<<

PHY401 - Nuclear and 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 ...... 5 1.2 Strangeness Oscillations ...... 5

2 Formulation of the Standard Model 6 2.1 and Leptons ...... 7 2.2 Content of ...... 8 2.3 Quark Content of ...... 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 Monsoon Semester 2020

In 1964, James Cronin and Val Fitch tested CP violations in neutral kaons. Neutral kaons can be represented as

|K0i = |sd¯ i, |K¯ 0i = |sd¯i. (1)

K¯ 0 is the anti-particle of K0.

They have opposite S and I3 quantum numbers

S|K0i = +1|K0i,S|K¯ 0i = −1|K¯ 0i, (2) 1 1 I |K0i = − |K0i,I |K¯ 0i = + |K¯ 0i. (3) 3 2 3 2

They look a lot like . 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 |K0i = −C|K0i = −|K¯ 0i, (5) CP |K¯ 0i = −C|K¯ 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 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 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 . 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¯ 0|K0i = 0, (10) 0 0 hK¯ |Hstrong|K i = 0. (11)

0 2 0 Hstrong|K i = mK0 c |K i, (12) ¯ 0 2 ¯ 0 Hstrong|K i = mK¯ 0 c |K i. (13)

We have

mK0 = mK¯ 0 ≈ 498 MeV. (14)

For the weak interactions we have

0 0 hK¯ |Hweak|K 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 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  |K1i = √ |K i − |K¯ i , 2 (19) 1 0 0  |K2i = √ |K i + |K¯ i . 2

Applying the CP operator to the K1 and K2 states, we can verify explicitly that

1 0 0  CP |K1i = √ CP |K i − CP |K¯ i 2 1 0 0  1 0 0  = √ −|K¯ i + |K i = √ |K i − |K¯ i = |K1i, (20) 2 2 and

1 0 0  CP |K2i = √ CP |K i + CP |K¯ i 2 1 0 0  1 0 0  = √ −|K¯ i − |K i = −√ |K i + |K¯ i = −|K2i. (21) 2 2

The two states |K1i and |K2i 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 |K i = √ (|K1i + |K2i) , 2 (22) 0 1 |K¯ i = −√ (|K1i − |K2i) . 2 In strong production processes, such as given in Eq. (6) only the eigenstates of the strong Hamiltonian are produced, namely |K0i or |K¯ 0i.

However, these states are superpositions of the |K1i and |K2i, 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 |K1i and |K2i states propagate in vacuum, both the |K1i and |K2i components decay away. 0 However, the state |K1i decays much faster than |K2i, and, after some time, the initial |K i or 0 |K¯ i will therefore be composed primarily of |K2i.

5 / 10 PHY401 - Nuclear and Particle Physics Monsoon Semester 2020

0 0 But |K2i has an equal admixture of |K i and |K¯ i, which means that, starting out either as a pure |K0i or as a pure |K¯ 0i state, any neutral 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. But whether you choose to analyze the process in terms of states of linear or circular polarization is largely a matter of taste.

2 Formulation of the Standard Model

Prior to the mid 1970s only a few low-mass were discovered. As the energies of accelerators increased, additional excited states of those particles, but with larger masses and higher spins, as well as particles with new flavors were found. In fact, even by the mid 1960s, there was a whole host of particles to contend with, and it was questioned whether they could all be regarded as fundamental constituents of matter. As we argued previously, even the lightest baryons, namely the proton and the neutron, show indirect evidence of substructure. For example, the large anomalous magnetic moments observed for these particles, especially dramatic for the neutron, imply a complex internal distribution of currents. From the pattern of the observed spectrum of hadrons, Gell-Mann and Zweig suggested inde- pendently, in 1964, that all such particles could be understood as composed of quark constituents. These constituents had rather unusual properties, and were initially regarded as calculational tools rather than as true physical objects. In the 1960s the data from the deep inelastic scattering experiments at SLAC revealed that it could be most easily understood if protons and neutrons were composed of point-like objects that 1 2 had charges − 3 e and + 3 e.

6 / 10 PHY401 - Nuclear and Particle Physics Monsoon Semester 2020

The presence of point-like quarks or partons was deduced from the characteristics of inelastically scattered electrons. By the early 1970s, it became quite apparent that hadrons were not fundamental point-like objects. In contrast, leptons still do not exhibit any evidence of structure, even at highest momentum transfers. It is natural therefore to regard leptons as elementary, but to regard hadrons as composed of more fundamental constituents. This line of thought - completely phenomenological in the beginning - merged the observations from electron scattering with those from particle spectroscopy and the , and culminated in the present Standard Model. The Standard Model incorporates all the known fundamental particles, namely, the quarks, leptons and the gauge bosons, and it provides a theory describing three of the basic forces of nature - the strong, weak and electromagnetic interactions.

2.1 Quarks and Leptons

We saw earlier that each charged lepton has its own neutrino. There are three families (flavors) of such leptons.

