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!

Having trouble with this normalization

  1. Mar 22, 2007 #1
    1.Question.

    the unnormalized excited state wavefuction of the H atom is:
    [tex]\psi = ( 2 - r/a_0 ) e^a[/tex]

    where a = [tex]-r/a_0[/tex]

    Normalize the function to one.

    2. My attempts.

    I tried 'integrating' the psi*psi, i.e. I squared the above wavefuction.

    [tex]N^2\int_{0}^{\infty} R^2e^{2a}\int_{0}^{\pi}sin \theta d\theta\int_{0}^{2\pi}d\phi =1[/tex]

    From here, I got confused with the examples. The second and third integrations were obviously 2 and [tex]2\pi[/tex], but the first integration I'm getting confused as to what to put as [tex]R^2[/tex] and I'm getting the same answer as the examples because when I done the calculation, the power to which R is the same with every question, which was [tex]a^3_0/4[/tex]

    Hence: [tex]a^3_0/4*2*2\pi[/tex] =[tex] 1/N^2[/tex]

    Hence I worked out N from this equation, but the answer I got was:

    [tex]\psi =(1/\pi a^3_0)^{1/2}e^{-r/a_0}[/tex]

    Which is the same as every other damn normalized wavefuction in the book.

    I think I'm having trouble actually understanding how [tex]R^2[/tex] is translated from the original wavefuction to the actual integration, hence my immense trouble to working out what R is. Can anyone help with this?
     
    Last edited: Mar 23, 2007
  2. jcsd
  3. Mar 22, 2007 #2

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I'm afraid I can't parse what you're saying. Could you try to clean it up?
     
  4. Mar 23, 2007 #3
    Sorry, Had a lot of trouble getting used to the Latex type setting. Updated it so you can read it a bit.. hopefully.
     
  5. Mar 23, 2007 #4

    Dr Transport

    User Avatar
    Science Advisor
    Gold Member

    The radial integral is
    [tex] \int_{0} ^{\infty} (4 -4a+a^{2}) e^{-2a} a^{2} da /a_{0} ^{2}[/tex]
    where [itex] a = r/a_{0} [/itex]. These are straightforward integrals that are found in any integral table.
     
    Last edited: Mar 23, 2007
  6. Mar 23, 2007 #5
    thank you!
     
  7. Mar 23, 2007 #6
    Wait, when I calculated the first integral, I got:

    [tex]\psi = (2 - a)e^{-a}[/tex]

    hence:

    First integral: [tex]\int_{0}^{\infty}(4-4a+a^2)(e^{-2a})a_0da[/tex]
    =[tex]a_0\int_{0}^{\infty}(4-4a+a^2)(e^{-2a})da[/tex]

    since: [tex]a = -r/a_0[/tex]

    [tex]da/dr = 1/a_0[/tex]
    [tex]da = (1/a_0)dr[/tex]
    [tex](a_0)da = dr[/tex]
    I'm not sure I understand how you got the [tex]a^2/a_0^2[/tex] Could you, if you have the time, care to elaborate?
     
    Last edited: Mar 23, 2007
  8. Mar 23, 2007 #7

    Dr Transport

    User Avatar
    Science Advisor
    Gold Member

    Change variables from [itex] r [/itex] to [itex] a = r/a_{0} [/itex], you have to remember that in sphereical coordinates you have [itex] r^{2} dr [/itex] for the integration over all r space. Check my original work because I forget to put in factors normalization until I check my units at the end. and I usually screw up early in the am.....
     
    Last edited: Mar 23, 2007
  9. Mar 23, 2007 #8
    Still having trouble with the first integration part.. I'll rewrite the original equation:

    [tex]\psi = (2-r/a_0)e^{-r/a_0}[/tex]
     
  10. Mar 23, 2007 #9

    Dr Transport

    User Avatar
    Science Advisor
    Gold Member

    Let [itex] \alpha = r/a_{0} [/itex], then

    [tex] |\psi |^{2} = (2-\alpha)^{2} e^{-2\alpha} [/tex] so the integral becomes

    [tex] \int_{0} ^{\infty} (4-4\alpha+\alpha^{2})e^{-2\alpha} a_{0} ^{3} \alpha^{2} d\alpha [/tex]

    From here you have yo do the integrals, the can be found in an integral table.... In general you have

    [tex] \int _{0} ^{\infty} x^{n} e^{-ax} dx = \frac{n!}{a^{n-1}} [/tex] for [itex] a > 0, n [/itex] an integer. From here you can find everything you need to calculate the normalization factor.
     
    Last edited: Mar 23, 2007
  11. Mar 24, 2007 #10

    cgw

    User Avatar

    I have nothing helpful to add. Just two (maybe stupid) questions as I try to follow along:
    In the original equation should the exp. of e be a/2?
    Should the integrals of the theta and Phi functions be included in the normalization?
     
  12. Mar 24, 2007 #11

    Dr Transport

    User Avatar
    Science Advisor
    Gold Member

    The problem may have been as written.

    Now as for the angular integrals, the normnalization of them are included in the integral in the form of factors of [itex] \pi [/itex] which up to this point have been ignored. The part of the original posters problem is that they are forgeting that the form of the volume element in spherical coordinates is [itex] r^{2} dr \cos(\theta) d \theta d \phi [/itex].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Having trouble with this normalization
Loading...