I am trying to show that $|\langle \psi|_{A} \otimes \langle \phi|_{B}|\theta\rangle_{AB}|^{2}<1$ given $|\theta\rangle$ is an entangled state, and as such has schmidt rank >1. Decomposing it, we get $$|\theta\rangle=\sum_{i}\lambda_{i}|i\rangle_{A}|i\rangle_{B}$$ so $$\langle \psi|_{A} \otimes \langle \phi|_{B}|\theta\rangle_{AB}=\sum_{i}\lambda_{i}\langle\psi|i\rangle_{A}\langle\phi|i\rangle_{B}$$ which means $$\begin{align}|\langle \psi|_{A} \otimes \langle \phi|_{B}|\theta\rangle_{AB}|^{2}&=\left(\sum_{i}\lambda_{i}\langle\psi|i\rangle_{A}\langle\phi|i\rangle_{B}\right)\left(\sum_{j}\lambda_{j}\langle\psi|j\rangle_{A}\langle\phi|j\rangle_{B}\right)^{*}\\&=\sum_{i,j}\lambda_{i}\lambda_{j}^{*}\langle\psi|i\rangle_{A}\langle j|\psi\rangle_{A}\langle\phi|i\rangle_{B}\langle j|\phi\rangle_{B}\end{align}$$

Now I am told to use the Cauchy Schwarz, and immediately it can be used on the 4 complex numbers resulting from the I.N. here, so we get $$\sum_{i,j}\lambda_{i}\lambda_{j}^{*}\langle\psi|i\rangle_{A}\langle j|\psi\rangle_{A}\langle\phi|i\rangle_{B}\langle j|\phi\rangle_{B}\le\sum_{i,j}\lambda_{i}\lambda_{j}^{*}$$

However, this last term is not strictly upper bound by 1. The only thing I can think of is that, due to the choice of the product state being free, you would choose them so the overlap with the product state assigned to the largest Schmidt coefficient would be 1, the modulus squared of which has to be less than 1. However, I am not sure how to present a tidy proof of this.


1 Answer 1


TL;DR: There is no need to use Schmidt decomposition. The non-strict variant of the inequality follows directly from Cauchy-Schwarz inequality and equality is ruled out because $|\theta\rangle_{AB}$ is entangled and therefore not a scalar multiple of a product state.

Recall that by Cauchy-Schwarz inequality for any vectors $|u\rangle$ and $|v\rangle$, we have

$$ |\langle u|v\rangle|^2 \le \langle u|u\rangle\langle v|v\rangle\tag1 $$

with equality if and only if $|u\rangle$ is a scalar multiple of $|v\rangle$ or $|v\rangle$ a scalar multiple of $|u\rangle$. Setting

$$ \begin{align} |u\rangle &:= |\psi\rangle_A\otimes|\phi\rangle_B\\ |v\rangle &:= |\theta\rangle_{AB} \end{align}\tag2 $$

and substituting into $(1)$, we obtain

$$ |\langle\psi|_A\otimes\langle\phi|_B|\theta\rangle_{AB}|^2\le\langle\psi|\psi\rangle_A\langle\phi|\phi\rangle_B\langle\theta|\theta\rangle_{AB}\tag3 $$

with equality if and only if $|\psi\rangle_A\otimes|\phi\rangle_B$ is a scalar multiple of $|\theta\rangle_{AB}$ or $|\theta\rangle_{AB}$ a scalar multiple of $|\psi\rangle_A\otimes|\phi\rangle_B$. Therefore, if $|\psi\rangle_A$, $|\phi\rangle_B$ and $|\theta\rangle_{AB}$ are all normalized, we have

$$ |\langle\psi|_A\otimes\langle\phi|_B|\theta\rangle_{AB}|^2 \le 1\tag4 $$

with equality if and only if

$$ |\theta\rangle_{AB}=\lambda\,|\psi\rangle_A\otimes|\phi\rangle_B\tag5 $$

for a non-zero $\lambda\in\mathbb{C}$. However, $(5)$ is satisfied if and only if $|\theta\rangle_{AB}$ is a product state. Since $|\theta\rangle_{AB}$ is entangled we see that $(5)$ does not hold and therefore $(4)$ becomes strict

$$ |\langle\psi|_A\otimes\langle\phi|_B|\theta\rangle_{AB}|^2 < 1\tag6 $$

as desired.

  • 1
    $\begingroup$ Thank you for this. $\endgroup$ Dec 15, 2021 at 12:08

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