- #1
maverick280857
- 1,789
- 4
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
[tex]-(\gamma\pi)^2 = \pi_{\mu}^2 - \frac{1}{2}e\sigma_{\mu\nu}F^{\mu\nu}[/tex]
where
[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]
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.
Thanks in advance.
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]
where
[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]
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.
Thanks in advance.
Last edited by a moderator: