# The order of excited states

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?

• Compare the energy levels. – Hanting Zhang Apr 9 at 3:02

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 $$1^2+2^2+2^2=9 < 3^2 +1^2+1^2=11$$.
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.