In basically every QFT book the Yang-Mills strength tensor $F_{\mu\nu}$ is defined as $$F_{\mu\nu}=[D_\mu,D_\nu]$$ where $D_\mu$ is the covariant derivative $$D_\mu=\partial_\mu-A_\mu$$ and $A_\mu$ is the Yang-Mills gauge field.

Explicitly working out the commutator most books obtain (see Peskin 15.15, Srednicki 69.14) $$F_{\mu\nu}=-\partial_\mu A_\nu +\partial_\nu A_\mu-[A_\mu,A_\nu]$$

However when I work out the commutator I get an extra term $$A_\nu\partial_\mu-A_\mu\partial_\nu$$

This term isn't mentioned in any of the resources I've come across and I don't know what to do with it. Obviously it vanishes somehow. So,

Question; Why does this term vanish?

  • $\begingroup$ If you like this question you may also enjoy reading this Phys.SE post. $\endgroup$
    – Qmechanic
    Jun 1, 2016 at 12:24

2 Answers 2


Note that, for example, \begin{align} [A_\mu,\partial_\nu]f&=A_\mu\partial_\nu f-\partial_\nu(A_\mu f)\\ &=A_\mu\partial_\nu f-\partial_\nu(A_\mu)f-A_\mu\partial_\nu f\\ &=-f\partial_\nu A_\mu\,. \end{align} So you don't get terms like $A_\mu\partial_\nu$.

  • $\begingroup$ This is fine in Peskin's case, where they consider $[D_\mu,D_\nu]$ acting on some other field, but in other texts it's just $[D_\mu,D_\nu]$. Do we always implicitly assume it's acting on some field/function? $\endgroup$
    – Okazaki
    Jun 1, 2016 at 15:44
  • 3
    $\begingroup$ @ryanp16 When you deal with derivatives in commutator, it is always assumed that is is acting on some kind of functions or fields. For example, after eq. (58.11) in Srednicki, he said that "We can also express the field strength in terms of the covariant derivative by noting that $$ [D_\mu,D_\nu]\Psi(x)=−ieF_{\mu\nu}(x)\Psi(x).$$ We can write this more abstractly as $$ F_{\mu\nu}=\frac{i}{e}[D_\mu,D_\nu],$$ where, again, the ordinary derivative in each covariant derivative acts on anything to its right." $\endgroup$
    – Frame
    Jun 1, 2016 at 23:54
  • $\begingroup$ Just to reiterate @Frame's point: always evaluate commutators with a test function, or you will make a mistake. (Nothing good has every happened to me any time I have tried to skip this step.) $\endgroup$
    – Andrew
    Mar 22, 2021 at 22:19

This is purely technical issue \begin{eqnarray} F_{\mu\nu} &=& \big[ D_\mu, D_\nu \big] \;,\\ &=&\big( \partial_\mu-A_\mu \big)\big( \partial_\nu-A_\nu \big)-\big( \partial_\nu-A_\nu \big)\big( \partial_\mu-A_\mu \big)\;,\\ &=& \partial_\mu \partial_\nu -\partial_\mu A_\nu -A_\mu \partial_\nu + A_\mu A_\nu\\ && -\partial_\nu \partial_\mu +\partial_\nu A_\mu +A_\nu \partial_\mu - A_\nu A_\mu\;,\\ &=&\big( \partial_\mu \partial_\nu-\partial_\nu \partial_\mu \big) -\big( \partial_\mu A_\nu -\partial_\nu A_\mu \big) -\big( A_\mu \partial_\nu- A_\nu \partial_\mu \big) +\big( A_\mu A_\nu- A_\nu A_\mu \big)\;,\\ &=&\big( \partial_\mu \partial_\nu-\partial_\nu \partial_\mu - A_\mu \partial_\nu+ A_\nu \partial_\mu \big)-\big( \partial_\mu A_\nu -\partial_\nu A_\mu \big) +\big( A_\mu A_\nu- A_\nu A_\mu \big)\;,\\ &=&\big( D_\mu \partial_\nu-D_\nu \partial_\mu \big)-\big( \partial_\mu A_\nu -\partial_\nu A_\mu \big) +\big( A_\mu A_\nu- A_\nu A_\mu \big)\;,\\ &=& \big(\Gamma_\mu{}^\lambda{}_\nu \partial_\lambda-\Gamma_\nu{}^\beta{}_\mu \partial_\beta \big) -\partial_\mu A_\nu +\partial_\nu A_\mu + \big[A_\mu, A_\nu\big]\;,\\ &=&-\partial_\mu A_\nu +\partial_\nu A_\mu + \big[A_\mu, A_\nu\big]\;. \end{eqnarray} Not sure why the last term is flips the sign.


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