Strong CP problem
According to quantum chromodynamics there could be a violation of CP symmetry in the strong interactions. However, no violation of the CP-symmetry is known to have occurred in experiments. As there is no known reason for it to be conserved in QCD specifically, this is a "fine tuning" problem known as the strong CP problem.
CP-symmetry states that the laws of physics should be the same if a particle were interchanged with its antiparticle (C symmetry, as charges of antiparticles are the negative of the corresponding particle), and then left and right were swapped (P symmetry). In particle physics, CP violation (CP standing for Charge+Parity) is a violation of the postulated CP-symmetry (or Charge conjugation Parity symmetry): the combination of C-symmetry (charge conjugation symmetry) and P-symmetry (parity symmetry).
How CP can be violated in QCD
QCD does not violate the CP-symmetry as easily as the electroweak theory; unlike the electroweak theory in which the gauge fields couple to chiral currents constructed from the fermionic fields, the gluons couple to vector currents. Experiments do not indicate any CP violation in the QCD sector. For example, a generic CP violation in the strongly interacting sector would create an electric dipole moment of the neutron which would be comparable to 10−18 e·m while the experimental upper bound is roughly one billionth that size.
This is a problem because at the end, there are natural terms in the QCD Lagrangian that are able to break the CP-symmetry.
For a nonzero choice of the θ angle and the chiral quark mass phase θ′ one expects the CP-symmetry to be violated. If the chiral quark mass phase θ′ can be converted to a contribution to the total effective θ angle, it will have to be explained why this effective angle is extremely small instead of being of order one; the particular value of the angle that must be very close to zero (in this case) is an example of a fine-tuning problem in physics. If the phase θ′ is absorbed in the gamma-matrices, one has to explain why θ is small, but it will not be unnatural to set it equal to zero.
If at least one of the quarks of the standard model were massless, θ would become unobservable; i.e. it would vanish from the theory. However, empirical evidence strongly suggests that none of the quarks are massless and so this solution to the strong CP problem fails.
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