- #1
dumbperson
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Hi, I'm trying to calculate the partition function for a certain system and I arrived at an expression for the partition function $Z$, and have been stuck here for two weeks at the least. This is not a homework problem. If this is the wrong place to post a question like this, could you please direct me as to where I could ask this? My goal is to express this partition function into an integral over some function, as long as I get the sigmas/spins out.
I will leave out the details of the system (it's not exactly physics), but I'll gladly provide more info if you want any, but it looks like the Ising model.
$$ Z = \sum_{\vec{G}}\exp{\left[\sum_{i} \left(\frac{J(M-1)}{2}\sum_{j}\sum_{\alpha}\sigma_{ij}^\alpha - \sum_{\alpha}\sum_{j}\sigma_{ij}^\alpha \frac{\theta_i + \theta_j}{4} + \frac{J}{4} \sum_{j}\left(\sum_{\alpha}\sigma_{ij}^\alpha \right)^2 - \frac{1}{2}\sum_{j}h_{ij}^0 - \frac{JMN}{4} \right) \right]}$$
where $$\sigma_{ij}^\alpha \in \{-1,1\}$$
and the summation over G means to sum over all possible configurations of $$\{\sigma_{ij}^\alpha\}$$
where $i<j$, and $j = 1,...,N$ and $\alpha = 1,...,M$
$$ \sum_{\vec{G}} = \sum_{\{\sigma_{ij}^\alpha \} = \pm 1} = \sum_{\sigma_{11}^1 = \pm 1 }\sum_{\sigma_{11}^2 = \pm 1 } ... \sum_{\sigma_{12}^1 = \pm 1 }\sum_{\sigma_{12}^2 = \pm 1 }... $$
$M$ are the amount of different values of $\alpha$ (so alpha goes from 1 to $M$), $N$ are the amount of different values for$i$ and $j, J$ and $\theta_i$ are (unknown) constants, and $h_{ij}^0$ is a function of $J, \theta_i, \theta_j$.Normally what is done with partition functions such as these is that the term inside the exponent is linearized by using a "Hubbard stratonovich" transformation:
$$ e^{-\frac{1}{2}Ks^2} = \left(\frac{K}{2\pi} \right)^{1/2}\int e^{-\frac{1}{2}Kx^2 - iKsx} dx $$
$$e^{\frac{1}{2}Ks^2} = \left(\frac{K}{2\pi} \right)^{1/2}\int e^{-\frac{1}{2}Kx^2 + Ksx} dx $$
or
$$ e^{\frac{1}{2}\sum_{ij}K_{ij}s_is_j} =\left(\frac{\det{K}}{(2\pi)^N} \right)^{1/2} \int^{\infty}_{-\infty}...\int^{\infty}_{\infty}\prod_{k=1}^Nd\phi_k \exp{\left[-\frac{1}{2}\sum_{ij}\phi_i K_{ij}\phi_j + \sum_{ij} s_iK_{ij}\phi_j\right]} $$
I have tried to use this here, first I tried to write the partition function as
$$ Z = \sum_{G}\exp{\left[\sum_i \left(\frac{J(M-1)}{2}m_i - \frac{\theta_i}{2} m_i + \frac{J}{4}m_i^2 - \frac{J}{2}\sum_{j<k}m_{ij}m_{ik} - \frac{1}{2}\sum_j h_{ij}^0 - \frac{JMN}{4} \right) \right]} $$
where
$$m_{ij} = \sum_{\alpha} \sigma_{ij}^\alpha , \qquad m_{i} = \sum_{j}m_{ij} $$
The parts of the exponent with $m_i , m_i^2$ i can write as an integral over some integration variable with subscript $i$, and then I'm left with the " cross terms" . I used that
$$m_{ij}m_{ik} = \frac{1}{2}m_{ij}^2 + \frac{1}{2}m_{ik}^2 + \frac{1}{2}(m_{ij}+m_{ik})^2 $$
and this transformation :
$$e^{-\frac{1}{2}Ks^2} = \left(\frac{K}{2\pi} \right)^{1/2}\int e^{-\frac{1}{2}Kx^2 - iKsx} dx $$
which leads to a final partition function that looks like (if I've done everything correctly...) that is enormous and I have no idea what to do with it
$$
Z = \exp{\left[- \frac{N^2K}{4}-\frac{1}{2}\sum_i \sum_j h_{ij}^0 \right]}\sum_{G}\left(\prod_i \left(\frac{K}{4\pi M}\right)^{1/2} \int \exp{\left[-\frac{K}{4M}x_i^2 + \frac{K}{2M} m_i x_i + \frac{K(M-1)}{2M}m_i - \frac{\theta_i}{2}m_i\right]}dx_i \right)
\\
\prod_{i}\prod_{k<j}\left( \left(\frac{K}{4\pi M} \right)^{1/2}\int \exp{\left[-\frac{K}{4M}x_{ij}^2 + \frac{K}{2M}m_{ij}x_{ij} \right]} dx_{ij} \left(\frac{K}{4\pi M} \right)^{1/2} \int \exp{\left[ -\frac{K}{4M}x_{ik}^2 + \frac{K}{2M}m_{ik}x_{ik} \right]} dx_{ik} \right)
\\
\left( \left(\frac{1}{2\pi} \right)^{1/2} \int \exp{\left[-\frac{1}{2}x_{ijk}^2 - i \sqrt{\frac{K}{M}}(m_{ij}+m_{ik})x_{ijk} \right]}dx_{ijk}\right) $$This last expression is really long and in the preview it looked separated over several lines (I used \\) but it doesn't seem to be separated now. How could I do that?
