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Basic Group Theory Question

  1. Aug 5, 2008 #1
    1. The problem statement, all variables and given/known data


    I'm trying to see the relation of the rotation of a vector in a plane to the generator of rotations...

    I want to see how [tex]e^{-i \theta J}[/tex] the rotation representation gives you the same result as acting on any vector with the rotation matrix say with the z direction fixed.

    [tex]
    \[ \left( \begin{array}{ccc}
    Cos(\theta) & -Sin(\theta) & 0 \\
    Sin(\theta) & Cos(\theta) & 0 \\
    0 & 0 & 1 \end{array} \right)\] = R_z [/tex]

    is [tex] R_z \textbf{v} = e^{-i \theta J_z^{(1)}} \textbf{v} [/tex]

    because a 3 dimensional vector has a spin one representation (right? because one full rotation gives the same vector back)

    with [tex] J_z^{(1)} = \[ \left( \begin{array}{ccc}
    1 & 0 & 0 \\
    0 & 0 & 0 \\
    0 & 0 & -1 \end{array} \right)\] [/tex]

    I get [tex] e^{-i \theta J_z^{(1)}} = \sum\frac{(-i \theta)^n}{n!}(J_z)^n = Cos(\theta)(J_z)^2-i J_z^{(1)} Sin(\theta) [/tex]

    [tex]
    e^{-i \theta J_z^{(1)}}= \[ \left( \begin{array}{ccc}
    Cos(\theta)-i Sin(\theta) & 0 & 0 \\
    0 & 0 & 0 \\
    0 & 0 & Cos(\theta)+i Sin(\theta) \end{array} \right)\] [/tex]

    when this matrix is applied to the vector [tex]\textbf{v}[/tex] it doesnt produce the same effect someone help finish the missing pieces thanks!
     
  2. jcsd
  3. Aug 5, 2008 #2
    ok I used the SO(3) matrix instead of SO(2) and that does give back the same matrix however if i still do it in 3-d i dont get the same matrix back

    [tex]
    R(d\theta) = I - i d\theta J \rightarrow J =
    \[ \left( \begin{array}{ccc}
    0 & -i & 0 \\
    i & 0 & 0 \\
    0 & 0 & 0 \end{array} \right)\]
    [/tex]

    if you expand [tex] e^{-i \theta J} = \[ \left( \begin{array}{ccc}
    Cos(\theta) & -Sin(\theta) & 0 \\
    Sin(\theta) & Cos(\theta) & 0 \\
    0 & 0 & 0 \end{array} \right)\]
    [/tex]
    with out the 1 in the bottom right (3,3)entry whys this?
     
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