# Exercise 4

## from Example Sheet 4

### Exercise 4.

Use the HSW theorem to find the product state capacity of the depolarizing channel, $$\Lambda$$, defined by $\Lambda(\rho) = p\rho + (1-p) \frac{I}{2}.$

Note we are only dealing with qubits (otherwise $$\Lambda$$ would not be trace-preserving). First, note that for any unitary $$U$$, we have that $\Lambda(U \rho U^*) = pU\rho U^* + (1-p) \frac{I}{2}$ and therefore $$\Lambda(U \rho U^*) = U \Lambda(\rho)U^*$$.

The HSW theorem gives that the product state capacity of $$\Lambda$$ is given by $\chi^*(\Lambda) = \max_{\{p_x, \rho_x\}} S( \Lambda(\sum_x p_x \rho_x)) - \sum_x p_x S(\Lambda(\rho_x)).$ We can reduce to pure state ensembles $$\{p_x,\psi_x\}$$. Then defining the average state $$\rho := \sum_x p_x \psi_x$$, we consider \begin{aligned} S( \Lambda(\rho)) - \sum_x p_x S(\Lambda(\psi_x)) \end{aligned} Since all pure states can be related by unitaries, we have that for each $$x$$ and $$y$$, there is a unitary $$U$$ such that $$\psi_x = U \psi_y U^*$$. Thus, $S(\Lambda(\psi_x)) = S( \Lambda( U \psi_y U^*)) = S(U \Lambda(\psi_y) U^*) = S(\Lambda(\psi_y))$ using that the entropy is unitarily invariant. Thus, $$S(\Lambda(\psi_x))$$ does not depend on $$x$$. To calculate its value, we can extend $$| \psi_x \rangle$$ to an orthonormal basis given by $$\{| \psi_x \rangle, | \psi_x \rangle^\perp\}$$. Then in this basis, $\Lambda(\psi_x) = p\psi_x + (1-p) \frac{I}{2} = \begin{pmatrix} p + \frac{1-p}{2} & 0 \\ 0 & \frac{1-p}{2} \end{pmatrix}.$ Thus, $$S(\Lambda(\psi_x)) = h\left( \frac{1-p}{2}\right)$$ where $$h$$ is the binary entropy. Since $$\sum_x p_x =1$$, we have $S( \Lambda(\rho)) - \sum_x p_x S(\Lambda(\psi_x)) = S( \Lambda(\rho)) - h\left( \frac{1-p}{2}\right).$ It remains to maximize the output entropy $$S( \Lambda(\rho))$$ over all states $$\rho$$. Since $$\Lambda$$ is unitarily invariant, we can work in the eigenbasis of $$\rho$$, in which case $$\rho = \begin{pmatrix} \lambda & 0 \\ 0 & 1- \lambda \end{pmatrix}$$ for some number $$0\leq \lambda\leq 1$$. In this basis, we have $\Lambda(\rho) = \begin{pmatrix} p \lambda + \frac{1-p}{2}& 0 \\ 0 & p(1- \lambda) + \frac{1-p}{2} \end{pmatrix}.$ Thus, $S(\Lambda(\rho)) = h\left(p \lambda + \frac{1-p}{2} \right) = h\left( \frac{1}{2} + p( \lambda - \frac{1}{2})\right)$ Since the binary entropy is maximized at $$\frac{1}{2}$$ at which point it takes the value $$\log_2 2 = 1$$ (since, e.g. the entropy is maximized on a uniform distribution), $$S(\Lambda(\rho))$$ is maximized when $$\lambda = \frac{1}{2}$$. Thus, $\chi^*(\Lambda) = 1- h\left( \frac{1-p}{2}\right)$ gives the product state capacity of the depolarizing (qubit) channel.