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The Lagrangian one would write down naively for QCD is invariant under CP, which is in agreement with all experiments.

Nevertheless, if we add the term

\begin{equation} \theta \frac{g^2}{32 \pi^2} G_{\mu \nu}^a \tilde{G}^{a, \mu \nu}, \end{equation}

where $\tilde{G}^{a,\mu \nu} = \frac{1}{2} \epsilon^{\mu \nu \rho \sigma} G^a_{ \rho \sigma}$ is the dual field strength tensor to the Lagrangian, QCD isn't CP invariant anymore. This is known as the strong CP problem

  • Why do we need to consider this term in QCD, and why is it never mentioned in weak- or electromagnetic interactions? (In the literature I was only able to find the nebulous statement that this is because of the topological structure of the QCD ground state)
  • This term isn't invariant under parity transformations, so why isn't there a strong P problem, too?
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You are asking four questions, whose answers are routinely provided in textbooks.

  • We consider it in QCD since there is no fundamental reason to exclude it, and topological configurations such as instantons, etc.. might well generate it in an effective low energy theory: the rule of thumb is that anything that is not prohibited has to emerge out of the workings of QFT.

  • In pure EM theory, you can quickly convince yourself it integrates out to a surface term--a total divergence.

  • Electroweak theory already violates P and CP at substantial levels, by dint of chiral couplings and the CKM matrix, so no further ultraweak effects due to such a term would stick out as they do in the strong interactions which lack such. If such effects are (so far) unobservable in QCD, despite sticking out like a sore thumb, they would be resolutely unobservable in the weak interactions, a fortiori. Nevertheless, speculative work, indeed a cottage industry, cf , has entertained the question. Finally, it is not clear the effects of such a topological term cannot be rotated/hidden into the CP-phases of the CKM matrix.

  • Who says there is no P violation in it? The EDM certainly breaks both P and T (~ CP).

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  • $\begingroup$ About your second point. I think even in QCD it is a total divergence. @CosmasZachos $\endgroup$ – SRS Jun 14 '18 at 15:07
  • $\begingroup$ Of course! It is always a topological term. I offered QED to let you see the point quickly. $\endgroup$ – Cosmas Zachos Jun 14 '18 at 15:50

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