Determining Energy Values for a One-Dimension Spin Chain

In summary, the conversation discusses determining the H_{ij} components of the Hamiltonian of a one dimension N site spin chain and finding the energy values. The Hamitonian is given by H=\sum_{k=1}^{N}[H_{z}(k)+H_{f}(k)] where H_{z}(k)=S^{z}(k)S^{z}(k+1) and H_{f}(k)=\frac{1}{2}[S^{+}(k)S^{-}(k+1)+S^{-}(k)S^{+}(k+1)]. The conversation also addresses the issue of finding the energy values using the eigenstates of H_f.
  • #1
PenKnight
11
0

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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  • #2
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]
 
Last edited:
  • #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:

FAQ: Determining Energy Values for a One-Dimension Spin Chain

What is a spin chain?

A spin chain is a model used in physics to study interacting systems of particles with spin. It consists of a one-dimensional chain of particles, where each particle has a spin value that interacts with its neighboring particles.

What are H eigenvalues in spin chains?

H eigenvalues are the eigenvalues of the Hamiltonian operator, which is used to describe the total energy of the spin chain system. They represent the possible energy states of the system and can be used to calculate the dynamics of the system.

How are spin chains used in research?

Spin chains are used in various fields of research such as condensed matter physics, quantum information theory, and statistical mechanics. They can be used to study the behavior of quantum systems, as well as for developing new technologies such as quantum computing.

What is the significance of H eigenvalues in spin chains?

H eigenvalues play a crucial role in understanding the behavior and properties of spin chain systems. They can reveal important information about the quantum states of the system and can be used to predict the behavior of the system under different conditions.

How do H eigenvalues change in a spin chain system?

The H eigenvalues can change in a spin chain system due to various factors such as external fields, temperature, and interactions between particles. Changes in the H eigenvalues can lead to different energy states of the system, resulting in changes in its dynamics and properties.

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