Exercise 3

from Revision class solutions

Exercise 3.

1. The Bell states $$| \Phi^+_{AB} \rangle$$, $$| \Phi^-_{AB} \rangle$$, $$| \Psi^+_{AB} \rangle$$, $$| \Psi^-_{AB} \rangle$$, can be characterized by two classical bits, namely the parity bit and the phase bit. Show that the latter are eigenvalues of two commuting observables.
2. The Bell states form an orthonormal basis of the two-qubit Hilbert space. It is referred to as the Bell basis. Let us denote it by $$B_1$$. A sequence of two operations can be used to convert states of the computational basis $$B_2 := \{| ij \rangle: i,j\in \{0,1\}\}$$ to the Bell states. State what these operations are. Can they also be used to convert states of $$B_1$$ to $$B_2$$? Justify your answer.
3. Prove that the Schmidt rank of a pure state cannot be increased by local operations and classical communication (LOCC), clearly stating any theorem that you use.
4. It is known that a matrix $$A$$ is doubly stochastic if and only if $$\vec x \prec \vec y$$ for all vectors $$\vec y$$, where $$x = A \vec y$$.
Let $$\rho\in \mathcal{D}(\mathcal{H})$$ be a state, where $$\dim \mathcal{H}=d$$, and let $$\Lambda:\mathcal{D}(\mathcal{H})\to \mathcal{D}(\mathcal{H})$$ be a unital channel. Let $$\vec r = (r_1,r_2,\dotsc,r_d)$$ and $$\vec s = (s_1,s_2,\dotsc,s_d)$$ respectively denote the vectors of eigenvalues of $$\rho$$ and $$\sigma = \Lambda(\rho)$$, arranged in non-increasing order. Using the above result, prove that $$\vec s\prec \vec r$$.
Note: the question originally said $$\vec r\prec \vec s$$, which was an error.
1. The parity bit is $$0$$ if it is in a $$| \Phi_{AB}^\pm \rangle$$ state, and $$1$$ if its in a $$| \Psi^\pm_{AB} \rangle$$ state. The phase bit is $$0$$ if its in a $$| \alpha^+ \rangle$$ state (for $$\alpha \in \{\Phi,\Psi\}$$), and $$1$$ if it’s in a $$| \alpha^- \rangle$$ state. Let $$X_{AB} = \sigma_x^{(A)}\otimes \sigma_x^{(B)}$$ where $$\sigma_x$$ is the Pauli $$x$$ matrix, and likewise $$Z_{AB} = \sigma_z^{(A)} \otimes \sigma_z^{(B)}$$. Then one can compute $$[X_{AB},Z_{AB}]= 0$$1. Moreover, \begin{aligned} X_{AB} | \alpha^+_{AB} \rangle &= | \alpha^+_{AB} \rangle \\ X_{AB} | \alpha^-_{AB} \rangle &= -| \alpha^-_{AB} \rangle \end{aligned} and thus the eigenvalues of $$X_{AB}$$ are the phase bit. Likewise, \begin{aligned} Z_{AB} | \Phi^{\pm}_{AB} \rangle &= | \Phi^{\pm}_{AB} \rangle\\ Z_{AB} | \Psi^{\pm}_{AB} \rangle &= - | \Psi^{\pm}_{AB} \rangle \end{aligned} so the eigenvalues of $$Z_{AB}$$ are the parity bit.

