Consider the following experiment: a double-slit set-up for firing electrons one at a time. Let's now add a second electron (orange), which is fired parallel to the first one, but in the opposite direction, and so as to pass closer to one slit than the other, and, importantly, above the plane in which interference occurs:

alt text

The firing of the two electrons is timed so that the orange electron is "closest" to the purple electron somewhere around point P, i.e. after the latter has passed through the two slits.

Presumably the electrons will repel due to having charge. Presumably an interference pattern will still be produced by the purple electron if the experiment is repeated thousands of times.

Now my question is, how exactly does the orange electron get deflected? If the experiment is repeated, would it always hit the same point on the opposite wall, or would it also start producing a sort of an interference pattern?

Thanks to user:Timwi for the diagram!

  • $\begingroup$ The interference pattern will disappear if you scatter another electron off it. $\endgroup$ Commented Jan 13, 2011 at 20:40
  • 1
    $\begingroup$ @Raskolnikov so you're saying that all the matter that makes up the slits doesn't "break" the pattern, but a single electron would? $\endgroup$ Commented Jan 13, 2011 at 20:47
  • 1
    $\begingroup$ The slits are what make the interference pattern possible in the first place. But if you try to determine which slit the electron went through by scattering off another electron, you will destroy the interference pattern. $\endgroup$ Commented Jan 13, 2011 at 20:54
  • 1
    $\begingroup$ @Raskolnikov I can imagine that if the scattered electron also produced a sort of an interference pattern that wouldn't exactly tell me which slit the purple electron went through. $\endgroup$ Commented Jan 14, 2011 at 17:17
  • $\begingroup$ Yes, if you made the orange electron go through slits as well, the overall result might well be such. But I'd have to compute it to check. But the end result would be that your perturbation would not be able to tell which slit either electron took. $\endgroup$ Commented Jan 14, 2011 at 17:42

2 Answers 2


This is a typical trade-off between the position properties of the purple electron and its wave properties.

If the orange electron is sufficiently far so that it doesn't influence the purple electron much, everything will continue as before: the purple electron will produce an interference pattern while the orange one will draw one point on the photographic plate.

However, you may try to bring the orange electron closer. As you're bringing it closer, you increase the likelihood that it is repelled by the purple electron. The deflection is pretty much dictated by the vertical position of the purple electron on the picture, so the orange electron is de facto measuring this position.

If most of the repulsive interaction occurs near the slits, the orange electron's motion is dominated by the which-slit information about the purple electron. The more accurately you measure it, by looking at the orange electron's path, the more you will disturb the interference pattern for the purple electron. The more you can read some "position information" from the orange electron's path, the less sharp will be the purple electron's interference pattern, and vice versa.

I want to emphasize that all such problems can be exactly calculated - just use the proper quantum mechanics for two particles. The wave function - in the non-relativistic picture - is psi(x1,y1,z1,x2,y2,z2,t) where 1,2 are labels of the two electrons. This wave function is antisymmetric under the exchange of x,y,z for 1 and 2 and it evolves according to a Schrodinger equation for 2 particles and its squared absolute value knows about the probability that both electrons will be observed at any pair of places.

One also has to be careful because the electrons are indistinguishable, so that if they get really close so that they could possibly get exchanged, one has to subtract (because of the Fermi statistics) the amplitudes in which they exchanged and one in which they have not, to keep their wave function antisymmetric.

I also want to emphasize that the electrostatic repulsion between two electrons in such experiments is tiny and should be more properly described as a "quantized" exchange of a single photon. Such a repulsive interaction always allows the possibility that additional photons are emitted (radiation from accelerating charges), and these extra photons may reduce the overall sharpness of the interference pattern further.

  • $\begingroup$ Such a calculation is certainly beyond my current abilities. Might there be a software package that requires less skill to use that might let me compute the exact pattern produced on both walls after repeating the experiment? It's ok if such a package requires programming. $\endgroup$ Commented Jan 14, 2011 at 17:16
  • $\begingroup$ Dear @romkyns, it's a relatively difficult problem but I just want to increase your self-confidence about QM. The motion of 2 quantum particles is described by Schrödinger's equation for 2 particles which is a straightforward differential equation, and the probability that a particle is absorbed somewhere is always proportional to $|\psi|^2$. You could do it if you wanted... $\endgroup$ Commented Feb 2, 2011 at 15:01
  • $\begingroup$ I probably could if I prepared for a couple of months... If I came across as "able" in physics then it is a complete accident; the last time I solved differential equations was 6 or 7 years ago, and I wouldn't have any clue how to put two of them together and how to set up initial conditions... And I'd definitely need a good tutor to help me along :) $\endgroup$ Commented Feb 2, 2011 at 19:58

I think that Luboš Motl has covered this rather well, but I get the idea that what you're looking for is a discussion on how to calculate the electron density for the combination of a single "free" electron and another electron that's gone through the interference. I'll give you the flavor of the calculation without actually doing it.

Let $\psi_p(x)$ be the wave function for the purple electron that went through the slit where $x$ means a 3-d point and I'm leaving off the time variable by assuming that there's no dependence on it. This is a complex function and when you evaluate it on the screen you'll find that its squared magnitude $|\psi_p(x)|^2$ gives you the interference pattern.

And let $\psi_o(x)$ be the wave function for the orange electron. Then the combined wave function is antisymmetrized:
$\psi(x_1,x_2) = \psi_o(x_1)\psi_p(x_2) - \psi_p(x_1)\psi_o(x_2)$

Then the probability that you get one electron at the point $x_1$ and the other at the point $x_2$ is proportional to $|\psi(x_1,x_2)|^2$. This calculation assumes that the electrons do not interact.

As Dr. Motl notes, so long as the two electrons don't get extremely close this will be quite accurate and the result will be the hole drilled by the orange electron and the scattering pattern from the other. To add the interaction you will have to use quantum field theory. In that theory you begin with the above calculation for $\psi(x_1,x_2)$ and add corrections to it. The first correction will be the addition of a single photon that is emitted by one electron and absorbed by the other.

The photon calculation is an example of a Feynman diagram. See http://en.wikipedia.org/wiki/Feynman_diagram
The pretty drawing is:
Feynman drawing

and the mathematics is kind of messy and should be the subject of a separate question.

  • $\begingroup$ your answer ended just as it was getting the most interesting :) If the electrons are close enough that they exchange one photon almost every time the experiment is repeated, what would happen to the patterns produced on the two screens? Or would answering this question require one to actually do that messy maths? $\endgroup$ Commented Feb 10, 2011 at 12:08
  • $\begingroup$ I'm going to guess that it would ruin the pattern completely and hope that if I'm wrong, Dr. Motl will correct me. $\endgroup$ Commented Feb 10, 2011 at 18:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.