Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
add comment

2 Answers

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 $[0, 1])$.

share|improve this answer
1  
@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 mathworld.wolfram.com/Zone.html. 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
add comment

Shameless self-publicity.

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

share|improve this answer
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.