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Spin Chains - H eigenvalues

  1. Oct 28, 2009 #1
    1. The problem statement, all variables and given/known data

    Same problem as this old post
    https://www.physicsforums.com/showthread.php?t=188714


    What I'm having problems with is determining the [tex]H_{ij}[/tex] components of the Hamiltonian of a one dimension N site spin chain. And then getting out somehow energy value to prove

    [tex]
    \lim_{n->\inf}\frac{E_{0}}{N} = \ln{2}+ \frac{1}{4}
    [/tex]

    2. Relevant equations

    The hamitonian of the spin chain

    [tex]
    \sum^{k=0}_{N-1}[H_{z}(k)+H_{f}(k)]
    [/tex]

    where

    [tex]
    H_{z}(k)=S^{z}(k)S^{z}(k+1)
    [/tex]

    [tex]
    H_{f}(k)=\frac{1}{2}[S^{+}(k)S^{-}(k+1)+S^{-}(k)S^{+}(k+1)]
    [/tex]

    The above can be gain from determing Sx and Sy from the rasing and lowering operators.

    3. The attempt at a solution

    I can see that at any site location within a state ,( a state is some configuration of site which hold either +- 1/2), the hamilotinan will pull out these eigenvalues.

    [tex]
    H|...\uparrow\uparrow... \rangle = S^{z}(k)S^{z}(k+1)|...\uparrow\uparrow... \rangle
    [/tex]

    [tex]
    = S^{z}(k)\frac{1}{2}|...\uparrow\uparrow ...\rangle =\frac{1}{4}|...\uparrow\uparrow ...\rangle
    [/tex]

    And for the other possible combinations

    [tex]
    H|...\uparrow\downarrow... \rangle = -\frac{1}{4}|...\uparrow\downarrow ...\rangle + \frac{1}{2}|...\downarrow\uparrow... \rangle
    [/tex]

    [tex]
    H|...\downarrow\uparrow ...\rangle = -\frac{1}{4}|...\downarrow\uparrow... \rangle + \frac{1}{2}|...\uparrow\downarrow... \rangle
    [/tex]

    [tex]
    H|...\downarrow\downarrow...\rangle = \frac{1}{4}|...\downarrow\downarrow...\rangle
    [/tex]

    But then finding the energy values ( taking the [tex]<\phi | H|\phi >[/tex] )
    will lead all states to having the same energy which is not correct. So either I've missed somthing or computing the H matrix incorrectly.
     
  2. jcsd
  3. Oct 28, 2009 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Surely you mean:

    [tex]H= \sum_{k=1}^{N}[H_{z}(k)+H_{f}(k)][/tex]

    Right?


    This makes no sense to me...All N particles can each be in the up or down state, and the Hamiltonian acts on all N particles... you can't just pick out two particles in a chain and calculate the effect of [itex]H[/itex] that way...You can however, say [itex]H(k)|...\uparrow\uparrow... \rangle = \frac{1}{4}|...\uparrow\uparrow... \rangle[/itex] and so on, for [itex]H(k)=H_z(k)+H_f(k)[/itex]
     
    Last edited: Oct 28, 2009
  4. Oct 28, 2009 #3
    Yep your right.

    I was trying to show what the spin operators pulls out for a particular pairing but mucked up in notation. I think going through a couple of examples may get my head around this.

    Here the first one with N = 3 and considering periodic boundary condition.

    [tex]
    H|\uparrow\downarrow\uparrow\rangle =
    H_z|\uparrow\downarrow\uparrow\rangle +
    H_f|\uparrow\downarrow\uparrow\rangle
    [/tex]

    [tex]
    =
    (-\frac{1}{4} -\frac{1}{4}+ \frac{1}{4} )|\uparrow\downarrow\uparrow\rangle +
    \frac{1}{2} ( |\downarrow\uparrow\uparrow\rangle +|\uparrow\uparrow\downarrow\rangle)
    [/tex]

    I'm not actually sure how to get the energy value with the [tex]H_f [/tex] terms eigenstates. Do i just add up all the co-efficient?

    Edit
    I think I've got it now. I'll be back if I can't get the eigenvalues from the Hamiltonian.
     
    Last edited: Oct 29, 2009
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