### Exercise 5.

Alice prepares a photon in one of two polarization states, given by the kets \(|a\rangle := |1\rangle\), and \(|b\rangle :=\sin \theta |0\rangle + \cos \theta |1\rangle\), depending on the outcome of a fair coin toss. If the outcome is heads, she prepares the state \(|a\rangle\). Otherwise she prepares the state \(|b\rangle\). Evaluate the Holevo \(\chi\) quantity for her ensemble of states. (Use the convention that \(|0\rangle =(1\,\,0)^T\) and \(|1\rangle = (0\,\,\,1)^T\)). For what value of \(\theta\) is the Holevo bound achieved? Explain why.

The average state is given by \[\begin{aligned} \rho &= \frac{1}{2} | 1 \rangle\langle 1 | + \frac{1}{2} | b \rangle \langle b |\\ &= \frac{1}{2} [ | 1 \rangle\langle 1 | + \sin^2\theta | 0 \rangle\langle 0 | + \cos^2\theta| 1 \rangle\langle 1 | + \sin \theta \cos \theta [ | 0 \rangle\langle 1 | + | 1 \rangle\langle 0 |] ]\\ &= \frac{1}{2} \begin{pmatrix} \sin^2 \theta & \sin \theta \cos \theta \\ \sin \theta \cos \theta & 1 + \cos^2 \theta \end{pmatrix} \end{aligned}\] which has eigenvalues \(\frac{1}{2} ( 1 \pm \cos \theta)\). Since the two states are pure, each has zero entropy. The Holevo quantity is therefore simply \(h(\frac{1}{2} ( 1 \pm \cos \theta))\) where \(h\) is the binary entropy. Since the binary entropy is maximized at \(\frac{1}{2}\), this quantity is maximized when \(\cos \theta =0\), i.e. \(\theta = \frac{\pi}{2} + \pi z\) for any \(z\in \mathbb{Z}\).