1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Variational Principle

  1. Mar 25, 2010 #1
    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]

    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.
  2. jcsd
  3. Mar 25, 2010 #2


    User Avatar
    Homework Helper
    Gold Member

    You need to minimize the expectation value of the whole Hamiltonian, not just the potential energy.


    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.
    Last edited: Mar 25, 2010
  4. Mar 27, 2010 #3
    okay but i still get integrals with r terms outside the exponential

    [itex] \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 )[/itex]

    and the normalisation on the denominator works out at [itex]\sqrt{2} \pi^{\frac{3}{2}}[/itex]

    any idea on how to proceed? thanks.
  5. Mar 27, 2010 #4


    User Avatar
    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?
  6. Mar 27, 2010 #5
    [itex]dV=r^2 \sin{\theta} dr d\theta d\phi[/itex]

    so we can pull out a [itex]4 \pi[/itex] 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

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

    the given formula doesn't apply here, does it?
  7. Mar 27, 2010 #6


    User Avatar
    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?
  8. Mar 27, 2010 #7

    [itex] \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 [/itex]

    i can't see why this is wrong...
  9. Mar 27, 2010 #8


    User Avatar
    Homework Helper
    Gold Member

    You are missing the normalization constant, but that can always be added later.
    I think the radial Laplacian term is incorrect. Can you show how you derived it?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Variational Principle
  1. Variational Principle (Replies: 3)

  2. Variational Principle (Replies: 4)