# Isn't the 'slit' in a double-slit experiment also a wave?

I'm new to QM so excuse my naivety. I was watching an online MIT QM course that described the double-slit experiment (with electrons) when it occurred to me that I have a question. In the video, the lecturer just drew a picture of a solid wall with two slits and then showed pictures of the interference patterns generated by shooting a single electron at the slits. Fine enough, the electron interferes itself, which is beautifully explained by the wave function.

However, aren't the atoms in the wall with the slits also described by a wave function? I mean, can we even meaningfully draw a picture of a 'slit' if it is roughly at the scale of an electron? Aren't we supposed to be dealing with a wave function there too? Looks to me the wall with the slits is treated as a 'classical' object (you can touch it, feel it, smell it) while electron is treated as a quantum object. But that simply cannot be the case.

Question: how does the wave function that describes the atoms around the slit 'know' to interact with the wave function of the electron? Does it collapse? The reason I ask this is because when the electron does not make it through the slit, it must have collided with one of the atoms: but wouldn't collision imply that two particles were at the same place at the same time? Doesn't that require wave function collapse? I'm confused ...

• Very perceptive, Lajos. Have you ever heard of the optical Fourier transform? Jul 19, 2015 at 8:02
• Very nice conclusion. In this answer I told the same and give a different answer about why we see intensity distributions behind edges. Jul 19, 2015 at 8:46
• The collision you talk about between a "slit electron" and your probe electron would be an exchange of virtual photons (the messenger particles of the EM force), causing the scattering. Electrons don't really collide classically (same place at the same time) just like you conclude. Jul 19, 2015 at 8:58
• Very similar to your point of view is my answer here physics.stackexchange.com/questions/234527/… Sep 15, 2016 at 18:25

To get a wave function one has to solve the quantum mechanical equation for the boundary conditions of the experiment:"electron impinging on two slits". In the usual description one is using approximations : the incoming electron is a plane wave, the effective potential of the electrons in the matter of the slits is high. Then the distance between slits and the width of the slits is chosen commensurate to the de Broglie wavelength of the incoming electron, so that the $\Delta x \Delta p \geq \hbar/2$ limit is binding and the quantum mechanical nature of the problem is apparent.

Question: how does the wave function that describes the atoms around the slit 'know' to interact with the wave function of the electron?

An effective potential approximates the collective behavior of the atoms around the slit.

Does it collapse?

Collapse is a bad terminology for interaction. The electron scatters off the collective potential of the atoms' electric field.

The reason I ask this is because when the electron does not make it through the slit, it must have collided with one of the atoms: but wouldn't collision imply that two particles were at the same place at the same time?

Do nor forget this is quantum mechanics. There exists a probability for the electron to scatter off or be absorbed by the atoms composing the slit. When the dimensions are chosen correctly, i.e. electron momentum, slit distances and width of slits, then the probability of going through is high and the interference pattern will be seen.

Doesn't that require wave function collapse?

Every measurements contributes to build the probability distribution for the problem at hand, i.e. the interference pattern is the complex conjugate square of the wavefunction for this set up . Each electron contributes a point on this distribution, showing how it interacted with the topology of the problem. The wavefunction is a general wavefunction for the setup and each electron gives one point by interacting with the recording screen. That is when the wave function "collapses" (I dislike the term, the wave function is not a balloon).

Just extending Anna's answer.

In order to observe the interference pattern which your probing electron(s) make, its associated de Broglie wavelength has to be several tens of nanometers at least. The wavelength of the electrons consisting the wall is approximately the distance between the atoms of the wall which is at least a few orders of magnitude lower than the wavelength of the probing electron, so they may be approximated as a solid, sharp-edged classical wall. Furthermore, their wavefunction oscillations are much faster and on average they can lower the coherence of the probing electron wavefunction only a bit.

Also, notice that the electrons of the wall exhibit a collective behavior as well, described in Solid state physics (see Bloch wave, Electronic band structure...) where their wavefunctions are smeared over the entire wall material, i.e. they are not so localized as you mention. Neither the probing electron may be localized well, since one has to well define its momentum/wavelength to observe its interference. The point is that you can't tell that a particular electron in the lattice was influenced by the scattered electron $-$ it's one against many.

Yes, to the notion that it is really interactions between charged particles with quantum natures. The "walls" of the slit are really the outer electrons of the surface atoms of the substance that are either covalently or metallically bonded to each other. Those bound electrons form a potential surface that is not exactly square, but when viewed from the perspective of energies of the electrons impinging on them would be close to a "square channel". The mass of the "walls" is so much higher than the mass of electrons that the walls can be considered fixed. the I wouldn't think of the wave function "collapsing" so much as being smoothly deformed.

Furthermore, the "real" situation is probably much more complex at the boundary. Since the potential is not infinite, there will be penetration into the "wall" and even absorption for some of the electrons, but they will also be re-emitted, possibly after traveling a distance along the length of the slit. It is rather amazing that such a simple model of this more complex process can yield such a high-fidelity answer. Almost as amazing as the quantum mechanical phenomena that are predicted. Might be fun to model different potentials applied to the slit material.

Yes, it is important to take in attention the surface electrons from edges. Electrons, shoot far enough to the edge, as well as photons, interact with their electric field with the electric field of the surface electrons. The difference is, tha the first intensity fringe from electrons makes the real shadow wider. In case of photons the first fringe makes the real shadow smaller than the geometrical shadow.

If recognize, that the common field is quantized, it is easy to explain, why single electrons (or photons) in interaction with a single edge over time produces intensity distribution.