Does a single atom have entropy? In statistical thermodynamics, the entropy is given by $S = k_B \ln{\Omega}$ where $\Omega$ is the number of possible microscopic states for a given macroscopic system. 
But the atom as a system can have several micro states corresponding to the same value of its energy. For example, the hydrogen atom in it's second energetic level can be in any of 4 states. So can we say that a hydrogen atom in a $n=2$ state has more entropy that another in a $n=1$ state? 
 A: Yes, a single atom could have entropy, if the state of the atom is defined as some density operator.  For example:
\begin{equation}\tag{1}
\rho = \frac{1}{2}| \, 2, 1, +1 \rangle \langle 2, 1, +1 \,| + \frac{1}{2} | \, 2, 1, -1 \rangle \langle 2, 1, -1 \,|,
\end{equation}
where $|\, n, l, m \rangle$ is the micro-state of energy level $E_{n, l, m}$, where
\begin{align}\tag{2}
n &= 1, 2, \ldots, \infty,
&l &= 0, 1, 2, \ldots n - 1,
&m &= 0, \, \pm 1, \, \pm 2, \ldots, \, \pm l.
\end{align}
According to (1), the atom has a probability $\frac{1}{2}$ to be in the state $|\, 2, 1, +1 \rangle$ and probability $\frac{1}{2}$ to be in a state $|\, 2, 1, -1 \rangle$.  Thus, the atom has statistical entropy $S = \ln{2}$.
This macro-state isn't the same as a quantum superposition.  For example:
\begin{equation}\tag{3}
| \, \psi \rangle = \frac{1}{\sqrt{2}} |\, 2, 1, +1 \rangle + \frac{1}{\sqrt{2}} |\, 2, 1, -1 \rangle
\end{equation}
is still a micro-state and it's entropy is 0.  In this case $\rho \ne |\, \psi \rangle \langle \psi |$.
The micro-state and macro-state depend on the way the atom has been prepared.  (1) and (3) could be the result of some interaction with another system in the past.
