Physics Help Forum Simplifying Partition Sum

 Jul 3rd 2017, 05:21 AM #1 Junior Member   Join Date: Jul 2017 Posts: 6 Simplifying Partition Sum 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: 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. Last edited by zemozamster; Jul 4th 2017 at 12:04 AM. Reason: Missing 't' inside exp() of modified Hamiltonian under method 2
 Jul 18th 2017, 10:41 AM #2 Physics Team   Join Date: Apr 2009 Location: Boston's North Shore Posts: 1,568 I'd love to help but all I see is a bunch of strange symbols which looks like Latex code. I don't read Latex code so as such I can't help. Is there a way for you to post it so I can see math symbols instead of code? Thanks.
 Jul 18th 2017, 12:40 PM #3 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 993 @PMB The Tex looks OK to me although I had to wait 23 seconds for mathjax to appear and another 20 for it to parse the Tex and display. Thanks Mash for the update. @zemozamster I take it you are dividing the quantity (normalised?) x between a reservoir, represented by q and some excitation levels, represented by p?
Jul 18th 2017, 12:55 PM   #4
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 Originally Posted by studiot @PMB The Tex looks OK to me although I had to wait 23 seconds for mathjax to appear and another 20 for it to parse the Tex and display.
What is mathjax? Could the problem be that I don't have something installed that I need to in order that I see it?

 Jul 18th 2017, 01:10 PM #5 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 993 Hi, PMB I am not an expert in mathjax. It is something used by the forum (and many others), whereby the tex or mathml script is sent to another site for translation into something that can be displayed and then sent back to this forum and displayed. It stopped working a week or so back and I complained so Mash fixed it. But it is very slow, I have noticed this slowness on other forums too. If it genuinely does not come up for you try refreshing the page and then waintin one minute. Meanwhile I have taken a screenshot of the OP. It should blow up well enough to be readable. Hope this helps Attached Thumbnails
 Jul 18th 2017, 01:13 PM #6 Physics Team   Join Date: Apr 2009 Location: Boston's North Shore Posts: 1,568 Ah ha! I see it now. You were right. All I had to do was wait a bit. Muchas gracias (and no, I don't speak Spanish)!
Aug 4th 2017, 06:01 AM   #7
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@studiot
Thank you for posting the output. My current workstation loads the equations instantly and thus, I was not aware of any issue.

 Originally Posted by studiot I take it you are dividing the quantity (normalised?) x between a reservoir, represented by q and some excitation levels, represented by p?
Quite similar to this idea.
You can interpret this as a problem of connectivity between two disjoint sets {a}s and {b}s. The system energy is defined by the maximum cost of existing links.

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