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First order perturbation for hydrogen

  1. Aug 13, 2013 #1
    1. The problem statement, all variables and given/known data
    Assume that there is a deviation from Coulomb’s law at very small distances, the Coulomb potential energy between an electron and proton is given by

    [itex]V_{mod}(r)=\begin{cases}
    -\frac{e^{2}}{4\pi\varepsilon_{0}}\frac{b}{r^{2}} & 0<r\leq b\\
    -\frac{e^{2}}{4\pi\varepsilon_{0}}\frac{1}{r} & r>b
    \end{cases}[/itex]

    (a) Specify the perturbation
    (b) Find the first order correction for the ground state
    (c) Show that the answer in (b) can be approximated by
    [itex]E_{1}^{(1)}\approx-\frac{4b^2}{a_0^2}E_R[/itex] where [itex]E_R=\frac{e^{2}}{8\pi\varepsilon_{0}a_0}[/itex] is the Rydberg energy.

    2. Relevant equations
    [itex]\psi_0=\frac{2}{a_0^{3/2}}e^{-r/a_0}[/itex]

    [itex]E_{1}^{(1)}=\int_{0}^{b}\left| \psi_0 \right|^2 \delta\hat{\textrm{H}} r^2\,\textrm{d}r[/itex]

    [itex]b \ll r[/itex]

    3. The attempt at a solution
    (a) [itex]\delta\hat{\textrm{H}}=-\dfrac{e^{2}}{4\pi\varepsilon_{0}}\left(\dfrac{b}{r^{2}}-\dfrac{1}{r}\right)[/itex]

    so with [itex]b \ll r[/itex] we have [itex]e^{-2b/a_0}\approx 1[/itex]
    (b)
    [itex]E_{1}^{(1)}=\dfrac{e^2 b}{2\pi \varepsilon_{0}a_0^2}[/itex]

    I think the above is correct, I just can't see how to get part (c).
     
    Last edited: Aug 13, 2013
  2. jcsd
  3. Aug 13, 2013 #2
    Sorry, I was stupid I have it now.
     
  4. Aug 17, 2013 #3
    Hi, what had you done wrong initially?
     
  5. Aug 20, 2013 #4
    I found the first order perturbation which included the exponential terms. To get part c what I should have done was to set the exponential to unity then perform the integration. So part b should have included the exponential terms.
    A better way I feel is to note that [itex]b/a_0 <<1[/itex] then take a 2nd order Taylor series and insert this into the integral in b.
     
    Last edited: Aug 20, 2013
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