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Radial Wave Function

  1. Sep 18, 2010 #1
    Radial Wave Function; Normalized Radial Wave Function

    5. Is the brute-force method the only way to show the Equation is satisfied? By that I mean differentiating R20 once, twice, and substituting it in? I tried that earlier, but it was very nasty.

    http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-18100956.jpg?t=1284823040 [Broken]

    http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-18100944.jpg?t=1284823023 [Broken]

    7. I'm not quite sure what I'm doing wrongly. I think I'm kind of close to the correct procedure.

    http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-18114302.jpg?t=1284828367 [Broken]

    http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-18114318.jpg?t=1284828434 [Broken]
     
    Last edited by a moderator: May 4, 2017
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  3. Sep 18, 2010 #2

    vela

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    Re: Radial Wave Function; Normalized Radial Wave Function

    I'm afraid you're going to have to revise your notion of nasty. The function is of the form c(1-x)e-x. It's pretty straightforward to show it satisfies the radial equation by direct substitution.
    You need a factor of r2 in your normalization integral. It comes from the volume element in spherical coordinates.
     
    Last edited by a moderator: May 4, 2017
  4. Sep 18, 2010 #3
    Re: Radial Wave Function; Normalized Radial Wave Function

    I'll look at it again. I must not have simplified it enough. But having to do the product rule so many times made it "nasty."

    Also, I'm only integrating with respect to r, not theta and phi. Do I still need r^2? I'm a bit rusty on my coordinate transformations.
     
    Last edited: Sep 18, 2010
  5. Sep 18, 2010 #4

    vela

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    Re: Radial Wave Function; Normalized Radial Wave Function

    Yes, you need the factor of r2. The complete wave function is ψnlm=Rnl(r)Ylm(θ,φ), and its normalization condition is

    [tex]1=\int\psi_{nlm}^*\psi_{nlm}r^2drd\Omega=\int R_{nl}^*R_{nl}r^2dr\int Y_{lm}^*Y_{lm}d\Omega[/tex]

    By convention, the individual pieces are normalized so that the radial and angular integrals are equal to 1 individually.
     
  6. Sep 18, 2010 #5
    Re: Radial Wave Function; Normalized Radial Wave Function

    So, d-omega is just d-theta, d-phi, right?

    For this problem I only need to show the radial function is normalized. How do I show the angular function is similarly normalized? Or is that even possible given that I do not know the magnetic quantum number?

    Of course, showing the radial function is normalized is quite easy in this case.
     
    Last edited: Sep 18, 2010
  7. Sep 18, 2010 #6

    vela

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    Re: Radial Wave Function; Normalized Radial Wave Function

    Not exactly. It's sin θ dθ dφ, which is sometimes written d(cos θ) dφ.
    You can't. Without knowing m because you don't know what Ylm is (though there's only three possibilities), but you're not asked about the angular part anyway. I was just showing you why the factor of r2 appears in the integral for normalizing the radial wave function.
     
  8. Sep 18, 2010 #7
    Re: Radial Wave Function; Normalized Radial Wave Function

    Ah. I forgot about those extra terms that appear when you transform to polar spherical coordinates. I think we covered how they got there in my vector analysis class last semester, but I forgot how to do that. And I'm not talking about simply using the little formulas that transform from Cartesian to polar cylindrical/spherical. We used tensor notation for all that stuff.
    That's what I figured. I always like to know the general case or context, so I'm glad you wrote everything out.
     
  9. Sep 18, 2010 #8
    Well, chief, it hasn't been pretty straightforward. Forgive my chicken scratch.

    http://i111.photobucket.com/albums/n149/camarolt4z28/1-1.jpg?t=1284850615 [Broken]

    http://i111.photobucket.com/albums/n149/camarolt4z28/2-1.jpg?t=1284850626 [Broken]

    http://i111.photobucket.com/albums/n149/camarolt4z28/3.jpg?t=1284850640 [Broken]
     
    Last edited by a moderator: May 4, 2017
  10. Sep 19, 2010 #9

    vela

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    Well, I'm not terribly interested in checking your algebra, but I can make some suggestions to make the algebra a bit less tedious. First, pull a factor of 2 out so you have

    [tex]R_{20}=\frac{1}{\sqrt{2}a_0^{3/2}}\left(1-\frac{r}{2a_0}\right)e^{-r/2a_0}[/tex]

    Note when you do this, r always appears over 2a0. Next, when you find R'20, simplify before proceeding. You should find

    [tex]R'_{20} = \frac{1}{\sqrt{2}a_0^{3/2}}\left[\frac{1}{2a_0}\left(\frac{r}{2a_0}-2\right)e^{-r/2a_0}\right][/tex]

    Keep the original normalization constant separate because it's going to be common to every term so you can cancel it out. Same with the exponential factor.

    Finally, the sum of the first two terms of the differential equation will be a polynomial multiplying an exponential and the normalization constant. That polynomial will factor. The product of one factor, the exponential times, and the normalization constant will be R20.
     
    Last edited: Sep 19, 2010
  11. Sep 19, 2010 #10
    Well, I took out the alpha and only differentiate the non-constant terms. Quite a few of them canceled. I think my algebra is correct up until the last image I posted. That's probably the only work that needs review. But let me incorporate your simplifications and see what I get.

    Okay. I got the same thing for R'20. I assume with R'', I can bring out the 1/2a0 to make things easier since it's just another constant.
     
    Last edited: Sep 19, 2010
  12. Sep 19, 2010 #11
    Here's what I got for R''20.

    http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-19135530.jpg?t=1284922635 [Broken]

    This is pissing me off. Nothing canceled this time, assuming I did R'' correctly.
     
    Last edited by a moderator: May 4, 2017
  13. Sep 19, 2010 #12

    vela

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    That matches what I got, so now you have

    [tex]R''(r)+\frac{2}{r}R'(r) = \frac{1}{\sqrt{2}a_0^{3/2}}e^{-r/2a_0}\left[-\frac{1}{(2a_0)^2}\left(\frac{r}{2a_0}-3\right)+\frac{2}{r}\frac{1}{2a_0}\left(\frac{r}{2a_0}-2\right)\right][/tex]

    Since r/2a0 appears to be a natural combination, I'd write the factor of 2/r in the differential equation in terms of it:

    [tex]\frac{2}{r} = 2\left(\frac{1}{2a_0}\right)\left(\frac{2a_0}{r}\right)[/tex]

    then you'll have a common factor of 1/(2a0)2 that you can factor out. You should be left with some combination of constants and powers of r/2a0.

    I misspoke earlier when I said you'd end up with a polynomial multiplying the exponential. You actually have to pull out a factor of 2a0/r before you get the polynomial you can factor.
     
    Last edited: Sep 19, 2010
  14. Sep 19, 2010 #13
    Okay. That's making sense. I don't know why I didn't factor out the common terms. Let me see how far I can get now.

    Besides the common terms, I'm getting -r/2a0 + 7 -6a0/r
     
    Last edited: Sep 19, 2010
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