Review of Statistical Mechanics

Transcription

1 Review of Statistical Mechanics 3. Microcanonical, Canonical, Grand Canonical Ensembles In statistical mechanics, we deal with a situation in which even the quantum state of the system is unknown. The expectation value of an observable must be averaged over: O = i w i i O i (3.) where the states i form an orthonormal basis of H and w i is the probability of being in state i. The w i s must satisfy w i =. The expectation value can be written in a basis-independent form: O = T r {ρo} (3.) where ρ is the density matrix. In the above example, ρ = iw i i i. The condition, wi =, i.e. that the probabilities add to, is: T r {ρ} = (3.3) We usually deal with one of three ensembles: the microcanonical emsemble, the canonical ensemble, or the grand canonical ensemble. In the microcanonical ensemble, 8

2 Chapter 3: Review of Statistical Mechanics 9 we assume that our system is isolated, so the energy is fixed to be E, but all states with energy E are taken with equal probability: ρ = C δ(h E) (3.4) C is a normalization constant which is determined by (3.3). The entropy is given by, S = ln C (3.5) In other words, ( ) S(E) = ln # of states with energy E (3.6) Inverse temperature, β = /(k B T ): Pressure, P : where V is the volume. First law of thermodynamics: β ( ) S E V ( ) P S k B T V E (3.7) (3.8) ds = S S de + E V dv (3.9) de = k B T ds P dv (3.) Free energy: Differential relation: F = E k B T S (3.) df = k B S dt P dv (3.) or, S = ( ) F k B T V (3.3)

3 Chapter 3: Review of Statistical Mechanics while P = ( ) F V T E = F + k B ( T S ) F = F T T = T F T T V (3.4) (3.5) In the canonical ensemble, we assume that our system is in contact with a heat reservoir so that the temperature is constant. Then, ρ = C e βh (3.6) It is useful to drop the normalization constant, C, and work with an unnormalized density matrix so that we can define the partition function: Z = T r {ρ} (3.7) or, Z = a e βea (3.8) The average energy is: E = E a e βea Z a = β ln Z = k B T T ln Z (3.9) Hence, F = k B T ln Z (3.) The chemical potential, µ, is defined by µ = F N (3.)

4 Chapter 3: Review of Statistical Mechanics where N is the particle number. In the grand canonical ensemble, the system is in contact with a reservoir of heat and particles. Thus, the temperature and chemical potential are held fixed and ρ = C e β(h µn) (3.) We can again work with an unnormalized density matrix and construct the grand canonical partition function: Z = N,ae β(ea µn) (3.3) The average number is: while the average energy is: N = k B T µ ln Z (3.4) E = β ln Z + µk BT µ ln Z (3.5) 3. Bose-Einstein and Planck Distributions 3.. Bose-Einstein Statistics For a system of free bosons, the partition function Z = e β(ea µn) (3.6) E a,n can be rewritten in terms of the single-particle eigenstates and the single-particle energies ɛ i : E a = n ɛ + n ɛ +... (3.7) Z = {n i }e β( i n iɛ i µ i n i) = i ( n i e β(n iɛ i µn i ) )

5 Chapter 3: Review of Statistical Mechanics = i e β(ɛ i µ) (3.8) n i = The chemical potential is chosen so that e β(ɛ i µ) (3.9) N = i = i n i e β(ɛ i µ) (3.3) The energy is given by E = i = i n i ɛ i ɛ i e β(ɛ i µ) (3.3) when N is increased by increasing µ (µ always). Bose-Einstein condensation occurs N > n i (3.3) i In such a case, n must become large. This occurs when µ =. 3.. The Planck Distribution Suppose N is not fixed, but is arbitrary, e.g. the numbers of photons and neutrinos are not fixed. Then there is no Lagrange multiplier µ and n i = e βɛ i (3.33) Consider photons (two polarizations) in a cavity of side L with ɛ k = hω k = hck and k = π L (m x, m y, m z ) (3.34) E = nmx,m y,m z m x,m y,m z ω mx,m y,m z (3.35)

6 Chapter 3: Review of Statistical Mechanics 3 We can take the thermodynamic limit, L, and convert the sum into an integral. Since the allowed k s are π L (m x, m y, m z ), the k-space volume per allowed k is (π) 3 /L 3. Hence, we can take the infinite-volume limit by making the replacement: Hence, f( k) = k ( f( k) ( k) 3 k) 3 k d 3 k f( k) (3.36) = L3 (π) 3 E = V = V = kmax kmax V k 4 B π ( hc) 3 T 4 d 3 k (π) 3 d 3 k (π) 3 hω k e β hω k hck e β hck x 3 dx e x β hckmax (3.37) For β hck max, and For β hck max, and E = V k4 B π ( hc) T 4 x 3 dx 3 e x C V = 4V k4 B π ( hc) T 3 x 3 dx 3 e x (3.38) (3.39) E = V k3 max 3π k B T (3.4) C V = V k3 maxk B 3π (3.4) 3.3 Fermi-Dirac Distribution For a system of free fermions, the partition function Z = e β(ea µn) (3.4) E a,n

7 Chapter 3: Review of Statistical Mechanics 4 can again be rewritten in terms of the single-particle eigenstates and the single-particle energies ɛ i : E a = n ɛ + n ɛ +... (3.43) but now n i =, (3.44) so that Z = {n i }e β( i n iɛ i µ i n i) = i = i e β(n iɛ i µn i ) n i = ( ) + e β(ɛ i µ) (3.45) n i = The chemical potential is chosen so that e β(ɛ i+µ) + (3.46) The energy is given by N = i E = i e β(ɛ i+µ) + ɛ i e β(ɛ i+µ) + (3.47) (3.48) 3.4 Thermodynamics of the Free Fermion Gas Free electron gas in a box of side L: ɛ k = h k m (3.49) with k = π L (m x, m y, m z ) (3.5)

