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3 total noise power of �2 after lowpass filtering. As long as no aliasing occurs, the total noise 12M power at the output of a compressor-by-M is the same as that of its input signal, so we know that the total noise power at the output of the system is Total noise power at output of system in OSB Figure 4.57 = �2...
https://ocw.mit.edu/courses/6-341-discrete-time-signal-processing-fall-2005/2e04b01fe7072772ba39f0c8dfb58306_lec06.pdf
ing a linear system to reduce the effects of quantization noise. Looking at OSB Figure 4.69 for further insight, the system works by first obtaining the error signal e[n], delaying it, and then using it to pre-compensate the input signal so that as long as e[n] is changing very slowly, it is mostly cancelled at y[n] b...
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is shown in OSB Figure 4.65, and a number of implementations of this type of cascaded system are discussed in OSB. Note that in practice, higher-order noise shaping systems are often designed to provide more detailed control of the resultant noise power spectrum. There are a number of practical issues associated wi...
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Table of Contents Lec 1 Introduction to Nanotechnology and Nanoscale Transport Phenomena. Microscopic Pictures of Heat Carriers Lec 2 Characteristic Time and Length, Simple Kinetic Theory, Characteristic Lec 3 Schrödinger Equation Lec 4 Quantum Wells, Harmonic Oscillators, Rigid Rotors, and Hydrogen Atoms Lec...
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1. Overview for nano sciences 1.1 Length scale 1.2 Examples in microtechnology 1.3 Examples in nanotechnology 1.4 Nano for energy (phonon, phonon, electron; wavelength, mean free path) 1.5 Nanoscale heat transfer in devices (e.g., CMOS) 1.6 Nano and microfabrication 1.7 Transport regimes 1.8 Overview of the boo...
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flux of radiation is evaluated as q =σ (T − T2 4 1 4 ) . T1 ε 1=1 T2 ε 2=1 2.2 Newton shear stress law The shear stress for the sketched one-dimensional flow is: τ = µ xy ∂ u ∂ y µ [N-s/m2] is dynamic viscosity. where: y τ xy x 2.57 Fall 2004 – Lecture 1 4 2.3 Fick’s diffusion law j = −ρD i dmi dx ...
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/ m , σ = 78mN / m into this equation, we get 3 2 r =1 m, γ = 8.4e4; r =1 mm, γ = 8.4e-2. mg 2.57 Fall 2004 – Lecture 1 5 4. Microscopic pictures of energy carriers 4.1 Heat 4.1.1 Heat conduction Gases: hotter air molecules (with larger kinetic energy) randomly pass their excess energy to cooler molecules. He...
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You can see similarities and why I said we can describe them in parallel. 2.57 Fall 2004 – Lecture 1 7 2.57 Nano-to-Macro Transport Processes Fall 2004 Lecture 2 5.1 Heat conduction Th Tc In last lecture, we describe electrons as free electron gas and lattice vibrations as phonon gas. Basically they are both...
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Lecture 2 1 8 5.4 Pressure and shear stress x As is shown in the figure, the velocities of gas molecules distribute randomly in all directions. Pressure is caused by their momentum changes normal to the wall. For one molecule, we have G G G F ma = = ( d mv ) dt or F = x m vx>0 − v ( x<0 = ) m∆v ∆t x . ∆...
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they travel? — How do they interact with each other? 6.1 How much energy/momentum can a particle have? Energy EKinetic Classical mechanics E = EKinetic+EPotential Quantum mechanics E is the eigenvalue of the Schrödinger equation E Translation = 2 mv 2 (for quantum case, I did not give answer but point out ...
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π = G | k | . 6.2 How many particles have the specified energy E? For a monoatomic ideal gas system, the only energy of each atom is their kinetic energy, m (v + v y + vz ). 2 x 2 2 E = 2 Statistical thermodynamics gives the probability density f(E), defined as the probability of finding the carriers at energ...
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dvx ∫ dvy ∫ -∞ -∞ ∞ 2 m (vx + v + vz 2 ) -∞ 2 2 y 3/2 ⎛ ⎞ m ⎟ exp ⎜ 2πκ T ⎠ ⎝ B ⎡ ⎢− ⎣⎢ 2 2 ( x m v + v2 + vz ) ⎤ y ⎥ dv 2κ BT ⎦⎥ z 3 = k T 2 B At room temperature (300 K), this average energy is 39 meV, or 6.21e-21 J. For He gas, m=6.4e-27 kg. Using 2 mv 2 = 3 2 B k T , we can calculate averag...
