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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Degeneracy and Symmetrization Requirement

**Physics Forums | Science Articles, Homework Help, Discussion**