Maximum overlap between an entangled state and a product state

Question

Say we have an entangled state $\left| \phi \right\rangle_{AB}$ and a product state $ \left| \psi \right\rangle_A \otimes \left| \psi \right\rangle_B$ , how can we show that their maximum overlap is strictly less than one, i.e. $$ \operatorname{max}_{\left| \psi \right\rangle_A, \left| \psi \right\rangle_B} \left| \left\langle \phi _{AB}\right | (\left| \psi \right\rangle_A \otimes \left| \psi \right\rangle_B) \right|^2 < 1.$$

This is intuitive, since an entangled state would not lie 100% in the direction of a product state.

We could do Schmidt decomposition on the entangled state, and recall that a state $\left| \phi \right\rangle_{AB}$ is entangled if and only if it has more than one nonzero Schmidt coefficients (i.e. it has Schmidt rank $r>1$). But I am not sure how to proceed from there.

Answer 1

This can be shown by contradiction; that is, we assume that the overlap of the two states is one, and show that then, $\lvert\phi\rangle_{AB}$ must be entangled:

If the overlap of two states is one, they are equal. This implies that $$ \lvert\phi_{AB}\rangle = \lvert\psi_A\rangle\otimes \lvert\psi_B\rangle\ . $$ Since this is is automatically also its Schmidt decomposition, it implies that $\lvert\phi_{AB}\rangle$ has only one non-zero Schmidt coefficient (i.e. Schmidt rank 1) and thus must be a product state.

Answer 2

The finite dimensional Hilbert space $\mathbb{C}^d$, as well as the related projective space $\mathbb{CP}^{d-1}$, are compact spaces. Thus, the maximum of $$ \left| \left\langle \phi _{AB}\right | (\left| \psi \right\rangle_A \otimes \left| \psi \right\rangle_B) \right|^2 $$ over $\left| \psi \right\rangle_A, \left| \psi \right\rangle_B$ exists indeed (in general only the supremum can be defined in other spaces). Thus, there must exist states $\left| \psi \right\rangle_A, \left| \psi \right\rangle_B$ for which $$ \left| \left\langle \phi _{AB}\right | (\left| \psi \right\rangle_A \otimes \left| \psi \right\rangle_B) \right|^2 = m, $$ where $m$ is the maximum. If $m$ is 1 then $|\phi _{AB}\rangle$ and $\left| \psi \right\rangle_A \otimes \left| \psi \right\rangle_B$ should be collinear (be the same in $\mathbb{CP}^{d-1}$), which is impossible, since one is entagled and other is not.

You can use the Schmidt decomposition of $|\phi _{AB}\rangle$ to actually find the maximal value $m$. It will be the largest Schmidt coefficient squared.