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Suppose a billiard ball moves in a straight line and the center of another billiard ball is at a distance from that line of less than the radius of a billiard ball. That distance determines which direction they bounce in an elastic collision.

A textbook (which I do not have before me, and the title and author of which are unknown to me) posits that a photon whose energy is 5000 electron-volts collides with a stationary electron and bounces in a certain direction. If photons and electrons, unlike billiard balls, lack extension in space, what is it about the physical conditions before the collision that determines what direction these particles move after the collision?

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Since the electron and photon are quantum mechanical objects, the angle at which the electron and photon move after the collision is probabilistic. As you have pointed out, they do not actually "collide" but instead they interact. You can think of this process as the electron absorbing the photon at one point, and then re-emitting one at another.

The scattering angle is therefore random, but the interaction does conserve energy and momentum. At the "instant" the interaction occurs, the scattered photon and electron are in a superposition of all possible states and all these possible states form a probability distribution, and this probability distribution can be computed using the appropriate quantum mechanical techniques, like QED and Feynman diagrams.

what is it about the physical conditions before the collision that determines what direction these particles move after the collision?

There are no exact specific conditions before the interaction that will tell you with certainty what direction the particles will travel. You are sort of thinking classically. Again, all quantum interactions are probabilistic. Even though we may calculate the differential cross-section as a function of angle (see the Klein-Nishina formula), the actual angle the particles take will be determined with certainty only after the measurement (collapsing the wave functions).

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  • $\begingroup$ In particular, even before the collision there is some variance in the particles' positions and momenta thanks to the uncertainty principle. $\endgroup$ – Carmeister Apr 14 at 16:14
  • $\begingroup$ I think your answer might be a little bit misleading due to wording. A particular result of an interaction might be random but collectively the angles show very meaningful distribution. I would really nitpick at "the actual angle the particles take will be determined with certainty only after the measurement" you are talking plural here, if you make sample large enough the collective result is not really random, you retrieve "classical" results. IMHO, for educational purposes, it's important to make a clear distinction between "a photon and an electron" and "photons and electrons". $\endgroup$ – luk32 Apr 14 at 22:33
  • $\begingroup$ When I use plural I'm talking about the particles, the photon and electron...two particles, therefore plural.Of course if we are talking about a large number of particles and many interactions, we approach a classical result. But thanks for pointing this out. $\endgroup$ – joseph h Apr 14 at 22:44
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The collision boundary differs from the classical mechanical billiard ball model in that there isn't a rigid boundary that defines whether the particles collide or not. In the billiard ball model, if "the center of another billiard ball is at a distance from that line of less than the radius of a billiard ball" then the balls clearly collide. If that distance is greater than the radius, then they simply miss.

The probability distribution, on the other hand, does not have a sharp point at which the probability drops to zero. If you attempt to understand this using the classical model, it's as if the balls were supposed to miss each other, but veered off anyway. But the quantum model asserts that there is a finite chance of the particles interacting even though they are further away than seems intuitively reasonable when based on classical mechanics.

Over a large number of interactions this behavior may average out to approximate classical behavior and to perhaps define a "usual" boundary at which they may bounce away from each other. But there will still be cases where the particles are closer and do not interact, or further away and do interact.

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    $\begingroup$ It’s also much too long to be a comment, so your reputation points don’t actually matter. Comments are limited to 600 characters. $\endgroup$ – Mike Scott Apr 14 at 20:59
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Electrons and photons do in fact have a spatial extension. It is not a spatial extension of the particles themselves though. They are point-like objects. On the contrary, the wavefunctions accompanying the electron and the photon are spatially extended. Because of this there is a continuous range of possible interaction (collision) outcomes. If the particles would behave classically (absence of a wavefunction) only one possible outcome of the collision would be observed. But because of the quantum Nature of the collision (interaction), i.e., the presence of a wavefunction, there is a multitude of possible outcomes. For each outcome one can calculate the chance of its occurrence and they all have to add up to one. As it has to be, because a chance that all possible events happen is bigger than one would imply that more things can happen than actually can happen.

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  • $\begingroup$ It's pushing things to claim that either a photon or an electron is in fact ever a "particle" or "object." We really can only say that in some circumstances their behavior mimics that of a particle. $\endgroup$ – Carl Witthoft Apr 14 at 13:43
  • $\begingroup$ @CarlWitthoft Are you denying that elementary particles are pointlike objects (I do, in fact, as I think they are non-string-like, extended structures, but that aside)? $\endgroup$ – Deschele Schilder Apr 14 at 14:02
  • $\begingroup$ Yes, I deny that anything can be point-like in the mathematical sense, and further that we are making a fundamental mistake in claiming that there is a physical structure in any Newtonian sense. $\endgroup$ – Carl Witthoft Apr 14 at 17:12
  • $\begingroup$ @CarlWitthoft And what about structure in a non-Newtonian sense? $\endgroup$ – Deschele Schilder Apr 14 at 17:21

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