I'm taking a QM class and I'm trying to work out a problem given to us by the professor.

We have two identical, non-interacting, spin-1 bosons that are located in a one dimensional harmonic oscillator potential with a characteristic frequency $\omega$. The problem to solve is to find the three lowest energy levels of this two-particle system, as well as their degeneracies.

I think that I need to use the Clebsch-Gordan table to combine the single-particle states into a combined state, and then act on this state with the hamiltonian to find the energies.

But in trying to do this I don't get very far. In earlier problems I always know the $m_{1}$ and $m_{2}$ quantum numbers ($m_{i} \in \{-s, -s+1,\dots,s\}$) that I need in order to use the table. This time the values of $m$ is not specified, so I am not sure how to proceed.

I\m thinking it might just be to write out a linear combination of all the combined states that I can get from every permutation of $m$'s, but that doesn't seem to right (plus I'm not getting anywhere when trying to do so). I assume I'm missing something quite elementary in understanding this.

Any help on how I should go about doing this would be great!


Since the two particles are bosons you need to analyse which combinations of the two spins are consistent with a total wave function that is even under particle exchange.

The total wave function is the product of the spatial wave function from the harmonic oscillator (a product state for non-interacting particles) and the spin wave function. For example, if the two bosons are both in the ground state of the harmonic oscillator their spatial wave function is a product of two Gaussians. This is clearly even under particle exchange, and therefore the total spin wave function should be even as well. This can only be the symmetric states with $S_{tot} = 0,2$ and therefore the total degeneracy of the ground state is 6.

  • $\begingroup$ Thank you for the response, I think a got a solution (of some kind anyway). I'm not quite sure how you get a degeneracy of 6. When I use Clebsch-Gordan and write out all the possible combinations for a given set of $m_1$ and $m_2$ I get 9 in total. Out of these I can only count 5 that would be even under particle exchange. How did you count to 6? @Praan $\endgroup$ – Bendik Oct 15 '15 at 20:10
  • $\begingroup$ @Bendik Note that you get $2S+1$ states for given $S$. That's how I got to six states in total. Did you consider the singlet $S=0$? To be sure, you could look up a table of the Glebsch-Gordan coefficients to check your results. $\endgroup$ – Praan Oct 15 '15 at 23:54
  • $\begingroup$ Well, that make sense to me, and I guess I can use this reasoning in my argument. What bothers me is that I'm unable to see the same results using the table. I guess I'm doing something wrong with reading the table, so I'll use the more intuitive argument you presented, and then ask the professor later how to do it "more rigorously". Thanks for the help in any case! $\endgroup$ – Bendik Oct 16 '15 at 5:50

In this case the particles are non-interacting and there is no magnetic field. Therefore the quantum numbers $m_i$ do not influence the energy of the system.

You can therefore calculate the energy eigenstates of the system while ignoring the quantum numbers $m_i$. You should however take the quantum numbers $m_i$ into account when you want to determine the degeneracy of the found states.


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