How to measure a qubit in a random basis

Question

Let a two dimensional system be in the state $\phi=|0\rangle\langle0|$, for any basis $M$ spanned by the orthogonal vectors $|\psi_0\rangle,|\psi_1\rangle$, we can measure $\phi$ in basis $M$ and obtain "0" and "1" with probabilities $p_0=\mathrm{tr}(|\psi_0\rangle\langle\psi_0|\phi)$, $p_1=1-p_0$.

My question is, if I want to select $M$ randomly, is there some commonly understood way of choosing a random basis? My second question is, what would be the distribution of $p_0,p_1$ resulting with this random selection of the basis?

Answer 1

Choosing uniformly distributed points on the two dimensional bloch sphere: $0 \leqslant \phi < 2\pi, 0 \leqslant \theta \leqslant \pi$, we can construct a random vector

$ |\psi_0\rangle = cos\frac{\theta}{2} |0\rangle + e^{i\phi} sin\frac{\theta}{2} |1\rangle $

In order to obtain a uniform distribution over the sphere's surface, $\phi$ should be uniformly distributed in the interval $[0, 2\pi)$ and $cos(\theta)$ should be uniformly distributed in $[-1, 1])$. To see that, please notice that in terms of the height of the unit sphere $z = cos\theta$, the surface element is uniform:

$ dS = sin(\theta) d\theta d\phi = -dz d\phi$

This is called Archimedes' spherical sampling theorem as it was known already to Archimedes.

The required expectation:

$ p = \mathrm{tr}(|\psi_0\rangle\langle\psi_0|\phi) = cos^2\frac{\theta}{2} = \frac{1+z}{2}$

Since $p$ is linear in $z$, it is uniformly distributed in $[0, 1])$.

Answer 2

Shameless self-publicity.

For random measurements of qutrits, ququarts, ..., qudits you can look in paper I co-authored: http://arxiv.org/abs/1010.4189