1.2: Example 2: The Harmonic Oscillator Revisited

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Suppose we guessed, instead, a trial wave function of the form:

\[\psi(x) = {1 \over x^2 + a^2} \equiv \psi(x;a)\]

The potential and this trial wavefunction are illustrated in the figure below:

figure=lec10_fig3.eps,height=4.0in,width=4.5in

We now regard $a$ as a variational parameter. Thus,

\[\langle \psi\vert\psi\rangle = \int_{-\infty}^{\infty}\;dx\;{1 \over (x^2+a^2)^2}\]

This integral can be evaluated easily by a trigonometric substitution:

$\displaystyle x$	$\textstyle =$	$\displaystyle a\tan\theta$
$\displaystyle dx$	$\textstyle =$	$\displaystyle a\sec^2\theta$

from which

$\displaystyle \langle \psi\vert\psi\rangle$	$\textstyle =$	$\displaystyle \int_{-\pi/2}^{\pi/2}\;{a\sec^2\theta d\theta \over a^4(1+\tan^2\theta)^2 }$

	$\textstyle =$	$\displaystyle {1 \over a^3}\int_{-\pi/2}^{\pi/2}\cos^2\theta\;d\theta = {\pi \over 2a^3}$

With a little algebra, it can be similarly shown that

\[\langle \psi\vert H\vert\psi\rangle = {\pi \hbar^2 \over 8ma^5} + {\pi m\omega^2 \over 4a}\]

Thus,

\[{\langle \psi\vert H\vert\psi\rangle \over \langle \psi\vert \rangle} = {\hbar^2 \over 4ma^2} +{1 \over 2}m\omega^2 a^2 = E(a)\]

Now, minimizing with respect to $\a$, we find

$\displaystyle E'(a) = {dE \over da}$	$\textstyle =$	$\displaystyle 0$
$\displaystyle -{\hbar^2 \over 2ma^3} + m\omega^2 a$	$\textstyle =$	$\displaystyle 0$
$\displaystyle a^2$	$\textstyle =$	$\displaystyle {\hbar \sqrt{2}m\omega}$

and the energy is obtained by

$\displaystyle E(a_{\rm min})$	$\textstyle =$	$\displaystyle {\hbar^2 \over 4m}{\sqrt{2}m\omega \over \hbar} + {1 \over 2}m\omega^2 {\hbar \over \sqrt{2}m\omega}$
	$\textstyle =$	$\displaystyle {\hbar \omega \over \sqrt{2}} > {\hbar \omega \over 2}$

The result is larger than the true ground state energy, as expected. The error made by this trial wavefunction is

\[E(a_{\rm min}) - E_0 \over E_0} ={\hbar \omega/\sqrt{2} -\hbar \omega/2 \over \hbar\omega/2} = \sqrt{2}-1 \approx 0.41 = 41 % \]

$\begin{displaymath} {E(a_{{\rm min}}) - E_0 \over E_0} = {\hbar \omega/\sqrt{2}... ...omega/2 \over \hbar\omega/2} = \sqrt{2}-1 \approx 0.41 = 41 \% \end{displaymath}$

which is a relatively large error.

\(\displaystyle x\)	\(\textstyle =\)	\(\displaystyle a\tan\theta\)
\(\displaystyle dx\)	\(\textstyle =\)	\(\displaystyle a\sec^2\theta\)

\(\displaystyle \langle \psi\vert\psi\rangle\)	\(\textstyle =\)	$\displaystyle \int_{-\pi/2}^{\pi/2}\;{a\sec^2\theta d\theta \over a^4(1+\tan^2\theta)^2 }$

	\(\textstyle =\)	\(\displaystyle {1 \over a^3}\int_{-\pi/2}^{\pi/2}\cos^2\theta\;d\theta = {\pi \over 2a^3}\)

\(\displaystyle E'(a) = {dE \over da}\)	\(\textstyle =\)	\(\displaystyle 0\)
\(\displaystyle -{\hbar^2 \over 2ma^3} + m\omega^2 a\)	\(\textstyle =\)	\(\displaystyle 0\)
\(\displaystyle a^2\)	\(\textstyle =\)	\(\displaystyle {\hbar \sqrt{2}m\omega}\)

\(\displaystyle E(a_{\rm min})\)	\(\textstyle =\)	$\displaystyle {\hbar^2 \over 4m}{\sqrt{2}m\omega \over \hbar} + {1 \over 2}m\omega^2 {\hbar \over \sqrt{2}m\omega}$
	\(\textstyle =\)	\(\displaystyle {\hbar \omega \over \sqrt{2}} > {\hbar \omega \over 2}\)

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