What does the Hamiltonian do in the Heisenberg picture? I know that the Hamiltonian is not supposed to change in time or from Schrodinger to Heisenberg picture, but when actually acting on a state with this Hamiltonian, isn't it true that the operators in the Hamiltonian are the Heisenberg ones that are functions of time? If this is true, how are these Heisenberg pictures operators even defined in the first place?
 A: Suppose you have a Hamiltonian $H$ with no explicit time dependence.  We can treat the Hamiltonian like any other observable operator: Start from the definition of the operator in the Schrödinger picture, then apply the unitary transformation $U = e^{-iHt/\hbar}$.  Normally this gives $A(t) = e^{iHt/\hbar} A e^{-iHt/\hbar}$ where $A$ was the time-independent operator in the Schrodinger picture and $A(t)$ the time-dependent operator in the Heisenberg picture.  But in this case $A$ is just $H$, which commutes with $e^{-iHt/\hbar}$, so we can cancel out the time dependent factors and end up with $H(t) = H$.
So we see the time dependence cancels out, and in fact this will still hold true even when you express $H$ in terms of other operators, which will themselves have time dependence in the Heisenberg picture.  For example, consider the harmonic oscillator Hamiltonian $H = \frac{p^2}{2m} + \frac{1}{2}m\omega^2x^2$.  In the Heisenberg picture one gets $x(t) = x(0) \cos(\omega t) + \frac{p(0)}{m \omega} \sin(\omega t)$ and $p(t) = p(0) \cos(\omega t) - m \omega x(0) \sin(\omega t)$.  So we have $$\frac{p^2}{2m} = \frac{p(0)^2}{2m}\cos^2(\omega t) + \frac{m \omega^2 x(0)^2}{2}\sin^2(\omega t) - \omega x(0) p(0) \cos(\omega t) \sin(\omega t)$$ and $$\frac{1}{2}m\omega^2x^2 = \frac{m \omega^2 x(0)^2}{2}\cos^2(\omega t) + \frac{p(0)^2}{2m}\sin^2(\omega t) + \omega x(0) p(0) \cos(\omega t) \sin(\omega t)$$ and thus $$H = \frac{p(0)^2}{2m}\cos^2(\omega t) + \frac{m \omega^2 x(0)^2}{2}\sin^2(\omega t) - \omega x(0) p(0) \cos(\omega t) \sin(\omega t) \\+ \frac{m \omega^2 x(0)^2}{2}\cos^2(\omega t) + \frac{p(0)^2}{2m}\sin^2(\omega t) + \omega x(0) p(0) \cos(\omega t) \sin(\omega t) \\ = \frac{p(0)^2}{2m} + \frac{m \omega^2 x(0)^2}{2}$$
So again the time dependence cancels out.  Although that's just one example, my point is that you can still have a time-independent Hamiltonian even if it's expressed in terms of operators that are themselves time-dependent.  I think that addresses your concern.
