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Regarding the Clifford algebra and spinors

  1. Feb 15, 2013 #1
    Hello! I´m currently taking a course in RQM and have some questions for which I didnt get any satisfactory answers on the lecture. All comments are appricieted!

    1. Is the gamma zero tensor some kind of metric in the space for spinors? When normalizing our solution to the Dirac equation it seems as we use them exactly as our metric in SR.

    2. As I understood it the gamma matricies are Lorentz invariant tensors. However when taking the norm of a spinor describing a particle moving in lets say the z-direction we get a answer that depends on the energy (only one component of our 4-momentum). In other words it seems as if our normalization is not invariant under Lorentz transformations and therefore not gamma 0?

    Thanks in advance!
     
  2. jcsd
  3. Feb 15, 2013 #2

    Bill_K

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    For a Lorentz 4-vector, x·x is not invariant. To get an invariant you have to insert a metric, x·η·x where η = (-1, 1, 1, 1).

    Likewise for spinors, ψψ is not an invariant, you must insert a 4x4 matrix that acts like a metric, and consider instead ψ†ηψ. Now if a Dirac spinor transforms like ψ → Lψ under a Lorentz transformation, then ψηψ → ψLηLψ, and so the condition that this quantity is a Lorentz scalar is LηL = η. The solution is η = γ0.
     
  4. Feb 15, 2013 #3
    Yes! This is what I mean when I say that it seems as we use the gamma 0 tensor as a metric. However ψγ0ψ = 1 - (p_z/E+m)2 for a particle moving in the z direction. Is the RHS really a Lorentz scalar?

    Is it mathematically correct to actually call γ0 our metric in the space of spinors? Or is this just a smiliarity of how we use an metric?
     
  5. Feb 16, 2013 #4
    Surely, if the electron is moving along the z axis then [itex]p_z = p[/itex] and then
    [tex]1 - (\frac{p}{(E + m)})^2 = \frac{E^2 + 2Em + m^2 - p^2}{(E + m)^2} = \frac{2m}{(E + m)}[/tex]
    I'm no expert, but this doesn't look invariant to me.
     
  6. Feb 16, 2013 #5

    Bill_K

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    The correctly normalized plane wave solution has an additional factor of ((E + m)/2m)½ in front. E.g. Bjorken and Drell vol I gives the plane wave solution as

    ψ(pz) = ((E + m)/2m)½(pz/(E + m), 0, 1, 0)

    Multiply your answer of 2m/(E + m) by this additional factor squared, and you get 1.
     
  7. Feb 16, 2013 #6
    for free particle spinors,the convention is
    u-u=2mS[itex]\dagger[/itex]S,where S is two component spinor normalized by
    S[itex]\dagger[/itex]S=1.
     
  8. Feb 16, 2013 #7
    Thanks Bill
     
  9. Feb 16, 2013 #8
    Yeah I agree with you, but dont you see the problem I´m pointing at? :)

    1) Our metric (γ0) defines an inner product which gives us a norm, |ψ|2 = ψηψ.
    2)When I calculate the norm of a spinor it is dependent of its energy.
    3) Energy is not a Lorentz scalar so the norm is not invariant under Lorentz transformations.

    So it seems to me that our metric is not Lorentz invariant. But I assume that there is some misunderstanding in my logic..
     
  10. Feb 16, 2013 #9

    Bill_K

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    No sorry, if that's the case, you're using improperly normalized wavefunctions! See above:
    The norm of this wavefunction is ψψ = 1, as it should be. The plane wave solution is derived by taking a solution at rest and Lorentz transforming it, so its norm is guaranteed to be Lorentz invariant, by construction.

    The full set of wavefunctions for given p consist of two positive energy solutions ur(p) and two negative energy solutions vr(p) where r = ±1 is the spin coordinate. The normalization is:
    ur(p)us(p) = δrs
    vr(p)vs(p) = - δrs
    ur(p)vs(p) = vr(p)us(p) = 0
     
  11. Feb 17, 2013 #10
    Yeah, Im with you now. Dont know why but I regareded your additional normalization factor as a dimensionless constant which only normalizes for one inertial frame. :)
     
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