[NOTES/QM-09004] Stationary States and Constants of Motion

We will assume that the Hamiltonian is independent of time.

 Stationary states
Let us consider time evolution of a system if it has a definite value of energy at an initial time $t_0$. The value of the energy then has to be one of the eigenvalues and the state vector will be the corresponding eigenvector. So $\ket{\psi t_0} = \ket{E_m}$, then at time $t$ the system will be in the state given by \begin{equation} \ket{\psi t} = U(t,t_0)\ket{E_m} = \exp(-iE_m(t-t_0)/\hbar) \ket{E_m}. \end{equation} It must be noted that the state vector at different times is equal to the initial state vector times a {\it numerical phase factor} $ ( \exp(-iE_m(t-t_0)/\hbar) )$. Therefore, the vector at time $t$ represents the same state at all times. Such states are called {\bf stationary states} because the state does not change with time. Every eigenvector of energy is a possible stationary state of a system. In such a state the average value of a dynamical variable, $\hat{X}$, not having time explicitly, is independent of time even if $\hat{X}$ does not commute with Hamiltonian. In fact the probabilities of finding a value on a measurement of the dynamical variable are independent of time.
Unless mentioned otherwise, we shall always assume that the Hamiltonian $H$ of the system under discussion is independent of time.

Constant of motion
The time evolution of average value of an operator \(\hat{F}\) is given by \begin{equation}\label{eq02} \dd{t}\average{\psi t}{\hat{F}} = \average{\psi t}{\Big(\pp[\hat{F}]{t}\Big)} + \frac{1}{i\hbar}\average{\psi t}{ [\hat{F}, \hat{H}]} \end{equation} This equation corresponds to the classical Poisson bracket equation of motion for (classical) dynamical variable \(F\) \begin{equation} \dd[F]{t} = \pp[F]{t} + \{F,H\}_\text{PB} \end{equation} If a dynamical variable $F$ does not contain explicit time dependence, then we have $ {\partial F\over \partial t}=0 $. If such an operator $\hat{F}$ commutes with the Hamiltonian operator $\hat{H}$, we will have \begin{equation} \boxed{ [\hat{F},\hat{H} ] =0 } \end{equation} and \eqref{eq02} shows that $$ {d\over dt}\, \bra{\psi t} \hat{F} \, \ket{\psi t} =0 $$ Therefore in an arbitrary state, the average value of $\hat{F}$ does not change with time. Such a dynamical variable will be called a {\bf constant of motion}.