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Identity in Schwinger's paper

  1. Sep 12, 2010 #1

    I'm working my way through Schwinger's paper (http://www.physics.princeton.edu/~mcdonald/examples/QED/schwinger_pr_82_664_51.pdf" [Broken]) and I came across the following identity

    [tex]-(\gamma\pi)^2 = \pi_{\mu}^2 - \frac{1}{2}e\sigma_{\mu\nu}F^{\mu\nu}[/tex]


    [tex]\pi_{\mu} = p_{\mu} - eA_{\mu}[/tex]

    [tex]F^{\mu\nu} = \partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}[/tex]

    [tex]\sigma^{\mu\nu} = \frac{i}{2}[\gamma^\mu,\gamma^\nu][/tex]

    (This is equation 2.33 of the paper, for those of you who refer to the pdf.)

    I am trying to prove this identity, but I ran into some problems. First of all, since his equation 2.4 states

    [tex]\frac{1}{2}\{\gamma_{\mu},\gamma_{\nu}\} = -\delta_{\mu\nu}[/tex]

    I'm guessing his sign convention for the metric is different. Also, shouldn't this be [itex]g_{\mu\nu}[/itex] on the RHS instead of the Kronecker delta?

    Returning to the identity, I know that

    [tex]\gamma^{\mu}a_{\mu}\gamma^{\nu}b_{\nu} = a\cdot b - i a_{\mu}\sigma^{\mu\nu}b_{\nu}[/tex]

    (\slashed doesn't work)

    In particular, setting [itex]a = b = \prod[/itex], this becomes

    [tex](\gamma \pi)^2 = \pi^2 - e\sigma^{\mu\nu}(\partial_{\mu}A_{\nu} + A_{\nu}\partial_{\mu})[/tex]


    1. How does one proceed from here?
    2. I seem to get no minus sign on the LHS. Is that because of Schwinger's metric?

    Any suggestions and inputs would be greatly appreciated. I've been stuck on this step for a few hours now.

    Thanks in advance.
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 13, 2010 #2
    Ideas, suggestions, anyone?
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