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ρ π D P k T B 2 πD PN A πD P R T u 2 Λ = = = , where the universal gas constant R = N k = 8.314 J mol K . / ⋅ u A B The ideal gas law can also be derived as following: In section 5.4, we derive P mn = 2 v 3 . In section 6.3, we get 2mv 2 = 3 2 B k T . Thus P = mn 2 v 3 = nk T = B N V B k T = ...
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theory In the figure, half of the carriers within vxτ can go across the interface before being scattered. Here vx is the x component of the random velocity of the heat carriers and τ is the relaxation time — the average time a heat carrier travels before it is scattered and changes its direction. So the net heat fl...
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Λ > d for thin films Λ=100 nm d=10~20 nm Disk drive Thin film k In example 2, we can further reduce the film thickness to enhance the size effects. With measured data for k and specific heat c, the mean free path in silicon can be estimated by k = cvΛ 3 , where v is sound velocity. The approximated mean free ...
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electron, D=1 mm, we have E = 2 n 8× 9.1 10 −31 × ( 6.6 ×10 −34 10−3 )2 ~ 10 −34 2 n << k T B = 4.14 ×10 −21 J at room temperature. 2) For D=1e-8 m, we calculate E k > B T . Further reducing D results in more observable E. 8.3 Fast transport For many materials, we have τ = 10−12 −10−11 s . Laser pulse can be ...
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sine and cosine functions, e.g. A e−i(ωt kx ) = A ⎡cos (ωt-kx ) − i sin (ωt kx )⎦⎤ . − ⎣ − 2.1.3 Standing wave x=0 x = π/k We can create a standing wave by superimposing two traveling waves along the positive and negative x-directions (assuming that the problem is linear such that the superposition principle ap...
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2.57 Fall 2004 – Lecture 3 16 To explain the blackbody radiation, Max Planck (1858-1948) introduced a radical hypothesis that the allowable energy of the electromagnetic field at a frequency ν is not continuous, but is a multiple of the following basic energy unit E = nhν , in which h is called the Planck constant...
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long wavelength. The first proof of the wave properties of particles came from the electron diffraction experiment, in which peak signals are observed in specified incident angles. Transmission Electron Microscope (TEM) A TEM uses thermal excitation or applying a high voltage to draw electrons from the tip end. Th...
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t = Ψ ∇ t i= − , 2 which is similar to heat conduction equation k T = ρ c ∇ 2 ∂ T ∂ t but the magic imaginary unit “i” really gives rise to wave behavior. 2.57 Fall 2004 – Lecture 3 18 Schrödinger himself did not come up with a correct explanation for the meaning of wavefunction. The right explanation was g...
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� 2 + U −= 2 = m2 + U = −= + U ⎜ ∂ 2 ⎝ ∂x + ∂ ∂y2 + ∂ ∂z 2 + U ⎞ ⎟ ⎟ ⎠ Note: in equation < Ω >= ∫ Ψ ΩΨ dV , you cannot switch Ψt gradient operator ∇ and the Laplace operator ∇ 2 . t * t * and Ψt if Ω contains the Standard deviation: Similar to x ∆ = n 1 − ∑ ( x − < > 1 n x i 1 = 2 ) , ...
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2 ⎢− Ψ ⎢⎣ 2m where E is a constant (eigenvalue) since Ψ depends on r only and Y depends on t only. Its meaning will be explained later ( H = E ). Solving for Y leads to ⎤ UΨ⎥ = i= ⎥ ⎦ 1 dY Y dt 2 + Ψ ∇ = E , Y C1 exp −i = E ⎤ ⎡ ⎥ = C1 exp [−iωt] . t ⎢⎣ = ⎦ And the governing equation for Ψ(r) is called the...
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principle states < ∆ >< ∆ >≥ x p = 2 ; < ∆ >< ∆ t E >≥ = 2 . 2.3 Example solutions: G G ( , ) ( ) by the boundary conditions and will not consider the u r t u r Here we determine case. 2.3.1 Free particles in 1D In this case, there are no constraints for the particles. The potential energy u=0 so that E ...
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Same as the free particles, the solution for first equation is still −ikx Ψ = Ae + Be ikx where k = , mE 2 2 = = 2 mE = = p = . The general boundary conditions are the continuity of the wave functions and their first derivatives at the boundaries. The latter derives from the continuity of particle flu...
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Finally we get A i= 1 2D . 2.3.2 Particle in a 2D box y D O U= 0 U = ∞ x D We establish a coordinate system as shown in the schematic above. Clearly, outside the potential well, we have Ψ = 0 because U = ∞ . We thus focus on the solution inside the potential well. The Schrödinger equation inside the well ...
