Solution to Problem Set 4

We consider a system of Ising spins (S_{i} = ±1) with the partition function Z[J] defined as,

Z[J] = Tre^{−(J_{ij}S_{i}S_{j}+J_{i}S_{i})} = e^{−W [J]}.  (1)

W [J] is the free energy of the system. We will follow the convention that when not mentioned explicitly, all repeated indices are to be summed.

The average magnetization is defined as,

\frac{∂W [J]}{∂J_{i}} = \frac{1}{Z} TrS_{i}e^{−(J_{ij}S_{i}S_{j}+J_{i}S_{i})} = ⟨ S_{i} ⟩ = m_{i}.  (2)

The free energy with fixed magnetization Γ[m] is defined by the Legendre tranform,

Γ[m] = W [J] − J_{i}m_{i}.  (3)

Using equations (2) and (3) we find that,

\frac{∂Γ[m]}{∂m_{i}} = \frac{∂W [J]}{∂J_{k}} \frac{∂J_{k}}{∂m_{i}} − J_{i} − m_{k} \frac{∂J_{k}}{∂m_{i}} = −J_{i}.  (4)

The term J_{ij} in equation (1) defines interaction between the spins. We want to treat this term perturbatively. For this we multiply J_{ij} by α, and expand relevant quantities in successive orders in α. Here α is a parameter which will be set equal to one, like the charge of the electron in DFT. The purpose of this exercise is to demonstrate that since magnetization is the interesting quantity for the spin system, it is meaningful to make a perturbative expansion of the thermodynamic quantity which keeps magnetization fixed, namely Γ[m]. Mathematically, as α is tuned from zero to one, from the point of view of Γ[m], J_{i}(α) has the role of a Lagrange multiplier which varies with α to satisfy the constraint ⟨ S_{i} ⟩ = m_{i}. This is the same principle that motivates DFT. In DFT, density of the electrons is the important quantity that must be kept fixed. The equivalent of J_{i} in DFT is the effective potential V (r), while the zeroth order approximation to it, which is equivalent to J_{i}(0) is the Kohn-Sham potential.

1. Zeroth Order

For α = 0 we are dealing with free Ising spins. The partition function factorizes and is given by,

Z_{0}[J] = ∏_{i} (e^{−J_{i}} + e^{J_{i}}) = ∏_{i} (2 cosh(J_{i})) .  (5)

The free energy W_{0}[J] is,

W_{0}[J] = − ln Z_{0} = − ∑_{i} ln [2 cosh(J_{i})] .  (6)

Since, m_{i} = ⟨ S_{i} ⟩ for all values of α, in particular this is true for α = 0. Then, using equation (2), the relation between m_{i} and J_{i}(0) is given by,

m_{i} = − tanh(J_{i}(0)).  (7)

We can use the above equation to express J_{i}(0) in terms of m_{i}. Using the definition of Γ[m] we can write,

Γ_{0}[m] = ∑_{i} \frac{1 + m_{i}}{2} ln(\frac{1 + m_{i}}{2} ) + ∑_{i} \frac{1 − m_{i}}{2} ln(\frac{1 − m_{i}}{2} )  (8)

This is nothing but the entropy of free Ising spins with average magnetization m_{i} at each site.

2. First Order

After introducing the parameter α, the partition function has the form

Z[J, α] = Tre^{−(αJ_{ij}S_{i}S_{j}+J_{i}S_{i})}.

We get,

\frac{∂W}{∂α} = −\frac{1}{Z} \frac{∂Z}{∂α} = J_{ij}⟨ S_{i}S_{j} ⟩,  (9)

and

\frac{∂Γ}{∂α} = \frac{∂W}{∂α} + \frac{∂W}{∂J_{i}} \frac{∂J_{i}}{∂α} − m_{i} \frac{∂J_{i}}{∂α} = \frac{∂W}{∂α} .  (10)

Then, (\frac{∂Γ}{∂α})_{α=0} = J_{ij}⟨ S_{i}S_{j} ⟩_{0}, where ⟨ ⋯ ⟩_{0} means average with Boltzman weight for free spins. We note that for i ≠ j, ⟨ S_{i}S_{j} ⟩_{0} = ⟨ S_{i} ⟩_{0}⟨ S_{j} ⟩_{0} = m_{i}m_{j}. While for i = j , ⟨ S_{i}^{2} ⟩ = 1 (for Ising spins). Thus, ignoring a constant term, we have,

Γ_{1}[m] = (\frac{∂Γ}{∂α})_{α=0} = ∑_{i≠j} J_{ij}m_{i}m_{j}.  (11)

We identify this term, together with the zeroth order term, as the free energy of the system within Weiss mean field approximation.

