Classical Phase Space Phase Space and Probability Density

Home , Phase space

Classical phase space Corresponding to a macro state of the system there are thus sets of micro states which belong to the ensemble Phase space and probability density with the probability ρ(P ) dΓ. ρ(P ) is the probability We consider a system of N particles in a d-dimensional density which satisﬁes the condition space. Canonical coordinates and momenta Z dΓ ρ(P ) = 1. q = (q1, . . . , qdN )

p = (p1, . . . , pdN ) The statistical average, or the ensemble expectation value, of a measurable quantity f = f(P ) is determine exactly the microscopic state of the system. Z The phase space is the 2dN-dimensional space {(p, q)}, hfi = dΓ f(P )ρ(P ). whose every point P = (p, q) corresponds to a possible state of the system. We associate every phase space point with the velocity A trajectory is such a curve in the phase space along ﬁeld µ ¶ which the point P (t) as a function of time moves. ∂H ∂H V = (q, ˙ p˙) = , − . Trajectories are determined by the classical equations of ∂p ∂q motion The probability current is then Vρ. The probability dq ∂H weight of an element Γ evolves then like i = 0 dt ∂p Z Z i ∂ dpi ∂H ρ dΓ = − Vρ · dS. = − , ∂t Γ0 ∂Γ0 dt ∂qi d S where

H = H(q1, . . . , qdN , p1, . . . , pdN , t) G 0 = H(q, p, t) = H(P, t) is the Hamiltonian function of the system. The trajectory is stationary, if H does not depend on Because Z Z time: trajectories starting from the same initial point P Vρ · dS = ∇ · (Vρ) dΓ, are identical. ∂Γ0 Γ0 Let F = F (q, p, t) be a property of the system. Now we get in the limit Γ0 → 0 the continuity equation dF ∂F ∂ = + {F,H}, ρ + ∇ · (Vρ) = 0. dt ∂t ∂t where {F,G} stands for Poisson brackets According to the equations of motion µ ¶ ∂H X ∂F ∂G ∂G ∂F q˙ = {F,G} ≡ − . i ∂p ∂q ∂p ∂q ∂p i i i i i i ∂H p˙i = − We deﬁne the volume measure of the phase space ∂qi we have YdN ∂q˙ ∂p˙ dqidpi −dN i i dΓ = = h dq1 ··· dqdN dp1 ··· dpdN . + = 0, h ∂qi ∂pi i=1 so we end up with the incompressibility condition Here h = 6.62608 · 10−34Js is the Planck constant. · ¸ X ∂q˙ ∂p˙ Note [dq dp] = Js, so dΓ is dimensionless. ∇ · V = i + i = 0. ∂q ∂p Note ∆0Γ = 1 corresponds to the smallest possible i i i volume element of the phase space where a point From the continuity equation we get then representing the system can be localized in accordanceR with the uncertainty principle. The volume ∆Γ = dΓ is ∂ρ 0 = + ∇ · (Vρ) then roughly equal to the number of quantum states in ∂t the part of the space under consideration. ∂ρ = + ρ∇ · V + V · ∇ρ The ensemble or statistical set consists, at a given ∂t moment, of all those phase space points which correspond ∂ρ = + V · ∇ρ. to identical macroscopic systems. ∂t

18 When we employ the convective time derivative We define the microcanonical ensemble so that its density distribution is d ∂ = + V · ∇ dt ∂t 1 µ ¶ ρE(P ) = δ(H(P ) − E). ∂ X ∂ ∂ ΣE = + q˙ +p ˙ , ∂t i ∂q i ∂p i i i Every point of the energy surface belongs with the same probability to the microcanonical ensemble. the continuity equation can be written in the form known The microcacnonical ensemble is stationary, i.e. ∂ρE = 0 as the Liouville theorem ∂t and the expectation values over it temporal constants. d The mixing flow is such an ergodic flow where the points ρ(P (t), t) = 0. dt of an energy surface element dΓE disperse in the course of time all over the energy surface. The points in the phase space move like an incompressible Ifρ ˆ (P, t) is an arbitrary non stationary density fluid which carries with it the constant probability E distribution at the moment t = t , then describing the ensemble. 0 1 lim ρÊ(P, t) = δ(H(P ) − E) = ρE(P ) Flow in phase space t→∞ ΣE The energy surface ΓE is the manifold determined by the equation and H(q, p) = E. Z

lim hfi = lim dΓρ Ê(P, t)f(P ) Since the energy is a constant of motion every phase t→∞ t→∞ i Z point P (t) moves on a certain energy surface ΓEi. The expectation value of the energy of the system = dΓ f(P )ρE(P ) Z E = hHi = dΓ Hρ i.e. the density describing an arbitrary (non equilibrium) state evolves towards a microcanonical ensemble. is also a constant of motion. The volume of the energy surface is Microcanonical ensemble and entropy Z Z If the total energy of a macroscopic system is known exactly its equilibrium state can be described by a ΣE = dΓE = dΓ δ(H(P ) − E). microcanonical ensemble. The corresponding probability density is The volume of the phase space is 1 Z Z ∞ ρE(P ) = δ(H(P ) − E). ΣE dΓ = dE ΣE. −∞ For a convenience we allow the energy to have some ”tolerance” and define Let us consider the element ∆ΓE of an energy surface. 1 Non ergodic flow: In the course of time the element ρE,∆E(P ) = θ(E + ∆E − H(P ))θ(H(P ) − E). ZE,∆E ∆ΓE traverses only a part of the energy surface ΓE. Here the normalization constant Ergodic flow: Almost all points of the surface ΓE are Z sometimes arbitrarily close to any point in ∆ΓE Z = dΓ θ(E + ∆E − H(P ))θ(H(P ) − E) ⇔ E,∆E The flow is ergodic if ∀f(P ), f(P ) ”smooth enough”, is the microcanonical state sum or partition function. ¯ ZE,∆E is the number of states contained in the energy f = hfiE slice E < H < E + ∆E (see the volume measure of the holds. Here f¯ is the time average phase space). In the microcanonical ensemble the Z probability is distributed evenly in every allowed part of 1 T the phase space. f¯ = lim dt f(P (t)) T →∞ T 0 Entropy and hfiE the energy surface expectation value We define the Gibbs entropy as Z 1 Z hfiE = dΓE f(P ). ΣE S = −kB dΓ ρ(P ) ln ρ(P ).

19 Let ∆Γi the volume of the phase space element i and ρi Quantum mechanical ensembles the average probability density in i. The state of the system is, with the probability Systems of identical particles Let H1 be a Hilbert space for one particle. Then the pi = ρi ∆Γi, Hilbert space for N identical particles is in the element i and HN = H1 ⊗ H1 ⊗ · · · ⊗ H1 . X | {z } pi = 1. N copies

1 We chooce the sizes of all elements to be smallest If, for example, |xii ∈ H is a position eigenstate the possible, i.e. ∆Γi = 1. Then N-particle state can be written as X X S = −kB ∆Γiρi ln ρi = −kB ρi∆Γi ln ρi∆Γi |Ψi = i i ZZ Z X ··· dx1 ··· dxN |x1,..., xN i ψ(x1,..., xN ), = −kB pi ln pi, i where since ln ∆Γi = 0. If ρ is constant in the range ∆Γ = W we have |x1,..., xN i = |x1i ⊗ |x2i ⊗ · · · ⊗ |xN i . 1 ρ = , There are two kinds of particles: W Bosons The wave function is symmetric with respect to so that Z 1 1 the exchange of particles. S = −kB ln dΓ. W W Fermions The wave function is antisymmetric with We end up with the Boltzmann entropy respect to the exchange of particles.

S = kB ln W. Note If the number of translational degrees of freedom is less than 3, e.g. the system is conﬁned to a two Here W is the thermodynamic probability: the number of dimensional plane, the phase gained by the many particle all those states that correspond to the macroscopical wave function under the exchange of particles can be properties of the system. other than ±1. Those kind of particles are called anyons. One can show that the entropy is additive, i.e. if the The Hilbert space of a many particle system is not the system is composed of two partial systems 1 and 2 its whole HN but its subspace: entropy is ½ N 1 1 S1+2 = S1 + S2. SH = S(H ⊗ · · · ⊗ H ) symm. H = N 1 1 If we require that the entropy has a maximum under the AH = A(H ⊗ · · · ⊗ H ) antisymm. condition Z dΓ ρ(P ) = 1, Dimension of space and statistics Let us consider two identical particles in an ρ takes the form n-dimensional Euclidean space En. ρ(P ) = ρ0 ∀P ∈ ΓE. We separate the center of mass and relative coordinates: The maximum principle of the entropy leads thus to the 1 X = (x1 + x2) ∈ En microcanonical distribution. 2 x = (x1 − x2) ∈ En. Entropy and disorder The maximum of entropy Since the particles are identical we identify the points ⇔ x = x − x Microcanonical ensemble 1 2 ⇔ −x = x2 − x1 Every microscopic state which satisﬁes in the space En of the relative motion. Let us call the E < H < E + ∆E, resulting space r(n, 2). The point  ∈ r(n, 2) is the singular point in this space. is present with the same probability, i.e. there is a complete lack of information Two dimensional space ⇔ The space r(2, 2) is a circular cone with the vertex Disorder is at maximum. aperture 60◦.

20 • corresponds to a curve connecting points x and −x - x in the original space En, i.e. corresponds to the exchange of the particles.

x • cannot be squeezed continuously to a point without crossing the singular point. A closed curve that does not circulate the vertex A closed curve that circulates the singular point twice

• corresponds in the original space En to a closed curve • corresponds to the double exchange. which connects a point x to the same point x. • can be squeezed continuosly to a point without • can be continuously squeezed to a point without crossing the singular point. crossing the singular point. The space r(3, 2) − {} is said to be doubly connected. A closed curve that goes around the vertex Quantization • corresponds in the original space En to a curve which connects points x and −x, i.e. corresponds to • The configuration space of two identical particles is particle exchange. flat with the exception of the singular point x = . • cannot be continuosly squeezed to a point without • In general, the configuration space of N identical crossing the singular point, no matter how many particles is flat with the exception of a finite number times the curve circulates the vertex. of singular points. The space r(2, 2) − {} is said to be infinitely connected. • The dynamics of classical systems is governed by local equations of motion; occasional singular points Three dimensional space have no effect. The vectors in the space r(3, 2) can be specified by telling their • Quantum mechanical description is global; the topology of the configuration space is essential. • length and • In the quantum mechanics of identical particles the • direction identifying, however, the opposite configuration space must be treated (somewhat) directions. warped. The space of the relative motion can be represented a the Proceeding formally product • At every point x we set a one dimensional Hilbert r(3, 2) − {} = (0, ∞) × P2, space hx. where P2 is a surface of a three dimensional hemisphere where the opposite points on the equator are identified. • The physical state of the system is described by the A closed curve that does not circulate the singular point vectors |Ψ(x)i ∈ hx. • is closed on the hemisphere. • In every Hilbert¯ E space hx we¯ specifyE the normalized ¯ ¯ base vector ¯χx . The set {¯χx } is called a gauge. • corresponds to a closed curve from the point x to the same point x in the original space E3. • A wave function ψ is the coordinate of a state vector with respect to the base: • can be squeezed continuosly to a point without ¯ E crossing the singular point. ¯ |Ψ(x)i = ψ(x) ¯χx .

C 2 ¯ E C 3 ¯ 0 • The transformation {¯χx } → {|χxi} from a base to C 3 another causes the gauge transformation

0 iφ(x) C 1 ψ(x) → ψ (x) = e ψ(x).

A closed curve that circulates the singular point once Physics is independent on the gauge. • is a curve on the sphere connecting opposite points • We employ a linear unitary operator P (x0, x), which and, consequently, opposite points on a hemisphere moves vectors from the space hx parallely to the with the equator passing through these points. space hx0 .

21 • The gauge can be chosen so that Density operator and entropy ¯ E ¯ E Let H be the Hilbert space of a many particle system. ¯ k ¯ The probability measure tells us the weight that a state P (x + dx, x) ¯χx = (1 + ibk(x)dx ) ¯χx+dx . |ψi ∈ H represents a system with given macroscopical properties. • Because the derivative operators The apriory probability: when there is no knowledge of the actual state of the system every state in H can taken ∂ Dk = − ibk(x) with equal weight. ∂xk We define the density operator ρ so that are invariant in the gauge transform the Hamiltonian N 1 X must be written using them. ρ = |ni hn| , N n=1 • The commutator where N = dim H and |ni ∈ H are the base vectors of H. ∂bl ∂bk fkl = i[Dk,Dl] = − The expectation value of an operator A is ∂xk ∂xl hAi = Tr ρA, is independent of the gauge. which is also called as the statistical expectation value. Due to the gauge invariance a vector potential b(x) shows Here Tr B is the trace of the operator B up in the Hamiltonian. XN • b(x) is a consequence of the topology. Tr B = hn|B|ni. n=1 • The force field related to the potential is fkl Now • We can set fkl = 0 everywhere in the configuration N space except the singular points. X 1 X X Tr ρ = hn|ρ|ni = hn|n0ihn0|ni N 0 • Moving an arbitrary vector |Ψ(x)i ∈ h along a n=1 n n x 1 X closed curve it = 1 = 1, N n – remains invariant provided that we are not circulating a singular point. so, for example, hIi = Tr ρI = 1 0 – maps to the vector |Ψ i = Px |Ψi ∈ hx if we and circulate a singular point. 1 hP i = , when P = |ni hn| . n N n • Because hx is one dimensional we must have Let |ψi ∈ H be an arbitrary normalized state. The iξ probability for the state |ψi is Px = e . X hPψi = Tr ρ |ψi hψ| = hn|ρ|ψihψ|ni • Since n X 0 0 −1 = hψ |nihn| ρ|ψi = hψ|ρ|ψi Px0 = P (x , x)PxP (x , x) = Px, | {z } n I the parameter ξ is independent on the point x; ξ is 1 X 1 X = hψ|nihn|ψi = |ψ |2 characteristic to the two particle system. N N n n n 1 • A route circulating a singular point once corresponds = . in the two particle configuration space to a curve N connecting points (x1, x2)(x2, x1): Px exchanges So, we can write 1 the particles. ρ = I. N • In two dimensional space there is no reason to restrict the values of ξ to 0 (bosons) or π (fermions). Ensemble

• In three dimensional space the extra condition Macrostate is the state determined by macroscopical 2 Px = 1 forces the condition ξ = 0 or π. parameters.

