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.

I'm looking at Section 2.4.1 of Nielsen and Chuang's Quantum Computation and Quantum Information were they derive the density operator versions of the evolution and measurement postulates of quantum mechanics and something is bugging me.

Let

$\{(p(i), | x_i \rangle) \colon i=1,2,...,n\}$

be an ensemble and suppose you perform a measurement on this ensemble that results in outcome $m$. Then the post-measurement ensemble is

$\{(p(i|m), | x'_i \rangle) \colon i=1,2,...,n\}$

where $p(i|m)$ is the probability that the state of the system was originally $| x_i \rangle$ given that the measurement outcome was $m$. It seems natural enough to use $p(i|m)$ for the probabilities of the post-measurement ensemble. However, mathematically, I do not see why that should be the case.

My question then is this: Given only the definition of what an ensemble is and the state vector version of the postulates of QM, is there a way to derive the rules to compute the post-measurement probabilities of an ensemble?

share|improve this question
    
Good question (and welcome to Physics Stack Exchange)! I'll have to think about this one but I'll see if I can post an answer later. Or perhaps someone else will answer you first. –  David Z Jul 22 '11 at 18:10
add comment

1 Answer

up vote 1 down vote accepted

This is just classical probability theory, note in particular the theorem

$p(m)*p(i|m) = p(i)*p(m|i)\;\;[=p(i\text{ and } m)]$.

Repeat the experiment an enormous number of times $N$. Out of these repetitions, $N*p(i)$ are the state $|x_i\rangle$. Out of these, $N*p(i)*p(m|i) = N*p(m)*p(i|m)$ result in the measurement outcome $m$. So out of the ensemble of $N*p(m)$ measurements that have outcome $m$, a fraction $p(i|m)$ started in state $|x_i\rangle$.

share|improve this answer
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.