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### Exercise 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 \(\mathcal{B}(\mathcal{H})\).
- Using the above identity, prove that \[ D(\rho,\sigma) = \max_P \operatorname{tr}(P(\rho -\sigma)) \] where the maximisation is over all projection operators \(P\in \mathcal{B}(\mathcal{H})\).
- 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.
- 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.

- State the Holevo-Schumacher-Westmoreland theorem.
- 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.
- 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 \(\mathcal{E}' = \{p_x,\Lambda(\rho_x)\}\).

- 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.*

- 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 \(p_x\geq 0\), \(\sum_x p_x = 1\).- 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 \(q =\frac{4p-1}{3}\).
- 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 \(\chi(\mathcal{E}')\leq \chi(\mathcal{E})\).

- This is the first half of Theorem 2 in Notes 14.

### Exercise 3.

- 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.
- 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.
- 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. - 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.

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 \([X_{AB},Z_{AB}]= 0\)

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.^{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.- Hadamard then CNOT. It is reversible.
- 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.
- 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 \(\vec s \prec \vec r\).

Takes a little time and expansion; this question was from an earlier iteration of the class where this was treated.↩

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