I am working with the covariant derivative and trying to show that the commutator of this derivative $[D_\mu , D_\nu]$ is proportional to the field $F_{\mu \nu}$. That is, I need the final term to be have $(\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu})$ contained in the answer (proportional to the field $F_{\mu\nu}$. Instead the commutator appears to be zero! When I work it all out I get

$\left[ { D }_{ \mu },{ D }_{ \nu } \right] =\left( { \partial }_{ \mu }-iq{ A }_{ \mu } \right) \left( { \partial }_{ \nu }-iq{ A }_{ \nu } \right) -\left( { \partial }_{ \nu }-iq{ A }_{ \nu } \right) \left( { \partial }_{ \mu }-iq{ A }_{ \mu } \right) ={ \partial }_{ \mu }{ \partial }_{ \nu }-{ q }^{ 2 }{ A }_{ \mu }{ A }_{ \nu }-iq\left( { \partial }_{ \mu }{ A }_{ \mu }+{ \partial }_{ \nu }{ A }_{ \mu } \right) -{ \partial }_{ \nu }{ \partial }_{ \mu }+{ q }^{ 2 }{ A }_{ \nu }{ A }_{ \mu }+iq\left( { \partial }_{ \nu }{ A }_{ \mu }+{ \partial }_{ \mu }{ A }_{ \mu } \right) =0$

I have checked this over and over but I cannot get it. The answer is in the book but I'm starting to think maybe the book is wrong? Other books show it is not, I am definitely missing something. If someone can point me to what I'm missing please?

  • 1
    $\begingroup$ You’re forgetting the (implied) thing that the covariant derivatives are taking the derivative of. You can’t take the derivative of nothing. $\endgroup$
    – G. Smith
    Nov 7, 2020 at 6:04
  • $\begingroup$ What do you mean? $\endgroup$
    – user276724
    Nov 7, 2020 at 6:08
  • $\begingroup$ This is a homework-like question, so you should not expect a complete answer that works it all out for you. Put a field, say a scalar $\phi$, after the commutator. Let each partial derivative operate on everything to the right of it. $\endgroup$
    – G. Smith
    Nov 7, 2020 at 6:13
  • $\begingroup$ i will try it, it i a homeowork question but i cant get it. any help appreciated. $\endgroup$
    – user276724
    Nov 7, 2020 at 6:15

1 Answer 1


You have assumed many terms in that expansion commute where they don't. Anyway, whenever you calculate any commutator, you should always check what happens if it operates on a function (field) or state etc. That is, calculate $[\mathrm{D}_\mu , \mathrm{D}_\nu] \psi$ for a field $\psi$. So

$[D_\mu , D_\nu] \psi = [\mathrm{D}_{\mu} \mathrm{D}_{\nu} - \mathrm{D}_{\nu} \mathrm{D}_{\mu}]\psi $

$ =(\partial_{\mu} -\mathrm{i} A_{\mu}) (\partial_{\nu} -\mathrm{i} A_{\nu}) \psi - (\partial_{\nu} -\mathrm{i} A_{\nu}) (\partial_{\mu} -\mathrm{i} A_{\mu}) \psi $

$= (\partial_{\mu} \partial_{\nu} -\mathrm{i} q \partial_{\mu} A_{\nu} -\mathrm{i} q A_{\nu} \partial_{\mu} - \mathrm{i} q A_{\mu} \partial_{\nu} - q^2A_{\mu} A_{\nu}) \psi$ $\large-$ [Same with $\mu$ and $\nu$ interchanged]$\psi$

Then you should get

$$[\mathrm{D}_{\mu},\mathrm{D}_{\nu}]\psi=-\mathrm{i} q (\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}) \psi=-\mathrm{i} q F_{\mu \nu} \psi$$

or in other words

$$F_{\mu \nu}=\frac{\mathrm{i}}{q} [\mathrm{D}_{\mu},\mathrm{D}_{\nu}]$$

  • $\begingroup$ thankx! that is correct answer! do all commutations on a state first. Got it! $\endgroup$
    – user276724
    Nov 7, 2020 at 6:18

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