measurement of an operator

M. next

when we have a certain state ψ(t)
and it is acted on by an operator A of eigenstates a, b, c and eigen vectors la>, lb>, lc>

does it mean that after measuring A ( if the result was 'a'), the state lψ(t)> becomes in state la>?

andrien

yes.

M. next

i will tell you what was the problem, we were given a state lpsi> the question was: let us carry out a set of two measurements where B is measured first and then, immediately A is measured. Find the probability of obtaining a value of 0 for B and a value of 1 for A.

the first part was easy, i found the eigen pvalues of operator B and eigenstates.
and then the probability as they asked.
when i moved to the second part, they said exactly, in the solution:
we deal now with the measurement of the other observablr A. the observables A and B do not have common eigenstates. After measuring B ( the result is b=0) the system is left in a state lphi> which can be found by projecting lpsi> onto the eigenvector of b=0!!!!
How come???? isnot it supposed to be as u claimed in your answer that it be directly in the eigenstate of b=0?

andrien

you have talked about only one operator in op,The second one is different.

M. next

then? can you elaborate more? and what is it that is different?

Jazzdude

Not quite. An operator acting on a state like A|psi> is not the same as |psi> being measured with A. The first is just what it should be, an application of the linear operator on the state, without any projections or probabilities going on. The latter is the application of the measurement postulate that says measuring |psi> with A gives back an eigenvector |a> of A with the probability |<a|psi>|^2.

M. next

yes, but it is not the case. please read my second post,

andrien

No.I said that if eigen value find is a,and the ket corresponding to it is|A> THEN by projecting |∅> into that state i.e calculating.<A|∅>,One can write |∅> for that state.

jfy4

Take a look at this: you start with a state [itex]|\Psi\rangle [/itex]. we can write out the operator [itex]\hat{B}[/itex] in terms of it's eigenstates
[tex]
\hat{B} = b'_{0} |b_0 \rangle\langle b_0| + b'_{1} |b_1\rangle \langle b_1| + ...
[/tex]
if we want to pull out an eigenvalue you can just hit this guy with any of the eigenstates we are interested in. Now, Dirac's number one rule: after measuring a system, it is left in an eigenstate of whatever was measured. So we measured the eigenvalue for, say, [itex]b'_0[/itex], so our system is in the state corresponding to that eigenvalue. So take the equation for [itex]\hat{B}[/itex] up above and hit it with [itex]|\Psi\rangle[/itex] on the right. Then [itex]|\langle b_0 |\Psi\rangle|^2[/itex] is the probability to get eigenvalue [itex]b'_0[/itex].

Now the system is in state [itex]|b_0\rangle[/itex], now write the operator [itex]\hat{A}[/itex] like before
[tex]
\hat{A}=a'_0 |a_0\rangle \langle a_0 |+ ...
[/tex]
so to compute [itex]\hat{A}|b_0\rangle[/itex] we just hit the above with the state [itex]|b_0\rangle [/itex] on the right. Then the probability for any of the n eigenvalues is [itex]| \langle a_n | b_0 \rangle |^2[/itex]. I hope this helps.