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]:
<br /> \begin{array}{lll}E_{113}&amp;=&amp;\frac{\pi^2\hbar^2}{2mL^2}\left((1)^2+(1)^2+(3)^2\right) \\ \\ \,&amp;=&amp;E_{0}\cdot11<br /> \end{array}<br />

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

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!