Question

Q8. Gaussian functions are easier to integrate numerically than exponential decays. For this reason, basis sets of Gaussian functions are frequently used for variational calculations. [4] Using the Gaussian trial wave function $\phi(r) = \exp(-cr^2)$ for the hydrogen atom gives $E' = \frac{3h^2}{8\pi^2\mu}c - \frac{e^2}{\sqrt{2\pi^{3/2}\epsilon_0}}\sqrt{c}$ a) Calculate the optimal values of c and the variational energy. b) Compare the variational energy to the exact energy $\mu e^4/8\epsilon_0^2h^2$ for the ground-state hydrogen atom.

          Q8.
Gaussian functions are easier to integrate numerically than exponential decays. For this
reason, basis sets of Gaussian functions are frequently used for variational calculations.
[4]
Using the Gaussian trial wave function $\phi(r) = \exp(-cr^2)$ for the hydrogen atom gives
$E' = \frac{3h^2}{8\pi^2\mu}c - \frac{e^2}{\sqrt{2\pi^{3/2}\epsilon_0}}\sqrt{c}$
a) Calculate the optimal values of c and the variational energy.
b) Compare the variational energy to the exact energy $\mu e^4/8\epsilon_0^2h^2$ for the ground-state
hydrogen atom.

Q8.
Gaussian functions are easier to integrate numerically than exponential decays. For this
reason, basis sets of Gaussian functions are frequently used for variational calculations.
[4]
Using the Gaussian trial wave function ϕ(r) = (-cr^2) for the hydrogen atom gives
E' = (3h^2)/(8π^2μ)c - (e^2)/(√(2π^3/2ϵ0))√(c)
a) Calculate the optimal values of c and the variational energy.
b) Compare the variational energy to the exact energy μ e^4/8ϵ0^2h^2 for the ground-state
hydrogen atom.

Added by Laura J.

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