Heisenberg Picture (Heisenberg, 1925)

Let observable $\hat{A}$ . The Heisen picture operator is

\tilde{A}(t)=U^\dagger (t)AU(t)

	Schrödinger Picture	Heisenberg Picture
State	$	\psi(t)\rangle = U(t)	\psi(0)\rangle$	$	\psi(0)\rangle$ constant
Observable	$A$ constant	$\tilde{A}(t) = U^\dagger(t)\, A\, U(t)$
Dynamics	$i\hbar\dfrac{d	\psi\rangle}{dt} = H(t)	\psi(t)\rangle$	$\dfrac{d\tilde{A}}{dt} = \dfrac{i}{\hbar}[\tilde{H}(t), \tilde{A}(t)]$

For Uniform dynamics, $\tilde{H}(t)=\tilde{H}$ . That means $U(t)\triangleq U(t,t_0)= e^{-iHt/\hbar}e^{iHt_0/\hbar}$

We know that something commutes with itself, so $[U(t), H]=0$

\Rightarrow\tilde{H}(t)=U^\dagger(t)HU(t)=H

Heisenberg equation of motion for $\tilde{A}(t)$

\frac{d}{dt}A(t)=\frac{d}{dt}(U^\dagger AU + U^\dagger A\frac{dU}{dt})

=-\frac{1}{i\hbar}U^\dagger H A U+\frac{1}{i\hbar}U^\dagger A H U

=-\frac{i}{\hbar}(U^\dagger HU U^\dagger A U - U^\dagger A U U^\dagger H U)

\boxed{\frac{d}{dt}\hat{A}(t)=\frac{i}{\hbar}[\hat{H}(t),\hat{A}(t)]}