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For two non-interacting particles, with eigenfunctions $\phi_{n1}(x1)$ and $\phi_{n2}(x2)$ in a one-dimensional potential well $V_{(x)}$ with n = 1,2,....

Consider two spinless non-identical particles:

What is the degeneracy of the ground state and first excited state?

I'm thinking it should be 0 for both.

Now Consider two spinless identical particles,

I'm thinking degeneracy for ground state is 0, and first excited state is 2. (since ground state both n = 0, and first excited state it can either be 1,0 or 0,1)

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    $\begingroup$ The state of two identical spinless particles must be symmetric w.r.t. permutations. So, the first excited state must be 1/sqrt(2) [|1,0> + |0,1>]. Interchaning the two particles does not yield a different state. $\endgroup$
    – Count Iblis
    Commented Jun 8, 2014 at 19:01
  • $\begingroup$ that's what I got as well. Since the energy is given by $E_n = E_{n1} + E_{n2} = (1 + n_1 + n_2)\hbar \omega $, is the degeneracy 2? This question is about two non-interacting particles in a potential well. $\endgroup$
    – user44840
    Commented Jun 8, 2014 at 19:24

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Spinless non-identical particles.
Ground state: $(0,0) \implies \text{non-degenerate}$
First excited state: $(0,1) \text{ and } (1,0) \implies \text{doubly degenerate}$

Spinless identical particles.
Ground state: $(0,0) \implies \text{non-degenerate}$
First excited state: $(0,1)+(1,0) \implies \text{non degenerate}$

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