y, the system is actually so complicated that this simple minded, single particle description itself is not really valid. (16.134) This is just a drop in the proverbial bucket of accelerated systems. Clearly, accelerated, relativistic systems have a much more involved structure than that described by the Lorentz transf...
I leave the rest of them to work out yourself. You should be able to show that: E′1 = E1 E′2 = γ(E2 − βB3) E′3 = γ(E3 + βB2) B′1 = B1 B′2 = γ(B2 + βE3) βE2) B′3 = γ(B3 − (16.171) (16.172) (16.173) (16.174) (16.175) (16.176) The component of the ﬁelds in the direction of the boost is unchanged, the perpendicular compon...
s rederive these results using only four–vectors and suitable scalar reductions. 17.1.2 The Elegant Way We can write the free particle Lagrangian using only scalar reductions of suitable 4–vectors: Lfree = − mc γ UαU α q (which is still mc2γ− − 1). The action is thus A = mc − τ1 τ0 q Z UαU αdτ. (17.20) (17.21) The vari...
ed the job of building a proper relativistic ﬁeld theory of electromagnetism. That is because we would like to be able to obtain all of the equations of motion (that is, physics) describing the system from a covariant action principle. We have done that for the particles in the ﬁelds, but what about the ﬁelds themselve...
eate photons (which must have quantized angular momenta), it is worthwhile to construct the symmetric stress tensor Θαβ = gαµFµλF λβ + 1 4 gαβFµλF µλ (17.85) in terms of which we can correctly construct a covariant generalization of the energy momentum conservation law ∂αΘαβ = 0 and the angular momentum tensor M αβγ = ...
causally connected retarded potential produced by a single charged particle as it moves in the absence of external potentials. This potential is “causal” in that the eﬀect (the potential ﬁeld) follows the cause (the motion of the charge) in time, where the advanced potential has the eﬀect preceding the cause, so to sp...
only care how that energy changes with distance. We will turn this into a rate equation by using the chain rule: Prad = e2 6πǫ0m2c3 dE dx dE dt dt dx (18.41) Thus the ratio of power radiated to power supplied by the accelerator Pacc = dE/dt is: Prad Pacc = e2 6πǫ0m2c3 1 v dE dx ≈ 1 6πǫ0 e2/mc2 mc2 dE dx where the latt...
of the dynamics. What is missing is radiation reaction. As charges accelerate, they radiate. This radiation carries energy away from the system. This, then means that a counterforce must be exerted on the charges when we try to accelerate them that damps charge oscillations. At last the folly of our ways is apparent. ...
), so that the energy that appears in the radiation ﬁeld is balanced by the work done by the radiation reaction force (relative to the total work done by an external force that makes the charge accelerate). We would like this force to vanish when the external force vanishes, so that particles do not spontaneously accel...
g to our discussion of dispersion but now with a physical model for why we expect there to be a damping term instead of a strictly phenomenological one. To simplify life, we consider a Lorentz “atom” to be an electron on a spring with constant k = mω2 0; a one–dimensional classical oscillator with a resonant frequency ...
α) , (4.45) where we have used the Maxwell equation ∂αF να = 0. The latter form shows that the diﬀerence between the two tensors has vanishing divergence, ∂ν∂α(AµF να) = 0, due to the antisymmetry of the ﬁeld-strength tensor. We can always add the term (4.45) to the conserved tensor in Eq. (4.44) to form (∂µAα − ∂αAµ) ...
.73) We see that for a photon gas the pressure is one-third of the energy density irrespective of the way that the energy is distributed over frequency (i.e., the radiation does not have to have a blackbody spectrum). 31 Part II Radiation of electromagnetic waves In this part we consider the production of electromagnet...
(6.9) Note that the ﬁrst contribution to the ﬁeld varies as 1/r2 while the second goes like 1/r. For a source oscillating at ω, the relative size of the 1/r part compared to the 1/r2 part is O(r/λ). We deﬁne the far-ﬁeld (sometimes known as the radiation zone or the wave zone) as the asymptotic region r λ. In the far-ﬁ...
A quantum description is necessary when the energy of an incident photon is comparable to the rest mass energy of the electron. In this limit, the scattered radiation undergoes a frequency shift due to recoil of the electron. 42 This is the eﬀective geometric cross-section of the electron to scattering. Note that it is...
(τ∗) as measured in the frame in which the charge is at rest at τ∗. Inspection of Eq. (8.26) reveals terms going like 1/R, that depend on the 4-acceleration, and a term going like 1/R2 that depends only on the 4-velocity of the charge. The 1/R terms describe radiation, while the 1/R2 term turns out to be the ﬁeld that ...
