Hi,
Embarrassing problem time. I'm actually tutoring this course in a few days time (Wednesday morning, so any help by then would be just fine), and I have totally forgotten how to do QM.
The problem concerns a particle in a infinite 3D well, with non symmetrical walls (so [imath]0 \leq x \leq a[/imath], [imath]0 \leq y \leq a[/imath] , [imath]0 \leq z \leq 3a[/imath].
So you solve your T.I.S.E, get your k values out
[math]k_x = \frac{n\pi}{a},\;\;\;\; k_y= \frac{n\pi}{a},\;\;\;\; k_z = \frac{n\pi}{3a}[/math]
And you have an energy
[math]E = \frac{\hbar^2 k^2}{2m} = \frac{\hbar^2 \pi^2}{2ma}(n_x^2 + n_y^2 + \frac{n_z^2}{9})[/math]
The problem is to calculate the energy of the different energy levels. The ground state is obvious [imath](n_x = n_y = n_z = 1)[/imath], but after that I can't see what order the [imath]n[/imath] values are going to increase in. I think the first excited state should be [imath](n_x = n_y = 1, n_z = 2)[/imath], but I can't remember why.
Any help will win you a prize.
Metaphorically, of course.
QM problem; nonsymmetrical 3D potential
Moderators: gmalivuk, Moderators General, Prelates
Re: QM problem; nonsymmetrical 3D potential
(1,1,1) has energy [imath]1^2 + 1^2 + \frac{1^2}{9} = 2.111[/imath]
to find the next energy level you have to plug possible values in
(1,2,1)=(2,1,1) has energy [imath]2^2 + 1^2 + \frac{1^2}{9} = 5.111[/imath]
(1,1,2) has energy [imath]1^2 + 1^2 + \frac{2^2}{9} = 2.444[/imath]
thus the first excited state would be the (1,1,2) state
to find the next excited state you need to look at (1,3,1) and (1,2,3), (1,1,3), etc and compare them to (1,2,1) and pick the lowest, and so on...
ground state (1,1,1)
1st (1,1,2)
2nd (1,2,1) (2,1,1)
3rd (1,2,2) (2,1,2)
4th (1,2,3)
to find the next energy level you have to plug possible values in
(1,2,1)=(2,1,1) has energy [imath]2^2 + 1^2 + \frac{1^2}{9} = 5.111[/imath]
(1,1,2) has energy [imath]1^2 + 1^2 + \frac{2^2}{9} = 2.444[/imath]
thus the first excited state would be the (1,1,2) state
to find the next excited state you need to look at (1,3,1) and (1,2,3), (1,1,3), etc and compare them to (1,2,1) and pick the lowest, and so on...
ground state (1,1,1)
1st (1,1,2)
2nd (1,2,1) (2,1,1)
3rd (1,2,2) (2,1,2)
4th (1,2,3)

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Re: QM problem; nonsymmetrical 3D potential
The first excited state will indeed be 1,1,2 because that has lower energy than any other state apart from 1,1,1 (clearly, increasing n_{x} or n_{y} increases the energy more than increasing n_{z}). From there, I find it easier to take out the 1/9, so you're working with the expression
[math]9n_x^2 + 9n_y^2 + n_z^2\ (*)[/math]
If I've calculated correctly (not guaranteed!), the first few levels are:
Configuration : value of (*)
1,1,1 : 19
1,1,2 : 22
1,1,3 : 27
1,1,4 : 34
1,1,5 : 43
1,2,1 / 2,1,1 : 46
etc.
The higher levels are made annoying by all the degeneracy (obviously there will be twofold degeneracy when n_{x}!=n_{y}, but you can get more; e.g. 1,1,6, 1,2,3, 2,1,3 are all degenerate). I can't see any easy way to characterise these.
Edit: ninja'd.
[math]9n_x^2 + 9n_y^2 + n_z^2\ (*)[/math]
If I've calculated correctly (not guaranteed!), the first few levels are:
Configuration : value of (*)
1,1,1 : 19
1,1,2 : 22
1,1,3 : 27
1,1,4 : 34
1,1,5 : 43
1,2,1 / 2,1,1 : 46
etc.
The higher levels are made annoying by all the degeneracy (obviously there will be twofold degeneracy when n_{x}!=n_{y}, but you can get more; e.g. 1,1,6, 1,2,3, 2,1,3 are all degenerate). I can't see any easy way to characterise these.
Edit: ninja'd.
Generally I try to make myself do things I instinctively avoid, in case they are awesome.
dubsola
dubsola
Re: QM problem; nonsymmetrical 3D potential
Thanks, that makes perfect sense. It's an introductory undergraduate course, so I don't think they have to worry about higher level degeneracy.
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