Yesterday it was stated in this site that a nucleus oscillating in a crystal lattice (may) produces a single photon. I thought that conservation of (angular) momentum requires the production of at least 2 photons, can someone please explain if that is true and how you produce one single photon, expecially considering the thermal radiation produced by the oscillation of a nucleus?


I realize that referring to crystal lattices hugely complicates matters, so I'll ask the general question: suppose we have an individual free charge and we make it oscillate at frequency k Hz on the z axis. How many photons can be emitted? if one is possible what happens to conservation of momentum? What determines the exact direction of the MF oscillation and propapation ?

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    $\begingroup$ Hint: If it is possible to absorb a single photon without violating any conservation laws. then it is also possible to emit one. $\endgroup$
    – Rococo
    Commented May 23, 2017 at 4:44
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    $\begingroup$ @bob: a nucleus in a lattice isn't isolated so it can transfer energy and momentum (including angular momentum) to the crystal as a whole. The restriction on creating two photons applies to e.g. particle annihilation where the annihilating particles are isolated and cannot transfer energy or momentum to anything but the photons they produce. $\endgroup$ Commented May 23, 2017 at 5:51
  • $\begingroup$ @JohnRennie, thanks, just suppose a body in thermal equilibrium where all nuclei have roughly some frequency, an electron hits one of them and discharges its KE which is turned into heat. The nucleus increases its KE and the electron's momentum/a is/are conserved and may be transferred to the lattice. Now , the question is: how is KE turned into thermal radiation? is it only the frequency of the oscillation that is adjusted or fresh photon/s is /are emitted? if no new photon is emitted, consider the limit case of near absolute 0: the nucleus is not oscillating and then emits thermal photon/s $\endgroup$
    – user137879
    Commented May 23, 2017 at 6:09
  • $\begingroup$ @bob: see my comment to your other question. $\endgroup$ Commented May 23, 2017 at 6:35
  • $\begingroup$ @JohnRennie, Thanks, I edited my question to pinpoint the issue at hand $\endgroup$
    – user137879
    Commented May 23, 2017 at 6:45

1 Answer 1


Let's take the concrete example of a heteronuclear diatomic molecule. This is probably as close as we can come to an ideal harmonic oscillator. To a good approximation the selection rule for the vibrational transition is:

$$ \Delta v = \pm 1 $$

(the anharmonicity in the internuclear potential means that other transitions do occur, but these have a low probability and can usually be ignored).

But in most cases a pure vibrational transition cannot occur precisely because of the conservation of angular momentum. There is a further selection rule:

$$ \Delta J = \pm 1 $$

That is, the transition is rovibrational so both the vibrational and rotational quantum numbers must change at the same time. This is because the angular momentum of the diatom molecule must change by the opposite of the photon spin to conserve angular momentum.

In combined electronic/rovibrational transitions the rotational quantum number can remain unchanged, i.e. $\Delta J = 0$ leading to the Q-branch in the spectrum, but only if the angular momentum of the excited state differs from the ground state by $\pm \hbar$. In this case the angular momentum of the electron state changes by the opposite of the photon spin.

I've chosen a specific example because your question:

Suppose we have an individual free charge and we make it oscillate at frequency k Hz on the z axis

is too vague to be answered. You need to consider what creates the potential within which the charge is oscillating, and what creates the potential will be involved in the conservation of angular momentum. In the case I describe it is the diatomic molecule that changes state to conserve angular momentum, and in the case of the lattice it is the lattice that conserves angular momentum.


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