This means that the radiation is observed to have Fourier components up to roughly γ3ω, and so a relativistic particle emits a broad frequency spectrum of radiation up to γ3 times the orbital frequency. 53 Part III Electromagnetism in media So far in this course we have considered isolated charges and currents in vacuu...
f an electron, and is also the z-component of the magnetic moment due to orbital motion of an electron with angular momentum Lz = ¯h.) We see that, even for a strong ﬁeld of 1 T, the induced magnetic moment is only O(10−5)µB. 10 Macroscopic Maxwell equations 10.1 Averaging the microscopic Maxwell equations In classical...
der in a/lsmooth and the internal relative velocities, but retaining the magnetisation can be important in some situations, for example magnetostatics where the magnetic ﬁeld induces magnetic dipole moments but no electric multipole moments. There is one loose end to tidy up – the neglect of the molecular velocities vn...
ergence-free, which requires k · E0 = 0 and k · B0 = 0. To satisfy the wave equations (11.2), ω and k must be related by the dispersion relation ω2 = v2k2 . Finally, to satisfy ∇ × E = −∂B/∂t, the polarization vectors must be related by If all components of the wavevector are real14 we have B0 = ˆk × E0/v. B0 = k × E0/...
ed radiation will be linearly polarized with its electric ﬁeld normal to the plane of incidence. There is a simple physical way to understand the condition θI + θT = π/2 required for no reﬂection. Consider an incident wave in vacuum. The reﬂected wave is really the radiation produced by the oscillating electric dipole ...
ependent, allowing a prism to split white light into its constituent colours. A further consequence of a frequency-dependent vp comes from considering localised wave pulses or packets. These can be formed by superposing waves with a range of frequencies. Since the individual frequency components propagate at diﬀerent (...
n Sec. 13.2 the consequences of this model for wave propagation in conductors. 13.1 Drude model In a metal, the outermost electrons of the constituent atoms form a delocalised “sea” of mobile electrons surrounding a lattice of ions (nuclei plus tightly-bound inner electrons). These electrons move in all directions and ...
ansition energies the condition on ω0 is not so obviously met. 90 13.2.3 Plasma oscillations We now return to Eq. (13.13), which we can rewrite as 0eﬀ r (ω)k · E0(ω) = 0 . (13.29) As noted earlier, if eﬀ r (ω) vanishes for some ω, there are solutions with k · E0(ω) = 0. Moreover, in this case, Eqs (13.14) and (13.15) g...
ernative take on covariant and contravariant components of 4-vectors: In 3D Euclidean space we are used to thinking of vectors as geometric objects, the archetypal example being displacement vectors that connect two points in space. Let us generalise this idea to spacetime so that the 4-vector dx connects two neighbour...
2h = 1 √ 2 hx d x. 13 A statistician will complain that I am confusing the average of a ﬁnite sample (a million, in this case) with the “true” average (over the whole continuum). This can be an awkward problem for the experimentalist, especially when the sample size is small, but here I am only concerned with the true ...
a f dg . Under the integral sign, then, you can peel a derivative off one factor in a product, and slap it onto the other one—it’ll cost you a minus sign, and you’ll pick up a boundary term. 18 An “operator” is an instruction to do something to the function that follows; it takes in one function, and spits out some oth...
J (a, t) − J (b, t), J (x, t) ≡ i 2m ∂∗ ∂ x − ∗ ∂ ∂ x . What are the units of J (x, t)? Comment: J is called the probability current, because it tells you the rate at which probability is “ﬂowing” past the point x. If Pab(t) is increasing, then more probability is ﬂowing into the region at one end than ﬂows out at the...
associated separation constant (E1, E2, E3, . . . = {En}); thus there is a different wave function for each allowed energy: 1(x, t) = ψ1(x)e−i E1t/, 2(x, t) = ψ2(x)e−i E2t/, . . . . Now (as you can easily check for yourself) the (time-dependent) Schrödinger equation (Equation 2.1) has the property that any linear comb...
2 a sin x . nπ a (2.31) 11 That’s right: ψ(x) is a continuous function of x, even though (x, t) need not be. 12 Notice that the quantization of energy emerges as a rather technical consequence of the boundary conditions on solutions to the time-independent Schrödinger equation. 13 Actually, it’s (x, t) that must be no...
ur amplitude is greater than a/2, go directly to jail.) (d) Compute p. (As Peter Lorre would say, “Do it ze kveek vay, Johnny!”) (e) If you measured the energy of this particle, what values might you get, and what is the probability of getting each of them? Find the expectation value of H . How does it compare with E1 ...
h energy less than zero, which (according to the general theorem in Problem 2.3) does not exist! At some point the machine must fail. How can that happen? We know that ˆa−ψ is a new solution to the Schrödinger equation, but there is no guarantee that it will be normalizable—it might be zero, or its square-integral migh...
asymptotic behavior is the standard ﬁrst step in the power series method for solving differential equations—see, for example, Boas (footnote 16), Chapter 12. 33 According to Taylor’s theorem, any reasonably well-behaved function can be expressed as a power series, so Equation 2.80 ordinarily involves no loss of genera...
k 2m t −ik x+ k 2m t . + Be (2.94) Now, any function of x and t that depends on these variables in the special combination (x ± vt) (for some constant v) represents a wave of unchanging shape, traveling in the ∓xdirection at speed v: A ﬁxed point on the waveform (for example, a maximum or a minimum) corresponds to a ﬁ...
