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Formula for expectation value of raidous in Hydrogen atom

  1. Dec 13, 2009 #1
    Let's consider eigenstates [tex] |nlm\rangle [/tex] of hamiltonian of hydrogen atom.
    Can anyone prove that
    [tex] \langle r \rangle = \langle nlm|r|nlm\rangle = \frac{a}{2}(3 n^2-l(l+1))[/tex].
    Where a - bohr radious.
    I've been trying to prove it using some property of Laguerre polynomials (which are
    radial part of eigenstates) but I stucked in it.

    I found this equation in "Princliples of QM" by Shankar
    (unfortunately with neither proof nor hint).
  2. jcsd
  3. Dec 13, 2009 #2


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    Here's an elementary argument that leads to the [itex]n^2[/itex] dependence and the dependence of a on fundamental constants, but not the dimensionless factor such as 3/2, or the l(l+1) term: http://www.lightandmatter.com/html_books/6mr/ch05/ch05.html#Section5.4 [Broken]

    The complete derivation of states in hydrogen is done in a lot of different textbooks. I'd think it would be lengthy for someone to transcribe a full derivation here. You can find one, for example, in Sakurai, Modern Quantum Mechanics, appendices A.5 and A.6.
    Last edited by a moderator: May 4, 2017
  4. Dec 13, 2009 #3


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    My advice is to take a look at the problems 6.29 and 6.30 from D.J. Griffiths's introductory text on QM.
  5. Dec 14, 2009 #4


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    Well it's a tricky integral to evaluate, but not too bad.. You'll need this:

    [tex]\int_0^{\infty}x^{\alpha+1} e^{-x} \left[L_n^{(\alpha)}\right]^2 dx= \frac{(n+\alpha)!}{n!}(2n+\alpha+1)[/tex]

    (which I'll freely admit to cribbing straight off Wikipedia. I don't bother memorizing these things! ;))
  6. Dec 14, 2009 #5


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    I searched for the formula you've written. It's on Wikipedia, indeed, but also on Wolfram which I trust more. Apparently there are at least 2 different ways to define the associated Laguerre polynomials. One of them is fomula (12) from http://mathworld.wolfram.com/LaguerrePolynomial.html" [Broken]. The other one can be found in various physics books, like for example L.Pauling and E. Wilson "Introduction to Quantum Mechanics".

    Pauling's definition reads:

    [tex] L_{r}^{s} (x)=\frac{d^{s}}{dx^{s}}\left[e^{x}\frac{d^{r}}{dx^{r}}\left(e^{-x}x^{r}\right)\right] [/tex]

    If you look at this formula and to the one on the Wolfram's page, I think you can see a difference.

    As to the OP, a full-blown derivation of [itex] \langle nlm|r|nlm\rangle [/itex] can be found in the splendid book by Bethe and Salpeter on the QM of one and 2-electron atoms. Pages 12 to 17.
    Last edited by a moderator: May 4, 2017
  7. Dec 18, 2009 #6
    Those two definitions are really different! They are polynomials of different order. What Pauling calls Lsr is a polynomial of degree r-s. What Wolfram calls Lsr is a polynomial of degree r.

    Which one is correct? Solving the hydrogen atom, it seems Wolfram's is correct. The two different definitions of associated Legendre polynomials don't give the same answer, and aren't even proportional.
    Last edited by a moderator: May 4, 2017
  8. Dec 18, 2009 #7
    Actually, Pauling's and Wolfram's definition are related:

    [tex]L_{l-m}^m (Wolfram) \mbox{ } \alpha \mbox{ } L_{l}^m (Pauling)[/tex]

    or more exactly:

    [tex](-1)^ml!L_{l-m}^m (Wolfram) \mbox{ } = \mbox{ } L_{l}^m (Pauling)[/tex]


    [tex](-1)^r(s+r)!L_{s}^r (Wolfram) \mbox{ } = \mbox{ } L_{s+r}^r (Pauling)[/tex]

    So I guess you can use either one, but the indices are different.

    Wolfram's definition only requires the bottom index to be a nonnegative integer and the top index to be greater than -1 (but not necessarily integral). Pauling's definition requires the bottom index to be a nonnegative integer and the top index to be a nonnegative integer (and if the top index is greater than the bottom index, the function is zero everywhere).