8 Chapter 3: Review of Statistical Mechanics 5 Then, taking into account the spin states, E = = V (3.5) ɛ mx,my,m z m x,m y,m z kmax d 3 k (π) 3 nmx,m y,m z h k m e β ( h k m µ ) + N = V kmax d 3 k (π) 3 ( e β h k m µ ) + (3.5) At T =, ( ( ) = θ µ h k ) (3.53) e β h k m µ m + All states with energies less than µ are filled; all states with higher energies are empty. We write k F = mµt =, ɛ F = µ T = (3.54) h N V = kf d 3 k (π) = k3 F (3.55) 3 3π For k B T e F, E V kf d 3 k h k = (π) 3 m = h kf 5 π m = 3 N 5 V ɛ F (3.56) 3 d 3 k (π) = m 3 π h 3 dɛ ɛ (3.57) N V = m 3 π h 3 = m 3 π h 3 µ dɛ ɛ e β(ɛ µ) + dɛ ɛ + m 3 π h 3 µ dɛ ɛ ( ) e β(ɛ µ) + + m 3 π h 3 µ dɛ ɛ e β(ɛ µ) +

9 Chapter 3: Review of Statistical Mechanics 6 = (m) 3 3 3π h 3 µ 3 m π h 3 3 µ 3 dɛ ɛ e β(ɛ µ) + + m π h 3 dɛ ɛ µ e β(ɛ µ) + k B T dx ( ) ( (µ + kb T x) e x (µ kb T x) ) + O e βµ + = (m) 3 3π h 3 µ 3 m + π h 3 = (m) 3 3 3π h 3 µ 3 (m) + π h 3 (k B T ) n µ 3 n Γ ( ) 3 n= (n )! Γ ( 5 dx xn n) e x + = (m) 3 3π h 3 µ ( ) kb T I + O ( T 4) (3.58) µ with We will only need Hence, (ɛ F ) 3 = µ 3 I k = + 3 dx I = π xk e x + (3.59) (3.6) ( ) kb T I + O ( T 4) (3.6) µ To lowest order in T, this gives: ( ) kb T µ = ɛ F I + O(T ) 4 ɛ F ( ) = ɛ F π kb T + O(T ) 4 (3.6) ɛ F E V = m 3 π h 3 dɛ ɛ 3 e β(ɛ µ) + = m 3 µ π h 3 dɛ ɛ 3 m 3 µ ( ) + π h 3 dɛ ɛ 3 e β(ɛ µ) + + m 3 π h 3 dɛ ɛ 3 µ e β(ɛ µ) + = (m) 3 3 5π h 3 µ 5 m µ π h 3 dɛ ɛ 3 e β(ɛ µ) + + m 3 π h 3 dɛ ɛ 3 µ e β(ɛ µ) + = (m) 3 3 5π h 3 µ 5 m k B T dx ( ) ( + π h (µ + 3 kb T x) 3 e x (µ kb T x) ) 3 + O e βµ + = (m) 3 3 5π h 3 µ 5 (m) + π h 3 (k B T ) n µ 5 n Γ ( ) 5 n= (n )! Γ ( 7 dx xn n) e x +

10 Chapter 3: Review of Statistical Mechanics 7 = (m) 3 5π h 3 µ 5 = 3 5 N V ɛ F + 5 ( ) kb T I + O(T ) 4 µ ( ) + 5π kb T + O(T ) 4 (3.63) ɛ F Hence, the specific heat of a gas of free fermions is: C V = π Nk k B T B (3.64) ɛ F Note that this can be written in the more general form: C V = (const.) k B g (ɛ F ) k B T (3.65) The number of electrons which are thermally excited above the ground state is g (ɛ F ) k B T ; each such electron contributes energy k B T and, hence, gives a specific heat contribution of k B. Electrons give such a contribution to the specific heat of a metal. 3.5 Ising Model, Mean Field Theory, Phases Consider a model of spins on a lattice in a magnetic field: H = gµ B B i S z i h i S z i (3.66) with S z i = ±/. The partition function for such a system is: The average magnetization is: Z = The susceptibility, χ, is defined by ( cosh h k B T S z i = tanh χ = ( h i S z i h k B T ) ) N (3.67) h= (3.68) (3.69)

11 Chapter 3: Review of Statistical Mechanics 8 For free spins on a lattice, χ = N k B T (3.7) A susceptibility which is inversely proportional to temperature is called a Curie sucseptibility. In problem set 3, you will show that the susceptibility is much smaller for a system of electrons. Now consider a model of spins on a lattice such that each spin interacts with its neighbors according to: H = JSi z Sj z (3.7) i,j This Hamiltonian has a symmetry S z i S z i (3.7) For k B T J, the interaction between the spins will not be important and the susceptibility will be of the Curie form. For k B T < J, however, the behavior will be much different. We can understand this qualitatively using mean field theory. Let us approximate the interaction of each spin with its neighbors by an interaction with a mean-field, h: H = i hs z i (3.73) with h given by h = i J S z i = Jz S z i (3.74) where z is the coordination number. In this field, the partition function is just cosh h k B T and S z = tanh h k B T (3.75) Using the self-consistency condition, this is: S z = tanh Jz Sz k B T (3.76)

12 Chapter 3: Review of Statistical Mechanics 9 S z i For k B T < Jz, this has non-zero solutions, S z which break the symmetry Si z. In this phase, there is a spontaneous magnetization. For k B T > Jz, there is only the solution S z =. In this phase the symmetry is unbroken and the is no spontaneous magnetization. At k B T = Jz, there is a critical point at which a phase transition occurs.