DrClaude said:
In a certain sense, yes.
As
@Vanadium 50 pointed out, one usually does not consider entropy when dealing with individual systems. Instead, we look at energy. The hydrogen atom exists because the bound state between a proton and an electron has a lower energy than a separated proton and electron.
But why do systems tend to the lowest energy state? It is because of entropy. Lower the energy of an isolated system usually corresponds an increase in entropy. If a photon is emitted when the electron and the proton combine, then the entropy of the bound atom and the photon is greater than the entropy of the unbounded proton and electron alone.
There are situations where this is not the case. There was a time period in the early universe where stable atoms could not exist because the universe was too hot. Heck, there was even a time when protons themselves could not form (look up quark-gluon plasma).
Well, it's a bit more subtle.
The entropy in the usual sense of (Gibbs) statistical mechanics is defined as (using natural units, where ##\hbar=c=k_{\text{B}}=1##)
$$S=-\mathrm{Tr} (\hat{\rho} \ln \hat{\rho}),$$
where ##\hat{\rho}## is the Statistical Operator of the system, representing the quantum state of this system.
Considering a system in a pure state, you have ##\hat{\rho}=|\psi \rangle \langle \psi|## you find ##S=0##. From an information-theoretical point of view this expresses the fact that within quantum theory a pure state represents a state of complete knowledge, i.e., you cannot specify a system "better" than by preparing it in a pure state.
Now consider a hydrogen atom. Here it depends on the level of sophistication you study it. Usually the first encounter is in the quantum-mechanics 1 lecture, where it is one of the few analytically solvable eigenvalue problem for the Hamilton operator taking into account only the Coulomb interaction between proton an electron. The energy eigenstates are the stationary states of a system and thus indeed of particular interest. That said makes immediately clear that when preparing a hydrogen atom (seen in this approximate way) in one of its energy eigenstates nothing happens to it, i.e., it stays in this state forever, i.e., as an isolated (closed) system it does not tend to relax to its ground state, i.e., the state of lowest energy.
Of course, this Hamiltonian is only a first approximation, which does not take into account many details, leading to the fine and hyperfine structure of its spectral lines. More important in our context is, however, that in fact the electromagnetic field itself is also a dynamical system, i.e., it is not merely providing the binding Coulomb potential for the electron and proton to a hydrogen atom but has to be quantized itself, and this enables the description of spontaneous emission of electromagnetic waves (or photons which is in this case really the adequate language, because spontaneous emission is the most obvious reason for the necessity of field quantization since it cannot be reliably described within semiclassical approaches). The quantization of the electromagnetic field leads to quantum fluctuations and the corresponding additional terms in a more complete Hamiltonian including the proton, electron, and the (quantized) electromagnetic field, which can be adequately treated in perturbation theory, leading to non-zero transition probabilities between the before calculated hydrogen energy eigenstates, and this implies that with some probability a hydrogen atom prepared in one of its excited states will emit a photon spontaneously bringing it to a lower energy eigenstate. Of course there are some selection rules concerning angular momentum, but the upshot is that indeed the hydrogen atom will relax to its ground state, which in this sense is the only really stationary state.
