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

1. Apr 15, 2014

### 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.

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.

2. Apr 15, 2014

### CAF123

Hi Rookiemonster, welcome to PF.
Yes, and this can be checked.

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?

3. Apr 16, 2014

### 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.

4. Apr 17, 2014

### CAF123

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.

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.

5. Apr 17, 2014

### Rookiemonster

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