Lecture 2: Meanfield Approximation, Variational Principle and Landau Expansion Naoki KAWASHIMA ISSP, U. Tokyo April 15, 2024 Statistical Mechanics I: Lecture 2 April 15, 2024 1 / 27 In this lecture, we see … ● Molecular-field theory yields the equation of the state, not the free energy. ● Gibbs-Bogoliubov-Feynman inequality give us a systematic and flexible framework for constructiong the mean-field-type approximations.
Molecular field theory revisited
● In the molecular-field theory, the effect of environment is replaced by an additional term. For the Ising ferromagnet, the many-body problem is reduced to a 1-body problem about a spin, say So,
where ( ij ) is the nearest neighbor pair of sites. ● It is also argued that the right choice of the effective field is Hmf = J } ● The uniformity condition m = 〈S i〉 (independent of i ) yields, m = tanh(β(H + zJm))   (z = (number of nearest-neighbors)) tatistical Mechanics I: Lecture April 15, 2024 3 / 27
Why should we complain?
● It is not generally obvious how we should choose the mean-field. (The order parameter may not have a simple form in terms of the model degrees of freedom.) ● In principle, we have multiple solutions of the self-consistent equation. ● The molecular-field theory does not tell us which solution we should choose. Is there any general framework for the mean-field approximation?
Theorem 1 (GBF inequality) For two Hamiltonians H ( S ) and H 0 ( S ) defined on the same space S ∈ Ω, Fv ≡ Fo + (H − Ho)o ≥ F, (1) where F  and F 0 are the free-energies of H and H 0 respectively and 〈···〉0 is the thermal average with respect to H 0. Variational calculation When H(S) is the Hamiltonian of the system that we want to study but is not solvable, by taking H 0 ( Λ, S ) that depends on a set of parameters Λ and is solvable for any Λ, the resulting F t (Λ) provides us with the computable upper bound of the correct free energy. To obtain the best upper-bound, we minimize it w.r.t.  Λ. April 15, 2024 5 / 27
GBF inequality from information-scientific view-point
Theorem 2 (Relation to Kullback-Leibler divergence) The “error” in the variational free-energy is proportional to the Kullback-Leibler divergence of the equilibrium distribution of H 0 relative to that of H. More precisely, Fv – F = kbT Dkl(polp) (2) where ρο ≡ e-βHo/Zo, and ρ ≡ e-βH/Z (3) and
(Z and Z 0 are partition functions of H and H 0 respectively.) Statistical Mechanics I: Lecture 2
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Statistical Mechanics I: Lecture 2 April 15, 2024 7 / 27 Kullback-Leibler information measure is positive
Remark This inequality together with Theorem 2 proves Theorem 1.
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● The theorems have been proved for classical systems. The corresponding quantum version of them can be also proved. ● For the extension, the KL divergence must be generaized to DkL(P|Q) ≡ Tr P(log P − log Q)
where P and Q are now density operators. (If a positive semi-definite operator X satisfies the normalization Tr X = 1, we call it a density operator.) ● The only non-trivial part in the proof of the quantum extension is the positivity of the KL information. The rest is straight-forward simply by replacing ∑ s by Tr. Statistical Mechanics I: Lecture 2 April 15, 2024 9 / 27 Positivity of quantum version of KL divergence (1/2) Theorem 3 (Positivity of quantum KL divergence) For any density operators P and Q, DkL(P|Q) ≡ Tr P log P – P log Q ≥ 0
Proof:
● Let us take the basis set in which P is diagonal, i.e., Pij = Piδij. ● Let U be the unitary operator U diagonalizing Q, i.e., △ ≡ U†QU is a diagonal operator. Then, Qij = Uikqku*jk· ● With this uij,
● Notice that aij ≥ 0, 2 θ ij = 1.
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Positivity of quantum version of KL divergence (2/2) ● By defining p ij = p i α ij ,    q ij = q j α ij , we can regard them as the “classical” distribution function in the squared Hilbert space H × H ≡ { (ij) | i, j ∈ H} because
● The two terms in (7) can be expressed as, respectively,
● By subtracting the second from the first, we obtain DKL(P|Q) = ∑ ij P ij log( p ij /q ij ), and this must be non-negative because the RHS is the classical KL divergence. (QED) tatistical Mechanics I: Lecture April 15, 2024 11 / 27 Some examples o  Ising model: Λ = (Γ)
(Г̸= 0 represents ferromagnetism.) ● Hubbard model: Λ = ({Λijμν})
where n i ≡ ∑ σ c iσ c iσ , and (a i 1 , a i 2 , a i 3 , a i 4 ) = (c i ↑ , c i ↓ , c i ↑ ( Λ i 1,j2̸= Λ i 4,j3 represents superconductivity.)
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Variational approximation to the Ising model (1/2) ●  For the target Hamiltonian H ≡ − J ∑ ( ij ) S i S j − H ∑ i S i , let us
    take the “trial”  Hamiltonian H 0 ≡ − Λ i S i .  where Λ is a variational parameter. ● Then, our variational free energy is Fv = Fo + {H − Ho}o = ⟨H⟩o − SoT where S 0 ≡ T − 1 ( H 0 ) 0 − F 0 ) is the entropy of the H 0 system ● By introducing m ≡ 〈Si〉0 = tanh βΛ,
Statistical Mechanics I: Lecture 2 April 15, 2024 13 / 27 Variational approximation to the Ising model (2/2) Variational free-energy density
– Hm – Tσ(m) ● Moreover, the GBF inequality tells us that we should take the solution with the smallest f v with respect to λ . Since f v depends on λ only through m, the stationary condition is ∂ƒ√/∂m = 0. ● From this, we obtain the same as the molecular-field approx.:
m = tanh βHmf   (Hmf ≡ zJm + H).

