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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?

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up vote 3 down vote accepted

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 $[0, 1])$.

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@Niel de Beaudrap The height $z$ is uniformly distributed, $\theta$ is not uniformly distributed. The surface area of all bands $z = z_0\pm\epsilon$ is equal $4 \pi \epsilon$. Please see This is the Archimedes' spherical sampling theorem – David Bar Moshe Mar 19 '13 at 13:53
Right, read too quickly. – Niel de Beaudrap Mar 19 '13 at 14:07
Thanks a lot, it was the answer I was looking for. – Ando Masahashi Mar 19 '13 at 15:35

Shameless self-publicity.

For random measurements of qutrits, ququarts, ..., qudits you can look in paper I co-authored:

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While your at it in self-publicity, could you please write a short summary in `laymens' terms so we don't have to search through the entire paper for an answer – Michiel Mar 20 '13 at 12:29

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