3. Second Order

To calculate the second order term we have to work a little more. But the strategy is similar to what we did for the first order. We note that,

\frac{∂^{2}W}{∂α^{2}} = J_{ij}J_{kl} [⟨ S_{i}S_{j} ⟩⟨ S_{k}S_{l} ⟩ − ⟨ S_{i}S_{j}S_{k}S_{l} ⟩] ,  (12)

and that

\frac{∂^{2}Γ}{∂α^{2}} = \frac{∂^{2}W}{∂α^{2}} + \frac{∂}{∂J_{i}} (\frac{∂W}{∂α} ) \frac{∂J_{i}}{∂α}
= J_{ij}J_{kl} [⟨ S_{i}S_{j} ⟩⟨ S_{k}S_{l} ⟩ − ⟨ S_{i}S_{j}S_{k}S_{l} ⟩]
+ J_{ij} [⟨ S_{i}S_{j} ⟩⟨ S_{k} ⟩ − ⟨ S_{i}S_{j}S_{k} ⟩] \frac{∂J_{k}}{∂α} .  (13)

In the above equation we know, in principle, how to calculate all the terms for α = 0, except for the term ∂J_{k}/∂α. We will calculate this term in two ways. The first method, though longer, sheds light on the mathematics of the formalism. For this we note that J_{i}(α) varies to satisfy the constraint m_{i} = ⟨ S_{i} ⟩. Thus ∂⟨ S_{i} ⟩/∂α = 0, and this condition should give how J_{i}(α) varies. We get,

(\frac{∂⟨ S_{i} ⟩}{∂α})_{α=0} = J_{kl} [⟨ S_{i} ⟩_{0}⟨ S_{k}S_{l} ⟩_{0} − ⟨ S_{i}S_{k}S_{l} ⟩_{0}] + [⟨ S_{i} ⟩_{0}⟨ S_{j} ⟩_{0} − ⟨ S_{i}S_{j} ⟩_{0}] (\frac{∂J_{j}}{∂α} )_{α=0} = 0.

(14)

We note that ⟨ S_{i} ⟩_{0}⟨ S_{j} ⟩_{0} − ⟨ S_{i}S_{j} ⟩_{0} = (m_{i}^{2} − 1)δ_{ij}. Then, from the above equation we have,

(\frac{∂J_{i}}{∂α} )_{α=0} = −\frac{1}{(m_{i}^{2} − 1)} ∑_{kl} J_{kl} [⟨ S_{i} ⟩_{0}⟨ S_{k}S_{l} ⟩_{0} − ⟨ S_{i}S_{k}S_{l} ⟩_{0}] .

In the above equation, the expression on the right hand side has contribution only for i = k ≠ l and for i = l ≠ k. Thus,

(\frac{∂J_{i}}{∂α} )_{α=0} = −\frac{1}{(m_{i}^{2} − 1)} ∑_{i≠j} 2J_{ij}(m_{i}^{2} − 1)m_{j}
= −2 ∑_{i≠j} J_{ij}m_{j}.  (15)

We can also derive the above equation very easily using equations (4) and (11). Thus,

(\frac{∂J_{i}}{∂α} )_{α=0} = − [\frac{∂}{∂α} (\frac{∂Γ}{∂m_{i}} )]_{α=0} = −\frac{∂}{∂m_{i}} (\frac{∂Γ}{∂α})_{α=0}
= −\frac{∂}{∂m_{i}} ∑_{i≠j} J_{ij}m_{i}m_{j}
= −2 ∑_{i≠j} J_{ij}m_{j}.

The above is nothing but an application of Maxwell relation. Note that we can write \frac{∂}{∂α} \frac{∂}{∂m_{i}} = \frac{∂}{∂m_{i}} \frac{∂}{∂α} because m_{i} is independent of α. Now we can evaluate the right hand side of equation (13). The first term, ⟨ S_{i}S_{j} ⟩_{0}⟨ S_{k}S_{l} ⟩_{0}− ⟨ S_{i}S_{j}S_{k}S_{l} ⟩_{0} is nonzero for i = l ≠ j = k. There are two such terms. Also, it is nonzero for i = k ≠ j ≠ l. There are four such terms. Thus,

J_{ij}J_{kl} [⟨ S_{i}S_{j} ⟩_{0}⟨ S_{k}S_{l} ⟩_{0} − ⟨ S_{i}S_{j}S_{k}S_{l} ⟩_{0}] = 2 ∑_{i≠j} J_{ij}^{2} (m_{i}^{2}m_{j}^{2} − 1)+4 ∑_{i≠j≠l} J_{ij}J_{jl} (m_{j}^{2} − 1) m_{i}m_{l}.

(16)

The second term in equation (13) gives,

J_{ij} [⟨ S_{i}S_{j} ⟩_{0}⟨ S_{k} ⟩_{0} − ⟨ S_{i}S_{j}S_{k} ⟩_{0}] (\frac{∂J_{k}}{∂α} )_{α=0}  (17)

= −4 ∑_{i≠j} J_{ij}m_{i} (m_{j}^{2} − 1) ∑_{k≠j} J_{jk}m_{k}

= −4 ∑_{i≠j≠l} J_{ij}J_{jl} (m_{j}^{2} − 1) m_{i}m_{l} − 4 ∑_{i≠j} J_{ij}^{2}m_{i}^{2} (m_{j}^{2} − 1) .  (18)

Using equations (13), (16) and (18) we finally get,

Γ_{2}[m] = (\frac{∂^{2}Γ}{∂α^{2}} )_{α=0}
= 2 ∑_{i≠j} J_{ij}^{2} (m_{i}^{2}m_{j}^{2} − 1) − 4 ∑_{i≠j} J_{ij}^{2}m_{i}^{2} (m_{j}^{2} − 1)
= −2 ∑_{i≠j} J_{ij}^{2} (1 − m_{i}^{2}) (1 − m_{j}^{2}) .  (19)

This is the famous Onsager correction to the mean field free energy.