Chapter 5

External field and the replica-symmetric solution

In this chapter we return to the SK model. One way to read Theorem 3.3 is that, if β < 1, we have

(5.1)  lim_{N→∞} E \frac{1}{N} log Z_{N} = lim_{N→∞} \frac{1}{N} log E Z_{N}.

It is always true, by Jensen’s inequality, that we have

E log Z_{N} ≤ log E Z_{N}

and the content of (5.1) is that this trivial bound gives the correct order. On the other hand (5.1) does not hold for β > 1, as Theorem 3.13 shows. This is of course a more interesting situation.

As it turns out, (5.1) represents an accidental feature rather than a general one. For several models of considerable interest, it does not occur, even at high temperature. In this chapter we investigate a typical case, the SK model with external field. With the notation (3.1), the Hamiltonian is given by

H_{N}(σ) = −\frac{1}{\sqrt{N}} ∑_{i<j} g_{ij}σ_{i}σ_{j} − h′ ∑_{i≤N} σ_{i}. (5.2)

The last term represents the influence of an external field (that is, created by an apparatus independent of our sample of matter). The reason for the unexpected ′ on the parameter h′ (which measures the strength of the external field) is that we find it convenient to consider β and h = βh′ as independent parameters.

In this chapter we prove the following theorem, where g is standard normal.

Theorem 5.1. There is a number β_{0} such that if β ≤ β_{0}, we have

(5.3)  lim_{N→∞} p_{N}(β, h) = log 2 + \frac{β^{2}}{4} (1 − q)^{2} + E log ch(βg\sqrt{q} + h)

where q is the unique solution of the equation

q = E th^{2}(βg\sqrt{q} + h) . (5.4)

Thus (to write the formulas of the physicists),

lim_{N→∞} p_{N}(β, h) = \frac{β^{2}}{4} (1 − q)^{2} + \frac{1}{\sqrt{2π}} ∫_{−∞}^{∞} e^{− \frac{t^{2}}{2}} log(2 ch(βt\sqrt{q} + h)) dt. (5.5)

On the other hand, it is very simple to show that

lim_{N→∞} \frac{1}{N} log E Z_{N} = log 2 + \frac{β^{2}}{4} + log ch h

and that (5.1) fails unless h = 0. (As we will show later, the value of q given by (5.4) minimizes the right-hand side of (5.3).) The remarkable character of a formula of the type (5.3) should be self apparent. This formula is not an accident, but reflects an underlying structure. It is the purpose of the present chapter to explain this structure and to prove (5.3). The formulae (5.3), (5.4) are called by physicists the “replica-symmetric” formulas.

The methods of this chapter are rather elementary. Considerably more powerful methods will be used in Chapter 6 and 7, and will provide shorter proofs of better results. Our motivation for presenting first an elementary (an leisurely) approach is that we feel that some of the important ideas are more apparent this way. These ideas have been very useful to the author.

If we think of the spins σ_{i} as functions on the probability space (Σ_{N}, G_{N}), the correlation of two spins is given by ⟨σ_{i}σ_{j}⟩ − ⟨σ_{i}⟩⟨σ_{j}⟩. The fundamental property behind (5.3) (and many other “replica-symmetric” formulas), is that in average over the disorder, these correlations are small, i.e.

lim_{N→∞} E(⟨σ_{i}σ_{j}⟩ − ⟨σ_{i}⟩⟨σ_{j}⟩)^{2} = 0. (5.6)

The purpose of the square is to prevent cancelation. By symmetry upon the sites, we can use any values for i, j.

Calculations are much simpler when one replaces products of integrals for Gibbs’ measure by single integrals (this is what makes replicas so useful), and we first explain how to do this in the present setting.

One first idea is to use a procedure called “symmetrization” to replace centering. (This idea is very useful in many areas of probability.) If X, Y are r.v., and (X′, Y′) is an independent copy of the pair (X, Y), we have

E(XY) − E(X) E(Y) = \frac{1}{2} E((X − X′)(Y − Y′)). (5.7)

We consider two replicas σ^{1}, σ^{2}, they are independent under G_{N}^{⊗2}, so that we can use (5.7) to write

2(⟨σ_{i}σ_{j}⟩ − ⟨σ_{i}⟩⟨σ_{j}⟩) = ⟨(σ_{i}^{1} − σ_{i}^{2})(σ_{j}^{1} − σ_{j}^{2})⟩.

The difference of two replicas σ^{1}, σ^{2} will occur many times so we will simplify notation by setting

σ˜ = σ^{1} − σ^{2}, (5.8)

and we see that (5.6) is equivalent to

lim_{N→∞} E⟨σ_{i}˜ σ_{j}˜ ⟩^{2} = 0. (5.9)

To remove this square, we bring in two new replicas σ^{3}, σ^{4} and we set

σ^{∗} = σ^{3} − σ^{4} (5.10)

so that, since ⟨σ_{i}^{∗}σ_{j}^{∗}⟩ = ⟨σ_{i}˜ σ_{j}˜ ⟩,

⟨σ_{i}˜ σ_{j}˜ ⟩^{2} = ⟨σ_{i}˜ σ_{j}˜ ⟩⟨σ_{i}^{∗}σ_{j}^{∗}⟩ = ⟨σ_{i}˜ σ_{j}˜ σ_{i}^{∗}σ_{j}^{∗}⟩. (5.11)

It is often useful to use symmetry among coordinates. We define

C_{N} = C_{N}(β, h) = E⟨(\frac{σ˜ · σ^{∗}}{N} )^{2}⟩ (5.12)

Writing

\begin{aligned}\frac{σ˜ · σ^{∗}}{N} & = \frac{1}{N} ∑_{i} σ_{i}˜ σ_{i}^{∗} \\ (\frac{σ˜ · σ^{∗}}{N} )^{2} & = \frac{1}{N^{2}} ∑_{i,j} σ_{i}˜ σ_{j}˜ σ_{i}^{∗}σ_{j}^{∗}, \end{aligned} (5.13)

we see that by symmetry upon the coordinates,

C_{N} = \frac{1}{N} E⟨σ_{1}^{ ˜ 2}σ_{1}^{∗2}⟩ + \frac{N − 1}{N} E⟨σ_{1}˜ σ_{2}˜ σ_{1}^{∗}σ_{2}^{∗}⟩. (5.14)

Thus, (5.6) is equivalent to lim_{N→∞} C_{N} = 0. One could say that when using

E⟨(\frac{σ˜ · σ^{∗}}{N} )^{2}⟩

one has a “global” point of view, while when using

E(⟨σ_{1}˜ σ_{2}˜ ⟩^{2}) = E⟨σ_{1}˜ σ_{2}˜ σ_{1}^{∗}σ_{2}^{∗}⟩

one has a “local” point of view (two spins involved). It seems fruitful to use both.

