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I have a problem with the $\theta$-term in the QCD-Lagrangian

$$ \mathcal{L} = \overline{q}(x)i\gamma_\mu{D^\mu} q(x)-\overline{q}(x)\mathcal{M}q-\frac{1}{4}F^a_{\mu\nu}F_a^{\mu\nu}+ \theta \frac{g^2}{32 \pi^2} F_{\mu \nu}^a \tilde{F}^{\mu \nu}_a$$

I guess the last term is gauge invariante by itselfe since it is proportional to the trace:

$$tr(F^{\mu\nu}\tilde{F}_{\mu\nu})=2\epsilon_{\mu\nu\rho\sigma}tr[(\partial^\mu A^\nu+igA^\mu A^\nu)(\partial^\rho A^\sigma +igA^\rho A^\sigma)] $$

But on the other hand the $\theta$-term is topological non-trivial and the integral

$$\int d^4x \ \ tr[F^{\mu\nu}\tilde{F}_{\mu\nu}] =: \int d^4x \ \ \partial_\mu K^\mu $$ is proportional to $n \in Z$ (the winding number) and not invariant under large gauge transformations.

Which step is wrong? And if the $\theta$-term is not gauge invariant how can we fix this problem in the Lagranian?

Alpha001
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2 Answers2

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The problem is that $K^{\mu}$ is not a true vector on the manifold. It depends on the gauge potential $A$ as well as the gauge field $F$. It is alright to use the Stokes theorem for these types of objects locally on coordinate pathches but not globally on the whole manifold.

Suppose that the integration is over $S^4$. We divide the integration into two parts, on the upper hemisphere and on the lower hemisphere, touching at the equator $S^3$. We know that on each hemisphere (I am using a coordinate free notation) $$\mathrm{tr}(F\wedge F) = d\mathrm{CS}(A)$$ Where $CS(A)$ is the Chern Simons class. Using the Stokes theorem $$\int_{S^4} \mathrm{tr}(F\wedge F) = \int_{D_{up}^4} d\mathrm{CS}(A_{up}) + \int_{D_{down}^4} d\mathrm{CS}( A_{down}) = \int_{S^3} \mathrm{CS}( A_{up}) - \mathrm{CS}( A_{ down })$$ The local gauge potentials $ A_{up}$ and $ A_{down}$ are connected by a transit gauge transformation $$A_{up} = gA_{down}g^{-1}+gdg^{-1}$$
It is known that the Chern-Simons classes are not gauge invariant, they change by a Wess-Zumino-Witten form under gauge transformation:

$$\mathrm{CS}( gA_{down}g^{-1}+ddg^{-1})- \mathrm{CS}( A_{ down }) = \mathrm{WZW}(g)$$ Thus we obtain: $$\int_{S^4} \mathrm{tr}(F\wedge F) = \int_{S^3}\mathrm{WZW}(g)$$ For nontrivial configurations like instantons, the transition gauge transformations are large gauge transformations, for which the Wess-Zumino-Witten is nonvanishing (It is the winding number of the transition group configurations).

Nikita
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David Bar Moshe
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It's exactly analogous to the following question. Consider a 2d manifold $M$ with $\Sigma = \partial M$ as its boundary. Consider the 2-form $F=dA$, and

$$ \frac{\theta}{2\pi} \int_M F = \frac{\theta}{2\pi} \int_M dA = \frac{\theta}{2\pi} \int_\Sigma A.$$

Under a gauge transformation, we have $A \to A + g^{-1}d g$, where $g\in \mathrm U(1)$. Of course, the leftmost term is gauge invariant. However, if you stare at the last term, it is not gauge invariant. If I choose $g$ such that the map $\Sigma\to g$ to have a nonzero winding number $n\in \Pi_1(\mathrm U(1))=\mathbb{Z}$, then it is a large gauge transformation, and $$ \frac{\theta}{2\pi}\int_\Sigma A \to \frac{\theta}{2\pi}\int_\Sigma A + n \theta. $$

How do we reconcile this? The answer is the large gauge transformation is only defined on $\Sigma$. You will have trouble embed it in $M$ without singularity. (The extension $\tilde g$ of a large gauge transformation $g$ onto $M$ has vortices.) This is consistent with the fact that no gauge transformation on $M$ can change the value of $\int_M F$.

To summarize, any gauge transformation that can alter the boundary action is not a legal gauge transformation in the bulk. This should also apply to your initial setup.

pathintegral
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