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Say I want to test if the following function is an eigenfunction of the momentum operator:

$$\psi(x,t) = A\exp{(-\alpha x^2})$$

$$\hat p [\psi(x,t)]= -i \hbar \frac{d}{dx}[A\exp{(-\alpha x^2})]$$

$$\hat p [\psi(x,t)] = -i \hbar (-\alpha 2x) (A\exp{(-\alpha x^2}))$$

$$\implies k = 2i\alpha x$$

This definitely tells me that this is in fact not an eigenfunction of the momentum operator, as this implies the wavenumber $k$ is imaginary and a function of $x$.

However, I don't have a good way of explaining why $k$ can't be imaginary and a function of $x$. Can someone motivate why this is?

sangstar
  • 3,192
  • I don't know. But your wave function is not just a stationary state, it's a super stationary state as you don't even need to square it to remove time. It is already time independent. The wave function itself is stationary. Your function is normalizable, but does it satisfy the Schrodinger equation? It seems like that might be an issue, but I don't know – DWade64 Apr 15 '18 at 23:43
  • Related: https://physics.stackexchange.com/q/98886/2451 , http://physics.stackexchange.com/q/221027/2451 and links therein. – Qmechanic Apr 16 '18 at 05:52

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An eigenstate of an operator $A$ is by definition a state such that $A\psi=c\psi$, where $c$ is a constant, called the eigenvalue. So not letting the eigenvalue depend on $x$ is simply part of the definition of an eigenvalue.

There is nothing wrong with having an imaginary eigenvalue of momentum. An example would be the wavefunction $\psi=A \exp(\beta x)$.