We will study C_{N} using the cavity method. This is simply induction upon N. One removes a spin to reduce to an (N −1)-spin system (thereby creating a cavity). This method is of fundamental importance, both in the rigorous work of mathematicians, and in the heuristic arguments of physicists.

Let us consider a sequence (g_{i})_{i≤N} of i.i.d. standard normal r.v., that is independent of all the g_{ij}. Consider σ_{N+1} ∈ {−1, +1}. We have

(5.15)  −βH_{N}(σ) + σ_{N+1}(\frac{β}{\sqrt{N}} ∑_{i≤N} g_{i}σ_{i} + h)

= \frac{β}{\sqrt{N}} ∑_{1≤i<j≤N+1} g_{ij}σ_{i}σ_{j} + h ∑_{i≤N+1} σ_{i}

where we set g_{i,N+1} = g_{i}. If we set

ϱ = (σ, σ_{N+1}) = (σ_{i})_{i≤N+1} ∈ Σ_{N+1}, (5.16)

we see that the right-hand side of (5.15) is

−β′ H_{N+1}(ϱ), (5.17)

where

β′ = β \begin{matrix}√ & \overline{N + 1} \\ √ & \overline{N} \end{matrix}. (5.18)

That is, (5.15) is the Hamiltonian of an (N + 1)-spin system at a slightly lower temperature. We denote by G_{N+1} = G_{N+1}(β′) the corresponding Gibbs’ measure, and by ⟨ · ⟩′ average with respect to this measure.

Consider now a function f on Σ_{N+1}. Using the notation (5.16), we can write either f = f(ϱ), ϱ ∈ Σ_{N+1} or f = f(σ, σ_{N+1}), σ ∈ Σ_{N}, σ_{N+1} = ±1. We then have the following algebraic identity

⟨f(ϱ)⟩′ = \frac{1}{Z} ⟨ Av_{σ_{N+1}=±1} f(σ, σ_{N+1})E⟩, (5.19)

where

E = E(σ, σ_{N+1}) = exp σ_{N+1}(\frac{β}{\sqrt{N}} ∑_{i≤N} g_{i}σ_{i} + h), (5.20)

and where

Z = ⟨ Av_{σ_{N+1}=±1}E(σ, σ_{N+1})⟩. (5.21)

(The reader should not confuse Z with the partition function Z_{N}.)

What these formulas mean is that we average first over σ_{N+1} = ±1; the resulting quantity depends on σ only. We then average over σ in G_{N}.

We will need a formula similar to (5.19), when f is a function on (Σ_{N+1}^{k}, G_{N+1}^{⊗k}). In that case, we have

⟨f(ϱ^{1}, …, ϱ^{k})⟩′ = \frac{1}{Z^{k}} ⟨Av f(ϱ^{1}, …, ϱ^{k})E_{k}⟩, (5.22)

where now

E_{k} = exp ∑_{l≤k} σ_{N+1}^{l}(\frac{β}{\sqrt{N}} ∑_{i≤N} g_{i}σ_{i}^{l} + h). (5.23)

The notation (5.22) uses two conventions, that we will be valid throughout the paper.

First, the relation between ϱ^{l} and (σ^{l}, σ_{N+1}^{l}) is as in (5.16),

ϱ^{l} = (σ_{1}^{l}, …, σ_{N}^{l}, σ_{N+1}^{l}). (5.24)

Second, the average is over σ_{N+1}^{1}, …, σ_{N+1}^{k} = ±1. The result of this averaging depends only upon σ^{1}, …, σ^{k}, and is then integrated for G_{N}^{⊗k}.

When one averages over σ_{N+1}^{1}, …, σ_{N+1}^{k} these are simply dumb variables, σ_{N+1}^{l} = ±1, so we will lighten notation by writing ε_{l} = σ_{N+1}^{l}, ε˜ = σ_{Ñ+1}, ε^{∗} = σ_{N+1}^{∗} inside such averages. Then we write of course (5.23) as

E_{k} = exp ∑_{l≤k} ε_{l}(\frac{β}{\sqrt{N}} ∑_{i≤N} g_{i}σ_{i}^{l} + h).

Proposition 5.2. There is a number β_{0} ≥ 1/3 such that if β ≤ β_{0}, then for all h ≥ 0, we have

NC_{N}(β, h) ≤ 32. (5.25)

Proof. From (5.14), we write

C_{N}(β, h) ≤ \frac{16}{N} + \frac{N − 1}{N} E⟨σ_{1}˜ σ_{1}^{∗}σ_{Ñ}σ_{N}^{∗}⟩,

and changing N in N + 1,

C_{N+1}(β′, h) ≤ \frac{16}{N + 1} + \frac{N}{N + 1} E⟨σ_{1}˜ σ_{1}^{∗}σ_{Ñ+1}σ_{N+1}^{∗}⟩′.

