Spherical Symmetric Harmonic Oscillator

[gasar8]

Homework Statement

An electron (S=1/2) is free in a spherical symmetric harmonic potential:
V(r)=\frac{1}{2}kr^2

a) Find energies and degeneracy of ground state and first excited state.
b) For these states find the l^2 and l_z basis.
c) How does these states split in a \vec{L} \cdot \vec{S} coupling?

3. The Attempt at a Solution
a) 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})
Ground state (l=0): E=\frac{3}{2}\hbar \omega, only one state - degeneracy 1
First excited state (l=1): E=\frac{5}{2} \hbar \omega, 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 | l, l_z \rangle?
Ground state (l=0, l_z=0): |00\rangle
First excited state: (l=1,l_z=\pm1,0): |1-1\rangle,|10\rangle,|11\rangle

c) \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
I get 0 for the ground state and \hbar^2 \beta for first excited state. Does that mean that ground state doesn't split and first excited in 0, \pm \hbar^2 \beta?

 

[blue_leaf77]

Is everything all right here?

Yes you are doing it right.

Do I only need to write |l,lz⟩|l,lz⟩| l, l_z \rangle?

You also need to specify the quantum number $n_r$.

Does that mean that ground state doesn't split and first excited in 0,±ℏ2β0,±ℏ2β0, \pm \hbar^2 \beta?

How did you get those splitting?
In part (c), a new degree of freedom which is spin is added into the system. Consequently, it's important to realize the degeneracy of each state in this system after taking spin into account but before considering the spin-orbit perturbation. For the first excited state, what is the degeneracy? Then upon taking spin-orbit into account, what are the possible value of $j$'s for the first excited state?

 

[gasar8]

You also need to specify the quantum number nrnrn_r.

In both states it is 0?

Ok, I calculated something wrong. So, from \frac{\beta \hbar^2}{2} [j(j+1)-l(l+1)-s(s+1)], I get:
Ground state: -\hbar^2 \beta, the ground state only lowers for this factor?
1. excited state: \frac{\hbar^2 \beta}{2} \ \textrm{for} \ j=3/2\\ -\hbar^2 \beta \ \textrm{for} \ j=1/2

For the first excited state, what is the degeneracy?

3, as I have calculated in a) part, but why do I need this now? nr and l are in 1. excited state always 0 and 1, respectively?

 

[blue_leaf77]

Ground state: −ℏ2β−ℏ2β-\hbar^2 \beta, the ground state only lowers for this factor?

Check again your calculation. For ground state, $l=0$ and $s=1/2$, what is $j$?

In both states it is 0?

1. excited state: \frac{\hbar^2 \beta}{2} \ \textrm{for} \ j=3/2\\ -\hbar^2 \beta \ \textrm{for} \ j=1/2

Yes. However, it's important to know that $\beta$ is often times a function of $r$. Therefore, instead of just $\beta$ I think it's more appropriate to write $\langle R_{01} |\beta(r)|R_{01}\rangle$.

3, as I have calculated in a) part, but why do I need this now?

After taking spin degree of freedom, the degeneracy should surely increase. Remember for spin 1/2 particle, each state will multiply two folds. Thus if initially a level has three degenerate states, what will it become after adding the spin?

 

[gasar8]

For ground state, l=0 and s=1/2, what is j?

Oh, I don't even know what I was calculating. j=l+s=0+{1 \over 2}= {1 \over 2}, so there is no change in final energy.

Ok, thank you. :)

Yes, so I must multiply every degeneracy by 2?
Thank you very much!

 

[blue_leaf77]

there is no change in final energy.

Yes.

must multiply every degeneracy by 2?

Yes. Before adding spin orbit there are 3x2=6 states in the first excited level, after being perturbed by spin-orbit these 6 states split into two groups: 4 of them belong to $j=3/2$ state and the rest 2 to $j=1/2$.

 

[gasar8]

Ok, thank you, I remember this from one of our other lectures. This is 2j+1 states for every degenerate state.