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Imaginary eigenvalues of gamma matrices

  1. Sep 13, 2012 #1
    Hi! I'm reading David Tong's notes on QFT and I'm now reading on the chapter on the dirac equation


    and I stumbled across a statement where he claims that

    [tex] (\gamma^0)^2 = 1 \ \ \Rightarrow \text{real eigenvalues}[/tex]


    [tex] (\gamma^i)^2 = -1 \ \ \Rightarrow \text{imaginary eigenvalues}.[/tex]

    I'm a bit rusty on my linear algebra and just wondered why this is necessarily true. Why does the square of a matrix being positive and negative respectively mean real and imaginary eigenvalues?
    Last edited: Sep 13, 2012
  2. jcsd
  3. Sep 13, 2012 #2
    Suppose x is an eigenvector of the matrix A with eigenvalue a, so that Ax = ax. Then A^2 x = a^2 x.

    Suppose A^2 x = x for all x (including the eigenvectors of A). Then in view of the above, all eigenvalues of A satisfy a^2 = 1. So the possible eigenvalues are +1 and -1.

    Now suppose A^2 x = -x for all x. Then all eigenvalues of A satisfy a^2 = -1. So the possible eigenvalues are +i and -i.
  4. Sep 13, 2012 #3
    Ah thanks a lot! :)
  5. Sep 13, 2012 #4

    Maybe you could help me with the next thing I'm stuck on? :)

    After obtaining that

    [tex] (S^{\mu \nu})^\dagger = \frac{1}4 [\gamma^\mu, \gamma^\nu]^\dagger = - \gamma^0 S^{\mu \nu} \gamma^0[/tex]

    D. Tong claims that

    [tex] \exp ( \frac{1}{2} \Omega_{\mu \nu} (S^{\mu \nu})^\dagger ) = \gamma^0 \exp( -\frac{1}{2} \Omega_{\mu \nu} S^{\mu \nu} ) \gamma^0.[/tex]

    I tried obtaining this by doing an exponential expansion and factorizing out the gammas, plus the property that
    [tex](\gamma^0)^2 = 1[/tex]

    but I get problems with the even terms in the expansion.
    Last edited: Sep 13, 2012
  6. Sep 13, 2012 #5
    What problems? Could you write out what you think the expansions of the left and right-hand sides of the equations are?
  7. Sep 13, 2012 #6

    [tex] \exp(\gamma^0 (-\frac{1}2 \Omega_{\mu \nu} S^{\mu \nu}) \gamma^0)
    = 1 + \gamma^0 A \gamma^0 + \frac{1}{2!} (\gamma^0)^2 A^2 (\gamma^0)^2 + \frac{1}{3!}(\gamma^0)^3 A^3 (\gamma^0)^3 + \ldots = \gamma^0 ( 1 + A + \frac{1}{2!} \gamma^0 A \gamma^0 + \frac{1}{3!} A^3 + \ldots )\gamma^0[/tex]

    where i have defined

    [tex]A = -\frac{1}2 \Omega_{\mu \nu} S^{\mu \nu}[/tex]
  8. Sep 13, 2012 #7
    OK, two things:

    1) Your expansion of the LHS isn't right because gamma-0 and A don't necessarily commute. If A and B are not-necessarily commuting matrices or operators, then

    exp(AB) = 1 + AB + (1/2!)ABAB + (1/3!)ABABAB + ...

    (recall that, for instance, (AB)^2 = (AB)*(AB)). But you seem to be assuming the incorrect form

    exp(AB) = 1 + AB + (1/2!)AABB + (1/3!)AAABBB + ...

    which is only equivalent to the correct form if A and B commute (because (AB)(AB) != AABB unless A and B commute). (Of course, you need to apply the equivalent statement for the exponential of the product of three matrices, not just two).

    2) Why are there some gamma-0's inside the parentheses in your expansion of the RHS?
  9. Sep 13, 2012 #8
    Ah, thank you again. The problem lied in the wrong expansion :) The gamma-0's were there due to this wrong expansion. When i tried to factor out gamma-0's to the left and right of the sum gamma-0's remained at the even terms.
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