Why stationary state wave functions are eigenfunctions of the Hamiltonian operator? can somebody please explain why stationary state wave functions are eigenfunctions of the Hamiltonian operator? 
 A: Assume a system which is in an eigenstate of the hamiltonian  $|\psi(0) \rangle \in \mathcal H$ at time $t=0$, for which by definition $\hat H|\psi(0) \rangle=E|\psi(0) \rangle$. The time evolution of this system by the Schrodinger equation is:
$$|\psi(t) \rangle = e^{-\frac {i}{\hbar}{\hat H}t}|\psi(0) \rangle=e^{-\frac {i}{\hbar}Et}|\psi(0) \rangle$$
You can easily see that $|\psi(t) \rangle$ only differs with $|\psi(0) \rangle$ by a physically irrelevant global phase factor. As a side note, also observe that the state vector at time $t$ is also an eigenstate of the hamiltonian; i.e. a system which is in an energy eigenstate will stay an energy eigenstate (as long as nothing is measured of course). Now for any observable $\hat \Omega$, the probability distribution of measuring the eigenvalues $\omega$ does not change with time. Explicitly, the probability distribution of the outcome of measuring the observable $\Omega$ , at time $t$ is:
$$\Pr(\Omega=\omega,t) = ||\hat P_{\omega}|\psi(t) \rangle||^2 $$
Where $\hat P_{\omega}$ is the projector onto the eigenspace of $\hat \Omega$ with eigenvalue $\omega$. Using the first equation for eigenstates of the hamiltonian, we get: 
$$\Pr(\Omega=\omega,t)=||e^{-\frac {i}{\hbar}Et}\hat P_{\omega}|\psi(0) \rangle||^2 =||\hat P_{\omega}|\psi(0) \rangle||^2 $$
Which means that:
$$\Pr(\Omega=\omega,t) = \Pr(\Omega=\omega,0)$$
Thus, the statistical distribution of the measurable values of any observable is stationary with time when the system is initially in an eigenstate of the hamiltonian, which is why we call it a stationary state. Note that this includes all statistical properties such as expectation values, etc.; as they can all be calculated from the probability distribution function.
