Hilbert space of two particles of spins $j_1$ and $j_2$, noninteracting versus interacting

Question

For a system of two noninteracting particles of spins $j_1$ and $j_2$, the joint Hilbert space $\mathcal{V}$ is the tensor product of the individual Hilbert spaces $\mathcal{V}_1$ and $\mathcal{V}_2$. Notationally, $\mathcal{V}=\mathcal{V}_1\otimes\mathcal{V}_2$ which is spanned by the $(2j_1+1)(2j_2+1)$ product states or their linear combinations.

• What is the Hilbert space when there is an interaction between the particles? Do the product states still serve as a basis which spans the space?

Answer 1

The Hilbert space is still the same, since you have not changed the number of degrees of freedom. The difference is that the eigenstates of the Hamiltonian are now superpositions of states within each of the subspaces. I wrote an answer here showing how this works. So yes, the product states would still be a basis, but they might not be as useful, depending on what the interaction is.

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Another way of seeing this is if you think of having $n$ qubits. The Hilbert space describing those is a product of the $n$ qubits and operating on them is equivalent to changing the Hamiltonian (i.e. introducing quantum gates) and time evolving the qubits. After a quantum computation, the space of states is still the same, you've only used the interactions to perform a computation.

Answer 2

There are two “naturally” useful bases for the same Hilbert space. The space ${\cal V}_1\otimes {\cal V}_2$ is reducible as a direct sum of $SU(2)$-invariant subspaces.

The first basis is the uncoupled basis with $\vert j_1m_1\rangle \vert j_2, m_2\rangle$ as basis states.

The second is the coupled basis $\vert JM\rangle$, with $j_1+j_2\ge J\ge \vert j_1-j_2\vert$, where states are expressed as linear combinations $$ \vert JM\rangle=\sum_{m_1m_2} C^{JM}_{j_1 m_1;j_2 m_2} \vert j_1m_1\rangle \vert j_2m_2\rangle\, . $$ Since states $\vert JM\rangle$ have “good angular momentum” quantum numbers $J$, one would expect that this basis is useful whenever you need to refer to the total angular momentum of the system. In this basis, the actual momentum operator act within each $SU(2)$-invariant subspace.

Contrariwise the uncoupled basis would be useful whenever single-particle properties are best to described the system.