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Physics 451 - Statistical Mechanics II - Course Notes

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Physics 451 - Statistical Mechanics II - Course Notes Physics 451 - Statistical Mechanics II - Course Notes David L. Feder January 8, 2013 Contents 1 Boltzmann Statistics (aka The Canonical Ensemble) 3 1.1 Example: Harmonic Oscillator (1D) . 3 1.2 Example: Harmonic Oscillator (3D) . 4 1.3 Example: The rotor . 5 1.4 The Equipartition Theorem (reprise) . 6 1.4.1 Density of States . 8 1.5 The Maxwell Speed Distribution . 10 1.5.1 Interlude on Averages . 12 1.5.2 Molecular Beams . 12 2 Virial Theorem and the Grand Canonical Ensemble 14 2.1 Virial Theorem . 14 2.1.1 Example: ideal gas . 15 2.1.2 Example: Average temperature of the sun . 15 2.2 Chemical Potential . 16 2.2.1 Free energies revisited . 17 2.2.2 Example: Pauli Paramagnet . 18 2.3 Grand Partition Function . 19 2.3.1 Examples . 20 2.4 Grand Potential . 21 3 Quantum Counting 22 3.1 Gibbs' Paradox . 22 3.2 Chemical Potential Again . 24 3.3 Arranging Indistinguishable Particles . 25 3.3.1 Bosons . 25 3.3.2 Fermions . 26 3.3.3 Anyons! . 28 3.4 Emergence of Classical Statistics . 29 4 Quantum Statistics 32 4.1 Bose and Fermi Distributions . 32 4.1.1 Fermions . 33 4.1.2 Bosons . 35 4.1.3 Entropy . 37 4.2 Quantum-Classical Transition . 39 4.3 Entropy and Equations of State . 40 1 PHYS 451 - Statistical Mechanics II - Course Notes 2 5 Fermions 43 5.1 3D Box at zero temperature . 43 5.2 3D Box at low temperature . 44 5.3 3D isotropic harmonic trap . 46 5.3.1 Density of States . 46 5.3.2 Low Temperatures . 47 5.3.3 Spatial Profile . 48 5.4 A Few Examples . 50 5.4.1 Electrons in Metals . 50 5.4.2 Electrons in the Sun . 50 5.4.3 Ultracold Fermionic Atoms in a Harmonic Trap . 51 6 Bosons 52 6.1 Quantum Oscillators . 52 6.2 Phonons . 53 6.3 Blackbody Radiation . 56 6.4 Bose-Einstein Condensation . 59 6.4.1 BEC in 3D . 59 6.4.2 BEC in Lower Dimensions . 60 6.4.3 BEC in Harmonic Traps . 62 Chapter 1 Boltzmann Statistics (aka The Canonical Ensemble) This chapter covers the material in Ch. 6 of the PHYS 449 course notes that we didn't get to last term. In particular, Sec. 1.1 here corresponds to Sec. 6.2.4 in the PHYS 449 course notes. 1.1 Example: Harmonic Oscillator (1D) Before we can obtain the partition for the one-dimensional harmonic oscillator, we need to find the quantum energy levels. Because the system is known to exhibit periodic motion, we can again use Bohr-Sommerfeld quantization and avoid having to solve Schr¨odinger'sequation. The total energy is p2 kx2 p2 m!2x2 E = + = + ; 2m 2 2m 2 p where ! = pk=m is the classical oscillation frequency. Inverting this gives p = 2mE − m2!2x2. Insert this into the Bohr-Sommerfeld quantization condition: I I p pdx = 2mE − m2!2x2dx = nh; where the integral is over one full period of oscillation. Let x = p2E=m!2 sin(θ) so that m2!2x2 = 2mE sin2(θ). Then I r Z 2π p 2E 2 2E 1 2πE pdx = 2 2mE cos (θ)dθ = 2π = = nh: m! 0 ! 2 ! So, again making the switch E ! n, we obtain h! = n = n¯h!: n 2π The full solution to Schr¨odinger'sequation (a lengthy process involving Hermite polynomials) gives 1 n = ¯h!(n + 2 ). Except for the constant factor, Bohr-Sommerfeld quantization has done a fine job of determining the energy states of the harmonic oscillator. 3 PHYS 451 - Statistical Mechanics II - Course Notes 4 Armed with the energy states, we can now obtain the partition function: X X Z = exp(−n=kBT ) = exp(−βn) = 1 + exp(−β¯h!) + exp(−2β¯h!) + :::: n n But this is just a geometric series: if I make the substitution x ≡ exp(−β¯h!), then Z = 1 + x + x2 + x3 + :::. But I also know that xZ = x + x2 + x3 + :::. Since both Z and xZ have an infinite number of terms, I can subtract them and all terms cancel except the first: Z − xZ = 1, which immediately yields Z = 1=(1 − x), or 1 Z = : (1.1) 1 − exp(−β¯h!) Now I can calculate the mean energy: 2 2 @ ln(Z) NkBT @Z 2 [1 − exp(−β¯h!)] ¯h! U = NkBT = = NkBT 2 (−1) 2 (−1) exp(−β¯h!) @T Z @T [1 − exp(−β¯h!)] kBT exp(−β¯h!) N¯h! = N¯h! = : 1 − exp(−β¯h!) exp(β¯h!) − 1 1 = N¯h!hn(T )i; where hn(T )i ≡ is the occupation factor: exp(¯h!=kBT ) − 1 At very high temperatures T 1, exp(¯h!=kBT ) ≈ 1 + (¯h!=kBT ), so hn(T )i ! kBT=¯h! and U(T 0) ! NkBT and CV (T 0) ! NkB: Notice that these high-temperature values are exactly twice those found for the one-dimensional particle in a box, even though the energy states themselves are completely different from each other. 