The Standard Model and Beyond: Problem Set 3

February 26, 2020 The Standard Model and Beyond: Problem Set 3 1. In this problem, we consider charged why charged pions decay preferentially into muons instead of electrons. The mechanism is known as \helicity suppression". µ µ (a) Transform the Dirac equation (γµ@ + m) = 0 into momentum space with coordinates p . Show that solutions to the momentum-space Dirac equation satisfy 0 i 0 ! µ 2 −piγ γ + imγ (p pµ + m ) (p) = 0 and (p) = ζ (p) pp2 + m2 0 0 with ζ = 1 if p > 0 and ζ = −1 if p < 0, where the pi are components of the spinor field’s three-momentum, i = 1; 2; 3, and jpj = ppipi. The former solutions are called particle solutions; the latter are antiparticle solutions. The helicity Λ of a particle is the projection of its angular momentum 3-vector in the direction i of the spatial momentum. Given that the Lorentz generators for spinors are Mµν = − 2 γµν = i 1 − 4 (γµγν −γνγµ) and the spatial angular-momentum 3-vector operator is Si = 2 ijkMjk, show that the helicity operator Λ can be written as p Λ = − 1 γ5γ0γi i (1) 2 jpj Hence, show that the helicity Λ of left- or right-chirality particles satisfies p 2 2 0 ! 9 ζ p + m L(p) − imγ R(p) > Λ L(p) = > 2jpj => 0 > 5 ζ = ±1 for p < 0 where γ L = ± L : p ! R R ζ p2 + m2 (p) − imγ0 (p) > Λ (p) = − R L > R 2jpj ;> (2) Hence, show that for massless or for ultrarelativistic spinor particles, chirality is related to helicity by ζ Λ L(p) ∼ L(p) 2 (3) ζ Λ R(p) ∼ − 2 R(p) : (b) Consider now a positively charged pion π+ at rest, decaying into a positively charged lepton + and the corresponding neutrino by exchange of a Wµ intermediate vector boson. An example is shown in Figure 1. 1 + u µ (p3) fπ W ¡ d¯ νµ(p4) Figure 1: Pion decay into a muon The problem is to explain the basic reason why the decay process shown in Figure 1 dominates over the analogous decay into a positron and its corresponding neutrino. In the minimal Standard Model, neutrinos are massless. Show that for the decay of a π+ at rest, the charged lepton and the neutrino decay products would have to have the same helicity if the lepton + were massless. Using the form of the Standard Model interaction between Wµ , the charged lepton and the neutrino, show that the decay of a pion via the process of Figure 1 would thus be impossible if the charged lepton were massless. For massive charged leptons, this helicity suppression is only partially effective, but it is stronger for lower mass charged leptons. Since the muon is about 200 times heavier than the electron, the helicity suppression for decay into muons is correspondingly weaker. 2. Now consider how to evaluate the relative rates of negatively charged pion decay into muons and electrons in the low-energy effective theory with a current-current interaction p +µ − µ Leff = 2 2GF (J Jµ + JZ JZµ) ; (4) ± where only the Jµ charged currents participate in charged pion decay. In the effective theory, for + decay of π into a charged lepton ` and its corresponding neutrino ν`, these are + 1 + Jµ = p c1fπ@µπ 2 (5) 1 J − = − ¯ γ (1 + γ ) µ 2 ν` µ 5 ` where fπ is the charged pion decay constant and c1 is a constant that needs to be obtained experimentally. The relevant term in the effective action is thus µ + ¯ Leff = −GF c1fπ@ π ν` γµ(1 + γ5) ` : (6) 2 One obtains momentum-space Feynman rules for a quantum spinor field by expanding in y P R d3p ip·x terms of bs(p), ds(p) annihilation and creation operators, (x) = s=± 2(2π)3p0 [bs(p)us(p)e + y −ip·x ds(p)vs(p)e ], where us(p) and vs(p) are c-number (i.e. ordinary commuting-number) spinor wavefunctions for incoming and outgoing particles respectively. The Dirac conjugatesu ¯s(p) and v¯s(p) are the wavefunctions for outgoing and incoming antiparticles. The tree level momentum- space decay amplitude for π+ is then µ + T = −iGF c1fπk π u¯ν` γµ(1 + γ5)v` (7) where v` is the wavefunction of the outgoing charged lepton,u ¯ν` is the wavefunction of the outgoing µ µ µ neutrino and the four-momenta of the pion, charged lepton, and neutrino are k , p3 and p4 as shown in Figure 1. Use then (−p=3 − im`)v` = 0 and −p=4uν` = 0 (remembering that the charged lepton and neutrino are outgoing) to show + T = −GF c1fππ mù¯ν` (1 − γ5)v` (8) so T is proportional to the charged lepton mass m`. Squaring the modulus of T and ignoring normalisation factors in the π+ wavefunction, one has 2 2 jT j = −(GF c1fπm`) u¯ν` (1 + γ5)v`v¯`(1 − γ5)uν` 2 = −(GF c1fπm`) tr (1 + γ5)v`v¯`(1 − γ5)uν` u¯ν` : (9) Then perform a sum over spins of the final states using the spin-sum rules X v`v¯` = −p=3 − im` (10) spins X uν` u¯ν` = −p=4 (11) spins to show that X 2 2 jT j = (GF c1fπm`) (−8p3 · p4) spins 2 2 2 = 4(GF c1fπm`) (mπ − m` ) (12) The decay rate Γ is given by 1 jp3j 2 Γπ = = 2 hjT j i : (13) τπ 8πmπ Show that the ratio of pion decay rates into electrons and into muons is consequently 2 2 2 2 me mπ − me Rπ = 2 2 : (14) mµ mπ − mµ −4 Inputting the experimental values of mπ, mµ and me, one obtains Rπ = 1:283 × 10 . 3.

Load more