sign in sign up
<<
Home , Canonical ensemble, Grand canonical ensemble, Grand potential

4. Classical quantum states in the part of the space under consideration. 4.1. Phase space and probability density The ensemble or statistical set consists, at a given We consider a system of N particles in a d-dimensional moment, of all those phase space points which correspond space. Canonical coordinates and momenta to a given macroscopic system. Corresponding to a macro(scopic) state of the system q = (q1,...,qdN ) there are thus a set of micro(scopic) states which belong

p = (p1,...,pdN ) to the ensemble with the probability ρ(P ) dΓ. ρ(P ) is the probability density which is normalized to unity: determine exactly the microscopic state of the system. The phase space is the 2dN-dimensional space (p, q) , dΓ ρ(P )=1, whose every point P = (p, q) corresponds to a possible{ } Z state of the system. A trajectory is such a curve in the phase space along and it gives the local density of points in (q,p)-space at which the point P (t) as a function of time moves. time t. Trajectories are determined by the classical equations of The statistical average, or the ensemble expectation motion value, of a measurable quantity f = f(P ) is dq ∂H i = f = dΓ f(P )ρ(P ). dt ∂p h i i Z dp ∂H i = , We associate every phase space point with the velocity dt ∂q − i field ∂H ∂H where V = (q, ˙ p˙)= , . ∂p − ∂q   H = H(q ,...,q ,p ,...,p ,t) 1 dN 1 dN The probability current is then Vρ. The probability = H(q,p,t)= H(P,t) weight of an element Γ0 evolves then like is the Hamiltonian function of the system. ∂ The trajectory is stationary, if H does not depend on ρ dΓ= Vρ dS. ∂t Γ − ∂Γ · time: trajectories starting from the same initial point P Z 0 Z 0 are identical. d S Trajectories cannot cross: if trajectories meet at a point, they must be identical!

Let F = F (q,p,t) be a property of the system. Now G 0 dF ∂F = + F, H , dt ∂t { } Because where F, G stands for Poisson brackets { } Vρ dS = (Vρ) dΓ, ∂Γ · Γ ∇ · ∂F ∂G ∂G ∂F Z 0 Z 0 F, G . continuity equation { }≡ ∂q ∂p − ∂q ∂p we get in the limit Γ0 0 the i i i i i → X   ∂ We define the volume measure of the phase space ρ + (Vρ)=0. ∂t ∇ · dN dqidpi dN According to the equations of motion dΓ= = h− dq dq dp dp . h 1 · · · dN 1 · · · dN i=1 Y ∂H q˙i = 34 Here h =6.62608 10− Js is the . (Often ∂pi dΓ has 1/N! in front· to remove degeneracy caused by ∂H p˙i = .) − ∂qi Note: [dq dp]= Js, so dΓ is dimensionless. Note: In classical dynamics the normalization is we have ∂q˙ ∂p˙ irrelevant. In quantum mechanics, it is natural to choose i + i =0, ∂q ∂p h = Planck constant. i i Note: ∆0Γ = 1 corresponds to the smallest possible so we end up with the incompressibility condition volume element of the phase space where a point (sourceless) representing the system can be localized in accordance ∂q˙ ∂p˙ with the QM uncertainty principle. The volume V = i + i =0. ∇ · ∂q ∂p ∆Γ = dΓ is then roughly equal to the number of i i i X   R 1 From the continuity equation we get then Ergodic flow: Almost all points of the surface ΓE are sometimes arbitrarily close to any point in ∆ΓE . ∂ρ 0 = + (Vρ) ∂t ∇ · ⇔ ∂ρ The flow is ergodic if f(P ), f(P ) ”smooth enough”, = + ρ V + V ρ ∀ ∂t ∇ · · ∇ ¯ ∂ρ f = f E = + V ρ. h i ∂t · ∇ holds. Here f¯ is the time average When we employ the convective time derivative 1 T f¯ = lim dt f(P (t)) d ∂ T T = + V →∞ Z0 dt ∂t · ∇ and f the energy surface expectation value ∂ ∂ ∂ E = + q˙ +p ˙ , h i ∂t i ∂q i ∂p i  i i  1 X f E = dΓE f(P ). h i ΣE the continuity equation can be written in the form known Z as the Liouville theorem We define the so that its density distribution is d ρ(P (t),t)=0. dt 1 ρE(P )= δ(H(P ) E). ΣE − Thus, the points in the phase space move like an incompressible fluid, and the local probability density ρ Every point of the energy surface belongs with the same remains constant during evolution. Different points probability to the microcanonical ensemble. ∂ρE naturally can have different ρ. The microcanonical ensemble is stationary, i.e. ∂t =0 and the expectation values over it temporal constants. 4.2. Flow in phase space The mixing flow is such an ergodic flow where the points of an energy surface element dΓ disperse in the The (constant) energy surface ΓE is the manifold E determined by the equation course of time all over the energy surface. Ifρ ˆE(P,t) is an arbitrary non stationary density distribution at the H(q,p)= E. moment t = t0, then 1 If the energy is a constant of motion, every phase point lim ρˆE(P,t)= δ(H(P ) E)= ρE(P ) i t ΣE − P (t) moves on a certain energy surface ΓEi, of dim. →∞ (2Nd 1). and The expectation− value of the energy of the system

lim f = lim dΓˆρE(P,t)f(P ) t h i t E = H = dΓ Hρ →∞ →∞ Z h i Z = dΓ f(P )ρE(P ) is also a constant of motion. Z = f The volume of the energy surface is h iE i.e. the density describing an arbitrary (non equilibrium) Σ = dΓ = dΓ δ(H(P ) E). state evolves towards a microcanonical ensemble under E E − Z Z mixing flow. The volume of the phase space is Liouville theorem (volume element(s) are conserved) + mixing: a phase space infinitesimal volume element

∞ becomes infinitely stretched and folded and is distributed dΓ= dE ΣE. over the full energy surface. Z Z−∞ Let us consider the time evolution of a surface element 4.3. Microcanonical ensemble and ∆ΓE of an energy surface. If the total energy of a macroscopic system is known Non-ergodic flow: In the course of time the element exactly and it remains fixed, the equilibrium state can be ∆ΓE traverses only a part of the whole energy described by a microcanonical ensemble. The surface ΓE. corresponding probability density is Examples: periodic motion; presence of other 1 constants of motion. ρE(P )= δ(H(P ) E). ΣE −

2 For a convenience we allow the energy to have some A) Let us now show that the maximisation of the Gibbs ”tolerance” and define entropy gives us microcanonical ensemble, when the energy of the system is restricted to (E, E + ∆E): 1 ρE,∆E(P )= θ(E + ∆E H(P ))θ(H(P ) E). ZE,∆E − − δS = k dΓ(δρ ln ρ + ρδ ln ρ) − B Here the normalization constant Z∆ΓE = k dΓδρ(ln ρ +1)=0. − B Z = dΓ θ(E + ∆E H(P ))θ(H(P ) E) ∆ΓE E,∆E − − Z Z Because δ1= dΓδρ = 0, above condition is satisfied if is the microcanonical state sum or partition function. ln ρ =const. or ρ(P ) =const., when P ∆Γ . R ∈ E ZE,∆E is the number of states contained in the energy S is indeed maximum: slice E

S = k ∆Γ ρ ln ρ = k ρ ∆Γ ln ρ ∆Γ − B i i i − B i i i i i i X X = k p ln p , − B i i i X since ln ∆Γi =0. If ρ is smooth in the range ∆Γ = W we have 1 ρ = , W so that 1 1 S = k ln dΓ. − B W W Z We end up with the Boltzmann entropy

S = kB ln W.

Here W is the thermodynamic probability: the number of all those states that correspond to the macroscopic state of the system.

3 5. Quantum mechanical ensembles Unitary operator U: UU † = U U = 1. Thus, • Uψ = ψ and U correspond† to a rotation in || || || || 5.1. Quantum mechanical concepts Hilbert space, n′ = U n . Unitary transformations correspond to a| changei | ofi eigenbasis. States and Hilbert space Trace of an operator: Quantum mechanical states (using Dirac notation, ψ ) of • a system form a Hilbert space , which is a linear vector| i Tr A = n A n = ak , space with inner (scalar) productH and associated norm. h | | i n k Linearity: if ψ , ψ′ , then cψ + c′ψ′ , for X X | i | i ∈ H | i ∈ H c,c′ C. where the first form is independent of the choice of The∈ inner product is a mapping C, with basis ketvn, and the second is obtained by choosing h·|·i H⊗H→ eigenvectors of A. ψ φ = φ ψ ∗ h | i h | i Tr AB = Tr BA, Tr U †AU = Tr A ψ αφ + α′φ′ = α ψ φ + α′ ψ φ′ h | i h | i h | i φ φ 0 h | i≥ If the quantum mechanical state of the system is ψ , φ φ =0 φ =0 • | i h | i ⇔ the projection operator corresponding to the state is ρ = ψ ψ . The norm is φ = φ φ 1/2. | i h | Time evolution|| || is definedh | i by Schr¨odinger equation: Now expectation values of observables are d A = Tr ρA . i¯h ψ(t) = H ψ(t) . h i dt | i | i Thus, trace in quantum mechanics corresponds to If H does not depend on t, the formal solution is phase space integral in classical mechanics: i ψ(t) = exp[ H(t t0)] ψ(t0) Tr A dΓA(q,p). | i −¯h − | i ⇐⇒ Z

