In my lecture notes for a course on QFT it is said that, also in non-abelian gauge theories, the identity $Z_3 = Z_\xi$ holds, where those renormalization parameters belong respectively to the following terms in a QCD Lagrangian in the Lorenz/$R_\xi$ gauge (repeated indices are summed over and $a$ corresponds to the gauge algebra index): $$\mathcal{L} = -\frac{Z_3}{2}(\partial_\mu A^a_\nu \partial^\mu A^\nu_a - \partial_\mu A^a_\nu \partial^\nu A^\mu_a) -\frac{Z_3}{2 Z_\xi \xi}(\partial_\mu A^\mu_a)(\partial^\nu A^a_\nu) + (\cdots)$$ The identity $Z_3 = Z_\xi$ is not justified further, but it is noted that if that was not the case, then the longitudinal component of the bosonic propagator would receive loop corrections, which is said to be "catastrophical" because the longitudinal degrees of freedom are gauge-dependent.

I understand how $Z_3 = Z_\xi$ ensures the longitudinal components of the propagator do not receive loop corrections, and I see why Ward identity imposes this constraint in QED. Nevertheless, in the present case I have two questions:

  1. What is exactly meant by saying that longitudinal degrees of freedom are gauge-dependent? Certainly a gauge such as the Lorenz one does impose a constraint on the longitudinal component of the field, without conditioning further the transverse components, but could I not find a gauge that constrains the transverse degrees of freedom instead?
  2. How is $Z_3 = Z_\xi$ guaranteed a priori in a non-abelian gauge theory? Is it only an a posteriori requirement to avoid the longitudinal component of the propagator to receive loop correction?

Edit: I am aware of the fact that physical massless bosons must have transverse polarizations, but I don't see how or why this implies gauge transformations would only constraint longitudinal degrees of freedom. I must confess, on the other hand, that any "gauge transformation" that I try to invent to constrain the transverse degrees of freedom happens to be non-linear in $A_\mu$, which I believe it means that this is not a gauge transformation (since any representation provides linear operators, and gauge transformations arise from representations of the corresponding gauge group).

  • $\begingroup$ If you used something other than an $R_\xi$ gauge, $Z_\xi$ wouldn't even be meaningful. $\endgroup$ Commented Aug 23, 2023 at 10:01
  • $\begingroup$ @ConnorBehan, do you mean by that that the statement about longitudinal d.o.f. is dependent on this gauge? Let me note, however, that I have often seen this $\xi$ prefactor be introduced in the Faddeev-Popov procedure, even before the gauge fixing functional was chosen to be $\mathcal{G}^a=\partial_\mu A^{a\mu}$. $\endgroup$
    – Albert
    Commented Aug 23, 2023 at 10:05
  • $\begingroup$ Besides, I have the impression that if the gauge fixing functional didn't depend on any parameter at all, then the propagator would automatically be fully transverse. Conversly, I have also the impression that introducing any parameter in the gauge fixing makes a part of the propagator longitudinal, in which case my question would probably hold with little change. Am I wrong? $\endgroup$
    – Albert
    Commented Aug 23, 2023 at 10:11

1 Answer 1


That's a good question.

  • In general there are 2 kinds of $Z$-factors in renormalization:

    1. A $Z$-factor $Z_{\phi}$ associated with wave function renormalization $\phi_0=Z_{\phi}^{1/2}\phi$.

    2. A $Z$-factor $Z_g$ associated with each coupling constant $g$. (A mass parameter $m$ and gauge parameter $\xi$ are here viewed as a coupling constant.)

  • The bare gauge-fixing term in $R_{\xi}$-gauge is more generally of the form $$ {\cal L}_{gf,0}~=~-\frac{1}{2\xi_0} \chi_0^2 ~=~-\frac{Z_{\xi}}{2\xi} \chi^2, $$ where $\chi$ is a gauge-fixing function.

  • Example: The bare gauge-fixing term in $R_{\xi}$ Lorenz gauge $\chi=\partial\cdot A$ reads $$ {\cal L}_{gf,0}~=~-\frac{1}{2\xi_0} (\partial\cdot A_0)^2 ~=~-\frac{Z_{\xi}}{2\xi} (\partial\cdot A)^2,$$ where $$\frac{1}{\xi_0}~=~\frac{Z_{\xi}}{\xi Z_A}. $$

  • If the gauge-fixing function $\chi$ is non-linear, the gauge-fixing term ${\cal L}_{gf}$ contains interaction terms. The corresponding (possibly generalized) Ward–Takahashi identities (WTI) may acquire additional terms. E.g. in QED, the naive WTI $$k_{\mu}\Pi^{\mu\nu}(k)~=~0$$ for the photon vacuum-polarization/self-energy may be modified by a non-zero RHS [1,2]. In particular, longitudinal and transversal parts may be deformed.

  • In the simplest cases (such as Lorenz gauge in QED and QCD), it is consistent to impose e.g. the renormalization condition [3,4] $$Z_{\xi}~=~1 \qquad\text{or}\qquad\xi_0~=~\xi,$$ but not always.


  1. R. B. Mann, G. McKeon & S.B. Phillips, Longitudinal contributions to the vacuum polarization in the 't Hooft–Veltman gauge, Can. J. Phys. 62 (1984) 1129.

  2. G. McKeon, S.B. Phillips, S.S. Samant & T.N. Sherry, Becchi–Rouet–Stora invariance in the 't Hooft–Veltman gauge, Can. J. Phys. 63 (1985) 1343.

  3. C. Itzykson & J.B. Zuber, QFT, 1985; Subsection 12-3-4, eq. (12-125). Be aware that I&Z use the notation $\lambda\equiv 1/\xi$.

  4. P. Ramond, Field Theory: A Modern Primer, 2nd ed, 1989; eqs. (8.2.55), (8.2.71) & (8.6.8). Be aware that Ramond uses the notation $\alpha\equiv \xi$.

  • $\begingroup$ Are you saying that the statement "longitudinal degrees of freedom are gauge-dependent" holds in the Lorentz gauge but not generally, and that the identity $Z_3 = Z_\xi$ in my post holds in QCD due to the generalize Ward-Takahashi identity? Because if that is the case, I am still quite lost, particularly when trying to see how the generalized W-T identity I know from Srednicki's Quantum field theory (ch. 22) implies that identifying both constants is consistent. $\endgroup$
    – Albert
    Commented Aug 24, 2023 at 15:47
  • $\begingroup$ What is mentionned in Srednicki, in his very symbolic notation, is that that the W-T identity derives from Schwinger-Dyson equation and reads $\partial_\mu \langle 0 | T\{j^\mu(x) \phi_{a_1}(x_1)\cdots\phi_{a_n}(x_n)\}|0\rangle = -i\sum_{j=1}^n \partial_\mu \langle 0 | T\{ \phi_{a_1}(x_1)\cdots\delta\phi_{a_j}(x)\delta(x-x_j)\cdots\phi_{a_n}(x_n)\}|0\rangle$ for a conserved current $j^\mu$ and $\delta \phi$ the variation of the field $\phi$ under the corresponding symmetry, for a collection of fields $\{\phi_j\}$. $\endgroup$
    – Albert
    Commented Aug 24, 2023 at 16:00
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Commented Aug 25, 2023 at 7:23

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