Consider the double-slit experiment done with a stream of electrons, sent individually toward the double-slit screen and detection plate. A moving electron is a moving electric charge, which by definition is an electric current, and moving electric current always generates an E-M field. Any piece of metal or any coil of wire in the laboratory, or 10 miles away on the moon, theoretically would be detecting the E-M field of the electron by some minute induced voltage. Most importantly, the E-M field induced voltage in some inadvertent “detector” would be greater or lesser for the electron going through the “Right-Hand” slit as opposed to going through the “Left-Hand” slit, since the moving electron would be closer to or further away from the “detector” depending on which slit it went through. Since the electron is always giving off an E-M signal as to “which-way” it travelled through the two slits, how can electrons ever be in a super-position state and provide the interference pattern on the detector screen, since moving electrons always “give-away” their presence?

  • $\begingroup$ The EM field of a moving electron is expressed in an exchange of a virtual photon between this electron and another charge (your detector). If this happens and you detect which slit the electron went through, then no interference on the screen, because the wave function of the electron is collapsed by this interaction and the electron cannot behave as a wave. $\endgroup$
    – safesphere
    Aug 20 '18 at 4:57
  • $\begingroup$ This is incorrect. The electron will always have interaction with the material of the slit system in the form of transient polarisation. This does not destroy the coherence and the interference. $\endgroup$
    – my2cts
    Aug 20 '18 at 9:20
  • $\begingroup$ The detector is typically/needs to be placed after the slit, so while initially the electron passed thru the slit and was diffracted it was later detected and its original wave function destroyed and a new one created. The new one interacts with the screen and does not produce a diffraction pattern. So you are both correct. $\endgroup$ Aug 20 '18 at 15:40

Since the electron is always giving off an E-M signal as to “which-way” it travelled through the two slits,

This is classical electrodynamics for charged particles. The electron is an elementary particle in the standard model of physics, and at the level of the interaction with the slits, it displays quantum mechanical behavior, i.e probability distributions can only be predicted, not trajectory signals through interaction with matter.

An electron will either interact before , and then it will be lost from the beam , or arrive at the slits and interact with them in a scattering experiment "electron of specific momentum scattering on specific width and distance slits", and the will have a complex wavefunction which when squared with its complex conjugate will predict the distribution to be seen at the screen.i.e. many electrons have to be scattered to get the distribution.

So at the level of electrons, they are not giving off electromagnetic signals. If they do, this means they are radiating photons (the signals) and the boundary conditions are not the ones of "electron of specific momentum scattering on specific width and distance slits".

  • $\begingroup$ If 1000 scientists each with 1000 identical double slit apparatuses measured the position of 1 and only 1 electron what would they find when they got together to compare/combine results? Would they still find the diffraction pattern? I think yes. $\endgroup$ Aug 20 '18 at 16:01
  • $\begingroup$ @PhysicsDave No. The boundary conditions have to be the same to have an experiment with electrons with the same slits. Different experiments mean different boundary conditions. $\endgroup$
    – anna v
    Aug 20 '18 at 16:14

All information that we have on the electron is contained in its wave function. If this wave function contains the information through which slit the electron went, by entanglement with some form of detector wave function, then the coherence is lost and no interference occurs.


You should understand that there is a continuum of options, not just "being in a superposition" and "having a collapsed wavefunction". So here are the two extreme cases: 1. If you put a huge detector after one of the slits, it clearly destroys the superposition and you effectively end up with a collapsed wave function. 2. In the perfectly idealized situation, nothing at all interacts with the electron and the wave function stays in a perfect superposition of the wave packets through the left and right slit.

Now every real situation lies somewhere in between these two, and depending on where in this continuum of possibilities it is, the coherence (visibility of the interference pattern) is higher or lower. The electromagnetic interaction you mention is one of the sources of problems in double-slit experiments. When people actually perform double-slit experiments with electrons (and note that this happened only in the 1970s, showing that it's difficult!), they are careful too isolate the electrons rather well. Never perfectly, of course. So if you do it well, the electrons still rather show the interference pattern.


The travelling electron is indeed emitting an EM field as it moves but as soon as you try and measure it or detect it (after the slit) you will alter or destroy it (the wave function, not the electron). When getting the diffraction pattern a detector has never been successfully employed. Also note that every electron that passes thru will get to the screen and create the bright spots, dark spots are areas where no electrons have landed. The interaction of the electron wave function (EM field) with the slit's EM field determines the pattern. Interference is a classical term used to explain dark areas before wave functions and QM etc were realized.


The electron travels as a wave, and when it travels through the slits, part of the electron's wave travel through both slits. These partial waves then interfere, and create constructive interference (bright pattern on the screen), and destructive interference (dark areas on the screen).

Now as soon as you try to put filters (which way experiment) on one slit, this filter will cause inelastic scattering on the electron, and this will create spherical waves, and these will be out of phase, so there will be no interference pattern.

Please see here:


This is because the electrons are shot one at a time, but because of the beam that creates the electrons is in phase (it creates the electrons so that the electrons are in phase), the electrons shot after each other will show a pattern. But if there is a filter on the slit, the electron's partial waves that are going through that slit will be out of phase, and will not interfere with the other parts of the wave.

Now in your case you say that the electron as it travels creates an EM field around it. If you want a detector to detect this EM field around the traveling electron, then you need the electron to interact with the detector.

This interaction can occur in three ways:

  1. elastic scattering, the electron will still keep its energy and phase, and create an interference pattern. This is always the case, since the electron is interacting with the air molecules, and elastically scattering off of the air atoms, and there is still a pattern.

  2. inelastic scattering, the electron will change energy and phase, and there will be no pattern. This is when the electron interacts with the detector.

  3. the electron interacts with the screen, and joins one of the atoms in the screen.

Now in your case, the EM field around the traveling electron will have to interact with the detector in a way that it keeps the electron's energy and phase. The electron will have to exchange virtual photons with the detector. Now this is the math that describes how the electron can interact with another particle, that is a part of the detector. This is called an EM interaction. This is the only way (through exchange of photons) an electron can have an EM interaction with another electromagnetically charged particle. Now since the electron is not accelerating, and it is free, it is not emitting real photons.

So the only way for a detector to detect the electron is through the exchange of virtual photons. There is no way that the virtual photons could be exchanged between the electron and the detector and create elastic scattering.

We simply cannot create a detector that will cause an elastic scattering on the electron and will keep the phase and energy of the electron to create a pattern.

All the detectors will create an inelastic scattering on the electron and will change the electrons energy and phase and there will be no pattern.

There would be another way to detect the electron, through gravitational interaction. Yes, the electron does have stress-energy, and has gravitational effects.

We simply cannot build detectors that would detect the gravitational effects of the traveling electron, as it exchanges virtual gravitons with the detector.

The same way there is not detector, that could use the weak force or the strong force to detect the electron.


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