We then use (5.22) on the last term to obtain

E⟨σ_{1}˜ σ_{1}^{∗}σ_{Ñ+1}σ_{N+1}^{∗}⟩′ = E \frac{1}{Z^{4}} ⟨Av σ_{1}˜ σ_{1}^{∗}ε ε̃^{∗}E_{4}⟩, (5.26)

where Av is over ε_{1}, …, ε_{4}. We observe that Av E(σ) ≥ 1, so that Z ≥ 1. Also

⟨Av σ_{1}˜ σ_{1}^{∗}ε ε̃^{∗}E_{4}⟩ = ⟨Av σ_{1}˜ ε Ẽ_{2}⟩^{2} ≥ 0

and thus,

E⟨σ_{1}˜ σ_{1}^{∗}σ_{Ñ+1}σ_{N+1}^{∗}⟩′ ≤ E⟨Av σ_{1}˜ σ_{1}^{∗}ε ε̃^{∗}E_{4}⟩. (5.27)

The previous argument does not take advantage of the fact that Z can be much bigger than one. But removing this denominator makes the computation of the expectation much easier. Before we start this computation we observe an important fact. Since ε_{l} ∈ {−1, +1}, if ε˜ ≠ 0, ε^{∗} ≠ 0 we have ε_{2} = −ε_{1}, ε_{4} = −ε_{3}, so that

ε ε̃^{∗}E_{4} = ε ε̃^{∗}E′

where

E′ = exp \frac{β}{\sqrt{N}} ∑_{i≤N} g_{i}(ε_{1}σ_{i}˜ + ε_{3}σ_{i}^{∗}).

Now

Av_{ε_{2}=±1} ε˜ = ε_{1}, Av_{ε_{4}=±1} ε^{∗} = ε_{3}

and thus

⟨Av σ_{1}˜ σ_{1}^{∗}ε ε̃^{∗}E_{4}⟩ = ⟨Av σ_{1}˜ σ_{1}^{∗}ε_{1}ε_{3}E′⟩. (5.28)

We denote by E_{g} the expectation in (g_{i})_{i≤N} only. Thus

E′′ := E_{g} E′ = exp \frac{β^{2}}{2N} ∑_{i≤N}((σ_{i}˜ )^{2} + (σ_{i}^{∗})^{2} + 2ε_{1}ε_{3}σ_{i}˜ σ_{i}^{∗}). (5.29)

Since the disorder in ⟨ · ⟩ is independent of the (g_{i})_{i≤N}, we have

E_{g}⟨Av σ_{1}˜ σ_{1}^{∗}ε_{1}ε_{3}E′⟩ = ⟨Av σ_{1}˜ σ_{1}^{∗}ε_{1}ε_{3} E_{g} E′⟩.

From (5.28) and (5.29), we then get that

(5.30)  E⟨Av σ_{1}˜ σ_{1}^{∗}ε ε̃^{∗}E_{4}⟩ = E⟨Av σ_{1}˜ σ_{1}^{∗}ε_{1}ε_{3}E′′⟩
= E⟨σ_{1}˜ σ_{1}^{∗} Av ε_{1}ε_{3}E′′⟩.

We then use symmetry among the sites to see that

E⟨Av σ_{1}˜ σ_{1}^{∗}ε_{1}ε_{3}E′′⟩ = E⟨(\frac{1}{N} ∑_{i≤N} σ_{i}˜ σ_{i}^{∗}) Av ε_{1}ε_{3}E′′⟩. (5.31)

In view of the notation σ^{1} · σ^{2} = ∑_{i≤N} σ_{i}^{1}σ_{i}^{2}, it is natural to write

∑_{i≤N}(σ_{i}˜ )^{2} = ∥σ˜ ∥^{2} ;  ∑_{i≤N}(σ_{i}^{∗})^{2} = ∥σ^{∗}∥^{2}. (5.32)

Thus the right-hand side of (5.31) is

\frac{1}{N} E⟨σ˜ · σ^{∗} Av ε_{1}ε_{3} exp \frac{β^{2}}{2N} (∥σ˜ ∥^{2} + ∥σ^{∗}∥^{2} + 2ε_{1}ε_{3}σ˜ · σ^{∗})⟩ (5.33)

= \frac{1}{N} E⟨σ˜ · σ^{∗} sh(\frac{β^{2}}{N} σ˜ · σ^{∗}) exp \frac{β^{2}}{2N} (∥σ˜ ∥^{2} + ∥σ^{∗}∥^{2})⟩.

5. External field and the replica-symmetric solution 45 Now we use the fact that for |x| ≤ 4, we have x sh β^{2}x ≤ \frac{x^{2}}{4} sh 4β^{2}, and that ∥σ˜ · σ^{∗}∥^{2}, ∥σ˜ ∥^{2}, ∥σ^{∗}∥^{2} ≤ 4N. From (5.31), (5.33) we obtain

E⟨Av σ_{1}˜ σ_{1}^{∗}ε_{1}ε_{3}E′′⟩ ≤ \frac{1}{4} sh 4β^{2} exp 8β^{2}⟨(\frac{σ˜ · σ^{∗}}{N} )^{2}⟩.

If we recall (5.25), we see that

C_{N+1}(β′, h) ≤ \frac{16}{N + 1} + \frac{1}{4}(\frac{N}{N + 1})^{2} sh 4β^{2} exp 8β^{2}C_{N}(β, h). (5.34)

We consider now β_{0} such that

\frac{1}{4} sh 4β_{0}^{2} exp 8β_{0}^{2} ≤ \frac{1}{2}

(e.g. β_{0} = \frac{1}{3}). Thus, if β′ ≤ β_{0} we have

\frac{1}{4} sh 4β′^{2} exp 8β′^{2} ≤ \frac{1}{2}

and

(N + 1)C_{N+1}(β′, h) ≤ 16 + \frac{1}{2}NC_{N}(β, h).

If we set

a_{N} = sup_{β≤β_{0}} NC_{N}(β, h),

since β′ > β we see that

a_{N+1} ≤ 16 + \frac{1}{2}a_{N}

and thus by induction over N that a_{N} ≤ 32. □

It should be pointed out that Proposition 5.2 is a very special case of a powerful result of [F-Z]; but this special proof we gave is particularly simple.

Let us now explore consequences of Proposition 5.2. We set

D_{N} = D_{N}(β, h) = E⟨(\frac{σ˜ · σ^{3}}{N} )^{2}⟩. (5.35)

Lemma 5.3. We have

(5.36)  C_{N} ≤ 4D_{N}
(5.37)  D_{N} ≤ \sqrt{C_{N}}.

Proof. To prove (5.36), using that (a + b)^{2} ≤ 2a^{2} + 2b^{2}, we have

C_{N} ≤ 2 E⟨(\frac{σ˜ · σ^{3}}{N} )^{2}⟩ + 2 E⟨(\frac{σ˜ · σ^{4}}{N} )^{2}⟩ = 4D_{N}.

