- #1

- 140

- 2

\begin{align}

{d \over dt} A(t)

= {i \over \hbar} H e^{iHt / \hbar} A e^{-iHt / \hbar} + e^{iHt / \hbar} \left(\frac{d A}{d t}\right) e^{-iHt / \hbar} + {i \over \hbar} e^{iHt / \hbar} A \cdot (-H) e^{-iHt / \hbar} \\

= {i \over \hbar} e^{iHt / \hbar} \left( H A - A H \right) e^{-iHt / \hbar} + e^{iHt / \hbar} \left(\frac{d A}{d t}\right) e^{-iHt / \hbar} \\

= {i \over \hbar } \left( H A(t) - A(t) H \right) + e^{iHt / \hbar} \left(\frac{d A}{d t}\right)e^{-iHt / \hbar} .

\end{align}[/tex]

I right [tex]A(t)[/tex] the Heisenberg picture of the operator(which depend of the time.). and [tex]A[/tex] the Schrödinger picture of the operator.

What permit me to say that : [tex]\frac{\partial A(t)}{\partial t} = e^{iHt / \hbar} \left(\frac{d A}{d t}\right)e^{-iHt / \hbar}[/tex] please?

I want to know if it's a Mathemtical definition or a physical definition.

Then I can say that with the classical Poisson brackets [tex]\{H, A\}[/tex] between the Hamiltonian and the Physical Quantity [tex]A[/tex] correspond to the [tex]\frac{i}{\hbar}[H, A][/tex] commutator between those 2 operator.

By the correspondance principle.

Thank you in advance and have a nice afternoon.