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SO(4) symmetry in the Coulomb potential

  1. Jan 31, 2016 #1

    ShayanJ

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    In chapter 4 of "Modern Quantum Mechanics" by Sakurai, in the section where the SO(4) symmetry in Coulomb potential is discussed, the following commutation relations are given:

    ## [L_i,L_j]=i\hbar \varepsilon_{ijk} L_k##
    ## [M_i,L_j]=i\hbar \varepsilon_{ijk}M_k##
    ## [M_i,M_j]=-i\hbar \varepsilon_{ijk} \frac 2 m H L_k ##

    Where H is the Hamiltonian ##H=\frac{\mathbf p^2}{2m}-\frac{Ze^2}{r} ## and ## \mathbf M=\frac{1}{2m} (\mathbf p \times \mathbf L-\mathbf L \times \mathbf p)-\frac{Ze^2}{r}\mathbf r ## and ## \mathbf L ## is the angular momentum operator.
    Then there is following paragraph:
    There are two things that aren't clear for me!

    1) Scattering states can also be eigenkets of the Hamiltonian so why do we need to consider the subspace of bound states only?

    2) Its true that for eigenkets of the Hamiltonian, we can just replace it with energy, but within the subpsace of bound states we still have superpositions of these bound states with different energies which are not eigenkets of the Hamiltonian so it seems to me, because of these superposition states, we can't simply replace Hamiltonian with energy.

    Thanks
     
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  3. Jan 31, 2016 #2

    vanhees71

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    Ad 1) Of course, you don't need to restrict yourselves to the eigenspaces with ##E<0## (for which you have the Lie algebra ##\mathrm{su}(2) \oplus \mathrm{su}(2)## or equivalently ##\mathrm{so}(4)##. For ##E=0## you have the Lie algebra of the Galilei group, and for ##E>0## that of the Lorentz group ##\mathrm{so}(1,3)##.

    Ad 2) Eeach eigenvalue defines a subspace in Hilbert space, and these can be characterized by the representations of the above mentioned Lie algebras. For the bound states you have the finite-dimensional representations of ##\mathrm{so}(4)##.
     
    Last edited: Jan 31, 2016
  4. Jan 31, 2016 #3

    ShayanJ

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    So different supspaces associated to different energies are just different finite dimensional representations of o(4)?
    And...I think you meant so(4) not o(4), right?

    Also, Do you know any book or paper that treats this more generally and thoroughly?

    Thanks again
     
  5. Jan 31, 2016 #4

    vanhees71

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    Yes, it's ##\mathrm{so}(4)##. I've corrected it in the original posting.

    The only book I know that treats this is Thirring's quantum mechanics book, vol. 3 in the four-volume series on mathematical physics (I'm not sure whether there's an English translation).
     
  6. Jan 31, 2016 #5

    dextercioby

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    So it was this Napolitano fellow who added this section in the original text by Sakurai. He couldn't copy-paste every word of Schiff (1968), so he reformulated this to make it a little less clear.

    Shiff (1968).JPG
     
  7. Jan 31, 2016 #6

    dextercioby

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    But I'm sure you know that Thirring's books have been translated into English (as two volumes, not four) and published by Springer. Anyways, his treatment is excellent from a mathematical perspective, but he misses the continuous spectrum or E=0, as you mention. A more extensive treatment is given by Arno Böhm, Ch. VI of his Quantum Mechanics book (Springer, 1979).
     
    Last edited: Jan 31, 2016
  8. Feb 1, 2016 #7

    strangerep

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    Heh, well, I have a private workfile but since I'm not the sole author I can't make it available. But I can probably fill in many of the missing pieces...

    The relevant dynamical symmetry group is constructed from a Hamiltonian and several conserved quantities (angular momentum ##L## and LRL vector ##M##). We make the standard assumption that these are to be represented on a Hilbert space as self-adjoint operators. Since ##L,M## commute with ##H## then, by the spectral theorem(s), there is a basis for the (rigged) Hilbert space in terms of eigenstates of ##H##. Also by the spectral theorem, any function ##f(H)## acting on an energy eigenstate can be represented in terms of ##f(E)## .

