Under what assumptions can we split a Hilbert space into subspaces?

Question

I was thinking about an apparently simple question about quantum mechanics, if I am looking at a quantum system described by a Hilbert space $\cal{H}$ under what hypothesis can I define A and B as subsystems whose union gives the full former system and decompose $\cal{H} = \cal{H_A}\otimes \cal{H_B}$. The other way, it is evident, if I join two quantum systems, their state will evolve in the tensorial products of the two, and the states are factorizable if there is no quantum correlation. But starting with the first joined system, I am puzzled by the meaning of this decomposition.

Thank you. And sorry if that question is a no brainer, it's just not that evident to me. =)

Answer 1

Without further specifications, you can do it always. Let $\psi_j$ ($j\in N_0$), where $N_0$ denotes the set of nonnegative integers, be a basis of the Hilbert space $H$. Let $I$ be a countable bijection from $N_0\times N_0$ to $N_0$, let $H_A$ be spanned by the $\psi_{I(j,0)}$, and $H_B$ be spanned by the $\psi_{I(0,k)}$. With the identification $\psi_{I(j,0)}\bigotimes\psi_{I(0.k)}=\psi_{I(j,k)}$, $H$ is the tensor product of $H_A$ and $H_B$.

On the other hand, if you have given two subsystems A and B of a bigger system, they are in a tensor product inside the big system if and only if A and B have no common observables (except for the c-numbers). In just this case there is inside the bigger system a subsystem composed of A and B described by this tensor product.

Answer 2

Decomposition of Hilbert states makes sense, when can write some operations as acting only on one part.

In particular, it happens with respect to particles, i.e. $\mathcal{H}=\mathcal{H}_A\otimes\mathcal{H}_B$.

But it is not only case; also it happens when you have a system of $n$ particles coupled to environment in the same way - then you can split state in the collective part and one related to individual particles.

For example, when you have $n$ spin$-\frac{1}{2}$ particles, then they can be decomposed with respect to total angular momentum (it's called Clebsch-Gordan decomposition or Schur-Weyl duality). To be more specific, for $n=4$ qubits you get $$\mathcal{H}_{1/2} \otimes \mathcal{H}_{1/2} \otimes \mathcal{H}_{1/2} \otimes \mathcal{H}_{1/2} =(\mathbb{C}^2\otimes\mathcal{H}_0) \oplus (\mathbb{C}^3 \otimes \mathcal{H}_1) \oplus (\mathbb{C}^0 \otimes \mathcal{H}_2),$$ where lower indices stand for total angular momentum. So for example for the triplet subspace (i.e. total angular momentum $1$) we get threefold degeneracy. Then permutation (i.e. changing order) of particles acts on $\mathbb{C}^3$, whereas interaction with external magnetic field - $\mathcal{H}_1$.

Such decomposition can be used to investigate so-called decoherence-free subsystems, where this $\mathbb{C}^k$ is not affected by decoherence.

Some more intro is in arXiv:1107.3786 (beginning of Sec. 2.).