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Gamma matrices

  1. Sep 30, 2011 #1
    I've just started a masters in physics after a 4-year break and am having some real trouble getting back into the swing of things! We have been asked to prove some properties of gamma matrices, namely:

    1. [itex]\gamma^{\mu+}=\gamma^{0}\gamma^{\mu}\gamma^{0}[/itex]

    2. that the matrices have eigenvalues +/- 1, +/- i

    3. The trace of [itex]\gamma^{\mu}[/itex] is zero

    4. if [itex]\gamma_{5} = -i\gamma^{0}\gamma^{1}\gamma^{2}\gamma^{3}[/itex] then [itex]\gamma_{5},\gamma^{\mu}[/itex]= 0, [itex]\gamma^{2}_{5}[/itex]=I, eigenvalues = +/- 1


    For #1 we are to use the Clifford algebra. We have not been given the definitions of the the gamma matrices -- I don't know if we are expected to know these or if they are irrelevant for the proof. We are also given that gamma 0 is equal to its conjugate transpose and gamma i (i = 1,2,3) is equal to its conjugate transpose times -1. I don't even know where to start on this one -- not quite clear on how the multiplication of the matrices works.

    #2 I can do if I take each matrix individually, but how do it do it for the 'general' case of [itex]\gamma^{\mu}[/itex]? By [itex]\gamma^{\mu}[/itex] does it mean I need to take all gammas at once as a set/group/4-vector (how can it be a vector if its components are matricies?), or does it mean for a general gamma mu, where mu = 0,1,2,3?

    #3 Same as above, fine if I take each matrix individually, but how to do it generally?

    # 4 I can do the eigenvalues and the gamma-five-squared = identity, but I'm not sure about the commutator? Again, do I use some 'general' gamma mu?


    I have never had to use tensors before so that whole area is still not quite clear to me -- I understand the concept but not really how tensor operations work.
     
  2. jcsd
  3. Sep 30, 2011 #2

    phyzguy

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    You will need to use the defining property of the gamma matrices, namely:
    [tex]\gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = 2 \eta^{\mu\nu}[/tex]
    Most of these should follow from this. For example, this tells you that
    [tex] (\gamma^0)^2 = 1, (\gamma^i)^2 = -1[/tex]
    What can you conclude about the eigenvalues from this?
     
  4. Oct 3, 2011 #3
    As they are unitary matrices, it means the eigenvalues are +/- i, +/- 1. Hooray!

    I am still stuck on how to show 1.

    Using the defining property I can generalize, ([itex]\gamma^{\mu}[/itex])[itex]^{2}[/itex]=[itex]\eta^{\mu\mu}[/itex]. I am given the conjugate transposes for 0 and i, putting those together I get

    [itex]\gamma^{\mu+}[/itex]=-[itex]\eta^{\mu\mu}\gamma^{\mu}[/itex]

    I can substitute ([itex]\gamma^{\mu}[/itex])[itex]^{2}[/itex] for [itex]\eta^{\mu\mu}[/itex]. For -1 I can substitute the square of gamma zero, giving

    [itex]\gamma^{\mu+}[/itex]=[itex]\gamma^{\mu}\gamma^{\mu}\gamma^{0}\gamma^{0}\gamma^{\mu}[/itex]

    How do I continue from here? I am stuck on how to rearrange this because the matrices are not commutative. I tried substituting a rearranged defining property for [itex]\gamma^{0}\gamma^{\mu}[/itex] but it didn't seem helpful.
     
  5. Oct 3, 2011 #4

    vela

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    Use the anticommutation relation to switch the order of [itex]\gamma^0[/itex] and [itex]\gamma^\mu[/itex] in [itex]\gamma^0\gamma^\mu\gamma^0[/itex]. It doesn't matter which [itex]\gamma^0[/itex] you use.
     
  6. Oct 5, 2011 #5
    Not sure how to do that. Everything I try seems to just cancel back out, e.g.

    [itex]\gamma^{0}\gamma^{\mu}\gamma^{0}[/itex] = [itex]\gamma^{0}[/itex](2[itex]\eta^{\mu 0}[/itex]-[itex]\gamma^{0}\gamma^{\mu}[/itex])

    =[itex]\gamma^{0}[/itex]2[itex]\eta^{\mu 0}[/itex]-([itex]\gamma^{0}[/itex])[itex]^{2}\gamma^{\mu}[/itex]

    = [itex]\gamma^{0}[/itex]([itex]\gamma^{\mu}\gamma^{0}+\gamma^{0}\gamma^{\mu}[/itex])+[itex]\gamma^{\mu}[/itex]

    =[itex]\gamma^{0}\gamma^{\mu}\gamma^{0}[/itex]+[itex]\gamma^{0}[/itex])[itex]^{2}\gamma^{\mu}[/itex]+[itex]\gamma^{\mu}[/itex]

    =[itex]\gamma^{0}\gamma^{\mu}\gamma^{0}[/itex]

    How does this help me?
     
  7. Oct 5, 2011 #6

    vela

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    What does the second line evaluate to when [itex]\mu=0[/itex] and when [itex]\mu=i[/itex]?
     
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