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. Applying these boundary conditions, we see that kx = nπ D (n = 1, 2,3 ") . Similarly, k y = lπ D Thus (l = 1, 2,3 ") . 2.57 Fall 2004 – Lecture 4 23 (A + n )π = 2 2 2 2 2 2mD E An = and ( A ,n=1,2,…) ⎛ Aπy ⎞ ⎛ nπx ⎞ . sin Ψ = C sin ⎜ ⎟ ⎜ ⎟ ⎝ D ⎠ ⎝ D ⎠ A...
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, however, does not tell the whole story on the quantum state of a particle. For example it cannot distinguish the spin of particles. For electrons, corresponding to each wavefunction obtained from the Schördinger equation, there are two quantum states (or two relativistic wavefunctions), which are usually denoted b...
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interaction between the nucleus and the orbiting electron is governed by the Coulomb force F = − 1 e2 4πεo r 2 = − du dr , where εo=1.124x10-10/4π [C2/(m2N)] is the electrical permittivity of the vacuum. It yields u r( ) = − 2e 4πε0r . Using separation of variables, we assume ΨnAm = RnA (r)YA energy level...
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orbitals 1s n=1 (- 13.6 eV) Note: the energy gap between different n values is much larger than the thermal energy (kBT~26 meV) and it is almost impossible to thermally excite electrons to a higher n level. A stable element is obtained only if all the orbitals for the highest n are completely filled, such as He....
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obtain from solving Schrodinger equation. Similarly, we can deal with electrons moving around the nucleus but changing the potential energy to u = e2 4πε0 xn . Finally we get E ∝ 1 n2 . Rigid rotation G G In classic mechanics, the angular momentum is expressed as r p . However, because of the uncertainty princ...
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n=1 U=0 x Energy has discrete levels, and we have one quantum number n. E = 2h n2 2 8m D (n=1,2,…) For 2D constraints, we have two quantum numbers n and l. In the discussions, the conception of “degeneracy” is introduced. 3. Spin 1 For electrons, we have talked about s = ± 2 , where s=1/2 is called spin up ...
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. The energy of 3d is lifted up above 4s because of the electron-electron interaction. For the element potassium (K), it has 19 electrons but the n=3 energy levels are not totally filled and one electron goes to the 4s orbit. 2.57 Fall 2004 – Lecture 5 29 2.3.5 Energy Quantization observation Absorption or emissi...
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(n f − ni );ν = k 1 2π m , where the reduced mass is m = m m2 = 1 m m2 +1 m1 ( m = m2 ), and the selection rule states 2 1 ni n f − = ± 1. Positive for absorption and negative for emission. We can measure the vibrational frequency from which, to deduce the spring constant. For hydrogen, the 2.57 Fall 2004 – ...
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=0,1,2, …). 2= I 2 hν = hB [ where B=1.8e12 Hz. Similarly we have A f − A i = ±1, 0 . As an example, A f − = 1 (absorption) gives νp = 2B l +1) . The normal wavelength for rotational energy is around 100 µm (far infrared regime), which is much larger than the emission wavelength at the room temperature (10 µm). A i ...
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of electrons. ( + ( 3.1.2 Possible approaches First we will review the problems solved in Lecture 4. (1) Free electron E p2 / 2m = 2 (2) Quantum well E ∝ n 2.57 Fall 2004 – Lecture 5 32 ENERGY AND WAVEFUNCTION U ∞ = n=3 n=2 n=1 U=0 x (3) Electron-nucleus system (hydrogen atom) En u r el = − 13.6 eV 2n ...
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let us go back to the quantum wells. Due to the finite depth, the wave functions will not be zero at the boundary like standing waves. Instead, they will decay exponentially from the interface. However, this indicates two wavefunctions overlap in the middle of the barrier, which conflicts with the Pauli exclusion pr...
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Born-von Karman periodic boundary condition. The Born-von Karman boundary condition deals with the end points of a crystal. Ordinarily, we would think that the two end points are different from the internal points. For many applications, however, distinguishing the boundary points from the internal points is not ne...
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same quantum state and should be counted only once. Thus, rather than plotting the energy eigenvalues for all the wavevectors, we can plot them in one period, as shown in the subsequent figure. This way of representation is called the reduced-zone representation. Often, only half of the band, [0,π /a], needs to be ...
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z i l a m r o N 0 -1 0 k / ( π /a) 1 2.57 Fall 2004 – Lecture 5 37 2.57 Nano-to-Macro Transport Processes Fall 2004 Lecture 6 Quick review of Lecture 5 ( , [x n ) ] = Ψ x ( ) ikn a +b) e In the last...