Operators, eigenvalues and trace Systems of identical particles Let 1 be a Hilbert space for one particle. Then the Physical observable quantum mechanical operator H • A, ψ A ψ . → Hilbert space for N particles is | i → | i N 1 1 1 Conjugated operator A†: = . • H H ⊗ H ⊗···⊗H N copies ψ A† φ = φ A ψ ∗ h | | i h | | i | {z } If, for example, x 1 is a position eigenstate the | ii ∈ H Eigenvalue and -vector: A a = a a . If A† = A N-particle state can be written as • (hermitean), a is real. Eigenvectors| i | i form a complete Ψ = (orthonormal) basis of the Hilbert space. | i dx dx x ,..., x ψ(x ,..., x ), Basis of : any vector can be written as a sum of · · · 1 · · · N | 1 N i 1 N • eigenvectorsH of some operator: Z Z Z where ψ = ψ n or ψ = daψ a | i n | i | i a | i x ,..., x = x x x . n 1 N 1 2 N X Z | i | i ⊗ | i⊗···⊗| i depending on whether the eigenvectors form a There are two kinds of particles: discrete or continuous set. Here we shall use the The wave function is symmetric with respect to former notation. the exchange of particles. Identity operator I = n n . • n | i h | The wave function is antisymmetric with Here Pn = n n is thePprojection operator to vector respect to the exchange of particles. n . | i h | | i Note: If the number of translational degrees of freedom Spectral representation of operator: if A n = an n , is less than 3, e.g. the system is confined to a two • | i | i dimensional plane, the phase gained by the many particle A = n a n . wave function under the exchange of particles can be | i n h | n other than 1. Those kind of particles are called . X ±

4 The Hilbert space of a many particle system is not the Occupation number operator of state ℓ is whole N but its subspace: | i H nˆℓ = aℓ†aℓ N = ( 1 1) symm. = SHN S H 1 ⊗···⊗H 1 with eigenvalues 0, 1, 2,... for bosons and 0, 1 for H = ( ) antisymm.  AH A H ⊗···⊗H fermions. Thus, antisymmetry of states Pauli exclusion principle. ⇒ For non-interacting systems the Hamilton operator can be Fock space enables us to describe many-particle quantum expressed as states with creation and annihilation operators. Formally, H = Eℓnˆℓ Fock space is the direct sum of all (anti)symmetrized Xℓ N-particle spaces: when the 1-particle states are eigenstates of H.

= (0) (1) . . . (N) . . . F Hp ⊕ Hp ⊕ Hp ⊕ 5.2. Density operator and entropy Let be the Hilbert space a many particle system. with (N) = (N) for bosons ( for fermions). Fock TheH probability measure tells us the weight that a state Hp SH A space wave function in coordinate representation is a ψ represents a system with given macroscopical vector |properties.i ∈ H The density operator ρ tells us the probability (1) (2) (2) Ψ = (C, ψ1(ξ1 ), ψ2(ξ1 , ξ2 ),...) of a given state, p = φ ρ φ . φ h | | i where C is a complex number, ξ = (x ,s ) (s labels the i i i i Ensemble spin and other internal degrees of freedom), and ψ is a N The quantum mechanical ensemble or statistical set can fully (anti)symmetric wave function. be defined in a similar fashion as in classical mechanics. Fock space normalized N-particle states can be written as Statistical macrostate is the state determined by macroscopic parameters, microstate is a particular QM n1,n2,...,nℓ,... | i state in Hilbert space. where n1,... are 1-particle state occupation numbers. In Pure state and mixed state N-particle state ℓ nℓ = N. In coordinate space representation these states can be written as If the quantum mechanical state of a system is fully P known it is in a pure state. In this case the density 1 operator is a (pure) projection operator ξ1,...,ξN n1,... = h | i N! n ! i i ρ = Ψ Ψ | i h | ǫP ξ1 ℓ1 .p . . ξNQℓN , × h | i h | i P (ξ1,...ξN ) of a pure state reduces into normal X quantum mechanics; for example A = Tr ρA = Ψ A Ψ h i h | | i where ℓ1,ℓ2,...ℓN contain the 1-particle indices of the Mixed state: only the probability pi that the system is N-particle state; i.e. it contains n1 times 1, n2 times 2 in state ψi is known. Density operator is, in orthonormal etc. Sum is over all permutations P of N coordinates, base, and ǫP is 1 for bosons, and 1 for fermions for even/odd ρ = ψ p ψ ± | ni n h n| permutations (Slater determinant). n X We can define creation and annihilation operators: Now Tr ρ = 1. Ensembe expectation values are a n ,...,n ,... = ( 1)Pℓ √n n ,...,n 1,... ℓ | 1 ℓ i − ℓ | 1 ℓ − i A = Tr ρA = pi ψi A ψi = pi Ai . P h i h | | i h i a† n ,...,n ,... = ( 1) ℓ √1 n n ,...,n +1,... i i ℓ | 1 ℓ i − ± ℓ | 1 ℓ i X X The a priori probability: if there is no knowledge of the with upper/lower sign for bosons/fermions, and P =0 ℓ actual state of the system every state in can taken with for bosons, and equal weight. Then H Pℓ = nk 1 Xk<ℓ ρ = n n , | i h | for fermions. N n The symmetry properties and normalization imply for X where = dim . bosons N H Properties of the density operator [a ,a ] = [a†,a† ]=0, [a ,a† ]= δ , l k l k l k l,k Density operator can be any operator with properties and for fermions ρ† = ρ a ,a = a†,a† =0, a ,a† = δ , { l k} { l k} { l k} l,k ψ ρ ψ 0 ψ h | | i ≥ ∀ | i ∈ H ([A, B] AB BA, A, B = AB + BA). Tr ρ = 1. ≡ − { }

5 The density operator associates with every normalized 1. S 0, because 0 pα 1. ψ the probability ≥ ≤ ≤ | i ∈ H 2. S = 0 corresponds to a pure state, i.e. α : p =1 ∃ α p = Tr ρP = ψ ρ ψ . and pα′ =0 α′ = α. ψ ψ h | | i ∀ 6 3. If the dimension of the Hilbert space is finite, Since ρ is hermitean there exists an orthonormal basis N H α for , where ρ is diagonal the entropy has a maximum when {| i} H 1 ρ = p α α . ρ = I α | i h | α N X 1 or pα = α . Then Here N ∀ | i ∈ H 0 pα 1 S = k ln . ≤ ≤ B N and Thus, in this case the entropy fully corresponds to pα =1. classical Boltzmann entropy S = kB ln W . X In this basis 4. The entropy is additive. If we have 2 (independent) systems, the total Hilbert space is A = Tr ρA = p α A α . h i αh | | i α = X H1+2 H1 ⊗ H2

The equation of motion and correspondingly Let us assume that the ensemble gives (and fixes) the ρ1+2 = ρ1 ρ2. probabilities pα corresponding to the states α . Now ⊗ | i (i) (i) (i) If ρi α = pα α , then ρ(t)= p α(t) α(t) . α | i h | α (1) (2) (1) (2) (1) (2) X ρ1+2 α ,β = pα pβ α ,β . Since the state vectors satisfy the Schr¨odinger equations E E

Now d i¯h α(t) = H α(t) Tr A = α(1),β(2) A α(1),β(2) , dt | i | i 1+2 h | | i d α,β i¯h α(t) = α(t) H, X − dt h | h | so that we end up with the equation of motion S = k Tr ρ ln ρ 1+2 − B 1+2 1+2 1+2 d = k p(1)p(2)(ln p(1) + ln p(2)) i¯h ρ(t) = [H,ρ(t)]. − B α β α β dt Xα,β = k p(1) ln p(1) k p(2) ln p(2) It is easy to show that − B α α − B β β α X Xβ d A ∂A 1 = S1 + S2. h i = + [A, H] . dt ∂t i¯h h i   The properties of the statistical entropy above are In a stationary ensemble the expectation values are equivalent to the thermodynamic entropy. This will be independent on time, soρ ˙ =0 or shown later. [H,ρ]=0. 5.3. Density of states Let us denote This is possible e.g. when ρ = ρ(H). H n = E n , | i n | i Entropy so that The entropy is defined by H = E n n . n | i h | n X S = k Tr ρ ln ρ. − B If the volume V of the system is finite the spectrum is discrete and the states can be conveniently normalized In a base where ρ is diagonal, like n m = δn,m. S = k p ln p . h | i − B α α α : X Entropy has the properties V and N → ∞ → ∞

6 so that N/V remains constant. If the particle has spin S, it has g =2S + 1 spin degrees The cumulant function of states is defined as of freedom. Then 2 p p V 2 J(E)= θ(E En), J1(E) = g dNkθ E = g 4π dp′p − − 2m h3 n Z   Z0 X V 4π = g p3. i.e. the value of J at the point E is the number of those h3 3 states whose energy is less than E. The density of states So we get (function) can be defined as 2 3/2 dJ(E) J1(E) = C1V E ω(E)= = δ(E En), 3 dE − 1/2 n ω (E) = C V E X 1 1 3/2 since dθ(x)/dx = δ(x). However, in macroscopic systems 2m C1 = 2πg . the energy levels are extremely closely spaced, and a more h2   sensible definition for the density of states is Example: 2. Maxwell-Boltzmann J(E + ∆E) J(E) Let us consider N free particles. The total energy is ω(E) = lim − ∆E 0 ∆E p2 → E = j 2m i.e. the number of states in the interval (E, E + ∆E), j where we assume that we do not allow ∆E to become X and the cumulant function smaller than the interval between energy levels. Density of states becomes a smooth function when the intervals JN (E) between energy levels 0 (volume ), as does J(E). 2 2 → → ∞ p1 pN We can also write, using the Hamilton operator, = dNk dNk θ E 1 · · · N − 2m −···− 2m Z Z   J(E) = Tr θ(E H) = dE dE ω (E ) ω (E ) − 1 · · · N 1 1 · · · 1 N ω(E) = Tr δ(H E). Z Z − θ(E E E ). × − 1 −···− N ω(E) corresponds to the volume ΓE of the energy surface of the classical phase space. Thus the corresponding density of states is Example: 1. Free particle dJ (E) ω (E) = N Let us consider a free particle in a box of size V = L3 N dE with periodic boundary conditions. The Hamiltonian is = dE dE ω (E ) ω (E ) 1 · · · N 1 1 · · · 1 N p2 Z H = . δ(E E E ). 2m × − 1 −···− N The eigenfunctions are the plane waves We define the Laplace transforms ∞ sE 1 ik r Ω1(s) = dE e− ω1(E) ψk(r)= e · , √V Z0 ∞ sE ΩN (s) = dE e− ωN (E). where the wave vector k = p/¯h can acquire the values 0 (periodicity) Z Now

