# Example Sheet 3

## March 17, 2018*

Example sheet 3
Example sheet 3 solutions
Example sheet 3 solutions (printing)

### Exercise 1.

Given two qubit states $$\rho$$ and $$\omega$$ with Bloch vectors $$\vec r$$ and $$\vec s$$ respectively, show that $\|\rho-\omega\|_1 = \|\vec r - \vec s\|_2.$ The two norm of a vector $$\vec v = (v_x,v_y,v_z)$$ is $$\|\vec v\|_2 = \sqrt{v_x^2 + v_y^2 + v_z^2}$$.

We have $\rho = \frac{1}{2}(I+ \vec r \cdot \vec \sigma), \qquad \omega = \frac{1}{2}(I+ \vec s \cdot \vec \sigma),$ so $\|\rho - \sigma\|_1 = \frac{1}{2}\| (\vec r - \vec s)\cdot \vec \sigma\|_1 =\frac{1}{2} \left\| \begin{pmatrix} v_z & v_x - i v_y \\ v_x + i v_y & - v_z \end{pmatrix}\right\|_1$ where $$\vec v = \vec r - \vec s$$, just as from exercise 1 from example sheet 2. We can use that for any self-adjoint matrix $$A$$, we have $$\|A\|_1 = \operatorname{tr}[|A|] = \sum_{j} |\lambda_j|$$ where $$\lambda_j$$ are the eigenvalues of $$A$$. Thus, we simply need to calculate the eigenvalues of the above $$2\times 2$$ matrix. We find the eigenvalues are $\lambda_\pm = \pm \sqrt{ v_x^2 + v_y^2 + v_z^2} = \pm \| \vec r - \vec s\|_2.$ Thus, $$\|\rho - \sigma\|_1 = \frac{1}{2}(|\lambda_-| + |\lambda_+|) = \|\vec r - \vec s\|_2$$.

### 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).
Show that for any two states $$\rho$$ and $$\sigma$$ and any unitary $$U$$, $F(U \rho U^\dagger,U \sigma U^\dagger) = F(\rho,\sigma)$ by using the fact that for any positive semi-definite operator $$A$$ and unitary $$U$$, we have $$\sqrt{U A U^\dagger} = U \sqrt{A} U^\dagger$$.
We have \begin{aligned} F(U\rho U^\dagger, U\sigma U^\dagger) &= \| \sqrt{U \rho U^\dagger}\sqrt{U \sigma U^\dagger}\|_1 \\ &= \| U \sqrt{\rho} U^\dagger U \sqrt{\sigma}U^\dagger\|_1\\ &= \| U \sqrt{\rho}\sqrt{\sigma}U^\dagger\|_1\\ &= \max_{V} | \operatorname{tr}[ V U \sqrt{\rho}\sqrt{\sigma}U^\dagger ]| \\ &= \max_{V} | \operatorname{tr}[ U^\dagger V U \sqrt{\rho}\sqrt{\sigma} ]| \\ &= \max_{UVU^\dagger} | \operatorname{tr}[ U^\dagger U V U^\dagger U \sqrt{\rho}\sqrt{\sigma} ]| \\ &= \max_{UVU^\dagger} | \operatorname{tr}[\sqrt{\rho}\sqrt{\sigma} ]| \\ &= \|\sqrt{\rho}\sqrt{\sigma}\|_1 \\ &= F(\rho,\sigma). \end{aligned} We use the characterization of the trace distance as a maximum over unitary operators $$V$$, and along the way proved that the trace norm was invariant under unitary conjugation. We also used that maximizing over all $$V$$ was the same as maximizing over all $$U V U^\dagger$$, since $\{U V U^\dagger: V \text{ unitary}\} = \{ V : V \text{ unitary}\}.$