1-D H.O. with exponential wavefunction

As a check of the procedure, take trial function \bgroup\color{black}$e^{-ax^2/2}$\egroup. This should give us the actual ground state energy.

\begin{displaymath}\bgroup\color{black} E'= { \int\limits^\infty_{-\infty} \psi^...
...]dx
\over {\int\limits^\infty_{-\infty}\psi^*\psi dx}} \egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black} = { \left\{ {-\hbar^2\over{2m}} \int\lim...
...\}
\over{ \int\limits^\infty_{-\infty} e^{-ax^2}dx }} \egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black} = \left[ {-a^2\hbar^2\over{2m}}+{1\over ...
...ty} e^{-ax^2} dx}}
+ \left[{\hbar^2 a\over{2m}}\right] \egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black} \int\limits^\infty_{-\infty} e^{-ax^2} dx =\sqrt{\pi\over a}
= \sqrt{\pi}a^{-1/2}\egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black} -\int\limits^\infty_{-\infty} x^2 e^{-ax^2} dx
= \sqrt{\pi}\left(-{1\over 2}\right)a^{-3/2} \egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black} \int\limits^\infty_{-\infty} x^2 e^{-ax^...
... \sqrt{\pi\over{a^3}} = {1\over{2a}}{\sqrt{\pi\over a}} \egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black}E'=\left[{-a\hbar^2\over{4m}}+{1\over{4a}...
...a\over{2m}}={1\over{4a}}m\omega^2 + {\hbar^2\over{4m}}a \egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black}{\partial E'\over{\partial a}}={-m\omega^2\over{4a^2}}+{\hbar^2\over{4m}}=0\egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black} 4a^2\hbar^2=4m^2\omega^2 \egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black}a={m\omega\over\hbar}\egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black}\psi=e^{-{m\omega\over{2\hbar}}x^2}\egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black}E'={m\omega^2\over 4}{\hbar\over{m\omega}...
...\hbar}
= {1\over 4}\hbar\omega + {1\over 4}\hbar\omega \egroup\end{displaymath}

OK.

Jim Branson 2013-04-22