# Spin Chains - H eigenvalues

1. Oct 28, 2009

### PenKnight

1. The problem statement, all variables and given/known data

Same problem as this old post

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

$$\lim_{n->\inf}\frac{E_{0}}{N} = \ln{2}+ \frac{1}{4}$$

2. Relevant equations

The hamitonian of the spin chain

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

where

$$H_{z}(k)=S^{z}(k)S^{z}(k+1)$$

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

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.

$$H|...\uparrow\uparrow... \rangle = S^{z}(k)S^{z}(k+1)|...\uparrow\uparrow... \rangle$$

$$= S^{z}(k)\frac{1}{2}|...\uparrow\uparrow ...\rangle =\frac{1}{4}|...\uparrow\uparrow ...\rangle$$

And for the other possible combinations

$$H|...\uparrow\downarrow... \rangle = -\frac{1}{4}|...\uparrow\downarrow ...\rangle + \frac{1}{2}|...\downarrow\uparrow... \rangle$$

$$H|...\downarrow\uparrow ...\rangle = -\frac{1}{4}|...\downarrow\uparrow... \rangle + \frac{1}{2}|...\uparrow\downarrow... \rangle$$

$$H|...\downarrow\downarrow...\rangle = \frac{1}{4}|...\downarrow\downarrow...\rangle$$

But then finding the energy values ( taking the $$<\phi | H|\phi >$$ )
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. Oct 28, 2009

### gabbagabbahey

Surely you mean:

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

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 $H$ that way...You can however, say $H(k)|...\uparrow\uparrow... \rangle = \frac{1}{4}|...\uparrow\uparrow... \rangle$ and so on, for $H(k)=H_z(k)+H_f(k)$

Last edited: Oct 28, 2009
3. Oct 28, 2009

### PenKnight

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.

$$H|\uparrow\downarrow\uparrow\rangle = H_z|\uparrow\downarrow\uparrow\rangle + H_f|\uparrow\downarrow\uparrow\rangle$$

$$= (-\frac{1}{4} -\frac{1}{4}+ \frac{1}{4} )|\uparrow\downarrow\uparrow\rangle + \frac{1}{2} ( |\downarrow\uparrow\uparrow\rangle +|\uparrow\uparrow\downarrow\rangle)$$

I'm not actually sure how to get the energy value with the $$H_f$$ 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