Does anyone have any ideas on how to proceed, or how I could try this differently? I'd greatly appreciate it! :)
I will leave out the details of the system (it's not exactly physics), but I'll gladly provide more info if you want any, but it looks like the Ising model.
$$ Z = \sum_{\vec{G}}\exp{\left[\sum_{i} \left(\frac{J(M-1)}{2}\sum_{j}\sum_{\alpha}\sigma_{ij}^\alpha - \sum_{\alpha}\sum_{j}\sigma_{ij}^\alpha \frac{\theta_i + \theta_j}{4} + \frac{J}{4} \sum_{j}\left(\sum_{\alpha}\sigma_{ij}^\alpha \right)^2 - \frac{1}{2}\sum_{j}h_{ij}^0 - \frac{JMN}{4} \right) \right]}$$
where $$\sigma_{ij}^\alpha \in \{-1,1\}$$
and the summation over G means to sum over all possible configurations of $$\{\sigma_{ij}^\alpha\}$$
where $i<j$, and $j = 1,...,N$ and $\alpha = 1,...,M$
$$ \sum_{\vec{G}} = \sum_{\{\sigma_{ij}^\alpha \} = \pm 1} = \sum_{\sigma_{11}^1 = \pm 1 }\sum_{\sigma_{11}^2 = \pm 1 } ... \sum_{\sigma_{12}^1 = \pm 1 }\sum_{\sigma_{12}^2 = \pm 1 }... $$
$M$ are the amount of different values of $\alpha$ (so alpha goes from 1 to $M$), $N$ are the amount of different values for$i$ and $j, J$ and $\theta_i$ are (unknown) constants, and $h_{ij}^0$ is a function of $J, \theta_i, \theta_j$.Normally what is done with partition functions such as these is that the term inside the exponent is linearized by using a "Hubbard stratonovich" transformation:
$$ e^{-\frac{1}{2}Ks^2} = \left(\frac{K}{2\pi} \right)^{1/2}\int e^{-\frac{1}{2}Kx^2 - iKsx} dx $$
$$e^{\frac{1}{2}Ks^2} = \left(\frac{K}{2\pi} \right)^{1/2}\int e^{-\frac{1}{2}Kx^2 + Ksx} dx $$
or
$$ e^{\frac{1}{2}\sum_{ij}K_{ij}s_is_j} =\left(\frac{\det{K}}{(2\pi)^N} \right)^{1/2} \int^{\infty}_{-\infty}...\int^{\infty}_{\infty}\prod_{k=1}^Nd\phi_k \exp{\left[-\frac{1}{2}\sum_{ij}\phi_i K_{ij}\phi_j + \sum_{ij} s_iK_{ij}\phi_j\right]} $$
I have tried to use this here, first I tried to write the partition function as
$$ Z = \sum_{G}\exp{\left[\sum_i \left(\frac{J(M-1)}{2}m_i - \frac{\theta_i}{2} m_i + \frac{J}{4}m_i^2 - \frac{J}{2}\sum_{j<k}m_{ij}m_{ik} - \frac{1}{2}\sum_j h_{ij}^0 - \frac{JMN}{4} \right) \right]} $$
where
$$m_{ij} = \sum_{\alpha} \sigma_{ij}^\alpha , \qquad m_{i} = \sum_{j}m_{ij} $$
The parts of the exponent with $m_i , m_i^2$ i can write as an integral over some integration variable with subscript $i$, and then I'm left with the " cross terms" . I used that
$$m_{ij}m_{ik} = \frac{1}{2}m_{ij}^2 + \frac{1}{2}m_{ik}^2 + \frac{1}{2}(m_{ij}+m_{ik})^2 $$
and this transformation :
$$e^{-\frac{1}{2}Ks^2} = \left(\frac{K}{2\pi} \right)^{1/2}\int e^{-\frac{1}{2}Kx^2 - iKsx} dx $$
which leads to a final partition function that looks like (if I've done everything correctly...) that is enormous and I have no idea what to do with it
$$
Z = \exp{\left[- \frac{N^2K}{4}-\frac{1}{2}\sum_i \sum_j h_{ij}^0 \right]}\sum_{G}\left(\prod_i \left(\frac{K}{4\pi M}\right)^{1/2} \int \exp{\left[-\frac{K}{4M}x_i^2 + \frac{K}{2M} m_i x_i + \frac{K(M-1)}{2M}m_i - \frac{\theta_i}{2}m_i\right]}dx_i \right)
\\
\prod_{i}\prod_{k<j}\left( \left(\frac{K}{4\pi M} \right)^{1/2}\int \exp{\left[-\frac{K}{4M}x_{ij}^2 + \frac{K}{2M}m_{ij}x_{ij} \right]} dx_{ij} \left(\frac{K}{4\pi M} \right)^{1/2} \int \exp{\left[ -\frac{K}{4M}x_{ik}^2 + \frac{K}{2M}m_{ik}x_{ik} \right]} dx_{ik} \right)
\\
\left( \left(\frac{1}{2\pi} \right)^{1/2} \int \exp{\left[-\frac{1}{2}x_{ijk}^2 - i \sqrt{\frac{K}{M}}(m_{ij}+m_{ik})x_{ijk} \right]}dx_{ijk}\right) $$This last expression is really long and in the preview it looked separated over several lines (I used \\) but it doesn't seem to be separated now. How could I do that?
Does anyone have any ideas on how to proceed, or how I could try this differently? I'd greatly appreciate it! :)
