Let the covariant derivative be $$D_\mu = \partial_\mu + \text ig\ A_\mu^a t^a,\quad a=1,\ldots,8$$where $g$ is the bare QCD coupling, $A_\mu^a$ are the eight gluon fields and $t^a=\tfrac{1}{2}\lambda^a$ are the generators of $SU(3)$. I am trying to figure out how the covariant derivative acts differently on quarks and gluons.

For quarks, the generators will act in their fundamental representation, as $3\times 3$ matrices proportional to the Gell-Mann matrices that act on a quark with three color components $i$: $$\begin{align}D_{\mu,ij} q_j &= \partial_\mu \delta_{ij} q_j + \text ig\ A_\mu^a (t^a)_{ij}\ q_j \\&=\partial_\mu q_i + \text ig\ A_\mu^a (t^a)_{ij}\ q_j \end{align}$$

For gluons, the generators should act via their adjoint representation, since the gluons themselves are part of the algebra (they are built up by the $t^a$). The adjoint representation is given by the $SU(3)$ structure constants, $(t^a)_{bc} = -\text i\ f_{abc}$. A generator $t^a$ in the adjoint representation acts on another element of the Lie algebra $t^b$ as $[t^a,t^b]$. $$ \begin{align}D_{\mu,ab}\ A_\nu &= D_{\mu,ab}\ A_\nu^c t^c\\&= \partial_\mu \delta_{ab} A_\nu^c t^c + \text ig\ A_\mu^dA_\nu^c\ [t^d,t^c]_{ab}\quad ?\end{align}$$

But then if I want to calculate the gluon field strength tensor, I don't get the correct result: $$ \begin{align} \frac{1}{\text ig}[D_\mu,D_\nu]X &= \frac{1}{\text ig}(\partial_\mu + \text ig A_\mu)(\partial_\nu+\text ig A_\nu)X-[\mu\leftrightarrow\nu]\\ &= \frac{1}{\text ig}\partial_\mu \partial_\nu X+\partial_\mu(A_\nu X)+ A_\mu \partial_\nu X+ \text ig A_\mu (A_\nu X)-[\mu\leftrightarrow\nu]\\ &= \frac{1}{\text ig}\partial_\mu\partial_\nu X + \partial_\mu A_\nu X + A_\nu \partial_\mu X+A_\mu \partial_\nu X\color{red}{+\text ig X [A_\mu,A_\nu]}+\text ig A_\mu A_\nu X\\ &\quad -\frac{1}{\text ig}\partial_\nu\partial_\mu X - \partial_\nu A_\mu X - A_\mu \partial_\nu X-A_\nu \partial_\mu X\color{red}{-\text ig X [A_\nu,A_\mu]}-\text ig A_\nu A_\mu X\\ &= (\partial_\mu A_\nu)X-(\partial_\nu A_\mu)X+\color{red}3\text ig [A_\mu,A_\nu]X \end{align} $$

The factor 3 should be 1. Terms in red should not be there (in case someone finds this question), thanks to @Toffomat!

How does the covariant derivative act on a gluon field $A_\mu = A_\mu^a\ t^a$? Where is the mistake in my calculations?


1 Answer 1


Indeed $\left[D_\mu,D_\nu\right]$ is not the same as $D_\mu A_\nu-D_\nu A_\mu$. In the end, this is because $D_\mu A_\nu$ is not a well-defined thing: $D_\mu$ maps covariant objects to covariant objects (in the same representation), and the gauge potential is not covariant (because of the inhomogeneous piece in the transformation).

What you need to do to compte the field strength is to consider the commutator acting on some arbitrary object $X$, $$\left[D_\mu,D_\nu\right] X\,.$$ When you expand that out in derivatives and gauge fields, you see that the result has no derivatives acting on $X$ (both the $\partial \partial X$ and the $\partial A\partial X$ contributions cancel), and you are left with something multiplying $X$, and that's (proportional to) the field strength.

Update for the updated question: Your mistake in the last equation is in the third equality sign. You expand (from second to third line) $$\text{i} g A_\mu\left(A_\nu X\right)\to\text{i}gX\left[A_\mu,A_\nu\right] + \text{i} g A_\mu X A_\nu\,,$$ whereas there really is not anything to expand, and the orders don't match either (if you e.g. take $\mu=\nu$). (Note also that you erroneously change the order on the second term, where $X\partial_\mu A_\nu$ should be $\partial_\mu A_\nu X$.)

Maybe what's confusing you is the "action by commutator". That's true in the sense that the structure constants basically form the adjoint representation, but it is really more instructive to take actual matrix representations. Then it becomes clear that when $X$ is in some representation (think column vector), $A_\nu X$ is in the same representation (think matrix times vector) , and $A_\mu A_\nu X$ is again in the same representation (matrix$\times$matrix$\times$vector), and no commutators are required.

  • $\begingroup$ I've edited my question to be more precise thanks to your input. Please take another look! $\endgroup$
    – ersbygre1
    Sep 28, 2018 at 5:02
  • $\begingroup$ That makes sense, thank you. So when does the "act as a commutator" come into play? $\endgroup$
    – ersbygre1
    Sep 28, 2018 at 11:16
  • 2
    $\begingroup$ The most direct way for that is that for e.g. $SU(N)$ you can build all representations from the fundamental, i.e. the $N$-component $\phi_i$ and its conjugate $\psi^j$. The adjoint would have one upper and one lower index, e.g. $\Sigma_i^{\phantom{i}j}$, and the group elements act from both sides ($\Sigma\to U\Sigma U^{-1}$), which translates to the commutator of generators. ($\Sigma$ is also traceless -- this is usually written as $\boldsymbol{N}\otimes\boldsymbol{\bar{N}}=\mathbf{adj}\oplus\boldsymbol{1}$.) But you can also choose the adjoint generators as matrices $T^a\sim f^{abc}$... $\endgroup$
    – Toffomat
    Sep 28, 2018 at 11:40
  • 1
    $\begingroup$ .... while taking care of upper/lower indices (Killing form and all that). Then the $T^a$ automatically have the right dimension, and the commutation relations are guaranteed by the Jacobi identity. $\endgroup$
    – Toffomat
    Sep 28, 2018 at 11:42

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