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1D Ising Correlation

  1. Apr 6, 2015 #1
    1. The problem statement, all variables and given/known data
    Compute correlation functions ##<\sigma_r \sigma_{r+l}>## for the 1D Ising model of length L with the follow BD conditions

    (i) Periodic
    (ii) Anti-Periodic
    (iii) ##\sigma_1 = \sigma_{L+1}=1##
    (iv) ##\sigma_1= -\sigma_{L+1}=1##

    2. Relevant equations

    ##<\sigma_r \sigma_{r+l}> = \displaystyle\frac{1}{Z}\sum_{\sigma_l=\pm 1}^{L-1}e^{K(\sum_{k=1}^{L-1}\sigma_k \sigma_{k+1} +\sigma_1 \sigma_{L+1})} \sigma_r \sigma_{r+l}##

    3. The attempt at a solution

    I know how to compute the partition function for the periodic case as its fairly common and I have a solution to computing the correlation function to here although I don't understand how he goes from 1.11 to 1.12.But these solutions use the Trace of the transfer matrix which I am pretty sure is unique to the periodic BD conditions. Any help on how to compute these in general would be appreciated as I missed the lecture on it.
     
  2. jcsd
  3. Apr 8, 2015 #2
    Did you understood the transfer matrix technique? the first n-1 spin sum just becomes the normal trace of the transfer matrix. But then you have [itex]\sigma_m[/itex], but the possible value of [itex]\sigma_m[/itex] are 1 and -1, with introduction of pauli matrix(+1 and -1 remember?), it just becomes a normal trace once again.
     
  4. Apr 10, 2015 #3
    Can you explain why this explicit calculation of the 1D model with 3 sites and periodic BDs is incorrect

    ##Z=\displaystyle\sum_{\sigma_1=\pm1}\sum_{\sigma_2=\pm1}\sum_{\sigma_3=\pm1}e^{K(\sum_{k=1}^2\sigma_k \sigma_{k+1} + \sigma_3 \sigma_1)}##

    ##=\displaystyle\sum_{\sigma_1=\pm1}\sum_{\sigma_2=\pm1}\sum_{\sigma_3=\pm1}e^{K(\sigma_1 \sigma_2 + \sigma_2 \sigma_3 + \sigma_3 \sigma_1)}##

    ##=\displaystyle\sum_{\sigma_1=\pm1}\sum_{\sigma_2=\pm1}e^{K(\sigma_1 \sigma_2 + \sigma_2 + \sigma_1)}+e^{K(\sigma_1 \sigma_2 + -\sigma_2 + -\sigma_1)}##

    ##=\displaystyle\sum_{\sigma_1=\pm1}e^{K(\sigma_1 + 1 + \sigma_1)}+e^{K(\sigma_1 -1 + -\sigma_1)}+e^{K(-\sigma_1 -1 + \sigma_1)}+e^{K(-\sigma_1+ 1 + -\sigma_1)}##

    ##=e^{K(1 + 1 + 1)}+e^{K(1 -1 + -1)}+e^{K(-1 -1 + 1)}+e^{K(-1+ 1 + -1)}+e^{K(-1 + 1 + -1)}+e^{K(-1 -1 + 1)}+e^{K(1 -1 -1)}+e^{K(1+ 1 + 1)}##

    ##=e^{K(3)}+e^{K(-1)}+e^{K(-1)}+e^{K(-1)}+e^{K(-1)}+e^{K(-1)}+e^{K(-1)}+e^{K(3)}##

    ##=2e^{3K}+6e^{-K}##

    nvm It's correct, I guess I don't understand how the transfer matrix works, I thought it was dependant on periodic BDs to get the trace
     
    Last edited: Apr 10, 2015
  5. Apr 10, 2015 #4
    Which boundary condition are you referring to?? If it's periodic then your calculation is right but if it's anti-periodic then your calculation is wrong.
     
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