If two neutrons collide in 3D space and we want to determine the final velocities of both nuetrons (3 components for each neutrons), we can use the conservation of momentum equations and the conservation of energy equation. Those equations are 4 while the number of unknowns we have is 6. There are 2 free variables.

For classical perfectly spherical objects, the solution is strait forward as explained here. However, neutrons because of their quantum nature can't be treated as classical perfectly spherical objects. We can't define a precise point of collision, so the trick we use for classical objects won't work for neutrons or any quantum level particle, or will it?

If it doesn't work, does that mean we can't avoid having free two degrees of freedom in this system??

NOTE: What is meant by a collision is a momentum-exchange event

  • $\begingroup$ Electrons don't interact by contact interactions but electromagnetically, so you haven't fully specified the classical case until you have given the impact parameter which supplies the two missing classical degrees of freedom. $\endgroup$ – dmckee --- ex-moderator kitten Sep 22 '13 at 17:51
  • $\begingroup$ I see in your link that the momentum transfer must be supplied. That applies equally to the electron-scattering case, only with electrons you can get $\vec{q}$ from $\vec{b}$ (the impact parameter). $\endgroup$ – dmckee --- ex-moderator kitten Sep 22 '13 at 18:11
  • $\begingroup$ @dmckee Thank you for comments, my question is about the collision process itself. Choosing electrons as an example was unfortunate choice. I changed it to neutrons instead $\endgroup$ – Gotaquestion Sep 22 '13 at 18:35
  • $\begingroup$ I understood. Look again at the question you linked. There is no answer until the momentum transfer ($\vec{q}$) is specified. The system (classical or quantum) is undetermined until you think up a way to provide it. Ted simply assumed it, and Mark avoided talking about it, but the problem exists there too. Supply $\vec{q}$ and the extra degrees of freedom are constrained, don't supply $\vec{q}$ and they are unconstrained. Independent of whether the system is classical or quantum. The difference in quantum mechanics is that $\vec{q}$ is subject to uncertainty and found probabilistically. $\endgroup$ – dmckee --- ex-moderator kitten Sep 22 '13 at 19:14
  • $\begingroup$ @dmckee Thank you very much, still there is something that is unclear to me. I completely understand that there is NO unique solution to this problem. What Mark did and Ted agreed with him is described in”Two- and three- dimensional” section of [1]. They made a choice of q that makes sense physically. In a quantum system such a choice can’t be made because there is no clear “line of collision” as it is called in Wikipedia. Am I correct about that? [1] en.wikipedia.org/wiki/Elastic_collision $\endgroup$ – Gotaquestion Sep 22 '13 at 21:20

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