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The energy level of electron in an infinite square well in three dimensions is given by $E_{n_1 n_2 n_3} =\frac{ \hbar^2 \pi^2}{2mL^2}(n_1^2 + n_2^2 +n_3^2)$. It is understood that $E_{111}$ represents the ground state. My question is how do we rank other excited states? The first excited state is surely any of $E_{211}$, $E_{121}$ and $E_{112}$. It seems $E_{221}$ is the second excited state and $E_{311}$ is the third. But why? I was thinking we can rank them based on the summation of indices but indices of $E_{221}$ and $E_{311}$ add up to the same number $2+2+1=3+1+1=5$. What is the justification for having $E_{311}$ as the third excited state?

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    $\begingroup$ Compare the energy levels. $\endgroup$ – Hanting Zhang Apr 9 at 3:02
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What has physical meaning is not the sum $n_1+n_2+n_3$ but the expression appearing in $E_{n_1 n_2 n_3}$: $n_1^2+n_2^2+n_3^2$. Therefore $1^2+2^2+2^2=9 < 3^2 +1^2+1^2=11$.

By writing systematically the increasing levels of energy one obtains the sequence of the first, second,... excited level of energy.

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I believe I figured it out myself. It is because we rank them based on the equation itself. So defining $E_0 = \frac{\hbar^2 \pi^2}{2mL^2}$ we have $E_{111} = E_0(1+1+1)=3E_0$. Similarly, the first excited state $E_{211} = E_0(2^2 +1^2+1^2)=6E_0$. For the second excited state we have $E_{221} = 9E_0$ and finally we get $E_{311} = 11E_0$ for the third excited state.

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    $\begingroup$ Correct. You order excited energy levels by their energy. $\endgroup$ – G. Smith Apr 9 at 3:40
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rank them according to their energies, put the different values of n and calculate the energy, we also can get degeneracy.

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