# Identity in Schwinger's paper

1. Sep 12, 2010

### maverick280857

Hi,

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

$$-(\gamma\pi)^2 = \pi_{\mu}^2 - \frac{1}{2}e\sigma_{\mu\nu}F^{\mu\nu}$$

where

$$\pi_{\mu} = p_{\mu} - eA_{\mu}$$

$$F^{\mu\nu} = \partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}$$

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

(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

$$\frac{1}{2}\{\gamma_{\mu},\gamma_{\nu}\} = -\delta_{\mu\nu}$$

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

Returning to the identity, I know that

$$\gamma^{\mu}a_{\mu}\gamma^{\nu}b_{\nu} = a\cdot b - i a_{\mu}\sigma^{\mu\nu}b_{\nu}$$

(\slashed doesn't work)

In particular, setting $a = b = \prod$, this becomes

$$(\gamma \pi)^2 = \pi^2 - e\sigma^{\mu\nu}(\partial_{\mu}A_{\nu} + A_{\nu}\partial_{\mu})$$

Questions:

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.