Tensor product postulate

Question

Non relativistic quantum mechanics assumes that a composite system should be described with the tensor product of the component systems. This is the tensor product postulate of quantum mechanics.

I think that the postulate originated in wave mechanics due to the following isomorphism: $\ L^2 \ (R\times R)=\ L^2 (R)\otimes \ L^2 (R) $. The lhs of the former equation is quite intuitive, the geometry of Hilbert spaces do the rest.

Now, disregarding position representation and taking for example two simple quantum systems (e.g a qubit and a qutrit) why, in principle, should we describe the composite system according to the tensor product postulate?

Answer 1

It comes down to linearity, and to be clear life isn't just a tensor product of two systems, it lives inside the tensor product.

If you think it reasonable to have a single particle state like $\psi_1$ for part of the system and to have a single particle state like $\psi_2$ for part of the system then it might seem reasonable to have a multiparticle state like $\psi_1\otimes\psi_2$ and then by linearity you would be forced to have at least the whole tensor product.

This really comes from having a basis for the tensor product made out of products of basis elements from the base spaces.

Now let's come down to reality. We merely said that it might seem reasonable to have a multi-particle state like $\psi_1\otimes\psi_2$ but if the states describe identical particles then $\psi_2\otimes\psi_1$ would be experimentally indistinguishable except in as far as this is a subsystem of a larger system. And what its also possible that nature could impose a superselection rule such as that a base state look like $\psi_1\otimes\psi_2+\psi_2\otimes\psi_1$ or $\psi_1\otimes\psi_2-\psi_2\otimes\psi_1$ and in fact the kicker is that nature does impose such superselection rules the former for bosons and the latter for fermions, and every particle is one or the other.

And then linearity gives you something provably smaller than the tensor product. So there isn't a tensor product rule. There is linearity and symmetry superselection and the former could have given you the full tensor product, but because of the latter it doesn't.

Answer 2

From a formal point of view, the reason for the tensor product rule is as follows:

For any quantum system the Hilbert space of states $\mathcal{H}$ is defined in terms of a complete set of degrees of freedom $S_1, S_2,…, S_n$ that can be measured simultaneously. The corresponding quantum states, characterized by sets of values $\{s_1,s_2,…,s_n\}$ and denoted $| s_1, s_2,…, s_n \rangle$, define both a basis in $\mathcal{H}$, and a complete set of mutually commuting, self-adjoint observables $\hat{S}_1$, $\hat{S}_2$, ..., $\hat{S}_n$, that have the states $| s_1, s_2,…, s_n \rangle$ as common eigenstates and the values $s_1, s_2,…, s_n$ as respective eigenvalues. The Hilbert space $\mathcal{H}$ so defined is isomorphically equivalently to the direct product $\mathcal{H}_1\otimes \mathcal{H}_2\otimes … \otimes \mathcal{H}_n$, where each $\mathcal{H}_i$ is spanned by states $| s_i\rangle$ corresponding to a single degree of freedom, and $| s_1, s_2,…, s_n \rangle$ is isomorphically equivalent to $\otimes_i| s_i\rangle$.

A similar construction applies when the total system is composed of a number of independent subsystems or different particles. In this case Hilbert spaces $\mathcal{H}_i^{(\alpha)}$ corresponding to different degrees of freedom of a subsystem $\alpha$ are naturally factors of the subsystem's Hilbert space $\mathcal{H}^{(\alpha)} = \otimes_i{\mathcal{H}_i^{(\alpha)}}$, and the total Hilbert space amounts to a direct product over subsystem spaces, $\mathcal{H} = \otimes_\alpha \mathcal{H}^{(\alpha)}$.

The case of identical particles is no different in this respect, although the set of states allowed is further restricted as required by indistinguishability and statistics. Supersymmetry does not supersede the tensor product rule, but reinforces it.