measurement of an operator


by M. next
Tags: measurement, operator
M. next
M. next is offline
#1
Nov22-12, 04:26 AM
P: 354
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>?
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andrien
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#2
Nov22-12, 04:28 AM
P: 987
yes.
M. next
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#3
Nov22-12, 05:01 AM
P: 354
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
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#4
Nov22-12, 05:11 AM
P: 987

measurement of an operator


you have talked about only one operator in op,The second one is different.
M. next
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#5
Nov22-12, 05:17 AM
P: 354
then? can you elaborate more? and what is it that is different?
Jazzdude
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#6
Nov22-12, 05:37 AM
P: 178
Quote Quote by M. next View Post
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>?
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
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#7
Nov22-12, 05:51 AM
P: 354
yes, but it is not the case. please read my second post,
andrien
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#8
Nov23-12, 07:11 AM
P: 987
Quote Quote by M. next View Post
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?
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
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#9
Nov23-12, 05:06 PM
jfy4's Avatar
P: 647
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.


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