2π 3 k = (n ,n ,n ), n Z, V = L . ΩN (s) L x y z i ∈ ∞ = dE dE ω (E ) ω (E ) The corresponding energy levels are 1 · · · N 1 1 · · · 1 N Z0 2 2 2 ∞ sE ¯h k p dE e− δ(E E1 EN ) ǫk = = . × − −···− 2m 2m Z0 ∞ sE1 sEN In the limit of large volume the summation can be = dE1 dEN ω1(E1)e− ω1(EN )e− · · · · · · Z0 transformed to the integration over the wave vector, using N = [Ω1(s)] . 2π dk = ∆k = . Since L Z n X X ∞ sE 1/2 3 3/2 Ω (s) = dE e− C V E = C V Γ( ) s− Thus, 1 1 1 2 Z0 1 3/2 V V = C V √πs− = dN = d3k = d3p. 1 2 k (2π)3 h3 Xk Z Z Z

7 we have N 3N/2 ΩN (s) = (C2V ) s− , s where 1 2πm 3/2 C = √πC = g . 2 2 1 h2   residues Performing the inverse Laplace transforms we get

1 N 3/2N 1 ωN (E)= 3 (C2V ) E − . Γ( 2 N)

Note: We ignored the permutation symmetry! Thus, for γ each N-particle states there are N! permutations which are physically equivalent (unless the particles are all different). We can correct the density of states by dividing this by N!, which gives us so-called Boltzmann counting. Using this we obtain the classical ideal gas or Maxwell-Boltzmann gas theory (will be discussed later):

1 N 3/2N 1 ωN (E)= 3 (C2V ) E − . N!Γ( 2 N) Note: This does not take into account the quantum 5.4. Energy, entropy and mechanical features of multiple occupation of 1-particle The density of states operator ρ enables us to calculate states (bosons, fermions). all thermodynamical properties of the system: for example the moments En and entropy S. We can also h i Sidebar: inverse Laplace transform define temperature T from properties of ρ. For this we Obviously, if f(t)= tα, then Laplace transform is need to define microcanonical ensemble, in analogy with the classical mechanics way done earlier:

∞ ts α α 1 ∞ y α α 1 dt e− t = s− − dye− y = s− − Γ(α + 1) Microcanonical ensemble Z0 Z0 We require that

Thus, if α =3N/2 1, we obtain the result for ωN . a) energy is restricted between (E, E + ∆E), and Many standard function− (inverse) Laplace b) the entropy is maximized. transformations can be found tabulated (Arfken). According to 5.2. this is satisfied when all states are More generally, inverse Laplace transforms can be equally likely, thus, the density of states operator is calculated using complex plane integral (see Arfken, for 1 example): ρE = θ(E + ∆E H)θ(H E), γ+i ZE − − 1 ∞ st ˆ f(t)= dse f(s) where 2πi γ i Z − ∞ where γ R is chosen so that the Laplace transform ZE = Tr θ(E + ∆E H)θ(H E) ∈ − − = Tr [θ(E + ∆E H) θ(E H)] ∞ st − − − fˆ(s)= dte− f(t) = J(E + ∆E) J(E) − Z0 is microcanonical partition function or the number of exists when s γ. It is easy to see that this is inverse states between (E, E + ∆E). When ∆E is small, we have Laplace: ≥

ZE ω(E) ∆E. γ+i ′ 1 ∞ st ∞ st ≈ dse dt′e− f(t′) 2πi γ i 0 Entropy is Z − ∞ Z γ+i ′ ∞ 1 ∞ s(t t ) = dt′f(t′) dse − SE = kBTr ρ ln ρ = kB ln ZE. 0 2πi γ i − Z Z − ∞ ′ ′ ∞ γ(t t ) 1 ∞ iy(t t ) Since ZE is a positive integer, SE 0 holds. Furthermore = dt′f(t′)e − dye − = f(t) we get ≥ 0 2π Z Z−∞ The integral can be closed to the left on complex s-plane, SE = kB ln[ω(E) ∆E] and calculated by finding residues within the contour. = kB ln ω(E)+ S0,

8 and we can write 6. Equilibrium distributions Microcanonical ensemble was discussed in the previous SE = kB ln ω(E), section. That is obtained by maximising entropy with the boundary condition H = E =constant, N constant. Now because ln∆E is non-extensive and negligible when V we shall discuss canonical and grand canonical ensembles, large. (This expression is slightly incorrect dimensionally; which are obtained by maximizing entropy with boundary we should use ln(Cω(E)), where C has dimensions of conditions H = E, N const. for (i.e. energy.) energy is allowedh i to fluctuate) and and H = E, Note: As a matter of fact Nˆ = N for (i.e.h i both energy handi particle number are allowed to fluctuate). ω = ω(E,V,N). 6.1. Canonical ensemble Temperature Let us now maximise the entropy under the constraints According to thermodynamics we have H = Tr ρH = E = constant h i 1 ∂S I = Tr ρ =1. = . h i T ∂E V,N   Using Lagrange multipliers, we require that In the microcanonical ensemble we define the δ(S λ H λ′ I )=0, temperature T so that − h i− h i

1 ∂ where λ are λ′ are multipliers. We get = k ln ω(E,V,N). T B ∂E δTr ( kB ρ ln ρ λρH λ′ρ)= Denoting − − − Tr ( kB ln ρ kBI λH λ′I)δρ =0. 1 − − − − β = , kBT Since δρ is an arbitrary variation, we can solve for ρ and end up (after relabelling the constants) with the canonical we have ∂ ln ω or Gibbs distribution β = . ∂E 1 βH ρ = e− , Example: Maxwell-Boltzmann gas Z Now 3/2N 1 where Z is the canonical sum over states (or partition ωN E − , ∝ function, which is determined from the condition Tr ρ = 1: so 3 ln ωN = N ln E + Z = Tr e βH = e βEn = dE ω(E)e βE. 2 · · · − − − n Z and X 3N β = The constant β is yet undetermined! We shall show below 2E that β =1/kBT . or we end up with the for 1-atomic ideal Note: In the canonical ensemble the number of particles gas: is constant, i.e. 3 E = k T N. 2 B Z = Z(β,V,N,...). The thermodynamics of a quantum mechanical system can be derived from the density of states ω(E,V,N). In The probability for the state ψ is practice the density of states of a microcanonical 1 βH ensemble (E and N constant) is difficult to calculate, due pψ = Tr ρPψ = ψ e− ψ . Z h | | i to the constraint in total E. Partcularly, in the case of an eigenstate of the Hamiltonian, H n = E n , | i n | i we have 1 βEn p = e− . n Z For one particle system we get

1 βǫν βǫν p = e− ; Z = e− . ν Z ν X

9 Here ǫν is the one particle energy. This is normalized dEP (E) = 1, and we get the correct expectation values Hn = dEEnP (E). Using canonical Entropy and temperature density ρ, we obtainhR i R Because in the canonical ensemble we have 1 βE P (E)= ω(E)e− . ln ρ = βH ln Z, Z − − When the volume is large, this typcially has a the entropy will be well-defined maximum at E = E¯ H . ≡ h i S = k Tr ρ ln ρ = k ln ρ w ( E ) e - b E − B − B h i = kB βE + kB ln Z. D E

Here E is the expectation value of the energy _ E 1 βH E E = H = Tr He− . h i Z Let us write the sum over states as

