
#1
Oct2209, 11:05 PM

P: 2

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
Oct2309, 12:41 PM

P: 383

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. 



#3
Oct2509, 11:14 PM

P: 2

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... 



#4
Dec1911, 09:26 AM

P: 11

Three Dimensional InfinitePotential Well Energies
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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