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condition is no longer valid when N becomes very small in nanomaterials. / o E E y g r e n E n o r t c e l E d e z i l a m r o N 15 10 5 0 - 1 0 1 k / (π/a) 2.57 Fall 2004 – Lecture 6 38 3.1.3 Consequences of solid energy levels we just obtained: (1) Electrons wave function extends through the whole...
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levels is small. Thus, these electrons can flow freely, making the materials good electrical conductors, which is the case for metals. 2.57 Fall 2004 – Lecture 6 39 Energy Energy Energy Energy Energy Energy Energy Energy Energy Energy Energy Energy Energy Energy Energy Energy Energy Energy Energy Conduction Co...
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trinsic Intrinsic Intrinsic Semiconductor Semiconductor Semiconductor n-type n-type n-type Semiconductor Semiconductor Semiconductor p-type p-type p-type Semiconductor Semiconductor Semiconductor If the valence electrons exactly fill one or more bands, leaving others empty, the crystal will be an insulator at zero tem...
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, x y , z talk about the electron filling in a 3D structure. (2) The shape of bands can be affected by heavily doping. (3) The lowest band starts from a nonzero energy, which is the consequence of the uncertainty principle. Impurities are added to most semiconductors and these impurities have energy levels somewher...
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) ( m = E k ) ( m + + 1 ∂ 2 E 2 ∂ k 1 =2 2 m * 2 |km (k − k )2 m (k − k )2 , m where the effective mass is defined as m * = 2= (∂ 2 E k ) |km / ∂ 2 . In differential geometry, the / ∂ 2 term 1/ ∂ 2 E k is just the curvature. Thus, effective mass is proportional to local curvature at band maxima or minim...
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Direct Eg k Eg k In above figures, we demonstrate the idea of direct and indirect band gaps. In a direct band structure, both the minimum in the conduction band and the maximum in the valence band occur at the same location of k (=0 for the example given). A good example for direct band gap is GaAs with Eg =1.4...
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between the nearest neighbors. Second, the interaction force between atoms is a harmonic force (which obeys Hook's law). This can be justified as we have done for harmonic oscillators. Now consider a typical atom j. The displacement of atom j from its equilibrium position o jx is 2.57 Fall 2004 – Lecture 6 43 ...
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2 K m sin ka 2 . Na (j-1)a ja (j+1)a Note: When k approaches zero for large wavelength, the frequency becomes a linear K ka m 2 function of the wavevector, i.e. ω≈ 2 . We can calculate the sound K m = ka velocity by v sound = dω dk . ω k π/a In last two lectures, we have derived the allowed k val...
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,2,3,…), in which ω = 2 K m sin ka 2 . The basic vibrational energy quanta, hν, is called a phonon. Comparison between electrons, phonons, and photons: (1) Electrons obey the Pauli exclusion principle, which says that each quantum state can only have at most one electron. Photons and phonons are not limited ...
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ons in 3D crystals a. Each direction is different b. Two transverse waves, one longitudinal wave Transverse Longitudinal c. For m atoms per basis, we have 3 acoustic waves and 3(m-1) optical waves. d. Each kx, ky, kz represents a normal mode. e. The energy dispersion (E-k relationship) can be totally different ...
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, i.e. crystal = lattice + basis. 2.57 Fall 2004 – Lecture 7 48 From a mathematical point of view, the location of each point can be described by a vector. Due to the periodic arrangement of lattice points, we can choose a basic set of vectors ...
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lattice points in such a rectangle---the center point plus the four corners, each of the latter is shared by four cells. Because the choices of primitive lattice vectors are not unique, there can be different ways to draw a primitive unit cell, as the two examples in the figure. One method to construct a unit cell u...
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6 3 5e −10 9 ) = e8 , which indicates the material should be similar to bulk crystal. (3) Miller index The Miller indices of crystal planes (hkl) are obtained in accordance with the following steps: (a) Find the intercepts of the crystal plane with the axes formed by the lattice vectors a1, a2, a3 in terms of the ...
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y x 2.57 Fall 2004 – Lecture 7 51 3.1.3 Bonding potential Interatomic potential Φ Repulsion x Attraction Recall the previous discussion about the interatomic forces. Here we will learn them in more details. The force interaction between atom...
