Question: In classical physics book of Kleppner, I read that

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> An atom can "jump" from one stationary state a to a lower b by emitting radiation with $E\_a-E\_b$. The frequency of the emitted "package of radiation" is $$\nu=\frac{E\_a-E\_b}{h}$$
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I remember accurately what I read on chemistry book of Bohr atomic model, I had read that electrons can jump from one state to another by emitting of absorbing radiation. So I think it ("atom") was a typo, wasn’t it? Or there's some other information what I am missing?

Answer: 
It is a useful-but-incorrect approximation to say that “the electron” changes state during an atomic transition, as if the behavior of the atomic nucleus were unchanged.

When an intro-quantum student solves the Schrödinger equation for the hydrogen atom, the first step is to turn the two-body system into a one-body mathematical problem. There are two ways to make this approximation, which are not quite equivalent:

1. You can assume that the nucleus of the atom is infinitely heavy, so that only the electron has any dynamics.
2. You can do a change-of-variable where, instead of the separate positions of the proton and the electron, your two position vectors are the electron-proton separation and the location of the atom’s center of mass. Then you choose a reference frame where the center of mass doesn’t move, and find the stationary states for the interparticle vector.

In the second approach, you admit that the electron and the proton are actually both moving about their “barycenter.” There is a computational difference: in the first approach, the mass parameter is the electron mass $m\_e$, while in the second approach the mass parameter is the “reduced mass” $\mu = (m\_e^{-1} + m\_p^{-1})^{-1} = \frac{m\_e }{1 + m\_e/m\_p} \approx m\_e$. But that’s the *only* computational difference. The second approach is more correct; it also generalizes more readily to exotic two-body systems where the interaction is electromagnetic but the mass ratio is less extreme, such as muonic atoms $(m\_\mu/m\_p \approx 1/10)$ or positronium $(m\_{e^-} / m\_{e^+} = 1).$

For positronium, it should be clear that you can’t say “the electron” has changed state during a transition. Which electron would you mean? The positive one or the negative one? Clearly they both have to change their orbits around the center of mass, which is halfway between them.

For multi-electron atoms it becomes more important to say “the atom” changes state, because you cannot in general assign any particular electron to any particular orbital. There are no simple multi-electron atoms; the approximation of a single-electron excitation is almost never useful except as a cartoon sketch at the very beginning of solving a problem.
The atom itself is said to be in an “excited state”, or higher energy state, when one of its electrons jumps to higher energy level. This can happen when an atom absorbs a photon, increasing its total energy. The converse, where an electron drops to a lower energy level via photon emission, we say the atom is now in a lower energy state and the atom decreases its total energy.

So using the words the *atom changes state* is accurate. However, to say the *electron has changed its state* is at best inaccurate.