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Lorentz Transformations and Spinors

  1. Apr 18, 2006 #1
    Hi,

    I have a question about spinors

    If [tex]\Lambda[/tex] is a Lorentz Transformation what is (and how do you show that it is) the spinor representation of the Lorentz group ? I think it has somnething to do with the equivalence transformation [itex]S\dagger{\gamma}S=\Lambda\gamma[/itex]

    But that is just a guess. Does anyone know the answer to this it has eluded me for a while. Also how do we show that it is the spinor representation...

    If someone can shed some light on this I would truely be greatful!

    Thanks
     
  2. jcsd
  3. Apr 19, 2006 #2

    George Jones

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    There is a 2-to-1 homomorphism between SL(2,C), the group of 2x2 complex matrices that have determinant +1, and the group of proper, orthochronous Lorentz transformations, i.e., the restricted Lorentz group. This means that every representation of the restricted Lorentz group is automatically a representation of SL(2,C), but not all representations of SL(2,C) are representations of the restricted Lorentz group.

    The terms "spinor representation" and "spinor" are not always used consistently in the literature.

    Here are a some possibilities for "spinor representation" of the restricted Lorentz group:

    1) any representation of SL(2,C);

    2) any representation of SL(2,C) that is not a representation of the restricted Lorentz group;

    3) any irreducible representation of SL(2,C).

    There are also other possibilities.

    Note that 1) includes all the "standard" tensor and vector representations of the restricted Lorentz group, but 2) and 3) do not.

    The equation that you gave is a way to get from a representation of SL(2,C) to a representation of the restricted Lorentz group. In this equation, S could be an element of SL(2,C), depending on the exact context of the equation. Note that whether S is used, or -S is used make no difference. This is an example of the 2-to-1 homomorphism between SL(2,C) and the group of restricted Lorentz transformations.

    If your equation is in the context of the Dirac equation, then S might not be an element of SL(2,C). In order to accomodate parity, the 4-dimensional space of Dirac spinors the direct sum of representation spaces of 2 inequivalent 2-dimesional representations of SL(2,C).

    Why do we need spinor representations? Why aren't the standard representations of SL(2,C) good enough for physical quantities?

    Because of arbitrary global phase factors in quantum theory. I can give a bit more detail on this if you or anyone else is interested.

    Regards,
    George
     
  4. Apr 20, 2006 #3

    dextercioby

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    I think you were trying to write

    [tex] U \gamma^{\mu} U^{-1} =\gamma' ^{\mu} [/tex]

    And somehow concluded that

    [tex] \gamma' ^{\mu} =\Lambda^{\mu}{}_{\nu} \gamma^{\nu} [/tex] (*)

    and then equaled terms of equalities.

    I must warn you that (*) is wrong. Gamma matrices are invariant under Lorentz transformation. Their elements are numbers x-independent. There's an explanation why they carry flat Greek index (and it is vector/one-form-type, since it can be lowered or raised using the flat metric), but that doesn't mean it automatically becomes a 4-vector/one-form.

    Daniel.
     
  5. Apr 20, 2006 #4

    SpaceTiger

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    Bear with me, I'm not an expert on this stuff. If the U matrices apply Lorentz transformations in spinor space, then wouldn't the relation you give have to be true in order for the gamma matrices to be Lorentz invariant? That is,

    [tex]\gamma^{\mu} -> (\Lambda^{-1})^{\mu}_{~\nu}U\gamma^{\nu}U^{-1}[/tex]

    [tex]=(\Lambda^{-1})^{\mu}_{~\nu}\Lambda^{\nu}_{~\alpha}\gamma^{\alpha}[/tex]

    [tex]=\delta^{\mu}_{~\alpha}\gamma^{\alpha}[/tex]

    [tex]=\gamma^{\mu}[/tex]


    In what ways are they unlike vectors/one-forms?
     
  6. Apr 20, 2006 #5

    CarlB

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    The gamma matrices act like coordinate vectors in the sense of determining directions in space and time. In Clifford algebra, they are called "canonical basis vectors" for CL(3,1) or CL(1,3) depending on your choice of signature. By "basis vector" is meant the same thing as any basis vector for a vector space.

    When you transform an object in the vector space, the basis vectors of the space are not transformed. For example, if an object is at the position (1,2,3) and you negate its coordinates, it will now have a position (-1,-2,-3), however, the [tex]\hat{x}, \hat{y}[/tex] and [tex]\hat{z}[/tex] basis vectors are still the same old (1,0,0), (0,1,0) and (0,0,1).


    Carl
     
  7. Apr 21, 2006 #6

    SpaceTiger

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    Ok, thanks. :smile:
     
  8. Apr 21, 2006 #7
    As far as I'm aware, the following is true.

    [tex]\Lambda^{-1}_{\frac{1}{2}}\gamma^{\mu}\Lambda_{\frac{1}{2}}=\Lambda^{\mu}_{\nu}\gamma^{\nu}[/tex]

    Which is the reason for the raised index, because it behaves like a Lorentz four-vector. ([itex]\Lambda_{\frac{1}{2}}[/itex] is the spinor representation of the Lorentz group.)
     
  9. Apr 21, 2006 #8

    SpaceTiger

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    Right, and couldn't we define our U matrices to boost in the opposite direction, making the relationships equivalent? If what you've quoted is right, then there exists a unitary matrix, U, for which that expression is true.
     
  10. Apr 21, 2006 #9
    Yeah, though I might have the two lambdas the wrong way around in what I mentioned, I was writing it off the top of my head.

    If S is the spinor representation of the Lorentz group, its adjoint isn't its inverse, as the generators of the Lorentz group are not Hermitian. That's why everyone subsequently has written the inverse of S where you've put the adjoint.
     
    Last edited: Apr 21, 2006
  11. Apr 21, 2006 #10

    SpaceTiger

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    Ah yes, I forgot this point. Notating the matrices as "U" implies that they're unitary, which is actually not the case. If we keep the OP's original notation, changing the hermitian conjugate to an inverse and taking the S matrices as [itex]\Lambda_{1/2}[/itex], the expression is correct.
     
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