In particle physics, CP stands for Charge+Parity or Charge-conjugation Parity symmetry: the combination of charge conjugation symmetry (C) and parity symmetry (P). According to the current mathematical formulation of quantum chromodynamics, a violation of CP-symmetry in strong interactions could occur. However, no violation of the CP-symmetry has ever been seen in any experiment involving only the strong interaction. As there is no known reason in QCD for it to necessarily be conserved, 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).
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 current experimental upper bound is roughly one billionth that size.
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 parity-violating chiral currents, the gluons of QCD couple to vector currents. The absence of any observed violation of CP-symmetry is a problem because 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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