Three Dimensional Infinite-Potential Well Energies

  1. Oct 22, 2009 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant equations
    -[tex]\frac{\hbar^{2}}{2m}[/tex][tex]\nabla^{2}[/tex][tex]\Psi[/tex] + V[tex]\Psi[/tex] = E[tex]\Psi[/tex]

    3. The attempt at a solution
    so I derived [tex]E[/tex] = [tex]\frac{\pi^{2}\hbar^{2}}{2mL^{2}}[/tex]([tex]n_{1}^{2}[/tex] + [tex]n_{2}^{2}[/tex] + [tex]n_{3}^{2}[/tex]) 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[tex]^{th}[/tex] and 5[tex]^{th}[/tex] 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...
  2. jcsd
  3. Oct 23, 2009 #2
    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 [tex]n_i[/tex] in your energy eigenvalue.

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

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

    where [tex]E_0=\pi^2\hbar^2/2mL^2[/tex]. So there is a lower energy in the [122] state than in the [113] state.
  4. Oct 25, 2009 #3
    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...
  5. Dec 19, 2011 #4
    I know this is a really old post but I had the exact same problem and just wanted to say thanks for the help!
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