# More than one Schmidt coefficient implies entanglement?

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.

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.