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?


1 Answer 1


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).

  • $\begingroup$ Thanks for your answer. For $tr(F\wedge F)=dCS(A)\neq 0$ we obtain a non-invariant Lagrangian under $SU(3)_c$ but how can one fix this issue? $\endgroup$
    – Alpha001
    Nov 8, 2017 at 13:47
  • $\begingroup$ We do not fix that. We consider the large gauge transformations as symmetries of the theory rather than gauge redundancies in the description of the system. (The group of large gauge transformations is discrete thus it does not reduce the infinite dimensional group of local gauge transformations by much). $\endgroup$ Nov 8, 2017 at 14:20
  • $\begingroup$ cont. Moreover, by adding the $\theta$ term we superficially break the large gauge symmetry because the Lagrangian is not invariant. However, since the equations of motion are invariant, this is a quasi-symmetry and after quantization, we can find a Hilbert space with an invariant action of the large gauge transformation. This is the Hilbert space built over the theta vacuum. $\endgroup$ Nov 8, 2017 at 14:20

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