    There is no symmetry transformation that can map between states of different energies. In particular, there is no symmetry transformation between the ##E<0##, ##E=0## and ##E>0## sectors of the Hilbert space. Therefore, the Hilbert space can be decomposed into a direct sum of 3 corresponding spaces. Then, on the spaces where ##E\ne 0##, we can introduce another operator in place of the LRL vector, taking the form: $$M' ~:=~ \frac{M}{\sqrt{2\mu |H|}} ~,$$where ##\mu## is the reduced mass of the system. Think of the messy denominator in terms of an ##f(H)## acting on energy eigenstates (of definite sign), then think about superpositions of those eigenstates which make up that sector of the Hilbert space.

    The commutation relations get modified to become: $$[M'_j, M'_k] ~=~ -\mbox{sign}(H) \, i\hbar \epsilon_{jk\ell} L_\ell ~,$$ hence there is a different Lie algebra for the +/- energy sectors. Indeed, one can analyze the +/- energy cases separately. In the -ve case, one derives bound states (after a lot of algebra). Some of this is done in Wybourne's textbook, but I'd caution that a few of his formulas seem to be wrong (imho), so one must check thoroughly everything he says.

    HTH.
     
  9. Feb 1, 2016 #8

    A. Neumaier

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    In the SO(4) case, the subspace spanned by the bound states is given by the direct sum of the appropriate representation spaces at fixed E<0, and in the SO(1,3) case, the subspace spanned by the scattering states is given by the direct integral of the appropriate representation spaces at fixed E>0. Their direct sum (and closure to get E=0) is the full Hilbert space.

    Barut and Raczka also treat the more general dynamical symmetry group SO(2,4), which allows to handle all energies simultaneously.
     
    Last edited: Feb 1, 2016
  10. Feb 2, 2016 #9

    dextercioby

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    @A. Neumaier 1. Barut's masterpiece is known to be plagued by typographical errors. 2. He's one of the creators of the theory of dynamical groups. I couldn't find the connection between H-atom and SO(4,2) in his book. :(
    @strangerep . Yes, Wybourne is the book to read on this particular topic, but it's horrendously difficult to digest it. Thanks for bringing it up.
    @vanhees71 in Post# 2. The SO(4) case and the bounded states are thoroughly discussed in the literature (there's even a book only on this restricted symmetry! https://www.amazon.com/Linearity-Sy...&qid=1454457823&sr=1-18&keywords=Frank+Singer). Where did you get the E=0 group?
     
    Last edited by a moderator: May 7, 2017
  11. Feb 2, 2016 #10

    strangerep

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    Unfortunately a number of his earlier papers have similar, or slightly different, errors. Check his definition(s) of the LRL vector... ?:) :oldgrumpy:

    Do you mean you couldn't locate the section? (If so, have a look at p383, and preceding sections.) Or do you mean you couldn't make sense of his calculations? (I couldn't either, even after substituting a correct LRL vector.)

    And he seems to quote B+R uncritically on some things (such as the SO(4,2) business, and also Barut's speculations about the role of the "conformal mass" in physics which have now mostly been consigned to the dustbin of history.

    I think it should probably be called a "Galilei-like" group. It arises because the components of the LRL vector commute in E=0 case, leaving a Galilei-like algebra. Come to think of it, I haven't actually tried the calculation in the classical case where one sets ##H=0##, i.e., $$\frac{p^2}{2\mu} ~=~ \frac{Y}{q} ~,$$(where ##Y## is a shorthand for some constants). I wonder what the LRL vector reduces to in that case?
     
  12. Feb 2, 2016 #11

    dextercioby

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    So the key word to look for in the whole book is Kepler, not Hydrogen. Perhaps we can make an error analysis someday of his chapter 12. :) It could be useful for a lot of people.
     