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– Lecture 7 52 What makes a crystal structure a favorable structure is that the total potential energy of the system ⎛ 1 B ⎜ U = ∑ ⎜ 12 2 i≠ j rij ⎝ − ⎞ ⎟ 1 A 6 ⎟ 2 rij ⎠ 12 ⎛ ⎞ ⎛ ⎜ ⎜ σ ⎟ = ∑ 4ε⎜ ⎟ ⎜ ⎜ ⎟ ⎜⎜ r i≠ j ⎝ ij ⎠ ⎝ − ⎛ ⎞ ⎜ σ ⎟ ⎟ ⎜ ⎜ ⎟ r ⎝ ij ⎠ 6 ⎞...
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R (r) = U e o U where B, ζ, and Uo are empirical constants determined from experimental data, such as the interatomic spacing and the binding energy. (Born-Mayer) Interatomic potential Φ Repulsion x Attraction 2.57 Fall 2004 – Lecture 8 54 Combining this attractive potential (van d...
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silicon, and germanium are all covalent crystals. Each atom has four electrons in the outer shell and forms a tetrahedral system of covalent bonds with four neighboring atoms, as indicated in the figure. 1 1 1 4 4 4 , , ) atom is shared by neighboring atoms. Note: For instance, in silicon the ( 2.57 Fall 2004 – L...
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i.e. t expanded into a Fourier series as, f ( + ) = f ( )t t T ), its can be ∞ ⎛ ⎛ 2πn ⎞⎞ ⎛ 2πn ⎞ f (t) = ∑ ⎜⎜ a n sin⎜ t ⎟⎟⎟ t ⎟ + bn cos⎜ T T ⎠ ⎠⎠ ⎝ ⎝ n = −∞ ⎝ = ∑( + n ' b e a n e ' −inωt inωt ∞ ) n =−∞ Here the angular frequency ω=2π/T is the Fourier conjugate of the time periodicity such that eiωT...
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In the above figure, the electron energy dispersion shows a period of 2π/a because of the periodic potential field ( + ) = u x ( ) (recall the Bloch theorem). u x a Note: (1) The discussed function f(x) can represent not only the charge density but also other periodically distributed properties. (2) The Born-von ...
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. For the one-dimensional case, we have G =2πn/a, r =x. With the above definitions, we can show that u(r) is indeed invariant with any translational lattice vector in the real space, T(=n1a1+n2a2+n3a3), where n1, n2 n3 integers, u( + ) = u e • + • ir G iT G i(r T+ •) G = r T ∑ G u e G ∑ G G ir G π n m n m ...
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an incident wave from the source along direction k. The incident wave is , where r is any point in the sample and rs is the location of the proportional to e source relative to the origin of coordinates. The wave scattered into the detector is then proportional to ( )r , where kf is the propagation direction of t...
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of the source, the sample, and the detector. The reciprocal lattice vectors G, and thus the crystal structure, can be determined from diffraction experiments. − f i θ a 2.57 Fall 2004 – Lecture 8 58 Consider the special set of crystal planes s...
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,…) For wavefunction Ψ ,n s (2) Harmonic oscillator The energy is , we have degeneracy g(n)=2 due to the spin. E = hν(n +1/ 2);ν = n 1 K 2π m (n=0,1,2…) The wavefunction is Ψ , and the degeneracy is g(n)=1. (3) Rigid rotation The energy eigenvalues are n E l = 2= I 2 A A + 1) = hB A A + 1) ( ( (for |m|≤ A ...
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10 5 0 -1 0 1 k / (π/a) ω Debye k π/a In the Debye approximation, we have energy dispersion as ω= vk = v 2 k x 2 + k y 2 k + z ; E n = hν(n + 1 2 ) , where v is sound velocity. Density of (quantum mechanical) states (DOS): (a) Electron ky π4 L π2 L 0 Volume of One Unit Cell k+dk dk π2...
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2m * k = ( 2 2π dE 2π = 1 2 2 )3/ 2 (E E ) c − 1/ 2 . E l e c t r o n D e n s i t y o f S t a t e s Ec EF E (b) Phonon According to the Debye model that assumes three modes (two transverse, one longitudinal) are identical, we have 2.57 Fall 2004 – Lecture 9 62 D( ) ω = 3 3 ∆N = 2 V ∆...
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� 1 ⎛ 3π2 ⎜ in which the electron density n=N/V. E f n = ∫ E c (E − E c ) ) ( Ed E D 3 / 2 3 / 2 = f , For nonzero temperatures, Ef is replaced by the chemical potential. Nanostructures: z D For a thin film with thickness d, we have the energy as ( k E x k , y n , ) = 2 2 = k xy m 2 * + n 2 = 2π2...
