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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
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1 Answers1

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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 $\psi=A \exp(\beta x)$.