The Physics Travel Guide

CP Symmetry

Prerequisites: Parity, Standard Model of Particle Physics. Full path (33 pages)

Intuitive

CP symmetry combines two operations. Charge conjugation C replaces every particle by its antiparticle. Parity P reflects space, which turns a left-handed spinning particle into a right-handed one. Each of these on its own is violated by the weak interaction. Neutrinos, for example, are only ever observed left-handed and antineutrinos only right-handed. Applying C alone to a left-handed neutrino gives a left-handed antineutrino, which does not exist in nature. Applying P alone gives a right-handed neutrino, which also does not take part in weak interactions. Applying both gives a right-handed antineutrino, which is exactly what we see. So for a while physicists believed that CP was an exact symmetry: a world in which all matter is replaced by antimatter and which is viewed in a mirror would behave exactly like ours.

This turned out to be almost, but not quite, true. In 1964 Cronin, Fitch and collaborators found that neutral kaons occasionally decay in a way that CP forbids. Later CP violation was also observed in the decays of B mesons. In the Standard Model all of this comes from a single complex phase in the way the three generations of quarks mix under the weak interaction.

The question matters because our universe contains matter but almost no antimatter. If the laws of physics treated matter and antimatter in exactly the same way, it would be hard to understand how such an imbalance arose from a hot early universe. CP violation is one of the necessary ingredients. The amount of CP violation in the quark sector of the Standard Model appears to be far too small to explain the observed imbalance, which is one reason to look for new sources, for example in the neutrino sector.

Concrete

Kaon Flavor Eigenstates:

∣K0⟩=∣sˉd⟩∣K‾0⟩=∣sdˉ⟩\begin{equation*} |K^0\rangle = |\bar{s}d \rangle \qquad |\overline{K}^0\rangle = |s \bar{d} \rangle \end{equation*} CP ⁣:  ∣K0⟩⟷∣K‾0⟩CP\!: \; |K^0\rangle \longleftrightarrow |\overline{K}^0 \rangle

Kaon CP Eigenstates:

∣KCP±⟩=12(∣K0⟩±∣K‾0⟩)\begin{equation*} |K_{CP \pm}\rangle = \frac{1}{\sqrt{2}} \left( |K^0\rangle \pm |\overline{K}^0 \rangle \right) \end{equation*}

Kaon Propagating Mass Eigenstates:

∣KL⟩∼∣KCP−⟩+ϵK∣KCP+⟩|K_L \rangle \sim |K_{CP-}\rangle + \epsilon_K |K_{CP+}\rangle

Observables:



CP Violation in the Neutrino Sector

Abstract

CP in a gauge theory. Write all fermions as left-chiral Weyl fields ψL\psi_L (every right-chiral field is written as the charge conjugate of a left-chiral one). CP maps ψL(t,x⃗)\psi_L(t,\vec x) to an expression in its Hermitian conjugate at (t,−x⃗)(t,-\vec x). The Hermitian conjugate of a left-chiral field transforms in the right-chiral representation, and this is why CP can be a symmetry of a theory with only left-chiral fields, where C and P separately are not. The most general CP transformation acts as (phases absorbed into UU) ψL(t,x⃗)→U γ0C ψˉL T(t,−x⃗),\psi_L(t,\vec x) \to U\, \gamma^0 C\, \bar\psi_L^{\,T}(t,-\vec x), with CC the charge conjugation matrix and UU a unitary matrix acting on flavour indices, together with a corresponding action on scalars and gauge fields. The gauge fields must be mapped by an automorphism of the gauge group under which every representation present in the theory goes into its complex conjugate. A theory is CP invariant if such a transformation exists for some choice of UU and of the automorphism. Because UU is free, the question is whether complex phases in the couplings can be removed by field redefinitions, which makes CP a property of the whole set of couplings rather than of any single term.

Quark sector of the Standard Model. With Yukawa mass matrices MuM_u and MdM_d, define Hu=MuMu†H_u = M_u M_u^\dagger and Hd=MdMd†H_d = M_d M_d^\dagger. The theory with NN generations has (N−1)(N−2)/2(N-1)(N-2)/2 physical phases in the CKM matrix VV, which is one phase for N=3N = 3. A basis-independent criterion for CP violation is det⁡ [Hu,Hd]≠0.\det\,[H_u, H_d] \neq 0 . This determinant is proportional to the Jarlskog invariant J=Im(VusVcbVub∗Vcs∗)J = \mathrm{Im}(V_{us}V_{cb}V_{ub}^*V_{cs}^*) times the product of all differences of squared masses among the up-type quarks and among the down-type quarks. CP violation therefore needs a nonzero phase and also non-degenerate masses in both sectors.

The θ\theta term. The term θ g232π2 GμνaG~aμν\theta\,\frac{g^2}{32\pi^2}\,G^a_{\mu\nu}\tilde G^{a\mu\nu} is a total derivative, odd under P and T, and even under C. It is therefore CP odd. Only the combination θˉ=θ−arg⁡det⁡(MuMd)\bar\theta = \theta - \arg\det(M_u M_d) (with the sign depending on conventions) is physical, because chiral rotations of the quarks shift θ\theta through the anomaly. Its experimental smallness is the strong CP problem.

Relation to T. By the CPT theorem, in a local Lorentz invariant quantum field theory CP violation is equivalent to violation of time reversal.

Why is it interesting?

A CP transformation transforms a particle into the corresponding antiparticle. An intact CP symmetry therefore means that particles and antiparticles (with mirrored spatial configuration) behave in exactly the same way. However, we know from experiments that CP is a broken symmetry since, for example, decays of neutral kaons and B mesons violate it. CP violation is also needed to explain why there are more particles than antiparticles in the visible universe.

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