6
$\begingroup$

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.

$\endgroup$
2
  • 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

7
$\begingroup$

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 $Z_1 = Z_2$.

$\endgroup$
5
$\begingroup$
  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.

References:

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.