# What would be a criteria to discover if a state is a tensor product or not?

In Quantum Mechanics states of composite systems are described by tensor products. As is known, if we have $\mathcal{H}_1$ and $\mathcal{H}_2$ (which of course could be the same space) in the tensor product $\mathcal{H}_1\otimes \mathcal{H}_2$ we have the so called factorizable tensors which are writen as $|\psi_1\rangle\otimes |\psi_2\rangle$ with $|\psi_i\rangle \in \mathcal{H}_i$ but we also have more general elements which cannot be factored this way.

Indeed if $|u_n(i)\rangle$ is a basis of $\mathcal{H}_i$ then $|u_n(1)\rangle \otimes |u_m(2)\rangle$ is a basis of $\mathcal{H}_1\otimes \mathcal{H}_2$ so that a general element is

$$|\psi\rangle = \sum_{nm} c_{nm}|u_n(1)\rangle \otimes |u_m(2)\rangle.$$

My question is: if we have one $|\psi\rangle$ and we want to determine whether or not it is a factorizable tensor, what would be a useful criteria to determine this?

• I think you're effectively asking whether subsystems 1 and 2 are entangled or not. If so, I believe the way to do this is by examining the reduced density matrices. I'm not an expert in this, so I can't write a full answer describing the technique, but maybe this will give you something to Google on. – Michael Seifert Apr 10 '17 at 1:02

Given a state $|\psi\rangle = \sum_{nm} c_{nm}|u_n(1)\rangle \otimes |u_m(2)\rangle$, you are asking whether there exist coefficients $a_n,b_n$ such that $$|\psi\rangle = \left(\sum_{m} a_n|u_n(1)\rangle \right) \otimes \left(\sum_{m} b_m|u_m(2)\rangle \right) = \sum_{nm} a_nb_m|u_n(1)\rangle \otimes |u_m(2)\rangle$$ In other words, you're asking whether or not there exist coefficients such that $c_{nm} = a_n b_m$. Equivalently, you are asking whether the $n \times m$ matrix $C$ whose entries are $c_{nm}$ can be written in the form $$C = \mathbf a \mathbf b^T$$ for vectors $\mathbf a = (a_1,\dots,a_n) \in \Bbb C^n, \mathbf b = (b_1,\dots,b_m)\in \Bbb C^m$.
In other words, a state $|\psi\rangle = \sum_{nm} c_{nm}|u_n(1)\rangle \otimes |u_m(2)\rangle$ is separable if and only if the matrix $C$ of coefficients $c_{nm}$ has rank 1.