22 Microstate is a particular state in a Hilbert space. Since ρ is hermitean there exists an orthonormal basis {|αi} for H, where ρ is diagonal Let us choose a set of identical macrostates. We perform X complete measurements whose results are the states ρ = pα |αi hα| . ψi, i = 1,..., M. We deﬁne the density operator of this α set, ensemble, as Here 0 ≤ p ≤ 1 M α 1 X ¯ ® ¯ ρ = ¯ψi ψi¯ . and X M M i=1 pα = 1. In this basis Then X Tr ρM = 1. hAi = Tr ρA = pαhα|A|αi. The ensemble expectation values of operators are α

XM The equation of motion 1 i i hAi = Tr ρ A = hψ |A|ψ i Let us fix the probabilities pα corresponding to the states M M i=1 |αi. Now X M ρ(t) = p |α(t)i hα(t)| . 1 X α = hAii , α M i=1 Since the state vectors satisfy the Schrödingerequations d where i¯h |α(t)i = H |α(t)i hAii = hψi|A|ψii dt ¯ ® d ¯ i −i¯h hα(t)| = hα(t)| H, is the expectation value of A in the quantum state ψ . dt In an ideal case there exists the limit we end up with the equation of motion M 1 X ¯ ® ¯ d ρ = lim ¯ψi ψi¯ , i¯h ρ(t) = [H, ρ(t)]. M dt M→∞ i=1 In a stationary ensemble the expectation values are defining the macrostate of the system. independent on time, soρ ˙ = 0 or Note In practice the method is unrealistic since it depends on the employed measurements. [H, ρ] = 0. Pure state: When the state of the system is known This is possible e.g. when ρ = ρ(H). ”quntum mechanically” accurately we can set Entropy ρ = |ψi hψ| . The entropy is defined by ¯ ® ¯ i In the corresponding ensemble every state ψ = |ψi. S = −kBTr ρ ln ρ. The statistical mechanics of a pure state reduces to ordinary quantum mechanics, e.g. In the base where ρ is diagonal, X S = −k p ln p . hAi = Tr ρA = hψ|A|ψi. B α α α Other states are known as mixed states. Entropy has the properties

1. S ≥ 0, because 0 ≤ pα ≤ 1. Properties of the density operator 2. S = 0 corresponds to a pure state, i.e. ∃α : pα = 1 0 and pα0 = 0 ∀α 6= α. ρ† = ρ 3. If the dimension N of the Hilbert space H is ﬁnite hψ|ρ|ψi ≥ 0 ∀ |ψi ∈ H the entropy has a maximum when Tr ρ = 1. 1 ρ = I The density operator associates with every normalized N |ψi ∈ H the probability 1 or pα = N ∀ |αi ∈ H. Then

pψ = Tr ρPψ = hψ|ρ|ψi. S = kB ln N .

23 4. The entropy is additive Now Let the total Hilbert space be J(E + ∆E) − J(E) ≈ ω(E) ∆E is the number of those states whose energy lies between H = H ⊗ H 1+2 1 2 (E,E + ∆E). We can also write and correspondingly J(E) = Tr θ(E − H)

ρ1+2 = ρ1 ⊗ ρ2. ω(E) = Tr δ(H − E). ¯ ® ¯ ® (i) (i) (i) If ρi ¯α = pα ¯α , then ω(E) corresponds to the volume ΓE of the energy surface ¯ E ¯ E of the classical phase space. When the system is large the ¯ (1) (2) (1) (2) ¯ (1) (2) energy spectrum is almost continuous and ω(E) can J(E) ρ1+2 ¯α , β = pα pβ ¯α , β . can be smoothened to continuos functions. Now Example 1. Free particle X The Hamiltonian is (1) (2) (1) (2) p2 Tr 1+2A = hα , β |A|α , β i, H = . α,β 2m The eigenfunctions are the plane waves so that

1 ik·r S1+2 = −kBTr 1+2ρ1+2 ln ρ1+2 ψk = √ e , X V (1) (2) (1) (2) = −kB pα pβ (ln pα + ln pβ ) α,β where the wave vector can acquire the values X X (1) (1) (2) (2) = −kB pα ln pα − kB p ln p 2π β β k = (n , n , n ), n ∈ I,V = L3. α β L x y z i = S1 + S2. The corresponding energies are

¯h2k2 p2 Density of states ²k = = . Let us denote 2m 2m H |ni = En |ni , In the limit of large volume the summation can be transformed to the integration over the wave vector, like so that X Z Z Z H = En |ni hn| . X V 3 V 3 n = dN = g d k = g d p, k (2π)3 h3 If the volume V of the system is ﬁnite the spectrum is k discrete and the states can be normalized like where g = 2S + 1 and S is the spin of the particle. Then

hn|mi = δn,m. Z µ 2 ¶ Z p p V 0 2 J1(E) = dNkθ E − = g 3 4π dp p Thermodynamic limit: 2m h 0 V 4π = g p3. V → ∞ and N → ∞ h3 3 so that N/V remains constant. So we get The state cumulant (function) is deﬁned as 2 X J (E) = C VE3/2 1 3 1 J(E) = θ(E − En), 1/2 n ω1(E) = C1VE µ ¶3/2 i.e. the value of J at the point E is the number of those 2m C1 = 2πg . states whose energy is less than E. h The state density (function) is deﬁned as Example 2. Maxwell-Boltzmann gas dJ(E) X Let us consider N free particles. The total energy is ω(E) = = δ(E − E ), dE n n X p2 E = j 2m since dθ(x)/dx = δ(x). j

24 and the state cumulant Note Neither the multiple occupation of bosons nor the Pauli exclusion principle have been taken into account. JN (E) Z Z µ ¶ p2 p2 Energy, entropy and temperature = dN ··· dN θ E − 1 − · · · − N k1 kN 2m 2m Z Z Microcanonical ensemble = dE1 ··· dEN ω1(E1) ··· ω1(EN ) We require, that

×θ(E − E1 − · · · − EN ). 1. with full certainty the energy lies between (E,E + ∆E). Thus the corresponding state density is 2. the entropy has its maximum. dJN (E) ωN (E) = Then the density operator is Z dE

= dE1 ··· dEN ω1(E1) ··· ω1(EN ) 1 ρE = θ(E + ∆E − H)θ(H − E), ZE ×δ(E − E1 − · · · − EN ). where (supposing ∆E > 0) We deﬁne the Laplace transformations ZE = Tr θ(E + ∆E − H)θ(H − E) Z ∞ −sE Ω1(s) = dE e ω1(E) = Tr [θ(E + ∆E − H) − θ(E − H)] 0 Z ∞ = J(E + ∆E) − J(E) −sE ΩN (s) = dE e ωN (E). 0 is the number of states between (E,E + ∆E). When ∆E is small, we have Now

ZE ≈ ω(E) ∆E. Ω (s) N Z ∞ Entropy is = dE1 ··· dEN ω1(E1) ··· ω1(EN ) 0 SE = kB ln ZE. Z ∞ −sE Since ZE is a positive integer, SE ≥ 0 holds. Furthermore × dE e δ(E − E1 − · · · − EN ) 0 we get Z ∞ −sE1 −sEN = dE1 ··· dEN ω1(E1)e ··· ω1(EN )e SE = kB ln[ω(E) ∆E] 0 N = kB ln ω(E) + S0, = [Ω1(s)] . so, we can write Since Z ∞ 1 √ SE = kB ln ω(E). Ω (s) = dE e−sEC VE1/2 = C V πs−3/2, 1 1 1 2 0 Note As a matter of fact we have N −3N/2 ω = ω(E,V,N). ΩN (s) = (C2V ) s , In the deﬁnition of the density operator we have applied where µ ¶ 1 √ 2πm 3/2 quantum mechanical ”ergodicity hyphothesis”: all allowed C = πC = g . 2 2 1 h2 states in the Hilbert space are equally probable.

Performing the inverse Laplace transformations we get Temperature According to thermodynamics we have 1 N 3/2N−1 ωN (E) = (C2V ) E . µ ¶ Γ( 3 N) 1 ∂S 2 = . T ∂E Note The permutation symmetry was ignored! An V,N approximative correction can be obtained when the state In the microcanonical ensemble we deﬁne the density is diveded by N!: temperature T so that

1 N 3/2N−1 1 ∂ ωN (E) = (C2V ) E . 3 = kB ln ω(E,V,N). N!Γ( 2 N) T ∂E

25 Denoting Equilibrium distributions 1 β = , kBT Canonical ensembles we have We maximise the entropy under the conditions ∂ ln ω β = . ∂E hHi = Tr ρH = E = constant Example Maxwell-Boltzmann gas hIi = Tr ρ = 1. Now 3/2N−1 So, we require that ωN ∝ E , 0 so δ(S − λ hHi − λ hIi) = 0, 3 ln ω = N ln E + ··· 0 N 2 where λ are λ are Lagrange multipliers. We get and 0 3N δTr (−kBρ ln ρ − λρH − λ ρ) = β = 0 2E Tr (−kB ln ρ − kB − λH − λ I)δρ = 0. or we end up with the equation of state for 1-atomic ideal Since δρ is an arbitrary variation, we end up with the gas: 3 canonical or Gibbs distribution E = kBTN. 1 2 ρ = e−βH , The thermodynamics of a quantum mechanical system Z can be derived from the density of states ω(E,V,N). In where Z is the canonical sum over states (or partition practice the state density of a microcanonical ensemble function) (E and N constant) is diﬃcult to calculate. X Z Z = Tr e−βH = e−βEn = dE ω(E)e−βE. n Note In the canonical ensemble the number of particles is constant, i.e. Z = Z(p, V, N, . . .). The probability for the state ψ is 1 p = Tr ρP = hψ|e−βH |ψi. ψ ψ Z Particularly, in the case of an eigenstate of the Hamiltonian, H |ni = En |ni , we have 1 p = e−βEn . n Z For one particle system we get Boltzmann distribution 1 X p = e−β²ν ; Z = e−β²ν . ν Z ν

Here ²ν is the one particle energy. Because in the canonical ensemble we have

ln ρ = −βH − ln Z,

the entropy will be

S = −kBTr ρ ln ρ = −kB hln ρi

= kBβE + kB ln Z. We recall that E is the expectation value of the energy 1 E = hHi = Tr He−βH . Z

26 The variation of the partition function is and ¡ ¢ δZ = Tr δ e−βH = −δβ Tr He−βH ln ω(E) − βE = = −δβ EZ. ln ω(E¯) − βE¯ =0, maximum µz }|¯ ¶{ The variation of the entropy is then ¯ 1 ∂S ¯ ¯ µ ¶ + ¯ − β (E − E) kB ∂E ¯ δZ ¯E=E δS = kB E δβ + β δE + 2 ¯ Z 1 ∂ S ¯ ¯ 2 + 2 ¯ (E − E) + ··· . = kBβ δE. 2kB ∂E E=E¯ ¯ According to thermodynamics the temperature will be At the point of maximum E = E we have ¯ µ ¶ ∂S ¯ 1 δE 1 k β = ¯ = T = = , B ¯ ¯ ∂E E=E¯ T (E) δS kBβ V,N 1 = . or average temperature 1 β = . So T is the average temperature. In the Taylor series kBT µ ¶ ∂2S ∂ 1 1 ∂T 1 2 = = − 2 = − 2 , Free energy ∂E ∂E T T ∂E T CV Since so Z ∂ 2 −βH ¯ − 1 (E−E¯) Z = −Tr e H = −Z hHi ¯ −βE 2k T 2C ∂β Z ≈ ω(E)e dE e| B {zV } . or normal distribution ∂ 2 ∂ ln Z As the variance of the normal distribution in the E = − ln Z = kBT , ∂β ∂T integrand we can pick up we can write (∆E)2 = k T 2C ∂ B V S = k (T ln Z) . B ∂T or p √ 2 The Helmholtzin free energy F = E − TS can be ∆E = kBT CV = O( N), expressed as because CV , as well as E, is extensive (O(N)). Thus the F = −kBT ln Z. ﬂuctuation of the energy is With the help of this the density operator takes the form ∆E 1 ∝ √ . E N ρ = eβ(F −H). Note Fluctuations can be obtained more straightforwardly from the free energy: Fluctuations D E ∂2(βF ) Let us write the sum over states as 2 (H − hHi) = − 2 . Z Z ∂β Z = dE ω(E)e−βE = dE e−βE+ln ω(E). Grand canonical ensemble We suppose that the function ω(E)e−βE has a sharp Let’s consider a system where both the energy and the maximum at E = E¯ and that ω(E) ≈ microcanonical number of particles are allowed to ﬂuctuate. The Hilbert state density. space of the system is then the direct sum w ( E ) e - b E H = H(0) ⊕ H(1) ⊕ · · · ⊕ H(N) ⊕ · · · D E and the Hamiltonian operator the sum

(0) (1) (N) _ E H = H + H + ··· + H + ··· . E ˆ Now We deﬁne the (particle) number operator N so that 1 ln ω(E) = S(E) Nˆ |ψi = N |ψi ∀ |ψi ∈ H(N). kB

27 We maximize the entropy S under constraints so E¯ N¯ S = − µ + k ln Z . hHi = E¯ = given energy T T B G D E Nˆ = N¯ = given particle number

hIi = 1. Grand potential In thermodynamics we deﬁned With the help of Lagrange multipliers we start with D E Ω = E − TS − µN, δ(S − λ hHi − λ0 Nˆ − λ00 hIi) = 0, so in the grand canonical ensemble the grand potential is and end up with the grand canonical distribution Ω = −kBT ln ZG. 1 ˆ ρ = e−β(H−µN). ZG With the help of this the density operator can be written as Here β(Ω−H+µNˆ) −β(H−µNˆ)) ρ = e . ZG = Tr e is the grand canonical partition function. In the base Note The grand canonical state sum depends on the where the Hamiltonian is diagonal this is varaibles T , V and µ, i.e. X X −β(E(N)−µN) ZG = ZG(T, V, µ). ZG = e n , N n where Fluctuations (N) (N) H |N; ni = H |N; ni = En |N; ni , Now

(N) 2 2 when |N; ni ∈ H is a state of N particles, i.e. ∂ ZG ∂ ˆ = Tr e−β(H−µN) ∂µ2 ∂µ2 Nˆ |N; ni = N |N; ni . D E −β(H−µNˆ) 2 2 2 2 = Tr e β Nˆ = ZGβ Nˆ ,

Number of particles and energy so Now D E D E (∆N)2 = (Nˆ − N¯)2 = Nˆ 2 − N¯ 2 ∂ ln ZG 1 −β(H−µNˆ) ˆ = Tr e βN ∂2 ln Z ∂N¯ ∂µ ZG 2 G ¯ D E = (kBT ) = kBT = O(N). ∂µ2 ∂µ = β Nˆ = βN¯ Thus the particle number fluctuates like and µ ¶ ∆N 1 ∂ ln ZG 1 ˆ = O √ . = − Tr e−β(H−µN)(H − µNˆ) N¯ ¯ ∂β Z N G D E = − hHi + µ Nˆ = −E¯ + µN,¯ A corresponding expression is valid also for the fluctuations of the energy. For a mole of matter the so fluctuations are ∝ 10−12 or the accuracy ≈ the accuracy of the microcanonical ensemble. ∂ ln ZG N¯ = kBT ∂µ Connection with thermodynamics ∂ ln Z ∂ ln Z E¯ = k T 2 G + k T µ G . Let us suppose that the Hamiltonian H depends on B ∂T B ∂µ external parameters {xi}:

Entropy H(xi) |α(xi)i = Eα(xi) |α(xi)i . According to the deﬁnition we have

S = −kBTr ρ ln ρ = −kB hln ρi . Adiabatic variation Now A system in the state |α(xi)i stays there provided that ln ρ = −βH + βµNˆ − ln ZG, the parameters xi(t) are allowed to vary slowly enough.

