7.1.2.2 So­lu­tion schrod­sol-b

Ques­tion:

For the one-di­men­sion­al har­monic os­cil­la­tor, the en­ergy eigen­val­ues are

\begin{displaymath}
E_n = \frac{2n+1}{2} \omega
\end{displaymath}

Write out the co­ef­fi­cients $c_n(0)e^{-{{\rm i}}E_nt/\hbar}$ for those en­er­gies.

Now clas­si­cally, the har­monic os­cil­la­tor has a nat­ural fre­quency $\omega$. That means that when­ever ${\omega}t$ is a whole mul­ti­ple of $2\pi$, the har­monic os­cil­la­tor is again in the same state as it started out with. Show that the co­ef­fi­cients of the en­ergy eigen­func­tions have a nat­ural fre­quency of $\frac 12\omega$; $\frac 12{\omega}t$ must be a whole mul­ti­ple of $2\pi$ for the co­ef­fi­cients to re­turn to their orig­i­nal val­ues.

An­swer:

The co­ef­fi­cients are

\begin{displaymath}
c_n(0) e^{-{\rm i}\frac{(2n+1)}{2}\omega t}
\end{displaymath}

Now if ${\omega}t$ is $2\pi$, the ar­gu­ment of the ex­po­nen­tial equals ${\rm i}$ times an odd mul­ti­ple of $\pi$. That makes the ex­po­nen­tial equal to mi­nus one. It takes un­til ${\omega}t$ $\vphantom0\raisebox{1.5pt}{$=$}$ $4\pi$ un­til the ex­po­nen­tial re­turns to its orig­i­nal value one.