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A question about SU(2)

  1. Nov 14, 2007 #1
    Any rotation about n(θ, φ) in SU(2) can be represent as
    u(n, ω) = I cos[ω/2] - i(σ.n)sin[ω/2],where I is the unit matrix and i is the complex number.Right?

    But can someone tell me why ω/2 rather than ω?

    Waiting for your response.Thank you.
  2. jcsd
  3. Nov 14, 2007 #2

    D H

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    A unit quaternion [itex]Q[/itex] rotates or transforms a 3-vector [itex]\vec x[/itex] via

    [tex]\bmatrix 0 \\ \vec x'\endbmatrix =

    Q \bmatrix 0 \\ \vec x\endbmatrix Q^*[/tex]

    where I have used [tex]\bmatrix 0 \\ \vec x\endbmatrix[/tex] to denote the pure quaternion (think pure imaginary) constructed from the 3-vector [itex]\vec x[/itex].

    If you work through the math you will see why the factor of 1/2 is needed.

    Note well: Swapping the quaternion and its conjugate also lead to a rotation or transformation. Both forms are in use.
  4. Nov 14, 2007 #3

    Chris Hillman

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    Spinorial double-covering

    Spinorial representations involve double coverings. Can you visualize a Moebius strip as a rectangle wrapped up with a twist so that it now has a single circular edge which is twice as long as the central circle in the band? Draw the vertical projections to see that two points on the edge correspond to each point at the center.

    This is easier to discuss with pictures!
  5. Nov 14, 2007 #4
    Another easy question:Does the Moebius strip pertain to the SU(2) group??
  6. Nov 15, 2007 #5

    Chris Hillman

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    That's what I just said! Consider the covering by SU(2) of SO(3). This is a higher dimensional analog of the covering of the circle by the circle which is given by the Moebius band (with a circular edge). See for example the chapter on spinors in MTW, Gravitation.
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