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In our introductory solid state lectures, the professor described von Laue's$^1$ diffraction conditions making the assumption of elastic scattering, which states that the incoming and the scattered radiation have the same wavelength. We also assumed scattering by each lattice point in all directions.

Now, don't these statements lead to violation of energy conservation, in the photon picture? (The corresponding wave picture makes sense though.) Because there was just one incoming photon while there were scattered photons in all the directions of the same wavelength. How to resolve this?

Or is that that I'm getting the photon picture of the elastic collision completely wrong? Maybe the correct photon picture is that a single lattice point scatters the incoming photon in a single but random direction. Please help!


$^1$Should it be von Laue or Von Laue?

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    $\begingroup$ Typically we don't capitalize the "von" or "van" -- see, e.g. Rembrandt. However, if a person was specifically named Van Whatever, that's different. The 'van' prefix is a now-obsolete way of indicating a person's geographic (or other affiliation) origin. $\endgroup$ – Carl Witthoft Aug 31 '20 at 12:00
  • $\begingroup$ @CarlWitthoft Thanks for noticing! And answering :) $\endgroup$ – Atom Aug 31 '20 at 12:23
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In elastic scattering one photon remains one photon. If you have an incident stream of photons (as is the case in most such experiments), then you will get scattered photons in all the directions.

However, thinking of photons as classical particles is rather misleading - mathematically they arise as excitation states when quantizing electromagnetic field, and as such already have a mode structure, i.e. like electromagnetic field a photon is already an incident and a scattered state. If one then tries to backtrack to the "intuitive" picture of ball-like particles scattered from an obstacle, one has to talk about photon wave packets, photon lifetime, etc. - it is quickly getting rather involved.

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  • $\begingroup$ Thanks, that makes sense! $\endgroup$ – Atom Aug 31 '20 at 8:25
  • $\begingroup$ "a photon is already an incident and a scattered state." Does this mean that the interaction with (say) the electron is just "collapsing" the photon into the scattered state? (Or is it a detector that puts the photon into the scattered state?) Are these two states probability distributions that map the location (trajectory?) of the photon? Also where is the energy for the scattering coming from? I am not a physicist but I am looking at scattering and trying to make sense of it. Your inputs would be much appreciated!! $\endgroup$ – Dunois Oct 2 '20 at 14:07

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