Obtain a variational estimate of the ground state energy of the hydrogen atom by taking as a trial function [itex]\psi_T(r) = \text{exp } \left( - \alpha r^2 \right)[/itex](adsbygoogle = window.adsbygoogle || []).push({});

How does your result compare with teh exact result?

You may assume that

[itex]\int_0^\infty \text{exp } \left( - b r^2 \right) dr = \frac{1}{2} \sqrt{ \frac{\pi}{b}}[/itex]

and that [itex] 1 \text{Ry} = \left( \frac{e^2}{4 \pi \epsilon_0} \right)^2 \frac{m}{2 \hbar^2}[/itex]

so i obviously need to minimise

[itex]E[r] = \frac{ \langle \psi \vline \hat{V} \vline \psi \rangle}{ \langle \psi \vline \psi \rangle}[/itex]

i think i'm getting the wrong answer because my V is wrong.

so i want a coulomb potential between a proton and a electron surely?

[itex]V=\frac{-e^2}{4 \pi \epsilon_0 r^2}[/itex]

clearly i'm going wrong since i have this [itex]r^{-2}[/itex] term that i don't know how to integrate (the only r dependence in hte given integral is in the exponent) and also i haven't used the rydberg thing (although i guess that might not be needed until the end of the question).

thanks for any help.

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# Homework Help: Variational Principle

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