Shine light on a plate carrying two narrow slits side by side and bands of light and dark appear on the screen beyond. The waves spreading from the two slits overlap, and it is bright where crest meets crest and dark where crest meets trough.
Send an electron beam into the same apparatus in place of the light. The same shape of fringes appears. The matter wave is not just something in the calculation but a wave that really does interfere.
An electron is a charged particle and can be counted one at a time. That such a particle does interference, which only a wave can do, is where this article really begins.
Weaken the beam until there is only one electron inside the apparatus at a time. Each arrival lights one point on the screen, so it is clear that it arrived as a particle. Where it lands is different every time.
Collect the points, though, and those positions that were supposed to be scattered line up in the shape of fringes. Ten points show nothing, a thousand show it faintly, and a hundred thousand make the fringes clear.
The fringes are not made by electrons running into one another, since there is only ever one inside the apparatus. A single electron has gone through both slits by itself.
Write the separation of the two slits as d and the distance to the screen as L. The two paths that reach a point x from the center of the screen differ in length by \dfrac{dx}{L}.
When this difference is a whole number of wavelengths, the two waves overlap crest on crest and it is bright. Since \dfrac{dx}{L} = n\lambda, the spacing of the bright lines is \Delta x = \dfrac{L\lambda}{d}.
Take the measured \Delta x and get \lambda out of this formula, and it agrees with \dfrac{h}{\sqrt{2meV}}. Measuring how fine the fringes are gives the wavelength of the electron directly.
Where one electron lands is not fixed. What is fixed is only the proportion, how many land where when many are sent.
It is the wave that fixes that proportion. Writing the size of the overlapped wave as \psi, the probability of landing at that point is proportional not to \psi itself but to the square of the size, |\psi|^2. A bright fringe is where the probability is high, a dark one where it is nearly 0.
Taking the square is the same as the brightness of light being the square of the amplitude. What differs is that for light it was the brightness of one wave, while here it is the chance of one particle arriving there.
Now look at which of the slits the electron went through. Put a detector at one of them and run the same experiment while checking whether the electron passed it.
The fringes vanish. What remains is two humps, one in front of each slit. The moment the path is fixed as that of a particle, the overlap as a wave is gone.
Stop looking and the fringes come back. Nothing about the apparatus was changed. What was changed is only whether it could be known which slit was taken. The two paths can overlap only while which one was taken is not fixed. That is what the double slit experiment shows.