# Seeming Mathematical Contradiction in Normalization of Momentum Eigenstates

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.
• Commented Jun 15, 2023 at 19:34
• see also e.g. physics.stackexchange.com/q/208596/50583 for other contradictions you can derive if you try to insist the "eigenstates" are actually inside the Hilbert space Commented Jun 15, 2023 at 19:35
• This explains how to deal with non-normalizable wave functions in the continuous spectrum and how not to make the mistakes most people make with them. Commented Jun 16, 2023 at 1:39