# 10.21: Variation Calculation on the 1D Hydrogen Atom Using a Trigonometric Trial Wave Function

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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:

$\Psi ( \alpha , x) := \frac{ \sqrt{12 \alpha ^3}}{ \pi} (x) sech( \alpha , x) \nonumber$

Evaluate the variational energy integral.

$E( \alpha ) := \int_{0}^{ \infty} \Psi ( \alpha , x) - \frac{1}{2} \frac{d^2}{dx^2} \Psi ( \alpha , x) dx + \int_{0}^{ \infty} \frac{-1}{x} \Psi ( \alpha , x)^2 dx |_{simplify}^{assume,~ \alpha > 0} \rightarrow \frac{1}{6} \alpha \frac{12 \alpha _ \alpha \pi^2 - 72 ln (2)}{ \pi ^2} \nonumber$

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, $$\alpha$$) $$\alpha$$ = 1.1410 E( $$\alpha$$) = -0.4808

The exact ground-state energy for the hydrogen atom is -.5 Eh. Calculate the percent error.

$| \frac{-.5 - E( \alpha )}{-.5}| = 3.8401 \nonumber$

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:

$\int_{0}^{a} \Psi ( \alpha , x)^2 dx = 0.95 \nonumber$

Find (a) = 2.8754

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

$\alpha := 1.1410~~~~ \frac{d}{dx}|\frac{ \sqrt{12 \alpha ^3}}{ \pi} (x) sech( \alpha , x)| = 0~|_{float,~3}^{solve,~x} \rightarrow 1.05 \nonumber$

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

$\int_{0}^{1.05} \Psi ( \alpha , x)^2 dx = 0.3464 \nonumber$

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

$T := \int_{0}^{ \infty} \Psi ( \alpha , x) \frac{-1}{2} \frac{d^2}{dx^2} \Psi ( \alpha , x) dx~~~T = 0.4808 \nonumber$

$V := \int_{0}^{ \infty} \frac{-1}{x} \Psi ( \alpha , x)^2 dx~~~ V = -0.9616 \nonumber$

$| \frac{V}{T}| = 2.0000 \nonumber$

Use the exact result to discuss the weakness of this trial function.

Eexact := -0.5

Using the virial theorem we know: Texact := 0.500 Vexact := -1.00

Calculate the difference between the variational results and the exact calculation:

E( $$\alpha$$) - Eexact = 0.0192

T - Texact = -0.0192

V - Vexact = 0.0384

The variational wave function yields a lower kinetic energy, but at the expense of a potential energy that is twice as unfavorable as the kinetic energy result is favorable.

Calculate the probability that tunneling is occurring.

Classical turning point:

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

Tunneling probability:

$\int_{2.08}^{ \infty} \Psi ( \alpha , x)^2 dx = 0.1783 \nonumber$

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