Okay, follow me here:
You have a free particle in a momentum eigenstate, meaning its wave function is as follows:
$$ \Psi(x,t) = e^{i(kx-\omega t)} $$
Of course, this wave function is not currently normalized as $ \int_{- \infty}^{\infty} ||\Psi(x,t)||^2dx = \infty $, so we need a normalization such that:
$$ \int_{- \infty}^{\infty} ||\Psi(x,t)||^2dx =1 $$
Further, given that $dP = ||\Psi(x,t)||^2dx$, the magnitude of the wave function over every integral of finite length must be zero, so it seems to me that $\Psi (x,t) =0$ would be the only function that satisfies this condition, meaning it can't satisfy the normalization criterion above.
Further, if we try to realize this wave function as a superposition of position eigenfunctions, i.e. $ \Psi (x,t) = \int_{D(x)}\hat\Psi(x,t) \cdot \delta(0-x)dx$, it seams that the value of $\hat \Psi$ for each function is zero, but again, we need that integral to add up to 1 over the whole domain.
So my question is, "can fix? and if so, how fix?"
The only thing that makes sense to me is to use a nonreal, infinitesimal number as a coefficient to either our original wave function and/or our position eigenstates.
So which is it?
- We can use the infinitesimal coefficient and this is a totally valid, physical thing to do.
- This is an honest-to-God mathematical contradiction, which means that no particle could ever truly be in a momentum eigenstate.
- We can use a infinitesimal coefficient to make the math work, but the problems noted above means this is not a physical system.