To prove (5.37), we write

σ˜ · σ^{3} = ∑_{i≤N} σ_{i}˜ σ_{i}^{3},

so that, using Cauchy-Schwarz twice,

D_{N} = \frac{1}{N^{2}} E ∑_{i,j≤N}⟨σ_{i}˜ σ_{i}^{3}σ_{j}˜ σ_{j}^{3}⟩

= \frac{1}{N^{2}} ∑_{i,j≤N} E(⟨σ_{i}˜ σ_{j}˜ ⟩⟨σ_{i}^{3}σ_{j}^{3}⟩)

≤ \frac{1}{N^{2}} ∑_{i,j≤N}(E⟨σ_{i}˜ σ_{j}˜ ⟩^{2})^{\frac{1}{2}} (E⟨σ_{i}^{3}σ_{j}^{3}⟩^{2})^{\frac{1}{2}}

≤ \frac{1}{N^{2}} ∑_{i,j≤N}(E⟨σ_{i}˜ σ_{j}˜ ⟩^{2})^{\frac{1}{2}}

≤ (\frac{1}{N^{2}} ∑_{i,j≤N} E⟨σ_{i}˜ σ_{j}˜ ⟩^{2})^{\frac{1}{2}}

= (\frac{1}{N^{2}} ∑_{i,j≤N} E⟨σ_{i}˜ σ_{j}˜ σ_{i}^{∗}σ_{j}^{∗}⟩)^{\frac{1}{2}}

= (E⟨(∑_{i≤N} \frac{1}{N} σ_{i}˜ σ_{i}^{∗})^{2}⟩)^{\frac{1}{2}}

= \sqrt{C_{N}}. □

Thus, it follows from (5.25) and Proposition 5.2 that D_{N} ≤ 8/\sqrt{N} for β ≤ β_{0}. It is in fact true, as we will show later, that D_{N} ≤ L/N. This however cannot be proved by general principles.

Challenge problem 5.4. Prove by an example that (5.37) is essentially sharp. That is, construct a random measure G′_{N} on {−1, 1}^{N} such that if we define C_{N} and D_{N} by replacing G_{N} by G′_{N}, then (5.37) is essentially optimal.

As a consequence, we will not be able to deduce that D_{N} ≤ L/N from (5.24) but we will have to make a specific analysis, along the lines of Proposition 5.2. It is unfortunately harder to relate D_{N} and D_{N−1} than C_{N} and C_{N−1}.

As simplicity is our main goal in the present chapter, we will proceed through (5.37). This does not give the correct rates of convergences, so we will not attempt to get any rate of convergence at all. The correct rates will be obtained later.

Lemma 5.5. We have

\frac{1}{4}D_{N} ≤ E⟨(R(σ, σ′) − ⟨R(σ, σ′)⟩)^{2}⟩ ≤ 4D_{N}. (5.38)

In particular, (5.6) is equivalent to

lim_{N→∞} E⟨(R(σ, σ′) − ⟨R(σ, σ′)⟩)^{2}⟩ = 0. (5.39)

What (5.39) means is that the function (σ, σ′) ↦ R(σ, σ′) on (Σ_{N}^{2}, G_{N}^{2}) is nearly constant. The value of the constant is quite naturally the integral of the function, namely ⟨R(σ, σ′)⟩. This quantity might depend upon the disorder.

Proof. First, we write

D_{N} = E⟨(\frac{σ˜ · σ^{3}}{N} )^{2}⟩
= E⟨(R(σ^{2}, σ^{3}) − R(σ^{1}, σ^{3}))^{2}⟩

≤ 2 E⟨(R(σ^{2}, σ^{3}) − ⟨R(σ^{2}, σ^{3})⟩)^{2}⟩ + 2 E⟨(R(σ^{1}, σ^{3}) − ⟨R(σ^{1}, σ^{3})⟩)^{2}⟩

= 4 E⟨(R(σ^{1}, σ^{2}) − ⟨R(σ^{1}, σ^{2})⟩)^{2}⟩.

Next we write

b = ⟨σ⟩ = (⟨σ_{i}⟩)_{i≤N}. (5.40)

We observe that by Jensen’s inequality

⟨(\frac{(σ^{1} − b) · σ^{3}}{N} )^{2}⟩ ≤ ⟨(\frac{σ˜ · σ^{3}}{N} )^{2}⟩. (5.41)

This is because the left-hand side relates to the right-hand side by averaging σ^{2} for G_{N} inside the square rather than outside. Similarly,

⟨(\frac{(σ^{1} − b) · b}{N})^{2}⟩ ≤ ⟨(\frac{σ˜ · σ^{3}}{N} )^{2}⟩. (5.42)

Let us also observe that

b · b = ∑_{i≤N}⟨σ_{i}⟩^{2} = ∑_{i≤N}⟨σ_{i}^{1}σ_{i}^{2}⟩ = ⟨σ^{1} · σ^{2}⟩. (5.43)

Thus

E⟨(\frac{σ^{1} · σ^{2}}{N} − ⟨\frac{σ^{1} · σ^{2}}{N} ⟩)^{2}⟩ = E⟨(\frac{(σ^{1} − b) · σ^{2}}{N} + \frac{(σ^{2} − b) · b}{N})^{2}⟩
≤ 4D_{N},

using (5.41), (5.42), and the fact that (a + b)^{2} ≤ 2a^{2} + 2b^{2}. □

In the use of the cavity method, we have met quantities such as

⟨exp \frac{t}{\sqrt{N}} ∑_{i≤N} g_{i}σ_{i}⟩ (5.44)

in (5.40) for t = βσ_{N+1}. These will occur often, so we write

∑_{i≤N} g_{i}σ_{i} = g · σ. (5.45)

The quantities in (5.40) are averages with respect to Gibbs’ measure, and as we do not yet understand Gibbs’ measure, we might fear that we do not understand them either. A fundamental consequence of (5.6) is that the quantities (5.44) essentially depend upon Gibbs’ measure only through the much simpler quantity b = ⟨σ⟩.

