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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.

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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 $

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