1D Harmonic Oscillator

Use

\begin{displaymath}\bgroup\color{black}\psi = \left(a^2-x^2\right)^2\qquad \vert x\vert\leq a\egroup\end{displaymath}

and \bgroup\color{black}$\psi=0$\egroup otherwise as a trial wave function. Recall the actual wave function is \bgroup\color{black}$e^{-m\omega x^2/2\hbar^2}$\egroup. The energy estimate is

\begin{displaymath}\bgroup\color{black}E'= { \left<\left(a^2-x^2\right)^2 \vert ...
...^2-x^2\right)^2 \vert \left(a^2-x^2\right)^2\right> }}. \egroup\end{displaymath}

We need to do some integrals of polynomials to compute

\begin{displaymath}\bgroup\color{black}E'={3\over 2}{\hbar^2\over{ma^2}}+{1\over{22}}m\omega^2a^2.\egroup\end{displaymath}

Now we optimize the parameter.

\begin{displaymath}\bgroup\color{black}{dE'\over{da^2}}=0={-3\over 2}{\hbar^2\ov...
...hbar^2\over{m\omega^2}}}=\sqrt{33}{\hbar\over{m\omega}} \egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black}E'={3\over 2}{\hbar\omega\over\sqrt{33}}
...
...\hbar\omega\left( {\sqrt{33}+\sqrt{33}\over{11}}\right) \egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black} = {1\over 2}\hbar\omega{\sqrt{4\cdot 3}\over\sqrt{11}}
= {1\over 2}\hbar\omega\sqrt{12\over{11}} \egroup\end{displaymath}

This is close to the right answer. As always, it is treated as an upper limit on the ground state energy.



Jim Branson 2013-04-22