I am struggling with question about possible outcomes of momentum measurement and their probability. I know I can calculate it with momentum operator, but a wavefunction is of form
$$\psi (x)=3\cos\pi x+\cos3\pi x$$
and I am unsure how to deal with it, as the derivative consists of sines.
I know that $\cos{kx}=\frac{e^{ikx}+e^{−ikx}}{2}$ and $p_{x}=\hbar k$, but does it mean that the momentum is sum of eigenvalues of individual exponentials?
I think your professor is regarding $\psi(x)$ as a sum $$ \frac 32 (|\pi\rangle +|{-}\pi\rangle)+ \frac 12 (|3\pi\rangle+|{-}3\pi \rangle), $$ with $|k\rangle$ as an eigenstate of momentum $\hbar k$. Thus he probably thinks that $p=\pm \hbar \pi$ and $p= \pm 3\hbar \pi$ are the only posible outcomes. However the exponential wavefunctions restricted to the finite box are not eigenstates of the momentum operator. (This is what Emilio is saying, I think) True momentum eigenstates are $\psi_k(x)=e^{ikx}$ for $x$ on the entire real line. If you first normalize you unnormalized wavefunction, and then take an inner product of these infinite plane waves with your localized wavefunction you will get a non-zero overlap for any value of $p=\hbar k$. So any-and-all momenta are possible outcomes.
