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Stupid question about Clebsch Gordan

  1. Feb 15, 2008 #1

    malawi_glenn

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    [SOLVED] Stupid question about Clebsch Gordan

    I know that

    [tex] \sum _{m_1,m_2}|<j_1m_1,j_2m_2|jm>|^2 = 1 [/tex]

    But is

    [tex] \sum _{m_1,m_2}|<j_1m_1,j_2-m_2|jm'>|^2 = 1 [/tex] ?

    I have tried to justify this by using a CG calculator and so on, but I just cant figure out why :S

    I think that doesent matter, since you sum over all m_2 and m_1 so that m_1 + m_2 = m'.

    Does anyone have a hint or clue?
     
    Last edited: Feb 15, 2008
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  3. Feb 15, 2008 #2

    George Jones

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    Maybe an example will help. Suppose m_1 = -1, 0 , 1. Write out the two sums over m_1 explicitly.
     
  4. Feb 15, 2008 #3

    pam

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    If you use -m_2, then you can only have m_1-m_2=m'.
    CG are not defined (or are zero) unless m_1+m_2=m.
     
  5. Feb 15, 2008 #4

    malawi_glenn

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    those things I know.

    Geroge Jones:

    [tex] \sum _{a=-1}^1a = \sum _{a=1}^{-1}(-a)
     
  6. Feb 15, 2008 #5

    malawi_glenn

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    those things I know.

    Geroge Jones:

    [tex] \sum _{a=-1}^1a = \sum _{a=1}^{-1}(-a) [/tex]
     
  7. Feb 15, 2008 #6

    kdv

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    As you say, since you are summing over all possible m_2 anyway, it does not matter at all if the label is [itex] - m_2 [/itex] or [itex] m_2 [/itex]. Unless I am missing something...
     
  8. Feb 15, 2008 #7

    malawi_glenn

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    Thats what I am wondering too :)

    I was looking in Greniers book "nuclear models" page 93 and 94, and saw that he must have uses this property. And this is a sub problem for my HW, so therefor I aksed here.
     
  9. Feb 15, 2008 #8

    kdv

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    The two expressions are equal. I was wondering if something had made you doubt it and if I was maybe not understanding the notation.
     
  10. Feb 15, 2008 #9

    malawi_glenn

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    No you have understand my question correct. I will mark this as solved now.
     
  11. Feb 15, 2008 #10

    George Jones

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    This is what I was trying to get at, but I accidentally wrote m_1 instead of m_2.
     
  12. Feb 15, 2008 #11

    pam

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    I'm sorry, but I'm still puzzled. If that isn't a misprint, you can't have
    <m_1,-m_2|m'> if m_1+m_2=m'.
     
  13. Feb 15, 2008 #12

    malawi_glenn

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    no of coruse not, But the other guys understood my question;)
     
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