    I really don't see the point in Pauling's definition. It seems worse in every respect.

    edit (conclusion):

    Actually, the relationship between Wolfram's and Pauling's formulas is just the well-known relation (for integral [tex]\nu [/tex]):

    [tex]L^{\nu}_{\eta}=(-1)^\nu \frac{d^\nu}{dx^\nu}L^{0}_{\eta+\nu} [/tex]

    between associated Laguerre and Laguerre (when the top of the associated Laguerre is zero, it becomes just Laguerre).
    Last edited: Dec 18, 2009
  9. Dec 19, 2009 #8
    It took me an hour to derive this, but the equation that will help you is not the one that Wikipedia says, but this one:

    [tex]\int^{\infty}_{0} x^{\alpha+2}e^{-x}(L^{\alpha}_{n})^2dx=

    where for integers, [tex]\Gamma(n)=(n-1)! [/tex]

    I've already checked out this formula for several different Laguerre polynomials, and it seems to be correct.

    Also, the Wikipedia article on the hydrogen atom ( http://en.wikipedia.org/wiki/Hydrogen_atom#Wavefunction ) I believe has the wrong normalization factor.
    The [tex][(n+\ell)!]^3[/tex] in the denominator should not be cubed. Unfortunately, in trying to make sure it was wrong, I checked several websites for the correct normalization factor, and a lot of them have it cubed. So either I'm wrong, or a lot of people are copying Wikipedia and getting it wrong. In fact, Wikipedia contradicts itself:


    where their normalization factor is not cubed. Could someone verify if it should be cubed or not, and change Wikipedia if it's wrong?
  10. Dec 20, 2009 #9


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    Condon & Shortley in their famous 1935 book on spectral lines theory mention both available definitions of the associated Laguerre. But they only use the one called above "Pauling". It's probably because the first solution of the H-atom using Schroedinger's equation was given by Schroedinger himself in 1926 and he used the "Pauling" definition. It's used throughout most physics books and the normalization factor with the 3rd power of the factorial comes out precisely if you use the "Pauling" definition.
  11. Dec 20, 2009 #10
    The formula suggested by the Wikipedia article:

    \int_0^{\infty}x^{\alpha+1} e^{-x} \left[L_n^{(\alpha)}\right]^2 dx= \frac{(n+\alpha)!}{n!}(2n+\alpha+1)

    when applied to the wavefunction without the 3rd power of the factorial (just the 1st power), shows that the wavefunction is correctly normalized. If it's really the 3rd power, then you can't use this relation:

    \int_0^{\infty}x^{\alpha+1} e^{-x} \left[L_n^{(\alpha)}\right]^2 dx= \frac{(n+\alpha)!}{n!}(2n+\alpha+1)

    Although most quantum mechanics book use the Pauling definition, I'm pretty sure most mathematical physics books use the Wolfram definition. The Wolfram is more natural in that it fits into the typical framework for classical orthogonal polynomials (in the Pauling definition you're taking derivatives of a Rodrigo formula; in the Wolfram scheme you have a Rodrigo formula).

    edit: added the solutions in terms of the two different forms

    Wolfram form:

    [tex] \psi_{n\ell m}(r,\vartheta,\varphi) = \sqrt {{\left ( \frac{2}{n a_0} \right )}^3\frac{(n-\ell-1)!}{2n[(n+\ell)!]} } e^{- \rho / 2} \rho^{\ell} L_{n-\ell-1}^{2\ell+1}(\rho) \cdot Y_{\ell}^{m}(\vartheta, \varphi )

    Pauling form:

    \psi_{n\ell m}(r,\vartheta,\varphi) = \sqrt {{\left ( \frac{2}{n a_0} \right )}^3\frac{(n-\ell-1)!}{2n[(n+\ell)!]^3} } e^{- \rho / 2} \rho^{\ell} L_{n+\ell}^{2\ell+1}(\rho) \cdot Y_{\ell}^{m}(\vartheta, \varphi )
    Last edited: Dec 20, 2009
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