Notices
 

[NOTES/QM-09002] Time Variation of Average Values

For page specific messages
For page author info

Category: 

Assuming time development of states to be given by  \[i\hbar \dd[\ket{\psi, t}]{t} = H \ket{\psi t}, \] an equation for time variation of average value of a dynamical variable is derived. Classical correspondence  is used to identify the generator of time evolution with Hamiltonian. A dynamical variable not depending explicitly on time is a constant of motion if it commutes with the Hamiltonian.

Time variation of average values}} The time evolution of a quantum system is governed by the Schrodinger equation \begin{equation} i \hbar \dd{t}\ket{\psi t}=\hat{H}\ket{\psi t}. \end{equation} We will obtain an equation for time development of averages of a dynamical variable $\hat{F}$. The result will turn out to have an obvious correspondence with the classical equation of motion for dynamical variable $F$. This then will suggest the identification of $\hat{H}$ as the operator representing the Hamiltonian of the system. Let $F(q,p,t)$ be an dynamical variable of the system and let $\hat{F}$ denote the corresponding operator. We are interested in finding out how the average value \begin{equation} \langle \hat{F} \rangle \equiv \bra{\psi t}\hat{F}\ket{\psi t} \label{E13} \end{equation} changes with time. The time dependence of the average value comes from dependence of the three objects, the operator$\hat{F}$, the bra vector $\bra{\psi t}$, and the ket vector $\ket{\psi t}$, present in \eqref{E13}. The equation conjugate to the Schrodinger equation \begin{equation}\label{eq03} i\hbar {d\over d t}\, \ket{\psi t} = \hat{H} \ket{\psi t} \end{equation} is given by \begin{equation} - i\hbar {d\over d t}\, \bra{\psi t} = \bra{\psi t} \hat{H}^\dagger \label{E14} \end{equation} Since the operator $\hat{H}$ is hermitian, the above equation takes the form \begin{equation} - i\hbar {d\over d t}\, \bra{\psi t} = \bra{\psi t} \hat{H} \label{E15} \end{equation} Therefore \begin{equation} \frac{d}{dt}\langle \hat{F} \rangle = \left( \frac{d}{dt}\bra{\psi t}\right)\hat{F}\ket{\psi t} + \average{\psi t}{\frac{d\hat{F}}{dt}} + \bra{\psi t}\hat{F} \bigg({d\over dt} \ket{\psi t}\bigg) \label{E16} \end{equation} Using \eqref{eq03} and \eqref{E15} in \eqref{E16} we get \begin{equation} {d\over dt}\langle \hat{F} \rangle = - {1\over i \hbar}\bra{\psi t}\hat{H} \hat{F}\ket{\psi t} + \bra{\psi t}{d\hat{F} \over dt}\ket {\psi t} +{1\over i \hbar} \bra{\psi t}\hat{F} \hat{H} \ket{\psi t}. \label{E17} \end{equation} The above equation is rearranged to give the final form \begin{equation} {d\over dt}\, \langle \hat{F} \rangle = \,\langle{\partial\over \partial t} \hat{F} \rangle + {1\over i\hbar} \langle\, [\hat{F},\hat{H} ]\, \rangle. \label{E18} \end{equation} This result is known as Ehrenfest theorem. Comparing \eqref{E18} with the equation of motion in classical mechanics for time evolution of dynamical variables \begin{equation} {d F\over d t} = {\partial F\over \partial t} + \{F,H\}_{PB} \label{E19} \end{equation} and remembering that the commutator divided by $i\hbar$ corresponds to the Poisson bracket in the limit $\hbar \rightarrow 0$, we see that $\hat{H}$ must be identified as the operator corresponding to the Hamiltonian $H$ of the system.

Exclude node summary : 

n

4727: Diamond Point, 4909: QM-HOME-I

700
0
 
X