How many photons are released upon de-excitation of an atom?

Question

In an single-electron atom, we consider the electron to be revolving around the nucleus, which remains at rest. But the nucleus should also be, in a way, revolving around the electron, however negligible this effect is. Both the nucleus and the electron would be following the Angular Momentum Quantization rule, therefore when the electron de-excites from a higher energy state, the nucleus must also do so. So my question is will this system release two photons or one?

I've read that the de-excitation of electron will release exactly one photon, regardless of the energy difference, but nothing is mentioned about its effect on the nucleus. Am I conceptually wrong in assuming that the nucleus de-excites from a "higher energy state"? Perhaps this can be better observed in an exotic atom where the orbiting particles are of comparable mass.

Answer 1

The full Schrödinger equation for an atom (let's say hydrogen), includes degrees of freedom for the electron (${\bf r_e}$) and the proton (${\bf r_p}$). They are factored into a center-of-mass motion and a separation:

$$ {\bf r} = {\bf r_e} - {\bf r_p}$$

with reduced mass:

$$ \mu = \frac{m_eM_P}{m_e + M_P} $$

The reduced mass accounts for the proton's complementary motion (which is somewhat classical sounds). Basically in an $L=1$ state, the $Y_1^m(\theta, \phi)$ has the angles defined with respect to ${\bf r}$, so yes: the proton is in a little spherical harmonic too.

Nevertheless, a nominal 1 photon transition occurs when the wave function with respect ${\bf r}$ changes from a higher energy state to a low energy one, and a single photon is emitted.