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PHY401 - Nuclear and Particle Physics

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PHY401 - Nuclear and Particle Physics PHY401 - Nuclear and Particle Physics Monsoon Semester 2020 Dr. Anosh Joseph, IISER Mohali LECTURE 27 Wednesday, October 28, 2020 (Note: This is an online lecture due to COVID-19 interruption.) Contents 1 CP Violation 1 1.1 CP Eigenstates of Neutral Kaons . .5 1.2 Strangeness Oscillations . .5 2 Formulation of the Standard Model 6 2.1 Quarks and Leptons . .7 2.2 Quark Content of Mesons . .8 2.3 Quark Content of Baryons . 10 1 CP Violation In 1956 Wu and collaborators conducted an experiment to look for P violation in cobalt-60 decays. The presence of P violation in this system suggested that there is something unique about the weak force. If P and C can be violated individually, how about the combination of the two? CP takes a physical particle state to a physical anti-particle state. Thus, invariance under CP is equivalent to having a particle-anti-particle symmetry in nature. However, the universe is known to be dominated by matter, with essentially no antimatter present. That is, there is a definite particle-anti-particle asymmetry in the universe. CP may not be a symmetry of all fundamental interactions. There are in fact processes in which CP is violated. CP violation implies that there are microscopic (sub-atomic) processes in which time has a unique direction of flow. PHY401 - Nuclear and Particle Physics Monsoon Semester 2020 In 1964, James Cronin and Val Fitch tested CP violations in neutral kaons. Neutral kaons can be represented as jK0i = jsd¯ i; jK¯ 0i = jsd¯i: (1) K¯ 0 is the anti-particle of K0. They have opposite S and I3 quantum numbers SjK0i = +1jK0i;SjK¯ 0i = −1jK¯ 0i; (2) 1 1 I jK0i = − jK0i;I jK¯ 0i = + jK¯ 0i: (3) 3 2 3 2 They look a lot like pions. But there is a subtle difference. Pions are superpositions of up and anti-up quarks, and down and anti-down quarks. π+ = ud;¯ π0 = uu¯ or dd;¯ (4) π− = du:¯ Pions are eigenstates of CP with an eigenvalue of −1. Neutral kaons, in contrast, are not eigenstates of CP but rather CP jK0i = −CjK0i = −|K¯ 0i; (5) CP jK¯ 0i = −CjK¯ 0i = −|K0i: In the above we used the fact that K-mesons are pseudo-scalars and consequently have odd intrinsic parities. Neutral kaons can be produced in the strong interaction processes such as K− + p ! K¯ 0 + n; K+ + n ! K0 + p; (6) π− + p ! Λ0 + K0: Since one is the anti-particle of the other, CPT theorem tells us that the two particles have identical masses and lifetimes. However, experimental data showed that there are two distinct lifetimes associated with both the K0 and K¯ 0. This can only be understood that if we assume that the K0 and K¯ 0 states consist of a superpo- sition of two distinct states with different lifetimes. There is a short-lived one, K1 and a long-lived one, K2. 2 / 10 PHY401 - Nuclear and Particle Physics Monsoon Semester 2020 The fact that K0 and K¯ 0 can decay through common channels K0 ! π0 + π0; (7) K¯ 0 ! π0 + π0; (8) suggests that these particles can mix through higher orders of the weak interaction. That is, in the presence of weak interactions, the two kaons share the same decay channels. This is a consequence of the fact that the weak interactions do not conserve strangeness. That is, we can have H H K0 −−−−!weak π0 + π0 −−−−!weak K¯ 0: (9) 0 0 Thus, the K and K¯ states, although eigenstates of the strong Hamiltonian Hstrong, cannot be eigenstates of the weak interaction Hamiltonian Hweak. Schematically, we can write hK¯ 0jK0i = 0; (10) 0 0 hK¯ jHstrongjK i = 0: (11) 0 2 0 HstrongjK i = mK0 c jK i; (12) ¯ 0 2 ¯ 0 HstrongjK i = mK¯ 0 c jK i: (13) We have mK0 = mK¯ 0 ≈ 498 MeV: (14) For the weak interactions we have 0 0 hK¯ jHweakjK i= 6 0: (15) Since the decay of K-mesons is a weak process, the observed K1 and K2 particles, with unique lifetimes, can be thought of as corresponding to the eigenstates of Hweak. If CP symmetry is perfectly conserved, then K1 and K2 could be distinguished by their decay products 0 0 −11 K1 ! π + π with τ = 9 × 10 s (16) and 0 0 0 −8 K2 ! π + π + π with τ = 5 × 10 s; (17) since an even number of pions will have a CP value of +1, and an odd number will have a CP of −1. Here, τ denotes the lifetime. 