- #1
gasar8
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Homework Statement
An electron (S=1/2) is free in a spherical symmetric harmonic potential:
[tex]V(r)=\frac{1}{2}kr^2[/tex]
a) Find energies and degeneracy of ground state and first excited state.
b) For these states find the [itex]l^2[/itex] and [itex]l_z[/itex] basis.
c) How does these states split in a [itex]\vec{L} \cdot \vec{S}[/itex] coupling?
3. The Attempt at a Solution
a) [itex]E = E_x+E_y+E_z=\hbar \omega (n_x+n_y+n_z+\frac{3}{2})=\hbar \omega (2 n_r + l+\frac{3}{2})[/itex]
Ground state (l=0): [itex]E=\frac{3}{2}\hbar \omega[/itex], only one state - degeneracy 1
First excited state (l=1): [itex]E=\frac{5}{2} \hbar \omega[/itex], three possible chances - degeneracy 3
Is everything all right here?
b) I am not sure what do I have to do here. Do I only need to write [itex]| l, l_z \rangle[/itex]?
Ground state ([itex]l=0, l_z=0[/itex]): [itex]|00\rangle[/itex]
First excited state: ([itex]l=1,l_z=\pm1,0[/itex]): [itex]|1-1\rangle,|10\rangle,|11\rangle[/itex]
c) [tex]\langle L S J M_J| \beta \vec{L} \cdot \vec{S} |L S J M_J\rangle = \frac{\beta \hbar^2}{2} \langle L S J M_J| j(j+1)-l(l+1)-s(s+1) |L S J M_J\rangle[/tex]
I get 0 for the ground state and [itex]\hbar^2 \beta[/itex] for first excited state. Does that mean that ground state doesn't split and first excited in [itex]0, \pm \hbar^2 \beta[/itex]?