Solution Let T = Tc(1 + τ) where Tc is such that νJ/(2kBTc) = 1. With these definitions, the Isingmodel magnetization m = 〈σ〉 becomes

m = tanh

(m

1 + τ

).

Suppose now that m� 1, and replace the hyperbolic tangent with its Taylor series, expanded to thirdorder, tanhx ' x− 1

3x3:

m =1

1 + τm− 1

3

(1

1 + τ

)3

m3.

Ignoring the m = 0 solution, we get the following:

m2 = 3 (1 + τ)3

(1

1 + τ− 1

)= 3

[(1 + τ)

2 − (1 + τ)3]' 3 [1 + 2τ − (1 + 3τ)] = −3τ,

⇒ m ∼ t1/2.

3. Landau Theory Consider the Landau’s theory of phase transition presented in the lectures.

(a) Calculate the low-field susceptibility:

χ0(T ) =

(∂m

∂h

)T

.

(b) What are the critical exponents γ and δ?

Solution

(a) Let φ(t,m) be the free energy, expanded in m and with coefficients from the Landau theory:

φ(t,m) = φ(t, 0) + r1tm2 + s0m

4. (2)

Legendre transforming to a free energy parametrized by the field h one has

φ(t, h) = φ(t,m)− hm = −hm+ φ(t, 0) + r1tm2 + s0m

4,

which is now minimized by m with h held constant,(∂φ

∂m

)h

= 0.

Straight-forward differentiation gives

−h+ 2r1tm+ 4s0m3 = 0 (3)

⇒ h ' −2r1tm ⇒ m(h) = − 1

2r1th.

And from that, one quite trivially gets the susceptibility(∂m

∂h

)t

=1

2r1t. (4)

(b) First, the exponent δ in m|t=0 ∼ h1/δ. From Eq. (3) with t→ 0:

h = 4s0m3 ⇒ m =

(h

4s0

)1/3

⇒ δ = 3.

Second, the exponent γ in (∂m/∂h) ∼ t−γ . This follows trivially from Eq. (4):

γ = 1.