28 Now E a ( x i ) X a = 1 a Tr H δPα = hβ| H (|αi hδα| + |δαi hα|) |βi = 0 β

= Eαδ hα| αi = 0, x i so Then the probabilities for the states remain constant and X δ hHi = Tr δρ(2)H − F δx . the change in the entropy i i i X S = −k p ln p Since B α α Z Z α X dE ω(E)f(E) = dE δ(E − Eα)f(E) is zero. Now α ¿ ¯ ¯ À X ∂E ∂ ¯ ∂H ¯ ∂ = f(E ), α = hα| H |αi = α¯ ¯α + E hα| αi α ∂x ∂x ¯ ∂x ¯ α ∂x α i ¿ i¯ ¯ À i i ¯ ∂H ¯ we can write the nonadiabatic term as = α¯ ¯α , ¯ ∂x ¯ X i (2) Tr δρ H = δpαEα since hα| αi = 1. Zα Let F be the generalized force i = dE ω(E)E δp(E). ¿ ¯ ¯ À ¯ ¯ ¯ ∂H ¯ ∂Eα Fi = − α¯ ¯α = − ∂xi ∂xi According to the deﬁnition the statistical entropy is X stat and δxi the related displacement. Then S = −kB pα ln pα. X α δ hHi = − F δx . i i Its variation is i =0 X Xz }| { δSstat = −k δp ln p − k δp Statistical study B α α B α α α Let us consider the density operator in an equilibrium X state ([H, ρ] = 0). In the base {|αi}, where the = −kB δpα ln pα Hamiltonian is diagonal, Zα

= −kB dE ω(E) δp(E) ln p(E). H |αi = Eα |αi , we have X In the microcanonical ensemble ρ = pαPα, 1 1 α p(E) ∝ ∝ , ZE,∆E ω(E) where Pα = |αi hα| . holds, so stat We divide the variation of the density operator into two −kB ln p(E) = kB ln ω(E) = S (E), parts: where Sstat(E) is the microcanonical entropy. The adiabatic nonadiabatic variation of the entropy can be written as Xz }| { Xz }| { Z δρ = pαδPα + δpαPα stat stat α α δS = dE ω(E)S (E) δp(E). = δρ(1) + δρ(2). We expand Sstat(E) as a Taylor series in a neighborhood Then of the point E = E¯:

δ hHi = Tr δρ H + Tr ρ δH Sstat(E) = Sstat(E¯) ¯ X stat (1) (2) ∂H ∂S (E)¯ = Tr δρ H + Tr δρ H + δxiTr ρ + ¯ (E − E¯) + ··· ∂x ¯ i i ∂E E=E¯ X X (2) E − E¯ = pαTr H δPα + Tr δρ H − Fiδxi. = Sstat(E¯) + + ··· . stat α i T (E¯)

29 Since In the thermodynamic equilibrium the entropy S has its Z X maximum dE ω(E) δp(E) = δpα = 0, 0 (0) (0) α S = S(E,X1 ,...,Xn ). we get Let us denote by Z (0) 1 xi = Xi − Xi δSstat = dE ω(E)E δp(E) T stat(E¯) deviations from the equilibrium positions. 1 The Taylor series of the entropy will be = Tr δρ(2)H T stat(E¯) 1 X S = S0 − k g x x + ··· , or X 2 B ij i j stat stat i,j δ hHi = T δS − Fiδxi. i where µ ¶¯ 2 ¯ This is equivalent to the first law of the thermodynamics, 1 ∂ S ¯ gij = ¯ . kB ∂Xi∂Xj {X(0)} δU = T thermδStherm − δW, i We use notation provided we identify   x1 hHi = E¯ = U = internal energy  .  x =  .  and g = (gij ). T stat = T therm xn Sstat = Stherm X Then Fiδxi = δW = work. − 1 xT gx f(x) = Ce 2 , i where p −n/2 Einstein’s theory of fluctuations C = (2π) detg. We divide a large system into macroscopical partial Correlation functions can be written as systems whose mutual interactions are weak. Z ˆ ⇒ ∃ operators {Xi} corresponding to the extensive hxp ··· xri ≡ dx f(x)xp ··· xr properties of the partial systems so that · ¸ ∂ ∂ = ··· F (h) , [Xî, Xˆj] ≈ 0 ∂hp ∂hr h=0 [Xî,H] ≈ 0. where

⇒ ∃ a mutual eigenstate |E,X1,...,Xni, which is one of dx = dx1 ··· dxn the macrostates of the system, i.e. corresponding to the and parameter set (E,X ,...,X ) there is a macroscopical 1 hT g−1h 1 n F (h) = e 2 . number of microstates. Let Γ(E,X1,...,Xn) be the number of the microstates corresponding to the macrostate |E,X1,...,Xni (the volume of the phase space). pVT-system The total number of the states is When studying the stability conditions of matter we X found out that Γ(E) = Γ(E,X ,...,X ) 1 n 1 X {Xi} ∆S = − (∆T ∆S − ∆p ∆V + ∆µ ∆N ). 2T i i i i i i i and the relative probability of the macrostate |E,X1,...,Xni Supposing that there is only one volume element in the system we get Γ(E,X ,...,X ) f(E,X ,...,X ) = 1 n . 1 n − 1 (∆T ∆S−∆p ∆V +∆µ ∆N) Γ(E) f = Ce 2kB T .

The entropy of the state |E,X1,...,Xni is We suppose that the system is not allowed to exchange particles, i.e. ∆N = 0. Employing the deﬁnitions of the S(E,X1,...,Xn) = kB ln Γ(E,X1,...,Xn) heat capacity and compressibility we can write or h i 1 1 1 CV 2 1 2 k S(E,X1,...,Xn) − (∆T ) + (∆V ) f(E,X ,...,X ) = e B . 2 k T 2 V kB T κT 1 n Γ(E) f(∆T, ∆V ) ∝ e B .

30 We can now read out the matrix g: Ideal systems

Ã TV ! System of free spins CV 1 T 2 0 Let us consider N particles with spin : g = kB T . 2 V 0 1 V kB T κT 1 Si = 2 ¯h 1 i = 1,...,N. The variances are then Siz = ± 2 ¯h ® k T 2 The z component of the total spin is (∆T )2 = B CV X ® 1 + − 2 Sz = Siz = ¯h(N − N ), (∆V ) = V kBT κT . 2 i where 1 Reversibel minimum work N + = + ¯h spin count Let x = X − X(0) be the ﬂuctuation of the variable X. 2 1 For one variable we have N − = − ¯h spin count. 2 − 1 gx2 f(x) ∝ e 2 . Sz determines the macrostate of the system. Now S = S(U, X, . . .) holds and Denoting Sz = ¯hν we have 1 dU = T dS − F dX −d¯W . N + = N + ν other 2 1 We get the partial derivative N − = N − ν 2 ∂S F = . and ∂X T 1 1 1 ν = − N, − N + 1,..., N. On the other hand we had 2 2 2 X Let W (ν) be the number of those microstates for which 0 1 S = S − kB gijxixj Sz = ¯hν, i.e. W (ν) tells us, how many ways there are to 2 + − i,j distribute N particles into groups of N and N + − + − 1 particles so that N + N = N and N − N = 2ν. = S0 − k gx2, 2 B From combinatorics we know that µ ¶ so N N! W (ν) = = ∂S N + N +!N −! = −kBgx ∂X N! and = 1 1 . ( 2 N + ν)!( 2 N − ν)! F = −kBT gx. When there is no action on X from outside, the deviation W (ν) the degeneracy of the state Sz = ¯hν. x ﬂuctuates spontaneously. Let us give rise to the same The Boltzmann entropy is deviation x by applying reversible external work: S = kB ln W (ν).

dU = −F dx = kBT gx dx. Using Stirling’s formula

Integrating this we get ln N! ≈ N ln N − N 1 (∆U) ≡ ∆R = k T gx2, we get rev 2 B ln W (ν) ≈ N ln N − N where ∆R is the minimum reversible work required for · ¸ 1 1 1 the ﬂuctuation ∆X. We can write − ( N + ν) ln( N + ν) − ( N + ν) ∆R 2 2 2 − k T · ¸ f(∆X) ∝ e B . 1 1 1 − ( N − ν) ln( N − ν) − ( N − ν) 2 2 2 1 N 2 1 N + ν = N ln − ν ln 2 . 1 2 2 1 2 4 N − ν 2 N − ν

31 We look for the extremum of W (ν): On the other hand we have

1 2 2 non extensive ∂ ln W (ν) 1 N − ν N 2 extensive z }| { 4 z }| { ³ ´ = N ¡ ¢2 2ν 1 π ∂ν 2 N 2 1 N 2 − ν2 ln W appr = N ln 2 + ln N 4 tot 2 2 1 2 N + ν = ln Wtot + non extensive − ln 1 2 N − ν 1 N − ν 1 N − ν + 1 N + ν −ν 2 2 ¡ 2 ¢ Energy 1 N + ν 1 2 2 2 N − ν Let’s put the system in the external magnetic ﬁeld 1 N + ν 2 B = µ0H, = − ln 1 = 0. 2 N − ν where We can see that ν = 0. H = Hzˆ Now ¯ ¯ is the magnetizing ﬁeld. ∂2 ln W (ν)¯ N ¯ The potential energy is ¯ = ¯ 2 ¯ 1 2 2 ¯ ∂ν ν=0 N − ν X X 4 ν=0 E = −µ µ · H = −µ H µ , 4 0 i 0 iz = − < 0, i i N where µ is the magnetic moment of the particle i. so ν = 0 is a maximum. i Now Let us expand ln W (ν) as a Taylor series in the vicinity of µ = γS, its maximum: where γ is the gyromagnetic ratio. For electrons we have 2 2 3 ln W (ν) = ln W (0) − ν + O(ν ), e N γ = 2γ = − , 0 m so W (ν) obeys the normal distribution e where γ0 is the classical value 2m . − 2 ν2 W (ν) ≈ W (0)e N , For electrons we can further write e whose deviation is µz = −µBσz = ∓µB.

1 √ Here σz is the Pauli spin matrix and ∆ν = N. 2 e¯h eV µ = = 5.79 · 10−5 In this distribution B 2m T

ln W (0) ≈ N ln 2 the Bohr magneton. Thus the energy is or X W (0) ≈ 2N . E = −µ0H µiz = −µ0γHSz = ²ν, where Total number of states ² = −¯hγµ0H We have exactly is the energy/particle. For electrons we have X µ ¶ N N W = = (1 + 1) ² = 2µ0µBH. tot N + N + Now = 2N . ∆E = ² ∆ν, According to the previous treatment we can write so from the condition approximatively ω(E) ∆E = W (ν) ∆ν X Z ∞ appr 2 2 2 2 − N ν − N ν Wtot ≈ W (0)e ≈ W (0) dν e we get as the density of states −∞ ν r µ ¶ π 1 E ≈ 2N N. ω(E) = W . 2 |²| ²

32 1) Microcanonical ensemble determine the thermodynamics of the system. Denoting 1 E = ²N, 2) Canonical ensemble 0 2 The canonical partition function is the total energy will lie between −E0 ≤ E ≤ E0. X −βEn With the help of the energy the degeneracy can be Z = e . written as n

2 Here 1 4E0 E E0 + E XN ln W (ν) = N ln 2 2 − ln 2 E0 − E ² E0 − E En = −µ0H µiz = ln ω(E) + ln |²|. i=1 the energy of a single microstate. As the entropy we get Denote 1 µiz = ¯hγνi, νi = ± . S(E) = kB ln ω(E) 2 · 2 ¸ 1 4E0 E E0 + E Now = NkB ln 2 2 − ln X P 2 E0 − E 2E0 E0 − E βµ0H µiz Z = e i +non extensive term. all microstates 1 1 The temperature was deﬁned like X2 X2 P βµ0hγH¯ νi = ··· e i 1 ∂S ν =− 1 ν =− 1 = , 1 2 N 2 T ∂E  1 N X2 so =  eβµ0Hγhν¯  = ZN , 1 N E0 + E 1 β(E) = = − ln . ν=− 1 kBT (E) 2E0 E0 − E 2

We can solve for the energy: where Z1 the one particle state sum

1 1 − βµ0Hγ βµ0Hγ βE0 Z1 = e 2 + e 2 E = −E0 tanh N µ0Hγ µ ¶ = 2 cosh . 1 µ0¯hγH 2kBT = − Nµ0¯hγH tanh . 2 2kBT The same result can be obtained using the degeneracy: The magnetization or the magnetic polarization means X Z = W (ν)e−βE(ν) the magnetic moment per the volume element, i.e. ν X 1 X = W (ν)e−β²ν M = µi. V ν i X µ ¶ N −β²(N +− 1 N) = e 2 The z component of the magnetization is N + N + 1 ¡ ¢N 1 ²ν 1 ¯hγµ0Hν − 2 β²N −β² Mz = − = = e 1 + e . V µ0H V µ0H 1 The free energy F is = γ¯hν. V F = F (T, H) = −k T ln Z · B ¸ Now µ Hγ¯h = −k TN ln 2 + ln cosh 0 . E = −µ0HVMz, B 2kBT so we get for our system as the equation of state The entropy is µ ¶ µ ¶ 1 µ0¯hγH ∂F M = ρ¯hγ tanh , S = − 2 2kBT ∂T " H where ρ = N/V is the particle density. µ0Hγ¯h = NkB ln 2 + ln cosh Note The relations derived above 2kBT # E = E(T,H,N) µ ¯hγH µ ¯hγH − 0 tanh 0 . M = M(T,H,N) 2kBT 2kBT

33 Diﬀerentiating the free energy with respect to the ﬁeld H When H → 0 we end up with Curie’s law we get C µ ¶ X χ = , ∂F 1 ∂ −β²ν T − = kBT W (ν)e ∂Hz Z ∂H T ν where µ ¶ X 2 1 −β²ν µ0ρ 1 = µ0γ¯h νW (ν)e C = ¯hγ . Z kBT 2 ν

= µ0γ¯h hνi = µ0VMz. Thermodynamical identifications Since the differential of the free energy is Earlier we identified

stat term dF = −S dT − µ0V M · dH, E ≡ E = hHi = U = internal energy, the magnetization is so µ ¶ 1 ∂F F = E − TS = F therm M = − µ0V ∂H = the Helmholtz free energy µT ¶ 1 µ ¯hγH = U − T S. = − ρ¯hγ tanh 0 . 2 2kBT Now This is identical with the result we obtained in the microcanonical ensemble. dF = −S dT − µ0V M · dH Also, the microcanonical entropy = the canonical entropy = dF therm = −S dT −d¯W, + a non extensive term. so Energy d¯W = µ0V M · dH. a) Another possibility Let us identify 1 ∂ E = hE(ν)i = ²ν¯ = − Z Z ∂β therm µ ¶ E = enthalpy = H = H. 1 1 = − N² tanh β² 2 2 Then = the energy of the microcanonical ensemble. F = E − TS = Htherm − TS = Gtherm b) According to thermodynamics = the Gibbs free energy = G

F = E − TS and or dG = −S dT − µ0V M · dH dH = T dS − µ V M · dH, ∂F 0 E = F + TS = F − T ∂T so ∂F ∂ = F + β = (βF ) ∂β ∂β G = G(T, H) ∂ H = H(S, H). = − ln Z ∂β = the energy given at a). In the thermodynamics we had for a pV T system dH = T dS + V dp,

Susceptibility from which we get the analogies According to the deﬁnition the susceptibility is

µ ¶ µ ¶ p ←→ −µ0H (intensive) ∂M 1 ∂2F χ = = − 2 V ←→ V M (extensive). ∂H T µ0V ∂H ¡ ¢2 µ ρ 1 ¯hγ On the other hand we had = 0 2³ ´. kBT cosh2 hγµ¯ 0H 2KB T U = H − pV

34 and S(E) dU = T dS − p dV = T dS −d¯W, b = 0 so now b > 0 b < 0 U = H + µ0V M · H T > 0 T < 0 and - | E 0 | + | E 0 | dU = T dS + µ V H · dM, 0 Now from which 1 ∂S N E + E0 β(E) = = − ln . d¯W = −µ0V H · dM. kB ∂E 2E0 E − E0 Example Adiabatic demagnetization Now e - b E S = ln 2 + ln cosh x − x tanh x, H-H NkB where µ0¯hHγ x = . -E 0 + E 0 + E 0 -E 0 2kBT Originally the maximum of ω(E)e−βE/Z is at a negative When T → 0, then x → ∞, so that value E. Reversing the magnetic ﬁeld abruptly E → −E 1 and correspondingly β → −β. ln cosh x = ln ex(1 + e−2x) 2 The temperature can be negative if the energy is bounded = x − ln 2 + e−2x + ··· both above and below. and Classical ideal gas (Maxwell-Boltzmann ex(1 − e−2x) tanh x = gas) ex(1 + e−2x) −2x = 1 − 2e + ··· . 2 Å Hence r S i → 2xe−2x + ··· . NkB When T → ∞, then x → 0, and S We deﬁne ri so that → ln 2. NkB The volume occupied by one molecule = S 4 V 1 N k 3 B H

b a or r 3 r = 3 . i 4πρ T Typically We decrease the ﬁeld adiabatically within the interval a → b. Now S = S(H/T ), so that • the diameter of an atom or a molecule d ≈ 2A.˚ µ ¶ µ ¶ Ha Hb • the range of the interaction 2–4A.˚ Sa = S = Sb = S Ta Tb ˚ or • the free path (collision interval) l ≈ 600A. Tb Hb = . • at STP (T = 273K, p = 1atm) ri ≈ 20A.˚ Ta Ha or d ¿ ri ¿ l Negative temperature 2 20 600 A˚ The entropy of the spin system is · 2 ¸ The most important eﬀect of collisions is that the system 1 4E0 E E0 + E thermalizes i.e. attains an equilibrium, which corresponds S(E) = NkB ln 2 2 − ln , 2 E0 − E 2E0 E0 − E to a statistical ensemble. Otherwise we can forget the where collisions. Let us consider a system of one molecule which can E0 = µ0µBHN ja − |E0| < E < |E0|. exchange energy (heat) with its surroundings. Then the

