# Two particles' spin Hamiltonian

1. Jul 20, 2017

### cacofolius

1. The problem statement, all variables and given/known data
Hi, I'm trying to familiarize with the bra-ket notation and quantum mechanics. I have to find the hamiltonian's eigenvalues and eigenstates.

$H=(S_{1z}+S_{2z})+S_{1x}S_{2x}$

2. Relevant equations
$S_{z} \vert+\rangle =\hbar/2\vert+\rangle$

$S_{z}\vert-\rangle =-\hbar/2\vert-\rangle$

$S_{x} \vert+\rangle =\hbar/2\vert-\rangle$

$S_{x} \vert-\rangle =\hbar/2\vert+\rangle,$

The states basis is $\vert++\rangle,\vert+-\rangle, \vert-+\rangle, \vert--\rangle$

3. The attempt at a solution

What I did was apply the hamiltonian to each basis ket

$H\vert++\rangle =(S_{1z}+S_{2z})\vert++\rangle + S_{1x}S_{2x}\vert++\rangle = \hbar/2\vert++\rangle + \hbar/2\vert++\rangle + \hbar/2\vert-+\rangle . \hbar/2\vert+-\rangle = \hbar/2\vert++\rangle$

$H\vert+-\rangle = 0$

$H\vert-+\rangle = 0$

$H\vert--\rangle = -\hbar/2\vert--\rangle$

My questions:
1) Is it right to consider $\vert-+\rangle . \vert+-\rangle = 0$, (since they're orthogonal states)? Because they're both ket vectors (unlike the more familiar $<a|b>$).

2) In that case, is the basis also the hamiltonian's, with eigenvalues $\hbar/2, -\hbar/2, 0$ (degenerate) ?

2. Jul 20, 2017

### gimak

Post this in the advanced physics homework section

3. Jul 24, 2017

### blue_leaf77

No, that's not right. Moreover, $S_{1x}S_{2x}|++\rangle \neq \hbar/2\vert-+\rangle . \hbar/2\vert+-\rangle$. It's like you are producing four electrons out of two electrons. The operator of the first particle only acts on the first entry of the ket and that of the second particle acts on the second entry.