I ask because I've always thought that the wave equation of the departing photon would be (barring nearby mirrors or absorbers) a sphere of wavelets growing out from the emitting atom at lightspeed. If this is the case, then the direction of the momentum isn't determined until the photon is absorbed, and that could be hundreds of years later. If the momentum is transferred upon emission, that would mean the photon somehow knows the direction of its eventual absorber? Or is the momentum only transferred upon absorption?

I imagine Heisenberg will be invoked, so I want the answer to deal with a concrete example: Imagine a powerful point light source at position A radiating in all directions. A basketball-court sized light shield (shout out to Webb) is at B, one light-second from A. The shield's material perfectly absorbs light, it is a perfect black. Another light shield of the same material, 4 times the area of the one at B, is at C, two light seconds from A and sqrt(2) light seconds from B (i.e. A-C is at right angles to A-B). The relative velocities of A, B and C are initially zero, and both light shields face A. At time t0 the light at A is turned on. Describe the momentum changes measured at A, B and C in the next 3 seconds.

  • $\begingroup$ Something something... spooky action at a distance... something something ... Bell's Theorem... something something... remains a superposition until you measure. Therefore something something, question is in classical terms, the answer is quantum. $\endgroup$
    – safkan
    Feb 5, 2022 at 13:13
  • $\begingroup$ Photons are quantum mechanical particles , described by the mathematics of relativistic quantum mechanics, and their wave nature is not in space-time, but in the probability of detection . My answer here may help physics.stackexchange.com/questions/628624/… $\endgroup$
    – anna v
    Feb 5, 2022 at 14:04
  • $\begingroup$ Related, though not an answer to your question - How can a red light photon be different from a blue light photon? $\endgroup$
    – mmesser314
    Feb 5, 2022 at 17:43
  • $\begingroup$ I presume A feels no net recoil for the first second after A is lit, because every photon emitted at A is in a superposition of going in all directions. My first thought was that during the second second, A would recoil away from B, because photons being absorbed at B have resolved their direction of momentum, while the others remain unresolved. But, thanks to everyone's responses, I now see that those others are partly resolved. They have a 'hole' (B's shadow) in their superposition sphere. Net, they produce a recoil exactly opposite to those absorbed at B. So A remains unaccelerated. $\endgroup$ Feb 6, 2022 at 3:36

1 Answer 1


The momentum is not definite until you measure it. It doesn’t matter if you measure it in the recoiling atom or the emitted photon. Either way gives you the information about the state of the system.

This is less about the uncertainty principle and more about no counter factual definiteness.

Your example uses a powerful light source, so it is classical and is not relevant to the question. But if you made it a very dim light source, that emits single photons, then the above applies. Until you measure the momentum it is not definite. And you can measure the momentum of any part of the system to get information.


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