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just like jumps at different Enl k . − nl ( kz (n) D(E ) kx bulk (3D) quantum well (2D) n=3 n=2 n=1 E ky Chapter 4 Statistical thermal & energy storage In Lecture 2, we have mentioned that matter tends to occupy the lowest energy levels. For an energy level Ei, we have given the Boltzmann factor for its ...
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applied to closed systems, while for an open system exchanging energy with the outside, the probability becomes P E Ni ) = Aexp[ −(E − µN ) / k T ] , where Ni is the particle number, chemical potential µ is the criteria for the equilibrium state of mass exchanging process with the outside, just as pressure for mech...
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ν k T B ) = 1 exp(− hν 1 B2k T 1− exp( − ) A A = exp( )(1 − exp( − hν 2k T B = 1. )) . ) hν k TB hν k T B And (P E) = exp( − )[1 − exp( − )] . nhν k T B hν k T B Then the average number of the phonons, or the occupancy of the quantum state is < >= ∑ nP E ( ) , n ∞ n n=0 = − exp( − [1 )]∑ n exp(...
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number of quantum states per unit interval of energy and per unit volume D E( ) = 1 ∆N V E ∆ = k 2 dk k 2 ∆k = 2π2 ∆E 2π2 dE . A factor that considers polarization of waves may be added (electron, spin up and down, thus a factor of 2, photon, two polarizations, phonons, 3 polarizations) For an energy level Ei,...
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= 1 k T ( E −µ ) / B i e , −1 where µ is again the chemical potential of the boson gas. The Bose-Einstein distribution changes the “plus one” in the denominator of the Femi- Dirac distribution into minus one. When E − µ  k T , we can ignore the ±1 term in the denominator. Both distributions reduce to the Boltzma...
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to the temperature. For constrained electrons, we can find that the ground state (quantum numbers n l = = = ) has much larger occupancy than any other energy level. This is more apparent at lower temperatures. Therefore, most molecules will go to the ground state when the temperature approaches 0 K. The phenomenon ...
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1 x2, P2 Quantum State Ω x3, P3 Microcanonical ensemble U, V, N fixed Isolated systems Canonical ensemble V, N, T fixed In contact with a thermal reservoir; isothermal. equal of 1/ iP = Ω Principle probability: Boltzmann principle gives S The entropy S is additive. S U V N , ) Ω = ln k B = ( ...
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ln = B becomes potential in this case. the thermal Grand canonical V, µ, T fixed isothermal system. Open, Exchanging both energy and particles with the reservoir. P E N ) , i i ( = ( − E i e − µ N i ) / k T B Z The numerator is the Gibbs factor and = ∑∑ ( − Z e k T B . N i E i µ...
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z )× exp ⎛ E − µ ⎞ ⎜ − ⎟ k T ⎝ ⎠ B ∑∑∑ ( x z ) E k k k , exp , y k x k y k z = ∑∑∑ exp ⎜ − ⎝ k z k k y x ⎛ E ⎞ ⎜ − ⎟ k T B ⎠ ⎝ , ⎛ E ⎞ ⎟ k T B ⎠ If the energy separation is very small (quasi-continuous), we can evaluate Z by integration instead of discrete summation, i.e. Z = ∑∑∑ exp ⎜ − ⎝...
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κBT/2 to the average energy of the system. The internal energy is 2.57 Fall 2004 – Lecture 10 71 u = k N T = R T , A B u 3 2 3 2 in which Ru is the universal gas constant. The specific heat is CV = ∂u ∂T 3 = Ru . 2 In the following figure we draw the specific heat of hydrogen gas. At low ...
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� dω 2.57 Fall 2004 – Lecture 10 73 2.57 Nano-to-Macro Transport Processes Fall 2004 Lecture 11 4.1.1 Photons (continue) First let us continue the discussion of photons. We have n = f ( ω T = ) , 1 =ω/ k TB e . −1 The internal energy is u = ∑∑∑ ωf ( = ω , k x k y k z T = ) ∞ ∫0 =ωf ( ω T D ( ) , ω dω...
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, we obtain uλ = dω dλ uω = 2πc λ 2 uω , which accounts for the power difference in above two expressions. Therefore, we have obtained the exact Planck’s blackbody radiation law. The radiation intensity, defined as energy flux per unit solid angle and normal area, is calculated as cu λ , 4π Iλ = where 4π i...
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3 2 2 π v D 3 ∫ exp 0 = ω3dω (= ) − ω/κ T 1 B . Note: For one monoatomic chain with N atoms, we have N quantum states. Similarly, for N atoms in a crystal, we have 3N quantum states (three acoustic branches) for phonons. This yields 1 V kx ω 3ω2 ∫ 2π vD 0 ∑∑∑ ∫ 1 = ω D D ) d ω ω ( ω 3 D 2 2π vD...
