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I am self-studying spin glasses and the first exercise I want to do is the following:

Characterize probability measures on $\{-1, +1\}^N$ that arise as Gibbs measures for a Hamiltonian.

I am not exactly sure what to do here. Since it’s the first exercise of a rather introductory book (and there is not much information before this exercise), I don’t expect the solution to be that complex (but that can be me being naive).

For some context, the Gibbs measure on $\{-1, +1\}^N$ with Hamiltonian $H_N$ is $$G_N(\boldsymbol{\sigma}) \triangleq \frac{\exp(-H_N (\boldsymbol{\sigma}))}{Z_N},$$ where $\boldsymbol{\sigma}$ is a spin configuration and $Z_N$ is the partition function.

I would appreciate a solution here. Thank you!

MathIsLife12
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    Consider a probability measure $\mu$ on ${-1,1}^N$ such that $\mu(\boldsymbol{\sigma})>0$ for all $\boldsymbol{\sigma}$. Can you see how to define a Hamiltonian $H_N$ so that $\mu(\boldsymbol{\sigma}) = C \exp(-H_N(\mathbf{\boldsymbol{\sigma}}))$ for some constant $C$ and all configurations $\boldsymbol{\sigma}$? – Yvan Velenik May 29 '24 at 07:37
  • @YvanVelenik I guess that’s my confusion, since I don’t understand why any Hamiltonian wouldn’t work here. As long as you have one, this Gibbs measure must exist, right? – MathIsLife12 May 29 '24 at 16:45
  • You need the measure $\mu$ to give positive probability to every configuration (so that you can take the logarithm and obtain the Hamiltonian). (I don't know what book you are using, but in general one does not allow the Hamiltonian to take infinite values, hence the condition.) – Yvan Velenik May 29 '24 at 16:47
  • @YvanVelenik Ah I see. I’d appreciate if you wrote something rigorous in the answer so I can give you a check, if you have time. – MathIsLife12 May 29 '24 at 16:49

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Given an arbitrary probability measure on $\{-1,1\}^N$ satisfying $$ \forall\boldsymbol{\sigma}\in\{-1,1\}^N, \qquad\mu(\boldsymbol{\sigma}) > 0, $$ one can define $$ H_N(\boldsymbol{\sigma}) = -\log\mu(\boldsymbol{\sigma}). $$ With this choice, we obviously obtain $$ \mu(\boldsymbol{\sigma}) = \exp(-H_N(\boldsymbol{\sigma})), $$ which shows that $\mu$ is indeed a Gibbs measure on $\{-1,1\}^N$.

The positivity condition is necessary, since one usually consider $H_N:\{-1,1\}^N \to \mathbb{R}$, thus forbidding infinite values.

Remark: Of course, this is really a necessary and sufficient condition, since $Z_N^{-1}\exp(-H_N(\boldsymbol{\sigma})) > 0$, for any Gibbs measure.

Remark: There are some contexts in which the restriction on the Hamiltonian is lifted, that is one considers $H_N:\{-1,1\}^N \to \mathbb{R}\cup\{+\infty\}$, as this allows modelling so-called hard-core interactions. In this case, of course, any probability measure on $\{-1,1\}^N$ is a Gibbs measure.

Yvan Velenik
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