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I am asked to derive the quadratic form of the Dirac equation in an electromagnetic field,

$\left[\left(i\hbar \partial - \frac{e}{c}A\right)^2 - \frac{\hbar e}{2c} \sigma^{\mu\nu} F_{\mu\nu} - m^2c^2\right] \psi = 0,$

where $F_{\mu\nu} =\partial_{\nu} A_{\mu} - \partial_{\mu} A_{\nu}$ is the usual e-m field tensor and $\sigma^{\mu\nu} = \frac{i}{2}[\gamma^{\mu}, \gamma^{\nu}].$

A hint in my textbook suggests that I left-multiply the Dirac equation,

$\left[-\gamma^{\mu} \left( i \hbar \partial_{\mu} - \frac{e}{c} A_{\mu} \right) + mc\right] \psi = 0,$

by the expression $\gamma^{\nu} \left( i \hbar \partial_{\nu} - \frac{e}{c} A_{\nu} \right) +mc$ and use the commutation relations for the gamma matrices. I have been unsuccessful to this end. I get

0 = $\left[-\gamma^{\nu} \gamma^{\mu} \left( i \hbar \partial_{\nu} - \frac{e}{c} A_{\nu} \right) \left( i \hbar \partial_{\mu} - \frac{e}{c} A_{\mu} \right) + m^2 c^2 \right] \psi = - \left[( \gamma^{\mu} \gamma^{\nu} + 2 i \sigma^{\mu\nu}) \left( i \hbar \partial_{\nu} - \frac{e}{c} A_{\nu} \right) \left( i \hbar \partial_{\mu} - \frac{e}{c} A_{\mu} \right) - m^2 c^2 \right] \psi = -\left[ -\left(i\hbar \partial - \frac{e}{c}A\right)^2 + 2 i \sigma^{\mu\nu} \left( i \hbar \partial_{\nu} - \frac{e}{c} A_{\nu} \right) \left( i \hbar \partial_{\mu} - \frac{e}{c} A_{\mu} \right) - m^2 c^2 \right] \psi, $

where in the last step I have used $(\gamma^{\mu})^2=-\mathbb{1}$.

From here I have tried expanding the expression $\left( i \hbar \partial_{\nu} - \frac{e}{c} A_{\nu} \right) \left( i \hbar \partial_{\mu} - \frac{e}{c} A_{\mu} \right)$ and rewriting the appropriate cross-term in terms of $F_{\mu\nu}$ but I don't know what to do with the other terms. I can't seem to extract the term $\frac{\hbar e}{2c} \sigma^{\mu\nu} F_{\mu\nu}$. Also, the sign of the $\left(i\hbar \partial - \frac{e}{c}A\right)^2$ term appears to be backwards. Can anyone help me with this?

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    $\begingroup$ What do you know about the symmetry of $\sigma^{\mu\nu}$? What does this tell you about which of those terms in your expansion should vanish? $\endgroup$ – Jerry Schirmer Dec 2 '13 at 4:00
  • $\begingroup$ Well certainly $\sigma^{\mu\nu}$ should be symmetric but I'm not sure how that helps. Also, I looked more closely at my work and the sign of the $\left(i \hbar \partial - \frac{e}{c} A \right)^2$ term appears to come out to be opposite what I'd expect. $\endgroup$ – mikefallopian Dec 2 '13 at 5:38
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    $\begingroup$ Re-examine your "well-certainly..." assertion; recall that $\sigma^{\mu\nu}$ is proportional to the commutator of $\gamma$ matrices. $\endgroup$ – joshphysics Dec 2 '13 at 5:43
  • $\begingroup$ Oh of course, I see what you're getting at. $\sigma^{\mu \nu}=-\sigma^{\nu \mu}$, so the diagonal terms of the expansion vanish $\endgroup$ – mikefallopian Dec 2 '13 at 5:54
  • $\begingroup$ so after a bit of algebra I get $\sigma^{\mu\nu} \left( i \hbar \partial_{\nu} - \frac{e}{c} A_{\nu} \right) \left( i \hbar \partial_{\mu} - \frac{e}{c} A_{\mu} \right) = \frac{i \hbar e}{2c} \sigma^{\mu \nu}F_{\mu \nu}$, which leads me to be off by a factor of 2 in the end. Also, the sign of the first term is still wrong... $\endgroup$ – mikefallopian Dec 2 '13 at 6:12
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You may write $\gamma^{\nu} \gamma^{\mu} = \frac{1}{2}(\gamma^{\nu} \gamma^{\mu} + \gamma^{\mu} \gamma^{\nu}) + \frac{1}{2}(\gamma^{\nu} \gamma^{\mu} - \gamma^{\mu} \gamma^{\nu}) \\= g^{\nu\mu} - i \sigma^{\nu\mu} \\= g^{\mu\nu} + i \sigma^{\mu\nu}$

Remember that $g^{\nu\mu}V_\mu V_\nu = g^{\mu\nu}V_\mu V_\nu = V^2$

Remember also that $F_{\mu\nu} =\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}$ (you made a sign inversion in your question)

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