Consider a free electron, with photon, that runs to electron under some angle(as everybody says). Compton scattering is happening, and electron instantly reemits photon in different angle.

First, how can we talk about angle between electron, and entity, that even not a point, but has no coordinates at all? And how we can say about the direction of moving?

My assumption- when electron emits photon (like during the Compton scattering), photon's direction is undefined- entangled. It gains an exact value when something absorbs photon, before - photon moves in all direction at the same time.

Wondering the next, I remember that photon's coordinate is not localized, hence according uncertainty principle, momentum of photon is 100% known.

The only mind, is that, by meaning that we 100% know momentum means that we know the speed, but not direction.

In the other hand, the thing that says, that photon propagates in all direction is this

enter image description here

People also represents this classic Coulomb's field ripple as em wave, or photon, as you can see, it propagates immediately in all directions. Furthermore, if you play with moving charge field simulators, you can notice, that there is no cases when em wave propagates in one direction, there always be the opposite propagation.

I'm confused.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – rob Sep 19 '18 at 7:33

You are confused because you are mixing incommensurable descriptions. Either you use a quantum mechanical view or you use a classical view. The idea that "a photon" is directly related to a ripple in a classical electromagnetic field as depicted in your question is incoherent (see this question for more on the relationship between the photon and a classical electromagnetic wave).

Next, a photon's "location" is a difficult notion to begin with, due to issues with relativistic notions of position operators (cf. this answer by Arnold Neumaier and references therein). What a photon does have is a probability to hit a detector (a surface) in a particular spot, but it does not have a probabilitiy to "be at a particular point". So your attempt to apply the uncertainty principle to create a paradox doesn't work either because the notion of position - and hence of "position uncertainty" - is inapplicable.

Finally, your idea that the photon and electron are entangled after scattering is certainly correct. Only when you measure either of them, the entanglement is broken and a value for the energy transferred and hence the scattering angle is determined. Note that the scattering angle does not need the position of the electron to be well-defined - it is simply the angle between the incoming and the outgoing photon rays, as determined by the location of emission and location of detection of the photon. The exact distribution for the different definite photon states with different angles in this entangled state are given by the Klein-Nishina formula, one of the first powerful results coming out of quantum electrodynamics in its early stages.

  • $\begingroup$ "it does not have a probabilitiy to "be at a particular point", can we say that probability, of photon being in any points, at any time are the same and equals zero? $\endgroup$ – user205628 Sep 18 '18 at 17:27
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    $\begingroup$ @Artur It does not have such a probability. It is not zero. It is not not zero. The question "At which point is the photon?" simply doesn't make any sense within the quantum relativistic framework. $\endgroup$ – ACuriousMind Sep 18 '18 at 17:36
  • $\begingroup$ Direction or rather momentum is actually $\dfrac{dx}{dt}$ but never mind. I've edited question with an ultimate question. Please, if You can, complete Your answer $\endgroup$ – user205628 Sep 18 '18 at 17:57
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    $\begingroup$ @Artur Please do not edit questions to ask different questions than they did before. If you have a new question, ask a new question. It is neither clear what exactly the question is you edited in, nor what its relation to original question is supposed to be. Please ask questions in proper text form instead of confusing pictures. $\endgroup$ – ACuriousMind Sep 18 '18 at 20:30

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