3 / 10 PHY401 - Nuclear and Particle Physics Monsoon Semester 2020 π+ K0 K¯ 0 π− π+ K0 π0 K¯ 0 π− Figure 1: Possible transformations of K0 and K¯ 0. Given the difference in decay times, a beam of K2 particles can be created simply by sending a beam of mixed kaons and observing a suitable distance away. Even at relativistic speeds, virtually all the K1 particles will decay after a few meters, and a beam of pure K2 particles will remain. Then the outgoing kaons should decay exclusively into three pions. But this is not what Cronin and Fitch observed. They observed the long-lived K2 (CP = −1) undergoing two pion decays, CP = (−1)(−1) = +1, thereby violating CP conservation. About 0.2% of the decays were to two pions. The goal was to search for 2π decays at the end, and thereby establish a more stringent upper limit on the 2π mode of K2. The short-lived component in the beam has a decay length of about 6 cm. It was expected to decay away by the end of the tube (part of their experimental apparatus). Another way of looking at the CP violation is that the reaction K0 $ K¯ 0 (18) is not symmetric. In 2001, CP violation in the B0 − B¯0 system was confirmed by the BaBar and the Belle exper- iments. Direct CP violation (CP violation through decay only) in the B0 − B¯0 system was reported by both the labs by 2005. Indirect CP violation - CP violation through mixing, or CP violation through a combination of mixing and decay. 4 / 10 PHY401 - Nuclear and Particle Physics Monsoon Semester 2020 1.1 CP Eigenstates of Neutral Kaons Using Eq. (5) we can define two linear orthonormal combinations of K0 and K¯ 0 that will be eigenstates of the CP operator, namely 1 0 0 jK1i = p jK i − jK¯ i ; 2 (19) 1 0 0 jK2i = p jK i + jK¯ i : 2 Applying the CP operator to the K1 and K2 states, we can verify explicitly that 1 0 0 CP jK1i = p CP jK i − CP jK¯ i 2 1 0 0 1 0 0 = p −|K¯ i + jK i = p jK i − jK¯ i = jK1i; (20) 2 2 and 1 0 0 CP jK2i = p CP jK i + CP jK¯ i 2 1 0 0 1 0 0 = p −|K¯ i − jK i = −p jK i + jK¯ i = −|K2i: (21) 2 2 The two states jK1i and jK2i do not carry unique strangeness. They can be defined as eigenstates of CP with eigenvalues +1 and −1, respectively. 1.2 Strangeness Oscillations We can invert the relations in Eq. (19) to obtain 0 1 jK i = p (jK1i + jK2i) ; 2 (22) 0 1 jK¯ i = −p (jK1i − jK2i) : 2 In strong production processes, such as given in Eq. (6) only the eigenstates of the strong Hamiltonian are produced, namely jK0i or jK¯ 0i. However, these states are superpositions of the jK1i and jK2i, which are eigenstates of the weak Hamiltonian. 0 0 At the time of production, the K and K¯ correspond to the superpositions of K1 and K2 given in Eq. (22). But as these specific mixtures of the jK1i and jK2i states propagate in vacuum, both the jK1i and jK2i components decay away. 0 However, the state jK1i decays much faster than jK2i, and, after some time, the initial jK i or 0 jK¯ i will therefore be composed primarily of jK2i. 5 / 10 PHY401 - Nuclear and Particle Physics Monsoon Semester 2020 0 0 But jK2i has an equal admixture of jK i and jK¯ i, which means that, starting out either as a pure jK0i or as a pure jK¯ 0i state, any neutral kaon will evolve into a state of mixed strangeness. This phenomenon is known as K0 - K¯ 0 strangeness oscillations. The question we can ask is: which then is the “real” particle? The above description refers to flavor (or strangeness) eigenstates and energy (or CP ) eigen- states. But which of them represents the “real” particle? What do we really detect in a laboratory? Quoting David J. Griffiths: The neutral Kaon system adds a subtle twist to the old question, ‘What is a particle?’ Kaons are typically produced by the strong interactions, in eigenstates of strangeness (K0 and K¯ 0), but they decay by the weak interactions, as eigenstates of CP (K1 and K2). Which, then, is the ‘real’ particle? If we hold that a ‘particle’ must have a unique lifetime, then the ‘true’ particles are K1 and K2. But we need not be so dogmatic. In practice, it is sometimes more convenient to use one set, and sometimes, the other. The situation is in many ways analogous to polarized light. Linear polarization can be regarded as a superposition of left-circular polarization and right-circular polarization. If you imagine a medium that preferentially absorbs right-circularly polarized light, and shine on it a linearly polarized beam, it will become progressively more left- 0 circularly polarized as it passes through the material, just as a K beam turns into a K2 beam.
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