Accordingly to Yang-Mills theories, after the introduction of a covariant derivative such that

$$D_\mu = \partial_\mu - igA_\mu, \tag1$$

you can built the kinetic term for the gauge potential $A_\mu$ as

$${\cal L}_A = -\frac{1}{2}tr\{F_{\mu \nu}F^{\mu \nu}\},\qquad F_{\mu \nu} = \frac{i}{g}[D_\mu, D_\nu]. \tag2$$

The action of the covariant derivative transforms under a local group tranformation $\Omega$ in the following way:

$$D_\mu\psi \rightarrow \Omega D_\mu\psi. \tag3$$

And the gauge field as

$$A_\mu \rightarrow \Omega A_\mu\Omega^\dagger + \frac{i}{g}\Omega\partial_\mu \Omega^\dagger . \tag4$$

Introducing Eq. (4) in Eq. (1) you get that under the group tranformations, the covariant derivate transforms as,

$$D_\mu \rightarrow \partial_\mu - ig\Omega A_\mu\Omega^\dagger + \Omega\partial_\mu\Omega^\dagger = \Omega D_\mu\Omega^\dagger + \partial_\mu . \tag5$$

Eq. (5) into definition of strength tensor in Eq. (2) gives

$$F_{\mu\nu} \rightarrow \Omega F_{\mu\nu}\Omega^\dagger + \frac{i}{g}(\partial_\mu(\Omega D_\nu\Omega^\dagger) - \partial_\nu(\Omega D_\mu\Omega^\dagger)) . \tag6$$

Taking $\Omega \simeq 1 - g\theta^iT^i$, with $\{T^i\}$ the set of generators of the group, $\theta^i \in \mathbb{R}$ and $f^{ijk}$ the structure constants:

$$F_{\mu\nu} \rightarrow \Omega F_{\mu\nu}\Omega^\dagger + \frac{i}{g}(\partial_\mu A_\nu^k - \partial_\nu A_\mu^k)T^k + ig^2f^{ajk}T^k[\partial_\mu(\theta^jA_\nu^a) - \partial_\nu(\theta^jA_\mu^a)] . \tag7$$

In Particle Physics books, it is said that $$F_{\mu\nu} \rightarrow \Omega F_{\mu\nu}\Omega^\dagger \tag{8},$$ but I don't get that in my calculus (Eq. (7)). What am I doing wrong? This result is important because if $F_{\mu\nu}$ doesn't transforms as books say, the kinetic term in Eq. (2) isn't gauge invariant.

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    $\begingroup$ Your equation (5) looks incredibly strange -- for example, for a trivial gauge transformation it says $D_\mu \to D_\mu + \partial_\mu$. $\endgroup$ – knzhou Jan 26 '19 at 16:59
  • $\begingroup$ Notice that by the very definition of covariant derivative, the second equality in the following transformaton law: $$D_\mu(A)\Psi \longrightarrow D_\mu(A^{\Omega}) \Omega \Psi = \Omega D^\mu (A) \Psi$$ holds. Your Eq. (5) is indeed wrong. $\endgroup$ – pppqqq Jan 26 '19 at 17:05
  • $\begingroup$ @pppqqq I don't understand what you mean. What you've written is the definition of covariant derivative that gives the gauge field transformation law, but how does it prove that my Eq. (5) is wrong? $\endgroup$ – Vicky Jan 26 '19 at 17:08
  • $\begingroup$ Actually you get the gauge field transformation law writting $D_\mu(A^\Omega) \Omega\psi = \Omega D_\mu\psi$ with $D_\mu(A^\Omega) = \partial_\mu - igA_\mu^\Omega$, so once you know $A_\mu^\Omega$ (Eq. (4)) you know the transformation law for covariant derivative, Eq(5) $\endgroup$ – Vicky Jan 26 '19 at 17:13
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    $\begingroup$ The issue is that you're freely changing what the derivative acts on. Sometimes, you take $\partial_\mu$ to only act on the right immediately to its right. Other times you take it to act on everything to the right. To get (5) you swapped the meaning, and in your previous comment you swapped them right back. $\endgroup$ – knzhou Jan 26 '19 at 18:20

Hint: OP's discrepancy seems spurred by inconsistent treatment of the derivative symbols, i.e. how far do the derivatives act to the right?/how many objects do the derivatives act on? See this Phys.SE post for a similar issue. E.g. OP's eq. (5) is right or wrong depending on the precise meaning of the derivatives therein.


Ok, I think I found the answer. Eq. (5) has to be written as:

$$D_\mu \rightarrow \partial_\mu + \Omega(\partial_\mu \Omega^\dagger) - ig\Omega A_\mu\Omega^\dagger \tag{A}$$

As you can deduce from Eq. (4) that is a consequence of imposing $D_\mu\psi \rightarrow \Omega D_\mu\psi$. Nevertheless, you can re-write Eq. (A) as:

$$D_\mu \rightarrow (\Omega\partial_\mu)\Omega^\dagger - ig\Omega A_\mu\Omega^\dagger = \Omega D_\mu\Omega^\dagger \tag{B}$$

After considering,

$$(\Omega\partial_\mu)\Omega^\dagger = \Omega(\partial_\mu \Omega^\dagger) + \Omega\Omega^\dagger\partial_\mu = \Omega(\partial_\mu \Omega^\dagger) + \partial_\mu \tag{C}$$

So with this, everything makes sense and the strength tensor is perfectly gauge-covariant, keeping in mind Eqs. (B)-(C) when interpreting OP's Eq. (8).

  • $\begingroup$ You should mark this as the accepted answer even though you solved it yourself. $\endgroup$ – Natanael Jan 28 '19 at 10:00
  • $\begingroup$ @Natanael There is a mandatory time you have to wait between self-answering and acceptance $\endgroup$ – Vicky Jan 28 '19 at 11:03

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