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