Problem 2.2 in Nielsen & Chuang - Properties of the Schmidt number

Question

This question has been asked here: "Problem 2.2 in Nielsen & Chuang - Properties of the Schmidt number", but no answer has been provided there yet, thus I move it here.

The problem is stated below. This is problem 2.2 (not exercise 2.2) in Nielsen & Chuang (Page 117 in the newest version).

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 $x$.

The link above provide some attempts, but not quite succeed. Any help is appreciated.

Answer 1

Following your attempt we have $$ \begin{align*} | \phi \rangle &= \sum_i \phi_i | a_i^{\phi} \rangle | b_i^{\phi} \rangle\\ | \gamma \rangle &= \sum_i \gamma_i | a_i^{\gamma} \rangle | b_i^{\gamma} \rangle, \end{align*} $$ $$ \rho \equiv \operatorname{tr}_B(| \psi \rangle \langle \psi |) = |\alpha|^2 \rho_{\phi \phi} + |\beta|^2 \rho_{\gamma \gamma} + \alpha \bar{\beta} \rho_{\phi \gamma} + \bar{\alpha} \beta \rho_{\gamma \phi}, $$ where $$ \rho_{\phi \phi} \equiv \operatorname{tr}_B(| \phi \rangle \langle \phi |) = \sum_i \phi_i^2 | a_i^{\phi} \rangle \langle a_i^{\phi} | \\ \rho_{\gamma \gamma} \equiv \operatorname{tr}_B(| \gamma \rangle \langle \gamma |) = \sum_i \gamma_i^2 | a_i^{\gamma} \rangle \langle a_i^{\gamma} | \\ \rho_{\phi \gamma}\equiv \operatorname{tr}_B(| \phi \rangle \langle \gamma |) = \sum_i \phi_i | a_i^{\phi} \rangle \sum_j \gamma_j \langle b_j^{\gamma} | b_i^{\phi} \rangle \langle a_j^{\gamma} | \\ \rho_{\gamma \phi}\equiv \operatorname{tr}_B(| \gamma \rangle \langle \phi |) = \sum_i \gamma_i | a_i^{\gamma} \rangle \sum_j \phi_j \langle b_j^{\phi} | b_i^{\gamma} \rangle \langle a_j^{\phi} |. $$ Now observe that for the following subspace of the first subsystem $$ H_\alpha = \text{span}\{| a_1^{\phi} \rangle, | a_2^{\phi} \rangle, ... ; | a_1^{\gamma} \rangle, | a_2^{\gamma} \rangle, ... \} $$ we have that $$ \rho_{\phi \phi} (H_\alpha^\perp) = \rho_{\gamma \gamma} (H_\alpha^\perp) = \rho_{\phi \gamma} (H_\alpha^\perp) = \rho_{\gamma \phi} (H_\alpha^\perp) = 0 $$ Thus $\rho(H_\alpha^\perp)=0$, which means that $\text{rank}(\rho)\leq \text{dim}(H_\alpha)$. Hence $$ \text{Sch}(\psi) \leq \text{Sch}(\phi) + \text{Sch}(\gamma) $$ This is just a triangle inequality. To obtain the necessary just apply the above inequality to $\alpha \rvert \phi \rangle = |\psi\rangle - \beta \rvert \gamma \rangle$ and $\beta \rvert \gamma \rangle = |\psi\rangle - \alpha \rvert \phi \rangle $

Answer 2

$\newcommand{\bra}[1]{\left<#1\right|} \newcommand{\ket}[1]{\left|#1\right>} \newcommand{\sch}[1]{\operatorname{Sch}\left( #1 \right)} \newcommand{\cases}[3]{ #1 = \begin{cases} #2 \hspace{1em} \text{if} \hspace{1em} i \le \sch{\gamma} \\ #3 \hspace{1em} \text{otherwise} \end{cases}} $If $\sch{\phi} = \sch{\gamma}$ then we are done because $\sch{\psi} \ge 0$ always holds for a pure state.

If $\sch{\phi} \ne \sch{\gamma}$ then let's assume that the statement of the problem does not hold, so we have: $$ \sch{\psi} \lt | \sch{\phi} - \sch{\gamma}| \tag{1}\label{1} $$ Without loss of generality we can assume that $\sch{\phi} > \sch{\gamma}$. From this we have: $$ \sch{\phi} > \sch{\gamma} + \sch{\psi} \tag{2}\label{2} $$ We can write: $$ \ket{\phi} = \frac{\beta}{\alpha} \ket{\gamma} - \frac{1}{\alpha} \ket{\psi} \tag{3}\label{3} $$ We can write the Schmidt decomposition of $\ket{\gamma}$ and $\ket{\psi}$ as: $$ \ket{\gamma} = \sum_{i=1}^{\sch{\gamma}} \gamma_i \ket{a_i^{\gamma}} \ket{b_i^{\gamma}} \\ \ket{\psi} = \sum_{i=1}^{\sch{\psi}} \psi_i \ket{a_i^{\psi}} \ket{b_i^{\psi}} $$ Plugging these into $\eqref{3}$, we have: $$ \ket{\phi} = \sum_{i=1}^{\sch{\gamma} + \sch{\psi}} \delta_i \ket{a_i^{\delta}} \ket{b_i^{\delta}} \tag{4}\label{4} $$ where $$ \cases{\delta_i }{\frac{\beta}{\alpha} \gamma_i}{-\frac{1}{\alpha} \psi_i}\\ \cases{a_i^{\delta}}{a_i^{\gamma}}{a_i^{\psi}}\\ \cases{b_i^{\delta}}{b_i^{\gamma}}{b_i^{\psi}} $$ Now we can invoke the result of Problem 2.2 (2) from Nielsen and Chuang. This says:

Suppose $\ket{\psi} = \sum_j \ket{\alpha_j}\ket{\beta_j}$ is a representation for $\ket{\psi}$, where $\ket{\alpha_j}$ and $\ket{\beta_j}$ are (un-normalized) states for systems $A$ and $B$, respectively. Prove that the number of terms in such a decomposition is greater than or equal to the Schmidt number of $\psi$, $\sch{\psi}$.

Applying this to \eqref{4}, we obtain: $$ \sch{\phi} \le \sch{\gamma} + \sch{\psi} \tag{5}\label{5} $$ This contradicts $\eqref{2}$ which is derived from $\eqref{1}$, so we must have $\sch{\psi} \ge | \sch{\phi} - \sch{\gamma}|$.