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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

    kuruman

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    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.
     
    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

    kuruman

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    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

    kuruman

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    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
    ok.

    [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

    kuruman

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    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?
     
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