Determining Energy Values for a One-Dimension Spin Chain

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Homework Statement



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]

Homework 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.

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.
 
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PenKnight said:
The hamitonian of the spin chain

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

Surely you mean:

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

Right?
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]...

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]
 
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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 =<br /> H_z|\uparrow\downarrow\uparrow\rangle +<br /> H_f|\uparrow\downarrow\uparrow\rangle[/tex]

[tex] =<br /> (-\frac{1}{4} -\frac{1}{4}+ \frac{1}{4} )|\uparrow\downarrow\uparrow\rangle +<br /> \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.
 
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