35 suitable ensemble is the canonical ensemble and the • The average of the speed distribution the Boltzmann distribution Z r ∞ 8k T 1 hvi = dv vF (v) = B . ρ = hl| ρ |li = e−β²l , πm l Z 0 where the canonical partition function is • The average of the square of the speed

X Z ∞ −β²l ® 3k T Z = e . v2 = dv v2F (v) = B . l 0 m

Since in the k-space the density of 1 particle states is Note constant, in the velocity space, where ¿ À ¿ À ¿ À µ ¶ 1 2 1 2 1 2 1 1 ¯h 3 mvx = mvy = mvz = kBT d3v = d3p = d3k, 2 2 2 2 m3 m and ¿ À ¿ À the density of states is also constant. 1 2 1 2 3 mv = 3 mv = kBT, Because the system is translationally invariant we have 2 2 x 2 i.e. the energy is evenly distributed among the 3 ¯h2k2 1 ² = hk| H |ki = = mv2, (translational) degrees of freedom: the equipartition of the k 2m 2 energy. so that the velocity distribution is

2 Partition function and thermodynamics − mv f(v) ∝ hk| ρ |ki = e 2kB T The single particle partition function is Z or −βE 2 Z (β) = dE ω(E)e − mv 1 f(v) = Ce 2kB T . Z X 2 2 p2 −β h¯ k V 3 − C can be determined from the condition = g e 2m = g d p e 2mkB T h3 Z ·Z ¸ k ∞ mv2 3 − x 3 2k T V 1 = f(v) d v = C dvxe B = g (2πmk T )3/2. −∞ h3 B µ ¶ 2πk T 3/2 = C B . Here g is the spin degeneracy. m When we denote the thermal de Broglie wave length by Thus the velocity obeys Maxwell’s distribution s h2 µ ¶ λT = 3/2 2 m − mv 2πmkBT f(v) = e 2kB T . 2πk T B we can write the one body partition function as From the relation V Z Z ∞ Z1(β) = g 3 . d3v = 4πv2dv λT 0 In the N particle system the canonical partition function we can obtain for the speed (the absolute value of the takes the form velocity v = |v|) the distribution F (v) X X 1 N −β(²k +···+²k ) ZN = g ··· e 1 1 F (v) = 4πv2f(v). N! k1 kN Ã !N 1 X F ( v ) = gN e−β²k N! k 1 N = Z1 . v N! Here N! takes care of the fact that each state • The most probable speed |k1,..., kN i r 2kBT is counted only once. Neither the multiple occupation nor vm = . m the Pauli exclusion principle has been taken into account.

36 Using Stirling’s formula ln N! ≈ N ln N − N the free The grand potential is energy can be written as βµ gV Ω(T, V, µ) = −kBT ln ZG = −kBT e 3 . FN = λT = −k T ln Z B · N ¸ Since N ¯ = Nk T ln − 1 − ln g + ln λ3 dΩ = −S dT − p dV − N dµ, B V T " # we get 2 N 3 3 h ∂Ω Ω βµ g = NkBT ln − ln T − 1 − ln g + ln . p = − = − = kBT e 3 V 2 2 2πmkB ∂V V λT and Since ¯ ∂Ω βµ gV pV dF = −S dT − p dV + µ dN, N = − = e 3 = ∂µ λT kBT the pressure will be or we end up with the ideal gas equation of state ∂F 1 ¯ p = − = Nk T pV = NkBT. ∂V B V i.e. we end up with the ideal gas equation of state Here P NzN Z pV = Nk T. ¯ PN N B N = hNi = N N z ZN With the help of the entropy 1 ∂Z ∂ ln Z = z G = G . Z ∂z ∂ ln z ∂F F 3 G S = − = − + Nk ∂T T 2 B Another way the internal energy is We distribute N particles among the 1 particle states so that in the state l there are nl particles. 3 U = F + TS = Nk T ²l 2 B 6 i.e. the ideal gas internal energy. The heat capacity is t nl = 1 µ ¶ ∂U 3 CV = = NkB. t t t t nl = 4 ∂T V,N 2 Comparing this with t t nl = 2 1 C = fk N V 2 B we see that the number of degrees of freedom is f = 3.

Grand canonical partition function According to the deﬁnition we have Now X X X X X N = nl and E = ²lnl. −β(E(N)−µN) N ZG = e n = z ZN , l l N n N The number of possible distributions is where βµ N! z = e W = W (n1, n2, . . . , nl,...) = . n1!n2! ··· nl! ··· is called the fugacity and ZN is the partition function of N particles. Since in every distribution (n1, n2,...) everyone of the N! So we get permutatations of the particles gives an identical state the partition function is X 1 N N zZ1 ZG = z Z1 = e −β(H−µNˆ) N! ZG = Tr e N · ¸ X∞ X∞ 1 −β(E−µN) βµ gV = ··· W e = exp e 3 . N! λT n1=0 n2=0

37 X∞ X∞ P 1 −β nl(²l−µ) Occupation number representation = ··· e l n !n ! ··· Let us consider a system of N non interacting particles. n =0 n =0 1 2 1 " 2 # Denote by ∞ Y X 1 |n , n , . . . , n ,...i = e−βn(²l−µ) 1 2 i n! l n=0 the quantum state where there are n particles in the one Y h i i = exp e−β(²l−µ) particle state i. Let the energy of the state i be ²i. Then Ã ! l " # X X H |n1, n2,...i = ni²i |n1, n2,...i −β(²l−µ) = exp e i l X £ ¤ N = n . = exp eβµZ i 1 i or exactly as earlier. We deﬁne the creation operator a† so that Now i † P∞ 1 −βn(² −µ) a |n , n , . . . , n ,...i = C |n , n , . . . , n + 1,...i ∂ ln Z −β n e l i 1 2 i 1 2 i G £ n=0 n! ¤ = Q P∞ 1 −βn(²l−µ) † ∂²l l n=0 n! e i.e. ai creates one particle into the state i. = −β hnli Correspondingly the destruction operator ai obeys: so the occupation number n¯l of the state l is 0 ai |n1, n2, . . . , ni,...i = C |n1, n2, . . . , ni − 1,...i , 1 ∂ ln ZG 1 ∂ n¯ = hn i = − = − e−β(²l−µ) l l i.e. ai removes one particle from the state i. β ∂²l β ∂²l The basis {|n1, n2,...i} is complete, i.e. = e−β(²l−µ). X |n , n ,...i hn , n ,...| = 1 The Boltzmann distribution gives a wrong result if the 1 1 2 1 2 {n } particle states are multiply occupied. Our approximation i is therefore valid only if and orthonormal or

0 0 n¯l ¿ 1 ∀l hn , n ,... | n , n ,...i = δ 0 δ 0 ··· . 1 2 1 2 n1n1 n2n2 or eβµ ¿ eβ²l ∀l. Bosons Now min ²l = 0, so that For bosons the creation and destruction operators obey eβµ ¿ 1. the commutation relations

† On the other hand [ai, aj] = δij † † N¯ [ai, aj] = [ai , aj] = 0. eβµ = λ3 , when g = 1 V T It can be shown that and √ N¯ 1 3 a |n , . . . , n ,...i = n |n , . . . , n − 1,...i = = , i 1 i i 1 i 3 † √ V vi 4πri ai |n1, . . . , ni,...i = ni + 1 |n1, . . . , ni + 1,...i . so we must have The (occupation) number operator λT ¿ ri. Now † s nˆi = ai ai h2 λT = obeys the relation 2πmkBT † is the minimum diameter of the wave packet of a particle nˆi |n1, . . . , ni,...i = ai ai |n1, . . . , ni,...i with the typical thermal energy (¯²l = kBT ) so in other = ni |n1, . . . , ni,...i words:

The Maxwell-Boltzmann approximation is valid when the and ni = 0, 1, 2,.... wave packets of individual particles do not overlap. An arbitrary one particle operator, i.e. an operator O(1), which in the conﬁguration space operates only on the

38 coordinates on one particle, can be written in the and ni = 0, 1. occupation number representation as One and two body operators take the same form as in the X D ¯ ¯ E case of bosons. ˆ(1) ¯ (1) ¯ † O = i¯ O ¯j ai aj. Note Since ai and aj anticommute one must be careful i,j with the order of the creation and destruction operators in O(2). A two body operator O(2) can be written as In the case of non interacting particles the Hamiltonian X D ¯ ¯ E operator in the conﬁguration space is ˆ(2) ¯ (2) ¯ † † O = ij¯ O ¯kl ai ajalak. X ijkl H = H1(ri), i Example Hamiltonian where 1 body Hamiltonian H1 is X 2 X ¯h 2 1 H = − ∇i + V (ri, rj) ¯h2 2m 2 H (r ) = − ∇2 + U(r ). i i6=j 1 i 2m i i takes in the occupation representation the form Let φj be eigenfunctions of H1 i.e. ¿ ¯ ¯ À X ¯ ¯h2 ¯ H = i¯ − ∇2 ¯j a†a H1φj(r) = ²jφ(r). ¯ 2m ¯ i j i,j X In the occupation space we have then 1 † † + hij| V |kli a a a a , X X 2 i j l k ˆ † ijkl H = ²jajaj = ²jnˆj j j where ¿ ¯ ¯ À Z and X X ¯ ¯h2 ¯ ¯h2 ˆ † ¯ 2 ¯ ∗ 2 3 N = ajaj = nˆj. i − ∇ j = − φi (r)∇ φj(r) d r ¯ 2m ¯ 2m j j and The grand canonical partition function is now X X P −β(Hˆ −µNˆ) −β nl(²l−µ) hij| V |kli = ZG = Tr e = ··· e l . Z n1 n2 ∗ ∗ 3 3 φi (r1)φj (r2)V (r1, r2)φk(r2)φl(r1) d r1 d r2. Bose-Einstein ideal gas Fermions In bosonic systems the occupations of one particle states The creation and destruction operators of fermions satisfy are nl = 0, 1, 2,.... The grand canonical state sum is the anticommutation relations X∞ X∞ P −β nl(²l−µ) † † † ZG,BE = ··· e l {ai, aj} = aiaj + ajai = δij n1=0 n2=0 † † " # {ai, aj} = {ai , aj} = 0. Y X∞ = e−βn(²l−µ) It can be shown that l n=0 Y 1 = . ai |n1, . . . , ni,...i = −β(²l−µ) ½ √ 1 − e Si l (−1) ni |n1, . . . , ni − 1,...i , if ni = 1 0, otherwise The grand potential is † h i ai |n1, . . . , ni,...i = X ½ √ −β(²l−µ) Si ΩBE = kBT ln 1 − e . (−1) ni + 1 |n1, . . . , ni + 1,...i , if ni = 0 0, otherwise l The occupation number of the state l is Here X X P Si = n1 + n2 + ··· + ni−1. 1 −β nm(²m−µ) n¯l = hnli = ··· nle m ZG The number operator satisﬁes n1 n2 1 ∂ ∂Ω nˆ |n , . . . , n ,...i = n |n , . . . , n ,...i = − ln ZG = , i 1 i i 1 i β ∂²l ∂²l

39 and for the Bose-Einstein occupation number we get The state sum in the grand canonical ensemble is

1 Z n¯ = . G,FD l β(² −µ) e l − 1 = Tr e−β(Hˆ −µNˆ) X1 X1 D ¯ ¯ E ¯ −β(Hˆ −µNˆ) ¯ = ··· n1n2 ··· ¯ e ¯n1n2 ··· Entropy n1=0 n2=0 Since dΩ = −S dT − p dV − N dµ we have X1 X1 P −β nl(²l−µ) µ ¶ = ··· e l ∂Ω S = n1=0 n2=0 ∂T ( ) µ Y X1 X h i −βn(²l−µ) −β(²l−µ) = e = −kB ln 1 − e l n=0 l Y h i X 1 −1 = 1 + e−β(²l−µ) . −k T (² − µ)e−β(²l−µ) . B −β(² −µ) l 2 1 − e l kBT l l The grand potential is Now X h i 1 −β(²l−µ) eβ(²l−µ) = 1 + ΩFD = −kBT ln 1 + e . n¯l l and The average occupation number of the state l is β(²l − µ) = ln(1 +n ¯l) − lnn ¯l, 1 −β(Hˆ −µNˆ) n¯l = hnli = Trn ˆle so ZG,FD µ ¶ 1 X1 X1 P X n¯ −β 0 nl0 (²l0 −µ) S = −k ln 1 − l = ··· nle l B ZG,FD n¯l + 1 n =0 n =0 l 1 2 X 1 ∂ ln ZG,FD ∂ΩFD +kB n¯l [ln(¯nl + 1) − lnn ¯l] = − = β ∂²l ∂²l l e−β(²l−µ) or = . X 1 + e−β(²l−µ) S = kB [(¯nl + 1) ln(¯nl + 1) − n¯l lnn ¯l] . l Thus the Fermi-Dirac occupation number can be written as 1 n¯l = . Fermi-Dirac ideal gas eβ(²l−µ) + 1 n The Hamiltonian operator is l k T 1 B X ˆ † H = ²lal al l

e l and the number operator m X The expectation value of the square of the occupation ˆ † N = al al. number will be l 2® 1 2 −β(Hˆ −µNˆ) nl = Trn ˆl e Now ZG,FD † 0 1 1 P {al, al0 } = δll 1 X X 2 −β 0 nl0 (²l0 −µ) = ··· nl e l and ZG,FD n1=0 n2=0 † † {a , a } = {a , a } = 0. 2 l l0 l l0 1 1 ∂ ZG,FD = 2 2 β ZG,FD ∂² The eigenvalues of the number operator related to the  l  state l, 1 1 Y h i †  −β(²l0 −µ)  nˆl = a a , = − 1 + e l l β ZG,FD l06=l are ∂ × e−β(²l−µ) nl = 0, 1. ∂²l

40 e−β(²l−µ) Bosonic systems = =n ¯l. 1 + e−β(²l−µ)

2 Bose condensate This is natural, since nl = nl. For the variance we get Number of particles ® 2 2 2 2 The avaerage number of particles is (∆nl) = nl − hnli =n ¯l − n¯l µ ¶ =n ¯l(1 − n¯l). ∂Ω N¯ = hNi = − n ∂µ T D l k T B X 1 = eβ(²l−µ) − 1 l

or X e l N¯ = n¯ . m l There are ﬂuctuations only in the vicinity of the chemical l potential µ. We denote The entropy is z = eβµ = fugacity, ∂Ω so S = − 1 ∂T n¯l = . X h i eβ²l z−1 − 1 = k ln 1 + e−β(²l−µ) B Let us consider a free non interacting gas. Then l =¯n l 2 2 2 z }| { ¯h kl pl X −β(²l−µ) ²l = = . 1 e 2m 2m + (²l − µ). T 1 + e−β(²l−µ) l Now β²l 1−n¯l −β(² −µ) 1 1 ≤ e < ∞. Now β(²l − µ) = ln and 1 + e l = , so n¯l 1−n¯l X Sincen ¯l ≥ 0, the fugacity is restricted to lie between S = −kB [(1 − n¯l) ln(1 − n¯l) +n ¯l lnn ¯l] . l 0 < z < 1

or µ < 0. We treat the state p = 0 separately, since the corresponding occupation numbern ¯0 can become macroscopic: z n¯ = → ∞. 0 1 − z z→1 We write the grand potential as

