Physics 262Topic 2: Foundations of Quantum Statistical Physics

Erel LevineSep 2015

1. QUANTUM-MECHANICAL ENSEMBLE THEORY: THE DENSITY MATRIX

Assume an ensemble ofN systems of N particles. Let∣∣ψk⟩ denote the (normalized) wavefunction characterizing the

state of system k at time t. Use a complete orthogonal set |n〉 to write∣∣ψk(t)⟩

=∑n

|n〉⟨n|ψk(t)

⟩≡∑n

akn(t) |n〉 . (1.1)

The wavefunctions akn evolve according to Schrodinger’s equation,

i~akn(t) =∑m

Hnmakm(t) .

The number∣∣akn(t)

∣∣2 represents the probability that a measurement at time t finds the kth system of the ensemble tobe in the particular state |n〉. Clearly, we must have∑

n

∣∣akn(t)∣∣2 = 1 (for all k).

Define now the density operator

ρmn(t) =1

N

N∑k=1

akm(t)ak∗n (t) (1.2)

The quantity ρnn(t) represents the probability that a system, chosen at random from the ensemble is found to be inthe state |n〉 at time t,

ρnn(t) =1

N

N∑k=1

∣∣⟨n|ψk⟩∣∣2 (1.3)

(both ensemble and QM averages), and of course∑n ρnn = Tr(ρ) = 1.

i~ρmn =1

N

N∑k=1

[i~(akm(t)ak∗n (t) + akm(t)ak∗n (t)

)]=

1

N

N∑k=1

[(∑l

Hmlakl (t)

)ak∗n (t)− akm(t)

(∑l

H∗nlak∗l (t)

)]=∑l

[Hmlρln − ρmlHln) = (Hρ− ρH)mn

i~ρ = [H, ρ] (1.4)

2

We thus conclude that an equilibrium ρ is diagonal in the energy representation. Otherwise ρ is symmetric. Thermalaverages are calculated as

〈G〉 =1

N

N∑k=1

⟨ψk∣∣G ∣∣ψk⟩

=1

N

N∑k=1

∑m,n

ak∗n akmGnm

=∑m,n

ρmnGnm =∑m

(ρG)mm

〈G〉 = Tr(ρG) (1.5)

[if the wavefunctions∣∣ψk⟩ are not normalized, then 〈G〉 = Tr(ρG)/Tr(ρ)].

2. MICRO-CANONICAL ENSEMBLE

In the energy basis (where ρ is diagonal) we postulate

ρnn = Γ−1δEn,E0. (2.1)

Pure state - all systems are in the same state |ψ〉: Γ = 1 and

ρ = |ψ〉 〈ψ| .

In another representation ρ may not even be diagonal. However, it is straightforward to show that in any represen-tation ρ2 = ρ.

Example: Assume a two-state system with energy eigenstates |1〉 and |2〉. In a pure state, all systems are in one state,say state |1〉. In the energy representation,

ρ =

(1 00 0

).

However if we choose an alternative representation |S±〉 = 1√2

(|1〉 ± |2〉) then

ρ =

(12

12

12

12

).

In this representation ρ is not diagonal, but it is symmetrical and it obeys ρ2 = ρ, as expected.

Mixed state - In this natural situation we must require

1. Equal a priory probability

2. Random a priory phases (equivalently: every∣∣ψk⟩ is an incoherent superposition of the basis |n〉.)

Then

ρmn =1

N

N∑k=1

akmak∗n =

1

N

N∑k=1

|a|2 ei(ϕkm−ϕ

kn) = c

⟨ei(ϕ

km−ϕ

kn)⟩

= cδmn .

Example: assume that an energy level is doubly degenerate, with eigenstates |R〉 and |L〉. We construct the ensembleby placing our systems in states ∣∣ψk⟩ = sin θ |R〉+ eiϕk cos θ |L〉 .

3

Then

ρk =

(cos2 θ eiϕk cos θ sin θ

e−iϕk cos2 θ sin θ sin2 θ

)and ρ =

(cos2 θ 0

0 sin2 θ

)and therefore we must require cos2 θ = sin2 θ = Γ−1 = 1/2 or θ = π/4.

