Three Dimensional Infinite-Potential Well Energies

[hADAR1]

Homework Statement

So the question asks me to find the energies of the 2nd, 3rd, 4th, and 5th excited states in a three dimensional cubical box and to state which are degenerate.

Homework Equations

-$$\frac{\hbar^{2}}{2m}$$$$\nabla^{2}$$$$\Psi$$ + V$$\Psi$$ = E$$\Psi$$

The Attempt at a Solution

so I derived $$E$$ = $$\frac{\pi^{2}\hbar^{2}}{2mL^{2}}$$($$n_{1}^{2}$$ + $$n_{2}^{2}$$ + $$n_{3}^{2}$$) for a cubical box. I think this is correct, so the derivation isn't where my question lies.
I'm having a little trouble knowing what n values to use for the different energy states. For ground state my book says it's [111]. It then says that for the first excited state it is either [112], [121], or [211] and goes into talking about degeneracy. I think I understand everything up until this point. But I am confused as to how I keep going to subsequent excited states...

Would the second excited state be [113], [131], and [311]? or would it be [122], [221], and [212]?
Would the third excited state be [114], [141], and [411]? or would it be [222]? or something even weirder like [123], [132], [213], [231], [312], and [321]

So on and so forth for the 4$$^{th}$$ and 5$$^{th}$$ excited states.
I just don't get when to increase what n.

Any help would be greatly appreciated! I looked around the forums and didn't find anyone else asking this question. I feel like I did the hard part (the derivation) correctly, but am stuck on the simple part. QM has its ways of being frustrating at times...

 

[jdwood983]

The next energy eigenvalues would be [122], [212], [221].

The easiest method of doing "Which excited state comes next" problems is to just put the numbers into the $$n_i$$ in your energy eigenvalue.

[113]:
$$\begin{array}{lll}E_{113}&=&\frac{\pi^2\hbar^2}{2mL^2}\left((1)^2+(1)^2+(3)^2\right) \\ \\ \,&=&E_{0}\cdot11 \end{array}$$

[122]:
$$\begin{array}{lll}E_{122}&=&\frac{\pi^2\hbar^2}{2mL^2}\left((1)^2+(2)^2+(2)^2\right) \\ \\ \,&=&E_{0}\cdot9 \end{array}$$

where $$E_0=\pi^2\hbar^2/2mL^2$$. So there is a lower energy in the [122] state than in the [113] state.

 

[hADAR1]

Oh ok, that makes a lot of sense. I feel stupid now lol.

Thanks a lot for the help though, I honestly just wasn't seeing it that way...

 

[Toni103]

I know this is a really old post but I had the exact same problem and just wanted to say thanks for the help!

 

[paranormal]

Dear student,

First of all, great job on deriving the expression for the energies in a three-dimensional infinite potential well! It looks correct to me.

To answer your question about the excited states, the principle quantum numbers n1, n2, and n3 represent the number of nodes in the x, y, and z directions, respectively. So for the ground state, [111] means there is one node in each direction. For the first excited state, [112] means there is one node in the x direction and two nodes in the y direction, and so on.

For the second excited state, it would be [113] for one node in x, one node in y, and three nodes in z. You can also think of it as [122] for two nodes in y and two nodes in z, or [221] for two nodes in x and two nodes in y. All of these combinations give the same energy, so they are degenerate.

For the third excited state, it would be [114], [123], [132], [213], [231], or [312], all of which have three nodes in total (one in each direction). This is because for the third excited state, the total number of nodes (n1 + n2 + n3) must be equal to three.

So for subsequent excited states, you just need to make sure that the total number of nodes is equal to the principle quantum number of that state. This will give you all the possible combinations for the quantum numbers n1, n2, and n3.

I hope this helps clear up your confusion. Good luck with your studies!