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Under the ordinary Pauli -Villars Regularisation one introduces a heavy mass ($\Lambda$) term $$\frac{1}{p^2-m^2+i\epsilon} \rightarrow \frac{1}{p^2-m^2+i\epsilon} - \frac{1}{p^2-\Lambda^2+i\epsilon}.$$

Now Kugo in his book "Gauge theories" says that the Pauli-Villars Regularisation term violates gauge invariance for non-Abelian case.

For the usual propagator term before regularisation one could argue that the gauge invariance is manifestly preserved since the propagator in its form is derived directly from the gauge-invariant Lagrangian.

The regularisation process doesn't take place on the Lagrangian level, so the gauge invariance can not be proved by simple gauge transformations of the fields. What is the argument for the gauge invariance violation of the $\Lambda$ - term. How can one "see" this? Or similarly, how can one see the gauge invariance of the regularisation in the abelian case?

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  • $\begingroup$ Looks like mathematics to me and mathematics are best suited for the math SE apparently. $\endgroup$ – user122066 Jul 10 '16 at 19:45
  • $\begingroup$ Our latest meta discussion concluded with the outcome that posts which show sufficient physical context, even though they may be asking an essentially mathematical question, are on topic here. I think this meets that description. $\endgroup$ – David Z Jul 11 '16 at 8:27
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In abelian case, the photon propagator under Pauli- Villars regularization goes as $$\frac{1}{k^2}\rightarrow\frac{1}{k^2}- \frac{1}{k^2-{\Lambda^2}}$$ which is basically $1/(k^2- \Lambda^{-2}k^4)$. The presence of the extra term in the propagator can be obtained by supplementing the Lagrangian with an extra term $$\frac{1}{2\Lambda^2}\Box A_\mu\Box A^\mu$$ which using Lorenz gauge can be converted into $$\frac{1}{2\Lambda^2}\partial^\nu F_{\nu\mu}\partial_\sigma F^{\sigma\mu}$$ as $$\partial_\sigma F^{\sigma\mu}=\partial_\sigma(\partial^\sigma A^\mu-\partial^\mu A^\sigma)$$ and using Lorenz gauge second term can be dropped. Now the expression $\frac{1}{2\Lambda^2}\partial^\nu F_{\nu\mu}\partial_\sigma F^{\sigma\mu}$ is U(1) invariant as $F^{\nu\mu}$ is U(1) invariant so Pauli -Villars regularization preserves gauge invariance in abelian case.

In Yang -Mills case, a term like $\frac{1}{2\Lambda^2}\Box A_\mu\Box A^\mu$ violates gauge invariance due to presence of simple derivative instead of covariant derivatives. It is not possible to convert this term by any gauge transformation to a form which has an explicit gauge $covariance$. Gauge invariance can only be obtained by modifying the Lagrangian appropriately for Yang-Mills case which does not follow from Pauli- Villars procedure.

The requirement of covariant derivative in non-Abelian case for gauge invariance comes form the fact that in non-Abelian case the Yang-Mills field strength transform homogeneously under gauge transformation as $F\rightarrow UFU^{-1}$ as opposed to the Abelian case where it is gauge invariant. Covariant derivative has the same transformation law as the field under gauge transformation as $D\rightarrow UDU^{-1}$, so the combination $D_\mu F_{\mu\nu}$ will transform homogeneously which will not be the case if we use a partial derivative instead of covariant derivative. Under a trace term in the Lagrangian which includes a couple of terms like this one can use cyclic property of the trace to show that the term (like the one mentioned below) is unaffected by gauge transformation.

A requirement of gauge invariant term in the Lagrangian will involve higher covariant derivative term like $tr(\frac{1}{\Lambda^2}D_\alpha F_{\mu\nu}D_\alpha F_{\mu\nu})$ .

This modified form of Pauli- Villars regularization was used by A. A. Slavnov in Nucl. Phys. B, 31, 301 (1971).

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  • $\begingroup$ Could you elaborate more on the point " simple partial derivative are appearing which are not covariant" or provide some link? $\endgroup$ – Statics Sep 2 '16 at 15:45
  • $\begingroup$ I have modified the answer. $\endgroup$ – ved Sep 3 '16 at 8:23

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