# Exercise 2

## from Example Sheet 3

### Exercise 2.

One Fuchs and van de Graaf inequality.

1. Show that for two pure qubit states $$\psi$$ and $$\phi$$, we have that $D(\psi,\phi) = \sqrt{1 - F(\psi,\phi)^2}.$ where $$D$$ is the trace distance, and $$F$$ the fidelity, defined in the lectures.
2. Using this, show that for any two mixed qubit states $$\rho$$ and $$\sigma$$, $D(\rho,\sigma) \leq \sqrt{1- F(\rho,\sigma)^2}.$
1. For two pure states, we have $$F(\psi, \phi) = |\langle \psi|\phi \rangle|$$. We may write $| \phi \rangle = \cos (\theta) | \psi \rangle + \sin(\theta) | \psi^\perp \rangle$(1) for some vector $$| \psi^\perp \rangle$$ orthogonal to $$| \psi \rangle$$. In these terms, $$F(\psi, \phi) = |\cos \theta|$$. To calculate the trace distance, we want to simplify the form of $$\psi - \phi = | \psi \rangle\langle \psi | -| \phi \rangle\langle \phi |$$. Using (1), we find $\phi = \cos^2(\theta)\, \psi + \sin (\theta)\cos(\theta) | \psi^\perp \rangle\langle \psi | + \cos (\theta)\sin(\theta) | \psi \rangle\langle \psi^\perp | + \sin^2 (\theta) \, \psi^\perp.$ In the basis $$\{| \psi \rangle,| \psi^\perp \rangle\}$$, we find $\psi - \phi = \begin{pmatrix} 1 -\cos^2(\theta) & - \sin (\theta)\cos (\theta) \\ - \sin(\theta)\cos(\theta) & - \sin^2(\theta)\end{pmatrix}$ which has eigenvalues $$\pm |\sin\theta|$$. Thus, $\frac{1}{2}\|\psi - \phi\|_1 = |\sin\theta|.$ Since $$\sqrt{1- \cos^2(\theta)} = |\sin\theta|$$, we therefore have $D(\psi,\phi) = \sqrt{1- \cos^2(\theta)} = \sqrt{1- F(\psi,\phi)^2}.$ Remark. In fact, this proof holds for qudits.
2. Let $$\rho = \rho_A$$ have purification $$\phi_{AR}$$ and $$\sigma = \sigma_A$$ have purification $$\psi_{AR}$$ such that $F(\rho_A,\sigma_A) = |\langle \phi_{AR}|\psi_{AR} \rangle|$ using Uhlmann’s theorem. Then $\frac{1}{2}\|\psi_{AR} - \phi_{AR}\|_1 = \sqrt{1 - F(\psi_{AR},\phi_{AR})} = \sqrt{1 - F(\rho_A,\sigma_A)}.$(2) Since the partial trace is a linear CPTP map, and therefore a quantum operation, we may use the monotonicity of the trace distance under quantum operations to find $\frac{1}{2}\|\psi_{AR} -\phi_{AR}\|_1 \geq \frac{1}{2}\|\rho_A-\sigma_A\|_1$ which concludes the proof by (2).