Degeneracy of the 3d harmonic oscillator

[mavyn]

Hi!

I'm trying to calculate the degeneracy of each state for 3D harmonic oscillator.
The eigenvalues are

En = (N + 3/2) hw

Unfortunately I didn't find this topic in my textbook.
Can somebody help me?

 

[StatMechGuy]

What does the "N" operator look like? Is it just a^\dagger a or something else?

 

[mavyn]

What does the "N" operator look like? Is it just a^\dagger a or something else?

Hi,
sorry I made a mistake in the formula. It should be

En = (n + 3/2) hw

 

[gulsen]

n is the sum of 3 quantum numbers, n_x, n_y, n_z; for the lowest rung, which is n=0, there's no degeneracy. For n=1, there's 3 fold degeneracy (one of the quantum numbers is 1, and rest is 0). For n=2, there's 6 fold degenertacy (one of them may be 2 and rest 0, or two of them may be 1) and so on. The result is, for nth energy level, there's 3n degeneracy.

 

[mavyn]

n is the sum of 3 quantum numbers, n_x, n_y, n_z; for the lowest rung, which is n=0, there's no degeneracy. For n=1, there's 3 fold degeneracy (one of the quantum numbers is 1, and rest is 0). For n=2, there's 6 fold degenertacy (one of them may be 2 and rest 0, or two of them may be 1) and so on. The result is, for nth energy level, there's 3n degeneracy.

Thanks for the answer!

 

[014137]

Watch out!

No that's not right.
You can't work it out for the first couple of cases and then presume the trend continues like that.
In fact, the degeneracy g(n) is:
g(n) = (1/2)(n+1)(n+2)

 

[nrqed]

No that's not right.
You can't work it out for the first couple of cases and then presume the trend continues like that.
In fact, the degeneracy g(n) is:
g(n) = (1/2)(n+1)(n+2)

And even this, I think, is nt completely right. There are some levels with additional accidental degeneracy which don't fit the pattern.

 

[014137]

Really? Which ones?
The formula can be derived like this:
n = n1 + n2 + n3
where 1,2,3 are three orthogonal directions.
Choose n1 then
n2 + n3 = n - n1
Can always pick n - n1 + 1 different pairs of n2, n3.
Sum over n1 from 0 to n:

Sum(n - n1 +1) = Sum(n - 1) - Sum(n1)
=(n - 1)*(n - 1) - (1/2)n(n + 1)
=(1/2)(n+1)(n+2)

Maybe there's a mistake somewhere?

 

[nrqed]

Really? Which ones?
The formula can be derived like this:
n = n1 + n2 + n3
where 1,2,3 are three orthogonal directions.
Choose n1 then
n2 + n3 = n - n1
Can always pick n - n1 + 1 different pairs of n2, n3.
Sum over n1 from 0 to n:

Sum(n - n1 +1) = Sum(n - 1) - Sum(n1)
=(n - 1)*(n - 1) - (1/2)n(n + 1)
=(1/2)(n+1)(n+2)

Maybe there's a mistake somewhere?

I was thinking about the cubic infinite square well which is more tricky since it involves the sum of squares, so never mind my comment about accidental degeneracy.

Your formula may be right but I don't understand how you got from (n - 1)*(n - 1) - (1/2)n(n + 1) to (1/2)(n+1)(n+2). Those two expressions are not equal. Moreover, the sum \sum_{n_1 = 0}^n gives (n+1), not (n-1), right?

EDIT: Actually, the formula does not seem to work. I may have missed some states, but I can't get it to work for n=10, say.

 

[014137]

sorry - i was being silly
Sum(n - n1 +1) = Sum(n + 1) - Sum(n1)
=(n + 1)*(n + 1) - (1/2)n(n + 1)
=(1/2)(n+1)(n+2)

 

[014137]

Does work for n = 10, look:
State
3 x 10, 0, 0
3 x 8, 1, 1
3 x 6, 2, 2
3 x 4, 3, 3
3 x 2, 4, 4
3 x 0, 5, 5
6 x 9, 1, 0
6 x 8, 2, 0
6 x 7, 3, 0
6 x 7, 2, 1
6 x 6, 4, 0
6 x 6, 3, 1
6 x 5, 4, 1
6 x 5, 3, 2
Degeneracy = 6 x 8 + 3 x 6 = 6 x 11 = 66

The formula for g(n) gives (1/2)*(11)*(12) = 66

So it works.

 

[nrqed]

Does work for n = 10, look:
State
3 x 10, 0, 0
3 x 8, 1, 1
3 x 6, 2, 2
3 x 4, 3, 3
3 x 2, 4, 4
3 x 0, 5, 5
6 x 9, 1, 0
6 x 8, 2, 0
6 x 7, 3, 0
6 x 7, 2, 1
6 x 6, 4, 0
6 x 6, 3, 1
6 x 5, 4, 1
6 x 5, 3, 2
Degeneracy = 6 x 8 + 3 x 6 = 6 x 11 = 66

The formula for g(n) gives (1/2)*(11)*(12) = 66

So it works.

Yes, you are completely right. I had missed one entry.
Good job. Thanks for the correction.

If you know a formula for the infinite cubic well, let me know...It's in that case that accidental degeneracy are a killer.

Regards

Patrick

 

[Xan*Sa*]

Thank you everyone.. i too was stuck in it and it helped me a lot