16
$\begingroup$

For example, Kaku's QFT pp. 214-215:

Massive vector theory with non-Abelian group is non-renormalizable.

Massive vector Abelian theory is renormalizable.

I heard about the following arguments, but I don't find them satisfactory.

  1. Someone will say the propagator of massive vector field is like $$ \frac{g_{\mu \nu} - k_{\mu} k_{\nu}/m^2}{k^2-m^2}. $$ In large $k$, it will not decay like $1/k^2$, so the power counting law breaks down. Certainly I admit that power counting law is violated, but why does a violation of power-counting have a relation with renormalizablity? And we already know that a massive $U(1)$ gauge field is still renormalizable even though it violates the power-counting.

  2. Someone will say mass term $m^2 \operatorname{tr} A^\mu A_\mu$ will break gauge invariance. But why is gauge invariance important? Guage invariance is not a symmetry, and it's nothing but a redundancy to describe the true physical degrees of freedom. Any theory without gauge invariance can be rewritten by the Stückelberg trick as a gauge theory which describes the same physics.

    For example, the massive Maxwell field $$\mathcal{L}= -\frac{1}{4}F^{\mu\nu}F_{\mu\nu}+\frac{1}{2}m^2 A^\mu A_\mu \tag{1}$$ under the replacement, $$A_\mu\rightarrow A_\mu +\partial_\mu \phi \tag{2}$$ becomes $$\mathcal{L}= -\frac{1}{4}F^{\mu\nu}F_{\mu\nu}+\frac{1}{2}m^2 (A_\mu +\partial_\mu \phi)^2 \tag{3}$$

    Now $(3)$ has local gauge invariance, $$\delta A_\mu =\partial _\mu \Lambda,\quad \delta \phi = -\Lambda \tag{4}$$

    It's obvious that $(3)$ with local gauge invariance $(4)$ describes the same theory $(1)$.

    Rescaling $\phi\rightarrow \frac{1}{m}\phi$, $$\mathcal{L}= -\frac{1}{4}F^{\mu\nu}F_{\mu\nu}+\frac{1}{2}m^2 A^\mu A_\mu + \frac{1}{2}\partial_\mu \phi \partial ^\mu \phi +m A_\mu \partial^\mu \phi\tag{5}$$ with local gauge invariance, $$\delta A_\mu = \partial_\mu \Lambda,\quad \delta \phi = -m \Lambda \tag{6}$$

    By the same way, any theory without gauge invariance, like massive non-abelian gauge field, can be rewritten as a gauge theory. So why is there a relation between gauge invariance and renormalizablity?

So it seems that the above two handwaving arguments are untenable. In general how to prove that massive abelian gauge theory is renormalizable but massive non-abelian gauge theory non-renormalizable?

Improve this question
$\endgroup$
11
  • 5
    $\begingroup$ Gauge invariance does not imply renormalizability. Think e.g. of the lagrangian $F_{\mu\nu}F^{\mu\nu}+[F_{\mu\nu}F^{\mu\nu}]^2+\ldots$ which is perfectly gauge invariant and not renormalizable. Gauge invariance is convenient only as bookkeeping, to highlight the relevant degrees of freedom (including those missing, such as the Higgs boson), and to write efficiently EFT with separation between masses and cutoff (possibly infinite separation of scales in the case of renormalizable theory). $\endgroup$
    – TwoBs
    Dec 16, 2017 at 23:38
  • 1
    $\begingroup$ I am not the specialist in HEP but if you are asking about relation of renormlizability and power counting it goes as follows. Renormalization group flow have to be closed in some finite dimensional subspace of particular kind, depending on mathematical structure of the theory. Every term in power series usually have physically meaning constant in front of it. Every such constant should be controlled by renormlization equations. If there is infinite number of such terms, theory still may be renormalizable, if terms depends on each other, or on finite number of physical parameters. $\endgroup$
    – kakaz
    Dec 17, 2017 at 9:08
  • 1
    $\begingroup$ Cont. Here, I suppose, it is not true, every term in nonabelian gauge is in important way different from another, so renormalization group flow occur in infinite dimensional space, and there are regions which always attract some terms to infinite values, making impossible to cancel it out. This is something I would expect, but of course what I was writing was only a handwaving :) $\endgroup$
    – kakaz
    Dec 17, 2017 at 9:09
  • 2
    $\begingroup$ @kakaz I strongly doubt that gauge invariance alone can make the theory renormalizable. Take the example I was suggesting above, $F^{\mu\nu}F_{\mu\nu}+a (F^{\mu\nu}F_{\mu\nu})^2+\ldots$. This is gauge invariant and arises, e.g., from integrating out a charged state of mass $m$ and charge $e$, so that $a\sim 1/(16\pi^2) e^4/m^4$. This theory is dominated at low-energy by the $F^4$ term, the others becoming important at $E\sim m 4\pi/e$. For $e\ll 1$, the theory is thus missing completely the presence of new states (the integrated-out charges), that give $o(1)$ correction to the amplitudes. $\endgroup$
    – TwoBs
    Dec 17, 2017 at 12:01
  • 1
    $\begingroup$ @coconut The $F^2$ term contributes zero to the scattering amplitude, being the kinetic term. The first non-zero contribution comes from $F^4$ which controls at low energy the non-trivial dynamics. In particular the amplitude at low energy goes like $e^4/(16\pi^2)\times (E/m)^4 $. The higher $F^n$ terms are suppressed by more powers of $(e E/m)^{n-4}$. $\endgroup$
    – TwoBs
    Feb 5, 2018 at 19:45

