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Variational principles for approximating thermodynamic potentials in the inverse Ising problem: How to go from double to single extremum?

Physics Asked by Gauss57 on July 4, 2021

I’m trying to wrap my head around section 2.2.6 (on variational principles) in the following paper (on the inverse Ising problem): https://arxiv.org/abs/1702.01522

Here the authors explain how to use variational principles to approximate the following thermodynamic potentials:
$$ F(boldsymbol{J}, vec{h}) = – ln{Z(boldsymbol{J}, vec{h})}, $$

$$ S(boldsymbol{chi}, vec{m}) = min_{boldsymbol{J}, vec{h}}left{{-sum_{i} h_i m_i -sum_{i < j} J_{ij} chi_{ij} – F(boldsymbol{J}, vec{h}) }right}, $$

$$G(boldsymbol{J}, vec{m}) = max_{vec{h}}left{ { sum_{i} h_i m_i + F(boldsymbol{J}, vec{h})} right}. $$

I understand that $F(boldsymbol{J}, vec{h})$ can be obtained by noting that for a distribution $q$,

$$D_{KL}(q||p) = langle H rangle_q + langle ln q rangle_q + ln Z(boldsymbol{J}, vec{h}),$$
so that, using the notation $U[q] equiv langle H rangle_q$, and $S[q] equiv langle ln q rangle_q$, we have:
$$F(boldsymbol{J}, vec{h}) = langle H rangle_q + langle ln q rangle_q – D_{KL}(q||p) longrightarrow F(boldsymbol{J}, vec{h}) = min_q{ U[q] – S[q]}.$$

However, the authors then go on to say that it is easy to see (using Lagrange multipliers) that the remaining two potentials can be expressed as:

$$G(boldsymbol{J}, vec{m}) = max_{vec{h}}left{ { sum_{i} h_i m_i + min_q{ U[q] – S[q]}} right} = min_{q in mathcal{G}}left{-sum_{i < j} J_{ij} langlesigma_i sigma_j rangle_q – S[q]right}, $$
and
$$S(boldsymbol{chi}, vec{m}) = min_{boldsymbol{J}, vec{h}}left{{-sum_{i} h_i m_i -sum_{i < j} J_{ij} chi_{ij} – min_q{ U[q] – S[q]} }right} = max_{q in mathcal{S}}{S[q]}.$$
Where $mathcal{G} equiv {q : langle sigma_i rangle_q = m_i},$ and $mathcal{S} equiv {q : langle sigma_i rangle_q = m_i & langle sigma_i sigma_j rangle_q = chi_{ij} }$.

My question is how do I, via Lagrange multipliers, go from these double extremum expressions to the single extremum expressions. Thanks in advance for any help I may recieve.

One Answer

I think I understand how to get the correct expressions now, however, I did not use Lagrange multipliers:

We want to show that

$$S(boldsymbol{chi}, vec{m}) equiv min_{boldsymbol{J}, vec{h}}left{{-sum_{i} h_i m_i -sum_{i < j} J_{ij} chi_{ij} - min_q{ U[q] - S[q]} }right},$$ can be expressed as $$S(boldsymbol{chi}, vec{m}) = max_{q in mathcal{S}}{S[q]}.$$

Note first that $-min_q {U[q] - S[q]} = max_q{-U[q] + S[q]}$. Let

$$S(boldsymbol{chi}, vec{m} ; boldsymbol{J}, vec{h}; q) = - sum_i h_i m_i -sum_{i < j} J_{ij} chi_{ij} - U[q] + S[q],$$ then, since $$U[q] equiv langle H rangle_q = langle -sum_i h_i sigma_i - sum_{i < j} J_{ij} sigma_i sigma_j rangle_q = -sum_i h_i langle sigma_i rangle_q - sum_{i < j} J_{ij} langle sigma_i sigma_j rangle_q,$$ if we restrict the distributions $q$ to distributions $$q in mathcal{S} equiv { q : langle sigma_i rangle_q = m_i text{ and } langle sigma_i sigma_j rangle_q = chi_{ij} },$$ we obtain begin{align*} S(boldsymbol{chi}, vec{m} ; boldsymbol{J}, vec{h}; q) &= - sum_i h_i m_i -sum_{i < j} J_{ij} chi_{ij} - Big( -sum_i h_i langle sigma_i rangle_q - sum_{i < j} J_{ij} langle sigma_i sigma_j rangle_q Big) + S[q] &= Big( - sum_i h_i m_i -sum_{i < j} J_{ij} chi_{ij} +sum_i h_i m_i + sum_{i < j} J_{ij} chi_{ij} Big) + S[q] &= S[q]. quad (q in mathcal{S}) end{align*} Thus, $S(boldsymbol{chi}, vec{m} ; boldsymbol{J}, vec{h}; q) = S(boldsymbol{chi}, vec{m}; q)$, i.e., there is no dependecy on $boldsymbol{J}$ and $vec{h}$. To get $S(boldsymbol{chi}, vec{m})$ then, we simply re-introduce the maximization over $q$ (with our condition that $q in mathcal{S}$):

$$S(boldsymbol{chi}, vec{m}) = max_{q in mathcal{S}} { S[q]}.$$

A similar argument yields the correct expression for $G(boldsymbol{J}, vec{m})$.

Correct answer by Gauss57 on July 4, 2021

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