How Does Symmetrization Affect Boson Spin Degeneracies?

[rmiller70015]

Homework Statement

Three identical bosons are in an infinite square well of length a. Find the ground state, first and second exited states, their energies, and the degeneracies.

Relevant Equations

$$S = \frac{1}{N!}\Sigma _{\alpha}P_{\alpha}$$
$$A = \frac{1}{N!}\Sigma _{\alpha} \epsilon _{\alpha} P_{\alpha}$$

My question is really about the degeneracies. I know that the symmetrization postulate says that there is only 1 unique ket in the subspace $Sym^{N}V$, but does this mean that if I know one unique spin configuration that is symmetric, say $|33>$ then is it correct to say the ground state wave function is $$C(\frac{2}{a})^{\frac{3}{2}}sin(\frac{\pi x_1}{a})sin(\frac{\pi x_2}{a})sin(\frac{\pi x_3}{a})|33>$$, where C is a scaling constant.

 

[kdv]

Homework Statement:: Three identical bosons are in an infinite square well of length a. Find the ground state, first and second exited states, their energies, and the degeneracies.
Relevant Equations:: $$S = \frac{1}{N!}\Sigma _{\alpha}P_{\alpha}$$
$$A = \frac{1}{N!}\Sigma _{\alpha} \epsilon _{\alpha} P_{\alpha}$$

My question is really about the degeneracies. I know that the symmetrization postulate says that there is only 1 unique ket in the subspace $Sym^{N}V$, but does this mean that if I know one unique spin configuration that is symmetric, say $|33>$ then is it correct to say the ground state wave function is $$C(\frac{2}{a})^{\frac{3}{2}}sin(\frac{\pi x_1}{a})sin(\frac{\pi x_2}{a})sin(\frac{\pi x_3}{a})|33>$$, where C is a scaling constant.

I am not sure what your S and A stand for here.
You wrote the correct spatial wavefunction. I am not sure what your ket $|33 \rangle$ stand for either.
You said that they are bosons, but their spin is not specified? If not, then, how many different spin states can you have in the ground state?

 

[rmiller70015]

I am not sure what your S and A stand for here.
You wrote the correct spatial wavefunction. I am not sure what your ket $|33 \rangle$ stand for either.
You said that they are bosons, but their spin is not specified? If not, then, how many different spin states can you have in the ground state?

S and A are the operators that project a ket from $V^{\otimes 3}$ into the subspaces of symmetric ($Sym^NV$) and antisymmetric ($Anti^NV$) configurations by way of the permutation operators. I wrote $|33>$ because it's the simplest bosonic spin configuration I could think of that would be symmetric, $|SM> = |s_1 m_1 s_2 m_2 s_3 m_3>$ so $|33> = |111111>$.

As far as how many spin configurations you can have in the ground state, that would be the dimension of $Sym^NV$, which is the number of symmetric spin configurations for the three bosons. Since the spatial wave function is symmetric under the exchange of particles, the spin part of the wave function also needs to be symmetric. I think if I am understanding correctly, that all but one of them are degenerate states.

 

[kdv]

Homework Statement:: Three identical bosons are in an infinite square well of length a. Find the ground state, first and second exited states, their energies, and the degeneracies.
Relevant Equations:: $$S = \frac{1}{N!}\Sigma _{\alpha}P_{\alpha}$$
$$A = \frac{1}{N!}\Sigma _{\alpha} \epsilon _{\alpha} P_{\alpha}$$

My question is really about the degeneracies. I know that the symmetrization postulate says that there is only 1 unique ket in the subspace $Sym^{N}V$, but does this mean that if I know one unique spin configuration that is symmetric, say $|33>$ then is it correct to say the ground state wave function is $$C(\frac{2}{a})^{\frac{3}{2}}sin(\frac{\pi x_1}{a})sin(\frac{\pi x_2}{a})sin(\frac{\pi x_3}{a})|33>$$, where C is a scaling constant.

S and A are the operators that project a ket from $V^{\otimes 3}$ into the subspaces of symmetric ($Sym^NV$) and antisymmetric ($Anti^NV$) configurations by way of the permutation operators. I wrote $|33>$ because it's the simplest bosonic spin configuration I could think of that would be symmetric, $|SM> = |s_1 m_1 s_2 m_2 s_3 m_3>$ so $|33> = |111111>$.

As far as how many spin configurations you can have in the ground state, that would be the dimension of $Sym^NV$, which is the number of symmetric spin configurations for the three bosons. Since the spatial wave function is symmetric under the exchange of particles, the spin part of the wave function also needs to be symmetric. I think if I am understanding correctly, that all but one of them are degenerate states.

Ah ok, so the P's are permutations.

I see that you assume that S=1. This is no stated in the question, so I think you will have to double check with your instructor because the question is ambiguous. S could be 0 (making the problem trivial) or 2, etc. The degeneracy will depend on that (unless your instructor wants an answer in terms of S, which can be worked out).
Note that if the spin is 1, the $| m_1 m_2 m_3 \rangle=|1,1,1\rangle$ is not the only state. You could also have $( |1,1,0\rangle + |1,0,1\rangle + |0,1,1\rangle )/\sqrt{3}$ and many other states. You have to count these and see the general formula in terms of S (again, if this is what your instructor wants).

 

[rmiller70015]

Ah ok, so the P's are permutations.

I see that you assume that S=1. This is no stated in the question, so I think you will have to double check with your instructor because the question is ambiguous. S could be 0 (making the problem trivial) or 2, etc. The degeneracy will depend on that (unless your instructor wants an answer in terms of S, which can be worked out).
Note that if the spin is 1, the $| m_1 m_2 m_3 \rangle=|1,1,1\rangle$ is not the only state. You could also have $( |1,1,0\rangle + |1,0,1\rangle + |0,1,1\rangle )/\sqrt{3}$ and many other states. You have to count these and see the general formula in terms of S (again, if this is what your instructor wants).

Ok, thanks, I was unclear on whether the permutations were degenerate or if the spin states were degenerate. However, if the spin states are not degenerate then there are like 27 of them for identical spin 1 boson and I'm not sure how many of those would be symmetric.