Does beta decay have an explanation similar to that of the atomic transitions? Atomic states are eigenstates of the atomic Hamiltonian. They are stationary states. Transitions from one atomic state to another are possible only in presence of an external perturbation e.g., the electromagnetic field.
Does the occurrence of radioactive beta decay have a similar explanation? More precisely, can we say that the states of the nucleus are eigenstates of the strong interaction Hamiltonian and beta decay happens due to weak interaction acting as the perturbation? If this is so, can someone give a reference?
 A: Your question points an important misunderstanding that creeps in to the study of the atom: the difference between our model of the system, and the actual system. We study the Coulomb potential as the model hamiltonian, and that gives us a beautiful and complete set of eigenstates with which to quantify the atom.
Eigenstates are indeed stationary states, and thus cannot transition without some sort of interaction. Enter the electromagnetic field. Since it is weak $(\alpha \approx 1/137)$, we can treat it as a external perturbation and use the eigenstates as approximations and calculate transition rates.
But the electromagnetic field cannot be turned on and off: it is always present. Transitions are always possible, and the orbitals are not stationary, and are not exact eigenstate of the hamiltonian. Exact analytic solutions to the fully interacting atom even in zero average field are not possible, so we proceed as described.
This is very much like the case with the neutron and proton. In the simplest strong-interaction picture, we ignore electromagnetism and the weak interaction, and set $m_u=m_d$. The neutron an proton are then identical particles, differing only in iso-spin. They form an $I=\frac 1 2$ doublet, with the proton (neutron) being the $I_3=\pm\frac 1 2$ eigenstate.
Note that this is very much like a magnetic moment in zero magnetic field: there are two states with $S_z=\pm \frac 1 2$ with equal energy. An external external field ($B_z\ne 0$) breaks the hamiltonians rotational symmetry and splits these two eigenstates. Likewise with the nucleon, isospin symmetry is not exact, and the neutron and proton masses are split.
In the strong interaction, the two states are connected by exchanging pseudoscalar bosons (the three $I=1$ pions), e.g.: $n + \pi^+ \rightarrow p$, but that's not under consideration now. We're talking about the weak interaction.
The weak interaction introduced a perturbation that connect the states via vector boson exchange ($W^{\pm}$); the $W^{\pm}$ also connects the $(\nu, e^-)$ doublet leading to:
$$ n \rightarrow p+e^-+\bar{\nu}_e $$
There is of course copious literature describing this process. A good place to state is the Fermi-interaction (https://en.wikipedia.org/wiki/Fermi%27s_interaction). This was long before electroweak unification, so it describes nucleon eigenstates in the presence of an ad hoc interaction that connects them, very much like the approximations used in atomic transitions.
