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Maggiore Book misunderstanding

  1. Feb 22, 2016 #1
    20160222_202843.gif

    Well, Look at the image.
    If T is a generator so VTV* (with V unitary) is another basis of the representation too, i am totally agree because it satisfy the structure equation. Now, he say that we can find V that set Gij = tr(TiTj) diagonal BUT when i try, i have :
    Gij = Tr(VTiV*VTjV*)
    = Tr(VTiTjV*) because V is unitary
    = Tr(TiTj) because Tr(AB) = Tr(BA)
    So V as no effect and it can't diagonize it. I don't understand why it don't work ?

    Thanks for all
     
  2. jcsd
  3. Feb 24, 2016 #2

    Demystifier

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    I think you are right. Indeed, the Killing form (Cartan metric) can be expressed in terms of structure constants, which clearly don't depend on V.
    https://en.wikipedia.org/wiki/Killing_form
    https://www.encyclopediaofmath.org/index.php/Killing_form

    To diagonalize ##G_{ij}## (for the case it is not already diagonal), the diagonalization matrix should act in the vector space in which ##G_{ij}## are components of a tensor, i.e. the diagonalization matrix should itself have the ##ij## components. It seems that the author of the book failed to distinguish different vector spaces, which is a mistake similar to that in
    https://www.physicsforums.com/threads/do-we-really-mean-hermitian-conjugate-here.858987/
     
    Last edited: Feb 24, 2016
  4. Feb 24, 2016 #3
    To conclude, i send an email to Maggiore himself, he said that the matrix V act directly on Gij, that seems logic but the sentence in the book is confusing because he speak of VTiV* implying that V act on the générators.

    Thanks for the answer.
     
  5. Feb 25, 2016 #4

    Demystifier

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    At the very least, I think he would need to rewrite this (small and inessential) part of the book.
     
  6. Feb 25, 2016 #5

    vanhees71

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    I'd say that's one of the most essential parts of any book on QT, because Lie algebras are at the heart of all QT :-).
     
  7. Feb 25, 2016 #6

    Demystifier

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    Then why books on non-relativistic QM (which is also a part of quantum theory) rarely mention Lie algebras? :wink:
    I'm sure every branch of theoretical physics can be expressed in terms of Lie algebras, but I think they are really essential only in Yang-Mills gauge theories.
     
  8. Feb 25, 2016 #7

    vanhees71

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    That speaks against the books. Already angular-momentum algebra is a (non-)abelian Lie algebra. Also, how do you motivate the commutation relations of the observables if not via the Lie algebra of the Galilei group? I think, you can not overstate the importance of Lie algebras and Lie groups in QT!
     
  9. Feb 25, 2016 #8

    Demystifier

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    Ask Heisenberg! :wink:
     
  10. Feb 25, 2016 #9

    Demystifier

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    I certainly can't, but you can. :biggrin:
     
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