I am encountering two-particle systems for the first time, and I am a little confused about something.(adsbygoogle = window.adsbygoogle || []).push({});

Consider a infinite square well with two noninteracting particles, both fermions (though, disregard spin degrees of freedom, this is a strictly 1-d scenario).

So I go to write down the states of the system corresponding to different energies. Consider the case:

[tex] \psi_{23} = \frac{1}{\sqrt{2}} \left[ \psi_2(x_1) \psi_3(x_2) - \psi_3(x_1) \psi_2(x_2) \right] [/tex]

Where

[tex] \psi_n (x) = \sqrt{\frac{2}{a}} \sin n \pi x / a [/tex]

This corresponds to energy

[tex] E = \frac{13 \hbar^2 \pi^2}{2 m a^2} [/tex]

This is all good. What I don't get is this: why isn't this energy degenerate? In other words, isn't the state:

[tex] \psi_{32} = \frac{1}{\sqrt{2}} \left[ \psi_3(x_1) \psi_2(x_2) - \psi_2(x_1) \psi_3(x_2) \right] [/tex]

Also an energy eigenfunction of the Hamiltonian, with the same eigenvalue. Aren't the two states distinct?

Now, the symmetrization requirement says:

[tex] \psi(x1,x2) = -\psi(x2,x1)[/tex]

for fermions, which is the case above. I feel like this is relevant. But I still see there being two separate states with the same energy. Isn't this the definition of degeneracy?

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# Degeneracy and Symmetrization Requirement

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