Spin-1 particle in uniaxial crystal field

[cscott]

Homework Statement

$$H = -D(S^z)^2$$ for cases D>0, D<0, where D<0 should remove the degeneracy in the ground state.

Homework Equations

$$H = -D(S^z)^2 = -D\hbar^2 (1 0 0; 0 0 0; 0 0 1)$$
(`;' separates rows)

$$det(H-1E)= 0$$, or by inspection...

The Attempt at a Solution

I get, for D>0,

$E_1 = -D \hbar^2$, $\psi_1=(1,0,0)$
$E_2 = -D \hbar^2$, $\psi_2=(0,0,1)$
$E_3 = 0$, $\psi_3=(0,1,0)$

So I get a degeneracy in the ground state, However I don't see how D<0 removes it.

 

[read]

For D<0 the ground state will be with m=0, i.e. the spin is perpendicular to the z-axis, whereas both m=+1 and -1 will have the same but larger energy.

 

[cscott]

Thanks! I definitely thought about that too hard haha