Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Spin-orbit coupling for s-orbital states

  1. Feb 10, 2013 #1
    In standard QM textbooks, when calculating the spin-orbit interaction term as a relativistic perturbation for hydrogenic atoms, it is said that the term gives 0 contribution for the s-orbitals (l = 0). This is apparently because the term has the form of S*L and L=0 for the s-orbitals.
    However this S*L is preceded by a factor proportional to 1/r^3 and the expectation value of 1/r^3 for the l=0 states gives ∞.
    So I think, strictly speaking, when one evaluates the 1st order perturbation caused by the spin-orbit coupling for the s-orbital states, one has to deal with ∞*0 and I guess it gives a finite contribution.
    But as far as I can see, I've never seen any textbooks/articles arguing this subtlety.
    Am I wrong or missing something?
    Any comments are welcome.
    Thanks.
     
  2. jcsd
  3. Feb 11, 2013 #2

    DrDu

    User Avatar
    Science Advisor

    The pre-factor 1/r^3 and thus the divergence of the spatial integral is an artifact of the non-relativistic approximation of the s-wavefunction, while the vanishing of L is exact on symmetry grounds.
     
  4. Feb 11, 2013 #3

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    In this case, it does not give a finite contribution. Mathematically, when you find expressions of the form ∞*0, you have to be very careful with the individual factors appearing in the limit that you're computing. So, for instance, if we want to compute [itex]\lim_{x\rightarrow 0} \sin x/ x[/itex], we note that [itex]\sin x \sim x[/itex] for small [itex]x[/itex] so that, in the limit, the expression becomes [itex]x/x =1[/itex].

    In your case, we can isolate the divergence in the radial integral by changing the region of integration to [itex]\epsilon \leq r < \infty[/itex]. The divergence would be obtained in the limit that [itex]\epsilon\rightarrow 0[/itex] and we can see that it takes the form [itex]\ln \epsilon[/itex]. This procedure is known as a regularization scheme and we can call [itex]\epsilon[/itex] a regulator for the divergence.

    However, the zero that we get when [itex]\hat{L}[/itex] acts on the [itex]l=0[/itex] is an exact operator statement, without any parameterization through a regulator. Since the angular momentum is quantized, it does not make any sense to consider an eigenvalue infinitesimally close to [itex]l=0[/itex]. So it is impossible to use the factor of [itex]\ln\epsilon[/itex] to somehow remove the zero. The 1st order perturbation is exactly zero.
     
  5. Feb 13, 2013 #4
    Thanks a lot for the comments.
    I agree that we must be careful when we take limits. But I wonder whether the limit you are talking about is of right kind, fzero. You take a limit in the region of the radial integration. However, as you say, the angular momentum part has nothing to do with it.

    In order just to evaluate the term purely mathematically (forgetting physics behind it), I think we could take [itex] l [/itex] as a real number and examine how it behaves as we take [itex] l \rightarrow 0[/itex].
    So let [itex] l = \epsilon [/itex]. Accordingly [itex] j = \epsilon+1/2 [/itex].
    Then the angular part becomes
    [tex]j(j+1)-l(l+1)-s(s+1) = \epsilon.[/tex]
    Obviously it approaches to zero as [itex] O(\epsilon) [/itex] when [itex] \epsilon \rightarrow 0 [/itex].
    On the other hand the radial part is given by (apart from an overall proportional constant)
    [tex]\int_0^\infty \mathrm{e}^{-r} r^{2\epsilon-1} ( L_{n+\epsilon}^{2\epsilon+1}(r))^2 dr,[/tex]
    where [itex] L_{n+\epsilon}^{2\epsilon+1}(r) [/itex] is the generalized associated Laguerre function with non-integer orders.
    This function can be expressed as the confluent hypergeometric function as
    [tex]L_{n+\epsilon}^{2\epsilon+1}(r) \propto F(-n+\epsilon+1;2\epsilon+2;r) = \Sigma_{k = 0}^\infty \frac{(-n+\epsilon+1)_k}{k!(2\epsilon+2)_k}r^k.[/tex]
    Let this be [itex]\Sigma_{k = 0}^\infty c_k r^k [/itex] for simplicity.
    By substituting this series into the above integral, we get
    [tex] \Sigma_{k = 0}^\infty \Sigma_{l = 0}^\infty c_k c_l \int_0^\infty \mathrm{e}^{-r} r^{k+l+2\epsilon-1} dr= \Sigma_{k = 0}^\infty \Sigma_{l = 0}^\infty c_k c_l \Gamma(k+l+2\epsilon),[/tex]
    where [itex] \Gamma(k+l+2\epsilon) [/itex] is the Gamma function.
    At the end of the day we've got a series of Gamma functions with some complicated pre-factors.

