# Spherical Symmetric Harmonic Oscillator

1. Aug 22, 2016

### gasar8

1. The problem statement, all variables and given/known data
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 allright 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 $0, \pm \hbar^2 \beta$?

2. Aug 22, 2016

### blue_leaf77

Yes you are doing it right.
You also need to specify the quantum number $n_r$.
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?

Last edited: Aug 22, 2016
3. Aug 22, 2016

### gasar8

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

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?

4. Aug 22, 2016

### blue_leaf77

Check again your calculation. For ground state, $l=0$ and $s=1/2$, what is $j$?
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 $\langle R_{01} |\beta(r)|R_{01}\rangle$.
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?

5. Aug 22, 2016

### gasar8

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!

6. Aug 22, 2016

### blue_leaf77

Yes.
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 $j=1/2$.

7. Aug 22, 2016

### gasar8

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

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted