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Discussions of conservation of momentum frequently use the metaphor of two billiard balls colliding. My impression is that this is not valid at the quantum scale - an illustration of the particles' trajectories should show the outgoing vectors with some uncertainty. Perhaps the total energy could still be conserved if the two particles were entangled in such a way that the imprecision of one particle's trajectory was balanced by the second particle's trajectory. Even if that was the case, I am not clear that the net vector would be as expected, which would therefore mean the momentum was not conserved.

An alternate way of viewing the problem: at the moment when two particles collide, the position is known very precisely (since they had to hit each other at the same place and time). Since momentum is complementary to position, this means the momentum has maximum uncertainty at that instant. While the momentum in a single collision may be perfectly conserved, perhaps the momentum being conserved is somewhat probabilistic such that over billions of interactions of billions of molecules (as the original force propagates) the original net momentum is not conserved.

I tried to look for answers to this question and here are some relevant ones. They seem to conclude that uncertainty does apply to single particles.

Does the Heisenberg uncertainty principle apply to the free particle?

Uncertainty principle: for an individual particle?

My question is prompted in part by a "tongue in cheek" video which shows a propeller in a closed box appearing to cause movement. The box is flimsy and the experiment is not meant to be definitive but made me wonder.

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  • $\begingroup$ as for the box, it is the pressure differential because of the pliability of the box seen. Rigid walls will bounce back the blown air. There should be a video with plexiglass $\endgroup$
    – anna v
    Apr 25, 2018 at 5:36

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Energy and momentum are exactly conserved at the microscopic level. They are not just conserved on the average. This was settled in 1926 by the Bothe-Geiger experiment, which disproved the BKS theory of quantum mechanics, in which conservation of energy was supposed to hold only on the average. (Exact conservation of energy implies exact conservation of momentum, if it holds in all frames of reference.)

Discussions of conservation of momentum frequently use the metaphor of two billiard balls colliding. My impression is that this is not valid at the quantum scale - an illustration of the particles' trajectories should show the outgoing vectors with some uncertainty.

Quantum mechanics doesn't say that all quantities are uncertain. The momentum of a particle can be exactly well defined.

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  • $\begingroup$ Thank you for answering. I did not mean that the momentum of a single particle could not be defined but just the suspicion that uncertainty may affect trajectories after a collision. $\endgroup$
    – PeterVermont
    Apr 21, 2018 at 23:29
  • $\begingroup$ The measurements of electron scattering in Compton and Simon's experiments were within 20 degrees of prediction which is not very precise. See books.google.com/… $\endgroup$
    – PeterVermont
    Apr 21, 2018 at 23:34
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    $\begingroup$ But can a scattering experiment also prove conservation in detail not just before and after, i.e., at all times, and if not then how do we know it is true so? $\endgroup$
    – hyportnex
    Apr 22, 2018 at 1:25
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    $\begingroup$ Excellent write-up on Bothe-Geiger experiment $\endgroup$
    – John Alexiou
    Apr 23, 2018 at 18:05
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If you initially knew the incident momentums, the sum of momentums will be preserved, but its difference almost certainly (it depends in the kind of interaction, but with probability 1) will have some uncertainty, thing that after the collision didn't happen.

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    $\begingroup$ I am sorry but have difficulty understanding your answer. Are you saying that while there is no uncertainty about the momentums be conserved, there is uncertainty as to whether or not they actually collided? $\endgroup$
    – PeterVermont
    Apr 21, 2018 at 23:49
  • $\begingroup$ Not only about if they collided, but also about how has they collided (at which angle will be deflected) $\endgroup$
    – Iliado Odiseo
    Apr 22, 2018 at 0:33
  • $\begingroup$ But essentially, yes. $\endgroup$
    – Iliado Odiseo
    Apr 22, 2018 at 0:33

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