Proposition 5.6. Under (5.7), if f is a bounded function on Σ_{N}, we have

lim_{N→∞} E|⟨f(σ) exp \frac{t}{\sqrt{N}} g · σ⟩ − ⟨f⟩ exp(\frac{t}{\sqrt{N}} g · b + \frac{t^{2}}{2} (1 − \overline{q}))| → 0 (5.46)

where

\overline{q} = ⟨R(σ, σ′)⟩ = \frac{∥b∥^{2}}{N} .

Strictly speaking the expression “f is a bounded function on Σ_{N}” is meaningless. This is a convenient way to express the following: for each N, we consider a function f_{N} on Σ_{N}, and we assume that sup_{N}|f_{N}| < ∞.

Proof. First let

X = ⟨f(σ) exp \frac{t}{\sqrt{N}} g · σ⟩
Y = ⟨f(σ) exp(\frac{t}{\sqrt{N}} g · b + \frac{t^{2}}{2} (1 − \overline{q}))⟩.

We will show that

E(X − Y)^{2} = E(E_{g}(X^{2} − 2XY + Y^{2})) → 0 (5.47)

where E_{g} denotes integration in g only. We use replicas to write

X^{2} = ⟨f(σ)f(σ′) exp \frac{t}{\sqrt{N}} g · (σ + σ′)⟩,
XY = ⟨f(σ)f(σ′) exp(\frac{t}{\sqrt{N}} g · (σ + b) + \frac{t^{2}}{2} (1 − \overline{q}))⟩,
Y^{2} = exp(\frac{2t}{\sqrt{N}} g · b + t^{2}(1 − \overline{q}))⟨f(σ)f(σ′)⟩,

so that, since ∥σ∥^{2} = ∥σ′∥^{2} = N, using (2.14),

E_{g} X^{2} = ⟨f(σ)f(σ′) exp t^{2}(1 + R(σ, σ′))⟩,
E_{g} XY = ⟨f(σ)f(σ′) exp(\frac{t^{2}}{2} (1 + 2 \frac{σ · b}{N} + \frac{∥b∥^{2}}{N} ) + \frac{t^{2}}{2} (1 − \overline{q}))⟩

= ⟨f(σ)f(σ′) exp t^{2}(1 + \frac{σ · b}{N})⟩,

E_{g} Y^{2} = exp(2t^{2} \frac{∥b∥^{2}}{N} + t^{2}(1 − \overline{q}))⟨f(σ)f(σ′)⟩

= exp t^{2}(1 + \overline{q}) ⟨f(σ)f(σ′)⟩.

Now, it follows from (5.39) that

E⟨(R(σ^{1}, σ^{2}) − \overline{q})^{2}⟩ → 0,

and from (5.42)and (5.43) that

E⟨(\frac{σ^{1} · b}{N} − \overline{q})^{2}⟩ → 0.

This implies that

E |E_{g} X^{2} − E_{g} Y^{2}| → 0 , E |E_{g} XY − E_{g} Y^{2}| → 0.

This proves (5.43) and Proposition 5.6. □

Not surprisingly, Proposition 5.6 will be very useful. But what is really behind it? Proposition 5.6 means that for the typical choice of g the law of (g·σ)/\sqrt{N} under G_{N} is approximately N ((g·b/\sqrt{N}), (1−\overline{q})). It is a kind of central limit theorem but it holds for the typical random coefficients g_{i} rather than for any choice. If it were true that G_{N} is a product measure (which it is certainly not) then for any sequence (a_{i})_{i≤N}, with max a_{i} ≪ (∑ a_{i}^{2})^{\frac{1}{2}} , the law under G_{N} of σ ↦ ∑ a_{i}σ_{i} would be approximatively

N (∑_{i≤N} a_{i}⟨σ_{i}⟩, ∑_{i≤N} a_{i}^{2}(1 − ⟨σ_{i}⟩^{2})).

When a_{i} = g_{i}/\sqrt{N}, one certainly expects that averaging takes place and that

∑_{i≤N} \frac{g_{i}^{2}}{N} (1 − ⟨σ_{i}⟩^{2}) ≃ 1 − \frac{1}{N} ∑_{i≤N}⟨σ_{i}⟩^{2} = 1 − \overline{q}.

Proposition 5.6 is much less surprising as part of a general theme: under (5.6), Gibbs’ measure somehow resembles a product measure. This is in fact the case as long as one considers only finitely many spins.

Proposition 5.7. Under (5.6), for each p, we have

lim_{N→∞} E(⟨σ_{1} … σ_{p}⟩ − ⟨σ_{1}⟩ … ⟨σ_{p}⟩)^{2} = 0. (5.48)

Thus, quite remarkably, the vanishing of the correlations of two spins implies a similar vanishing on any given number of spins.

Proof. We proceed by induction over p. This is true for p = 2. For the induction from p − 1 to p, since |⟨σ_{i}⟩| ≤ 1, it suffices to prove that

lim_{N→∞} E(⟨σ_{1} … σ_{p}⟩ − ⟨σ_{1} … σ_{p−1}⟩⟨σ_{p}⟩)^{2} = 0,

or, with the notation

σ̇_{i} = σ_{i} − ⟨σ_{i}⟩, (5.49)

that

lim_{N→∞} E⟨σ_{1} … σ_{p−1}σ̇_{p}⟩^{2} = 0. (5.50)

We observe that, by symmetry upon the sites, we have

M E⟨σ_{1} … σ_{p−1}σ̇_{p}⟩^{2} ≤ E ∑_{i_{1},...,i_{p}}⟨σ_{i_{1}} … σ_{i_{p−1}} σ̇_{i_{p}}⟩^{2} (5.51)

where M = N(N − 1) … (N − p + 1), and where the summation is over all choices i_{1}, …, i_{p} ≤ N. Now, using replicas,

(5.52) ∑_{i_{1},...,i_{p}}⟨σ_{i_{1}} … σ_{i_{p−1}} σ̇_{i_{p}}⟩^{2} = ⟨ ∑_{i_{1},...,i_{p}} σ_{i_{1}} … σ_{i_{p−1}} σ̇_{i_{p}}σ′_{i_{1}} … σ′_{i_{p−1}}σ̇′_{i_{p}} ⟩

= N^{p}⟨(\frac{σ̇ · σ̇′}{N} ) R(σ, σ′)^{p−1}⟩.

