# Exercise 5

## from Example Sheet 4

### 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 $$z\in \mathbb{Z}$$.