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# 436: Gaussian Trial Wavefunction for the Hydrogen Atom

A Gaussian function, exp(‐αr2), is proposed as a trial wavefunction in a variational calculation on the hydrogen atom. Determine the optimum value of the parameter α and the ground state energy of the hydrogen atom. Use atomic units: h = 2π, me = 1, e = ‐1.

$\Phi (r, \beta ) := ( \frac{2 \beta}{ \pi})^{ \frac{3}{4}} exp(- \beta r^2)$

$T = \frac{-1}{2r} \frac{d^2}{dr^2} (r \blacksquare )$

$V = \frac{1}{r}$

$\int_{0}^{ \infty} \blacksquare 4 \pi r^2 dr$

a. Demonstrate the wave function is normalized.

$\int_{0}^{ \infty} \Psi (r, \beta )^2 4 \pi r^2 dr |_{simplify}^{assume,~ \beta >0} \rightarrow 1$

b. Evaluate the variational integral.

$E ( \beta ) := \int_{0}^{ \infty} \Psi (r, \beta ) [ \frac{-1}{2r} \frac{d^2}{dr^2} (r \Psi (r, \beta ))] 4 \pi r^2 dr ... |_{simplify}^{assume,~ \beta >0} \rightarrow \frac{1}{2} \frac{3 \pi^{\frac{1}{2}} \beta - (4) 2^{ \frac{1}{2}} \beta^{ \frac{1}{2}}}{ \pi^{ \frac{1}{2}}} + \int_{0}^{ \infty} \Psi (r, \beta ) \frac{-1}{r} \Psi (r, \beta ) 4 \pi r^2 dr$

c. Minimize the energy with respect to the variational parameter $$\beta$$.

$$\beta$$ := 1 $$\beta$$ := Minimize (E, $$\beta$$) $$\beta$$ = 0.283 E( $$\beta$$) = -0.424

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

$\frac{-.5 - E( \beta )}{-.5} = 15.117$

e. The differences between the Gaussian and Slate type wavefunctions are illustrated with the surface plots shown below.

N := 50 b := 5 i := 0..N j := 0..N $$y_{i} := -b + \frac{2bi}{N}$$ $$x_{j} := -b + \frac{2bj}{N}$$

$Gauss_{i,~j} := ( \frac{2 \beta}{ \pi})^{ \frac{3}{4}} exp[- \beta [ (x_{i})^2 + (y_{j})^2]]$

$Slater_{i,~j} := \frac{1}{ \sqrt{ \pi}} exp [ - \sqrt{ (x_{i})^2 + (y_{j})^2}]$ f. These wavefunctions can also be compared to their radial distribution functions:

r := 0, .1 .. 6

$G(r) := ( \frac{2 \beta}{ \pi}) ^{ \frac{3}{4}} exp( - \beta r^2)$

$S(r) := \frac{1}{ \sqrt{ \pi}} exp( -r )$ • Was this article helpful?