Consider the Lagrangian (renormalised + counterterm) of QED:

$$\mathcal{L} = -\frac{1}{4} F_{\mu \nu}F^{\mu \nu} - \frac{1}{2 \xi}(\partial_{\mu} A^{\mu})^2 + \bar{\psi}(i \displaystyle{\not} D - m)\psi -\frac{\delta Z_A}{4} F_{\mu \nu}F^{\mu \nu} - \frac{\delta Z_A}{2 \xi}(\partial_{\mu} A^{\mu})^2 + i\delta Z_{\psi} \bar{\psi}\displaystyle{\not} \partial \psi + iZ_1e\bar{\psi}\displaystyle{\not} A \psi -i\delta Z_2 m\bar{\psi}\psi.\tag{1}$$

where the $Z_1$ counter-term includes the counter-terms coming from the electron charge, and the field renormalisations of the photon anf the fermion. Gauge invariance implies $Z_1 = Z_{\psi}$ so when renormalising the coupling, we only need the field strength renormalisation of the photon $\delta Z_A$. However, this argument apparently doesn't work in Yang Mills theory as an explicit calculation reveals they are not equal:

$$ \delta Z_{\psi}= -\frac{g^2}{8\pi^2} \frac{1}{\epsilon} C_F$$ and $$\delta Z_1= -\frac{g^2}{8\pi^2} \frac{1}{\epsilon} (C_F + C_A)$$

where $C_F$ is the Casimir of the fermion and $C_A$ is the casimir of the gluons. Wouldn't this break gauge invariance? I can see that the difference comes due to the self interaction of gluons but I don't see how it resolves my issue as under a gauge transformation, the coupling term between the fermion and the gluon shouldn't change under. Does the answer have to do with BRST symmetry? Thanks for the help in advance.

  • 1
    $\begingroup$ An observation about why the guage field strength renormalization does not have to be gauge independent in a non-Abelian theory: In the Abelian (QED) theory, the charge current is a gauge-invariant observable on its own, so the $e$ that appears in in must have a gauge-invariant value. However, the non-Abelian current is not an observable, because it carries a group index and thus transforms under gauge transformations, so it is not necessary that the renormalized $g$ appearing in the current has to be defined in a gauge-invariant way. $\endgroup$
    – Buzz
    Commented Apr 10, 2023 at 2:43
  • $\begingroup$ Thank you, that’s very helpful! $\endgroup$
    – emir sezik
    Commented Apr 10, 2023 at 9:41

2 Answers 2


To renormalize, we rescale $$ A \to \sqrt{Z_3} A , \qquad \psi \to \sqrt{Z_2} \psi , \qquad c \to \sqrt{Z_{3c}} c , \qquad g \to Z_g g $$ We then define $$ Z_1 \equiv \sqrt{Z_3} Z_g Z_2 , \qquad Z_{1c} \equiv \sqrt{Z_3} Z_g Z_{3c} \quad \implies \quad \frac{Z_1}{Z_2} = \frac{Z_{1c}}{Z_{3c}} . \tag{1} $$ Now, in QED, the gauge group is Abelian so the ghost field completely decouples from the theory and the RHS of the above is simply 1 and we get $Z_1 = Z_2$.

This decoupling does NOT generally happen in non-Abelian gauge theories so the general formula shown above actually holds. Indeed, using the explicit formulas from Schwartz's QFT book (section 26.5.3) \begin{align} \delta_1 &= \frac{1}{\epsilon} \frac{g^2}{16\pi^2}[ - 2 C_F - 2 C_A + 2 ( 1 - \xi) C_F + \frac{1}{2} ( 1 - \xi ) C_A ] + {\cal O}(g^3) , \\ \delta_2 &= \frac{1}{\epsilon} \frac{g^2}{16\pi^2} [ - 2 C_F + 2 ( 1 - \xi ) C_F ]+ {\cal O}(g^3) , \\ \delta_{1c} &= \frac{1}{\epsilon} \frac{g^2}{16\pi^2} [ - C_A + (1-\xi) C_A ]+ {\cal O}(g^3) , \\ \delta_{3c} &= \frac{1}{\epsilon} \frac{g^2}{16\pi^2}[ C_A + \frac{1}{2} ( 1 - \xi ) C_A ] + {\cal O}(g^3) \end{align} We can then easily check that $$ \delta_1 - \delta_2 = \delta_{1c} - \delta_{3c} + {\cal O}(g^3) . $$ which is the linearized version of (1).

This was of course evaluated in $R_\xi$-gauge. If we chose to work in axial gauge where the ghost decouples (just like in QED), then we would find $Z_1 = Z_2$.

  1. It is true that gauge invariance in QED implies the Ward–Takahashi identity, which in turn implies $Z_1=Z_{\psi}$, but it cannot be directly read off from the action (1) as OP seems to suggest.

  2. In contrast, the non-abelian Yang-Mills case is governed by the Slavnov-Taylor identities.


  1. M. Srednicki, QFT, 2007; chapters 68 + 73. A prepublication draft PDF file is available here.

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