In the double-slit experiment with electrons, if a detector is placed in one of the paths to show where the electron is passing, the wave nature of the particle is lost and you have no interference.

Why don't the slits themselves cause the wave function to collapse? After all a slit interacts physically with the particle to cause the interference.

Conversely if the slit is acting in the "wave domain", leaving the wave nature of the particle unchanged, wouldn't it be possible to build a detector by the same principle? Like for example a third slit in front of the first one, showing an asymmetric interference at that side?

(Note: of course I'm not doubting the experiment and the physics, only trying to understand it better.)

  • $\begingroup$ Your Q is a good one. Just you push to much the reasoning when you apply it to the first grid/slits.. It causes diffraction, and as such it detects electrons. But it does say nothing to their path through a specific opening. The rest is interesting and I ll follow this Q. $\endgroup$
    – Alchimista
    Commented Feb 25, 2019 at 13:39
  • $\begingroup$ Really closely related question here. (Just substitute the word "Stern-Gerlach machine" with "slit".) $\endgroup$
    – knzhou
    Commented Feb 25, 2019 at 17:49

2 Answers 2


Your question is an excellent one, and puzzled me when I was studying for my PhD in quantum theory.

The answer is that the intermediate screen in which the slits are formed can indeed collapse the wave functions of electrons projected towards it; but they are the electrons that make it through neither slit, being instead absorbed or scattered by the intermediate screen. (You can think of them, if you like as electrons that are poorly aimed and pass far to the left or right of the two slits, but that is a simplification.) The effect is very easy to see with two-slit experiments performed with light, as you will see the intermediate screen illuminated by the photons hitting it- only the photons that do not get absorbed or reflected by the intermediate screen pass through the slits to make the interference pattern.

More generally, a collapse of the wave function occurs when a particle's position becomes localised as a consequence of some physical interaction. Usually we refer to such an interaction as a 'measurement', but that label is an unfortunate one as it can be misinterpreted as meaning that the collapse doesn't take place except when there is a deliberate attempt to measure.

In the case you mention, the electron can either be localised by hitting the intermediate screen, or, if that does not happen, by hitting the detector beyond it. It is only the electrons which make it to the detector which contribute to the interference pattern.

Of course, the equations we use to model the two slit effect are simplifications of reality. We solve a one-electron Schrodinger equation with the intermediate screen modelled as a potential barrier with sharp edges where the slits are, ignoring the fact that the potential barrier is much more complicated in reality, and in fact its impact is mediated by photon exchange with the inbound electron. But the simplified model does a good enough job of predicting the outcome of the experiment.


The electron is a quantum mechanical entitity, i.e. it obeys solutions of quantum mechanical equations.

The quantum mechanical problem to be solved is : electron of momentum p impinges on two slits of width $w$ each, a $d$ distance apart. What is the probability distribution of seeing the electron on a screen after it has left the slits?

The probability will be given by the $Ψ^{*}Ψ$ of the solution of the above quantum mechanical problem.

Here is the experiment:

enter image description here

So truly the wave function justifies its name by showing a probability distribution with interference patterns expected from the solutions of a wave equations.

Why don't the slits themselves cause the wave function to collapse?

Because the fringe electrical fields of the slits are part of the boundary conditions which will define the wavefunction of the particular experiment.

Conversely if the slit is acting in the "wave domain", leaving the wave nature of the particle unchanged

The wave nature is not an attribute of the particle, the way the spin, the mass and the charge are. The wave nature is part of the particular solution of the quantum mechanical equation with its set up and boundary conditions, which will allow to see it or not.

A single electron leaves a bubble chamber track defining it as a particle. Its wave nature appears in the probability of hitting a hydrogen atom and ionizing it, there is a probability distribution for this.

elect elect2

The curly line was produced by an electron that was struck by one of twelve passing beam particles in a liquid hydrogen bubble chamber. It curves in an applied magnetic field and loses energy rapidly, spiraling inwards.

The curly line is from the tail of the interaction of the charged Kaon beam traversing the chamber, a high momentum electron kicked off. All the dots are small energy electrons kicked off the hydrogen atoms of the chamber. It is all about probabilities, and the wave nature becomes evident only in the special experiments where interference in the accumulated distribution of individual scatters can be seen, as above with the single electron double slit.

This is an interesting experiment, showing how changing the boundary conditions affects the interference patterns.

So the answer to the title question :

In the double-slit experiment, why don't slits destroy interference pattern?

is that:

the slits are part of the boundary conditions that generate the specific wavefunction which gives to the probability distribution the interference patterns.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.