3. THE CANONICAL ENSEMBLE

Anticipating Pn ∼ exp(−βEn) the density matrix in the energy representation should take the form

ρmn =1

ZN (β)e−βEnδmn (3.1)

ρ =∑n

|n〉 1

ZN (β)e−βEn 〈n| = 1

ZN (β)e−βH

∑n

|n〉 〈n| = e−βH

Tr(e−βH)(3.2)

In this ensemble the thermal quantum average is given by

〈G〉N =Tr(Ge−βH)

Tr(e−βH)(3.3)

4. THE GRAND CANONICAL ENSEMBLE

Again using the analogous classical ensemble we can define the density operator

ρ =e−β(H−µN)

Tr(e−β(H−µN))(4.1)

and the corresponding thermal (and quantum) average

〈G〉 =Tr(Ge−β(H−µN))

Z=

∑N ζ

N 〈G〉N ZN∑N ζ

NZN(4.2)

5. SIMPLE EXAMPLES

5.1. Particle in a potential well

The problem of a single particle in a box will serve us soon as the initial step towards constructing a theory ofquantum ideal gases.

H = − ~2

2m∇2

〈x|k〉 =eik·x√V

E(k) =~2k2

2m

Z1 = Tr(ρ) =∑k

exp

(−β ~

2k2

2m

)' V

∫d3k

(2π)2exp

(−β ~

2k2

2m

)=

V

vQ

〈x′| ρ |x〉= vQV

∫V d3k

(2π)3e−ik·(x−x

′)

Vexp

(−β ~

2k2

2m

)=

1

Vexp

[−m(x− x′)2

2β~2

]=

1

Vexp

[−π(x− x′)2

vQ

]P (x) = 〈x| ρ |x〉 =

1

V

4

The transition from a sum to an integral corresponds to taking the classical limit. Note that the consistency with thefundamental assumption of statistical mechanics.

5.2. Harmonic oscillator

One more important and recurring problem.

H =

(N +

1

2

)~ω

Tr(e−βH) =∑n

〈n| e−βH |n〉 =∑n

e−β~ω(n+1/2) =e−β~ω/2

1− e−β~ω

U = 〈H〉 = TrHρ = − ∂

∂βln Tr e−βH =

1

2~ω coth(β~ω/2)

〈q| ρ |q′〉 =

[mω tanh(β~ω/2)

π~

]1/2exp

[−mω

4~((q + q′)2 tanh(β~ω/2) + (q − q′)2 coth(β~ω/2)

)]P (q) = 〈q| ρ |q〉 =

[mω tanh(β~ω/2)

π~

]1/2exp

[−mω

~(q2 tanh(β~ω/2)

)]Gaussian distribution, σ2

q =~

2mω tanh(β~ω/2).

High T : σ2q '

kBT

mω2Thermal fluctuations

Low T : σ2q '

~2mω

Quantum fluctuations

5

6. MANY-PARTICLE QUANTUM SYSTEMS

We denote by P a permutation operator, such that

P |ψ(1, 2, ..., N)〉 = |ψ(P1, P2, ..., PN)〉 .

Since the particles are identical we require

|ψ|2 = |Pψ|2

and we therefore distinguish between two types of particles,

• Bosons correspond to

P |ψ〉 = |ψ〉

• Ferminos correspond to

P |ψ〉 = (−1)P |ψ〉

where (−1)P is +1 for an even permutation and (-1) for and odd permutation.

Now assume a single particle HamiltonianHi with eigenstates |k〉 and energies ε(k). The total Hamiltonian is givenby

H =∑i

Hi .

We can construct the N -particle states as follows:

• The product space is obtained from

|ψ〉0 = |k1〉 |k2〉 ... |kN 〉

but these do not obey any of the symmetry requirements.

• Bosonic space obtained by

|ψS〉 =1√N+

∑P

P |ψ〉0 . (6.1)

Here some (or all) ki may appear more than once, and one can specify the state by the numbers nk such that∑nk = N .

• Fermionic space obtained by

|ψA〉 =1√N−

∑P

(−1)PP |ψ〉0 , (6.2)

where all the ki’s are different.

Note that the |ψA〉 can be written in terms of a Slater determinant

|ψA〉 =1√N−

∣∣∣∣∣∣∣∣∣ui(1) ui(2) · · · ui(N)uj(1) uj(2) · · · uj(N)· · · · · ·· · · · · ·

uk(1) uk(2) · · · uk(N)

∣∣∣∣∣∣∣∣∣ . (6.3)

The numbers N+ and N− should be chosen such that the state function is normalized. For fermions this is simplyN− = N ! since all particles must be at different states. For bosons it is given by N+ = N !/

∏k nk!.

