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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?

  1. We can use the infinitesimal coefficient and this is a totally valid, physical thing to do.
  2. This is an honest-to-God mathematical contradiction, which means that no particle could ever truly be in a momentum eigenstate.
  3. We can use a infinitesimal coefficient to make the math work, but the problems noted above means this is not a physical system.
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Whenever you have a continuous spectrum part of an observable, the eigenfunctions live in Rigged Hilbert space. i.e. not normalisable.

A real quantum system has to live in Hilbert space, i.e. normalisable.

This means that

  1. This is an honest-to-God mathematical contradiction, which means that no particle could ever truly be in a momentum eigenstate.

is true. That is not a problem—the real quantum system is described by wavepackets, superpositions of these eigenfunctions. It is still guaranteed that any Hilbert space element can be expanded in terms of these Rigged Hilbert space stuff.

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