What is meant by the stationary states in quantum mechanics? This question came up as an exercise in my quantum text?
I typically find these "word" questions quite difficult. I want to answer:
"What is meant by the stationary states?" as concisely as possible.
I would say something along the lines of " Normalizable/(sufficiently smooth) solutions to the time- independent Schrödinger equation". But perhaps there is a better answer ( perhaps more physical)...
Any suggestions? Many thanks!
 A: Stationary states are eigenstates of the Hamiltonian, since they time evolve by just a phase. If $H|n\rangle= E_n |n \rangle$, then the time evolution of the state is given by
$$e^{i H t} | n \rangle= e^{i E_n t} | n \rangle, $$
which is just a phase.
A: The stationary states are those states which do not evolve as time passes - in this sense they are stationary. One needs to be careful however, and not forget that what we mean here is that the states do not have any observable changes over time. Thus, if a state evolves by just a global phase factor over time, this change is not observable, and thus such a state is a stationary state. You should have seen a characterisation of such states in terms of the Hamiltonian of the system in your course by now.
A: Let us say, that in the position basis, we study the evolution of the components $\psi(x,t)$.  By taking into account only those wave functions that have a property of being separable, i.e. $\psi(x,t)=\phi(x)\phi(t)$. This limits the choice of solutions to a specific class of wave function (eigen-functions) out of a wide variety of possibilities.
The term stationary state is used for those solutions of the T.I.S.E (time independent Schrödinger equations) for which the solutions are the eigen-functions (standing wave). These solutions would always be in a confining potential i.e. having some boundary conditions. The potential term in such a case is constant in time and is just a function of space i.e. V(x). The exponential terms of the plane wave get cancelled out when measuring the probability density and one gets $|\Psi(x)|^2\,\times (\text{Amp.})^2$ with no time dependence factor attached. In the energy basis, these solutions to the T.I.S.E would give the solutions having definite energy i.e. they are eigenvalues of the Hamiltonian.
However, what it does not mean is that electron is at rest!