Alternatively, one can consider the projections $P_{\text{parity}} = | \Psi^+_{AB} \rangle\langle \Psi^+_{AB} |+| \Psi^-_{AB} \rangle\langle \Psi^-_{AB} |$ and $P_\text{phase} = | \Phi_{AB}^- \rangle\langle \Phi_{AB}^- | + | \Psi_{AB}^- \rangle\langle \Psi_{AB}^- |.$ Then the eigenvalues of these operators are exactly the phase and parity bit, in the sense that $P_{\text{parity}} | \Psi^\pm_{AB} \rangle = | \Psi^\pm_{AB} \rangle, \qquad P_{\text{parity}} | \Phi_{AB}^\pm \rangle = 0,$ and for $$\alpha\in\{\Phi,\Psi\}$$, $P_{\text{phase}} | \alpha^- \rangle = | \alpha^- \rangle, \qquad P_{\text{phase}} | \alpha^+ \rangle = 0.$ Moreover these projections commute.
2. Hadamard then CNOT. It is reversible.
3. Nielsen’s Majorization Theorem states that a bipartite state $$| \psi_{AB} \rangle$$ can be converted to $$| \phi_{AB} \rangle$$ if and only if $\lambda_\psi \prec \lambda_\phi$ where $$\lambda_\psi$$ and $$\lambda_\phi$$ are the vectors of eigenvalues of $$\psi_A := \operatorname{tr}_B | \psi_{AB} \rangle\langle \psi_{AB} |$$ and $$\phi_A = \operatorname{tr}_B | \phi_{AB} \rangle\langle \phi_{AB} |$$ and $$\prec$$ is the majorization pre-order. Let $$n(\psi)$$ and $$n(\phi)$$ denote the Schmidt ranks of two pure states $$| \psi_{AB} \rangle$$ and $$| \phi \rangle_{AB}$$, such that $$| \psi \rangle_{AB} \xrightarrow{\text{LOCC}} | \phi_{AB} \rangle$$. Assume $n(\psi) < n(\phi).$(1) Let $$\lambda_\psi = (\nu_1,\dotsc,\nu_d)$$ and $$\lambda_\phi = (\mu_1,\dotsc,\mu_d)$$, where $$d = \dim \mathcal{H}_A$$ is the dimension of the Hilbert space for system $$A$$. The assumption (1) implies that there exists some integer $$m\leq d$$ such that $$\mu_m \neq 0$$ but $$\nu_m = 0$$. Hence $\sum_{i=1}^{m-1}\nu_i = 1, \qquad\text{but}\qquad \sum_{i=1}^{m-1} \mu_i < 1.$ This contradicts that $$\lambda_\psi \prec \lambda_\phi$$. Therefore, (1) cannot hold, and therefore the Schmidt rank cannot increase by LOCC operations.
4. Let us write the eigendecompositions \begin{aligned} \rho &= \sum_{i=1}^d r_i | e_i \rangle\langle e_i |\\ \sigma &= \sum_{i=1}^d s_j | f_j \rangle\langle f_j | \end{aligned} where the eigenvalues are arranged in non-increasing order. We have that \begin{aligned} s_j &= \operatorname{tr}[ \sigma | f_j \rangle\langle f_j |]\\ &= \operatorname{tr}[\Lambda(\rho) | f_j \rangle\langle f_j |] &= \sum_i r_i \operatorname{tr}[\Lambda(| e_i \rangle\langle e_i |)| f_j \rangle\langle f_j |]. \end{aligned} Now define a matrix $$D$$ with entries $$D_{ji} = \operatorname{tr}[\Lambda(| e_i \rangle\langle e_i |)| f_j \rangle\langle f_j |]$$. Then we’ve shown $s_j = \sum_i D_{ji} r_i.$ Thus, it remains to show that $$D$$ is doubly-stochastic, by the result quoted in the question. We have \begin{aligned} \sum_j D_{ji} &= \sum_j \operatorname{tr}[\Lambda(| e_i \rangle\langle e_i |)| f_j \rangle\langle f_j |] \\ &= \operatorname{tr}[\Lambda(| e_i \rangle\langle e_i |) I] = \operatorname{tr}[\Lambda(| e_i \rangle\langle e_i |)] = 1 \end{aligned} since $$\Lambda$$ is trace-preserving. Next, \begin{aligned} \sum_i D_{ji} &= \sum_i \operatorname{tr}[\Lambda(| e_i \rangle\langle e_i |)| f_j \rangle\langle f_j |] \\ &= \operatorname{tr}[\Lambda( I)| f_j \rangle\langle f_j |] = \operatorname{tr}[| f_j \rangle\langle f_j |] = 1 \end{aligned} since $$\Lambda$$ is unital. Therefore, $$D$$ is doubly stochastic. Since $$\vec s= D \vec r$$, we have $$\vec s \prec \vec r$$.
1. Takes a little time and expansion; this question was from an earlier iteration of the class where this was treated.