Quantum Mechanics - Identical and Non-Identical Spin 1 particles

[Tangent87]

Hi, I am doing question 32D on page 18 here:

http://www.maths.cam.ac.uk/undergrad/pastpapers/2005/Part_2/PaperII_3.pdf

and I am stuck on the second paragraph where we have to explain how to construct the two-particle states of lowest energy for (i). identical spin-1 particles with combined total spin J=1, and (ii). non-identical spin-1 particles with combined total spin J=1.

I found in the first part of the question that all three of the J=1 spin states are antisymmetric, and thus for (i) since we have identical spin 1 particles the total state must be symmetric overall therefore the only possible state is \psi_1(A)\psi_1(B). But I'm not sure about this because I don't think it takes into account the fact that we're in a total spin J=1 state.

For (ii), I said that since the particles are non-identical there is no exchange symmetry and so we can have any of the three states: \psi_1(A)\psi_1(B)|1 M> for M=-1,0 or 1. Is that correct?

[ideasrule]

I found in the first part of the question that all three of the J=1 spin states are antisymmetric, and thus for (i) since we have identical spin 1 particles the total state must be symmetric overall therefore the only possible state is \psi_1(A)\psi_1(B). But I'm not sure about this because I don't think it takes into account the fact that we're in a total spin J=1 state.

If the total state must be symmetric and the spin states are all antisymmetric, that means the wavefunction must be antisymmetric. \psi_1(A)\psi_1(B) is not antisymmetric.


For (ii), I said that since the particles are non-identical there is no exchange symmetry and so we can have any of the three states: \psi_1(A)\psi_1(B)|1 M> for M=-1,0 or 1. Is that correct?

Yes, and all three states have identical energy.

[Tangent87]

If the total state must be symmetric and the spin states are all antisymmetric, that means the wavefunction must be antisymmetric. \psi_1(A)\psi_1(B) is not antisymmetric.


So in order to have the wavefunction being antisymmetric would I need to have something like \frac{1}{\sqrt{2}}(\psi_1(A)\psi_2(B)-\psi_2(A)\psi_1(B))?

The only trouble I have with this is that the wavefunction now involves terms from the second energy level whilst we're only dealing with the lowest energy level, is this a problem?

[ideasrule]

So in order to have the wavefunction being antisymmetric would I need to have something like \frac{1}{\sqrt{2}}(\psi_1(A)\psi_2(B)-\psi_2(A)\psi_1(B))?


Yes, that's correct.

The only trouble I have with this is that the wavefunction now involves terms from the second energy level whilst we're only dealing with the lowest energy level, is this a problem?

The wavefunction involves second-energy-level one-particle states, while the problem asks for the lowest-energy two-particle state. For fermions, the lowest-energy two-particle state indeed includes excited one-particle states.

[Tangent87]

Ah okay, that clears everything up, thanks again.