Double slit experiment

We are familiar with the double-slit experiment using light, and the interference pattern it produces. We consider the same experiment but with a physical particle (usually an electron) in the low-intensity limit. Experimentally we see that even though only a single electron at a time passes through the system, an interference pattern still appears.

This means that the underlying de Broglie wave of the particle interferes with itself, even though we consider only a single particle to be in the system at a time.

Consider a setup like shown, where a laser is shined at the particle as it passes through one of the slits. A microscope records the scattered photons and can determine which slit the particle passed through depending on the scattering angle.

The $\color{red}\text{red}$ waves show photons, while the black waves show the incident particle. The dashed lines show the path to a far-off screen where one would expect interference.

What we see experimentally is that if we attempt to measure the slit passed through in this way, the interference pattern disappears. Shining a laser on the particle’s path at all, even if the scattering is not recorded, destroys the interference pattern.

Consider lowering the intensity of the laser such that on the order of a single photon is emitted at a time. What we see is that we recover a partial interference pattern albeit with incomplete contrast.

The black line shows an interference pattern with complete contrast, scaled from $0$ to $1$. The blue line has incomplete contrast, the lows don’t reach $0$ and the highs don’t reach $1$.

Instead, we might lower the frequency of the photons so they carry less energy and interact weakly with the particle. If the frequency is low enough we again recover the interference pattern, but when $\lambda \sim d$ we lose the optical resolution to determine which slit a photon was scattered from.

Any way the information is available, we cannot know both the position of the particle and maintain an interference pattern.

Consider the wave functions at the two slits $\psi_1 = A e^{i \phi_1}$ and $\psi_2 = A e^{i \phi_2}$, with $\phi_1 = \phi_2$ initially. At the wall (call it point $D$) where we expect interference, the wave function $\psi_D = A e^{i k L_1} + A e^{ikL_2}$ where $k$ is the wavenumber and $L$ is the length traveled by each wave. This gives us probability

Now consider the behavior when a photon interacts with the particle.

When the photon and electron collide the photon imparts some momentum upon the electron. This corresponds to a change in phase of wave vector such that $\psi_i \to \psi_i \,e^{i (\mathbf k_\mathrm{in} - \mathbf k_\mathrm{sc}) \cdot \mathbf x_i}$. Here $x_1,x_2$ are the positions of the slits; we take the dot product to adjust for the phase difference of the incident laser. This gives us the probability

We see that the only non-constant term is $\mathbf k_\mathrm{sc}$. Since the scattering angle is random, the phase shift will be random too. These random phases cause the interference patterns to disappear. However, if we know the scattering angle of the photon that collided with a given electron, we can reconstruct the interference pattern by adjusting for the phase difference.

This also explains the behavior the incident wavelenght is high. $k=\frac{2\pi}{\lambda}$, so when $\lambda$ is large, $\mathbf k_\mathrm{in} - \mathbf k_\mathrm{sc}$ is small, as is the phase shift.