# Revision class solutions

## May 22, 2018*

Revision sheet
Revision sheet solutions
Revision sheet solutions (printing)

### Exercise 1.

1. Define the trace distance $$D(\rho,\sigma)$$ between two states $$\rho,\sigma\in \mathcal{D}(\mathcal{H})$$ and prove that it can be expressed in the form: $D(\rho,\sigma) = \frac{1}{2}(\operatorname{tr}Q + \operatorname{tr}R)$ where $$Q$$ and $$R$$ are suitably defined positive semi-definite operators in $$\mathcal{B}(\mathcal{H})$$.
2. Using the above identity, prove that $D(\rho,\sigma) = \max_P \operatorname{tr}(P(\rho -\sigma))$ where the maximisation is over all projection operators $$P\in \mathcal{B}(\mathcal{H})$$.
3. Further, prove that $D(,\rho,\sigma) = \max_T \operatorname{tr}(T(\rho-\sigma))$ where the maximisation is over all positive semi-definite operators $$T\in \mathcal{B}(\mathcal{H})$$ with eigenvalues less than or equal to unity.
4. Let $$\rho$$ be a quanutm state and $$\Lambda$$ be a linear completely positive trace-preserving map. Prove that $F_e (\rho,\Lambda) \leq ( F(\rho,\Lambda(\rho)))^2$ where $$F_e(\rho,\Lambda)$$ denotes the entanglement fidelity, and $$F(\rho,\Lambda(\rho))$$ denotes the fidelity of the states $$\rho$$ and $$\Lambda(\rho)$$.

### Exercise 2.

1. State the Holevo-Schumacher-Westmoreland theorem.
1. Use it to obtain the product-state classical capacity of a qubit depolarizing channel $$\Lambda$$ defined as follows: $\Lambda(\rho) = p \rho + \frac{1-p}{3}(\sigma_x\rho\sigma_x + \sigma_y \rho \sigma_y + \sigma_z \rho \sigma_z),$ where $$\sigma_x$$, $$\sigma_y$$, and $$\sigma_z$$ are the Pauli matrices.
2. Consider an ensemble of quantum states $$\mathcal{E}= \{p_x,\rho_x\}$$ and let $$\chi(\mathcal{E})$$ denote its Holevo quantity. Let $$\Lambda$$ be a quantum channel. Prove that $\chi(\mathcal{E}') \leq \chi(\mathcal{E})$ where $$\mathcal{E}' = \{p_x,\Lambda(\rho_x)\}$$.
2. Consider a memoryless quantum information source characterized by $$\{\pi,\mathcal{H}\}$$, where $$\pi\in \mathcal{D}(\mathcal{H})$$. Suppose on $$n$$ uses, the source emits a signal state $$| \Psi_k^{(n)} \rangle\in \mathcal{H}^{\otimes n}$$ with probability $$p_k^{(n)}$$, the index $$k$$ labelling the different possible signal states. State the Typical Subspace Theorem, and use it prove that for such a source there exists a reliable compression-decompression scheme of rate $$R>S(\pi)$$ where $$S(\pi)$$ denotes the von Neumann entropy of the source.

Note: we won’t cover part (2) in the revision class, since it’s just bookwork.