! ! ! ν ν ν e , µ , and τ . (23) e− µ− τ − In the above, we have used the convention introduced previously in connection with strong symmetry, namely, the higher member of a given multiplet carries a higher electric . The quark constituents of hadrons also come in three families ! ! ! u c t , , and . (24) d s b

1 The numbers are B = 3 for all the quarks, and the charges are

2 Q = Q = Q = + e, (25) u c t 3 1 Q = Q = Q = − e. (26) d s b 3

Although the fractional nature of their electric charges was deduced indirectly from electron scattering for only the u and d quarks, phenomenologically, such charge assignments also provide a natural way for classifying the existing hadrons as bound states of quarks. Quarks also appear to have flavor quantum numbers. For example, because we defined the strangeness of the K+ as +1, we will see shortly that the will have to be assigned a strangeness of −1.

7 / 10 PHY401 - Nuclear and Particle Physics Monsoon Semester 2020

The , top and the bottom quarks, correspondingly, carry their own flavor quantum num- bers. Of course, each quark has its own antiquark, which has opposite and other internal quantum numbers such as strangeness and charm.

2.2 Quark Content of Mesons

The quarks, like the leptons, are point-like fermionic particles. 1 In other words, they have spin angular momentum of 2 ~. This suggests that, since mesons have integer spin, then, if they are bound states of quarks, they can only consist of an even number of these particles. In fact, every known can be described as a bound state of a quark and an antiquark. Thus, for example, a π+ meson, which has spin zero and electric charge +1, can be described as the bound state π+ = ud.¯ (27)

It follows, therefore, that the π− meson, which is the antiparticle of the π+, can be described as the bound state π− =ud. ¯ (28)

The π0 meson, which is charge neutral, can, in principle, be described as a bound state of any quark and its antiquark. However, other considerations, such as the fact that all three π-mesons belong to a strong-isospin multiplet, and should therefore have the same internal structure, lead to a description of the π0 meson as 1 π0 = √ (uu¯ − dd¯). (29) 2 The strange mesons can similarly be described as bound states of a quark and an antiquark, where one of the constituents is strange. Thus, we can identify the following systems

K+ = us,¯ (30) K− =us, ¯ (31) K0 = ds,¯ (32) K¯ 0 = ds.¯ (33)

It is quite easy to check that not only are the charge assignments right, but even the strangeness quantum numbers work out to be correct if we assign a strangeness quantum number S = −1 to the s-quark. Because there are quarks with higher mass and new flavor quantum numbers, phenomenologi- cally, on the basis of the quark model, we would also expect new kinds of mesons.

8 / 10 PHY401 - Nuclear and Particle Physics Monsoon Semester 2020

Many such mesons have already been found. For example, the charge-neutral J/ψ meson, whose discovery in 1974 (by two independent groups) suggested first evidence for the existence of the charm quark, can be described as a bound state of charmonium (named in analogy with positronium)

J/ψ = cc.¯ (34)

This is a “normal” meson, in the sense that the quantum numbers of charm add up to zero, but its properties (decays) cannot be explained using only the older u, d and s quarks. There are, of course, mesons that contain open charm, such as

D+ = cd,¯ (35) D− =cd, ¯ (36) D0 = cu,¯ (37) D¯ 0 =cu. ¯ (38)

We can think of such mesons as the charm analogs of the K mesons, and the properties of these mesons have by now been studied in great detail. In analogy with the K+, the D+ meson is defined to have charm flavor of +1, which then defines the charm quantum number for the c-quark to be +1. There are also mesons that carry both strangeness and charm quantum numbers, two of these are denoted as

+ Ds = cs,¯ (39) − Ds =cs. ¯ (40)

Finally, there is also extensive evidence for hadrons in which one of the constituents is a bottom quark. For example, the B mesons, analogous to the K mesons, have structure of the form

B+ = u¯b, (41) B− =ub, ¯ (42) 0 ¯ Bd = db, (43) ¯0 ¯ Bd = db. (44)

The charge-neutral states involving b and s quarks are particularly interesting because, just like the K0 − K¯ 0 system, they exhibit CP violation in their decays

0 ¯ Bs = sb, (45) ¯0 Bs =sb. ¯ (46)

9 / 10 PHY401 - Nuclear and Particle Physics Monsoon Semester 2020

Extensive studies of B decays are currently being pursued at particle colliders, to search for any discrepancies with expectations from the Standard Model.

2.3 Quark Content of Baryons

Just as mesons can be thought of as bound states of quarks and antiquarks, so can baryons be considered as constructed out of these constituents. But because baryons carry half-integral spin angular momenta (they are fermions), they can be formed from only an odd number of quarks. Properties of baryons are most consistent with being composed of only three quarks. Thus, we can think of the proton and the neutron as corresponding to the bound states

p = uud, (47) n = udd. (48)

Similarly, the , which carry a strangeness quantum number, can be described by

Λ0 = uds, (49) Σ+ = uus, (50) Σ0 = uds, (51) Σ− = dds. (52)

Also, the cascade particles, which carry two units of strangeness, can be described as

Ξ0 = uss, (53) Ξ− = dss. (54)

Since all baryons have of unity, it follows therefore that each quark must carry 1 a baryon number of 3 . Furthermore, since a meson consists of a quark and an antiquark, and since an antiquark would 1 have a baryon number − 3 , we conclude that mesons do not carry baryon number.

References

[1] Dave Goldberg, The Standard Model in a Nutshell, Princeton University Press (2017).

[2] David Griffiths, Introduction to Elementary Particles, 2nd edition, Wiley-VCH (2008).

[3] A. Das and T. Ferbel, Introduction To Nuclear And Particle Physics, World Scientific (2003).

10 / 10