£ βµ¤ ΩBE = kBT ln 1 − e X h 2 2 i βµ −β h¯ k +kBT ln 1 − e e 2m . k6=0

Let us deﬁne functions gα(z) so that

X∞ zn g (z) = . α nα n=1

Then V kBT ΩBE = kBT ln(1 − z) − 3 g5/2(z). λT For the number of particles we get

¯ V N =n ¯0 + 3 g3/2(z). λT

41 When the temperature is high or the density low, the Pressure g (z) n¯0 3/2 With the help of the grand potential the pressure is term V is negligible as compared with λ3 , i.e. T µ ¶ ¯ ∂Ω ΩBE N 3 p = − = − λ = g3/2(z). ∂V V V T T,N kBT Now g (z) is a positive monotonically increasing = g (z), 3/2 λ3 5/2 function and T µ ¶ so N¯ λ3 = g (0) = 0  V T 3/2 above the z=0  kB T g (z) µ ¶  λ3 5/2 N¯ T critical point λ3 = g (1) = ζ(3/2) = 2.612. p = T 3/2  V z=1  below the  kB T kB T λ3 g5/2(1) = 1.342 λ3 T T critical point. N 3 V l T p transition border 2 . 6 1 2 T 1 > T 2 > T 3 T 1

T 2 z 1 T N¯ 3 Let us choose the density ρ = V and T so that V ¯ We are dealing with a 1st order phase transition. N 3 C λ = 2.612, V V T and z = 1. If we still increase the density or decrease the N¯ 3 temperature the increase of the term V λT must originate n¯0 3 from V λT , since z ≤ 1, i.e. T T C N¯ 3 N¯ 3 V λT = g3/2(z), when V λT < 2.612 4He liquid N¯ 3 n¯0 3 N¯ 3 V λT = V λT + g3/2(1), when V λT ≥ 2.612. A second order phase transition to a super liquid state at the temperature T = 2.17K. The expression given above, When c V λ3 ≥ 2.612 , 2π¯h2 ³ ρ ´2/3 T N¯ T = , c mk 2.612 the state p = 0 will be occupied macroscopically forming B the Bose-Einstein condensate. The formation starts when results Tc = 3.13K. C the temperature is less than the critical temperature V µ ¶ H e I I 2π¯h2 ³ ρ ´2/3 H e I Tc = mkB 2.612 or the density greater than the critical density T T C µ ¶3/2 mkBT This is called a λ-transition. ρc = 2.612 . 2π¯h2 Two liquid modell 4 When T < Tc, the relative fraction of the condensate is When T < Tc, we suppose that He is composed of two µ ¶ components: super and normal components. Then n¯ 2.612 V T 3/2 0 = 1 − = 1 − . ¯ 3 ¯ N λT N Tc ρ = ρs + ρn j = js + jn n 0 N . 1 .

ρs n¯0 When T → 0, then ρ → 1, but N¯ →∼ 0.1. This is due to the fact that 4He is not an ideal liquid: T T 4 C between He atoms there is

42 • a strong repulsion at short distances, Here k is the wave vector and ½ L, left • an attraction at longer distances. λ = R, right Black body radiation (photon gas) is the polarization. The photon is a relativistic massles boson, whose spin is Photons obey the Bose-Einstein statistics S = 1, so g = 2S + 1 = 3. In the vacuum only transversal Let’s consider n photons each with the angular velocity polarization exisits, so g = 2. ω. The total energy of this system is The energy of a photon is ²n(ω) = n¯hω, p 2 2 2 ²(p) = (m0c ) + (pc) so the system is equivalent with a single harmonic = pc = ¯hkc. ocillator, 1 With the help of the frequency f or of the angular En = (n + )¯hω = n¯hω + 0-point motion. velocity ω the energy is 2 Thus we can consider a system of one harmonic oscillator ² = ¯hω = ¯h2πf = hf. which is allowed to exchage energy with its surroundings. So we can set µ = 0. Since the wave length λ is The Hamiltonian of the system is X † 2π Hˆ = (¯hck)a a . λ = , kλ kλ k k,λ we have According to the Bose-Einstein distribution the c occupation of the energy state ²(ω) is f = λ 1 n¯(ω) = . ω = ck. eβhω¯ − 1 The total energy is Z Density of states ∞ ¯hω Employing the periodic boundary conditions the wave E = dω f(ω) . eβhω¯ − 1 vector is 0 2π k = (n , n , n ), The energy density will be L x y z Z ∞ 3 so the number of states in the vicinity of k is E ¯h ω = e(T ) = 2 3 dω βhω¯ V π c 0 e − 1 µ ¶3 Z L ∞ dNk = g dk = dω e(ω, T ), 2π 0 V 2 where the energy density at the given angular velocity = g 3 4πk dk. (2π) obeys Planck’s law of radiation With the help of the angular velocity this is ¯hω3 e(ω, T ) = 2 3 βhω¯ . V ω2 dω π c (e − 1) dN = dN = g 4π k ω (2π)3 c2 c e ( w ,T) V ω2dω = g . w 2π2c3 m a x

T 3 We denote now T 2 < T < dNω = f(ω) dω, 1 w so We can see that the maximum of the intensity follows the ω2 dω f(ω) dω = V . Wien displacement law π2c3 The sum over quantum states can be replaced by the ωmax = constant × T. integration like At the long wave length limit, λ À hc or ω ¿ kB T , the kB T h¯ X X Z ∞ energy density obeys the Rayleigh-Jeans law ··· = ··· = dω f(ω) ··· . 0 2 l k,λ e(ω, T ) = vakio × ω T.

43 At a given temperature the energy density will be Z ∞ ¯hω3 e(T ) = dω π2c3(eβhω¯ − 1) 0 Z ¯h 1 ∞ x3 = 2 3 4 dx x π c (β¯h) 0 e − 1 4 4 4 We can think that the emitting surface is a hole on a kBT π = . hollow container ﬁlled with isotropic black body π2c3¯h3 15 radiation. The radiation power can be determined by Thus the energy density obeys the Stefan-Boltzmann law counting the number of photons escaping through the hole per time interval. 4 e(T ) = σT 4, c q where σ is the Stefan-Boltzmann constant

π2k4 W σ = B = 5.67 · 10−8 . q 60¯h3c2 m2K4 l ( ) Now In the time interval τ the photons escaping into the Ω = F − µN = F, direction θ originate from the region whose depth is since µ = 0. Thus the free energy is `(θ) = cτ cos θ. Z ∞ £ −βhω¯ ¤ F = kBT dω f(ω) ln 1 − e The total energy of photons landing into the space angle 0 Z ∞ element dΩ at the direction θ is V 2 £ −βhω¯ ¤ = 2 3 kBT dω ω ln 1 − e π c 0 dΩ Z ∞ e(T )Acτ cos θ . V kBT 2 £ −x¤ 4π = 2 3 3 dx x ln 1 − e π c (β¯h) 0 Thus the total energy of the radiation is 4 4 4 kBT π Z π/2 = −V 3 , dΩ π2c3¯h 45 E = e(T )Acτ cos θ rad 4π or θ=0 4 σ 1 Z π/2 F = − VT 4 = − E. 1 3 c 3 = e(T )Acτ dθ sin θ cos θ 2 0 Here 1 = Ae(T )cτ. E = e(T )V 4 is the total energy. The radiation power per unit area is The entropy is ∂F Erad 1 S = − P = = ce(T ) ∂T Aτ 4 or = σT 4. 16 σ S = VT 3. 3 c The pressure is Absorption and intensity of radiation ∂F p = − ∂V or A 4 σ p = T 4. 3 c We see that the photon gas satisﬁes the relation 1 When the radiation arrives from a given direction its pV = E. intensity is 3 E Acτe(T ) I = = Aτ Aτ Radiation of a black surface = ce(T )

44 or Since we are at a equilibrium this force must be zero. I = 4σT 4. Thus the linear term in the series expansion of U vanishes. The absorption power to a perpendicular surface is IA. Up to the second order we are left with Phonons U = U eq + U harm, where U eq is the potential energy of the equilibrium and Classical harmonic lattice We let the ions of a crystal to oscillate in the vicinity of 1 X U harm = [u (R) − u (R0)]φ (R − R0) their equilibrium position. We suppose that 4 µ µ µν RR0 1. At the average equilibrium position the crystal is a µ,ν=x,y,z 0 Bravais lattice. With every point R of the lattice we ×[uν (R) − uν (R )] can thus associate an atom. The vector R, however, ∂2φ(r) reperents only the average position of the ion. φµν (r) = . ∂rµ∂rν 2. Typical deviations from the equilibrium positions are If we are not interested in the quantities related to the small as compared with the interatomic distances. equilibrium of the crystal ( total energy, total volume, According to the hypothesis 1 the atoms of the crystal total compessibility, ...) we can forget the term U eq. The can be identified with the Bravais lattice points R; e.g. harmonic potential is usually written more generally as r(R) stands for the actual position of the ion associated 1 X with the lattice point R. If u(R) is the deviation of the U harm = u (R)D (R − R0)u (R0). 2 µ µν ν ion R from its equilibrium then RR0 µν r(R) = R + u(R). The former expression can be obtained by setting Let φ(r) be the potential energy of two ions separated by X 0 00 0 the distance r. The energy of the whole lattice is then Dµν (R − R ) = δRR0 φµν (R − R ) − φµν (R − R ). 1 X R00 U = φ(r(R) − r(R0)) 2 RR0 The heat capacity of classical lattice 1 X = φ(R − R0 + u(R) − u(R0)). The volume element of the the 3N dimensional classical 2 RR0 phase space formed by the N ions of the lattice is When we use the notation P(R) for the momentum of the Y 1 Y 1 dΓ = du(R) dP(R) = du (R)dP (R) ion R the total Hamiltonian is h h µ µ R R,µ X P2(R) H = + U. 2m and the canonical partition sum R Z Z = dΓ e−βH . Harmonic approximation Since the evaluation of the total potential U starting from The total energy E is then the actual pair interactions is hopeless we approximate it Z resorting to the hypothesis 2 (u(R) is small). The first 1 ∂ E = dΓ e−βH H = − ln Z. terms in the Taylor series of the potential U are Z ∂β N X U = φ(R) When we change variables, 2 R X u(R) = β−1/2u¯(R) 1 0 0 + (u(R) − u(R )) · ∇φ(R − R ) −1/2 2 P(R) = β P¯(R), RR0 1 X + [(u(R) − u(R0)) · ∇]2φ(R − R0) the partion function can be written as 4 · µ ¶¸ RR0 Z X 2 P(R) eq harm 3 Z = dΓ exp −β + U + U +O(u ). 2M Z eq Y 1 In the equilibrium the total force due to other ions = e−βU β−3N du¯(R) dP¯(R) × affecting the ion R is h R X · ¸ 0 X ¯ 2 X F = − ∇φ(R − R ). P(R) 1 0 0 exp − − u¯µ(R)Dµν (R − R )¯uν (R ) R0 2M 2

45 Since all dependence on the temperature is outside of the Substituting the exponential trial into the equation of integral the energy can be calculated easily motion we see that the angular velocity ω must satisfy r r ∂ −βUeq −3N 2K(1 − cos ka) K 1 E = − ln(e β × vakio) ω(k) = = 2 | sin ka|. ∂β M M 2 eq = U + 3NkBT. The solutions represent a wave advancing in the ring with The heat capacity is the phase velocity c = ω/k and with the group velocity v = ∂ω/∂k. If the wave length is large or the wave vector ∂E C = = 3Nk . k small then the disperssion relation v ∂T B Ã r ! This expression for heat capacity, due to the lattice K ω = a k vibrations, is known as Dulong-Petit’s law. M Experimentally is linear and the phase and group velocities equal. • at low temperatures the heat capacity is smaller than One dimensional lattice with base the one obtained from the Dulong-Petit law. When We suppose that in the primitive cell there are two we approach the temperature T = 0 the heat atoms. Let the equilibrium positions of the ions to be na capacity tends to zero. and na + d, where d ≤ a/2. We denote the deviations of • even at higher temperatures the measured heat the ions these equlibrium positions by u1(na) and u2(na). capacities do not approach the Dulong-Petit limit. For the simplicity we suppose that the masses of the atoms are equal. The harmonic interaction due to the nearest neighbours is Normal modes of the harmonic crystal K X One dimenional Bravais lattice U harm = [u (na) − u (na)]2 If the separation of the lattice points in the one 2 1 2 n dimensional Bravais lattice is a the lattice points are na, X G 2 n an integer. Every lattice point na is associated with + [u2(na) − u1((n + 1)a)] , 2 one atom. n We suppose that in this one dimensional lattice only the where K describes the interaction of the ions na and nearest neighbours interact. Using the notation na + d, and G the interaction of na + d and (n + 1)a. K = φ00(x), The classical equations of motion are ∂U harm the harmonic potential of the lattice is Mu¨ (na) = − 1 ∂u (na) X 1 harm 1 2 U = K [u(na) − u((n + 1)a)] . = −K[u1(na) − u2(na)] 2 n −G[u1(na) − u2((n − 1)a)] The classical equations of motion are ∂U harm Mu¨ (na) = − 2 ∂u (na) ∂U harm 2 Mu¨(na) = − = −K[u (na) − u (na)] ∂u(na) 2 1 −G[u (na) − u ((n + 1)a)]. = −K[2u(na) − u((n − 1)a) − u((n + 1)a)]. 2 1

We suppose that the N points of the lattice form a ring, Again we look for a solution of the form i.e. the deviations satisfy the boundary conditions i(kna−ωt) u1(na) = ²1e i(kna−ωt) u((N + 1)a) = u(a); u(0) = u(Na). u2(na) = ²2e .

We seek solutions of the form Substituting these into the equations of motion we end up with the linear homogenous simultaneous equations u(na, t) ∝ ei(kna−ωt) 2 −ika [Mω − (K + G)]²1 + (K + Ge )²2 = 0 To satisfy the boundary conditions we must have ika 2 (K + Ge )²1 + [Mω − (K + G)]²2 = 0. eikNa = 1. This system has a non trivial solution only if the We see that the allowed values for k are coeﬃcient determinant vanishes. From this we obtain 2π n K + G 1 p k = , n integer. ω2 = ± K2 + G2 + 2KG cos ka. a N M M

46 The ration of the amplitudes is Case 4. K = G Now we have a Bravais lattice formed by single atoms ika ²1 K + Ge with the primitive cell length a/2. = ∓ ika . ²2 |K + Ge | Three dimensional Bravais lattice of single atoms For every allowed wave vector k (counting N) we get two Using the matrix notation the harmonic potential can be solutions. Alltogether the number of the normal modes is written more compactly now 2N. 1 X We consider couple of limiting cases. U harm = u(R)D(R − R0)u(R0). 2 Case 1. k ¿ π/a RR0 The angular velocities of the modes are now Independent on the interionic forces the matrix r 0 2(K + G) D(R − R ) obeys certain symmetries: ω = − O((ka)2) 1. D (R − R0) = D (R0 − R) M µν νµ s This property can be verified by exchanging the order of KG ω = (ka). differentiations in the definitions of the elements of D: ¯ 2M(K + G) 2 ¯ 0 ∂ U ¯ Dµν (R − R ) = 0 ¯ . Since the latter dispersion relation is linear the ∂uµ(R)∂uν (R ) u= correspondingp mode is called acoustic. In the former 2. D(R) = D(−R) mode ω = 2(K + G)/M, when k = 0. Since at the long Let’s consider a lattice where the displacements from the wave length limit this mode can couple with equilibrium are u(R). In the corresponding reversal electromagnetic radiation it is called the optical branch. lattice the displacements are −u(−R). Since every At the long wave length limit, when k ≈ 0, the Bravais lattice has the inversion symmetry the energies of amplitudes satisfy both lattices must be equal, no matter what the ²1 = ∓²2 deviations u(R) are, i.e. the upper sign corresponding to the optical mode and the 1 X lower sing to the acoustic mode. U harm = u(R)D(R − R0)u(R0) 2 Case 2. k = π/a RR0 At the border of the Brillouin zone the modes are 1 X = (−u(−R))D(R − R0)(−u(−R0)) r 2 2K RR0 ω = , optical branch 1 X M = u(R)D(R0 − R)u(R0), r 2 2G RR0 ω = , acoustical branch. M for an arbitrary u(R). This can be valid only if Correspondingly for the amplitudes D(R − R0) = D(R0 − R).