1.2 Example: Harmonic Oscillator (3D) By analogy to the three-dimensional box, the energy levels for the 3D harmonic oscillator are simply nx;ny ;nz = ¯h!(nx + ny + nz); nx; ny; nz = 0; 1; 2;:::: Again, because the energies for each dimension are simply additive, the 3D partition function can 3 be simply written as the product of three 1D partition functions, i.e. Z3D = (Z1D) . Because 3 almost all thermodynamic quantities are related to ln (Z3D) = ln (Z1D) = 3 ln (Z1D), almost all quantities will simply be mupltiplied by a factor of 3. For example, U3D = 3NkBT = 3U1D and CV (3D) = 3NkB = 3CV (1D). One can think of atoms in a crystal as N point masses connected to each other with springs. To a first approximation, we can think of the system as N harmonic oscillators in three dimensions. In fact, for most crystals, the specific heat is measured experimentally to be 2:76NkB at room temperature, accounting for 92% of this simple classical picture. It is interesting to consider the expression for the specific heat at low temperatures. At low temperature, the mean energy goes to U ! 3N¯h! exp(−¯h!=kBT ), so that the specific heat approaches 3N¯h! ¯h! ¯h! 2 ¯h! CV ! − 2 (−¯h!) exp − = 3NkB exp − : kBT kBT kBT kBT PHYS 451 - Statistical Mechanics II - Course Notes 5 This expression was first derived by Einstein, and shows that the specific heat falls off exponentially at low temperature. It provided a tremendous boost to the field of statistical mechanics, because it was fully consistent with experimental observations of the day. Unfortunately, it turns out to be 3 wrong: better experiments revealed that CV / T at low temperatures, not exponentially. This is because the atoms are not independent oscillators, but rather coupled oscillators, and the low-lying 3 excitations are travelling lattice vibrations (now known at phonons). Actually, even CV / T is wrong at very low temperatures! The electrons that can travel around in crystals also contribute to the specific heat, so in fact CV (T ! 0) / T . 1.3 Example: The rotor Now let's consider the energies associated with rotation. In classical mechanics, the rotational kinetic energy is 1 T = ~! · I · ~!; 2 where I is the moment of inertia tensor and ~! is the angular velocity vector. In the inertial ellipsoid, this can be rewritten L2 L2 L2 T = x + y + z ; 2Ixx 2Iyy 2Izz where Lj is the angular momentum along direction ^and Ijj is the corresponding moment of inertia. Suppose that we have a spherical top, so that Ixx = Iyy = Izz = I: 1 L2 T = L2 + L2 + L2 = : 2I x y z 2I In the quantum version, the kinetic energy is almost identical, except now the angular momentum is an operator, denoted by a little hat: L^2 T = : 2I The eigenvalues of this operator are `(` + 1)¯h2=2I, where ` = −L; −L + 1; −L + 2;:::;L − 1;L so that ` can take one of 2L + 1 possible values. For a linear molecule (linear top), the partition function for the rotor can then be written as 1 L 1 X X `(` + 1)¯h2 X L(L + 1)¯h2 Z = exp − ≈ (2L + 1) exp − ; 2IkBT 2IkBT L=0 `=−L L=0 where the second term assumes that the contributions from the different ` values are more or less equal. This assumption should be pretty good at high temperatures where the argument of the exponential is small. In this case, there are simply 2L + 1 terms for each value of L. Again, because we are at high temperatures the discrete nature of the eigenvalues is not important, we can approximate the sum by an integral: Z 1 L(L + 1)¯h2 Z ≈ (2L + 1) exp − dL: 0 2IkBT PHYS 451 - Statistical Mechanics II - Course Notes 6 We can make the substitution x = L(L + 1) so that dx = (2L + 1)dL, which is just term already in the integrand. So the partition function becomes Z 1 2 ¯h x 2IkBT Z = exp − dx = 2 : 0 2IkBT ¯h Again, I can calculate the mean energy 2 2 2 @ ln(Z) NkBT @Z 2 ¯h 2IkB U = NkBT = = NkBT 2 = NkBT: @T Z @T 2IkBT ¯h This is exactly the contribution that we expected from the equipartition theorem: there are two ways the linear top can rotate, so there should be two factors of (1=2)NkBT contributing to the energy. For a spherical top, each of the energy levels is (2L + 1)-fold degenerate. The partition function for the rotor can then be written as 1 L 2 1 2 X X `(` + 1)¯h X 2 L(L + 1)¯h Z = (2L + 1) exp − ≈ (2L + 1) exp − ; 2IkBT 2IkBT L=0 `=−L L=0 where the second term assumes that the contributions from the different ` values are more or less equal.
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