# Two electros in one harmonic oscillator

Suppose a one dimensional harmonic oscillator with two spin $\frac{1}{2}$ fermions. The state of each fermion can be $|n\rangle|\pm\rangle$ with $n\in\{0,1,2,\dots\}$.

My question is: which states compose the first excited level?

My (possibly incomplete) answer is: I know that, because of the Pauli exclusion principle, if I were to measure the four quantum numbers for the system I would get one of the following configurations: These are the possible configurations but not the possible states, because fermions have antisymmetric states, and none of the previous states is antisymmetric.

In order to build antisymmetric states I usually take my "possible configurations" and plug them into Slater determinants. This has been working fine when I have only one quantum number to worry about, for example the fundamental state of the current system. But this technique does not seem to be working here, I don't know how to find the states (nor how many states there should be).

I was told by my professor that I should find 8 states (I don't know why 8). Using Slater determinants I have found the states $|a\rangle, |b\rangle, |c\rangle$ and $|d\rangle$, and my professor made me notice that $|e\rangle$ and $|f\rangle$ are also possible states: but another two questions arise:

1. What about the states $|g\rangle$ and $|h\rangle$ ?
2. Which is the algorithmic method to find all the states?

Thanks in advance for any help. The exam is tomorrow :(

• Observe that $\left|e\right>=\frac{\left|d\right>-\left|c\right>}{\sqrt{2}}$ and $\left|f\right>=\frac{\left|d\right>+\left|c\right>}{\sqrt{2}}$ so the states you wrote are not linear independent. – eranreches Dec 11 '17 at 23:50
• Try Gram-Schmidt orthonormalization. It works the same on finite-dimensional Hilbert spaces as it does on ordinary vector spaces. – probably_someone Dec 12 '17 at 0:01
• If you use second quantization notation, you don't need all this antisymmetric stuff. – DanielSank Dec 12 '17 at 3:04