²1 = ∓²2. In addition, according to the symmetry 1, we have

0 0 Case 3. K À G Dµν (R − R ) = Dνµ(R − R ), The dispersion relations are now so the matrix D is symmetric. r · µ ¶¸ P 2K G 3. R D(R) =  ω = 1 + O M K We move every ion R to R + d. This is equivalent with r · µ ¶¸ translating the whole lattice by the amount d. The 2G 1 G ω = sin ka 1 + O , potential energies of the original and the translated M 2 K lattices are equal; in particular at the equilibrium 0, i.e. X 0 and the amplitudes satisfy 0 = dµDµν (R − R )dν RR0 ² ≈ ∓² . µν 1 2 Ã ! X X The frequency of the optical branch is now independent = Ndµdν Dµν (R) . on the wave vector. Its magnitude corresponds to the µν R vibration frequency of a molecule of two atoms with equal Since the vector d is arbitrary we must have masses and coupled with the spring constant K. X On the other hand, the acoustical branch is the same as D(R) = . in the case of the linear chain. R

47 The classical equations of motion As the eigen values of a real and symmetric matrix λs(k) are real and the eigenvectors ² (k) can be harm X s ∂U 0 0 orthonormalized, i.e. Mu¨µ(R) = − = − Dµν (R − R )uν (R ), ∂uµ(R) R0ν 0 ²s(k) · ²s0 (k) = δss0 , s, s = 1, 2, 3. or in the matrix notation X The polarizations of three normal modes are ²s(k) and Mu¨(R) = − D(R − R0)u(R0) the angular velocities correspondingly R0 r λs(k) form a system of 3N equations. Again we seek solutions ωs(k) = . M of the form u(R, t) = ²ei(k·R−ωt). Let us suppose now that the mutual interaction of the ions decreases rapidly with the increasing separation. Here the polarisation vector ² tells us the direction of the Strictly speaking we suppose that motion of the ions. Furthermore we require that for every primitive vector ai the solutions satisfy the Born-von lim D(R) = O(R−5). Karman boundary conditions R→∞ Then, at long wave length, i.e. when k ≈ , we have u(R + Niai) = u(R),

2 1 1 2 when the total number of primitive cells is N = N1N2N3. sin ( k · R) ≈ ( k · R) These conditions can be satisﬁed only if the wave vector 2 2 k is of form and 2 X k 2 n1 n2 n3 D(k) ≈ − (kˆ · R) D(R). k = b1 + b2 + b3. 2 N1 N2 N3 R ˆ 2 Here bi are vectors in the reciprocal lattice and ni Let cs(k) be the eigenvalues of the matrix integers. 1 X We see that we get diﬀerent solution only if k is restricted − (kˆ · R)2D(R). 2M into the 1st Brillouin zone, i.e. there are exactly N R allowed values for the wave vector. We substtitute the trial into the equations of motion and We see that at small wave vectors the frequency is end up with ω (k) = c (kˆ)k. Mω2² = D(k)², (∗) s s where X Thus the dispersion of all three modes is a linear function −ik·R ˆ D(k) = D(R)e of k so all three modes are acoustical. In general cs(k), ˆ R together with ωs(k), depend also on the direction k of the is so called dynamical matrix. For every allowed k we propagation in addition to the mode s. have as the solution of (∗) three eigen values and vectors. The number of normal modes is therefore 3N. Three dimensional lattice with base Employing symmetry properties of D(R) we can rewrite We proceed exactly like in the case the one dimensional the dynamical matrix as lattice with base. We suppose that there are p ions in the primitive cell. Every ion in the primitive cell adds one 1 X degree of freedom so the total number of modes at a D(k) = D(R)[e−ik·R + eik·R − 2] 2 given wave vector k is 3p. The corresponding frequences R i X are ωs(k), where now s = 1, 2, 3 and i = 1, 2, . . . , p. The = D(R)[cos(k · R) − 1]. corresponding displacements are R i i i(k·R−ωi (k)t) u (R, t) = ² (k)e s . Thus the dynamical matrix is s s X 1 The polarizations are no more orthogonal but satisfy D(k) = −2 D(R) sin2( k · R). 2 Xp R i ∗ i ²s (k) · ²s0 (k) = δss0 . We see that D(k) is a real and symmetric function of k. i=1 Since D(R) is symmetric D(k) is also symmetric. We rewrite the equation (∗) as Analogically with one dimensional lattice 3 of the modes are now acoustical and the rest 3(p − 1) modes optical.

D(k)²s(k) = λs(k)²s(k).

48 Quantum mechanical treatment ² (k)eik·R s r Let us consider the harmonic Hamiltonian −i X ¯hMω (k) P(R) = √ s (a − a† ) × X 1 ks −ks Hharm = P (R)2 N 2 2M ks R ik·R ²s(k)e . 1 X + u(R)D(R − R0)u(R0) 2 The Hamiltonian is now RR0 X † 1 H = ¯hωs(k)(aksaks + ). describing the lattice. Let ωs(k) and ²s(k) be the 2 ks frequences and polarizations in the corresponding This is simply the Hamiltonian of the system of 3N classical lattice. We deﬁne the operator aks so that independent harmonic oscillators whose energies are X 1 −ik·R correspondingly aks = √ e ²s(k) · N R X 1 "r s # E = (nks + )¯hωs(k). Mω (k) 1 2 s u(R) + i P(R) . ks 2¯h 2¯hMωs(k) Here nks the eigenvalues of the occupation number operatorn ˆ = a† a , i.e. n = 0, 1, 2,.... † ks ks ks ks The Hermitean conjugate aks of the operator aks is X Einstein’s model † 1 ik·R Let us suppose that every ion of the crystal moves in a a = √ e ²s(k) · ks N R similar potential well. Then "r s # X † 1 Mωs(k) 1 H = ¯hω (a a + ), u(R) − i P(R) . E ks ks 2 2¯h 2¯hMωs(k) ks

† where the parameter The operator aks is called the phonon creation operator and a the phonon destruction operator. kBTE ks ωE ≡ We employ the canonical commutation relations for the ¯h position and momentum is the Einstein frequency common for all 3N oscillators and T the corresponding Einstein temperature. 0 E [uµ(R),Pν (R )] = i¯hδµν δRR0 The partition function of one single harmonic oscillator is 0 0 [uµ(R), uν (R )] = [Pµ(R),Pν (R )] = 0, X∞ −βhω¯ (a†a+ 1 ) −βhω¯ (n+ 1 ) Z (ω) = Tr e 2 = e 2 the identities harm ½ n=0 X X − 1 βhω¯ ik·R 0, k is not a reciprocal vector − 1 βhω¯ ¡ −βhω¯ ¢n e 2 e = = e 2 e = N, k is a reciprocal vector 1 − e−βhω¯ R n 1 and X = . ik·R 1 e = 0, R 6= 0 2 sinh( 2 β¯hω) k Since the number of modes is 3N the canonical partition together with the property of an orthogonal vector set function is ∞ ∞ ∞ P X X X 3N 1 3 −βhω¯ E (nj + ) X Z = ··· e j=1 2 [²s(k)]µ[²s(k)]ν = δµν . n1=0 n2=0 n3N =1 s=1 3N ∞ Y X 1 −βhω¯ E (n+ 2 ) 3N One can straightforwardly show that the creation and = e = Zharm(ωE) annihilation operators obey the commutation relations j=1 n=0 · µ ¶¸−3N † 0 TE [aks, ak0s0 ] = δkk0 δss = 2 sinh . † † 2T [aks, ak0s0 ] = [aks, ak0s0 ] = 0. The heat capacity is With the help of the creation and destruction operators µ ¶ ∂E ∂ ∂ ∂ ∂ ln Z the operators u(R) and P(R) can be written as C = = − ln Z = k T 2 V ∂T ∂T ∂β ∂T B ∂T s X 2 2 1 ¯h † ∂ (TE/2T ) √ = kBT T ln Z = 3NkB . u(R) = (aks + a−ks) × 2 2 N 2Mωs(k) ∂T sinh (TE/2T ) ks

49 C V The canonical partition function is 3 N k B D ulong-Petit 1 ∞ ∞ ∞ P X X X 1 experim ental −β hω¯ s(k)(n + ) Z = ··· e ks ks 2 E i n s t e i n n1=0 n2=0 n3N =0 − 1 βhω¯ (k) Y e 2 s T = , 1 − e−βhω¯ s(k) ks Debye’s model To get the exact solution we should evaluate the partition from which we can derive as the free energy function h i P X 1 X † 1 −βhω¯ s(k) −β hω¯ s(k)(a a + ) F = ¯hωs(k) +kBT ln 1 − e Z = Tr e ks ks ks 2 2 ∞ P ks ks X 1 | {z } −β hω¯ s(k)(n + ) = e ks ks 2 , 0-point energy {n =0} ks or which in turn would require the knowledge of of the Z 9N ∞ ¡ ¢ dispersions ω (k). In practice we have to be satisﬁed 2 −βhω¯ s F = F0 + kBT 3 dω ω ln 1 − e . with, normally quite realistic, Debye’s model: ωD 0 • At low temperatures only the contribution of the low ∂F ∂S ∂2F Since S = − ∂T and CV = T ∂T , is CV = −T ∂T 2 , so we energetic phonons is prominent, so have µ ¶ TD – we take into account only the acoustic modes: 2 CV = 3NkBfD . T tranversal and 1 longitudinal. – we take only the phonons associated with small Here Z 3 x y4ey k, so we can employ the linear dispersions fD(x) = 3 dy y 2 x 0 (e − 1) ωl(k) = clk is the so called Debye function. ωt(k) = ctk. Typical Debye temperatures

• We cut the spectra at the Debye frequency TD k T Au 170 ω ≡ B D , Cu 315 D ¯h Fe 420 where TD is the corresponding Debye temperature. Cr 460 B 1250 In each mode j the density of states is C (diamond) 1860 µ ¶ L 3 V dN (ω) = 4πk2dk = ω2dω. Note The higher T the stiﬀer, harder crystal. j 2π 2π2c3 D j Behaviour of CV : Thus the total density of states is T → ∞ Since µ ¶ Z x V 2 1 2 3 2 dN(ω) = + ω dω. fD(x) → dy y = 1, 2 3 3 x→0 x3 2π ct cl 0 Since the total number of states is we have Z µ ¶ CV → 3NkB, ωD V 2 1 3N = dN(ω) = + ω3 , 2 3 3 D or we end up with the Dulong-Petit heat capacity. ω=0 6π ct cl T → 0 where N is the number of primitive cells, we get as the Since Debye temperature Z ∞ 4 y µ ¶ 3 y e constant −1 fD(x) → dy = , N 2 1 x→∞ 3 y 2 3 3 2 x 0 (e − 1) x ωD = 18π 3 + 3 . V ct cl we get Correspondingly the state density is 4 µ ¶3 3 12π T 9N 2 CV (T ) → vakio × T = NkB . dN(ω) = 3 ω dω (ω < ωD). 5 TD ωD

50 C V The vibrational degrees of D e b y e freedom of the separation d of nuclei correspond at small a T amplitudes to a linear harmonic e l e c t r o n i c contribution oscillator. T † a T 3 nˆ = a a = 0, 1, 2,... Each energy level is non Diatomic ideal gas degenerate

Hel = electronic energies

d • jumping of electrons from an orbital to another

We classify molecules of two atoms as • ionization > 4 • homopolar molecules (identical atoms), e.g. H2,N2, • energies ∼1eV ≈ kB10 K O , ..., and 2 • in normal circumstances • heteropolar molecules (diﬀerent atoms), e.g. CO, these degrees of freedom are NO, HCl, ... frozen and can be neglected. When the density of the gas is low the intermolecular interactions are minimal and the ideal gas equation of state holds. The internal degrees of freedom, however, energies corresponding to Hnucl = change the thermal properties (like CV ). nucleonic degrees of freedom When we suppose that the modes corresponding to the In normal circumstances only the internal degrees of freedom are independent on each nuclear spins are interesting. The other, we can write the total Hamiltonian of the molecule spin degeneracy is as the sum gy = (2I1 + 1)(2I2 + 1), where I and I are the spins of H ≈ Htr + Hrot + Hvibr + Hel + Hnucl. 1 2 the nuclei Here Energy terms do not couple appreciably, i.e. the energy p2 E of the state i is Htr = = kinetic energy i 2m m = mass of molecule Ei ≈ Etr + Erot + Evibr, L2 Hrot = = rotational energy so the partition sum of one molecule is 2I L = angular momentum X X∞ X∞ I = moment of inertia Z1 = gy(2l + 1) × p l=0 n=0

X p2 2 m1m2 −β −β h¯ l(l+1)−βhω¯ (n+ 1 ) I = m x2 = d2 e 2m 2I v 2 i i m + m i 1 2 = ZtrZrotZvibrZnucl, Example H2-molecule d = 0.75A˚ i.e. the state sum can be factorized. p Above L = ¯h l(l + 1), l = 0, 1, 2,...