6

7. QUANTUM IDEAL GAS

We’ll construct the theory of an ideal gas moving from a single particle, for which we’ve already seen that

〈x′| ρ |x〉 =1

Vexp

[−m(x− x′)2

2β~2

], (single particle)

to two particles, and then generalize.

Two particles

Following the above discussion, the energy states of this ensemble are

|E〉 =1√2

(|E1E2〉 ± |E2E1〉)

so the density matrix in the space representation is

〈x′| ρ |x〉 =1

Z

∑E1,E2

e−β(E1+E2) 〈x′|E〉 〈E|x〉

with

〈x|E〉 =1

V

1√2

(ei(k1·x1+k2·x2) ± ei(k1·x2+k2·x1)

),

or

〈x′| ρ |x〉 =1

Z

1

4

1

V

∑k1,k2

e−β~2

2m (k21+k22)(ei(k1·x′1+k2·x′2) ± ei(k1·x′2+k2·x′1)

)(e−i(k1·x1+k2·x2) ± e−i(k1·x2+k2·x1)

)We now use

1

L3/2

∑k

· · · = 1

(2π)3

∫d3ke−αk

2

eikx =1

(4πα)3/2e−|x|

2/4α

to get

〈x′| ρ |x〉 =1

Z

1

2v2Q

[e− m

2β~2 (|x1−x′1|2+|x2−x′2|

2) ± e−m

2β~2 (|x1−x′2|2+|x2−x′1|

2)],

so the diagonal elements are

P (x) = 〈x| ρ |x〉 =1

Z

1

2v2Q

[1± e−

mβ~2 (|x1−x2|2)

]Z =

1

2v2Q

∫d3x1d

3x2[ 1︸︷︷︸O(V 2)

± e−mβ~2 (|x1−x2|2)︸ ︷︷ ︸O(V )

] ' V 2

2v2Q

P (x) =1

V 2

[1± e−

mβ~2 (|x1−x2|2)

](7.1)

ΒuH

xL FD

BE0.5 1

x�Λ

-0.5

0.5

1One could capture this behavior of a non-interactingquantum gas in terms of an interacting classical gas.In this case, the Boltzmann factor e−βu(x) should bethe same as the term in the square brackets in P (x)

7

above; that is1

u(x) = −kBT ln

[1± exp

(−2πx2

v2/3Q

)]. (7.2)

The strong repulsive interaction at close distances between fermions is due to Pauli. Notice however that Bosonsexperience ”statistical attraction”. Both interactions vanish as the inter-particle distance becomes comparable withthe quantum wavelength, that is within a few nanometers.

N particles

We now write the N -particle matrix element

〈1, 2, ..., N | e−βH |1′, .., N ′〉 =1

N !

1

vQ

3N/2∑P

δP f(|xP1 − x′1|)f(|xP2 − x′2|) · · · f(|xPN − x′N |)

with

f(x) = exp

(− m

2β~2x2).

In each term of the sum, particles that are not permuted contribute f(0) = 1. We can therefore order the permeationsin the sum by their order, and rewrite it as∑

P

δP f(|xP1 − x′1|)f(|xP2 − x′2|) · · · f(|xPN − x′N |) = 1±∑i<j

fijfji +±∑i<j<k

fijfjkfki + ...

which can be viewed as an expansion in vQ/V :

ZN =1

N !

(V

vQ

)N [1± N2

2

vQ23/2V

+ ...

]. (7.3)

To lowest order, we need only to consider the non-permuted (Boltzmann) state, and get the classical partitionfunction. Note that the Gibbs correction factor N ! appears automatically. In the subleading term the factorN2/2 ' N(N − 1)/2 is just the number of such permutations.

We can now write F = −kBT lnZ and P = − ∂F∂V to obtain

βP = ρ∓ ρ2 vQ25/2

(7.4)

which looks like a virial expansion, with a virial coefficient B2 = ∓ vQ25/2

. Indeed,

B2 = ∓ vQ25/2

, B2 = − 1

2V

∫dq1dq2f12 =⇒ f12 = ± exp

(−2πq2

v2/3Q

)

f12 = e−βu(q) − 1 = ± exp

(−2πq2

v2/3Q

)=⇒ u(q) = −kBT ln

[1± exp

(−2πq2

v2/3Q

)]

as seen above.

