## QM problem; non-symmetrical 3D potential

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seladore
Posts: 586
Joined: Thu Mar 20, 2008 8:17 pm UTC
Location: Tumbolia

### QM problem; non-symmetrical 3D potential

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
$k_x = \frac{n\pi}{a},\;\;\;\; k_y= \frac{n\pi}{a},\;\;\;\; k_z = \frac{n\pi}{3a}$

And you have an energy
$E = \frac{\hbar^2 k^2}{2m} = \frac{\hbar^2 \pi^2}{2ma}(n_x^2 + n_y^2 + \frac{n_z^2}{9})$

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.

KyleOwens
Posts: 73
Joined: Thu Oct 23, 2008 4:21 am UTC

### Re: QM problem; non-symmetrical 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)

Ended
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Joined: Fri Apr 20, 2007 3:27 pm UTC
Location: The Tower of Flints. (Also known as: England.)

### Re: QM problem; non-symmetrical 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 nx or ny increases the energy more than increasing nz). From there, I find it easier to take out the 1/9, so you're working with the expression
$9n_x^2 + 9n_y^2 + n_z^2\ (*)$
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 two-fold degeneracy when nx!=ny, 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

seladore
Posts: 586
Joined: Thu Mar 20, 2008 8:17 pm UTC
Location: Tumbolia

### Re: QM problem; non-symmetrical 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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