2 X p2 tr −β V ¯h Z = e 2m = = 85.41K λ3 2IkB p T eigenvalues s h2 ¯h2 λ = l(l + 1) T 2πmk T 2I B X∞ are (2l + 1)-fold degenerated rot − Tr l(l+1) Z = (2l + 1)e T l=0 ¯h2 vibr 1 H = ¯hωv(ˆn + ) = vibration energy Tr = 2 2IkB

51 X∞ vibr −βhω¯ (n+ 1 ) Rotation Z = e v 2 Typical rotational temperatures n=0 · ¸ Gas Tr T −1 = 2 sinh v H2 85.4 2T N2 2.9 ¯hωv NO 2.4 Tv = kB HCl 15.2 nucl Cl2 0.36 Z = gy = (2I1 + 1)(2I2 + 1). We see that Tr ¿ the room temperature. Approximatively (neglecting the multiple occupation of T ¿ Tr states) the state sum of N molecules is Now X∞ 1 N rot − Tr l(l+1) −2 Tr Z = Z , Z = (2l + 1)e T ≈ 1 + 3e T , N N! 1 l=0 where 1/N! takes care of the identity of molecules. We so the corresponding free energy is associate this factor with the translational sum. The free energy rot −2 Tr F ≈ −3NkBT e T F = −kBT ln ZN can be divided into terms and the internal energy · ¸ µ rot ¶ tr 1 tr N rot 2 ∂ F −2 Tr F = −k T ln (Z ) U = −T ≈ 6Nk T e T . B N! ∂T T B r " # µ ¶ 3 N 1 2πmk T 2 = −k T ln V B Rotations contribute to the heat capacity like B N! h2 µ ¶2 · ¸ rot Tr −2 Tr V 3 3 2πm C ≈ 12NkB e T → 0. = −k TN ln + 1 + ln k T + ln V T T →0 B N 2 B 2 h2 ( ) ∞ X T À Tr rot − Tr l(l+1) F = −NkBT ln (2l + 1)e T Now · l=0 ¸ Z ∞ rot − Tr l(l+1) vibr Tv Z ≈ dl (2l + 1)e T F = NkBT ln 2 sinh 0 2T ,∞ Fnucl = −NkBT ln gy. T − Tr l(l+1) T = − e T = , Tr Tr The internal energy is 0 ∂F so the free energy is U = F + TS = F − T ∂T µ ¶ rot T ∂ F F ≈ −NkBT ln = −T 2 , Tr ∂T T and the internal energy so the internal energy corresponding to tranlational rot degrees of freedom is U ≈ NkBT. µ ¶ ∂ F tr 3 The contribution to the heat capacity is U tr = −T 2 = N k T ∂T T 2 B 1 Crot ≈ Nk = f rot Nk , and V B 2 B tr 3 CV = NkB rot 2 or in the limit T À Tr there are f = 2 rotational so we end up with the ideal gas result. degrees of freedom. Since only F tr depends on volume V the pressure is Precisely: r o t C V tr ∂F ∂F NkBT N k p = − = − = , B ∂V ∂V V i.e. we end up with the ideal gas equation of state T T pV = NkBT. r

52 Vibration We consider these two cases: Typical vibrational temperatures: I = 1 I = 0 Gas Tv Iz = −1, 0, 1 Iz = 0 H2 6100 triplet singlet N2 3340 NO 2690 orthohydrogen parahydrogen spin spin wave O2 2230 HCl 4140 wavefunctions function We see that T À the room temperature. symmetric: antisymmetric: v |1 1i = |↑↑i T ¿ Tv 1 |1 0i = √ (|↑↓i + |↓↑i) |0 0i = √1 (|↑↓i − |↓↑i) The free energy is 2 2 |1−1i = |↓↓i h i Tv Tv Space wave Space wave vibr 2T − T F = NkBT ln e (1 − e ) function function

1 − Tv antisymmetric: symmetric: ≈ Nk T − Nk T e T , l l 2 B v B (−1) = −1 (−1) = 1 so The corresponding partition functions are µ ¶2 T vibr Tv − v X Tr C ≈ Nk e T . − T l(l+1) V B T Zortho = (2l + 1)e l=1,3,5,...

X Tr T À Tv − T l(l+1) Zpara = (2l + 1)e Now the free energy is l=0,2,4,...

Tv and the partition function associated with rotation is F vibr ≈ Nk T ln B T rot Z = 3Zortho + Zpara. and the internal energy correspondingly When T À Tr collisions cause conversions between ortho vibr U ≈ NkBT, and para states so the system is in an equilibrium. In addition Zorto ≈ Zpara, so all 4 spin states are equally so the heat capacity is probable. < When T ∼Tr the gas may remain as an metastable mixture vibr of ortho and para hydrogens. In the mixture the ratio of C ≈ NkB. V the spin populations is 3 : 1. Then we must use the partion sum We see that in the limit T À Tv two degrees of freedom 3N N rot 4 4 are associated with vibrations like always with harmonic ZN = ZortoZpara. oscillators (E = hT i + hV i = 2 hT i). The internal energy is now r o t C V r o o m N k B tem perature rot 3 orto 1 para i o n i z a t i o n U = U + U dissociation 4 4 e t c . and the heat capacity correspondingly

T T r T v 3 1 Crot = Corto + Cpara. 4 4 Rotation of homopolar molecules The symmetries due to the identity of nuclei must be taken into account. Example H2-gas: The nuclear spins are

1 I = I = , 1 2 2 so the total spin of the molecule is

I = 0, 1.

53 Fermionic systems and can be discarded. If φ(²) regular enough in the vicinity of ² ≈ µ we can Electron gas expand The ideal Fermi gas is a good approximation for example 0 1 000 3 for the conducting electrons in a metal. φ(µ + ²) − φ(µ − ²) ≈ 2φ (µ)² + 2 φ (µ)² + ··· . 2 2 3! When the single particle energies are ² = h¯ k the k 2m Now density of states is Z ∞ φ(²) µ ¶ d² 3/2 β(²−µ) gs 2m √ 0 e + 1 ω1(²) = V ² Z 4π2 ¯h2 µ ≈ d² φ(²) µ ¶3/2 2m √ 0 Z = V 2πgs ². ∞ h2 0 2 z +2φ (µ)(kBT ) dz z 0 e + 1 As the density we get Z ∞ 3 000 1 4 z +2φ (µ) (kBT ) dz µ ¶3/2 Z √ 3! ez + 1 N¯ g 2m ∞ ² 0 ρ = = s d² . + ··· 2 2 β(²−µ) V 4π ¯h 0 e + 1 and we end up with Sommerfeld’s expansion The energy per particle will be Z ∞ Z µ 2 φ(²) π 2 0 R 3/2 d² ≈ d² φ(²) + (k T ) φ (µ) ∞ ² β(²−µ) B E d² β(²−µ) 0 e + 1 0 6 ²¯ = = R0 e +1 . ¯ ∞ ²1/2 4 N d² 7π 4 000 0 eβ(²−µ)+1 + (k T ) φ (µ) + ··· . 360 B

Degenerated Fermi gas Temperature T = 0 Suppose that kBT ¿ µ. Now Let’s write n¯(²) = θ(µ − ²) 1 and h(x) = 0. = θ(µ − ²) + h(² − µ), eβ(²−µ) + 1 The Fermi energy is 2 2 where ¯h kF 1 ²F = µ = . h(x) = sign(x) . 2m eβ|x| + 1 The Fermi momentum is The function h(x) deviates from zero only at the narrow < domain |x|∼kBT ¿ µ. pF = ¯hkF . h ( x ) The density is

µ ¶3/2 Z µ gs 2m √ ρ = 2 2 d² ² x = e - m 4π ¯h 0 µ ¶3/2 gs 2m²F = 2 Let’s evaluate the integral 6π2 ¯h

Z ∞ or φ(²) gs 3 d² ρ = kF . β(²−µ) 6π2 0 e + 1 Z ∞ The spin degeneracy factor of electrons is = d² φ(²)[θ(µ − ²) + h(² − µ)] 1 gs = 2 · 2 + 1 = 2, so Z0 Z µ ∞ k3 = d² φ(²) + d² h(²)[φ(µ + ²) − φ(µ − ²)] F ρ = 2 . 0 0 3π Z ∞ + d² h(²)φ(µ − ²). For the energy per particle we get µ R 1 3/2 0 dx x 2/5 ²¯ = ²F R = ²F The last term is of the order 1 1/2 2/3 0 dx x 1 −µ/kB T 3 h(µ) = ≈ e = ²F . eµ/kB T + 1 5

54 The total energy is Now µ ¶ 2 µ ¶2 3 3 ¯h2 6π2ρ 2/3 ¯h 1.92 3.69 ²F = 2 = 2 13.6eV. E = ²F N = N . 2ma rs r 5 5 2m gs | {z 0} s binding Since energy of E = constant × N 5/3V −2/3, hydrogen we have µ ¶ Since ∂E 2 E −p = = − 1eV = 11604kBK, ∂V N 3 V we have µ ¶ or 1.92 2 2 T = 13.6 · 11604K. pV = E. F r 3 s For aluminium the Fermi temperature is TF = 136 000K. In general, the metals satisfy Metallic electron gas When we write the density as T ¿ TF , ¯ N 1 so the metallic electron gas is strongly degenerated. ρ = = 4 3 V 3 πri Speciﬁc heat Let now T > 0, but T ¿ TF . N¯ We need µ = µ(T ), when V = ρ is known. and deﬁne the dimensionles number With the help of the Sommerfeld expansion we get ri rs = , 2 µ 2 ¶3/2 Z ∞ 1/2 a 2 3/2 4π ¯h ² 0 ² = ρ = d² F β(²−µ) 3 gs 2m 0 e + 1 where a is the Bohr radius 0 2 π2 1 ≈ µ3/2 + (k T )2 √ + ··· 2 3 12 B µ 4π²0¯h ˚ a0 = 2 = 0.529A, me so we can write we can see that · 2 ¸ 2 3/2 π 2 1 2 3/2 µ 1 + (kBT ) + ··· = ² . 1.613 · 1030 1 3 8 ²2 3 F ρ = . F r3 m3 s From this we get for the chemical potential the expression For metals we have " µ ¶ # π2 k T 2 µ(T ) = ² 1 − B + ··· . < < F 1.9∼rs∼5.6. 12 ²F

The Fermi wave vector will become Employing again the Sommerfeld expansion we get

r Z ∞ 1 9π 1.92 ²3/2 3 d² kF = = . β(²−µ) a0rs 4 a0rs 0 e + 1 2 2 5/2 π 2√ The Fermi velocity is ≈ µ + (kBT ) µ + ··· 5 " 4 # µ ¶2 pF ¯hkF 1.92¯h 2 5/2 5 2 kBT vF = = = = ²F 1 + π + ··· . m m ma0rs 5 12 ²F 4.2 · 106 m = . Now the energy/particle is rs s R ∞ ²3/2 For example in aluminium d² β(²−µ) ²¯(T ) = R0 e +1 ∞ ²1/2 km c 0 d² eβ(²−µ)+1 vF = 2029 = . " µ ¶ # s 148 3 5 k T 2 = ² 1 + π2 B 5 F 12 ² The Fermi temperature or the degeneracy temperarure TF F is deﬁned so that 3 π2 k2 T 2 = ² + B . kBTF = ²F . F 5 4 ²F

55 The heat capacity which can be written as The Fermi wave vectors can be determined from the conditions 2 2 ∂N²¯ π kB CV = = N T 2 2 ∂T 2 ²F ¯h kF + + µBB = µ π2 T 2m = Nk , 2 B 2 T ¯h k2 F F − − µ B = µ. 2m B is small when compared e.g. with the speciﬁc heat of the 3 Maxwell-Boltzmann gas (CV = NkB 2 ). This is Since the number density is understandable since the number of those particles that g can be excited with the thermal energy ∼ k T in Fermi s 3 B ρ = 2 kF , gas is much less than in MB or BE gases due to the Pauli 6π exclusion principle. the spin population densities are

Pauli’s paramagnetism k3 ρ = F + The magnetic moment of the electron is + 6π2 3 e kF − µ = − s ρ− = . m 6π2 or If the strength of the magnetic field is µz = −µBσz, ²F where B0 = , µB e¯h −5 eV µB = = 5.66 · 10 2m T the magnetic energy is of the same order as the Fermi 5 and energy. For metals ²F ≈ 5eV, so B0 ≈ 10 T. So the 2 σz = sz = ±1. realistic magnetic fields are ¿ B0 and we can work at the ¯h small B limit. Let us denote In an external magnetic field the energy of an electron is kF ± = kF ± δkF , p2 ² = ² = − µ B = ² + µ Bσ pσz p± 2m z p B z so when the kinetic energy is 2 2 2 2 2 ¯h kF ± ¯h kF ¯h kF ± µBB = ± δkF ± µBB p2 2m 2m m ²p = . ¯h2k2 2m = µ = F . 2m We still treat electrons as non interacting so the grand canonical partition function is as before, provided that we From this we get replace ² → ² + µBBσz. p p mµB The occupation numbers of the states are now δkF = − 2 B ¯h kF 1 n¯pσz =n ¯p± = . and eβ(²p+µB Bσz −µ) + 1 3 2 kF kF ρ± = ± δkF n p s 6π2 2π2 3 n k kF mµB p - = F ∓ B. 6π2 2π2¯h2 n p + The relative polarization is e p ρ − ρ 3mµ Since the metallic electron gas is strongly degenerated r ≡ + − = − B B 2 2 (T ¿ TF ), we can restrict to the temperature T = 0. ρ+ + ρ− ¯h kF e k 3µB e k - e k + = − B. 2²F = e F m The magnetization per volume element is k k k N F- F + M = hµ i = −ρµ hσ i = −ρµ r V z B z B

56 or The eigenenergies are the discrete Landau levels 2 3 µB µ ¶ M = ρ B. 1 2 ²F En = n + ¯hωc, n = 0, 1, 2,.... The susceptivity is, according to its deﬁnition, 2 ∂M ∂M Choosing the Landau gauge χ = = µ0 . ∂H ∂B A = (0, Bx, 0) Pauli’s paramagnetic susceptivity is then the single particle Hamiltonian is 3 µ2 " # B µ ¶2 χ = µ0ρ 1 eB 2 ²F H = p2 + p + x . 0 2m∗ x y c provided that T ¿ TF and µBB ¿ ²F . In aluminium the electron density is The eigenfunctions are

29 −3 µ ¶ 2 2 x − X ρ = 1.82 · 10 m iky y −(x−X) /2` φnX = e e 0 Hn , `0 and the Fermi energy µ ¶ where the center of the oscillatory motion is given by 1.92 2 ²F = 13.6eV = 11.7eV. 2 2.07 X = −ky`0.

The susceptivity Conﬁne the system in a rectangular cell z 3 (5.66 · 10−5)2 χ = · 4π · 10−7 · 1.82 · 1029 · 6 2 11.7 Vs 1 (eV)2 - y Am m3 T2eV µ ¶ B © eV Vs m2 2 = 9.4 · 1013 x 4 6 Am Vs Lx = 9.4 · 1013 · 1.6 · 10−19 = 1.5 · 10−5 - A is now small since only the electrons very close to the Fermi surface can be polarized magnetically. ©

Two dimensional electron gas Ly - The Hamiltonian for a free electron in the magnetic ﬁeld

B = ∇ × A is given by Using periodic boundary conditions we have 1 ³ e ´2 H0 = −i¯h∇ + A . 2πny 2m∗ c ky = , ny = 0, ±1, ±2,... Ly Convenient unit of and 2πny 2 • the energy for non-interacting electrons ishω ¯ c, where X = − `0, 0 ≤ X < Lx. ∗ Ly ωc = eB/m c is the cyclotron frequency.

2 The number of allowed values of ny, i.e. the degeneracy • the energy for interacting electrons is e /²`0, where of each Landau level, is 1 2 • the length is `0 = (¯hc/eB) , the magnetic length. LxLy e Φ Ns = 2 = Φ = , Consider electrons 2π`0 hc Φ0 • confined to xy-plane. where Φ0 = hc/e is the flux quantum. Thus, on each Landau level there is exactly one state for • subjected to a perpendicular magnetic field Bkzˆ. each flux quantum and for each spin polarization.

57 When Ne is the number of electrons in an area and Ns Quantum Hall states the number of flux quanta we define the filling fraction as Consider an experiment like µ ¶ Ne n0 T ν = = 4.136 15 −2 . Ns 10 m B To treat the spin we note that

• there should be the Zeeman coupling term jx B HZeeman = µ · B = −gµBBsz 6 in the Hamiltonian. Here g is the Lande factor and µB the Bohr magneton. + − • in addition to the Zeeman term there are no spin dependent terms in the Hamiltonian, not even in the interacting many body system. + Ey − - • the problem can be solved disregarding the spin. At + − V later stages we can add the total Zeeman energy L

EZeeman = gµBBSz. + −

The conductivity σ and the resistivity ρ are deﬁned by j = σE, E = ρj. Classically the diagonal and Hall conductivities are

2 n0e τ 1 σxx = 2 m 1 + (ωcτ) n0ec σxx σxy = − + , B ωcτ

where τ is the relaxation time. In particular ρxy = −B/n0ec. Experimentally the resistivities behave like

58 The Hall conductivity can be written in the form n ec σ = − 0 + ∆σ , xy B xy where, according to the Kubo formula, the contribution from a localized state |αi to ∆σxy is

f(E )ec ∆σα = α . xy B Here f(E) is the Fermi distribution function. When the number of electrons changes we observe (at T = 0) that

• as long as the Fermi level lies within the localized states, σxy remains constant. • if all states below the Fermi level are localized, the terms in σxy cancel exactly and σxy = 0. • for QHE to exist there must be extended states in Landau levels.

