1-D H.O. with exponential wavefunction

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

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

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

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

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

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

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

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

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

$$\bgroup\color{black} 4a^2\hbar^2=4m^2\omega^2 \egroup$$

$$\bgroup\color{black}a={m\omega\over\hbar}\egroup$$

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

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

OK.