Three Dimensional Infinite-Potential Well Energies

  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. Chegg
    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.

    [113]:
    [tex]
    \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}
    [/tex]

    [122]:
    [tex]
    \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}
    [/tex]

    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. 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. 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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