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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 $x = +\infty$ ?

Emile Couzin
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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.

rob
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  • Thanks for your answer ! So if I understood you well: the uncertainty on k $(\Delta k)\psi = 0$ $\Longrightarrow$ the uncertainty of the momentum $(\Delta p)\psi = 0$ $\Longrightarrow$ because of Heisenberg relation $\Delta x \Delta p \ge \hbar/2$, you have the uncertainty on the position $\Delta x = \infty$ $\Longrightarrow$ the neutron can be anywhere $\Longrightarrow$ this is not a physical evenment. Is this it ? – Emile Couzin May 29 '22 at 14:47
  • And waht does the waves of the wavepacket represent in this context, since you don't have a "packet" but just on single particle ? – Emile Couzin May 29 '22 at 14:50
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    It’s a little more fundamental than that: the relation $p = \hbar k = h/\lambda$ essentially is the uncertainty principle. As for the “waves in the wavepacket,” remember that the complex plane wave $\psi \sim e^{ikx}$ has constant amplitude, which sloshes around between its real and imaginary parts. The probability density $\psi^*\psi$ undoes all of this sloshing. An interesting homework problem for you would be to construct a Gaussian wavepacket and its corresponding probability density. – rob May 29 '22 at 15:58
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    You might also be interested in this answer about single-neutron interferometry. – rob May 29 '22 at 15:58