Moreover, since |R(σ, σ′)| ≤ 1, by Cauchy-Schwarz,

E⟨(\frac{σ̇ · σ̇′}{N} )R(σ, σ′)^{p−1}⟩ ≤ (E⟨(\frac{σ̇ · σ̇′}{N} )^{2}⟩)^{\frac{1}{2}} ≤ CN^{\frac{1}{2}}, (5.53)

where in the last inequality we have used Jensen’s inequality as in (5.41). To finish the proof, we observe that (5.50) follows from (5.51) to (5.53). □

Theorem 5.8. Under (5.6), given any number p, σ_{1}, …, σ_{p} are asymptotically independent under G_{N}.

Proof. The proof will provide a precise statement. Denote by G_{N,p} the law of σ_{1}, …, σ_{p} under G_{N}, or equivalently, the projection of G_{N} onto {−1, 1}^{p}. Let us denote by μ the (random) product measure on {−1, 1}^{p} such that

∫ σ_{i} dμ(σ_{1}, …, σ_{p}) = ⟨σ_{i}⟩ = ∫ σ_{i} dG_{N,p}(σ_{1}, …, σ_{p}).

Then E |G_{N,p} − μ| → 0, where |G_{N,p} − μ| is the total variation distance. To prove this, it suffices to show that given σ_{1}, …, σ_{p},

E |G_{N,p}({(σ_{1}, …, σ_{p})}) − μ({(σ_{1}, …, σ_{p})})| → 0. (5.54)

Now, if (η_{1}, …, η_{p}) ∈ {−1, 1}^{p},

1_{{(σ_{1},...,σ_{p})}}(η_{1}, …, η_{p}) = 2^{−p} ∏_{i≤p}(1 + η_{i}σ_{i}) = 2^{−p} ∑_{I} η_{I}σ_{I}

where the summation is over I ⊂ {1, …, p}, and σ_{I} = ∏_{i∈I} σ_{i}, η_{I} = ∏_{i∈I} η_{i}. Thus, to prove (5.54), it suffices to prove that for each I

E |∫ σ_{I} dG_{N,p}(σ_{1}, …, σ_{p}) − ∫ σ_{I} dμ(σ_{1}, …, σ_{p})| → 0, (5.55)

i.e.

E |⟨σ_{I}⟩ − ∏_{i∈I}⟨σ_{i}⟩| → 0,

which follows from Proposition 5.7. □

The remarkable feature of Theorem 5.8 is of course that the conclusion looks stronger than the hypothesis.

At this point the reader should be convinced of the importance of (5.6). With the exception of Proposition 5.2, the conclusions we have drawn from (5.6) are quite general, and rely only upon the symmetry between sites. The physicists say that a system satisfying (5.6) is in a pure equilibrium state. For all the systems where the author has been able to prove at high temperature the validity of the replica symmetric formulas, proving that the system is in a pure state has always been the first (and most difficult) step.

Lemma 5.5 tells us that the overlap of two generic configurations is independent of the particular choice of these configurations, although it might depend upon the disorder. Of course, we would like it to be in fact independent of the disorder. That is, we would like

⟨R(σ, σ′)⟩ = \frac{1}{N} ∑_{i≤N}⟨σ_{i}σ′_{i}⟩ = \frac{1}{N} ∑_{i≤N}⟨σ_{i}⟩^{2} = \overline{q}

to be independent of the disorder. Before we prove this, we collect a few simple facts, the first of which, the integration by parts formula, is of fundamental importance.

Proposition 5.9. If g is centered normal, for any smooth function f such that (1 + x^{2})^{−k}f(x) is bounded for some k then

E(gf(g)) = E(g^{2}) E(f′(g)). (5.56)

Proof. This is, indeed, integration by parts. If E(g^{2}) = τ^{2},

\frac{1}{τ\sqrt{2π}} ∫_{−∞}^{+∞} tf(t) exp(−\frac{t^{2}}{2τ^{2}} ) dt = τ^{2} \frac{1}{τ\sqrt{2π}} ∫_{−∞}^{+∞} f′(t) exp(−\frac{t^{2}}{2τ^{2}} ) dt. □

We consider

(5.57) Ψ(x) = E th^{2}(βg\sqrt{x} + h) = \frac{1}{\sqrt{2π}} ∫ th^{2}(βg\sqrt{t} + h) exp(−\frac{t^{2}}{2} ) dt.

This notation will be used in Chapters 5 and 6. (The function ψ of Chapter 4 will be not used any more).

Lemma 5.10. We have

Ψ′(x) = β^{2} E \frac{1 − 2 sh^{2}(βg\sqrt{x} + h)}{ch^{4}(βg\sqrt{x} + h)}

and in particular |Ψ′| ≤ 2β^{2}, and |Ψ′| ≤ \frac{1}{2} if β ≤ \frac{1}{3}.

Proof. We have

Ψ′(x) = E (\frac{βg}{\sqrt{x}} \frac{th}{ch^{2}} (βg\sqrt{x} + h))

and the result by (5.56), since

(\frac{th(x)}{ch^{2}(x)})′ = \frac{1 − 2 sh^{2}(x)}{ch^{4}(x)}.  □

It follows, in particular, that (5.4) has indeed as unique solution q.

Theorem 5.11. If β < β_{0}, for each h we have

lim_{N→∞} E(\overline{q} − q)^{2} = 0. (5.58)

Proof. We set

Q_{N} = Q_{N}(β, h) = E(\overline{q} − q)^{2} = E(\frac{1}{N} ∑_{i≤N}⟨σ_{i}⟩^{2})^{2} − 2q E \frac{1}{N} ∑_{i≤N}⟨σ_{i}⟩^{2} + q^{2}.

By expanding the square and using symmetry between the sites, we have

Q_{N} ≤ \frac{1}{N} + E⟨σ_{1}⟩^{2}⟨σ_{N}⟩^{2} − q E⟨σ_{1}⟩^{2} − q E⟨σ_{N}⟩^{2} + q^{2}, (5.59)

and, changing N into N + 1,

(5.60)  Q_{N+1}(β′, h) ≤ \frac{1}{N + 1}
+ E ⟨σ_{1}⟩′^{2}⟨σ_{N+1}⟩′^{2} − q′ E ⟨σ_{1}⟩′^{2} − q′ E ⟨σ_{N+1}⟩′^{2} + q^{′2},

where q′ is the solution of (5.4) when β′ replaces β. It should be obvious that

|q − q′| = o(1), (5.61)

where o(1) is a quantity that goes to zero as N → ∞.