Let us now calculate ∂S/∂E. Note that the parameter β βE βE+ln ω(E) very well can depend on E! Thus, using the Z = dE ω(E)e− = dE e− . thermodynamical identity we can define T : Z Z Now we can use the saddle-point method to calculate the 1 ∂S ∂ ln Z sharply peaked integral: expand the exponent to second = = k β + k Eβ′ + k T ∂E B B B ∂E order  V,N βH = kB (β + Eβ′ β′Tr He− /Z) ln ω(E) βE = − − = kB (β + Eβ′ β′ H )= kBβ ln ω(E¯) βE¯ − h i − =0, maximum where β′ = ∂β/∂E. Thus, we obtain the familiar relation ∂ ln ω + β (E E¯) 1 ∂E − − β = . z }|E=E¯ { kBT 2 1 ∂ ln ω 2 + (E E¯) + . 2 2 − · · · ∂E E=E¯ Free energy ¯ The partition function is the central quantity, and all Linear term must vanish at E = E: thermodynamic properties can be derived from it: ∂ ln ω 1 ∂S(E) 1 β = = = ¯ ∂E E=E¯ kB ∂E E=E¯ kB T (E) ∂Z βH = Tr e− H = Z H = ZE ∂β − − h i − where we used the microcanonical entropy ¯ or S(E)= kB ln ω(E) to obtain the microcanonical T (E). ∂ ∂ ln Z Thus, the temperature of the heat bath in canonical E = ln Z = k T 2 . −∂β B ∂T ensemble = T = T (E¯), the temperature of the microcanonical ensemble at energy E¯ = H . Actually, Using the expression above we can write h ican. ∂ ρ = dE P (E,β) ρ S = k (T ln Z) . β,can. E,microcan. B ∂T Z and canonical ensemble can be thought as an ensemble of Using the definition for the microcanonical ensembles. The 2nd order term in the F = E TS, we get − Taylor series is F = k T ln Z. 2 − B ∂ S ∂ 1 1 ∂T 1 2 = = 2 = 2 , ∂E ∂E T −T ∂E −T CV With the help of this the density operator takes the form   so 1 ¯ 2 β(F H) βE¯ 2 (E E) ρ = e − . Z ω(E¯)e− dE e− 2kB T CV − . ≈ Z normal distribution Note that F is a number, H operator. Thus, we find the variance of the| normal{z distribution:} 2 2 Fluctuations (∆E) = kB T CV The probability distribution of the energy E is or P (E)= δ(H E) = Tr ρδ(H E). ∆E = k T 2C = (√N), h − i − B V O p 10 because CV , as well as E, is extensive ( (N)). Thus the ending up with the grand canonical distribution fluctuation of the energy is O 1 β(H µNˆ ) ∆E 1 ρ = e− − . ZG E ∝ √N Note: More straightforward way to calculate the width is (again relabeling constants). Here to use β(H µNˆ) ZG = Tr e− − (∆E)2 = (H H )2 = H2 H 2, − h i h i − h i is the grand canonical partition function. D E βH which is, using F = k T ln Tr e− , The trace can be split in N-sectors (effectively using − B eigenstates of Nˆ): ∂2 (∆E)2 = (βF ) −∂β2 β(H µNˆ ) βµN βH(N) ZG = Tr N e− − = e Tr N e− 2 ∂ 2 ∂ F N N = kBT T X X − ∂T ∂T T N 2 = z ZN , 3 ∂ F 2 N = kBT 2 = kBT CV X − ∂T V,N   where ZN is the canonical partition function with N Note: Alternative (and perhaps more rigorous) way to particles, and fugasity z eβµ , and H(N) is a N-particle derive the density operator for canonical ensemble would Hamiltonian. ≡ be to start from a very large microcanonical system, and This directly gives the probability distribution of particle divide this into 2 pieces: part I, “system” and II, “heat number: bath”, where volume of II volume of I. Using only ≫ 1 N microcanonical total density operator, it is P (N) δ(N Nˆ) = Tr ρNˆ = z ZN straightforward to derive the canonical density for I: ≡ h − i ZG

βII HI i.e. the grand canonical ensemble is equivalent to the sum ρ e− I ∝ of N-particle canonical ensembles where the weight of N where βII =1/kBTII . each N sector is given by z . In the base where the Hamiltonian is diagonal the 6.2. Grand canonical ensemble partition function is Let us consider a system where both the energy and the β(E(N) µN) number of particles are allowed to fluctuate. The Hilbert ZG = e− n − , n space of the system is then the Fock space, direct sum XN X = (0) (1) (N) where H H ⊕ H ⊕···⊕H ⊕··· (N) (N) and the Hamiltonian operator the sum of 1,2,. . . -particle H N; n = H N; n = En N; n , | i | i | i Hamiltonians: when N; n (N) is a state of N particles, i.e. H = H(0) + H(1) + + H(N) + . | i ∈ H · · · · · · Nˆ N; n = N N; n . We define the (particle) number operator Nˆ so that | i | i Nˆ N = N N Particle number and energy | i | i Now for eigenstates of Nˆ. Grand canonical ensemble can derived by allowing both ∂ ln ZG 1 β(H µNˆ ) = Tr e− − βNˆ E and N to fluctuate, and requiring that the expectation ∂µ ZG values are fixed: = β Nˆ = βN¯ ¯ H = E = given energy D E h i and Nˆ = N¯ = given particle number ∂ ln ZG 1 β(H µNˆ ) ˆ DTrEρ = I = 1 probability normalization = Tr e− − (H µN) h i ∂β −ZG − We can now demand that the entropy S = kBρ ln ρ is = H + µ Nˆ = E¯ + µN,¯ maximized with the above constraints. Thus,− using − h i − D E Lagrange multipliers we obtain so that, using β =1/(kBT ), ˆ ∂ ln Z 0 = δ(S + λ H + λ′ N + λ′′ I ) N¯ = k T G h i h i B ∂µ D E ˆ = δTr (kB ρ ln ρ λρH λ′ρN λ′′ρ) ∂ ln Z ∂ ln Z − − − E¯ = k T 2 G + k T µ G . = Tr δρ(k ln ρ + k λH λ′Nˆ λ′′) B B B B − − − ∂T ∂µ

11 (this is assuming that β =1/(kBT ), which we have not manner. Let us now study the relation closer. We shall really yet shown!) assume that the Hamiltonian H and eigenstates α | i depend on external parameters xi : Entropy and { } According to the definition of entropy we have H(xi) α(xi) = Eα(xi) α(xi) . | i | i

S = kBTr ρ ln ρ = kB ln ρ . Now the change in energy of α : − − h i | i

Now ∂Eα ∂ ∂H ∂ ˆ = α H α = α α + Eα α α ln ρ = βH + βµN ln ZG, ∂x ∂x h | | i ∂x ∂x h | i − − i i  i  i so that ∂H 1 1 = α α , S = E¯ µ N¯ + ln Z . ∂xi G   kBβ − β

In thermodynamics we defined the grand potential because α α = 1 (Feynman-Hellman theorem). h | i Ω= E TS µN, Adiabatic variation − − In quantum mechanics it can be shown that, if the which agrees with the above expression if we identify variation of the parameters x(t) is slow enough and the β =1/(kBT ) (as expected) and initial state of the system is α , the system will remain in the eigenstate α(x(t)) and| therei are no transitions to | i Ω= kB T ln ZG. other Hamiltonian eigenstates (note that this is only true − for eigenstates of H). Thus, the density operator can be written as E a ( x i ) ˆ ρ = eβ(Ω H+µN). a = 1 − a = 0 Note: The grand canonical partition function depends on variables T , V and µ, i.e. x i Z = Z (T,V,µ). This implies that the probabilities for the states remain G G constant and the change in the entropy

Fluctuations S = kB pα ln pα − α Now X 2 ∂ β(H µNˆ) β(H µNˆ ) 2 2 is zero. Tr e− − = Tr e− − β Nˆ ∂µ2 Let us consider the density operator in an equilibrium state ([H,ρ] = 0). We assume N is constant. In the base 2 ˆ 2 = ZGβ N , α , where the Hamiltonian is diagonal, {| i} D E so H α = E α , | i α | i (∆N)2 = (Nˆ N¯)2 = Nˆ 2 N¯ 2 − − we have D ∂2 lnEZ D E ∂N¯ ρ = p P , P = α α . = (k T )2 G = k T = (N¯), α α α | i h | B 2 B α ∂µ ∂µ O X We divide the variation of the density operator into two ¯ because only N is extensive in the last expression. Thus parts: the particle number fluctuates like adiabatic nonadiabatic ∆N 1 = . N¯ O √N¯ δρ = pαδPα + δpαPα   α α Xz }| { Xz }| { A corresponding expression is valid also for the = δρ(1) + δρ(2). fluctuations of the energy. For a mole of matter the 12 fluctuations are 10− or the accuracy the accuracy ∝ ≈ The first part corresponds to adiabatic variation (in of the microcanonical ensemble. statistical sense), because the probabilities of the eigenstates remain fixed. 6.3. Relation with thermodynamics Let us define Fi, the generalized force conjugate to the In thermodynamics the entropy is a state variable related generalized displacement xi as to the exchange of the heat energy (dQ = T dS) between the system and the environment. However, the statistical ∂H ∂H Fi = Tr ρ = . Gibbs entropy kBρ ln ρ is constructed in a very different − ∂x − ∂x − i  i 

12 For example, F = p, x = V , and F δx the work done by Isolated system and microcanonical ensemble the system. If the system is microcanonical, then Eα = E for all Then the change in energy is states. However, when we change x, E(x) can vary! In this case δ H = Tr δρ H + Tr ρ δH h i δ H = δE = δpαEα Fiδxi (1) (2) ∂H h i − = Tr δρ H + Tr δρ H + δx Tr ρ α i i ∂x X X i i X = Fiδxi , (2) − = p Tr H δP + Tr δρ H F δx . i α α − i i X α i X X which is equivalent to thermodynamical work done by an The first term vanishes, because isolated system

δU = δW = Fiδxi Tr H δPα = β H ( α δα + δα α ) β − − h | | i h | | i h | | i i Xβ X = Eαδ α α =0, h | i 6.4. Einstein’s theory of fluctuations so that the change in energy becomes Let us divide a large system into macroscopic parts with weak mutual interactions. (2) δ H = Tr δρ H Fiδxi = δpαEα Fiδxi. operators Xˆ , which correspond to the extensive h i − − ⇒ ∃ { i} i α i properties of the partial systems so that X X X The definition of the statistical Gibbs entropy is [Xˆ , Xˆ ] 0 i j ≈ [Xˆi, H] 0. Sstat = k Tr ρ ln ρ = k p ln p , ≈ − b − B α α α mutual eigenstates E,X ,...,X , which are X 1 n macrostates⇒ ∃ of the system,| i.e. for eachi set of the and its variation is parameters (E,X1,...,Xn) there is a macroscopic =0 number of microstates. Let Γ(E,X1,...,Xn) be the number of the microstates corresponding to the state δSstat = k δp ln p k δp − B α α − B α E,X1,...,Xn (the volume of the phase space). α α | i X zX}| { The total number of the states is = k δp ln p − B α α α Γ(E)= Γ(E,X1,...,Xn) X Xi = kBβ δpαEα, {X} α and the relative probability of the state (E, X ) X { } where in the last stage we used the canonical ensemble Γ(E,X ,...,X ) f(E,X ,...,X )= 1 n . 1 n Γ(E) 1 βEα p = e− α Z The entropy of the state E,X ,...,X is | 1 ni stat Thus, if we denote β =1/(kBT ), we obtain S(E,X1,...,Xn)= kB ln Γ(E,X1,...,Xn) or stat stat 1 δ H = T δS Fiδxi. 1 k S(E,X1,...,Xn) h i − f(E,X1,...,Xn)= e B . i X Γ(E) This is the probability of state X with fixed energy E. This is equivalent to the first law of the thermodynamics, In thermodynamic equilibrium the entropy S is maximized: δU = T thermδStherm δW, − 0 (0) (0) S = S(E,X1 ,...,Xn ). provided we identify Let us denote by H = E¯ = U = (0) xi = Xi Xi hstati therm − T = T deviations from the equilibrium positions, and expand the Sstat = Stherm entropy to 2nd order around extremum: the Taylor series

