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?

  • $\begingroup$ What is the question? $\endgroup$ – Andrei Geanta Oct 10 '17 at 8:40
  • $\begingroup$ Those cosines are sums of complex exponentials, and those have a simpler relationship with the momentum operator. $\endgroup$ – Emilio Pisanty Oct 10 '17 at 8:43
  • $\begingroup$ 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? $\endgroup$ – M.B. Oct 10 '17 at 8:53
  • $\begingroup$ This is hard to answer in its current form without access to the precise wording of the question you were given. It's important to note that the wavefunction you were given does not have a well-defined momentum (and neither do its individual components $3\cos(\pi x)$ and $\cos(3\pi x)$); that doesn't mean that you can't say anything useful about its momentum properties but it's important to know exactly which property you're being asked about. $\endgroup$ – Emilio Pisanty Oct 10 '17 at 10:49
  • $\begingroup$ The particle is contained in 1D infinite potential well, bounded in the range -1<x<1. The question stated is: what are the possible outcomes of this measurement ($p_{x}$) and what is the probability of each outcome. $\endgroup$ – M.B. Oct 10 '17 at 12:00

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.

  • $\begingroup$ The momentum operator in the finite square well is notoriously problematic, cf. e.g. this and this answers. I agree that the lecturer likely meant that superposition, but interpreting states in the infinite well through the lens of the full-line momentum operator is a rather more subtle question. $\endgroup$ – Emilio Pisanty Oct 10 '17 at 16:02
  • 1
    $\begingroup$ @Emilio. Yes I agree: the "momentum operator" on a finite interval is the classic example of a symmetric but not self-adjoint operator. My thought is that the OP's question can be formulated more physically by asking what happens in finite depth square well on the entire line. Then a self-adjoint momentum operator exists (but does not commute with the Hamiltonian) and the overlaps do give the possible momenta, One can then explore what happens as the depth increases. $\endgroup$ – mike stone Oct 10 '17 at 17:01
  • $\begingroup$ @ mike that's a nice way to look at things. $\endgroup$ – Emilio Pisanty Oct 10 '17 at 17:10

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