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k T , which is consistent with C = 3 B N V B result is similar to the ideal gas case. A common Cv-T curve is drawn as following. V k . This CV Cv ~T3 CV → const Temperature In many sources, people use the specific heat per unit mass instead of per unit volume. This normally causes a factor difference. In general...
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y ⎞ ⎛ dkz ⎞ µ,T ) ⎟ ⎜ ⎝ 2 /π L ⎠ ⎝ 2 /π L ⎠ ⎝ 2 /π L ⎠ ⎜ ⎟ ⎟ ⎜ ∞ , 0 ) f E ( , , )µ T D E dE ( = ∫ where N is the total number of electrons, and we use density of state to rewrite the summation in an integral form. The chemical potential µ is solved if N/V is given. The specific heat is derived as T 1 N 2 ...
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77 ) . K g k / J ( T A E H C F C E P S I I 103 102 101 100 10-1 10-2 10-3 1 PHONON ELECTRON 100 10 TEMPERATURE (K) 1,000 Chapter 5 Energy transport by waves Consider energy transported between two points. The net energy transfer rate is ( q = 12 E f T E D E ...
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. Based on the Schrödinger equation 2∇ Ψ + (U − E)Ψ = 0 , − = 2m we have Ψ = Ae− ω ( i i t k ) − 1x Ψ = Be− ω ( i t k 1 ) + x r Ψ = Ce− ω i t k x ( − 2 ) t (incoming wave), (reflected wave), k = 1 (transmitted wave), k = 2 2mE 2= 2 ( m E u − ) 2 = . r i x=0+ The boundary conditions are applied ...
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) 2 A k1 , where we use the fact that k1 is a real number in the last step. 2.57 Fall 2004 – Lecture 11 79 2.57 Nano-to-Macro Transport Processes Fall 2004 Lecture 12 5.1 Plane waves & their interface reflection (continue) Transmission wave Reflection wave Incoming wave Energy b...
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e (ω − * ) t k x 1 | ⎞ x=0 ⎟ ⎠ * ( 1 − 1 ) x i k k | x=0 ) = = = = m = m = m 2 A * 1 Re(k e ) * k1 A 2 Re( 2 A k1 , where we use the fact that k1 is a real number in the last step. Similarly, we have 2.57 Fall 2004 – Lecture 12 80 2B k1 (negative ...
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occur for all the three types of carriers (phonons, electrons, photons) and the descriptions of these phenomena are also similar. JJG An electromagnetic wave in vacuum is characterized by an electric field vector, E , and a magnetic field vector, H . Consider a pair of charged particles placed in an electrical fiel...
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which is superimposed onto the external magnetic field. A measure of the total magnetic field inside the medium is called magnetic induction, B (N.s m-1.C-1), B=µH where µ is the magnetic permeability. The propagation of an electromagnetic wave is governed by the following Maxwell equations: (1) ∇ × E = − ∂B ...
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� 2E ∂t ∂E ∂t denotes the current. Without µσ ∂E ∂t , the equation ∇2E=µε ∂ 2E ∂t 2 is just a regular wave function. The additional term µσ ∂E ∂t corresponds to damping and is also called dissipation term. By solving equations (1)-(5), we obtain the following results for EM waves E(r,t) = Eoexp[-i(ωt- k • ...
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2 = εµω can be rewritten as 2 2 2 2 ω ω c0 ω 2 N , 2 = = k = 2 2 c 0 2 c 2 c 0 2 c or k = Nω co . One can prove that the electromagnetic wave is a transverse wave, and that the electrical and magnetic fields are perpendicular to each other, i.e., E⊥H⊥k 2.57 Fall 2004 – Lecture 12 83 ...
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along direction ki (wave vector direction) and meets an interface with norm nˆ . The reflected wave and refracted wave propagates along the kr and kt directions, respectively. We call the plane formed by ki and nˆ as the plane of incidence, and the angle formed between nˆ and ki as the angle of incidence. n1 n2 E...
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⎞⎤ ⎟⎥ ⎥ ⎠⎦ , λ o ⎛ E// t exp ⎢−i ⎜ωt − n2 2π ⎝ ⎡ ⎢ ⎣ x sinθt + z cos θt ⎞⎤ ⎟⎥ , ⎠⎥ ⎦ λo where in the second equation the negative sign before z cos θr indicates different propagation direction (upward in the z direction) from the incoming and refraction waves. The subscript “//” means that the electric field...