As a function of the density the conductivities behave like

We observe that D(E)

• the Hall resistivity develops plateaus with e 2 h

x y n 0 e C

h s ρ = , n = 1, 2, 3,.... - B xy ne2 This quantization condition is obeyed with extreme x x

accuracy. In fact, the current ISO standard for s resistivity defines 02 p l 2 n 1 25812.807 0 0 ρ = Ω. Noting that xy n Ne 2 n0 ν = = 2π`0n0 ∝ , Ns B • at the same time the diagonal resistivity practically vanishes. decreasing magnetic field corresponds to increasing filling factor, i.e. decreasing the magnetic field is equivalent to For the moment we suppose that the electrons are increasing the number of electrons. polarized. If the current carrying electrons fill up exactly Increasing the magnetic field (i.e. reducing the electron 2 n Landau levels, it can be shown that ρxy = h/ne and density) furthermore one finds resistivities to behave like ρxx = 0. The plateaus can be explained by noting that

• in an ideal pure 2DEG the density of states is a series of δ-peaks separated by ¯hωc.

• In a real impure system the δ-peaks are spread and between the Landau levels there are localized states. w h c w h c D(E) D(E)

E Filled Empty E E F

59 The plateaus in the Hall resistivity and the minima in the The corresponding energies are longitudinal resistivity correspond to ﬁlling fractions 1 p En,m = (2n + |m| + 1 − m)¯hωc. ν = , 2 q In particular, in the lowest Landau level (n = 0, m ≥ 0), where the wave functions are

· ¸ 1 µ ¶ < 2 m • p and q are small integers (∼11). 1 z −|z|2/4`2 ψ (z) = e 0 , m 2π`22mm! ` • q is an odd integer. 0 0 where we have written This behaviour is called as the Fractional Quantum Hall Eﬀect (FQHE) as opposed to the previous Integer z = re−iφ = x − iy. Quantum Hall Eﬀect (IQHE). Regarding the IQHE we note that It is easy to show, that the quantum number m can take the values • the plateaus correspond to full Landau levels, m = 0, 1,...,Ns − 1, • the Landau levels are energetically far from each where other as compared to typical electron-electron A < Ns = 2 interaction energies (at least when ν∼5). 2π`0 • the mutual electronic interactions play practically no is the degeneracy of a Landau level. In the lowest Landau role. level the wave functions are therefore of the form

2 Ns−1 While this single particle picture is suﬃcient in the IQHE 1, z, z , . . . , z times Gaussian. it cannot explain the FQHE where The great idea of Laughlin was to propose the Jastrow • the Landau levels are only partially ﬁlled, so that type function

Ne Ne • there is room for the Coulomb intra level interaction. Y Y 2 2 m −|zj | /4` ψm = (zj − zk) e 0 It turns out that the correlations due to the electron j

2 • it can be mapped to a charge neutral two p 1 1 ∗ 2 2 H = + ωcpφ + m ω r . dimensional classical plasma, which makes it possible 2m∗ 2 8 c to use classical statistical mechanics to evaluate e.g. The Schr¨odingerequation takes the form the energy.

· µ ¶ ¸ < ¯h2 1 ∂ ∂ψ 1 ∂2ψ • small systems (∼12 particles) can be solved exactly. − r 2m∗ r ∂r ∂r r2 ∂φ2 Comparisions with Laughlin’s wave function show µ ¶ that it is practically the exact solution of the many 1 ∂ψ 1 − i¯hω + m∗ω2r2 − E ψ = 0. body problem 2 c ∂φ 8 c

Its solutions can be written as

· ¸ 1 n! 2 2 2 −imφ−r /4`0 ψn,m(r, φ) = 2 m e 2π`02 m! µ ¶|m| µ 2 ¶ r |m| r × Ln 2 . `0 2`0

60 Spin polarization • H0 is the single particle Hamiltonian, We consider the ﬁlling fraction ν = 1, i.e. the lowest Landau level is fully occupied. We turn on the • He-e is the Coulomb interaction between an electron-electron interaction and note that electron and all other electrons and their images summed over all electrons, • typically the Landau level separation ¯hωc is (much) larger than the characteristic Coulomb • He-im is the interaction between an electron and its 2 interaction energy e /²`0. images,

• if the electrons remain polarized the interaction • He-b is the electron-background interaction, cannot do much: all energetically favorable states are already occupied. • Hb-b is the background-background interaction. So, we let electrons to ﬂip their spins. However, The Zeeman coupling is treated afterwards. We • according to Hund’s rule the repulsive interaction is the smaller the larger the total spin S. 1. restrict to the lowest Landau level.

• In the absence of the Zeeman coupling all possible Sz 2. work in occupation representation. There states are degenerate. X † H = wjajσajσ • the Zeeman coupling gµBBSz tends to polarize the jσ system, although the Lande g-factor is rather small X + A a† a† a a , (in GaAs g ≈ 0.5). j1j2j3j4 j1σ1 j2σ2 j3σ3 j4σ4 j σ j σ We conclude that the ground state at ν = 1 is polarized. 1 1 2 2 j3σ3 j4σ4 The diagonalization method † We will work in rectangular geometry with periodical where operators ajσ (ajσ) create (destroy) an boundary conditions. electron with spin σ in a single particle state j. y 3. fix Ns, the number of flux quanta (≈ 10). This is 6 - also the number of allowed single particle states. Lx 4. fix N , the number of electrons. At full Landau level e e e e e e e e e e (ν = 1) Ne = Ns. e e e e e e e e 5. fix thee polarization Sz and the total momentum since they are preserved by Coulomb interaction. ue e 6 e e u u e e 6. form the basis by constructing all possible non-interacting states satisfying the above conditions. ue e L ue e y ue 7. represente the Hamiltonian as a matrix in the basis constructed? above. e e e e e e e e e e e e 8. diagonalize the matrix. As a result we have the e e e energy spectrum and corresponding eigenvectors. e e e 9. for each eigenstate find its total spin S. Since [H,S] = 0, we know that these eigenstates are eigenstates of spin, too. - x

hSzi We now have the spectrum E0,E1,E2,... for the Our Hamiltonian is interacting many particle system. To calculate the polarization we note that H = H0 + He-e + He-im + He-b + Hb-b, where • the energies Ei are associated with other quantum numbers like the total spin Si and its z-component • we suppose a homogenous positive background, Szi.

61 2 1/3 • since there are no spin dependent term in the When kF = (3π ρ) is of the order kc, the relativistic Hamiltonian all states with quantum numbers corrections must be taken into account. The (Ei,Si,Szi = −Si), (Ei,Si,Szi = −Si + 1), ..., corresponding density is (Ei,Si,Szi = +Si) are degenerate. So, the 3 kc 35 1 expectation value of Sz would be 0. ρ = = 5.87 · 10 c 3π2 m3 • the Zeeman interaction must be turned on. The ≈ 106 × density of metallic electron gas energies will shift like We have an ultrarelativistic electron gas when kF À kc or ²i = Ei − gµBBSzi. correspondingly ρ À ρc. Let us consider cold relativistic material, i.e. let us It turns out that, as expected, the total spin in the ground suppose T ¿ TF . The total energy is state is S0 = Ne/2 (supposing Ne to be even). Due to the R p Zeeman coupling the ground state is polarized at T = 0. kF 2 2 2 dk k c¯h k + kc The spins of the excited states, however, have all the E = N²¯ = N 0 R , kF 2 possible values 0, 1,...,Ne/2. So, we expect the 0 dk k polarization to decrease with increasing temperature. where The dependence on temperature is evaluated in the R √ canonical ensemble as kF /kc dx x2 1 + x2 ²¯ = mc2 0 R 1 X kF /kc dx x2 −(Ei−gµB BSzi)/kB T 0 hSzi = Szie , R £√ ¤ Z kF /kc 2 i dx x 1 + x2 − 1 = mc2 + mc2 0 R kF /kc 2 where Z is the canonical partition sum 0 dx x X is the average electronic energy. Z = e−(Ei−gµB BSzi)/kB T . At the non relativistic limit we have i " R # kF /kc dx x2[ 1 x2 + ···] ²¯ ≈ mc2 1 + 0 R 2 kF /kc 2 Relativistic electron gas 0 dx x " µ ¶ # The rest energy of an electron is 3 k 2 = mc2 1 + F + ··· , mc2 = 0.511keV 10 kc and the relativistic total energy from which our earlier results can be derived, provided p that the rest energy of electrons is taken into account. 2 2 2 ²p = (mc ) + (cp) At the ultrarelativistic limit kF À kc we get 2 R k /k 2 p F c 3 = mc + + ··· . 2 0 dx x 3 2m ²¯ ≈ mc R = c¯hkF . kF /kc 2 4 0 dx x Denote by mc 12 −1 Thus the energy density is kc = = 2.59 · 10 m ¯h E 3 = (3π2)1/3c¯hρ4/3 the Compton wave vector of an electron and by V 4

2π −12 and the pressure µ ¶ λc = = 2.43 · 10 m ∂E kc p = − ∂V its Compton wave length. N Since p = ¯hk, we have at the ultrarelativistic limit p 1 E 1 2 1/3 4/3 2 2 p = = (3π ) c¯hρ . ²k = c¯h k + kc . 3 V 4 Periodic boundary conditions are the same as in the non White dwarf relativistic case i.e. In a properly functioning star the energy released in 2π nuclear reactions (mainly 2H → He) and the collapsing k = (n , n , n ), L x y z gravitational force are in balance. When the nuclear fuel is comsumed the start collapses. If the mass of the star is so we have k3 large enough all material will become ionized. Depending ρ = F . 3π2 on the mass of the star the ﬁnal state can be for example

62 • white dwarf, if the pressure of the degenerated Because in nuclei there are roughly as many neutrons as electronic plasma prohibits further compression. protons, and, on the other hand, there are as many protons as electrons, we have • neutron star if the electronic pressure is not enough to compensate the gravitational force. The matter ρm(r) ≈ 2mpρ(r). compresses further to neutrons and their degeneracy pressure prohibits further collapse. Here m = 1.673 · 10−27kg Typical properties of a white dwarf: p is the proton mass and ρ(r) the number density of the 4 • the diameter of the star 2R ≈ 10 km. electrons. As a good approximation the electron density of a star • the total number of nuclei N ≈ 1057. N can be taken as a constant, ρ say. Then 30 • the mass M ≈ 10 kg ≈ M¯, where 8 30 3 M = 1.989 · 10 kg is the mass of the sun. M(r) = πmpρr ¯ 3 10 −3 6 • the mass density ρm ≈ 10 kg m is about 10 × the and thus the total mass density of the sun or of the earth. 8 M = πm ρR3, • the number density of electrons ρ ≈ 1036m−3. Then 3 p kF ≈ kc, so the electron gas is only moderately relativistic. In inner parts the gas can be much when R is the radius of the star. The pressure must now denser and thus ultrarelativistic. satisfy the diﬀerential equation dp 16 • the pressure p ≈ 1022Pa ≈ 1017atm. = − πm2ρ2Gr dr 3 p 7 • the temperature in inner parts T ≈ 10 K ≈ T¯. with the boundary condition that the pressure vanishes at 10 Since the Fermi temperature is TF ≈ 10 K À T we, the surface, i.e. however, have a cold electron gas. p(R) = 0.

Let p(r) be the pressure at the distance r from the center Integrating the diﬀerential equation we get for the of the star, g(r) the corresponding gravitational pressure at the center acceleration and ρm(r) the density. 8π p A p = Gm2ρ2R2. 3 p p + d p r + d r Since the electron gas is not quite ultrarelativistic we p r ( p + d p ) A calculate more accurately than before. The average r m A d r g electronic energy is The condition for the balance of hydrostatic mechanical R √ kF /kc dx x2 x2 + 1 forces is ²¯ = mc2 0 R dp kF /kc dx x2 = −g(r)ρ (r). 0 dr m R £ ¤ kF /kc dx x3 1 + 1 1 + ··· = mc2 0 2 x2 Now R k /k GM(r) F c dx x2 g(r) = , · 0 ¸ r2 3 k 3 k = mc2 F + c + ··· . where M(r) is the mass inside of the radius r and 4 kc 4 kF

2 From this we can get for the pressure −11 Nm G = 6.673 · 10 2 kg ¯hc p = (k4 − k2k2 + ···) 12π2 F c F is the gravitational constant. We get the pair of equations · ¸ 1 m2c2 = (3π2)1/3¯hcρ4/3 1 − + ··· . dp(r) M(r)ρ (r) 4 ¯h2(3π2ρ)2/3 = −G m dr r2 dM(r) This is the equation of state of the relativistic electron = 4πr2ρ (r). dr m gas.

63 We require that the pressures obtained from the equation Neutron star of state and from the hydrodynamic balance conditions When the mass of a star exceeds the Chandrasekar limit are equal in the center, i.e. the Fermi pressure of the electrons is not enough to cancel the gravitational force. The star continues its 8π 2 2 Gmpρ R = collapse. The star forms a giant nucleus where most 3 electrons and protons have transformed via the reaction · 2 2 ¸ 1 2 1/3 4/3 m c (3π ) ¯hcρ 1 − 2 + ··· . − 4 ¯h (3π2ρ)2/3 p + e → n + νe

When we substitute the electron density (as a function of to neutrons. The radius of the star is the mass and radius) R ≈ 10km, 3M ρ = 8πm R3 the nucleon count p 57 NN ≈ 10 we get the condition and the mass density µ ¶2/3 µ ¶2 µ ¶2/3 M R Mc 18 −3 = 1 − , ρm ≈ 10 kg m . Mc Rc M The pressure acting against the gravitation is mostly due where to the pressure of the Fermi gas and to the strong, at µ ¶1/2 µ ¶3/2 short distances very repulsive nuclear forces. 9π ¯hc 57 Mc = mp 2 ≈ 0.52 · 10 mp 512 Gmp Quark matter µ ¶1/3 µ ¶1/3 ¯h 9π Mc When nuclear matter is compressed 2–10 times denser Rc = ≈ 4700km. than in atomic nuclei the nucleons start to ”overlap” and mc 8 mp their constituent quarks form a quark plasma. For the radius of the star we get 3 " # Liquid He µ ¶1/3 µ ¶1/3 1 M M The nucleus is p+p+n and the nuclear spin 2 . R = Rc 1 − . Mc Mc At low temperatures the nuclear spin determines the statistics, i.e. 3He atoms are Fermions. We see that the white dwarf has the maximum mass The Fermi temperature corresponding to the normal M = Mc. A more careful calculation shows that the mass density is ²F of a white dwarf cannot exceed Chandrasekhar’s limit, TF = ≈ 5K. kB about 1.4M¯, without collapsing to a neutron star or a black hole. Since the mutual interactions between 3He atoms are considerable the 3He matter forms an interacting Fermi Other Fermionic systems liquid. The 3He liquid has two super phases (A and B). These are in balance with the normal phase at the critical Nuclear matter point T The mass density of heavy nuclei is T ≈ 2.7mK < F . c 1000 17 −3 ρm ≈ 2.8 · 10 kg m .

When we assume that the proton and neutron densities are equal the Fermi wave vectors of both gases are

15 −1 kF ≈ 1.36 · 10 m and the Fermi energies

²F ≈ 38MeV.

2 Since mnc = 938MeV, the nuclear matter is non relativistic. The attractive nucleon interactions cancel the pressure due to the kinetic energy.