I'm looking for the simple argument to prove that a separable pure bipartite quantum state is in fact a product state. This question comes from a statement in Wikipedia on separable states: In the special case of pure states the definition simplifies: a pure state is separable if and only if it is a product state.

On one hand, $|\psi\rangle$ has a Schmidt decomposition $|\psi\rangle = \sum_i \lambda_i |a_i\rangle \otimes |b_i\rangle$, where $\lambda_i \in \mathbb{R}_{\geq 0}, \forall i$, and $\sum_i \lambda_i^2 = 1$ (to ensure the normalisation of the state). This decomposition is not unique since the sets of orthogonal vectors (Schmidt bases) $\{|a_i\rangle\}$ and $\{|b_i\rangle\}$ are not unique. However, the set of Schmidt coefficients are.

Now, suppose that the state is separable (but not a product state). Then, $|\psi\rangle = \sum_i p_i |A_i\rangle \otimes |B_i\rangle$, where $p_i \in \mathbb{R}^+, \forall i$ (more than one value), and $\sum_i p_i = 1$. From the definition of being separable, there is no assumption on the sets $\{|A_i\rangle\}$ and $\{|B_i\rangle\}$. From the existence of such decomposition can I find a contradiction? For example, based on the uniquess of the Schmidt coefficients?

Would it be simpler to use the density matrix definition? The density matrix of a pure state is (i) a positive semi-definite Hermitian matrix, and (ii) a projection. Thus, its eigenvalues are non-negative real (Hermitian). Furthermore, since it is a projection, it has only one eigenvalue (equal to 1). The rank is therefore equal to 1. Now, if the state is separable, the rank of $\sum_i p_i |A_i\rangle \langle A_i| \otimes |B_i\rangle\langle B_i|$ has to be one also. But technically, the sum of two matrices of rank one can also be of rank one. So no chance to conclude that only one density matrix in the summation is not null.


1 Answer 1



By definition, a bipartite quantum state $\rho$ is a product state if $\rho=\sigma\otimes\tau$. By another definition, $\rho$ is separable if it can be written as a convex combination of product states $$ \rho=\sum_{i=1}^r p_i \sigma_i\otimes\tau_i\tag1 $$ where $p_i>0$ and $\sum_{i=1}^rp_i=1$. By yet another definition, $\rho$ is pure if $\rho=|\psi\rangle\langle\psi|$ for some $|\psi\rangle$.

Proof that every separable pure state is a product state

Consider the overlap of both sides of $(1)$ with $|\psi\rangle$ $$ \sum_{i=1}^r p_i \langle\psi|\sigma_i\otimes\tau_i|\psi\rangle=\langle\psi|\rho|\psi\rangle=1.\tag2 $$ This can only be the case if $\langle\psi|\sigma_i\otimes\tau_i|\psi\rangle=1$ for all $i=1,\dots,r$ which means that $\sigma_i\otimes\tau_i$ is the rank-one projector onto $|\psi\rangle$ for every $i$. In particular, $$ \sigma_i\otimes\tau_i=\sigma_j\otimes\tau_j\tag3 $$ for all $i,j$. By tracing out each subsystem in turn, we see that $\sigma_i=\sigma_j$ and $\tau_i=\tau_j$ for all $i,j$. But then $$ \rho=\sum_{i=1}^r p_i \sigma_1\otimes\tau_1=\sigma_1\otimes\tau_1\tag4 $$ so $\rho$ is a product state. $\square$

  • $\begingroup$ I found probably an equivalent approach by using the density matrix formulation of the separability criteria. Wlog this formulation can restrict the separable states to be convex combinations of pure states. The density $\rho$ of a pure quantum state is such that $Tr(\rho^2) = 1$. Now, assuming that $\rho = \sum_i p_i \rho_{A,i} \otimes \rho_{B,i}$, the trace of $\rho^2$ is smaller than or equal to 1 by using the Schwarz inequality. It would be equal to 1 only if all the pure states defining $\rho_{A,i}$ and $\rho_{B,i}$ are all equivalent, and thus can be reduced to one product state. $\endgroup$
    – JMark
    Mar 11 at 16:08

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