Example II: The harmonic oscillator revisited

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

Figure 2:

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:

$$x = a\tan\theta$$

$$dx = a\sec^2\theta$$

from which

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

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

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,

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

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

$$E'(a) = {dE \over da} = 0$$

$$-{\hbar^2 \over 2ma^3} + m\omega^2 a = 0$$

$$a^2 = {\hbar \sqrt{2}m\omega}$$

and the energy is obtained by

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

$$= {\hbar \omega \over \sqrt{2}} > {\hbar \omega \over 2}$$

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

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

which is a relatively large error.