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Ising Model Spin configurations

  1. Oct 11, 2015 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    Consider the below network of spins, the spins numbered on the nodes of the diagram. The spins interact via the following Hamiltonian ##\mathcal H = \sum_{\langle i j \rangle} \sigma_i \sigma_j##, where the sum ##\langle i j \rangle## is over nearest neighbours and ##\sigma_i = \pm 1##. Call ##K = \beta k##.

    Compute the partition function for the network of spins. Use the fact that ##e^{K\sigma \sigma'} = \cosh(K)[1+\sigma \sigma' \tanh(K)]## which holds for Ising spins.

    2. Relevant equations
    In section 1

    3. The attempt at a solution
    $$Z = \sum_{\sigma} e^{-\beta \mathcal H} = \sum_{\sigma} e^{k\beta \sum_{\langle i j \rangle} \sigma_i \sigma_j} = \sum_{\sigma} \prod_{\langle i j \rangle} e^{K \sigma_i \sigma_j} = \sum_{\sigma} \prod_{\langle i j \rangle}(\cosh K(1+\sigma_i \sigma_j \tanh K))$$ My question is how to interpret the product here. I thought it meant taking each node in the network, writing out the links which are its nearest neighbours, repeating for all the other nodes and multiplying them all together. So for example, for node 1, we would get a term ##(1+\sigma_1 \sigma_5 \tanh K + \sigma_1 \sigma_2 \tanh K)## and similarly for the others. But if I do this for the rest and multiply them all together, I'll never get a ##\tanh^6 K## term present in the answer.

    Answer given is ##\cosh^6K(1+2\tanh^2 K + \tanh^6 K)##

    So how am I misinterpreting the product?

    Many thanks.
     

    Attached Files:

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  3. Oct 11, 2015 #2

    TSny

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    I believe the product over ##<ij>## is over all pairs of indices ##i## and ##j## for which ##i## and ##j## are nearest neighbor pairs. So, one possible value of ##<ij>## is ##<23>##, but the value ##<35>## is not allowed.

    In the answer, should the ##\tanh^2K## be ##\tanh^3K##? Does the answer have an overall numerical factor that you left out?

    Is there a missing numerical factor in the Hamiltonian ##\mathcal H = \sum_{\langle i j \rangle} \sigma_i \sigma_j##?
     
    Last edited: Oct 11, 2015
  4. Oct 12, 2015 #3

    CAF123

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    Hi TSny,
    I see, so is the product like $$(1+(\sigma_3 \sigma_4+\sigma_4 \sigma_3)\tanh K)(1+(\sigma_3 \sigma_2+\sigma_2 \sigma_3)\tanh K)(1+(\sigma_2 \sigma_4+\sigma_4 \sigma_2)\tanh K)(1+(\sigma_1 \sigma_2+\sigma_2 \sigma_1)\tanh K)(1+(\sigma_1 \sigma_5+\sigma_5 \sigma_1)\tanh K)(1+(\sigma_2 \sigma_5+\sigma_5 \sigma_2)\tanh K)?$$

    Using this, I get terms that look like $$1+8\tanh^3 K(\sigma_3 \sigma_4 \sigma_3 \sigma_2 \sigma_2 \sigma_4 + \sigma_2 \sigma_5 \sigma_1 \sigma_5 \sigma_1 \sigma_2) + 64 \tanh^6 K (\sigma_1 \sigma_5 \sigma_3 \sigma_4 \sigma_3 \sigma_2 \sigma_2 \sigma_4 \sigma_1 \sigma_2 \sigma_2 \sigma_5),$$ where all other terms can be reduced to the the form ##\sigma_i \sigma_j, i \neq j## or ##\sigma_i \sigma_j \sigma_k## with ##i \neq j \neq k## using the fact that ##\sigma_i^2=1##. Those terms vanish under the sum over ##\left\{\sigma\right\}##.

    I would continue by saying that the sum now brings in a factor of 2^5 (there are 2^5 possible ways that ##\sigma_1 \sigma_5 \sigma_3 \sigma_4 \sigma_3 \sigma_2 \sigma_2 \sigma_4 \sigma_1 \sigma_2 \sigma_2 \sigma_5## can be 1, but this seems to not give the correct numerical factors. Is there a mistake in what I got before I took the sum?

    Yes sorry it was a typo and there is no numerical factor in the answer given, but I suppose there is a factor 2^5 coming in when we sum over the ##\left\{\sigma\right\}## at the end?
    Yup it should have a -k outside the sum.
     
    Last edited: Oct 12, 2015
  5. Oct 12, 2015 #4

    TSny

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    I don't think that's right. The Hamiltonian is the sum of ##\sigma_i \sigma_j## for each nearest neighbor pair ##<ij>##. Each ##\sigma_i \sigma_j## yields a factor ##e^{K\sigma_i \sigma_j}##. But I'm getting an answer that's off by an overall factor of 25. So maybe I'm missing something.
     
    Last edited: Oct 12, 2015
  6. Oct 12, 2015 #5

    CAF123

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    Ah, then I guess I was double counting the links previously. Is the product then $$\cosh^6K(1+\sigma_3 \sigma_4 \tanh K)(1+\sigma_3 \sigma_2 \tanh K)(1+\sigma_2 \sigma_4 \tanh K)(1+\sigma_2 \sigma_1 \tanh K)(1+\sigma_1 \sigma_5 \tanh K)(1+\sigma_2 \sigma_5 \tanh K).$$

    The factor 2^5 might be correct (at least I can see how it possibly arises) so the answer given might just be in error up to this factor.
     
  7. Oct 12, 2015 #6

    TSny

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    I believe so.
     
  8. Oct 12, 2015 #7

    CAF123

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    Ok thanks. I have reduced it down to $$\cosh^6 K \sum_{\sigma} (1+ (\sigma_4^2 \sigma_2^2 \sigma_3^2 + \sigma_1^2 \sigma_5^2 \sigma_2^2)\tanh^3K + (\sigma_1^2 \sigma_2^2 \dots \sigma_5^2)\tanh^6K).$$

    I have kept only terms where all the sigmas are squared. The rest go to zero when I sum over all sigma. Each prefactor to the ##\tanh^3## could be 1 in ##2^3## ways yes? Similarly, in the ##\tanh^6## term, the prefactor can be 1 in ##2^5## ways. But this does not give me the same numerical factors as in the answer. Am i thinking about this in the right way?
     
  9. Oct 12, 2015 #8

    TSny

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    The prefactor of ##\tanh^3## will be (1+1) for each possible state of the system.
     
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