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● We have also obtained the explicit expression for the free energy. ● Since its behavior near m ≈ 0 is most important for the critical phenomena, let us expand f v with respect to m.
● From the condition (coefficient of m²) = 0, we obtain kBTc = zJ. ● Near T ~ T c , we can get a generic form of Landau expansion: △fv ≈ −Hm + tm² + um²
where △f v ≡ f v − f v , f v ≡ −zJ log 2, t ≡ (kBT − zJ)/2 and u ≡ zJ/12. ● It is also clear we should choose non-zero m solution below T c . tatistical Mechanics I: Lecture 2 April 15, 2024 15 / 27 Exercise 2.1:  Verify (8) and (9).

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Statistical Mechanics I: Lecture 2 April 15, 2024 17 / 27 Exercise 2.2:   At the point where we have arrived at (10), m is just the magnetization of the variational Hamiltonian. How can we interpret it as the magnetization of the target Hamiltonian within the mean-field approximation? Within the mean-field approximation, the variational free energy F v , not F 0 , with the optimal value of A is our approximation to the target free energy. Therefore, for the variational minimum F* = min Λ F ν (Λ), if we can prove that Nm = −∂F*/∂H, then it would establish the interpretation. We can do this as follows:
because (∂ƒ*/∂m)н = 0.
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Exercise 2.3:  Based on the variational principle, propose an improvement on the simplest mean-field approximation of the Ising model discussed in the lecture, and obtain its equation of state. Specifically, propose a variational Hamiltonian that would be closer to H than H 0 discussed in the lecture but still solvable. For example, we may consider the Hamiltonian of non-interacting pairs instead of the Hamiltonian of non-interacting single spins considered in the lecture. Specifically, Ho = ∑(ij) Hij with Hij ≡ − Λ 1 (S i + S j ) − Λ 2 S i S j where ∑(jj) is the summation over all mutually non-overlapping nearest-neighbor pairs of spins that covers the whole system. Similarly, we can improve the approximation systematically by considering bigger local clusters: H 0 = ∑ c H c where ∑ c is the summation over all mutually non-overlapping clusters of spins that covers the whole system, and Hc is some parameterized function of spins in the cluster. (The required computational effort would increase very fast, though, as we increase the cluster size.) Statistical Mechanics I: Lecture 2 April 15, 2024 19 / 27 Exercise 2.4:   Obtain the Landau expansion of the 3-state Potts model by taking the trial Hamiltonian H 0 ≡ − Λ ∑ i δ s i ,1 with S i ＝ 1,2,3. What is the symmetry-breaking order-parameter and what is the essential difference from the Ising case? By defining m ≡ 〈δ s , , 1〉 − , we obtain
From this we obtain the Landau free energy similar to that of the Ising model discussed in the lecture. However, the important difference lies in the existence of the 3rd order term —vm³ with a positive constant v. This makes the transition the 1st-order one, in contrast to the Ising model.
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Exercise 2.5:   Consider the S=1 Ising model:
Propose a trial Hamiltonian for this model, and, by the variational approximation with it, obtain the T — D phase diagram. Let the variational Hamiltonian be
Or, equivalently, let us denote the probability of having S i = 1,0, —1 under this trial Hamiltonian as p, r, q respectively.  Then, m ≡ ⟨ S i ⟩ o ＝ p − q, bi )o = p + q. Statistical Mechanics I: Lecture 2 April 15, 2024 21 / 27 It follows that
Obviously, we can regard ( m, n ) or ( p, q ) as our variational parameters, instead of (Г, Л). Now, the expectation valur of the target Hamiltonian becomes
As for the entropy, using Shanon's entropy S' = − Ca Pa log pa,
Therefore, the variational free energy per spin is, in the unit of zJ,
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From the first equation, we obtain the transition temperature as a function of n, tc = n By replacing n by t in the second equation, we obtain the equation that determines the critical temperature tc as a function of g:
Solution of this equation is plotted in the figure. So far, we have been assuming that the transition is of the second order, which means that the order parameter is continuous function of the state parameter, i.e., lim t→t c +0 m( t ) = lim t→t c −0 m( t ). However, this is not the case with the first order transition. As a result, some part of the phase boundary defined by (15) is taken over by the first order transition and not realized. The first-order transition line is determined by the condition that the m = 0 solution and the m̸= 0 solution to (13) and (14) have the same free energy. The detailed analysis is given in papers; Capel, Physica 32 966 (1966); Blume, Phys. Rev. 141 517 (1966). Statistical Mechanics I: Lecture 2 April 15, 2024 25 / 27 Figure:  The transition line is separated by the point located at (T/z.J, D/z.J) = ( ² log ), ² ) ≈ (−0.462028 · · · , 0.33333 · · ), which is marked by filled red circle. Above this point, the purple curve represents the critical (the 2nd order transition) line whereas the red curve below is the first-order transition line. The