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Spinors, vectors and quaternions

  1. Aug 13, 2010 #1
    I am interested in using hypercomplex numbers and not using tensors.

    Therefore a question about the difference between spinors and vectors.
    I read that they both can be written as quaternions.

    Vq = ix + jy + kz

    Sq = ix + jy + kz

    So what is the difference between them?
    Can anyone tell me?

  2. jcsd
  3. Aug 13, 2010 #2


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    Science Advisor

    Can you give some references or link where you read this, for the people who are not so familiar with this stuff but would like to help? :)

    I'm not into quaternions, but the short answer to your question is that spinors are representations of SU(2) and vectors are representations of SO(3), where SU(2) is the double cover of SO(3).
  4. Aug 13, 2010 #3
    I think it goes something like this. If [itex]\sigma_x , \sigma_y , \sigma_z[/itex] are the Pauli matrices, then [itex]\sigma_0 , i\sigma_x , i\sigma_y , i\sigma_z[/itex] act like the unit quaternions, where [itex]\sigma_0[/itex] is the 2x2 identity matrix and [itex]i=\sqrt{-1}[/itex]. A vector v can be written [itex]v = v_x\sigma_x + v_y\sigma_y + v_z\sigma_z[/itex] and a spinor can be written [itex]\psi = \alpha\sigma_0 + \beta i\sigma_x + \gamma i\sigma_y + \delta i\sigma_z[/itex] where [itex]\alpha, \beta, \gamma, \delta[/itex] are all real. You get back to the idea of spinors as 2-component complex vectors by writing
    [tex]|\psi\rangle = \left(\begin{array}{c}
    \alpha +i\delta \\
    -\gamma + i\beta\end{array}\right)[/tex]

    The set [itex]\sigma_0 , \sigma_x , \sigma_y , \sigma_z , i\sigma_x , i\sigma_y , i\sigma_z , i\sigma_0[/itex] forms a basis for the Clifford algebra.

    This is in 3-d, no idea about 4-d.

    At least I think some of that is right...
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