5. External field and the replica-symmetric solution 53 Lemma 5.12. We have E |⟨σ_{1}⟩′^{2} − ⟨σ_{1}⟩^{2}| → 0 (5.62)

E |⟨σ_{N+1}⟩′^{2} − th^{2}(\frac{β}{\sqrt{N}} g · b + h)| → 0. (5.63)

Proof. To prove (5.62), from (5.19) we have

⟨σ_{1}⟩′ = \frac{⟨Av σ_{1}E⟩}{Z} ,

for

Z = ⟨Av E⟩
E = exp σ_{N+1}(\frac{β}{\sqrt{N}} ∑_{i≤N} g_{i}σ_{i} + h) .

It follows from Proposition 5.6 that

E |Z − exp \frac{β^{2}}{2} (1 − \overline{q}) ch(\frac{β}{\sqrt{N}} g · b + h)| → 0,

E |⟨Av σ_{1}E⟩ − ⟨σ_{1}⟩ exp \frac{β^{2}}{2} (1 − \overline{q}) ch(\frac{β}{\sqrt{N}} g · b + h)| → 0.

Since Z ≥ 1, this implies

E |⟨σ_{1}⟩′ − ⟨σ_{1}⟩| → 0

from which (5.62) follows. The proof of (5.63) is similar. □ End of the proof of Theorem 5.11. If we combine (5.60) to (5.63), we see that

(5.64) Q_{N+1}(β′, h) ≤ E(⟨σ_{1}⟩^{2} th^{2}(\frac{β}{\sqrt{N}} g · b + h))

− q E⟨σ_{1}⟩^{2} − q E th^{2}(\frac{β}{\sqrt{N}} g · b + h) + q^{2} + o(1).

Since

E_{g}(\frac{1}{\sqrt{N}} g · b)^{2} = \frac{1}{N} ∥b∥^{2} = \overline{q},

we have

E_{g} th^{2}(\frac{β}{\sqrt{N}} g · b + h) = Ψ(\overline{q}). (5.65)

Moreover, since ⟨σ_{1}⟩^{2} does not depend upon g, we have

E(⟨σ_{1}⟩^{2} th^{2}(\frac{β}{\sqrt{N}} g · b + h)) = E(⟨σ_{1}⟩^{2} E_{g} th^{2}(\frac{β}{\sqrt{N}} g · b + h))
= E(⟨σ_{1}⟩^{2}Ψ(\overline{q})).

With (5.64) this implies

Q_{N+1}(β′, h) ≤ E(⟨σ_{1}⟩^{2}Ψ(\overline{q})) − q E⟨σ_{1}⟩^{2} − q E Ψ(\overline{q}) + q^{2} + o(1) (5.66)

and, using the symmetry between sites,

(5.67)  Q_{N+1}(β′, h) ≤ E(\overline{q}Ψ(\overline{q})) − q E(\overline{q}) − q E Ψ(\overline{q}) + q^{2} + o(1)

= E((\overline{q} − q)(Ψ(\overline{q}) − q)) + o(1)

= E((\overline{q} − q)(Ψ(\overline{q}) − Ψ(q))) + o(1)

≤ \frac{1}{2} E((\overline{q} − q)^{2}) + o(1)

= \frac{1}{2}Q_{N}(β, h) + o(1).

Here, we have used Lemma 5.10 in the fourth line. Iteration of (5.67) yields (5.58). □

We have now improved (5.39) into

lim_{N→∞} E⟨(R(σ, σ′) − q)^{2}⟩ = 0 (5.68)

where q is given by (5.4).

We are now ready to start the proof of (5.3). We first remove some of the mystery of this formula. Fixing h once and for all, we denote by SK(β, q) the right-hand side of (5.3), when we think of β, q as independent variables. We then have

\frac{∂ SK}{∂q} = −\frac{β^{2}}{2} (1 − q) + E(\frac{βg}{2\sqrt{q}} th(βg\sqrt{q} + h)),

and using integration by parts (Proposition 5.9),

\frac{∂ SK}{∂q} = −\frac{β^{2}}{2} (1 − q) + \frac{β^{2}}{2} E \frac{1}{ch^{2}(βg\sqrt{q} + h)}.

Since th^{2}(x) = 1 − \frac{1}{ch^{2}(x)}, (5.4) means that q = q(β, h) is such that

\frac{∂ SK}{∂q}(β, q) = 0.

It follows that

\frac{d}{dβ} SK(β, q(β, h)) = \frac{∂ SK}{∂β}(β, q(β, h)). (5.69)

Now,

\frac{∂ SK}{∂β} = \frac{β}{2} (1 − q)^{2} + E(g\sqrt{q} th(βg\sqrt{q} + h))

= \frac{β}{2} (1 − q)^{2} + βq E \frac{1}{ch^{2}(βg\sqrt{q} + h)},

using Proposition 5.9 again. Now, by (5.4),

E \frac{1}{ch^{2}(βg\sqrt{q} + h)} = 1 − q,

so that

\frac{∂ SK}{∂β} = \frac{β}{2} (1 − q^{2}). (5.70)

We note for β = 0

p_{N}(0, h) = log(2 ch h) = log(e^{h} + e^{−h})

so that (5.3) is obviously true. So to prove (5.3) we will show that

lim_{N→∞} |\frac{∂p_{N}}{∂β} (β, h) − \frac{β}{2} (1 − q^{2})| = 0, (5.71)

where q is given by (5.4). This, of course, removes much of the mystery of (5.3). Lemma 5.13. We have

\frac{∂p_{N}}{∂β} = \frac{β}{2} (1 − E⟨R(σ, σ′)^{2}⟩).

