Struggling with (I think pretty trivial) eigenfunction work for QM.

[Rookiemonster]

This is the sort of thing that should be easy but my brain has blanked due to be overloaded in other areas of Physics.

An operator \check{A} , corresponding to an observable A, has two normalized eigenfunctions \varphi₁ and \varphi₂ , with distinct eigenvalues a₁ and a₂ . An operator \check{B} , corresponding to an observable B, has normalized eigenfunctions \chi₁ and \chi₂ , with distinct eigenvalues
b₁ and b₂ . The eigenfunctions are related by

\varphi₁=\frac{2\chi_{1} + 3\chi_{2}}{\sqrt{13}}

\varphi₂=\frac{3\chi_{1} - 2\chi_{2}}{\sqrt{13}}

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 \frac{97}{169}.

I've tried a great deal of different approaches but I'll walk though the one that led closest.

I assumed that \hat{A} and \check{B} where both hermitian so there operators were orthogonal. Therefore the bra-ket of \varphi_{1} and \varphi_{2} is zero.

This yields two options. Either \chi_{1}=3 and \chi_{2}=-2 or \chi_{1}=2 and \chi_{2}=-3. I used the one that didn't make \varphi_{1} vanish.

Then my plan was to go from \varphi_{1} to \chi_{1} and then back again. So take bra-kets squared and then do the same for \chi_{2}. So I'm assuming the wave function is in the state \varphi_{1} and then working out the probability of it going to either of the B states and then the probability of it going back to \varphi_{1}.

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.

 

[CAF123]

Hi Rookiemonster, welcome to PF.

I assumed that \hat{A} and \check{B} where both hermitian so there operators were orthogonal. Therefore the bra-ket of \varphi_{1} and \varphi_{2} is zero.

Yes, and this can be checked.

Then my plan was to go from \varphi_{1} to \chi_{1} and then back again. So take bra-kets squared and then do the same for \chi_{2}. So I'm assuming the wave function is in the state \varphi_{1} and then working out the probability of it going to either of the B states and then the probability of it going back to \varphi_{1}.

Yes, that is the right idea. You have some general wavefunction describing a quantum system. You are told that when the observable corresponding to the operator A is measured, you obtain the value a₁. So, the wavefunction has collapsed into an eigenstate of A, that being $\varphi_1$. Now, assume we measure B right away (we are not given that $\varphi_1$ is a stationary state of some Hamitonian, so the system might naturally evolve if we leave it long enough, but that need not concern us.) To obtain the possible values of B in the state $\varphi_1$, you need to express this state in terms of the eigenfunctions of B. Happily, the question did that for you. So, what are the possible outcomes and their corresponding probabilities?

 

[Rookiemonster]

so the bra-ket of \varphi_{1} and \varphi_{2} should be equal to zero.

Therefore, either \varphi_{1} or \varphi_{2} needs to be zero in total.

Therefore, \chi_{1}=3 and \chi_{2}=-2 or \chi_{1}=2 \chi_{2}=3

The problem is putting these values into a squared bra-ket of \varphi_{1} and \chi_{1} yields a probability greater than one.

 

[CAF123]

so the bra-ket of \varphi_{1} and \varphi_{2} should be equal to zero.

Therefore, either \varphi_{1} or \varphi_{2} needs to be zero in total.

I am not sure what you mean by that - the bra-ket being zero is due to the orthogonality of the states, i.e $\varphi_1, \varphi_2$ are eigenstates of A with different eigenvalues, so their inner product vanishes.

Therefore, \chi_{1}=3 and \chi_{2}=-2 or \chi_{1}=2 \chi_{2}=3

The $\chi_i$ and $\varphi_i$ need not be identically equal to a constant always. I'll start you off on the right track: You were given that A is measured and the result $a_1$ found. This means your system is now in the eigenstate $\varphi_1$. Now you measure B while the system is in this state. To find the possible values of B, you must decompose $\varphi_1$ into the eigenstates of B (the $\chi_i$). So $\varphi_1 = c_1 \chi_1 + c_2 \chi_2$. If the state is normalized, then $|c_1|^2 + |c_2|^2 = 1$ and so the interpretation is that the $c_i$ are the possible outcomes of B in the state $\varphi_1$ and $|c_i|^2$ is the corresponding probability. Since you must measure some value of B, these probabilities must add up to 1, as you can also check.

 

[Rookiemonster]

Are you are a hero. I hadn't thought to use the probability of coefficients.