1 But many authors have a strong objection to this idea.

8

8. BOSE-EINSTEIN AND FERMI-DIRAC DISTRIBUTIONS

Quantum gases are much easier to handle in the grand canonical ensemble. We therefore write the grand partitionfunction

Z(T, V, µ) =∑{nk}

W ({nk})e−β[∑k nkεk−µ

∑nk] =

∑{nk}

∏k

W (nk)e−β[nkεk−µnk] =∏k

Zk(T, V, µ)

where Zk(T, V, µ) =∑nkW (nk)e−β[nkεk−µnk] =

∑nkW (nk)znke−βnkεk is the k-state partition function.

The combinatorial prefactor W (nk) dependes on the way we constructed the ensemble:

W (nk) =

1 BEδnk,0 + δnk,1 FD

1/nk! MB(8.1)

and thus the partition function

Zk =

[1− ze−βεk ]−1 BE1 + ze−βεk FD

exp(ze−βεk) MB(8.2)

We therefore obtain for the equation of state

βPV = lnZ =∑k

1

aln(1 + aze−βεk) (8.3)

where a = −1, 0,+1 for BE, MB and FD, resp.

〈nk〉 =1

z−1eβεk + a

U =∑k

εknk =∑k

εkz−1eβεk + a

8.1. Ideal Bose gas

Before we deal with the ideal bose gas, it is useful to introduce the density of state, a(ε). What we like to do is toreplace the sum over states

∑k with an integral over a ”continuous” state space, weighted appropriately by the

density of states in each volume. In other words we want to make the approximation∑n

↔∫d3n↔ 4π

∫n2dn↔

∫a(ε)dε .

For example, for the energy levels of an free particle in a box

εn =h2n2

2mL2

we have

a(ε) = 4πn(ε)2dn

dε

hence

a(ε)dε = (2πV/h3)(2m)3/2ε1/2dε .

9

With this density of state we can immediately find

βP = −2π(2mkBT )3/2

h3

∫ ∞0

x1/2 ln(1− ze−x)dx =1

vQg5/2(z) (8.4)

N −N0

V=

2π(2mkBT )3/2

h3

∫ ∞0

x1/2

z−1ex − 1dx =

1

vQg3/2(z) . (8.5)

Here we took care not include the number of particles in the ground state in the integral, since we understand thatin moving from a sum to an integral we are missing this term. In general, this is just one state so we might this itis negligible. But since N0

V = z1−z , and we expect that at high densities (or low temperatures) z should approach 1,

we’d better not dismiss this term too quickly.

The Bose-Einstein function gν(z) is defined as

gν(z) =1

Γ(ν)

∫ ∞0

xν−1dx

z−1ex − 1= z +

z2

2ν+z3

3ν+ · · ·

and has the property

g′ν(z) =1

zgν−1(z) .

Use this in the expressions above to get a virial expansion

βP = ρ− vQ

4√

2ρ2 −

(2

9√

3− 1

8

)v2Qρ

3 − · · ·≡∞∑`=0

a`ρ`+1v`Q (8.6)

Since gν(z) is only defined for z ≤ 1 the number of particles in the excited statesNe is bound from above by VvQg3/2(1)

with g3/2(1) = ζ(3/2) ' 2.612. If N is smaller than this maximum, then practically Ne ' N . However at higherdensities

N0 = N − V

vQζ(3/2)

and therefore

z =N0

1 +N0' 1− 1

N0.

This leads to a macroscopic occupation of the ground state, known as the Bose Einstein Condensation (BEC). Thecondition for BEC is therefore

N < V T 3/2

(2πmkBh2

)3/2

ζ(3/2) (8.7)

which can be seen either as a condition on N (keeping V and T fixed) or a condition on the temperature (holding Nand V ).

As T → Tc from below

NoN

= 1−(T

Tc

)3/2

' 3

2(1− t)

with the reduced temperature t ≡ T/Tc.Below Tc, the pressure is just

P (T ) =kBT

vQζ(5/2)

10

(the correction is order 1/N ), which is independent of V , implying infinite compressibility. From this expression wefind

The energy of the system is

U = − ∂

∂βlnZ = −∂β(βPV ) = −V g5/2(z)∂βv

−1Q = V g5/2(z)∂β(2πm/h2β)3/2 =

3

2kBT

V

vQg5/2(z) =

3

2PV (8.8)

From this, we are interested in calculating

cV =

(∂u

∂T

)V,N

.

At cold temperatures, T < Tc this is straightforward:

cV =1

N

3

2kBV ζ(5/2)∂T (T/vQ) =

kBN

15

4ζ(5/2)

V

vQ(T ≤ Tc) ∼ T 3/2.