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��⎦ o r r = cosθt E// t exp ⎢−iω⎜ t − 2 ⎡ ⎢ ⎣ ⎛ ⎝ n x sinθ ⎞⎤ t ⎟⎥ ⎠⎥ ⎦ co where the above equation is valid only when the exponents are equal because x can take any value. Thus we have 2.57 Fall 2004 – Lecture 12 85 i r 2 1 n sinθ = n sinθ = n sinθ t 1 which ...
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2θ + sin 2θ t t t , Note: (1) For an incident light shown as below (e.g. from air to water), in this case the refraction light will be bended and the object in the second media will looks higher if the observer is in the first media. (2) If n2<0, n1>0, the refraction light will be bended as the right figure. Howev...
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k1 + k 2 r − = t = = Ψ r Ψi 2k1 k1 + k 2 R = −J / J = r 2 i B A Re (k2 ) )k1 * * Jt = Ji Re ( T = 2 = k k2−1 k1 + k2 R// = 2 t T// = Normal incidence onto an interface − n2 + n1 n2 + n1 E// r E// i r// = = t // = 2n1 n2 + n1 Sr z Sr , = Si z Si , St z = Re ⎜ , Si z , = 2 r// ⎛ n2 cos θ ...
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angle, the Snell law gives, sinθ = t n1 sinθi > 1, n2 and thus, ⎛ n1 sin θi ⎟ ⎞ cosθt = 1− sin θt = i ⎜ n2 ⎠ ⎝ 2 2 −1 = ai . In the wave function of the transmitted wave, the imaginary cos θt leads to an exponential decay wave ⎡ ⎛ E// t exp ⎢ −iω⎜ t − ⎢ ⎝ ⎣ 2 n x sinθ + n z cos θt ⎞⎤ ⎟⎥ co ⎠⎥⎦ 2 t E=...
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adding layers that have a low refractive index than the core. 2.57 Fall 2004 – Lecture 13 89 Electrode Cladding Layer Core Cladding Layer Quantum Wells Electrode 2) Brewster angle π 2 When θ θi+ = t , we have the electric wave r// = 0 , but the magnetic wave r⊥ ≠ 0 . This incident angle θ is called the Bre...
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the phase Refraction wave factor. The energy flow is still JK JJK * S = Re ( E × H ). 1 2 Incident wave Amplitude contours 4) Acoustic waves n Recall the Newton’s law K JK , F ma = JK K K = ⋅ where the force is F n K u x= − x0 . We have K K K d x du v = . dt dt = σ, acceleration is a = . Denote u ...
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wave When the incident wave is polarized in the direction parallel to the plane of incidence, one longitudinal wave and one transverse wave are excited for both reflection and transmission waves. If the materials are anisotropic, one more transverse is excited for both reflection and transmission waves. Incident w...
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of the electric field E ( ) and y-component of ( ) on the interface are related by 2×2 matrix A, 2×2 matrix M1, 2 the magnetic field H y ×1 matrix B. And we obtain ⎛ Ei ⎞ ⎜ ⎟ ⎝ Er ⎠ = A M BEt . −1 1 x z 2.57 Fall 2004 – Lecture 13 93 (0) ⎞ ⎛ E x ⎜ ⎜ H (0) ⎝ y ⎛ Ei ⎞ ⎟ E r ⎠ ⎟⎟ = A⎜ ⎝ ⎠ x (0) ⎞ ⎛ E...
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2 12 r 23 r 23 cos ϕ2 r 2 12 2 2 r r cos ϕ2 + 2 23 , 12 in which the cosine function indicates periodicity of R. This is just the interference effect. Discussions: (1) Periodic variation in R R φ Note: In microfabrication, the color of a thin film will change periodically according to the thickness, ...
https://ocw.mit.edu/courses/2-57-nano-to-macro-transport-processes-spring-2012/2e4ecaa5cf55f03bcefbc8ccce79aed6_MIT2_57S12_lec_notes_2004.pdf
4 ( U m 2 o − d E ) / =] = τ or ≈ τ o − E) exp[− 2 16 ( U E 2U o ( U m 2 ) o − d E 16 / =] = ( o − E) − k 2 2 U E e 2 Uo d . 2.57 Fall 2004 – Lecture 13 95 2.57 Nano-to-Macro Transport Processes Fall 2004 Lecture 14 Review of last lecture d R n1 n2 n3 φ In lecture 13, we talked abou...
https://ocw.mit.edu/courses/2-57-nano-to-macro-transport-processes-spring-2012/2e4ecaa5cf55f03bcefbc8ccce79aed6_MIT2_57S12_lec_notes_2004.pdf