### 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.