Two Particles' Spin Hamiltonian Analysis?

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cacofolius
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Homework Statement


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}##

Homework 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) ?
 
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cacofolius said:
Is it right to consider |−+⟩.|+−⟩=0\vert-+\rangle . \vert+-\rangle = 0,
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.