B.3 HEISENBERG Picture

In the HEISENBERG representation state vectors are defined as

\begin{displaymath}\begin{array}{l}\displaystyle \displaystyle \vert\Psi_\mathrm...
...{i\hat{H}t/\hbar} \vert\Psi_\mathrm{S}(t)\rangle \ .\end{array}\end{displaymath} (B.10)

Its time derivative may be combined with (B.3) to yield $ i\hbar\partial_t\vert\Psi_\mathrm{H}(t)\rangle = 0$, which shows that $ \vert\Psi_\mathrm{H}(t)\rangle$ is time-independent. Since an arbitrary matrix element in the SCHRÖDINGER picture can be written as

\begin{displaymath}\begin{array}{l}\displaystyle \langle\Psi_\mathrm{S}^{'}(t) \...
...-i\hat{H}t/\hbar}\vert\Psi_\mathrm{H}(t)\rangle \ , \end{array}\end{displaymath} (B.11)

a general operator in the HEISENBERG picture is given by

\begin{displaymath}\begin{array}{l}\displaystyle \hat{O}_\mathrm{H}(t) \ = \ e^{...
...H}t/\hbar}\hat{O}_\mathrm{S}e^{-i\hat{H}t/\hbar}\ . \end{array}\end{displaymath} (B.12)

Equation (B.12) can be rewritten in terms of the interaction picture operators

\begin{displaymath}\begin{array}{l} \displaystyle \hat{O}_\mathrm{H}(t) \display...
...(t) e^{i\hat{H}_0 t/\hbar} e^{-i\hat{H}t/\hbar} \ , \end{array}\end{displaymath} (B.13)

or in terms of the operator $ \hat{S}$ derived in the next section

\begin{displaymath}\begin{array}{l}\displaystyle \hat{O}_\mathrm{H}(t) \ = \ \hat{S}(0,t) \hat{O}_\mathrm{I}(t) \hat{S}(t,0)\ . \end{array}\end{displaymath} (B.14)

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