3
$\begingroup$

While reading Peskin and Schroeder (page 64) I come across this

Although any relativistic field theory must be invariant under the proper orthocronous Lorentz group, it need not be invariant under P, T, or C. What is the status of these symmetry operations in the real world? From experiment, we know that three of the forces of Nature the gravitational, electromagnetic, and strong interactions are symmetric with respect to P, C, and T. The weak interactions violate C and P separately, but preserve CP and T. But certain rare processes (all so far observed involve neutral K mesons) also show CP and T violation. All observations indicate that the combination CPT is a perfect symmetry of Nature.

The question is this. As is stated above all known interactions preserve T, yet it is said that there are processes that violate it. How is this possible? Via which interactions do these processes happen?

$\endgroup$
  • $\begingroup$ Weren't there measurements of CP violating weak decays via mixing? $\endgroup$ – Hydro Guy Jul 22 '14 at 17:11
  • 6
    $\begingroup$ Have you checked the Wikipedia article on CP violation? There's a paper on experimental measurement of T violation here. $\endgroup$ – John Rennie Jul 22 '14 at 17:13
  • 1
    $\begingroup$ I am not going to even attempt to give a full answer because I'm still a newbie with the SM. But as far as I know, the CP violation is not a result of an explicit CP violation in terms of the modulus of a single Feynman diagram, but of interference of two Feynman diagrams (two "processes" or "channels") by a non-CP invariant phase difference. As the whole Lagrangian is $CPT$-invariant, $CP$ acts the same as $T^{-1}$, so identical $T$-violation is implied if $CPT$ is true. So yes, explicit experimental violation of time-reversal also has to be shown $\to$ John Rennie's link. $\endgroup$ – Void Jul 23 '14 at 8:35
3
$\begingroup$

The sentence in Peskin's and Schroeder's book that "the weak interactions preserve CP and T" is a bit misleading but there is a sense in which it is right.

Experimentally, CP and T is known to be violated and CPT is always a symmetry. Theoretically, CPT is always a symmetry, too – it's proven by the CPT theorem. The CPT transformation is effectively a rotation by $\pi$ in the $t_Ez$ plane in the Euclideanized spacetime which is a symmetry due to the Lorentz symmetry and analyticity of the theory (the charge conjugation is automatically added because by sending the particles backwards in time, they become antiparticles, so the geometric operation is physically interpreted as CPT and not just PT). Theoretically, we know that CP and T may be violated. Note that because CPT is always a symmetry, a theory is symmetric under CP if and only if it is symmetric under T.

There are various potential physical phenomena that violate CP and T – including the "theta angle" of QCD (which would mean that the strong force also breaks CP and T) – but the only experimentally observed source of CP violation is the "complex phase in the CKM matrix".

The CKM matrix is a unitary transformation that transforms the upper-type quark mass eigenstates to the $SU(2)_W$ upper partners of the down-type quark mass eigenstates. All the quark masses are generated by the Higgs mechanism – from the Yukawa couplings and the vev – and all the Higgs-related things may be incorporated into the "weak interaction". But at least up to some extent, the known breaking of the CP occurs due to mass terms which are not "interactions at all" (they are quadratic terms in the Lagrangian while interactions have to be higher-order). In this sense, the CP and T violation isn't caused by the "weak interaction" only – only by a subtle combination of the weak interaction and subtleties in the mass matrices that wouldn't matter separately if there were no interactions.

At any rate, the CP and T are known to be violated much more "unambiguously" than the formulation in the Peskin-Schroeder textbook seems to suggest. The breaking of this symmetry is there; its effects are just a few orders of magnitude weaker (and less qualitative) than the effects of the breaking of C and P. The latter symmetry violations are "immediately obvious" when we consider the weak force – for example because the observed neutrinos are always left-handed (and antineutrinos are right-handed).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.