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Correlation function, 1d polymer

  1. Apr 25, 2007 #1
    1. The problem statement, all variables and given/known data
    1D polymer, fixed segment length a
    If the angle between segment j and j+1 is 0, the energy is 0
    If the angle is pi the energy is +2J.

    Compute the correlation function [tex]<s_i s_{i+n}>[/tex], where [tex]s_j = \pm 1[/tex] denotes the direction of segment j

    Find the persistence length Lp, defined through
    [tex]<s_i s_{i+n}> = e^{-|n|a/Lp} [/tex]

    Find an expression for the end-to-end distance [tex]S(N) = <(x_N - x_0)^2>^{1/2}[/tex] as a function of temperature and the number of links N

    2. Relevant equations
    ?


    3. The attempt at a solution
    [tex]<s_i s_{i+n}> = \frac{ Tr s_i s_{i+n} e^{-\beta H} }{ Tr e^{-\beta H} }[/tex]

    But I don't know any hamiltonian? Or even what sort of trace to do.
    The problem sort of reminds me of the 'XY'-modell for spins on a 1d lattice, but I don't really understand how to make any use of that.
     
    Last edited: Apr 25, 2007
  2. jcsd
  3. Apr 26, 2007 #2

    nrqed

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    You can write down a Hamiltonian.
    It will be the sum of the energy of all the adjacent pairs, taking into account the rule they give. You want an expression that gives zero when [itex]s_j = s_{j+1}[/itex] and which gives 2J when [itex]s_j = -s_{j+1}[/itex]. This is simply [itex]J(1-s_j s_{j+1})[/itex].
     
  4. Apr 29, 2007 #3
    So, starting with a Hamiltonian like
    [tex]H = J \sum_{j=1}^{N-1} (1-s_j s_{j+1}) [/tex]
    and proceeding to calculate the partition function much the same way as for the 1d ising chain,
    [tex]Z = Tr e^{-\beta H} = \dots = 2 e^{-\beta J (N-1)} [2 cosh(\beta J)]^{N-1}[/tex]
    then noteing I can write
    [tex] <s_j s_{j+n}> = \frac{1}{Z \beta^n} \frac{d^n Z}{d J^n} [/tex]
    I end up with
    [tex] <s_j s_{j+n}> = tanh(\beta J)^n [/tex]
    is this really right? feels like I missed something

    thanks for the help!
     
    Last edited: Apr 29, 2007
  5. May 11, 2008 #4
    the last equation should be (tanh(\beta J))^n
     
  6. Oct 30, 2011 #5
    how can we calculate $$S^2(N) = <(x_N-x_0)^2>$$ in above case?
     
  7. Nov 1, 2011 #6
    How one can calculate

    S(N)2 = <(xN-x0)2>

    thanx
     
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