# Variational Principle

Obtain a variational estimate of the ground state energy of the hydrogen atom by taking as a trial function $\psi_T(r) = \text{exp } \left( - \alpha r^2 \right)$

How does your result compare with teh exact result?

You may assume that

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

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

so i obviously need to minimise

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

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?

$V=\frac{-e^2}{4 \pi \epsilon_0 r^2}$
clearly i'm going wrong since i have this $r^{-2}$ 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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kuruman
Homework Helper
Gold Member
You need to minimize the expectation value of the whole Hamiltonian, not just the potential energy.

<H>=<T>+<V>

where <T> is the expectation value of the kinetic energy term, so you have to do two integrals. Furthermore, the Coulomb potential goes as 1/r not 1/r2.

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okay but i still get integrals with r terms outside the exponential

$\langle \psi | \hat{H} | \psi \rangle = - \int_V \frac{\hbar^2}{m} \text{exp} ( - 2 \alpha r^2) ( 2 \alpha r^2 - \alpha ) dV - \int_V \frac{e^2}{4 \pi \epsilon_0 r} \text{exp} ( - 2 \alpha r^2 )$

and the normalisation on the denominator works out at $\sqrt{2} \pi^{\frac{3}{2}}$

any idea on how to proceed? thanks.

kuruman
Homework Helper
Gold Member
Are you saying you don't know how to do the integrals? What is dV and what are your limits of integration?

$dV=r^2 \sin{\theta} dr d\theta d\phi$

so we can pull out a $4 \pi$ from the theta and phi integrals. then the r integral is from 0 to infinity since dV is all space.

so, for example how do we integrate

$\int_0^\infty -\frac{4 \pi \hbar^2}{m} 2 \alpha r^2 \text{exp} ( - 2 \alpha r^2) dr$

the given formula doesn't apply here, does it?

kuruman
Homework Helper
Gold Member
No it does not. However, before we go into that, can you write the complete expression for the integral giving the expectation value for the kinetic energy?

ok.

$\langle \psi | \hat{T} | \psi \rangle = \int_0^\infty \int_0^\pi \int_0^{2 \pi} e^{- \alpha r^2} -\frac{\hbar^2}{2m} (2 \alpha r^2 - \alpha) e^{- \alpha r^2} r^2 \sin{\theta} dr d \theta d \phi = -\frac{\hbar^2}{2m} \int_0^\infty \int_0^\pi \int_0^{2 \pi} r^2 (2 \alpha r^2 - \alpha) e^{- 2 \alpha r^2} \sin{\theta} dr d \theta d \phi$

i can't see why this is wrong...

kuruman