Fiδxi = δW = work. of the entropy will be i 1 X S = S0 k g x x + , − 2 B ij i j · · · i,j X

13 where We can naturally also calculate correlations of other 2 1 ∂ S quantities, for example gij = . −kB ∂Xi∂Xj X(0)   i ∂S ∂S { } ∆S∆V = ∆T ∆V + (∆V )2 We use vector-matrix notation h i ∂T h i ∂V h i  V  T x ∂p 1 = V kB TκT . ∂T V x = . and g = (gij ).    .  1 − xn ∂V ∂T   = V kBTκT   − " ∂p T ∂V p# Then the probability becomes     = kB V T αp 1 xT gx f(x)= Ce− 2 , where we used the Maxwell relation ∂S = ∂p . ∂V T ∂T V where Note that if we were to origially take other set of  n/2  C = (2π)− detg. variables, say, S, V as the independent ones, the matrix g = 1/k ∂2S/∂~x2 would look different (in this case p B Correlation functions non-diagonal).− Correlation functions can be written as 6.5. Reversible minimum work x x xf(x)x x Another way to calculate the fluctuation probability is to h p · · · ri ≡ D p · · · r Z consider the concept of reversible minimum work. ∂ ∂ (0) = F (h) , Let x = X X be the fluctuation of the variable X. ∂h · · · ∂h −  p r h=0 For one variable we have 1 gx2 where f(x) e− 2 . ∝ x = dx dx D 1 · · · n Let us assume that X is generalized displacement; thus and the generating function is dU = T dS F dX dW¯ . − − other − 1 xT gx+ 1 (hT x+xT h) 1 hT g 1h F (h)= C xe− 2 2 = e 2 . If we write S = S(U,X,...) we get the partial derivative D Z ∂S F = . SVN-system ∂X T When studying the stability conditions of matter we On the other hand we had found out that the 1 S = S0 k g x x − 2 B ij i j 1 i,j ∆S = (∆Ti∆Si ∆pi∆Vi + ∆µi∆Ni). X −2T − 1 i = S0 k gx2, X − 2 B Let us consider now only one volume element: so ∂S 1 (∆T ∆S ∆p ∆V +∆µ ∆N) = kBgx f = Ce− 2kB T − . ∂X − and in the limit of small displacement We assume that the system is not allowed to exchange F = k T gx. particles, i.e. ∆N = 0. Taking now T and V as our − B independent variables we can employ the definitions of When there is no action on X from outside, the deviation the heat capacity and compressibility: x fluctuates spontaneously. Let us give rise to the same deviation x by applying reversible adiabatic external 1 CV 2 1 2 (∆T ) + V k T κ (∆V ) f(∆T, ∆V ) e− 2 kB T B T . work: ∝ h i dU = F dX = k T g x dx. − B We can now read out the matrix g: Integrating this we get

1 2 T V (∆U)rev ∆R = kB T gx , CV ≡ 2 T 2 0 g = kB T . V 0 1 where ∆R is the minimum reversible work required for V kB TκT ! the fluctuation x = ∆X. We can write ∆R The variances are then f(∆X) e− kB T . ∝ 2 2 kBT Note that this relates a fluctuation of X at constant 2 (∆T ) = δ S/kB CV U = E (f(x) e ) to the adiabatic work done to 2 ∼ (∆V ) = V kBTκT . achieve same change of X (δS = 0).

14 1 1 1 7. Ideal equilibrium systems ( N ν) ln( N ν) ( N ν) In ideal systems there are no interactions between the − 2 − 2 − − 2 −   particles/degrees of freedom. If we know the solution for 2 1 1 N 2 N + ν 1-particle state, N-particle state can be solved by taking = N ln ν ln 2 1 N 2 ν2 − 1 N ν into account the correct statistics. 4 − 2 − 1 1 1+2ν/N = N ln2+ N ln ν ln 2 1 4ν2/N 2 − 1 2ν/N 7.1. System of free spins − − Let us consider N particles with spin 1 . The ν2 2 N ln 2 2 z-component of the spins are ≈ − N

1 where in the last step we have taken the approximation Siz = ¯h i =1,...,N. ± 2 ν N. The maximum of W is clearly at ν = 0. | | ≪ The z component of the total spin is Thus, the binomial distribution can be approximated by a normal distribution: 1 + Sz = Siz = ¯h(N N −), 2ν2/N N 2 − W (ν) W (0)e− , W (0) 2 . i X ≈ ≈ where The mean deviation ∆ν √N/2 N, i.e. large values of ν are very unlikely. ≈ ≪ 1 N + = + ¯h spin count Note that the total number of states 2 N 1 Wtot = W (ν)=2 N − = ¯h spin count. −2 ν X Sz determines the macrostate of the system. Thus, a correct normalization of the gaussian distribution Denoting Sz =¯hν we have would be

2 1 W = C dνe 2ν /N = C Nπ/2=2N N + = N + ν tot − 2 Z p 1 N N − = N ν which implies C =2 / Nπ/2. However, the correction 2 − in this case is non-extensive and can be neglected in the p and large N limit: 1 1 1 ν = N, N +1,..., N. non-extensive −2 −2 2 extensive 1 Let W (ν) the number of those microstates for which ln C = N ln 2 ln(Nπ/2) − 2 Sz =¯hν, i.e. W (ν) tells us, how many ways there are to z }| { + z }| { distribute N particles into groups of N and N − + + Energy particles so that N + N − = N and N N − =2ν. From combinatorics we know that this is− given by the Let us put the system in an external magnetic field binomial distribution: H = Hzˆ, with magnetic flux density

N N! B = µ0H. W (ν) = = N + N +!N !   − The potential energy is N! = 1 1 . ( N + ν)!( N ν)! E = µ0 µ H = µ0H µiz, 2 2 − i · − − i i X X W (ν) is the degeneracy (=number of states) of the where µ is the magnetic moment of the particle i. macrostate S =¯hν. i z The magnetic moment of a particle is proportional to its The Boltzmann entropy is spin: µ = γS, S = kB ln W (ν). where γ is the gyromagnetic ratio. In classical Using Stirling’s formula electrodynamics, with a homogeneously charged solid body of mass m, total charge q and angular momentum S ln N! N ln N N , this has the value ≈ − q γ = . we get 0 2m However, for elementary particles this is different, and for ln W (ν) N ln N N ≈ − we have 1 1 1 ( N + ν) ln( N + ν) ( N + ν) e − 2 2 − 2 γ 2γ0 = ,   ≈ −m

15 Usually this is given in terms of Bohr magneton The magnetization or the magnetic polarization is the magnetic moment per the volume element, i.e. e¯h 5 eV µ = =5.79 10− . B 2m · T 1 M = µ . V i The energy of the system is i X E = µ H µ = µ γHS = ǫν, The z component of the magnetization is − 0 iz − 0 z X 1 ǫν 1 ¯hγµ Hν where M = = 0 z −V µ H V µ H ǫ = ¯hγµ0H 0 0 − 1 = γ¯hν. is the energy/particle. For electrons we have V

ǫ =2µ0µBH. Now E = µ HVM , Now the change in energy is − 0 z so we get for our system as the equation of state ∆E = ǫ ∆ν, 1 µ ¯hγH Changing the variable ν E, and using the condition M = ρ¯hγ tanh 0 , ↔ 2 2k T (Jacobi!)  B  ω(E) ∆E = W (ν) ∆ν | | | | where ρ = N/V is the particle density. Note: The we get as the density of states relations derived above

1 E ω(E)= W . E = E(T,H,N) ǫ ǫ | |   M = M(T,H,N)

determine the thermodynamics of the system. 1) Microcanonical ensemble Let us first consider the spin system at fixed energy. 2) Canonical ensemble Denoting The canonical partition function is 1 E0 = ǫN, 2 βEn Z = e− . the total energy will lie between E0 E E0. n With the help of the energy W can− be≤ written≤ as X Here 1 4E2 E E + E N ln W (ν) = N ln 0 ln 0 2 2 En = µ0H µiz 2 E0 E − ǫ E0 E − − − i=1 = ln ω(E) + ln ǫ . X | | the energy of a single microstate. As the entropy we get Denote 1 µiz =¯hγνi, νi = . S(E) = kB ln ω(E) ±2 1 4E2 E E + E = Nk ln 0 ln 0 Now B 2 E2 E2 − 2E E E  0 − 0 0 −  βµ H µiz +non extensive term. Z = e 0 i all microstatesX P The temperature was defined like 1 1 2 2 βµ0hγH¯ νi 1 ∂S = e i = , 1 · · · 1 ν = νN = T ∂E 1X− 2 X− 2 P 1 N so 2 1 N E0 + E = eβµ0Hγhν¯ = ZN , β(E)= = ln .   1 kBT (E) −2E0 E0 E ν= 1 − X− 2 We can solve for the energy:   where Z1 the one particle state sum βE0 E = E tanh 1 1 0 βµ0hHγ¯ βµ0hHγ¯ − N Z1 = e− 2 + e 2 1 µ0¯hγH µ0¯hHγ = Nµ0¯hγH tanh . = 2cosh . −2 2kBT 2k T   B

16 The same result can be obtained using W (ν): Thus the differential of the free energy is

βE(ν) Z = W (ν)e− dG = S dT µ V M dH, − − 0 · ν X βǫν so that the magnetization is = W (ν)e− ν 1 ∂G X M = N βǫ(N + 1 N) = e− − 2 −µ0V ∂H T N +   N + 1 µ ¯hγH X   0 1 N = ρ¯hγ tanh . 2 βǫN βǫ −2 2k T = e− 1+ e− .  B  From the partition function we can determine This is identical with the result we obtained in the thermodynamic potential, free energy. However, what is microcanonical ensemble. the “correct” thermodynamic potential in this case? The microcanonical entropy = the canonical entropy + a Naively we might identify non extensive term. Energy F (T, H)= k T ln Z − B a) Energy E(T, H) () can be calculated directly as the Helmholtz free energy (as is done in many from partition function textbooks). This is (mostly) OK for thermodynamics, 1 ∂ and the simple choice, but it is not fully consistent: E = E(ν) = ǫν¯ = Z h i −Z ∂β Note that H is generalized force (“p”), M is displacement 1 1 (“V ”). Thus, when our set of independent variables is (T , = Nǫ tanh βǫ −2 2 H), the right thermodynamic potential is the Gibbs free   energy = the energy of the microcanonical enesmble.

