# Eigenvalues of Hamiltonian

1. Sep 16, 2008

### Krischi

1. The problem statement, all variables and given/known data Consider two Ising spins coupled together
−βH = h(σ1 + σ2) + Kσ1σ2,
where σ1 and σ2 commute and each independently takes on the values ±1.
What are the eigenvalues of this Hamiltonian? What are the degeneracies of the states?

3. The attempt at a solution Four possible combinations for (σ1,σ2): (1,1), (1,-1), (-1,1) and (-1,-1).
Therefore H=(-h/β)*(σ1 + σ2) + K/β*σ1σ2 can be written in a 2×2 matrix. And the eigenvalues λ are obtained by det(H-Eλ)=0.

it follows: [(-2h/β)-(K/β)-λ)][(-2h/β)-(K/β)-λ)]-(2K/β)=0

and so: λ1,2=-((2h-K)/β)±sqrt[(2h-K)22)-((2h-K)22-(2K/β)]

and: λ1,2=-((2h-K)/β)±sqrt[2k/β]

Are these really the eigenvalues of the hamiltonian? I dont gain any physical insight by this solution and therefore I doubt my calculation. I dont know how to go on and clculate the degeneracies of the states.

Krischi

2. Sep 18, 2008

### genneth

You should use a 4x4 matrix for the Hamiltonian --- the system has 4 basis states (which you listed). Find the eigenvalues of that matrix.

3. Sep 18, 2008

### Krischi

Really, a 4$$\times$$4 matrix? If there are 4 base states, why can't I use a 2$$\times$$2 matrix? 4 states fit into a 2$$\times$$2 matrix, right? I tried this and calculated 2 eigenvalues, $$\lambda$$, but I am not sure, if the result is correct, since it "looks" to complicated (see my 1st reply)

4. Sep 18, 2008

### yaychemistry

genneth is right, for 4 base states you need a 4x4 matrix.
Consider for a moment a Hamiltonian for a single spin that can be +1 or -1. We need to know how the Hamiltonian operator acts on the particle if its spin is +1 and also how the Hamiltonian acts on the particle if its spin is -1. Thus, we need a basis state for each state of the particle.

In your case you have two particles. So you have two particles with two states each 2*2 = 4 basis states. You need to know how the Hamiltonian acts on each individual configuration of spins and there are 4 possible configurations.

Hope that helps