I Are Eigenstates of operators always stationary states?

No. The eigenstates of an Hermitian operator are stationary only if that operator commutes with the Hamiltonian of the system.

 

Does the Hamiltonian always commute with itself?

 

Let's work in the Schrödinger picture. Then the state vectors carry the entire time evolution, while the essentially self-adjoint (Hermitean is not sufficient!) operators that represent the observables are time-independent (we leave out the somwhat more complicated case of explicitly time-dependent observables). Then the state vector obeys
$$\mathrm{i} \hbar \partial_t |\psi(t) \rangle = \hat{H} |\psi(t) \rangle.$$
Obviously the formal solution solution of this equation is
$$|\psi(t) \rangle = \exp \left (-\frac{\mathrm{i}}{\hbar} \hat{H} t \right ) |\psi(0) \rangle.$$
Now suppose at the initial time ##t=0## the system has been prepared in an energy eigenstate ##|\psi(0) \rangle=|u_E \rangle##. Then you have [corrected in view of #6]
$$|\psi(t) \rangle = \exp \left (-\frac{\mathrm{i}}{\hbar} \hat{H} t \right ) |u_E \rangle =
\exp \left (-\frac{\mathrm{i}}{\hbar} E t \right ) |u_E \rangle,$$
which means that the time evolution of the state is just given by multiplying ##|u_E \rangle## with a phase factor.

Now since the physical content of the state ket is just given by Born's rule, i.e., if you have a complete set of compatible observables ##A_i## (##i \in \{1,\ldots,n \}##) and ##|a_1,\ldots,a_n \rangle## a complete set of common eigenvectors of the corresponding essentially self-adjoint operators ##\hat{A}_i##, then the probability (distribution) to measure at time ##t## the values ##(a_i)## is given by
$$P_{\psi}(t,a_1,\ldots,a_n) = |\langle a_1,\ldots a_n |\psi(t) \rangle|^2,$$
a phase factor doesn't play any role, i.e., for the above case that an energy eigenstate is the initial state of the system, the probability distribution is time-independent
$$P_{\psi}(t,a_1,\ldots a_n)=|\langle a_1,\ldots a_n |u_E \rangle|^2.$$
Thus, the eigenvectors of the Hamiltonian represent stationary states of the quantum system.

 

Perhaps you meant:

$$|\psi(t) \rangle = \exp \left (-\frac{\mathrm{i}}{\hbar} \hat{H} t \right ) |u_E \rangle = \exp \left (-\frac{\mathrm{i}}{\hbar} E t \right ) |u_E \rangle,$$

 

An operator ##H## commutes with itself if ##HH = HH##

That doesn't leave a lot of room for any operator to get out of that one!

 

So, for a non-self-commuting operator, H, we have:

##HH \ne HH##?

 

I don't see any difference between my formula and yours, or am I overlooking something?

 

You just missed out the ##\exp## from the last expression.

 

Argh, yes, you are right. I'll correct it. Thanks.

 

Well, trivially any operator commutes with itself. The commutator of two operators is defined by
$$[\hat{A},\hat{B}]=\hat{A} \hat{B}-\hat{B} \hat{A}.$$
Now set ##\hat{A}=\hat{B}##, and you immediately get 0!

 

If the Hamiltonian is explicitely time dependent in the Schrödinger picture, then H(t₁) will not strongly commute with H(t₂). This is actually the root cause for Dyson's series.

 

If the Hamiltonian is explicitly time-dependent it would generally not make any sense to talk about stationary states anyway.