G = G(T, H)= kBT ln Z b) Alternatively, using thermodynamic Legendre − transformation from enthalpy to Gibbs potential µ0Hγ¯h = kBTN ln2 + lncosh . − 2kBT G = E TS   − Furthermore, in this case we should identify our energy E or as the enthalpy of the system: ∂G E = G + TS = G T E = µ0HVM = U H (µ0VM) − ∂T − − × ∂G ∂ = G + β = (βF ) which is related to the true internal energy U by a ∂β ∂β Legendre transform. In this case internal energy vanishes: ∂ = ln Z U =0. −∂β

True internal energy U should not depend on external Susceptibility force like H, only on internal properties like M. However, According to the definition the susceptibility is in this case there is no H-independent energy! 2 Thus, the entropy is ∂M 1 ∂ G χ = = 2 ∂H −µ0V ∂H  T   ∂G 2 S = 1 ¯hγ − ∂T µ0ρ 2  H = . kB T cosh2 hγµ¯  0H µ0Hγ¯h 2KB T = NkB ln2 + lncosh   " 2kBT When H 0 we end up with Curie’s law → µ ¯hγH µ ¯hγH 0 tanh 0 . C − 2k T 2k T χ = , B B # T H Differentiating the free energy with respect to the field where we get µ ρ 1 2 C = 0 ¯hγ . k T 2 ∂G 1 ∂ βǫν B = k T W (ν)e−   − ∂H B Z ∂H  z T ν X Thermodynamic identifications 1 βǫν = µ0γ¯h νW (ν)e− Earlier we identified Z ν X = µ γ¯h ν = µ VM . Estat E = Hˆ = enthalpy, 0 h i 0 z ≡ D E 17 so that or T H b = b . G = E TS = Gtherm T H − a a = the Can be serialised to reach even smaller T . = U TS µ V M H. − − 0 · The entropy of the spin system is Now using the differentials we get 2 1 4E0 E E0 + E dU = dG + d(TS)+ d(µ0V M H) S(E)= Nk ln ln , · B 2 E2 E2 − 2E E E = T dS + µ V H dM  0 − 0 0 −  0 · = T dS dW¯ where − E = µ µ HN ja E 0 b < 0 Example: Adiabatic demagnetization is used to T > 0 T < 0 achieve cooling of atomic spins to nanokelvin . This is due to very large degeneracy of - | E 0 | + | E 0 | spin states, which means large entropy down to very Now 1 ∂S N E + E small temperatures. Removing entropy with magnetic β(E)= = ln 0 . fields can then reduce temperature very effectively. kB ∂E −2E0 E E0 − Now S - b E = ln2+lncosh x x tanh x, e Nk − B H-H where µ ¯hHγ x = 0 . 2kBT -E 0 + E 0 + E 0 -E 0 When T 0, then x , so that βE → → ∞ Originally the maximum of ω(E)e− /Z is at a negative 1 x 2x value E. Reversing the magnetic field abruptly E E ln cosh x = ln e (1 + e− ) →− 2 and correspondingly β β. 2x →− = x ln2+ e− + The temperature can be negative if the energy is bounded − · · · both above and below. and x 2x 7.2. Classical ideal gas e (1 e− ) tanh x = x − 2x (Maxwell-Boltzmann gas) e (1 + e− ) 2x Kinetic gas theory, (1860): gas of = 1 2e− + . − · · · atoms/molecules. Explained many properties of real Hence , supporting the atomic hypothesis of matter. S 2x At standard temperature and (STP) most gases 2xe− + . NkB → · · · are dilute: When T , then x 0, and → ∞ → 2 Å S ln 2. r i NkB → Let us cool down a system by some means at a large field Ha, reaching point a: We define ri so that the volume occupied by one molecule S N k B = H 1

3 3 ri = . T 4πρ r We decrease the field adiabatically within the interval Typically a b. Now S = S(H/T ), so that → the diameter of an atom or a molecule d 2A.˚ H H • ≈ S = S a = S = S b a T b T the range of the interaction 2–4A.˚  a   b  •

18 the mean free path (collision interval) l 600A.˚ From the relation • ≈ at STP (T = 273K, p = 1atm) r 20A.˚ 3 ∞ 2 • i ≈ d v = 4πv dv or Z Z0 d ri l we can obtain for the speed (the absolute value of the ≪ ≪ 2 20 600 A˚ velocity v = v ) the distribution function F (v) | | The most important effect of collisions is that the system F (v)=4πv2f(v). thermalizes i.e. attains an equilibrium, which corresponds to a statistical ensemble. Otherwise we can forget the collisions and describe the gas as non-interacting F ( v ) molecules. Let us consider a system of one molecule which can exchange energy (heat) with its surroundings. Then the suitable ensemble is the canonical ensemble and v the distribution the Boltzmann distribution

1 βǫl pl = l ρ l = e− , The most likely speed (maximum of the distribution) h | | i Z • where the canonical partition function is 2k T v = B . m m βǫl r Z = e− . l The average speed X • We recall that in k-space the density of 1-particle states ∞ 8kBT is constant: in volume V = L3 the wave function is v = dv vF (v)= . h i 0 r πm ik x Z ψk(x) e · ∝ The average of the square of the speed • where, using periodic b.c. k =2πn/L, n Z3. Thus, in ∈ ∞ 3kBT volume element (k, k + ∆k) the number of states is v2 = dv v2F (v)= . m constant independent of k. Z0

For convenience, we shall switch to velocity space, with Note: v = p/m =¯h/mk. However, everything would work also in k-space too. 1 1 1 1 mv2 = mv2 = mv2 = k T The volume element in velocity space is 2 x 2 y 2 z 2 B       1 ¯h 3 and d3v = d3p = d3k, 1 1 3 m3 m mv2 =3 mv2 = k T,   2 2 x 2 B     i.e. the density of states is also constant. The energy of i.e. the energy is evenly distributed among the 3 k-state is (translational) degrees of freedom: the equipartition of the 2 2 energy. ¯h k 1 2 ǫk = k H k = = mv , h | | i 2m 2 Partition function and thermodynamics so that the velocity distribution becomes The single particle partition function is

mv2 βE Z (β) = dE ω(E)e− f(v) k ρ k = e− 2kB T 1 ∝ h | | i Z 2 2 p2 β h¯ k V 3 or = g e− 2m = g d pe− 2mkB T mv2 3 2k T h f(v)= Ce− B . Xk Z V C can be determined from the condition = g (2πmk T )3/2. h3 B 2 3 ∞ mvx 3 2k T 1 = f(v) d v = C dvxe− B Here g is the spin degeneracy. When we denote the thermal de Broglie wave length by Z Z−∞  3/2 2πkBT = C . 2 m h   λT = s2πmkB T Thus the velocity obeys Maxwell’s distribution we can write the 1 body partition function as 3/2 m mv2 f(v)= e− 2kB T . V Z1(β)= g . 2πkBT 3   λT

19 In the N particle system the canonical partition function the the equation of state, we obtain µ in terms of the takes the form natural variables for the Gibbs ensemble: 2 1 N β(ǫk + +ǫk ) 5 3 h Z = g e− 1 ··· 1 µ(p,T )= T ln p ln kBT + ln ln g . N N! · · · − 2 2 2πm − k kN   X1 X N 1 N βǫ = g e− k Grand canonical partition function N! ! According to the definition we have Xk 1 N β(E(N) µN) N = Z . Z = e− n − = z Z , N! 1 G N n XN X XN Here N! takes care of the fact that each state where z = eβµ k ,..., k | 1 N i is called the fugacity and ZN is the partition function of is counted only once. Neither the multiple occupation nor N particles. Thus we get the Pauli exclusion principle has been taken into account.

