# Exercise 1

## from Example Sheet 3

### 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, $$\|\rho - \sigma\|_1 = \frac{1}{2}(|\lambda_-| + |\lambda_+|) = \|\vec r - \vec s\|_2$$.