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SO(3),SU(2) connection

  1. Apr 5, 2009 #1
    1. The problem statement, all variables and given/known data
    I'm supposed to make a connection between SU(2) and SO(3), by using
    [tex]U=\frac{a_0+i\vec{\sigma}\vec{a}}{a_0+i\vec{\sigma}\vec{a}}[/tex], where [tex]\sigma[/tex] are the usual 2 dimensional Pauli matrices.
    I need to find the corresponding axis and angle of rotation in three dimensions, in terms of [tex]a_0[/tex] and [tex]\vec{a}[/tex]


    3. The attempt at a solution
    Well I can normalize this and get something like
    [tex]A(1+B \vec{\sigma}\vec{a})[/tex]
    But how can you relate this to a three dim. rotation? It's pretty clear that the axis will be [tex]\vec{a}[/tex], but then the usual generators of SO(3) are 3x3 matrices.. Any hints please?
     
  2. jcsd
  3. Apr 5, 2009 #2
    Just realized it's problem 2 of Sakurai 1985, ch 3. Any thoughts? Do you have to relate each matrix, U_z(a/2) to R_z(a), etc, or is there an easier way?
     
  4. Apr 6, 2009 #3

    Avodyne

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    Science Advisor

    The numerator and denominator in your expression for U are the same, so I can't tell what the starting point is supposed to be.
     
  5. Apr 8, 2009 #4
    Oops, the bottom one was with a minus.
     
  6. Apr 9, 2009 #5
    There is a general formula connecting the Unitaries from SU2 and the orthogonals from SO3. This is:

    [tex]F_{kl}=\frac12 \text{Tr}(U\sigma_l U^{-1}\sigma_k)[/tex]

    So the kl-th component of the SO3 form corresponding two can be expressed with the trace of the k, and l, th pauli matrices.

    Try to derive this formula (its not that hard).

    Now if the axis of rotation is [tex]\vec n[/tex] and the angle of rotation is [tex]\varphi[/tex], then the general form of this orthogonal matrix that is the SO3 form is:

    [tex]F_{kl}=\cos\varphi\delta_{kl}+n_k n_l(1-\cos\varphi)+\sin\varphi\sum_{m=1}^{3}\varepsilon_{kml}n_m[/tex]

    You can deduce this, from a diagram....

    Now putting this together with the first formula you get for the SU2 form:

    [tex]U_{kl}=\cos\left(\frac{\varphi}{2}\right)\cdot \delta_{kl} -i\sin\left(\frac{\varphi}{2}\right)(\vec{n}\sigma)_{kl}[/tex]

    The derivation is quite tedious but the end is so nice... :D
     
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