Nuclear and Particle Physics - Lecture 15 Meson Decays and the CKM Matrix

Nuclear and Particle Physics - Lecture 15 Meson decays and the CKM matrix 1 Introduction We know that six of the nine ground state mesons have different flavour quarks and antiquarks 0 and so cannot decay by the strong or EM force. These are the π, K and K0, K . We will now look at their decays and see what this tells us about how quarks interact with the weak force. In general, any meson decay can be classified in one of three ways, depending on the final state: 1. Leptonic: meaning a purely leptonic final state, 2. Semi-leptonic: meaning a mixture of leptons and hadrons in the final state, 3. Hadronic: meaning a purely hadronic final state. We will use these classifications as we study the meson decays. 2 Charged pion decay The main decays of the charged pion are leptonic − − − − π ! e νe; π ! µ νµ + + In all that follows, the charged conjugate decays, i.e. here π ! e νe, etc, are also implied. This decay requires the quarks, e.g. du in the case of a π−, to have annihilated each other as there are (by definition for a leptonic decay) no quarks left. This therefore means the u and d quarks must have a weak interaction vertex just like the e and νe d W- u Note the d quark charge is −e=3 and the u quark charge is +2e=3 so they are different by e, just right to emit a W −. Also, the u and d quarks do not have the same mass, just like the e and νe. Note also that this interaction does not conserve flavour; the weak interaction allows us to change the types of quarks. The Feynman diagram for pion decay is then straightforward The ratio of the number of decays to electrons versus muons also demonstrates a property of the weak interaction. I have stated several times the muon is just a heavier version of the electron but is otherwise identical. Specifically, it has identical charges and here, this means it has the same gW weak charge. We in fact saw the result of this in tau decays, where the electron and muon branching fractions were almost equal (and were each one third of the hadron rate). This means the electron and muon might be expected to have the equal amplitudes for pion decay 1 d l- W π- g cosθ g W C W ν u l and hence have equal rates here too. However, the muon mass, 105.7 MeV/c2, is not that much lower than the pion mass, 139:6 MeV/c2, so there is somewhat limited energy available for the muon decay and it would therefore be expected that the electron rate may be higher than the muon rate. A calculation shows this would lead to a difference in the rates by a factor of around three. However, the muon rate is not only bigger than the electron rate, but is around 8000 times bigger! This is impossible to explain without considering the way the weak force interaction works. We know the weak force only couples to LH fermions and RH antifermions. Also, in the relativistic limit, these correspond to h = −1=2 and h = +1=2 respectively and when not at that limit, they have (1 v=c)=2 of each state. The antineutrino is almost massless compared to the charged leptons, so it must be at the relativistic limit to a very good approximation. Hence, as it is RH, it must be h = +1=2. The pion is spin 0, so to balance angular momentum, the charged lepton must also be h = +1=2. - - ν p ν π p µ µ- sν sµ h = +1/2 h = +1/2 RH (1+v/c)/2 RH, (1-v/c)/2 LH If the charged lepton were also at the relativistic limit, this would not be allowed as it would be a RH fermion and so the decay could not occur. However, the charged leptons are not quite at v = c and so when h = +1=2, they have a little LH state included, namely (1 − v=c)=2. Hence, there is a suppression factor which gets smaller as the velocity gets closer to c. What is the velocity of the charged leptons? The standard two-body decay formula gives 2 2 − 2 mπ + ml mν 2 El = c 2mπ which, in the approximation that the neutrinos are massless is 2 2 mπ + ml 2 El = c 2mπ The momentum is then given by 2 − 2 2 − 2 4 mπ ml plc = El ml c = c 2mπ q 2 so that the velocity is 2 − 2 vl plc mπ ml = βl = = 2 2 c El mπ + ml Putting in the relevant masses, this evaluates to βe = 0:99997 and βµ = 0:272, which is a ratio of (1 − βµ)=(1 − βe) = 28000. This explains the major part of the relative branching fractions for muons and electrons. Because of the small mass difference between the pion and the muon, the restricted energy available to the muon means it has a relatively low velocity. However, this means there is a lot more LH state mixed in and hence the muon rate is much higher. 2 The charged pion mass, mπ = 139:6 MeV/c , is slightly higher than the neutral pion mass, 2 mπ0 = 135:0 MeV/c , so that mπ > mπ0 + me. This means there is also a semi-leptonic decay − 0 − possible, namely π ! π e νe e- W ν d e π- u u π0 u However, as the energy available is so limited, the branching fraction is minute. 3 Charged kaon decay The mere fact that kaons decay tells us something about the weak force and quarks which is new. The u and d quarks form the first generation of quarks, like the e and νe form the first generation of leptons, and the weak force allows interactions between them. Hence, since c and s are the next generation, like µ and νµ, we would expect weak interactions between these two quarks also. However, c quarks are heavier than s quarks, so if s could only be turned to c, then the charged kaon, which is su (not sc) could not decay; all the hadrons with c are much heavier than the kaon. However, the kaon has leptonic decay modes just like the pion − − − − K ! e νe; K ! µ νµ and indeed had a ratio of these decays of around 40000 more to muons than to electrons, which is what we would expect from using the kaon mass rather than the pion mass from the previous calculation. Hence, we have to believe the s and u quarks are annihilating so there must be a weak interaction vertex 3 s W- u This means the decay diagram is just like in the pion s l- W - g sinθ g K W C W ν u l This means there must be a cross-generational coupling for quarks, even though it is absent for leptons. Given that this exists, the other decays of the kaon are easy to understand; the − − semileptonic decays are K ! l νl(nπ) l- W ν s l K- u u nπ u where the quarks can emerge as two or three pions. Here, there is no helicity issue with the leptons due to the pions also being present and so the ratio of e to µ decays is much more similar, being 1.5 more electrons than muons. There are also hadronic decays to pions with diagrams 4 d W s u K- u u u where again the quarks can emerge as two or three pions. This is in fact the decay which was previously mentioned as being parity violating when two pions are produced. The ratio of decays to two and three pions is around three, where the two pion decay is higher as would be expected from the extra energy available, in the absence of parity constraints. Hence, it acts as if parity was not an issue at all. 4 Charm mesons decays We have found there is a cross-generational coupling for quarks, where the u quark can connect to either the d or the s. The obvious question is whether the c quark can do the same thing, which would mean each could connect to the other two of the different EM charge. Looking at charm meson decays shows this is indeed possible. The charmed D+ meson, cd, has been seen to + 0 + 0 decay (amoungst other decays modes) semi-leptonically to both l νlK and also l νlπ , which are given by diagrams like l+ l+ ν ν c l c l D+ D+ d s d d 0 K π0 d d Hence, for quarks, any +2e=3 quark (u or c) can connect to any −e=3 quark (d or s) in a weak interaction. The relative weak force strengths are usually described by the Cabibbo matrix Vij Vud Vus 0:975 0:222 cos θC sin θC = − = − Vcd Vcs ! 0:222 0:975 ! sin θC cos θC ! ◦ where the \Cabibbo angle", θC ≈ 13 . Hence, it is close to, but not exactly, a unit matrix. This is in fact a rotation matrix and means every ud vertex has a multiplicative factor of Vud = cos θC , etc. In fact, looking at the third generation of b and t quarks, it turns out this matrix has to be extended so again any of u, c and t can connect to any of d, s and b.

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