[NOTES/QM-09006] Heisenberg Picture of quantum mechanics

The Heisenberg picture state vector is defined by \begin{equation}\label{eq05} \ket{\psi t}_H \stackrel{\text{def}}{\equiv} e^{iHt/\hbar}\ket{\psi t}_S = \ket{\psi_0}. \end{equation} The Heisenberg state vector is independent of time and coincide with the state vector in the Schr\"{o}dinger picture at initial time. The time development of the Heisenberg picture operators is defined so that the average value of any dynamical variable at time \(t\) in the Schr\'{o}dinger and Heisenberg pictures coincide. Thus we demand \begin{equation}\label{eq06} _H\average{\psi t}{X_H(t)}_H = \,\, _{S\hspace{-1pt}}\average{\psi t}{X_S}_S \end{equation} Substituting \eqref{eq05} in \eqref{eq06} gives \begin{equation}\label{eq07} \average{\psi_0}{X_H(t)} = \average{\psi_0}{e^{iHt/\hbar}\,X_S\,e^{-iHt/\hbar}}. \end{equation} We, therefore, define the Heisenberg picture operators by \begin{equation}\label{eq08} X_{H}(t) = {e^{iHt/\hbar}\,X_S\, e^{-iHt/\hbar}}. \end{equation}
Equation of Motion
In Heisenberg picture the state vector does not evolve with time. So how do we describe the time development of a system? The answers is that in the Heisenberg picture the operators carry the entire time dependence. So for a point particle, the position operator, the momentum operator, in fact all dynamical variables become time dependent. This is parallel to the classical description where the time evolution of the state is carried by the position and momentum. The equations of motion are then the equations telling us how the a given dynamical variable will change with time. The equation of motion is easily derived from \eqref{eq08} and we compute \begin{eqnarray}\label{eq09} \dd[X_H]{t} &=& \dd{t}\Big[{e^{iHt/\hbar}\,X_S\, e^{-iHt/\hbar}}\Big]\\ &=& \dd{t}\Big({e^{iHt/\hbar}\Big)\,X_S\, e^{-iHt/\hbar}} + {e^{iHt/\hbar}\,\Big(\dd{t}X_S\Big)\, e^{-iHt/\hbar}}\nonumber + {e^{iHt/\hbar}\,X_S\, \dd{t}\Big(e^{-iHt/\hbar} \Big)} \\\label{eq10}\\ &=&\frac{iH}{\hbar} e^{iHt/\hbar}\,X_S\, e^{-iHt/\hbar} + e^{iHt/\hbar}\,\Big(\pp{t}X_S\Big)\, e^{-iHt/\hbar}\nonumber + e^{iHt/\hbar}\,X_S\,e^{-iHt/\hbar} \frac{-iH}{\hbar}\\ \label{eq11}\\ &=& \frac{i}{\hbar} H X_H + \pp{t}X_H - X_H \frac{i}{\hbar} H \label{eq12} \end{eqnarray} Here we have used \begin{equation} \dd{t}e^{iHt/\hbar} = \big(\tfrac{iH}{\hbar}\big)e^{iHt/\hbar} = e^{iHt/\hbar} \big(\tfrac{iH}{\hbar}\big). \end{equation} Thus we arrive at the final form of equations of motion in the Heisenberg picture
\begin{equation}\label{eq13} \boxed{\dd[X_H]{t} = \pp[X_H]{t}+ \frac{1}{i\hbar}\big[X_H, H\big]_-.} \end{equation}

• [\NoteThisPoint] Recalling that the ($\frac{1}{i\hbar} \times$ commutator) has correspondence with the Poisson bracket, we have an obvious correspondence with the Poisson bracket form of equations of motion in classical mechanics.

The steps \eqref{eq09}- \eqref{eq12}, leading to the final result \eqref{eq13}, require some explanation and care as explained in  Notes and Comments section at the end.