We’re happy to find that cV → 0 when the temperature approaches 0, as required by the third law of thermodynam-ics.

At high temperatures, we can take two alternative approaches. One is to use this expression for U(z), express z interms of ρ, and then take the derivative:

cV = ∂T

(∂U

∂N

)V,N

=3

2∂T

(kBTβP

ρ

)ρ

=3

2∂T kBT

∞∑`=0

a`(ρvQ)` =3

2kB∑`

2− 3`

2a`(vQρ)` =

3

2kB

[1 +

vQ

8√

2ρ+ · · ·

],

(8.9)with the classical limit at small ρ or high temperature (namely, small vQ) and positive correction at higher densities.Together with our result for low temperature, we conclude the cV must have a maximum at Tc.

The other approach is to keep working with z, but remember that z is in fact a function of T and N :

CVNkB

=∂

∂T

(3

2Tg5/2(z)

g3/2(z)

)Above Tc we have g3/2(z) = 1

vQNV ∼ T

3/2 and therefore(∂g3/2(z)

∂T

)v

=3

2Tg3/2(z) and z

∂

∂zg3/2(z) = g1/2(z) =⇒ 1

z

(∂z

∂T

)v

= − 3

2T

g3/2(z)

g1/2(z)

Another way to get dzdT is to realize that in the canonical ensemble dN

dT = 0, and therefore

dN

dT= 0 = V

d

dT

g3/2(z)

vQ=

V

vQ

[3

2Tg3/2(z) +

dz

dT

∂

∂zg3/2(Z)

]=

V

vQ

[3

2Tg3/2(z) +

dz

dT

1

zg1/2(Z)

]=⇒ 1

z

dz

dT= − 3

2T

g3/2(z)

g1/2(z)

CVNkB

=3

2

g5/2

g3/2− 9

4

g3/2

g1/2

∂

∂z

g5/2

g3/2=

15

4

g5/2(z)

g3/2(z)− 9

4

g3/2(z)

g1/2(z)(T ≥ Tc). (8.10)

At z = 1 the expression here coincides with the T ≤ Tc expression at T = Tc, so CV is continuous at Tc, but its firstderivative is not.

11

188 Chapter 7 . Ideal Bose Systems

Equation (33) now gives

CV

Nk= 15

4

g5/2(z)

g3/2(z)� 9

4

g3/2(z)

g1/2(z); (37)

the value of z, as a function of T , is again to be determined from equation (26). In thelimit z ! 1, the second term in (37) vanishes because of the divergence of g1/2(z), whilethe first term gives exactly the result appearing in (32). The specific heat is, therefore, con-tinuous at the transition point. Its derivative is, however, discontinuous, the magnitude ofthe discontinuity being

✓

@CV

@T

◆

T=Tc�0�

✓

@CV

@T

◆

T=Tc+0= 27Nk

16⇡Tc

⇢

⇣

✓

32

◆�2

' 3.665NkTc

; (38)

see Problem 7.6. For T > Tc, the specific heat decreases steadily toward the limiting value✓

CV

Nk

◆

z!0= 15

4� 9

4= 3

2. (39)

Figure 7.4 shows all these features of the (CV ,T)-relationship. It may be noted that itwas the similarity of this curve with the experimental one for liquid He4 (Figure 7.5) thatprompted F. London to suggest, in 1938, that the curious phase transition that occurs inliquid He4 at a temperature of about 2.19K might be a manifestation of the Bose–Einsteincondensation taking place in the liquid. Indeed, if we substitute, in (24), data for liquidHe4, namely m = 6.65 ⇥ 10�24 g and V = 27.6cm3/mole, we obtain for Tc a value of about3.13K, which is not drastically different from the observed transition temperature of theliquid. Moreover, the interpretation of the phase transition in liquid He4 as Bose–Einsteincondensation provides a theoretical basis for the two-fluid model of this liquid, which wasempirically put forward by Tisza (1938a,b) to explain the physical behavior of the liquidbelow the transition temperature.

According to London, the N0 particles that occupy a single, entropyless state (" = 0)

could be identified with the “superfluid component” of the liquid and the Ne particlesthat occupy the excited states (" 6= 0) with the “normal component.” As required in the

00 1 2 3

1.5

C v~

T3/

2

1.925

(T /Tc)

(Cv/

Nk)

FIGURE 7.4 The specific heat of an ideal Bose gas as a function of the temperature parameter (T/Tc).

