Dirac equation with an Abelian symmetry can be written as $$(\gamma^{\mu}D_{\mu} - m)\psi = 0$$ where $$D_{\mu}\psi = (\partial_{\mu} - iqA_{\mu})\psi$$ Then how do we get this second order equation $$(D_{\mu}D^{\mu} - \frac{1}{2}iq\gamma^{\mu \nu}F_{\mu \nu} - m^2)\psi = 0\,\,?$$ Also here $[D_{\mu}, D_{\nu}]=-iqF_{\mu \nu}$
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$\begingroup$ Act on the first equation with $\gamma^\nu D_\nu + m$ $\endgroup$– PraharCommented Dec 12, 2014 at 14:46
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$\begingroup$ That is what I am doing, but I cannot get the 1/2 factor in front of the field strength. $\endgroup$– MarionCommented Dec 12, 2014 at 14:52
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Act on the first equation with $\gamma^\nu D_\nu + m$. We find \begin{align} (\gamma^\mu D_\mu + m )( \gamma^\nu D_\nu - m ) \psi &= (\gamma^\mu \gamma^\nu D_\mu D_\nu - m^2 ) \psi \\ &= (\frac{1}{2} \left\{ \gamma^\mu ,\gamma^\nu \right\} D_\mu D_\nu + \frac{1}{4} [ \gamma^\mu , \gamma^\nu ] [ D_\mu , D_\nu] - m^2 ) \psi \\ &= (D_\mu D^\mu - \frac{ i q }{4}[ \gamma^\mu , \gamma^\nu ] F_{\mu\nu} - m^2 ) \psi \\ \end{align} Assuming you are defining $$ \gamma^{\mu\nu} := \frac{1}{2} [ \gamma^\mu , \gamma^\nu ] $$ we are done!
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$\begingroup$ Alright thanks. I could not make the second line as simple a you have so I have to understand this bit. $\endgroup$– MarionCommented Dec 12, 2014 at 15:35
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$\begingroup$ I write $\gamma^\mu \gamma^\nu = \frac{1}{2} \{ \gamma^\mu , \gamma^\nu \} + \frac{1}{2} [ \gamma^\mu , \gamma^\nu ]$. For the first part, I use the Clifford algebra. The second part is contracted with $D_\mu D_\nu$. However, since $\frac{1}{2} [ \gamma^\mu , \gamma^\nu ]$ is anti-symmetric, I can replace $D_\mu D_\nu$ with $\frac{1}{2} [ D_\mu , D_\nu ]$. $\endgroup$– PraharCommented Dec 12, 2014 at 15:36