  13. Feb 3, 2016 #12

    A. Neumaier

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    Yes, one has to check his calculations independently to make sure one has a correct result. I haven't done it completely for the SO(2,4) case. However, the classical kepler problem also has an SO(2,4) dynamical symmetry, so the general features must be right, and it is only a question of making the algebra fully correct. Unfortunately, Wybourne's SO(2,4) treatment doesn't seem to be much better.
     
  14. Feb 3, 2016 #13

    vanhees71

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    For the subspace ##E=0## (vanishing energy eigenvalue) you get the commutation relations of the Galilei group (see #1 in this thread): The ##M_j## commute with the ##L_j## (angular momentum operators) as vector operators and commute among themselves in this subspace.
     
    Last edited by a moderator: May 7, 2017
  15. Feb 8, 2016 #14

    ShayanJ

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    I have one more question.
    In the classical Kepler problem, its easy to say that the conservation of the LRL vector means that the shape of the bound orbit doesn't change.
    But in the quantum mechanical case, the LRL vector becomes a vector symmetry operator. So the state of the system goes to an equivalent state under the action of the LRL vector operator. But what is this action? What does the LRL vector operator do to a state?
    Thanks
     
  16. Feb 8, 2016 #15

    strangerep

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    I suspect you'll have a few more... :oldwink:

    [Edit: In the classical case, observables commute. So one may speak of both the angular momentum and LRL eccentricity of an orbit simultaneously. But the quantum operators for angular momentum and LRL do not commute in general, hence we cannot characterize states in terms of simultaneous eigenstates of these operators.]

    That's not simple. It turns out that we have 2 Casimirs, being the squares of ##J_\pm := (L\pm M')/2##. But for this problem, it also turns out that ##L \cdot M = 0##, which has the consequence that these Casimirs are equal. (You'll have to crunch through the computations to verify this.) So I'll just call them both ##J##.

    Although ##J## obeys so(3) commutation relations, it is not the physical angular momentum ##L##. Nevertheless, ##J^2## has the usual eigenvalues ##j(j+1)## (i.e., half-integral ##j##). Just keep in mind that this ##j## is not the same as the physical angular momentum eigenvalues. We're just working in an abstract Hilbert space to find the ##J^2## eigenvalues.

    Since ##J## commutes with ##H##, we can use it to find the detailed spectrum for given energy. Indeed, it turns out that $$ E \propto \frac{1}{(2j+1)^2}$$which is usually written as $$ E \propto \frac{1}{n^2}$$where ##n := 2j+1## is the more familiar principal quantum number for the H-atom.

    Now, ##2J^2 = (L^2 + M'^2)/2##, hence ##L^2 = 4J^2 - M'2##. Denoting the eigenvalues of ##L^2## by ##\ell(\ell+1)##, this implies that ##\ell(\ell+1) \le 4j(j+1)## (since ##M'## is Hermitian). And since ##n := 2j+1##, we then deduce ##\ell \le n-1##, which is the well-known formula constraining possible angular momenta for each value of the principal quantum number ##n##. One then constructs states as ##|n\ell m\rangle##, where ##m## is the usual eigenvalue for the z-component of angular momentum.

    Unfortunately, ##[L, M'] \ne 0## in general, so the action of ##M'## on these states is complicated. Authors such as Barut and Wybourne attempt to construct operators in the larger group SO(4,2) which can ladder between the various ##|n\ell m\rangle## states. But this is really messy, and I've so far been unable to verify their claimed results.

    IIRC, they also offer alternate constructions of H-atom states (i.e., in a different basis) in terms of the ##J## operators, but those computations also become rather messy. Note however, that ##M'## does not commute with the ##J## operators in general, so it's action on eigenstates of ##J^2## is messy.

    That's probably more than you wanted to know, right? :biggrin:
     
    Last edited: Feb 8, 2016
  17. Feb 9, 2016 #16

    ShayanJ

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    Well, in a way yes. Its more than I wanted to know but is not a superset of what I wanted to know!
    On the other hand, its surely valuable information which I do want to know.
    Well, I guess its too messy to be answered here.
    Thanks
     
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