Q:

Why do electrons in an atom occupy only the stationary states?

When we talk about the elementary problems in quantum mechanics like particle in a box, we first calculate the energy eigen-function. Then we say that the most general state is the linear combination or superposition of these basis eigen-functions. But when we go to the atoms, say hydrogen atom, we end up in calculating energy eigen-functions and say the electrons  occupy these stationary states starting from least energy state (ground state). I have seen this in solid state physics too.  For instance in nearly free electron model, we calculate energy eigen-functions with eigenvalues 
$$E=\frac{h^2k^2}{2m}$$ where $k=n\pi/a$, 
$a$=length of the sample and assume that electrons are going to occupy these states. Here too there is no discussion of linear combination of these states.
So my question is why don't we talk about the state functions here that may be the linear combination of two or more than two stationary states in atoms or are the conditions under which electrons occupy only the stationary states ? 

A:

Why do electrons in an atom occupy only the stationary states?

This isn't true. An electron in an atom can be in any superposition of states. This is one of the basic postulates of quantum mechanics: linearity.
For example, say an atom has a ground state 1 and an excited state 2, and let's say we're able to prepare it in a pure state 2. It will decay electromagnetically to state 1. This decay is represented mathematically by a process in which the wavefunction becomes a mixture of states 1 and 2, with the amplitude of 2 decaying exponentially and the amplitude of 1 growing correspondingly.
Energy is special here only because many of the measuring devices we use to study atoms are energy-sensing devices. When we measure using one of these devices, we always get a definite energy. Take the two-state example again for simplicity. In the Copenhagen interpretation (CI), this is because of wavefunction collapse. In the many-worlds interpretation (MWI), the measuring device becomes a superposition, but it's a superposition of a state in which the device measured a single energy and another state in which the device measured the other energy. You can also discuss this in terms of decoherence.

A:

Depending on specific situations, I can think of two independent answers to the question "why we mostly only care about stationary states":

decoherence by photon interaction (I won't focus on this aspect);
all (effectively nondegenerate) systems becomes (improper) ensembles of stationary states IF we have uniform ignorance of time (I will focus on explaining this idea).

First, I assume you have basic knowledge about density matrix. (I think I have to do so, since otherwise my answer would be too long.)
Given an arbitrary initial density matrix $\rho_0$, we construct a time ensemble based on uniform distribution of time. Why do we want such time ensemble? Say we set the initial state, and then we wait while we don't record the time. Not recording time roughly means we have an uniform ignorance of time. Given that uniform ignorance of time, the density matrix describing the system is the time ensemble $\rho_{\text{time}}$. Say $U_t$ is the time evolution operator.
$$\rho_{\text{time}} =\lim_{t \rightarrow \infty} \frac{1}{t} \int^t_0 \mathrm{d}t \ U_t\rho_0 U_t^\dagger$$
Say the initial state is in a pure state superposition $\rho_0=\left(\sum_m  C_m \left|m \right>\right) \left( \sum_n C_n^\dagger \left< n \right|\right)$
$$\rho_{\text{time}} =\lim_{t \rightarrow \infty} \frac{1}{t} \int^t_0 \mathrm{d}t \ U_t\rho_0 U_t^\dagger = \lim_{t \rightarrow \infty} \frac{1}{t} \int^t_0 \mathrm{d}t \sum_{m,n} e^{-i (E_m-E_n)t} C_m C_n^\dagger \left| m \right> \, \left< n \right|$$
Taking the limit first gives the delta function $\delta_{m,n}$
$$\rho_{\text{time}}=\sum_{m,n} \delta_{m,n} C_m C_n^\dagger \left| m \right> \left< n \right|=\sum_{n} \left|C_n \right|^2 \left|n \right> \left< n \right|$$
Here we go! If we have an initial state and an uniform ignorance of time, all phase information is lost and the density matrix is diagonal in energy so it's an (improper) ensemble of energy states. For thermodynamic properties which are independent of time, such time ensemble capture all information about the system. Therefore, only energy states and the occupation probabilities of the states matter. The ideas of superpositon of energy states and interference between energy states become non important roughly speaking. (The above derivation requires non-degeneracy. The reason why it's often reasonable is that there is rarely perfect symmetry. I won't elaborate this more.)

A:

One reason we focus on energy eigenstates is that atoms spend almost all of their time in an energy eigenstate, and their spectrum is a result of transitions between them.  
Another reason is pedagogical:  to peel back the onion one layer at a time.  But before too long, many courses do include examples of systems that are not in an energy eigenstate.  One popular example is the harmonic oscillator with the system in a Gaussian state that is displaced from the origin.  All energy eigenstates are centered on the origin; the displaced Gaussian is a superposition of all energy eigenstates.  When one calculates the time evolution one finds that the state oscillates back and forth about the origin ... simple harmonic motion.

