Why is the plane wave $\tilde{A}e^{ikx}$ representing an isolated neutron not physically acceptable?

Question

I consider a plane wave $Ae^{ikx-i\omega t}$ where $A \in \mathbb{C}$, $k = \frac{2\pi }{\lambda} \in \mathbb{R}$, $\omega = 2 \pi \nu = \frac{2\pi c}{\lambda} \in \mathbb{R}$ ($\lambda$ is the wave length) and $t = t_0 \in \mathbb{R}$ such as: $$\psi(x)|_{t=t_0} = Ae^{ikx}e^{-i\omega t_0} \equiv \tilde{A}e^{ikx}$$ A plane wave is physically acceptable ($\equiv$ represents a physical evenment) when: $$\psi(x) \in L^2(\mathbb{R}, \mathbb{R}) \Longrightarrow ||\psi(x)||^2 = \int_\mathbb{R}|\psi(x)|^2 \ dx \equiv I \in \mathbb{R}_0 \ \mathrm{and \ such \ as} \ I = 1$$ (or at least $\exists \phi \in L^2(\mathbb{R}, \mathbb{R})$ with $||\phi(x)||^2 = 1$ such as $\phi(x) \equiv \frac{\psi(x)}{||\psi(x)||}$)

Now, let's consider a neutron $n$ travelling in a vacuum space. I believe its wave function is define by $\psi(x) = \tilde{A}e^{ikx}$ such as $\lambda = \lambda_0 := 100 \ \mathrm{pm}$ for instance. My problem is: this wave function does not belong to $L^2$.

Indeed: $$\int_\mathbb{R}|\psi(x)|^2 \ dx = \int_\mathbb{R} \psi^\star(x)\psi(x) \ dx = \tilde{A}\int_\mathbb{R} \exp{(ikx - ikx)} \ dx = \tilde{A} \int_\mathbb{R} 1 \ dx \notin \mathbb{R}_0$$ So my question is: Why this neutron, which obviously exists, has the wave function of a non-physical evenment ? Is it perhaps because the neutron is isolated (thus not trapped in any kind of potential) and can therefore be anywhere between $x = -\infty$ and $x = +\infty$ ?

Answer 1

Your favorite derivation of the Uncertainty Principle is based on the properties of plane waves. If you have a neutron with definite wavevector $\vec k$, it is equally likely to be found anywhere is the universe. To model a neutron which is “somewhere” along its flight path, you have to allow some distribution of momenta to construct a wavepacket.

This restriction applies to all three components of $\vec k$. If the neutron’s momentum has a well-defined direction, that’s the same as setting the two perpendicular components of $\vec k$ to exactly zero, which makes the neutron unlocalizable perpendicular to the beam.

The plane wave can still be an excellent and useful approximation. An interesting system for exploring these limits is a pulsed neutron source where a monochromator crystal sends one wavelength out to a long beamline. Only “one energy” has the correct diffraction angles to enter the beamline from the monochromator, and only “one orientation” is sufficiently parallel to the beamline to reach your experiment at the end of it. But at a pulsed source, the neutrons with a particular momentum all reach the end of the beamline at a particular time. The angular divergence is small, but a clever experiment can separate out the position and angle correlations and take a kaleidoscopic picture of the brightness of the monochromator. The experimental uncertainties associated with a “monochromatic, unidirectional” neutron source are much bigger than the fundamental uncertainties associated with your non-normalizable wavefunction.