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

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?
• Also, this might answer your question too ? Jun 25, 2020 at 16:24
• Interactions don't change the Hilbert space, only the hamiltonian. Furthermore, by the dimension theorem of linear algebra, any two finite-dimensional, complex Hilbert spaces of the same dimension are isomorphic, so any interacting system whose Hilbert space dimension $n$ you know has a Hilbert space isomorphic to $\mathbb C^n$. Jun 26, 2020 at 0:14

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.

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.

• When the particles are coupled by interaction, it possible to define $\mathcal{V}_1$ and $\mathcal{V}_2$ separately? Jun 25, 2020 at 16:53
• @mithusengupta123 I'm not sure what you mean by "define them separately". The total hilbert space is still the tensor product of the two. Each one of $\mathcal{V}_{1}$ and $\mathcal{V}_{2}$ is still a Hilbert space on its own. Jun 25, 2020 at 19:27
• The space of possible states for two particles is the SAME regardless of interaction (at least in the nonrelativistic regime, the proof of this that I know depends on the schrodinger equation). The dynamics of states (how states change over time) is different depending on the interaction. I suggest Asher Peres' amazing book, Quantum Theory: Concepts and Methods, for the nonrelativistic theory Jun 26, 2020 at 4:04

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.