# 10.22: Variation Calculation on the 1D Hydrogen Atom Using a Gaussian Trial Wavefunction

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The energy operator for this problem is:

$\frac{-1}{2} \frac{d^2}{dx^2} \blacksquare - \frac{1}{x} \blacksquare \nonumber$

The trial wave function is:

$\psi (x, \alpha ) = 2 ( \frac{2 \alpha}{ \pi^{ \frac{1}{3}}}) (x) exp( - \alpha x^2) \nonumber$

Evaluate the energy integral.

$E( \alpha ) = \int_{x}^{ \alpha} - \frac{1}{2} \frac{d^2}{dx^2} \psi (x, \alpha ) dx + \int_{0}^{ \infty} \frac{-1}{x} \psi (x, \alpha )^2 dx |_{simplify}^{assume,~ \alpha > 0} \rightarrow \frac{-1}{2 \pi^{ \frac{1}{2}}} [ (-3) \pi^{ \frac{1}{2}} \alpha + (4) 2^{ \frac{1}{2}} \alpha ^{ \frac{1}{2}}] \nonumber$

Minimize the energy with respect to the variational parameter $$\alpha$$ and report its optimum value and the ground-state energy.

$$\alpha$$ = 1 $$\alpha$$ = Minimize (E, $$\alpha$$) $$\alpha$$ = 0.2829 E ( $$\alpha$$) = -0.4244

The exact ground state energy for the hydrogen atom is -0.5 Eh. Calculate the percent error.

$\left| \dfrac{-0.5 - E( \alpha )}{-0.5} \right| = 15.1174 \nonumber$

Plot the optimized trial wave function and the exact solution, $$\Phi (x) = 2(x) exp(-x)$$.

Find the distance from the nucleus within which there is a 95% probability of finding the electron.

$$\alpha$$ = 1. Given:

$\int_{0}^{a} \psi (x, \alpha )^2 dx = .95 \nonumber$

Find (a) = 2.6277

Find the most probable value of the position of the electron from the nucleus.

$\alpha = 0.2829 \frac{d}{dx} | \psi (x, \alpha )| = 0 |_{float,~3}^{solve,~x} \rightarrow {\begin{pmatrix} -1.33 \\ 1.33 \end{pmatrix}} \nonumber$

Calculate the probability that the electron will be found between the nucleus and the most probable distance from the nucleus.

$\int_{0}^{1.33} \psi ( \alpha , x)^2 dx = 0.3584 \nonumber$

Break the energy down into kinetic and potential energy contributions. Is the virial theorem obeyed?

\begin{align*} T &= \int_{0}^{ \infty} \psi (x, \alpha ) \frac{1}{2} \frac{d^2}{dx^2} \psi (x, \alpha ) dx \\[4pt] &= 0.4244\end{align*}

\begin{align*} V &= \int_{0}^{ \infty} \frac{-1}{x} \psi (x, \alpha )^2 dx \\[4pt] &= -0.8488 \end{align*}

$| \frac{V}{T}| = 2.00 \nonumber$

Calculate the probability that tunneling is occurring.

Classical turning point:

$E( \alpha ) = \frac{-1}{x}|_{float,~3}^{solve,~x} \rightarrow 2.36 \nonumber$

Tunneling probability:

$\int_{2.36}^{ \infty} \psi (x, \alpha )^2 dx = 0.0978 \nonumber$

This page titled 10.22: Variation Calculation on the 1D Hydrogen Atom Using a Gaussian Trial Wavefunction is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Frank Rioux via source content that was edited to the style and standards of the LibreTexts platform.