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:

Possible 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:

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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 :(

  • $\begingroup$ 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. $\endgroup$ – eranreches Dec 11 '17 at 23:50
  • $\begingroup$ Try Gram-Schmidt orthonormalization. It works the same on finite-dimensional Hilbert spaces as it does on ordinary vector spaces. $\endgroup$ – probably_someone Dec 12 '17 at 0:01
  • $\begingroup$ If you use second quantization notation, you don't need all this antisymmetric stuff. $\endgroup$ – DanielSank Dec 12 '17 at 3:04

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