Using Stirling’s formula ln N! N ln N N the free 1 N N zZ1 βµ gV ≈ − ZG = z Z1 = e = exp e 3 . energy can be written as N! λT XN   FN = The grand potential is = k T ln Z − B N βµ gV N 3 Ω(T,V,µ)= kBT ln ZG = kBTe . = NkBT ln 1 ln g + hλ 3 V − − T − − λT   N 3 3 h2 Because dΩ= S dT p dV N¯ dµ, we get = Nk T ln ln k T 1 ln g + ln . − − − B V − 2 B − − 2 2πm " # ∂Ω Ω βµ g p = = = kB Te 3 − ∂V T,µ −V λT Because dF = S dT p dV + µ dN, the pressure will be   − − and ∂F 1 ¯ ∂Ω βµ gV pV p = = NkBT N = = e 3 = , − ∂V V − ∂µ λT kB T  T,N and we end up again with the ideal gas equation of state i.e. we end up with the ideal gas equation of state pV = Nk¯ BT. pV = NkBT. Here With the help of the entropy NzN Z N¯ = N = N N h i zN Z ∂F F 3 P N N S = = + Nk − ∂T − T 2 B 1 ∂ZG ∂ ln ZG  V,N = z P = . ZG ∂z ∂ ln z the internal energy is Another way 3 We distribute N particles among the 1 particle states so U = F + TS = NkBT that in the state l there are n particles. 2 l ǫ l 6 i.e. the ideal gas internal energy. The heat capacity is t ∂U 3 nl =1 C = = Nk . V ∂T 2 B  V,N t t t t nl =4 Comparing this with

1 t t nl =2 C = fk N V 2 B we see that the number of degrees of freedom is f = 3. We can also calculate the Gibbs function G = F + pV = µN, and substuting N/V = p/(kBT ) using

20 Now so we must have N = n and E = ǫ n . λ r . l l l T ≪ i l l X X Now The number of possible distributions is h2 λ = N! T 2πmk T W = W (n ,n ,...,n ,...)= . s B 1 2 l n !n ! n ! 1 2 · · · l · · · is the minimum diameter of the wave packet of a particle Since in every distribution (n1,n2,...) each of the N! with the typical thermal energy (¯ǫl = kBT ) so in other permutatations of the particles gives an identical state words: the partition function is The Maxwell-Boltzmann approximation is valid when the wave packets of individual particles do not overlap. ∞ ∞ 1 β(E µN) Z = We− − G · · · N! n =0 n =0 7.3. Diatomic ideal gas X1 X2 ∞ ∞ 1 β nl(ǫl µ) = e− l − · · · n !n ! n =0 n =0 1 2 X1 X2 · · · P ∞ 1 βnl(ǫl µ) = e− − d n ! "n =0 l # Yl Xl β(ǫl µ) = exp e− − We classify molecules of two atoms to l Y h i homopolar molecules (identical atoms), e.g. H , N , • 2 2 β(ǫl µ) O , . . ., and = exp e− − 2 " # Xl heteropolar molecules (different atoms), e.g. CO, βµ • = exp e Z1 NO, HCl, . . . or exactly as earlier.  When the density of the gas is low the intermolecular Validity range of MB gas law interactions are minimal and the ideal gas equation of Now state holds. The internal degrees of freedom, however, change the thermal properties (like CV ). When we 1 βn(ǫl µ) ∂ ln ZG β ∞ n e− − assume that the modes corresponding to the internal = − n=0 n! ∂ǫl 1 βnm(ǫm µ) degrees of freedom are independent on each other, we can P∞ e− − m nm=0 nm! write the total Hamiltonian of the molecule as the sum = Qβ nhPl i − h i H Htr + Hrot + Hvibr + Hel + Hydin. so the occupation number n¯l of the state l is ≈ Here 1 ∂ ln Z 1 ∂ 2 G β(ǫl µ) p n¯l = nl = = e− − Htr = = kinetic energy h i −β ∂ǫl −β ∂ǫl 2m

β(ǫl µ) = e− − . and m = mass of molecule.

The Boltzmann distribution gives a wrong result (for real L2 ¯h2l(l + 1) Hrot = = = rotational energy gases) if 1-particle states are multiply occupied. Our 2I 2I approximation is therefore valid if with L = angular momentum and I = moment of inertia. If xi are distances of atomic nuclei from the center of n¯l 1 l ≪ ∀ mass, 2 m1m2 2 or I = mixi = d βµ βǫl m1 + m2 e e l. i ≪ ∀ X Now min ǫl = 0, so that Example: H2-molecule d =0.75A,˚ L =¯h l(l + 1), l =0, 1, 2,.... Now eβµ 1. ≪ p ¯h2 On the other hand = 85.41K 2IkB ¯ βµ N 3 e = λT , when g =1 is the characteristic temperature of rotational d.o.f. V 2 Eigenvalues h¯ l(l + 1) have (2l+1) -fold degeneracy. and 2I N¯ 1 3 vibr 1 = = 3 , H =¯hωv(ˆn + ) = vibration energy V vi 4πri 2

21 The vibrational degrees of freedom of the separation d of can be divided into terms nuclei correspond at small amplitudes to a linear tr 1 tr N harmonic oscillator.n ˆ = a†a =0, 1, 2,... Each energy F = kBT ln (Z ) level is non-degenerate. − N!   3 2 N 1 2πmkB T el = k T ln V H = orbital energy − B N! h2 "   # Corresponds to jumping of electrons from an orbital to V 3 3 2πm = k TN ln +1+ ln k T + ln another and ionization. Characteristic energy levels are − B N 2 B 2 h2 > 4   1eV kB10 K, thus, in normal circumstances these ∼ ≈ nucl rot ∞ Tr l(l+1) degrees of freedom are frozen and can be neglected. H F = NkBT ln (2l + 1)e− T corresponds to energy related to nucleonic spin − ( ) Xl=0 interactions. The spin degeneracy is T F vibr = Nk T ln 2 sinh v gy = (2I1 + 1)(2I2 + 1), where I1 and I2 are the spins of B 2T the nuclei   F = Nk T ln g . Energy terms are effectively decoupled at low nucl − B y temperatures, i.e. the energy Ei of the state i is The internal energy is

E E + E + E , ∂F i ≈ tr rot vibr U = F + TS = F T − ∂T so the partition sum of one molecule is ∂ F = T 2 , − ∂T T ∞ ∞   Z = g (2l + 1) 1 y so the internal energy corresponding to tranlational p × l=0 n=0 degrees of freedom is X X2 X p h¯2 1 β 2m β 2I l(l+1) βhω¯ v (n+ 2 ) e− − − tr tr rot vibr nucl tr 2 ∂ F 3 = Z Z Z Z , U = T = N kB T − ∂T T 2   i.e. the state sum can be factorized. Above and 3 2 tr tr β p V CV = NkB Z = e− 2m = 2 λ3 p T so we end up with the ideal gas result. X Since only F tr depends on volume V the pressure is h2 λ = T tr s2πmkBT ∂F ∂F NkBT p = = = , ∞ T −∂V − ∂V V rot r l(l+1) Z = (2l + 1)e− T l=0 i.e. we end up with the ideal gas equation of state X 2 ¯h pV = Nk T. Tr = B 2IkB

∞ 1 vibr βhω¯ v(n+ ) Z = e− 2 n=0 Rotation X 1 Typical rotational temperatures Tv − = 2 sinh Gas Tr 2T   H2 85.4 ¯hωv N2 2.9 Tv = kB NO 2.4 nucl Z = gy = (2I1 + 1)(2I2 + 1). HCl 15.2 Cl2 0.36 Approximatively (neglecting the multiple occupation of We see that Tr the room temperature. ≪ states) the state sum of N molecules is T Tr Now≪ 1 N ZN = Z , 1 rot ∞ Tr l(l+1) 2 Tr N! Z = (2l + 1)e− T 1+3e− T , ≈ where 1/N! takes care of the identity of molecules. We Xl=0 associate this factor with the tranlational sum. The free so the corresponding free energy is energy rot 2 Tr F = k T ln Z F 3Nk Te− T − B N ≈− B

22 and the internal energy Now the free energy is

rot rot 2 ∂ F 2 Tr vibr Tv U = T 6Nk T e− T . F NkB T ln − ∂T T ≈ B r ≈ T   Rotations contribute to the heat capacity like and the internal energy correspondingly 2 U vibr Nk T, rot Tr 2 Tr B C 12NkB e− T 0. ≈ V ≈ T T→0   → so the heat capacity is

T Tr ≫ Cvibr Nk . Now V ≈ B

rot ∞ Tr l(l+1) We see that in the limit T Tv two degrees of freedom Z dl (2l + 1)e− T ≫ ≈ are associated with vibrations like always with harmonic Z0 ∞ oscillators (E = T + V =2 T ). Tr T l(l+1) T h ri o t h i h i = e− T = , C V N k r o o m −Tr , Tr B tem perature 0 i o n i z a t i o n dissociation so the free energy is e t c .

rot T T F NkBT ln T T ≈− Tr r v and the internal energy Rotation of homopolar molecules The symmetry due to the identical nuclei must be taken U rot Nk T. into account. ≈ B Example H2-gas: The contribution to the heat capacity is The nuclear spins are

rot rot 1 1 CV NkB = f NkB, I = I = , ≈ 2 1 2 2 or in the limit T T there are f rot = 2 rotational ≫ r so the nuclei are fermions and the total wave function degrees of freedom. must be antisymmetric. The total spin of the molecule is Precisely: I =0, 1. We consider these two cases: r o t C V

N k B I ↑↑=1 I ↑↓=0 Iz = 1, 0, 1 Iz =0 triplet− singlet T T r orthohydrogen parahydrogen spin spin wave Vibration wavefunctions function Typical vibrational temperatures: symmetric: antisymmetric: Gas Tv 11 = | i |↑↑i H2 6100 10 = 1 ( + ) 00 = 1 ( ) | i √2 |↑↓i |↓↑i | i √2 |↑↓i − |↓↑i N2 3340 1 1 = NO 2690 | − Spacei wave|↓↓i Space wave O2 2230 function function HCl 4140 antisymmetric: symmetric: We see that Tv the room temperature. ( 1)l = 1 ( 1)l =1 T T ≫ − − − ≪ v The free energy is The corresponding partition functions are

T T v v Tr vibr T T l(l+1) F = NkB T ln e 2 (1 e− ) Z = (2l + 1)e T − ortho − l=1,3,5,... 1 h Tv i T X NkBTv NkBTe− , Tr l(l+1) ≈ 2 − Zpara = (2l + 1)e− T so l=0,2,4,... 2 X vibr Tv Tv C NkB e− T . and the partition function associated with rotation is V ≈ T   T T Zrot =3Z + Z . ≫ v ortho para

