# Quantum Mechanics - Measurements and Observables

1. Jun 23, 2009

### Gray

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. The attempt at a solution
First off I realised 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

2. Jun 25, 2009

### Gray

Ugh I started working on this problem again and realised I was being stupid earlier (as usual :/). For some reason I thought psi was a combination of the states a1 and b2 but these aren't states at all, they are the measurements of the observables A and B and so just numbers (eigenvalues to be precise). (I guess I'm also assuming the second measurement is made sufficiently long after the first measurement that the wavefunction psi has spread out again).
So this means
P(a1,b2) = |<b2|psi>|^2.|<a1|psi>|^2
and similarly
P(b2,a1) = |<a1|psi>|^2.|<b2|psi>|^2
right?
Again if a1=b2 then
P(a1,b2)/P(b2,a1) = 1 trivially (i.e the measurement sequences have equal probability since the measured values are the same in each case),
and in general
P(a1,b2)/P(b2,a1) = |<b2|psi>|^2.|<a1|psi>|^2 / |<a1|psi>|^2.|<b2|psi>|^2
= (<b2|psi>*<b2|psi>.<a1|psi>*<a1|psi>) / (<a1|psi>*<a1|psi>.<b2|psi>*<b2|psi>)
= (b2^2<psi*|psi>.a1^2<psi*|psi>) / (<a1^2<psi*|psi>.b2^2<psi*|psi>)
= 1 again? which I don't think it should...

Any help/advice for this and my previous thread would be greatly appreciated..

3. Jun 25, 2009

### Gray

Woops that "properties" is meant to be "probabilities".

4. Jun 26, 2009

### Matterwave

Ok, you are going on the right track. BUT you CANNOT allow the wave functions to "spread out" again in order to do this problem because you are not given the Hamiltonian of the system. You don't know how the system would evolve in time, and so if you allowed the wave function to spread out, you can't say anything about probabilities because you no longer know the psi's (in this case, the psi's become unknown and you can't get quantities such at <b|psi>).

I will give a hint, and hopefully you can figure it out from there:

We know that:

$$\psi = c_1|A_1>+c_2|A_2>+c_3|A_3>+...$$

for some coefficients c's (and A's are the eigen vectors of A) since the eigen-vectors of A must span the space. Similarly, you can also express:

$$\psi = d_1|B_1>+d_2|B_2>+d_3|B_3>+...$$

for some coefficients d's (and B's are the eigen vectors of B) since the eigen-vectors of B must also span the space.

With this knowledge, and the knowledge that making the observation on A collapses the state psi into an eigenstate should be sufficient for you to complete this problem.

5. Jun 26, 2009

### LiorE

My attempt was: We know that: $$p(A = a1) = |<a1 | \psi>|^2$$ and $$p(B = b2) = |<b2 | \psi>|^2$$. Also, the probability of measuring a1 and then b2 is: $$p(a1,b2) = |<b2 | a1><a1 | \psi>|^2$$, and similarly: $$p(b2,a1) = |<a1 | b2><b2 | \psi>|^2$$. However, what I don't understand is what does the fact that a1 happens to equal b2 have to do with this? I mean, if these are two equal eigenvalues, it doesn't mean that they share the same eigenstate too, right? (That's because simultaneous eigenstates are possible only if the operators commute)

Even if you check this specifically this doesn't make sense. Suppose we're in the spin 1/2 space, and the state is $$|\psi > = a|+> _z+ b|->_z,\;|a|^2+|b|^2 = 1$$. Then:

$$p(S_x = \hbar/2,S_y = \hbar/2) = \frac{|1+i|^2 |a+b|^2}{8}$$,

and:

$$p(S_x = \hbar/2,S_y = \hbar/2) = \frac{|1+i|^2 |a+ib|^2}{8}$$

clearly they're not equal.

6. Jun 26, 2009

### Matterwave

Your work looks fine to me, but in general you don't want to solve the problem for the poster. Give them hints so they can solve it themselves. Doing their homework for them isn't going to help them become better.

7. Jun 27, 2009

### LiorE

Thanks, and I'm sorry if I wrote more than I should have, but what I'm saying is that I have a counterexample to what he's trying to show, and that seems really weird to me. What am I missing here?

8. Jun 27, 2009

### Matterwave

I don't see why it would be the case that the two probabilities are equal as well.

The values a1 and b1 are just the measurements, and they don't have any bearing on the probabilities themselves. So, if one A=Sx and B=Sy the eigenvalues themselves won't have any bearing on the outcome of the results, only the eigenstates and the coefficients for "how much that state contains those eigenstates" will matter.

The only way what was typed would make sense to me is if it were instead |a1>=|b1>

which is not possible for non-commuting operators (or else you could violate the HUP).

9. Jun 27, 2009

### LiorE

Right, that's exactly what I was thinking.