1. The HSW theorem says that the product state classical capacity of a quantum channel $$\Lambda$$ is given by $C^{(1)}(\Lambda) = \chi^*(\Lambda)$ where $$\chi^*(\Lambda)$$ is the Holevo capacity, defined as $\chi^*(\Lambda) := \max_{\{p_x,\rho_x\}} \chi(\{p_x,\Lambda(\rho_x)\})$(1) where $$\chi(\cdot)$$ is the Holevo $$\chi$$-quantity, defined by $\chi(\{p_x,\Lambda(\rho_x)\}) := S(\Lambda(\sum_x p_x \rho_x)) - \sum_x p_x S(\Lambda(\rho_x))$ and the maximum in (1) is taken over all ensembes $$\{p_x,\rho_x\}$$ of possible input states $$\rho_x$$ of the channel, and $$p_x\geq 0$$, $$\sum_x p_x = 1$$.
1. Since $\frac{I}{2} = \frac{1}{4}[ \rho + \sigma_x \rho \sigma_x + \sigma_y \rho \sigma_y + \sigma_z \rho \sigma_z]$ we find that \begin{aligned} \Lambda(\rho) &= p \rho + \frac{1-p}{3} (2 I - \rho)\\ &= \rho [ p - \frac{1-p}{3}] + 4 \left( \frac{1-p}{3} \right) \frac{I}{2}\\ &= q \rho + (1-q) \frac{I}{2} \end{aligned} where $$q = p - \frac{1-p}{3} = \frac{4p-1}{3}$$. Let $$\mathcal{E}= \{ p_j, | \psi_j \rangle\langle \psi_j | \}$$ be an input ensemble of pure states. Then $\Lambda( | \psi_j \rangle\langle \psi_j |) = q | \psi_j \rangle\langle \psi_j | + (1-q)\frac{I}{2}$ which has eigenvalues $$\frac{1\pm q}{2}$$, regardless of which pure state was input. Thus, $$S(\Lambda( | \psi_j \rangle\langle \psi_j |)) = h(\frac{1+q}{2})$$ where $$h(\cdot)$$ denotes the binary Shannon entropy. Thus, since we can restrict the maximum in (1) to be only over pure states, we have \begin{aligned} C^{(1)}(\Lambda) &= \chi^*(\Lambda) \\ &= \max_{\{p_j, \rho_j\}} S(\sum_j p_j \rho_j) - \sum_j p_j h( \frac{1+q}{2})\\ &= \max_{\{p_j, \rho_j\}} S(\sum_j p_j \rho_j) - h(\frac{1+q}{2})\\ &\leq \log 2 - h(\frac{1+q}{2})\\\\ &= 1-h(\frac{1+q}{2}) \end{aligned} since the von Neummon entropy of a qubit is bounded by $$\log 2$$. By choosing the ensemble to be $$p_1 = p_2 = \frac{1}{2}$$ and $$| \psi_1 \rangle = | 0 \rangle$$, $$| \psi_2 \rangle = | 1 \rangle$$, we have $$\sum_j p_j \rho_j = \frac{I}{2}$$ which has von Neummon entropy $$\log 2=1$$. Since this upper bound is achievable, we thus have $C^{(1)}(\Lambda) = 1-h(\frac{1+q}{2})$ where $$q =\frac{4p-1}{3}$$.
2. We have that \begin{aligned} \chi(\mathcal{E}) &= S(\sum_x p_x \rho_x) - \sum_x p_x S(\rho_x) \\ &= \sum_x p_x S(\rho_x \| \rho) \end{aligned} where $$\rho = \sum_x p_x \rho_x$$. We can verify this by expanding the second form: \begin{aligned} \sum_x p_x S(\rho_x \| \rho) &= \sum_x p_x \operatorname{tr}[\rho_x (\log \rho_x - \log \rho)]\\ &=\sum_x p_x ( -S(\rho_x) - \operatorname{tr}[\rho_x \log \rho ]\\ &= - \sum_x p_x S(\rho_x) - \sum_x p_x \operatorname{tr}[\rho_x \log \rho ]\\ &= - \sum_x p_x S(\rho_x) - \operatorname{tr}[\rho \log \rho ]\\ &= S(\rho) - \sum_x p_x S(\rho_x) \end{aligned} as desired. Likewise, $$\chi(\mathcal{E}') = \sum_x p_x S(\Lambda(\rho_x) \| \Lambda(\rho))$$. But by the monotonicity of the relative entropy under CPTP maps (i.e. data-processing), we have $S(\Lambda(\rho_x) \| \Lambda(\rho)) \leq S(\rho_x \| \rho)$ for each $$x$$. Multiplying by $$p_x$$ and summing over $$x$$ yields the result that $$\chi(\mathcal{E}')\leq \chi(\mathcal{E})$$.
2. This is the first half of Theorem 2 in Notes 14.

### 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.