# Modified double-slit experiment - two electron sources instead of two slits

In the famous double-slit experiment, if I inserted two identical sources (i.e electron guns) at the locations of the slits, would I still see an interference pattern on the screen?

I would allow only one particle to be fired out of the two electron guns in turns. In short, I replace the double slits with electron guns.

• What do you think? What would happen if you made it an optical experiment with two extremely narrow band laser sources tuned to almost the same frequency? If you are not comfortable with quantum phenomena, try reasoning trough the same scenario with water waves. Can you get interference from two sources creating water waves at slightly different frequencies? May 14, 2015 at 10:05
• Why do you only fire one electron at a time? May 14, 2015 at 10:13
• @innisfree to prevent interference like an electron "bumps" against each other May 14, 2015 at 10:15
• @CuriousOne since a particle can behave like wave doesn't mean it must have interference pattern so I think not, however your analogy suggest otherwise which puzzles me. I thought there is something tricky about the slits but it seems the properties of quantum is even trickier. I'm still thinking how a wave can interfere with itself... May 14, 2015 at 10:35
• Particles don't behave like waves. That's a poorly reasoned interpretation of quantum mechanics that has gone out of fashion eighty years ago. Bad ideas die hard, though, so this one is still around among laymen and some physicists are still shopping them around for no good reason. The sooner you get it out of your mind, the better for you. There is no way around learning proper QM, if you want to know how these things really work. It's not that hard, either, the math is certainly a lot easier than in general relativity. May 14, 2015 at 19:10

The situation is entirely different from the double slit experiment!

In the double slit experiment, one electron propagates through the slits, its parts interfere, thus we have a density matrix like (this prepares a pure state $\lvert\psi\rangle = \frac{1}{\sqrt{2}} (\lvert 1 \rangle + \lvert 2 \rangle)$):

$$\rho = \frac 1 2 \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$$

This has coherent terms, therefore it will be possible to observe interference effects.

The case with two electron guns. Here you have one electron from one source, this can be achieved best by a low emission rate (and then neglecting the few events with multiple electrons). This amounts to a density matrix of:

$$\rho = \frac 1 2 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$

This cannot show interference as the different electrons are incoherent. This is the picture for electron guns (which are macroscopic, therefore distinguishable, and so on).

Now if you build clever "electron guns" such that they will emit one electron, but it is not sure where (not classically, but in a quantum sense), then sure, you will have interference. (So the answer will get arbitrarily complicated, if you specify the situation in more detail).

# Addendum: The two-particle case in detail

Here I assume very clever electron guns each emitting one electron.

The single-particle density (which is measured on the screen) is given by (with the shorthand for arguments: $1 := \vec r_1, \sigma_1$, $\int d1 := \sum_{\sigma_1} \int d^3r_1$):

$$n(\vec r_1) = \sum_{\sigma_2} \int d1\, \sum_\sigma \lvert\psi(1, 2)\rvert^2$$

I neglect electron-electron interaction, as they will be far from each other.

## Same spin

If the emitted electrons have the same spin, the real-space part of the wave function must be anti-symmetric due to the Pauli exclusion principle. The spatial part of the wave function is then given by:

$$\phi(\vec r_1, \vec r_2) = \phi_1(\vec r_1)\phi_2(\vec r_2) - \phi_1(\vec r_2)\phi_2(\vec r_1)$$

\begin{align*} n(\vec r) &= \int d^3r'\, \left| \phi_1(\vec r)\phi_2(\vec r') - \phi_1(\vec r')\phi_2(\vec r) \right|^2\\ &= \int d^3r'\, \Big( \big\lvert \phi_1(\vec r)\phi_2(\vec r') \big\rvert^2 + \big\lvert \phi_1(\vec r')\phi_2(\vec r) \big\rvert^2 - 2 \Re \phi_1(\vec r)\phi_2(\vec r')\phi_1^*(\vec r')\phi_2^*(\vec r) \Big) \\ &= \big\lvert \phi_1(\vec r) \big\rvert^2 + \big\lvert \phi_2(\vec r) \big\rvert^2 - 2 \Re \phi_1(\vec r) \phi_2^*(\vec r) \underbrace{\int d^3r' \phi_1^*(\vec r')\phi_2(\vec r')}_{=\,0}. \end{align*}

The integral is zero, because the states $\phi_1, \phi_2$ must be orthogonal (because the initial states are orthogonal).

That means, we do not observe an interference pattern.

## Opposite spin

We will not observe interference either (as the seperate spin channels do not interfere).

The general (multi-electron, density matrix preparation) case is left as an exercise for the reader ;).

• It's not quite that simple. Electron guns are only incoherent because their energy dispersion is much wider than that of coherent photon sources. If you do the experiment with two lasers that have very narrow spectra, you will see (classical and quantum) interference overlaid by a beat. Indeed, the major problem in that experiment would be to actually keep the two lasers from mode-locking trough reflected photons! The exact same would be true for electron guns, if we could make their energy dispersion small enough. Technologically we can't, but that doesn't mean there is no interference. May 14, 2015 at 20:50
• It just occurred to me that there may be a way to actually do this experiment. If the electron sources consist of two laser traps which cool electrons to a very low temperature before they release them adiabatically, then one can make single electron sources that will get two electron wave-functions to overlap in space in time and with long enough coherence length to actually observe the interference. That would be a very cool (literally!) quantum bowling experiment! May 14, 2015 at 20:58
• Well something like this I meant by clever electron guns ;) May 14, 2015 at 21:16
• I still can't wrap my head around the two electron case. Intuitively, I think it will make a difference from the photon case, that they are fermions and we cannot construct a coherent state out of them. I'll try to do the math for the two electron case and report. May 14, 2015 at 21:22
• Not only that, but we will have an effect of the anti-symmetrization of total wave function (as when scattering identical fermions). May 14, 2015 at 22:56