I have a Hamiltonian of a system as \(\displaystyle H(x\in X) = \max\limits_{a,b} \left( p_{ab} + q\frac{ n_b}{ n_a} \right) x_{ab}\)

where \(\displaystyle n_a=\sum_b x_{ab}, n_b=\sum_a x_{ab}\), and \(\displaystyle X = \{ x = [x_{ab}]_{a\in A}^{b\in B} |x_{ab}\in\{0,1\}, n_a\geq 1 \}\). Here, \(\displaystyle p_{ab}\) is a random variable with gamma distribution and \(\displaystyle q\) is a constant.

I need to simplify/find close-form expression for Partition sum \(\displaystyle Z = \sum_{x\in X} e^{-\beta H(x) }\).

My attempts:

Modify Hamiltonian as \(\displaystyle H(x,t) = t\) with additional constraints \(\displaystyle \left( p_{ab} + q\frac{ n_b}{ n_a} \right) x_{ab} \leq t\) for all \(\displaystyle a,b\).

Then I think the partition sum should be \(\displaystyle Z = \sum_{x\in X} \int_{0}^{\infty} e^{-\beta H(x,t) } dt\).

I have no clue how to simplify this due to having mixed integer and linear parameters.

Modify Hamiltonian as \(\displaystyle H(x) = \frac{1}{t} \ln \frac{1}{AB} \sum_{a,b} \exp \left( t \left( p_{ab} + q\frac{ n_b}{ n_a} \right) x_{ab} \right)\).

As \(\displaystyle t\to\infty\), I can obtain the original Hamiltonian.

Then,

\(\displaystyle Z = \sum_{x\in X} e^{-\beta H(x) } = \prod\limits_{a,b}\sum\limits_{x_{a,b}=0,1} \sum\limits_{n_a\geq1} \sum\limits_{n_b\geq 0} e^{-\beta H(x) }\)

\(\displaystyle Z = \sum\limits_{x_{a,b}=0,1} \sum\limits_{n_a\geq1} \sum\limits_{n_b\geq 0} e^{-\beta H(x) } \prod\limits_{a}\delta_{n_a,\sum_b x_{ab}} \prod\limits_{b}\delta_{n_b,\sum_a x_{ab}}\)

Substitute \(\displaystyle \delta_{n_a,\sum_b x_{ab}} = \int_{0}^{2\pi}\frac{d\lambda}{2\pi}e^{\imath \lambda (n_a-\sum_b x_{ab})}\) to decouple \(\displaystyle x_{ab}\) variables.

I have used this method for a simpler form of Hamiltonian such as \(\displaystyle H=p_{ab}x_{ab} - q n_a - r n_b\) in which after above step I could separate the variables \(\displaystyle x_{ab}\) to a product term where I could substitute \(\displaystyle x_{ab}=0,1\) and carryout the integrals.

For above choice of Hamiltonian, I cannot decouple \(\displaystyle x_{ab}\) variables.

I would really appreciate if any of you could guide me/provide alternate method to simplify the partition sum.

Thanks.

where \(\displaystyle n_a=\sum_b x_{ab}, n_b=\sum_a x_{ab}\), and \(\displaystyle X = \{ x = [x_{ab}]_{a\in A}^{b\in B} |x_{ab}\in\{0,1\}, n_a\geq 1 \}\). Here, \(\displaystyle p_{ab}\) is a random variable with gamma distribution and \(\displaystyle q\) is a constant.

I need to simplify/find close-form expression for Partition sum \(\displaystyle Z = \sum_{x\in X} e^{-\beta H(x) }\).

My attempts:

**Method 1:**Modify Hamiltonian as \(\displaystyle H(x,t) = t\) with additional constraints \(\displaystyle \left( p_{ab} + q\frac{ n_b}{ n_a} \right) x_{ab} \leq t\) for all \(\displaystyle a,b\).

Then I think the partition sum should be \(\displaystyle Z = \sum_{x\in X} \int_{0}^{\infty} e^{-\beta H(x,t) } dt\).

I have no clue how to simplify this due to having mixed integer and linear parameters.

**Method 2:**Modify Hamiltonian as \(\displaystyle H(x) = \frac{1}{t} \ln \frac{1}{AB} \sum_{a,b} \exp \left( t \left( p_{ab} + q\frac{ n_b}{ n_a} \right) x_{ab} \right)\).

As \(\displaystyle t\to\infty\), I can obtain the original Hamiltonian.

Then,

\(\displaystyle Z = \sum_{x\in X} e^{-\beta H(x) } = \prod\limits_{a,b}\sum\limits_{x_{a,b}=0,1} \sum\limits_{n_a\geq1} \sum\limits_{n_b\geq 0} e^{-\beta H(x) }\)

\(\displaystyle Z = \sum\limits_{x_{a,b}=0,1} \sum\limits_{n_a\geq1} \sum\limits_{n_b\geq 0} e^{-\beta H(x) } \prod\limits_{a}\delta_{n_a,\sum_b x_{ab}} \prod\limits_{b}\delta_{n_b,\sum_a x_{ab}}\)

Substitute \(\displaystyle \delta_{n_a,\sum_b x_{ab}} = \int_{0}^{2\pi}\frac{d\lambda}{2\pi}e^{\imath \lambda (n_a-\sum_b x_{ab})}\) to decouple \(\displaystyle x_{ab}\) variables.

I have used this method for a simpler form of Hamiltonian such as \(\displaystyle H=p_{ab}x_{ab} - q n_a - r n_b\) in which after above step I could separate the variables \(\displaystyle x_{ab}\) to a product term where I could substitute \(\displaystyle x_{ab}=0,1\) and carryout the integrals.

For above choice of Hamiltonian, I cannot decouple \(\displaystyle x_{ab}\) variables.

I would really appreciate if any of you could guide me/provide alternate method to simplify the partition sum.

Thanks.

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