### Exercise 2.

**One Fuchs and van de Graaf inequality.**

- 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. - Using this, show that for any two mixed qubit states \(\rho\) and \(\sigma\), \[ D(\rho,\sigma) \leq \sqrt{1- F(\rho,\sigma)^2}. \]

- 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. - 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).