# Three Dimensional Infinite-Potential Well Energies

 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 -$$\frac{\hbar^{2}}{2m}$$$$\nabla^{2}$$$$\Psi$$ + V$$\Psi$$ = E$$\Psi$$ 3. 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...
 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 $$n_i$$ 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.