In the considerations in many QM books the usual simple scheme of a slit is represented and a plane wave (e.g. particles of almost fixed velocity) are falling on it. After the slit one has (if the slit is small) a spherical wave and the px component of the impulse can take values from almost – infinity to + infinity (Fourier theorem shows this but at least px take values from –p to +p where p is the initial impulse.
Now I wonder how this happens? [In the meaning of: what is the physical and not mathematical reason (like Fourier and commutators etc.). I am pretty sure how this happen in math.] I consider three scenarios. All three fell.

  1. I suppose that p just changes direction after interaction with long range forces from the slit walls. That means that px increases for the expenses of py. But that assumes that in y direction (along the initial velocity) py can change from p to 0. Now one has no slit along y direction so the wave packet does not contract in that direction and Fourier theorem will not give spread for py. So I conclude that this can not happen. So 1 is wrong.
  2. Then it is left the option that the particle receives impulse in x direction by the atoms of the walls of the slit. This is also not possible. If the T (temperature) of the walls was absolute zero than they would not give up dp to the falling particles. But Fourier theorem doesn’t care (or does it?) Even if T is not 0 every particle will cool down the walls and after a while T goes to 0 and the spread of px will stop. So 2 is wrong.
  3. @ClaudioSuspinsky (as far as I get it) answered in another post that p doesn’t change after the slit but one changes p by looking at it. E.g. applying an electron microscope (as in Heisenberg example in QM) in order to assure the particle went through the slit. 3A. In the example with the slit however there is no need for a microscope. Each time there is a particle behind the slit it is 100% sure that it went through the slit. A detector movable in x (or a CCD camera) placed after the slit registers every particle and calculates its px from the x position or simply measures them. It must show that px varies at least [-p,p], which means change of px happened in the slit. 3B. Also if there is no slit and we subject the falling particles to observation via an electron microscope the microscope will also cause spread in px. (I think it would be the same magnitude as with slit.) So in my opinion it is not the real reason about the spread after the slit. So I conclude that the reason is not in the electron microscope itself. 3 is wrong too. It must be the walls. But than 1. and 2. show that this is not true. What really happens in the slit and behind?
  • $\begingroup$ Is it fair to say that in essence, your question is, "how does momentum/energy conservation hold in the process of the particle passing past the slit because it looks like it gains all these high-momenta modes after passing through the slit which weren't present initially and thus, should have come from somewhere for the conservation of momentum/energy?"? $\endgroup$
    – user87745
    Sep 2, 2021 at 12:58
  • $\begingroup$ What are the frequency components of a plane wave perpendicular to the direction of propagation? Are those considered part of the momentum along the direction of propagation? Why is any transfer of energy or momentum needed between the wave and the slit? $\endgroup$
    – Jon Custer
    Sep 2, 2021 at 13:03
  • $\begingroup$ @DvijD.C. yes that's the essence. Plus 3 propositions which are not satisfactory. $\endgroup$
    – Mercury
    Sep 2, 2021 at 15:31

1 Answer 1


Since we are talking about waves, I suggest you consider what happens in analogous cases- for example, consider a water wave approaching a barrier in which there is a narrow opening. When the wave passes through the opening, it spreads in a circular front. That is a fundamental property of wavelike motion on a surface- a tendency to spread. The linear nature of the wavefront before it passes through the opening can be modelled as the result of a semicircular propagation of the wave from every point along the wavefront- the semicircular components spreading from every point interfere with each other in a way that yields a single long straight wave. The effect the gap in the barrier has is to isolate the portion of the wave passing through the barrier from the interference effects from points all along the length of the wave, so that small portion can now spread radially.

I hope you can see from the analogy that a long straight water wave that is heading in a y-direction say, meeting a long barrier across the x-direction, initially has momentum only in the y direction. When a portion of the wave makes it through the gap in the barrier, it spreads radially, and thus has momentum components to either side in the x direction. You would not assume that the wave has somehow acquired those components of momentum from its interaction with the barrier- you would simply assume that the wave will naturally try to spread in all directions in a semicircular front.

I suggest you consider the wave-function of particles aimed at a slit in a screen to behave in the same way as the water wave. Approaching the screen, you have a propagating wavefront- from each point on that wavefront you have a semicircular spread, with the semicircular disturbances interfering to produce the main wavefront. When the wave passes through the slit, the interference pattern that yields the main wavefront is suppressed- the short portion of the wavefront that passes through the slits can then spread in a semi-circular fashion without interference.

  • $\begingroup$ Yes you are right completely but for material waves. The good old Hyugens principle. In the case of QM the wave attached to the particle is just probabilistic. It can not interact with wall like the material wave. In fact in just vanishes meeting a wall without any effect in Energy impulse etc, But all properties of the particles must be in all points of the wave and nevertheless they vanish to nil. Very disappointed for me is that one is left with no interaction in this picture. The particle changes its direction and nothing happens to the wall like the 3 Newton's Law would afford. $\endgroup$
    – Mercury
    Sep 3, 2021 at 19:45

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