- #1
Rookiemonster
- 3
- 0
This is the sort of thing that should be easy but my brain has blanked due to be overloaded in other areas of Physics.
I've tried a great deal of different approaches but I'll walk though the one that led closest.
I assumed that [itex]\hat{A}[/itex] and [itex]\check{B}[/itex] where both hermitian so there operators were orthogonal. Therefore the bra-ket of [itex]\varphi_{1}[/itex] and [itex]\varphi_{2}[/itex] is zero.
This yields two options. Either [itex]\chi_{1}[/itex]=3 and [itex]\chi_{2}[/itex]=-2 or [itex]\chi_{1}[/itex]=2 and [itex]\chi_{2}[/itex]=-3. I used the one that didn't make [itex]\varphi_{1}[/itex] vanish.
Then my plan was to go from [itex]\varphi_{1}[/itex] to [itex]\chi_{1}[/itex] and then back again. So take bra-kets squared and then do the same for [itex]\chi_{2}[/itex]. So I'm assuming the wave function is in the state [itex]\varphi_{1}[/itex] and then working out the probability of it going to either of the B states and then the probability of it going back to [itex]\varphi_{1}[/itex].
None of this worked and I got nonsense out. Any help is appreciated, I recognise that I'm being very simple and apologies if I haven't posted correctly.
An operator [itex]\check{A}[/itex] , corresponding to an observable A, has two normalized eigenfunctions [itex]\varphi[/itex]1 and [itex]\varphi[/itex]2 , with distinct eigenvalues a1 and a2 . An operator [itex]\check{B}[/itex] , corresponding to an observable B, has normalized eigenfunctions [itex]\chi[/itex]1 and [itex]\chi[/itex]2 , with distinct eigenvalues
b1 and b2 . The eigenfunctions are related by
[itex]\varphi[/itex]1=[itex]\frac{2\chi_{1} + 3\chi_{2}}{\sqrt{13}}[/itex]
[itex]\varphi[/itex]2=[itex]\frac{3\chi_{1} - 2\chi_{2}}{\sqrt{13}}[/itex]
A is measured and the value a1 is obtained. If B is then measured and then A again,
show that the probability of obtaining a1 a second time is [itex]\frac{97}{169}[/itex].
I've tried a great deal of different approaches but I'll walk though the one that led closest.
I assumed that [itex]\hat{A}[/itex] and [itex]\check{B}[/itex] where both hermitian so there operators were orthogonal. Therefore the bra-ket of [itex]\varphi_{1}[/itex] and [itex]\varphi_{2}[/itex] is zero.
This yields two options. Either [itex]\chi_{1}[/itex]=3 and [itex]\chi_{2}[/itex]=-2 or [itex]\chi_{1}[/itex]=2 and [itex]\chi_{2}[/itex]=-3. I used the one that didn't make [itex]\varphi_{1}[/itex] vanish.
Then my plan was to go from [itex]\varphi_{1}[/itex] to [itex]\chi_{1}[/itex] and then back again. So take bra-kets squared and then do the same for [itex]\chi_{2}[/itex]. So I'm assuming the wave function is in the state [itex]\varphi_{1}[/itex] and then working out the probability of it going to either of the B states and then the probability of it going back to [itex]\varphi_{1}[/itex].
None of this worked and I got nonsense out. Any help is appreciated, I recognise that I'm being very simple and apologies if I haven't posted correctly.