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To check if these wave functions are normalized to 1

  1. Nov 3, 2007 #1
    [SOLVED] to check if these wave functions are normalized to 1

    I need to check if the following radial functions are properly normalized to unit probability

    R(1,0) (r) =


    2(1/ao)^3/2 e^(-r/ao)

    R(2,1) (r) =


    (1/2*ao)^3/2 *[ r/ sqrt(3)*a0] e^(-r/2ao)

    We do know that


    ∫ u ^n e^(-u) du = factorial(n)
    0

    To normalize the wave function in the following way



    ∫ [tex] \phi ^2[/tex] = 1
    -∞

    Now applying the same for R(1,0)


    ∫ [2(1/ao)^3/2 e^(-r/ao)]^2 dr
    -∞

    =


    ∫ 4(1/ao)^3 e^(-2r/ao) dr
    -∞


    How do I proceed further.

    Regards.
    QP
     
  2. jcsd
  3. Nov 3, 2007 #2

    malawi_glenn

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    Hint: Radial wave functions are only defined from r = 0 to +infinty.
     
    Last edited: Nov 3, 2007
  4. Nov 3, 2007 #3
    Yes.Thats useful thanks.How do I still perform the integration.I dont know how to integrate it.
     
  5. Nov 3, 2007 #4

    malawi_glenn

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    but have the general formular for it, just do a substitution.

    let 2r/ao = u

    what is then n? Well n = 3. And what is 1/ao if 2r/ao = u ?

    Have you done calculus classes?

    Now try do this, and show me/us what you get.
     
    Last edited: Nov 3, 2007
  6. Nov 3, 2007 #5
    As I understand if we integrate it with respect to r then we dont need to apply in that general formulae at all since everrything else is a constant except e raised to -r/ao. Suppose we integrate with respect to ao then this formulae can be applied.What should we integrate with respect to ao or r.

    Regards
    QP.
     
  7. Nov 3, 2007 #6

    malawi_glenn

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    ao is just a constant, infact it is the bohr radius. r is the guy that you should integrate over.

    And in the R(2,0) case, you must use it. And now also, i dont know if your professor had said this, but the radial part of the wave function, there are "two ways" to have a radial wave function.

    When you solve the radial Schrödinger eq, you do this ansatz:
    [tex] R(r) = u(r)/r [/tex]

    And then you solve the radial part with u(r) as your radial wave function; and the total wave function is normalized according to:

    [tex] \int ^{\infty} _0 |R(r)|^2 r^2 dr = 1 [/tex]

    due to the [tex] r^2 [/tex] that comes from the volume integration. [tex]d\vec{r} = r^2 sin \theta drd\theta d\phi [/tex]

    And this then implies that:
    [tex] \int ^{\infty} _0 |u(r)|^2 dr = 1 [/tex]

    So what is your R(1,0) , R(2,0)? is it the R or u that I denoted here?
     
  8. Nov 3, 2007 #7
    It is R which you denoted here so an r(square) term must be introduced when I integrate with respect to r is it not.

    So it this should be finally integrated is it not

    ∫ 4(1/ao)^3 e^(-2r/ao) r^2 dr
    0
     
  9. Nov 3, 2007 #8

    malawi_glenn

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    Now I have hard to understand what you are saying, but if the R(1,0) is the R(r) i wrote, then yes, you should do the integration that you wrote. Now try to do the substitution i wrote eariler, and see if this becomes unity. If it doesn't, try figure out why =)
     
  10. Nov 3, 2007 #9
    I solved it no problems.It comes to one.All that needs to be done is substitution. The most crucial part of this problem is that additional factor r^2.Without that we cannot proceed anywhere.Thanks a lot for the hints.

    How do I find the most probable value of radius r in the same question?
     
  11. Nov 3, 2007 #10

    malawi_glenn

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    where the wave-function has maximum. Just as you do with an "ordinary" function. The wave function represents probablity right? =)
     
  12. Nov 3, 2007 #11
    I still dont understand how to proceed.

    This is what I need to compute.The radius r needs to be found for both radial wave functions.
     
  13. Nov 4, 2007 #12

    malawi_glenn

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    You dont know how to find maximum for a function?

    First let me first show you the definitions again:

    [tex] | \psi (x') |^2 dx' [/tex] is the probabilty to find the particle between x' and x' + dx', if the wave function [tex] \psi (x) [/tex] is normalized to unity.

    Now for a 3-d wave function:

    [tex] | \Phi (\vec{r'}) |^2 d\vec{r'} [/tex] is the probability to find the particle inside the infinitesimal volume element. (if it is normalized to unity of course).

    Now the radial wave function comes from the ansatz of separation of variables when we solve the 3D shrödinger Equation.
    [tex] \Phi (\vec{r}) = R(r) \cdot h(\theta , \phi ) [/tex]

    Now what do you think you must do to find the most probable radius for the R(1,0) and R(2,0) ? Remember the volume element.

    This is a good link: http://www.physics.gatech.edu/gcuo/lectures/ModernPhysicsLectures/MP16HydrogenAtom.ppt
    OBS it is a power point file, so it can take some time to load. Slide 31 shows you the wave function vs. probability.

    Now good luck =)
     
    Last edited: Nov 4, 2007
  14. Nov 4, 2007 #13
    Its solved now.

    Thanks a lot.

    Regards.

    QP.
     
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