3
$\begingroup$

I'm studying Quantum Information Theory by Mark M. Wilde, and I got stuck in solving the exercise 3.8.2 in page 100. The exercise is to show that a pure bipartite state is entangled if and only if it has more than one Schmidt coefficient. The hint below the problem says that

\begin{equation*} \max_{|φ\rangle_A,|ψ\rangle_B} |\langleφ|_A ⊗ \langleψ|_B |ϕ\rangle_{AB}|^2 < 1 \end{equation*}

if there is more than one Schmidt coefficient for the state $|ϕ_{AB}\rangle$. Also, the hint suggests using the Schwarz inequality. It is clear that one Schmidt coefficient implies a product state, but I cannot figure out how to show the strict inequality above. I appreciate any help.

$\endgroup$
1
$\begingroup$

You're asking how to prove \begin{equation*} \max_{|φ\rangle_A,|ψ\rangle_B} |\langleφ|_A ⊗ \langleψ|_B |ϕ\rangle_{AB}|^2 < 1 \end{equation*} as opposed to actually answering the question?

To prove this, consider the Schmidt decomposition of $|ϕ\rangle_{AB}=\sum_i\lambda_i|\phi^i_A\rangle|\phi^i_B\rangle$, and let $$ |\gamma_A\rangle=\sum_i\lambda_i|i\rangle \langle\varphi|\phi^i_A\rangle,\qquad |\gamma_B\rangle=\sum_i|i\rangle\langle\psi|\phi^i_B\rangle. $$ Note that $|\gamma_B\rangle$ has norm 1, $\langle\gamma_B|\gamma_B\rangle=1$ by the completeness relation of the orthonormal basis $|\phi^i_B\rangle$. On the other hand, $|\gamma_A\rangle$ only has norm 1 if $|\phi_{AB}\rangle$ has one Schmidt coefficient, since otherwise $\sum_i\lambda_i^2|\langle\varphi|\psi^i_A\rangle|^2<\sum_i|\langle\varphi|\psi^i_A\rangle|^2=1$.

Bringing this together for the case of more than one Schmidt coefficient, $$ |\langle\varphi\psi|\phi_{AB}\rangle|^2=|\langle\gamma_A|\gamma_B\rangle|^2\leq \langle\gamma_A|\gamma_A\rangle\langle\gamma_B|\gamma_B\rangle<1 $$ having used the Schwartz inequality.

| improve this answer | |
$\endgroup$
1
$\begingroup$

I'd ignore that hint and instead try to prove the contrapositives: show that a pure state is a product state iff it has one (nonzero) Schmidt coefficient.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.