# Derivation of the quadratic form of the Dirac equation

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?

• 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? Dec 2, 2013 at 4:00
• 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. Dec 2, 2013 at 5:38
• Re-examine your "well-certainly..." assertion; recall that $\sigma^{\mu\nu}$ is proportional to the commutator of $\gamma$ matrices. Dec 2, 2013 at 5:43
• Oh of course, I see what you're getting at. $\sigma^{\mu \nu}=-\sigma^{\nu \mu}$, so the diagonal terms of the expansion vanish Dec 2, 2013 at 5:54
• 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... Dec 2, 2013 at 6:12

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)

I had the same problem and after many hours i figured out the right solution that provide the correct quadratic Dirac equation:

$$\lbrace \left( i\gamma^\mu\partial_\mu-e\gamma^\mu A_\mu\right)\left( i\gamma^\nu\partial_\nu-e\gamma^\mu A_\nu\right)-m^2 \rbrace \psi =0$$

$$=\gamma^\mu\gamma^\nu \left( i\partial_\mu-e A_\mu\right)\left( i\partial_\nu-e A_\nu\right)\psi-m^2 \psi =$$

$$=\gamma^\mu\gamma^\nu \left( i\partial_\mu-e A_\mu\right)\left( i\partial_\nu-e A_\nu\right) \psi -m^2\psi=$$

$$=\left(g^{\mu\nu}-i\sigma^{\mu\nu}\right) \left( i\partial_\mu-e A_\mu\right)\left( i\partial_\nu-e A_\nu\right) \psi -m^2\psi=$$

$$=g^{\mu\nu} \left( i\partial_\mu-e A_\mu\right)\left( i\partial_\nu-e A_\nu\right)-i\sigma^{\mu\nu}\left( i\partial_\mu-e A_\mu\right)\left( i\partial_\nu-e A_\nu\right) \psi -m^2\psi=$$

$$= \left( i\partial^\nu-e A^\nu\right)\left( i\partial_\nu-e A_\nu\right)-i\sigma^{\mu\nu}\left( i\partial_\mu-e A_\mu\right)\left( i\partial_\nu-e A_\nu\right) \psi -m^2\psi=$$ $$=\lbrace\left( i\partial-e A\right)^2+\frac{1}{i}\sigma^{\mu\nu}\left( i\partial_\mu-e A_\mu\right)\left( i\partial_\nu-e A_\nu\right)-m^2\rbrace\psi=$$ Now we write the same equation after index exchanging and apply the property $$\sigma^{\mu\nu}=\frac{i}{2}[\gamma^\mu,\gamma^\nu]=\frac{-i}{2}[\gamma^\nu,\gamma^\mu]=-\sigma^{\nu\mu}$$ $$=\lbrace\left( i\partial-e A\right)^2+\frac{1}{i}\sigma^{\nu\mu}\left( i\partial_\nu-e A_\nu\right)\left( i\partial_\mu-e A_\mu\right)-m^2\rbrace\psi=$$ The equations above are the same of the half of the sum of the two added, the one with the original indices and the other with the switched indices and so appears the commutator and all the numeric factors are correct. $$=\lbrace\left( i\partial-e A\right)^2+\frac{1}{2i}\sigma^{\mu\nu}\left[ i\partial_\mu-e A_\mu, i\partial_\nu-e A_\nu\right]-m^2\rbrace\psi=0$$ Evaluating the commutator we get that is exactly the same as the electromagnetic tensor write in a covariant form, thus the final result come up $$=\lbrace\left( i\partial-e A\right)^2 + \frac{1}{2i} \sigma^{\mu\nu}F_{\mu \nu}-m^2\rbrace\psi=0$$