23 When T Tr collisions cause conversions between ortho Bosons and para≫ states so the system is in an equilibrium. In For bosons the creation and annihilation operators obey addition Zorto Zpara, so all 4 spin states are equally the commutation relations probable. ≈ < When T Tr the gas may remain as an metastable mixture [ai,aj†] = δij of ortho∼ and para hydrogens. In the mixture the ratio of [a ,a ] = [a†,a†]=0. the spin populations is 3 : 1. Then we must use the i j i j partion sum It can be shown that 3N N rot 4 4 ZN = ZortoZpara. a n1,...,ni,... = √ni n1,...,ni 1,... The internal energy is now i | i | − i ai† n1,...,ni,... = √ni +1 n1,...,ni +1,... . 3 1 | i | i U rot = U orto + U para 4 4 The (occupation) number operator and the heat capacity correspondingly nˆi = ai†ai 3 1 Crot = Corto + Cpara. obeys the relation 4 4 nˆ n ,...,n ,... = a†a n ,...,n ,... i | 1 i i i i | 1 i i = n n ,...,n ,... 7.4. Occupation number representation i | 1 i i Let us recall the Fock space states, which are vectors in and n =0, 1, 2,.... Hilbert space i An arbitrary one particle operator, i.e. an operator O(1), = (0) (1) . . . (N) . . . which in the configuration space operates only on the F Hp ⊕ Hp ⊕ Hp ⊕ coordinates on one particle, can be written in the (N) occupation number representation as with p is N-particle Hilbert space. Here actually (N)H (N) (N) = is symmetrised for bosons and (1) (1) p Oˆ = i O j a†a . HantisymmetrisedSH for fermions. AH i j i,j D E Denote by X n ,n ,...,n ,... (2) | 1 2 i i A two body operator O can be written as the quantum state where there are ni particles in the 1 ˆ(2) (2) particle state i. Let the energy of the state i be ǫi. Then O = ij O kl ai†aj†alak. ijkl D E X

H n ,n ,... = n ǫ n ,n ,... Example: Hamiltonian | 1 2 i i i | 1 2 i i ! X 2 ¯h 2 1 N = ni. H = + V (r , r ) −2m ∇i 2 i j i i i=j X X X6

We define the creation operator ai† so that takes in the occupation representation the form

2 ai† n1,n2,...,ni,... = C n1,n2,...,ni +1,... ¯h 2 | i | i H = i j a†a − 2m ∇ i j i,j   i.e. a† creates one particle into the state i. X i 1 Correspondingly the annihilation operator a obeys: + ij V kl a†a†a a , i 2 h | | i i j l k Xijkl ai n1,n2,...,ni,... = C′ n1,n2,...,ni 1,... , | i | − i where i.e. ai removes one particle from the state i. 2 2 ¯h 2 ¯h 2 3 The basis n1,n2,... is complete, i.e. i j = φ∗(r) φ (r) d r {| i} − 2m ∇ −2m i ∇ j  

n1,n2,... n1,n2,... =1 | i h | and ni {X} ij V kl = and orthonormal or h | | i 3 3 φi∗(r1)φj∗(r2)V (r1, r2)φk(r2)φl(r1) d r1 d r2. n′ ,n′ ,... n1,n2,... = δn n′ δn n′ . h 1 2 | i 1 1 2 2 · · · Z

24 1 Fermions 2 -integer, the particle is fermion with antisymmetric wave The creation and annihilation operators of fermions function. satisfy the anticommutation relations In bosonic systems the occupations of one particle states are nl =0, 1, 2,.... Only the grand canonical ensemble is a ,a† = a a† + a†a = δ { i j } i j j i ij simple to calculate. The grand canonical state sum is ai,aj = ai†,aj† =0. β(Hˆ µNˆ ) { } { } ZG,BE = Tr e− −

It can be shown that ∞ β(Hˆ µNˆ) = n n e− − n n 1 2 · · · 1 2 · · · n =0 ai n1,...,ni,... = D E | i {X} Si ∞ ( 1) √ni n1,...,ni 1,... , if ni =1 β nl(ǫl µ) = e− l − 0−, | − i otherwise  n =0 {X} P ai† n1,...,ni,... = | i ∞ Si βn(ǫl µ) ( 1) √ni +1 n1,...,ni +1,... , if ni =0 = e− − − | i "n=0 # 0, otherwise Yl X  1 = . Here 1 e β(ǫl µ) l − − Si = n1 + n2 + + ni 1. Y − · · · − Note: In Maxwell-Boltzmann we had The number operator satisfies W ( n )= 1 N! ; here we effectively have { } N! n1!n2!... nˆi n1,...,ni,... = ni n1,...,ni,... W ( n ) = 1, which is the correct way to symmetrise | i | i multiple{ } occupation numbers. and ni =0, 1. The grand potential is One and two body operators take the same form as in the β(ǫl µ) Note: ΩBE = kB T ln 1 e− − . case of bosons. Since ai and aj anticommute one − must be careful with the order of the creation and Xl h i annihilation operators in O(2). The occupation number of the state l is In the case of non-interacting particles the Hamiltonian 1 β nm(ǫm µ) operator in the configuration space is n¯ = n = n e− m − l h li Z · · · l G n n X1 X2 P H = H1(ri), 1 ∂ ∂Ω i = ln ZG = , X −β ∂ǫl ∂ǫl where 1-particle Hamiltonian H1 is and for the Bose-Einstein occupation number we get 2 ¯h 2 1 H1(ri)= i + U(ri). n¯l = . −2m ∇ eβ(ǫl µ) 1 − − Let φj be eigenfunctions of H1 i.e. Entropy H1φj (r)= ǫjφ(r). Because dΩ= S dT p dV N dµ we have − − − In the occupation space we have then ∂Ω S = ∂T ˆ  µ,V H = ǫjaj†aj = ǫjnˆj β(ǫl µ) j j = kB ln 1 e− − X X − − l h i and X 1 β(ǫl µ) 1 ˆ kBT (ǫl µ)e− − − . N = a†a = nˆj . β(ǫl µ) 2 j j − 1 e− − − kB T j j l − X X X The grand canonical partition function is now Now β(ǫl µ) 1 e − =1+ β(Hˆ µNˆ ) β nl(ǫl µ) n¯l ZG = Tr e− − = e− l − . · · · and n1 n2 X X P β(ǫ µ) = ln(1 +n ¯ ) lnn ¯ , l − l − l so 7.5. Bose-Einstein (BE) ideal gas n¯l In relativistic QM it can be shown that the spin of the S = kB ln 1 − − n¯l +1 particle is connected to the statistics: if the spin is l X   integer, s =0, 1,..., the particle is and the wave +kB n¯l [ln(¯nl + 1) lnn ¯l] function is symmetric wrt. permutations. If the spin is − Xl

25 n or l k T 1 B S = k [(¯n + 1)ln(¯n + 1) n¯ lnn ¯ ] . B l l − l l Xl

7.6. Fermi-Dirac ideal gas e l m The Hamiltonian operator is The expectation value of the square of the occupation ˆ number will be H = ǫlal†al l X 2 1 2 β(Hˆ µNˆ ) nl = Trn ˆl e− − and the number operator ZG,FD 1 1 ˆ † N = al al. 1 2 β ′ n ′ (ǫ ′ µ) = n e− l l l − l Z · · · l X G,FD n =0 n =0 X1 X2 P Now 2 1 1 ∂ ZG,FD a ,a†′ = δ ′ l l ll = 2 2 { } β ZG,FD ∂ǫl and

al,al′ = al†,al†′ =0. 1 1 β(ǫ ′ µ) { } { } = 1+ e− l − −β ZG,FD  ′  The eigenvalues of the number operator related to the l =l Y6 h i state l,   ∂ β(ǫl µ) nˆl = al†al, e− − ×∂ǫl are β(ǫl µ) e− − nl =0, 1. = =n ¯l. β(ǫl µ) 1+ e− − The state sum in the grand canonical ensemble is 2 This is natural, since nl = nl. For the variance we get ZG,FD β(Hˆ µNˆ ) 2 2 2 2 = Tr e− − (∆nl) = nl nl =n ¯l n¯l 1 1 − h i − β(Hˆ µNˆ ) =n ¯l(1 n¯l). = n n e− − n n − · · · 1 2 · · · 1 2 · · · n1=0 n2=0 D E D n l X X k T 1 1 B β nl(ǫl µ) = e− l − · · · n =0 n =0 X1 X2 P 1 βn(ǫl µ) = e e l − − m (n=0 ) Yl X There are fluctuations only in the vicinity of the chemical β(ǫl µ) = 1+ e− − . potential µ. Yl h i The entropy is The grand potential is ∂Ω S = β(ǫl µ) −∂T ΩFD = kBT ln 1+ e− − . β(ǫl µ) − = k ln 1+ e− − Xl h i B Xl h i The average occupation number of the state l is =¯nl

1 β(Hˆ µNˆ) β(ǫl µ) n¯l = nl = Trn ˆle− − 1 e− − + (ǫl µ). h i ZG,FD β(ǫl µ) T z1+ e−}| − { − 1 1 Xl 1 β n ′ (ǫ ′ µ) l′ l l = nle− − 1 n¯l β(ǫl µ) 1 Z · · · Now β(ǫl µ) = ln − and 1+ e− − = , so G,FD n =0 n =0 n¯l 1 n¯l X1 X2 P − − 1 ∂ ln ZG,FD ∂ΩFD = = S = k [(1 n¯ ) ln(1 n¯ )+¯n lnn ¯ ] . −β ∂ǫl ∂ǫl − B − l − l l l l e β(ǫl µ) X = − − . β(ǫl µ) 1+ e− − Thus the Fermi-Dirac occupation number can be written as 1 n¯l = . β(ǫl µ) e − +1

26