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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.

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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.

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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.

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    $\begingroup$ Indeed, the hint in the book is very bad. Instead, the book should encourage students to train their basic logic skills. It is enough to show the implication "product state" $\Rightarrow$ "Schmidt rank is 1" (which is trivial). The negation of this is "Schmidt rank $\neq$ 1" $\Rightarrow$ "Not a product state". And you're done. $\endgroup$ Oct 23 '20 at 7:54

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