    I don't argue if this series itself is converging. It is pretty complicated.
    But at least if we look at the first term for [itex] k = l = 0 [/itex], being [itex] \Gamma(2\epsilon) [/itex], this approaches to [itex] \Gamma(0) [/itex] as [itex] \epsilon \rightarrow 0 [/itex].
    Because the Gamma function has a pole of order 1 at 0, this term is of order [itex] 1/O(\epsilon) [/itex].
    All the other higher terms [itex] \Gamma(k+l+2\epsilon) (k+l \ge 1) [/itex] gives a finite value approaching to [itex](k+l-1)![/itex].

    So unless the series as whole cancel the first term (and I suspect it does't), the radial part is [itex] 1/O(\epsilon) [/itex].
    This seems to cancel [itex] O(\epsilon) [/itex] from the angular part, giving a finite contribution!

    Of course my treatment of taking the limit presented above is by no means rigorous. There could be easily a flaw.
    Any further comments are well appreciated!
    Thanks
     
  6. Feb 13, 2013 #5

    DrDu

    User Avatar
    Science Advisor

    Why don't you perform this analysis with the eigenfunctions of the Dirac equation instead?
     
  7. Feb 13, 2013 #6
    If you apply Dirac's theory to the hydrogen atom, you get an exact solution with an exact eigen energy. Everything is fine. You don't get those individual perturbative terms like the spin-orbit term, Darwin term etc.

    On the other hand my interest is on the validity of the non-relativistic approximation, or in other words the relativistic correction of Schrodinger equation very commonly found in many textbooks.

    As I've been trying to explain, there seems to be some subtlety regarding the convergence of the spin-orbit term for the s-orbital states. However I've never seen any literature, even not a single textbook, arguing this. All the textbooks I've looked into more or less swiftly say this term is 0 for l=0 because l=0.

    So my first thought is that I should be wrong, because this problem should've been established long long time ago (in 1920's?) and if there were really such a subtlety it should've been well known among physicist and many texts would be written more carefully.
    But anyway I want to see a reasonable account about my doubt even if whichever case is correct.
     
  8. Feb 13, 2013 #7

    Bill_K

    User Avatar
    Science Advisor

    Bethe-Salpeter, "Quantum Mechanics of One and Two-Electron Atoms", in their discussion of spin-orbit coupling on p185, has the following footnote:
    The "first term in (40.3)" is like r1-3 k1·S. (And perversely k1 is the notation they use for the orbital angular momentum.) :uhh:
     
  9. Feb 13, 2013 #8
    Oh thanks a lot for the info!
    I will look into that book.
     
  10. Feb 13, 2013 #9

    DrDu

    User Avatar
    Science Advisor

    The problem has extensively been treated in relativistic quantum chemistry, see, e.g.

    http://www.springerlink.com/index/P6J1535L1K61J106.pdf
    http://theochem.chem.rug.nl/publications/pdf/ft231.pdf
    http://theochem.chem.rug.nl/publications/PDF/ft245.pdf
    http://dare2.ubvu.vu.nl/bitstream/handle/1871/23031/123236.pdf?sequence=1

    Note especially the now regular expression for spin orbit effects in eq. 42 in van Lenthe et al 1993.
    Proofs of regularity of the hamiltonian can be found in
    http://iopscience.iop.org/1402-4896/34/5/007
     
    Last edited: Feb 13, 2013
  11. Feb 14, 2013 #10
    Fantastic...
    It seems to be a well-known problem in some community.
    I'll dig into there papers.
    Cheers
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Spin-orbit coupling for s-orbital states
  1. Spin orbit coupling (Replies: 1)

  2. Spin-Orbit Coupling (Replies: 2)

  3. Spin-Orbit Coupling (Replies: 10)

Loading...