1.1: Example 1: The Harmonic Oscillator

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We will use the harmonic oscillator Hamiltonian in order to illustrate the procedure of using the variational theory. The Hamiltonian we wish to consider, therefore, is

$\begin{displaymath} H = -{\hbar^2 \over 2m}{d^2 \over dx^2} + {1 \over 2}m\omega^2 x^2 \end{displaymath}$

Suppose that we do not know the exact ground state solution of this problem, but, using intuition and knowledge of the shape of the potential, we postulate the shape of the wavefunction:

figure=lec10_fig2.eps,height=4.0in,width=4.5in

and postulate a form for the ground state wave function as

$\begin{displaymath} \psi(x) = e^{-\alpha x^2} \equiv \psi(x;\alpha) \end{displaymath}$

We view $\alpha$ as a variational parameter with respect to which we can minimize $\langle H\rangle$.

Thus, we compute

$\begin{displaymath} E(\alpha) = {\langle \psi\vert H\vert\psi\rangle \over \langle \psi\vert\psi\rangle } \end{displaymath}$

Clearly,

$\begin{displaymath} \langle \psi\vert\psi\rangle = \int_{-\infty}^{\infty}\;dx\;e^{-2\alpha x^2} = \sqrt{\pi \over 2\alpha} \end{displaymath}$

The quantity

$\begin{displaymath} \langle \psi\vert H\vert\psi\rangle = \int_{-\infty}^{\infty... ...2 \over dx^2} + {1 \over 2}m\omega^2x^2\right] e^{-\alpha x^2} \end{displaymath}$

can be easily shown to be

$\begin{displaymath} \langle \psi\vert H\vert\psi\rangle = \left({\alpha \hbar^2 ... ...m} + {m\omega^2 \over 8\alpha}\right) \sqrt{\pi \over 2\alpha} \end{displaymath}$

There, the ratio of these gives

$\begin{displaymath} E(\alpha) = \left({\alpha \hbar^2 \over 2m} + {m\omega^2 \over 8\alpha}\right) \end{displaymath}$

We then compute the best approximation to $E_0$ by minimizing $E(\alpha)$ with respect to $\alpha$:

$\displaystyle E'(\alpha) = {dE \over d\alpha}$	$\textstyle =$	$\displaystyle 0$
$\displaystyle {\hbar^2 \over 2m} - {m\omega^2 \over 8\alpha^2}$	$\textstyle =$	$\displaystyle 0$
$\displaystyle \alpha^2$	$\textstyle =$	$\displaystyle {m^2 \omega^2 \over 2\hbar^2}$
$\displaystyle \alpha$	$\textstyle =$	$\displaystyle {m\omega \over 2\hbar}$

Then the energy is obtained from

$\begin{displaymath} E(\alpha_{\rm min}) = {\hbar^2 \over 2m}{m\omega \over 2\hba... ... \over 8}{2\hbar \over m\omega} = {\hbar \omega \over 2} = E_0 \end{displaymath}$

In this case, the exact ground state energy is obtained because we assumed the correct functional form for the trial wave function. Thus, the ground state wavefunction is clearly given by

$\begin{displaymath} \psi_0(x) = \left({m\omega \over \pi\hbar}\right)^{1/4}e^{-m\omega x^2/ 2\hbar} \end{displaymath}$

\(\displaystyle E'(\alpha) = {dE \over d\alpha}\)	\(\textstyle =\)	\(\displaystyle 0\)
\(\displaystyle {\hbar^2 \over 2m} - {m\omega^2 \over 8\alpha^2}\)	\(\textstyle =\)	\(\displaystyle 0\)
\(\displaystyle \alpha^2\)	\(\textstyle =\)	\(\displaystyle {m^2 \omega^2 \over 2\hbar^2}\)
\(\displaystyle \alpha\)	\(\textstyle =\)	\(\displaystyle {m\omega \over 2\hbar}\)

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