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Clebsch-Gordan coefficients symmetry relation

  1. Feb 23, 2016 #1
    Why are ##<j_1j_2m_1m_2|jm>## and ##<j_2j_1m_2m_1|jm>## negative of each other when ##j_1+j_2-j## is odd as given below?
    Screen Shot 2016-02-23 at 8.07.15 pm.png

    I would expect ##<j_1j_2m_1m_2|jm>## and ##<j_2j_1m_2m_1|jm>## to always have the same sign since nature doesn't care which particle we label as particle 1 and which as particle 2.

    Screen Shot 2016-02-23 at 8.06.49 pm.png

    From (6.274), isn't it true that
    ##|j_1j_2m_1m_2>\,=|j_1m_1>|j_2m_2>\,=|j_2m_2>|j_1m_1>\,=|j_2j_1m_2m_1>##?

    And hence
    ##<j_1j_2m_1m_2|=\,<j_2j_1m_2m_1|##?

    And so ##<j_1j_2m_1m_2|jm>## and ##<j_2j_1m_2m_1|jm>## should always have the same sign.

    ##j## is the general angular momentum quantum number and ##m## is the associated magnetic quantum number.
     
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  3. Feb 23, 2016 #2

    Vanadium 50

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    Think of a cross product: a x b is - b x a, even though nature doesn't care which particle we call a and which we call b.
     
  4. Feb 23, 2016 #3
    Why does the order in the direct product matter?

    Screen Shot 2016-02-23 at 8.35.06 pm.png

    From (6.276),
    ##J_z \psi_{j_2 j_1 m_2 m_1}=(m_1+m_2)\hbar\psi_{j_2 j_1 m_2 m_1}##

    It seems like the order doesn't matter. Could you give an example where the negative sign is necessary when the order is flipped around?
     
  5. Feb 23, 2016 #4

    Vanadium 50

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    The important thing the C-G coefficients is their relative sign. That's where the SU(2) algebra is captured. You will need the sign only when you add terms.
     
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