em is designed to guide you through a “proof” of Plancherel’s theorem, by starting with the theory of ordinary Fourier series on a ﬁnite interval, and allowing that interval to expand to inﬁnity. (a) Dirichlet’s theorem says that “any” function f (x) on the interval [−a, +a] can be expanded as a Fourier series: ∞ f (x)...
ro, and that’s why dψ/d x is ordinarily continuous. But when V (x) is inﬁnite at the boundary, this argument fails. In particular, if V (x) = −αδ(x), Equation 2.116 yields dψ d x = − 2mα 2 ψ(0) . For the case at hand (Equation 2.125), dψ/d x = −Bκe−κ x , for (x > 0) , dψ/d x = +Bκe+κ x , for (x < 0) , so dψ/d x so dψ/d...
e wave functions. (c) What are the bound state energies in the limiting cases (i) a → 0 and (ii) a → ∞ (holding α ﬁxed)? Explain why your answers are reasonable, by comparison with the single delta-function well. ∗∗ Problem 2.28 Find the transmission coefﬁcient, for the potential in Problem 2.27. 2.6 THE FINITE SQUARE ...
CHAPTER 2 Time-Independent Schrödinger Equation (a) What is the probability that it will “reﬂect” back, if E = V0/3? Hint: This is just like Problem 2.34, except that the step now goes down, instead of up. (b) I drew the ﬁgure so as to make you think of a car approaching a cliff, but obviously the probability of “bounc...
ound outside the well (x > a)? Answer: 0.542, so even though it is “bound” by the well, it is more likely to be found outside than inside! ∗∗∗ Problem 2.49 (a) Show that (x, t) = 1/4 mω π exp ! − mω −2iωt + it m " − 2x0xe−iωt satisﬁes the time-dependent Schrödinger equation for the harmonic oscillator potential (Equati...
r combination of two (probably unfamiliar) functions. Plot each of them, for (−15 < z < 5). One of them clearly does not go to zero at large z (more precisely, it’s not normalizable), so discard it. The allowed values of (and hence of E) are determined by the condition ψ(0) = 0. Find the ground state 1 numerically (in ...
but others seem to be more general, and it would be nice to prove them once and for all (the uncertainty principle, for instance, and the orthogonality of stationary states). The purpose of this chapter is to recast the theory in more powerful form, with that in mind. There is not much here that is genuinely new; the i...
int) of an operator ˆQ is the operator ˆQ† such that g for all f and g ˆQg ˆQ† f . = f (3.20) A hermitian operator, then, is equal to its hermitian conjugate: ˆQ = ˆQ†. ∗ g Problem 3.3 Show that if ˆQ f are equivalent). Hint: First let h = f + g, and then let h = f + ig. h for all f and g (i.e. the two deﬁnitions of “h...
(b) Check that f (x) = exp (x) and g(x) = exp (−x) are eigenfunctions of the operator d2/d x 2, with the same eigenvalue. Construct two linear combinations of f and g that are orthogonal eigenfunctions on the interval (−1, 1). Problem 3.8 (a) Check that the eigenvalues of the hermitian operator in Example 3.1 are real....
function: 1 = \| = 2! cn fn = n n n c∗ ncnδnn = n " ! "3 cn fn = n c∗ ncn = n \|cn\|2 . n n fn fn c∗ ncn (3.47) (3.48) Similarly, the expectation value of Q should be the sum over all possible outcomes of the eigenvalue times the probability of getting that eigenvalue: Q = n qn \|cn\|2 . Indeed, Q = ˆQ = 2! cn fn n ! " ˆQ n...
mmutator identities: ˆA, ˆC # ˆA + ˆB, ˆC # = # # $ $ $ # $ + $ ˆB, ˆC # , $ ˆA ˆB, ˆC = ˆA ˆB, ˆC + ˆA, ˆC ˆB. (b) Show that (c) Show more generally that x n, ˆp = inx n−1. f (x), ˆp = i d f d x , for any function f (x) that admits a Taylor series expansion. (d) Show that for the simple harmonic oscillator # $ ˆH , ˆa...
width of the mass plot, using the uncertainty principle as input. However, the point is valid, even if the logic is backwards. Moreover, if you assume the is about the same size −23 sec is roughly the time it takes light to cross the particle, and it’s hard to as a proton imagine that the lifetime could be much less th...
on for the inner product, α\|β, into two pieces, which he called bra, α\|, and ket, \|β (I don’t know what happened to the c). The latter is a vector, but what exactly is the former? It’s a linear function of vectors, in the sense that when it hits a vector (to its right) it yields a (complex) number—the inner product. (W...
(x, t) d x . Finally we recognize the integral as ( p, t) (Equation 3.54). Problem 3.30 Derive the transformation from the position-space wave function to the “energy-space” wave function (cn(t)) using the technique of Example 3.9. Assume that the energy spectrum is discrete, and the potential is time-independent. 124 ...

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