Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 3 / 33B (Principles of Quantum Mechanics)

For distinguishable particles with one-particle Hilbert spaces ${\mathcal{H}}_1,\dots ,{\mathcal{H}}_N$, the multiparticle quantum state belongs toA product of one-particle states gives a product vector, and arbitrary linear combinations give entangled states. A noninteracting Hamiltonian is the sum of one-particle Hamiltonians acting on their respective tensor factors, so product-state energies add.

For two identical particles, the particle exchange operator $P_{12}$ swaps every degree of freedom. Since $P_{12}^2=I$, its eigenvalues are $\pm 1$. Particles whose allowed total states satisfyare bosons, while those satisfyingare fermions. Thus bosonic states are symmetric and fermionic states antisymmetric under interchange. The same condition applies to every transposition in a system of more than two identical particles.

Spin-one particles are bosons, so their total two-particle state must be symmetric. Write $|m,n⟩=|m{⟩}_1|n{⟩}_2$ for $m,n\in \{1,0,-1\}$. A basis of symmetric spin states isDenote these six states by ${\chi }_1^{+},\dots ,{\chi }_6^{+}$. A basis of antisymmetric spin states isThis is the six-plus-three decomposition of two identical spin-one bosons.

In the ground orbital both particles occupy $n=0$, so the spatial wavefunctionis symmetric. The complete list of ground states is therefore

At the first excited energy, one particle occupies $n=0$ and the other $n=1$. The normalized symmetric and antisymmetric spatial states areOverall bosonic symmetry permits symmetric spatial times symmetric spin and antisymmetric spatial times antisymmetric spin. Hence all first excited states areThe ground and first excited degeneracies are therefore six and nine.

Write the equally spaced one-particle spectrum asThe two-particle level labelled by $N$ has energyand receives contributions from orbital pairs $(n_1,n_2)$ with $n_1+n_2=N$.

For every unequal unordered pair $n_1<n_2$, there is one symmetric and one antisymmetric spatial combination. The symmetric one combines with the six-dimensional symmetric spin space; the antisymmetric one combines with the three-dimensional antisymmetric spin space. Thus each unequal pair contributes nine states.

If $N$ is odd, all pairs are unequal and there are $(N+1)/2$ unordered pairs. Therefore

If $N$ is even, there are $N/2$ unequal pairs and the additional equal pair $n_1=n_2=N/2$. The latter has only a symmetric spatial state and therefore contributes six spin states. HenceThis is the degeneracy of two identical spin-one bosons with equally spaced orbital levels.