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Stationary States

An eigenstate of the energy operator $H\equiv{\rm i} \hbar \partial/\partial t$ corresponding to the eigenvalue $E_i$ satisfies
\begin{displaymath}
{\rm i} \hbar \frac{\partial \psi_E(x,t,E_i)}{\partial t} = E_i \psi_E(x,t,E_i).
\end{displaymath} (293)

It is evident that this equation can be solved by writing
\begin{displaymath}
\psi_E(x,t,E_i) = \psi_i(x) {\rm e}^{-{\rm i} E_i t/\hbar},
\end{displaymath} (294)

where $\psi_i(x)$ is a properly normalized stationary (i.e., non-time-varying) wavefunction. The wavefunction $\psi_E(x,t,E_i)$ corresponds to a so-called stationary state, since the probability density $\vert\psi_E\vert^{ 2}$ is non-time-varying. Note that a stationary state is associated with a unique value for the energy. Substitution of the above expression into Schrödinger's equation (137) yields the equation satisfied by the stationary wavefunction:
\begin{displaymath}
\frac{\hbar^2}{2 m} \frac{d^2 \psi_i}{d x^2} =
\left[V(x)-E_i\right]\psi_i.
\end{displaymath} (295)

This is known as the time-independent Schrödinger equation. More generally, this equation takes the form
\begin{displaymath}
H \psi_i = E_i \psi_i,
\end{displaymath} (296)

where $H$ is assumed not to be an explicit function of $t$. Of course, the $\psi_i$ satisfy the usual orthonormality condition:
\begin{displaymath}
\int_{-\infty}^\infty \psi_i^\ast \psi_j  dx = \delta_{ij}.
\end{displaymath} (297)

Moreover, we can express a general wavefunction as a linear combination of energy eigenstates:
\begin{displaymath}
\psi(x,t) = \sum_i c_i \psi_i(x) {\rm e}^{-{\rm i} E_i t/\hbar},
\end{displaymath} (298)

where
\begin{displaymath}
c_i = \int_{-\infty}^{\infty} \psi_i^\ast(x) \psi(x,0) dx.
\end{displaymath} (299)

Here, $\vert c_i\vert^{ 2}$ is the probability that a measurement of the energy will yield the eigenvalue $E_i$. Furthermore, immediately after such a measurement, the system is left in the corresponding energy eigenstate. The generalization of the above results to the case where $H$ has continuous eigenvalues is straightforward.

If a dynamical variable is represented by some Hermitian operator $A$ which commutes with $H$ (so that it has simultaneous eigenstates with $H$), and contains no specific time dependence, then it is evident from Eqs. (297) and (298) that the expectation value and variance of $A$ are time independent. In this sense, the dynamical variable in question is a constant of the motion.



Subsections
next up previous
Next: Exercises Up: Fundamentals of Quantum Mechanics Previous: Continuous Eigenvalues
Richard Fitzpatrick 2010-07-20