I started reading the Nielsen & Chuang's Quantum Computation and Quantum Information. I got stuck by the last question of Problem 2.2. I got the other problems, but I can't see this one. I guess it's not really difficult, but as I am new in this field, some help will be nice. :)
Problem statement:
Suppose $|\psi\rangle $ is a pure state of a composite system with components $A$ and $B$, such that: $$|\psi\rangle = \alpha \rvert \phi \rangle + \beta \rvert \gamma \rangle$$ Prove that: $$ \operatorname{Sch}(\psi) \geq | \operatorname{Sch}(\phi) - \operatorname{Sch}(\gamma)|$$ where $\operatorname{Sch}(x)$ is the Schmidt number of the pure state labeled $x$.
My attempt:
Here is what I tried. Let's assume that $ \operatorname{Sch}(\phi) > \operatorname{Sch}(\gamma)$. If we write the Schmidt decomposition of $\phi$ and $\gamma$: \begin{align*} | \phi \rangle &= \sum_i \phi_i | a_i^{\phi} \rangle | b_i^{\phi} \rangle\\ | \gamma \rangle &= \sum_i \gamma_i | a_i^{\gamma} \rangle | b_i^{\gamma} \rangle, \end{align*} we can then calculate the partial trace of $\psi$ regarding component $A$: $$ \rho \equiv \operatorname{tr}_B(| \psi \rangle \langle \psi |) = |\alpha|^2 \rho_{\phi \phi} + |\beta|^2 \rho_{\gamma \gamma} + \alpha \bar{\beta} \rho_{\phi \gamma} + \bar{\alpha} \beta \rho_{\gamma \phi}, $$
where:
$$ \rho_{\phi \phi} \equiv \operatorname{tr}_B(| \phi \rangle \langle \phi |) = \sum_i \phi_i^2 | a_i^{\phi} \rangle \langle a_i^{\phi} | \\ \rho_{\gamma \gamma} \equiv \operatorname{tr}_B(| \gamma \rangle \langle \gamma |) = \sum_i \gamma_i^2 | a_i^{\gamma} \rangle \langle a_i^{\gamma} | \\ \rho_{\phi \gamma}\equiv \operatorname{tr}_B(| \phi \rangle \langle \gamma |) = \sum_i \phi_i | a_i^{\phi} \rangle \sum_j \gamma_j \langle b_j^{\gamma} | b_i^{\phi} \rangle \langle a_j^{\gamma} | \\ \rho_{\gamma \phi}\equiv \operatorname{tr}_B(| \gamma \rangle \langle \phi |) = \sum_i \gamma_i | a_i^{\gamma} \rangle \sum_j \phi_j \langle b_j^{\phi} | b_i^{\gamma} \rangle \langle a_j^{\phi} |. $$
It can be seen that $\operatorname{Im}(\rho_{\phi \gamma})$ is a subspace of $\operatorname{Im}(\rho_{\phi})$, and similarly that $\operatorname{Im}(\rho_{\gamma \phi})$ is a subspace of $\operatorname{Im}(\rho_{\gamma})$. We define $ P_{\gamma} $ the projector onto $\operatorname{Im}(\rho_{\gamma})$ and $$ P_{\gamma}^{\perp} = I - P_{\gamma} $$ the projection onto the orthogonal complement of $\operatorname{Im}(\rho_{\gamma})$. As $\operatorname{Sch}(\gamma) < \operatorname{Sch}(\phi)$, the subspace corresponding to $P_{\gamma}^{\perp}$ is not reduced to the zero vector. We end up with: $$ \rho = P_{\gamma}( |\alpha|^2 \rho_{\phi \phi} + |\beta|^2 \rho_{\gamma \gamma} + \alpha \bar{\beta} \rho_{\phi \gamma} + \bar{\alpha} \beta \rho_{\gamma \phi} ) + P_{\gamma}^{\perp} ( |\alpha|^2 \rho_{\phi \phi} + \alpha \bar{\beta} \rho_{\phi \gamma} ). $$
As the subspaces corresponding to the two defined projectors are orthogonal: \begin{align*} \operatorname{rank}(\rho) &= \operatorname{rank}(P_{\gamma}( |\alpha|^2 \rho_{\phi \phi} + |\beta|^2 \rho_{\gamma \gamma} + \alpha \bar{\beta} \rho_{\phi \gamma} + \bar{\alpha} \beta \rho_{\gamma \phi} ))\\ &+ \operatorname{rank}(P_{\gamma}^{\perp} ( |\alpha|^2 \rho_{\phi \phi} + \alpha \bar{\beta} \rho_{\phi \gamma} )) \\ &\geq \operatorname{rank}(P_{\gamma}^{\perp} ( |\alpha|^2 \rho_{\phi \phi} + \alpha \bar{\beta} \rho_{\phi \gamma} )), \end{align*} I was hoping to conclude by claiming that: $$ \operatorname{rank}(P_{\gamma}^{\perp} ( |\alpha|^2 \rho_{\phi \phi} + \alpha \bar{\beta} \rho_{\phi \gamma} )) \geq \operatorname{Sch}(\phi) - \operatorname{Sch}(\gamma), $$ but it is not correct, as I find counterexamples. Therefore I don't think that I am following the right track.