12

8.2. Ideal Fermi gas

Following the same steps as for the Bose gas, we find

βP =g

vQf5/2(z) (8.11)

N

V=

g

vQf3/2(z) (8.12)

The FD function

fν(z) =1

Γ(ν)

∫ ∞0

xν−1dx

z−1ex + 1= z − z2

2ν+z3

3ν− · · ·

has the property zf ′(z) = fν−1(z). Thermodynamic properties are readily obtained

U = − ∂

∂βlnZ = kT 2 ∂

∂TlnZ =

3

2kBT

gV

vQf5/2(z) =

3

2NkBT

f5/2(z)

f3/2(z)(8.13)

P =2

3

U

V(8.14)

cV /kB =15

4

f5/2(z)

f3/2(z)− 9

4

f3/2(z)

f1/2(z). (8.15)

The classical limit, corresponding to high-T or low density, is obtained through the limit z � 1, where fν(z) ' z. Inthis case the results above yield as expected

U =3

2kBT, cV =

15− 9

4kB =

3

2kB , P = NkBT/V .

We could of course take more terms in the expansion, the same as we did it for the Bose gas, and obtain a virialexpansion.

In the limit T → 0 the occupation number

〈ns〉0 = limβ→∞

1

eβ(εs−µ) + 1= Θ(µ− ε)

where Θ(x) is the Heaviside function, Θ(x) = 1 for x > 0 and zero otherwise. In this limit

N0 =

∫ εF

0

a(ε)dε =gV

6π2

(2m

~2

)3/2

ε3/2F =⇒ εF = (6π2n/g)2/3

~2

2m

U0

N0=

1

N

∫ εF

0

εa(ε)dε =3

5εF

P0 =2

3

U

V=

2

5nεF ∼ n5/3 .

At low but finite temperature we take large but finite z. We use an asymptotic expansion of fν(z) in terms of (ln z)−1:

fν(z) =(ln z)ν

Γ(ν + 1)

[1 + ν(ν − 1)

π2

6(ln z)−2 + ν(ν − 1)(ν − 2)(ν − 3)

7π4

360(ln z)−4 + · · ·

](8.16)

which yields (to first order)

N

V=

4πg

3

(2m

h2

)3/2

(kBT ln z)3/2[1 +

π2

8(ln z)−2 + · · ·

](8.17)

13

and

µ = kBT ln z = εF

[1− π2

12

(kBT

εF

)2]. (8.18)

With this results we can obtain

U

N=

3

5εF

[1 +

5π2

12

(kBT

εF

)2

+ · · ·

]

P =2

3

U

V=

2

5nεF

[1 +

5π2

12

(kBT

εF

)2

+ · · ·

]

Thus cV grows linearly for T � TF before saturating at the classical limit cV /kB = 3/2.

8.3. Fermi gas in the presence of a magnetic field – Pauli paramagnetism

Assuming spin-1/2 particles, the energy is given by

ε =p2

2m− µBBσz

with σz = ±1. At T = 0

M = µB(N+ −N−) =4πV

3h3(2m)3/2

[(εF + µBB)3/2 − (εF − µBB)3/2

]More generally,

E =∑p

[(p2

2m− µBB

)n+p +

(p2

2m+ µBB

)n−p

]=∑p

p2

2m(n+p + n−p )− µBB(N+ −N−)

ZN =

N∑N+=0

eβµBB(2N+−N)[Z

(0)N+Z

(0)N−N+

]1

NlnZN = −βµBB +

1

Nln∑N+

exp[2βµBBN+ − βF (0)

N+ − βF (0)N−N+ ]

∂N+ = 0 =⇒ µ(0)(⟨N+⟩)− µ(0)(

⟨N −N+

⟩) = 2µBB

And finally for small fields we expand

µ(0)(⟨N±

⟩) ' µ(0)(N± = N/2)+∂N±µ

(0)(N± = N/2)(N±−N/2) = µ(0)(N± = N/2)±∂N+µ(0)(N± = N/2)(N±−N/2)

to get

M = µB⟨N+ −N−

⟩=

2µ2BB

∂µ(0)(N+)∂N+

∣∣∣N+=N/2

.

We can now use the expansions given above for high and low temperatures, to find

χ =M

VB=

3

2

nµ2B

εF

[1− π2

12

(kBT

εF

)2]

low T

=nµ2

B

kBT

[1− nvQ

25/2

]high T.