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Lie algebra su(2)

  1. Dec 14, 2008 #1
    hi ,i see from a book su(2) has the form
    U[tex]\left(\hat{n},\omega\right)[/tex]=1cos[tex]\frac{\omega}{2}[/tex]-i[tex]\sigma[/tex]sin[tex]\frac{\omega}{2}[/tex]
    in getting the relation with so(3),why we choose [tex]\frac{\omega}{2}[/tex],how about changing for [tex]\omega[/tex]?
    thank you
     
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  3. Dec 14, 2008 #2

    Fredrik

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    The ½ is there because [itex]J_i=\frac 1 2 \sigma_i[/itex] satisfies the commutation relations [itex][J_i,J_j]=i\epsilon_{ijk}J_k[/itex].
     
  4. Dec 14, 2008 #3
    thanks a lot,but i still cannot understand it ,because the book just gives the equation ,where can i find a detailed derivation?
     
  5. Dec 14, 2008 #4

    George Jones

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    Note that U is an element of Lie group SU(2), not an element of the Lie algebra su(2).

    Roughly, there is a factor of 1/2 because of the 2 to 1 relationship between the groups SU(2) and SO(3).
     
  6. Dec 14, 2008 #5

    Fredrik

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    I don't think that's correct, but it's possible that I'm wrong. I think the only point of using SO(3) instead of SU(2) is that it guarantees that we can find an actual representation (with U(R')U(R)=U(R'R) for all R) instead of a projective representation. (If we take R and R' to be members of SO(3), there will sometimes be a minus sign in front of one of the U's).

    I think the 1/2 appears only because a rotation operator is always [itex]1-i\theta^iJ_i[/itex] to first order in the parameters, with the [itex]J_i[/itex] satisfying the usual commutation relations.
     
  7. Dec 14, 2008 #6

    George Jones

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    There is quite a lot of interesting stuff going on here, and I don't have time to tex it right now, but I stand by my statement. Note that what I wrote doesn't negate anything that you wrote; there are often a number of (somewhat equivalent) ways to look at the same thing.

    Maybe in a couple of days I'll write a much longer post.
     
  8. Dec 15, 2008 #7
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