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Perturbation first order hydrogen - Coulomb model

  1. May 1, 2016 #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{q^{2}}{4\pi\varepsilon_{0}}\frac{b}{r^{2}} & 0<r\leq b\\
    -\frac{q^{2}}{4\pi\varepsilon_{0}}\frac{1}{r} & r>b
    \end{cases}[/itex]

    (a) Find the first order correction for the ground state using the standard integrals

    2. Relevant equations

    Ground State - [itex]\psi_0=\frac{2}{a_0^{3/2}}e^{-r/a_0}[/itex]

    To find the first order correction - [itex]E_{1}^{(1)}[/itex]=<ψ0| [itex]\delta\hat{\textrm{H}} [/itex]|ψ0>

    Perturbation - [itex]\delta\hat{\textrm{H}}=-\dfrac{q^{2}}{4\pi\varepsilon_{0}}\left(\dfrac{b}{r^{2}}-\dfrac{1}{r}\right)[/itex] for 0<r≤b
    Standard integrals
    [itex]\int_{0}^{x}[/itex] e-u du = 1 - e-x
    -[itex]\int_{0}^{x}[/itex] u e-u du = 1 - e-x - xe-x

    3. The attempt at a solution
    Using [itex]E_{1}^{(1)}[/itex]=[itex]\int_{-∞}^{∞}[/itex]ψ(0) * [itex]\delta\hat{\textrm{H}} [/itex] ψ(0)
    [itex]E_{1}^{(1)}=\dfrac{1}{\pi a_{0}^{3}}\dfrac{q^{2}}{4\pi\varepsilon_{0}}\int_{0}^{b}(e^{\dfrac{-2r}{a_{0}}}) (\dfrac{b}{r^{2}}-\dfrac{1}{r})[/itex] dr

    So I've taken the constants out and applied a 'sandwich integral' with the perturbation Hamiltonian... This problem is I have no idea where to start making this easier for me, ive been staring at it for aleast and hour. Can someone help please?
     
    Last edited: May 1, 2016
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  3. May 1, 2016 #2

    blue_leaf77

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    Where is the volume element for the integration ##d^3\mathbf{r}##?
     
  4. May 1, 2016 #3
    r is radius not a vector in this question. So integration is only on radius. I added dr
     
  5. May 1, 2016 #4

    blue_leaf77

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    That's not the point of my question. Any integral must take the form of ##\int f(x) \, dx##, where is your ##dx## in your calculation of the integral?
     
  6. May 1, 2016 #5
    I added dr
     
  7. May 1, 2016 #6

    blue_leaf77

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    That's still not completely correct, you are integrating over the entire volume which means the integration element must be that of a volume element. Find the correct form of ##d^3\mathbf{r}## (sometimes also written as ##dV##) in spherical coordinate, for example take a look at https://en.wikipedia.org/wiki/Spherical_coordinate_system.
     
  8. May 1, 2016 #7
    But the perturbation hamilton that I have is independent of Φ,θ
     
  9. May 1, 2016 #8

    blue_leaf77

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    The form of the volume element is entirely determined by the coordinate system being used, not by the integrand.
     
  10. May 1, 2016 #9
    [itex]E_{1}^{(1)}=\dfrac{1}{\pi a_{0}^{3}}\dfrac{q^{2}}{4\pi\varepsilon_{0}} \int_{0}^{2pi} \int_{0}^{pi}\int_{0}^{b}(e^{\dfrac{-2r}{a_{0}}}) (\dfrac{b}{r^{2}}-\dfrac{1}{r})[/itex] dr dθ dΦ
     
  11. May 1, 2016 #10

    blue_leaf77

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    Still something missing, look for ##dV=?## in the wikipedia link above.
     
  12. May 1, 2016 #11
    [itex]E_{1}^{(1)}=\dfrac{1}{\pi a_{0}^{3}}\dfrac{q^{2}}{4\pi\varepsilon_{0}} \int_{0}^{2pi} \int_{0}^{pi}\int_{0}^{b}(e^{\dfrac{-2r}{a_{0}}}) (\dfrac{b}{r^{2}}-\dfrac{1}{r})[/itex] r2sin θ dr dθ dΦ =
    [itex]E_{1}^{(1)}=\dfrac{1}{\pi a_{0}^{3}}\dfrac{q^{2}}{4\pi\varepsilon_{0}} \int_{0}^{2pi} \int_{0}^{pi}\int_{0}^{b}(e^{\dfrac{-2r}{a_{0}}}) (b-r)[/itex] sin θ dr dθ dΦ
     
  13. May 1, 2016 #12

    blue_leaf77

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    Yes, that's right. Now all ##r##'s will be in the numerator and the integral should be analytically solvable.
     
  14. May 1, 2016 #13
    hmm i'm still not seeing how to make use of the standard integrals given
     
  15. May 1, 2016 #14

    blue_leaf77

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    I mean this equation:
    $$
    E_{1}^{(1)}=\dfrac{1}{\pi a_{0}^{3}}\dfrac{q^{2}}{4\pi\varepsilon_{0}} \int_{0}^{2pi} \int_{0}^{pi}\int_{0}^{b}e^{\frac{-2r}{a_0}} (b-r) dr d\theta d\phi
    $$
    Do the integrals over the angles first if the appearance makes obscure your mind. You will then be left with the integral over ##r##, which takes the form of the formula you posted under the "Relevant equations" in your starting post.
     
  16. May 1, 2016 #15
    [itex]\int_{0}^{b}(e^{\dfrac{-2r}{a_{0}}}) (b-r) dr = \int_{0}^{b}-r(e^{\dfrac{-2r}{a_{0}}})+b(e^{\dfrac{-2r}{a_{0}}}) [/itex]

    This is hurting me, I cant see how these work with standard integrals?? Everything im trying on my paper is not working :/ Have I gone wrong somewhere else and thats why I cant see it
     
    Last edited: May 1, 2016
  17. May 1, 2016 #16

    Integrate via a substitution using u=2r/a
     
  18. May 1, 2016 #17

    blue_leaf77

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    One more time, don't leave the integration element behind whenever you write an integral.
    As to the method to solve, do you know integration by substitution? It's the same as the one someone above suggested.
     
  19. May 2, 2016 #18
    I have done the subs after a very long day of staring at this and I think its correct but its incredibly messing and hard to simplify.
    For the first integral...
    = ((a0b) /2) x (1-e-2b/a0)
    Second
    = (a02 / 4) x ( 1 - e - 2b/a0 - 2b/a0 e-2b/a0)
    So I'll add them to complete.
    I have Just to double check that the limits are from 0 to 2b/a0?
     
  20. May 2, 2016 #19

    blue_leaf77

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    Yes, add them.
    Why do you still want to double check again when you have computed the integral?
     
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