
#1
Nov2212, 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>? 



#2
Nov2212, 04:28 AM

P: 987

yes.




#3
Nov2212, 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? 



#4
Nov2212, 05:11 AM

P: 987

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




#5
Nov2212, 05:17 AM

P: 354

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




#6
Nov2212, 05:37 AM

P: 178





#7
Nov2212, 05:51 AM

P: 354

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




#8
Nov2312, 07:11 AM

P: 987





#9
Nov2312, 05:06 PM

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. 


Register to reply 
Related Discussions  
quantum entanglement and measurement operator  Advanced Physics Homework  0  
Actual earth measurement contradicts measurement predicted by special relativity  Special & General Relativity  93  
Eigenfunctions of translation operator and transposed operator property proof  Calculus & Beyond Homework  1  
operator and measurement  Quantum Physics  17  
about time measurement vs measurement of movement  Special & General Relativity  21 