1 Answer 1

Reset to default
11
+100
$\begingroup$

Certainly I admit that power counting law is violated, but why does violation of power-counting have relation with renormalizablity?

As per the Dyson-Weinberg power-counting theorem (see Ref.1, chapter 12 and Ref.2, chapter 8-1), a diagram is convergent if and only if each of its subdiagrams has a negative superficial degree of divergence $\omega$. The latter is defined as \begin{equation} \begin{aligned} \omega&=\phantom{-}\left(\text{# of factors of momentum in the numerator}\right)\\ &\,\phantom{-}-\,\left(\text{# of factors of momentum in the denominator}\right)\\ &\phantom{-}\,+d\cdot \left(\text{# of independent momentum variables}\right) \end{aligned} \end{equation} where $d$ is the number of spacetime dimensions. Therefore, the number of powers of momentum in the numerator is an essential ingredient in the analysis of potentially divergent diagrams in a certain theory. This number contributes with a plus sign to $\omega$, so the higher the former the higher the latter, and there are more diagrams that are superficially divergent. In fact, it is possible to argue that the number of divergent diagrams is finite if and only if all the interactions are power-counting renormalisable. This is why power-counting is a key part of renormalisability.

And we already know that a massive $U(1)$ gauge field is still renormalizable even though it violates the power-counting.

This is, to some extent, a happy coincidence. Recall that each vertex carries a factor of the current ($J^\mu\sim e\bar\psi\gamma^\mu\psi$). In (massive or massless) QED the current is conserved, and therefore the factors of $k^\mu$ in the numerators don't contribute to scattering amplitudes. Formally speaking, the propagator behaves as $\mathcal O(k^{-2})$ for high momentum (instead of being $\mathcal O(1)$, as one may expect). Therefore, even if massive QED violates power-counting renormalisability, the theory is in fact renormalisable.

In non-abelian gauge theories the current is not conserved (but covariantly conserved instead). Therefore, the factors of $k^\mu$ do contribute and the general analysis of power-counting renormalsiability applies: the propagator is $\mathcal O(1)$ for high momentum, and the theory is non-renormalisable. Only in the massless case do the longitudinal modes decouple, and therefore the propagator effectively behaves as $\mathcal O(k^{-2})$. Renormalisability is thus recovered (and only because the ghosts cancel the longitudinal contribution; without them, the longitudinal part contributes as well).

So why is there relation between gauge invariance and renormalizablity?

This is a matter of semantics. Any divergence, in any local theory, can be eliminated by introducing counter-terms. When the counter-terms have the same form as the original Lagrangian, we say that the theory is renormalisable. Therefore, if the initial theory is gauge invariant, then the theory will be renormalisable only if the counter-terms are gauge-invariant, by definition. If you happen to need a non-gauge-invariant counter-term, that means you need a counter-term that was not initially in the Lagrangian, and the theory is non-renormalisable.

In the case of Yang-Mills, one can prove that the counter-terms are indeed gauge-invariant, and they have the same form as the terms originally present in the initial Lagrangian (see Ref.2, chapter 12-4 and Ref.3, chapter 23). Therefore, the theory is renormalisable. The proof is rather non-trivial and is best understood in the context of Batalin-Vilkovisky (based on the early work of Zinn-Justin). In the case of naïve Quantum Gravity, one can prove that the counter-terms are gauge invariant as well but they are not of the same form as the original Lagrangian (cf. this PSE post). Therefore, the theory is non-renormalisable. Finally, in a gauge-theory where anomalies are present, the counter-terms are not gauge-invariant and the theory is non-renormalisable.

In general how to prove that massive abelian gauge field is renormalizable but massive non-abelian gauge field nonrenormalizable?

You can find an explicit proof of the renormalisability of massive QED in Ref.2, chapter 8-4. You can find the discussion of the non-renormalisability of massive non-abelian gauge theories in Ref.2, chapter 12-5-2 (the essential points are summarised in this PSE post).

References.

  1. Weinberg's QFT, Vol.1.

  2. Itzykson & Zuber's QFT.

  3. DeWitt's The Global Approach to QFT, Vol.1.

Improve this answer
$\endgroup$
1
  • $\begingroup$ To fix: the number of divergent diagrams is not (necessarily) finite. Rather, the number of external-leg-structures is finite. $\endgroup$
    – AccidentalFourierTransform
    May 27, 2018 at 15:03

Your Answer

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

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