Proof. This fundamental formula is again a consequence of integration by parts. We write (5.72) p_{N}(β, h) = \frac{1}{N} E log Z_{N} = \frac{1}{N} E log ∑ ; σ B(σ), where

B(σ) = B_{N}(β, h, σ) = exp(\frac{β}{\sqrt{N}} ∑_{i<j} g_{ij}σ_{i}σ_{j} + h ∑_{i≤N} σ_{i})

and thus

(5.73)  \frac{∂B(σ)}{∂β} = \frac{1}{\sqrt{N}} (∑_{i<j} g_{ij}σ_{i}σ_{j})B(σ)
(5.74)  \frac{∂B(σ)}{∂g_{ij}} = \frac{β}{\sqrt{N}} σ_{i}σ_{j}B(σ).

Thus, from (5.71),

(5.75)  \frac{∂}{∂β} p_{N}(β, h) = \frac{1}{N^{3/2}} E ∑ ; σ \frac{∑_{i<j} g_{ij}σ_{i}σ_{j}B(σ)}{Z_{N}}

= \frac{1}{N^{3/2}} ∑_{i<j} ∑ ; σ σ_{i}σ_{j} E \frac{g_{ij}B(σ)}{Z_{N}}.

We now use integration by parts using the fact that the g_{ij} are independent, so that we can use Proposition 5.9 conditionally upon the (g_{i′j′}), (i′, j′) ≠ (i, j) and we get, using (5.74)

E(g_{ij} \frac{B(σ)}{Z_{N}}) = \frac{β}{\sqrt{N}} (σ_{i}σ_{j} E(\frac{B(σ)}{Z_{N}}) − E \frac{B(σ)(∑σ_{′} σ′_{i}σ′_{j}B(σ′))}{Z_{N}^{2}})

and (5.75) yields

\frac{∂}{∂β} p_{N}(β, h) = \frac{β}{N^{2}} (\frac{N(N − 1)}{2} − E ∑_{i<j}⟨σ_{i}σ_{j}σ′_{i}σ′_{j}⟩)

= \frac{β}{2N^{2}} (N^{2} − E ∑_{i,j}⟨σ_{i}σ_{j}σ′_{i}σ′_{j}⟩)

= \frac{β}{2} (1 − \frac{1}{N^{2}} E⟨(∑_{i} σ_{i}σ′_{i})^{2}⟩)

= \frac{β}{2} (1 − E⟨R(σ, σ′)^{2}⟩). □

Combining Theorem 5.11, Lemma 5.13 and (5.70) proves (5.71), and hence Theorem 5.1.

Thus, it seems quite appropriate to say that the replica symmetric formula (5.3) is neither so mysterious nor quite central. The central fact is (5.68).

Theorem 5.8 means that “locally” (on finitely many spins), Gibbs’ measure is determined by the sequence b = (⟨σ_{i}⟩)_{i≤N}, and Proposition 5.6 says that this is even true “globally”, in some respects. So, how does this sequence ⟨σ_{i}⟩_{i≤N} behave?

Theorem 5.14. For β ≤ β_{0}, given any number p, ⟨σ_{1}⟩, …, ⟨σ_{p}⟩ are asymptotically i.i.d. Their common limit law is the law ν of th(βg\sqrt{q} + h) where g is N (0, 1) and q is given by (5.4).

Proof. By induction over p, we show that if f : ℝ^{p} → ℝ is continuous with bounded support,

E f(⟨σ_{1}⟩, …, ⟨σ_{p}⟩) − ∫ f(x_{1}, …, x_{p}) dν(x_{1}) … dν(x_{p}) → 0

uniformly over β ≤ β_{0}. This is true for p = 0 (nothing to prove) so we proceed to the induction step from p − 1 to p. Since this requires use of the cavity method, we will change N into N + 1 and, using the symmetry among sites, we will prove that

E f(⟨σ_{1}⟩′, …, ⟨σ_{p−1}⟩′, ⟨σ_{N+1}⟩′) − ∫ f(x_{1}, …, x_{p}) dν′(x_{1}) … dν′(x_{p}) → 0 (5.76)

uniformly in β′ ≤ β_{0}, where β′ and β are related by β′ = β (1 + \frac{1}{N} )^{1/2}, and where ν′ corresponds to β′ rather than β. We observe from Lemma 5.12 that

E |⟨σ_{N+1}⟩′ − th(\frac{β}{\sqrt{N}} g · b + h)| → 0

E |⟨σ_{i}⟩′ − ⟨σ_{i}⟩| → 0, for all i, (5.77)

so that

E f(⟨σ_{1}⟩′, …, ⟨σ_{p−1}⟩′, ⟨σ_{N+1}⟩′) − E f(⟨σ_{1}⟩, …, ⟨σ_{p−1}⟩, th(\frac{β}{\sqrt{N}} g · b + h)) → 0. (5.78)

Consider the function

u(x_{1}, …, x_{p−1}, y) = E f(x_{1}, …, x_{p−1}, th(βg + h)),

where g is N (0, y). It is obviously continuous. In (5.78), g is independent of ⟨ · ⟩, and the law of (g · b)/\sqrt{N} given b is N (0, \overline{q}). It follows that

E f(⟨σ_{1}⟩, …, ⟨σ_{p−1}⟩, th(\frac{β}{\sqrt{N}} g · b + h)) = E u(⟨σ_{1}⟩, …, ⟨σ_{p−1}⟩, \overline{q}).

Since we have shown that E |\overline{q} − q| → 0, we have

E u(⟨σ_{1}⟩, …, ⟨σ_{p−1}⟩, \overline{q}) − E u(⟨σ_{1}⟩, …, ⟨σ_{p−1}⟩, q) → 0.

Now the function u(x_{1}, …, x_{p}, q) has bounded support, so by induction hypothesis (and since β ≤ β′) we have

E u(⟨σ_{1}⟩, …, ⟨σ_{p−1}⟩, q) − ∫ u(x_{1}, …, x_{p−1}, q) dν(x_{1}) … dν(x_{p−1}) → 0.

Moreover

u(x_{1}, …, x_{p−1}, q) = ∫ f(x_{1}, …, x_{p−1}, x_{p}) dν(x_{p}).

Combining these estimates we have proved

E f(⟨σ_{1}⟩′, …, ⟨σ_{p−1}⟩′, ⟨σ_{N+1}⟩′) − ∫ f(x_{1}, …, x_{p}) dν(x_{1}) … dν(x_{p}) → 0

from which (5.76) follows. □