- #1
Gray
- 10
- 0
Homework Statement
Consider a state |psi>, and two non-commuting observables A and B. Now study the following chain of measurements:
(i) On |psi> a A [sic] measurement gives the result a1, and a subsequent measurement of B gives the result b2.
(ii) On |psi> a B measurement gives the result b2, and a subsequent measurement of A gives the result a1.
Express the result of (i) as a product of two relevant properties. Do the same for (ii). Call these P(a1,b2) and P(b2,a1). Now show that if a1=b2, then for this "accidental" case of two non-commuting observables sharing the same eigenvalue,
P(a1,b2)/P(b2,a1)=1
Provide a general expression for the above ratio when a1 =/= b2.
Further discuss this setup by taking A=J_x and B=J_y; specifically remark on the equality a1=b2 in this context.
Homework Equations
The Attempt at a Solution
First off I realized that state psi must be a combination of the states a1 and b2 so
|psi> = c1|a1> + c2|b2> where c1*.c1, c2*.c2 are the probabilites of the respective states.
I then figured that the total probability of case (i) = prob of result a1 times prob of result b2 after a1, so
P(a1,b2) = |<b2|a1>|^2.|<a1|a1>|^2
and similarly
P(b2,a1) = |<a1|b2>|^2.|<b2|b2>|^2
If a1=b2 then these expressions are the same and P(a1,b2)/P(b2,a1) obviously = 1.
What about a general expression? My lecturer doesn't just mean |<b2|a1>|^2.|<a1|a1>|^2 / |<a1|b2>|^2.|<b2|b2>|^2 surely, so is there some way that this expression simplifies? I can't see how...
For the last part regarding J_x, J_y, if a1=b2 does this mean that then even though they do not commute, state psi is a simultaneous eigenstate of J_x and J_y and so they would have the same eigenvalue j? I thought being non-commutative prevented this possibility? Or does this work because psi is a linear combination?
Thanks,
Gray