# Exercise 2

## from Revision class solutions

### 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.