The Standard Model and Beyond: Problem Set 3

Home , Chirality (physics), Lepton, Pion

February 26, 2020

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)ψ = 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 )ψ(p) = 0 and ψ(p) = ζ ψ(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 |p| = 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 |p|

Hence, show that the helicity Λ of left- or right-chirality particles satisﬁes

p 2 2 0 !  ζ p + m ψL(p) − imγ ψR(p)  ΛψL(p) =  2|p|  0 > 5 ζ = ±1 for p < 0 where γ ψ L = ±ψ L . p ! R R ζ p2 + m2ψ (p) − imγ0ψ (p)  Λψ (p) = − R L  R 2|p|  (2) Hence, show that for massless or for ultrarelativistic spinor particles, chirality is related to helicity by ζ ΛψL(p) ∼ ψL(p) 2 (3) ζ ΛψR(p) ∼ − 2 ψR(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 .Nwcnie o oeaut h eaiertso eaieycagdpo ea nomosand muons into decay pion charged negatively of rates relative the evaluate to how consider Now 2. h eeattr nteeetv cini thus is action effective the in term relevant The experimentally. where interaction current-current a with theory effective low-energy the in electrons ea of decay the only where h uni bu 0 ie eve hnteeeto,tehlct upeso o ea into decay for suppression helicity the weaker. electron, Since correspondingly the leptons. than is charged heavier muons mass times lower 200 helicity for about this stronger is is leptons, muon thus it charged the would but massive 1 effective, For Figure partially of massless. only process were is the lepton suppression via charged pion the a lepton if of decay the impossible between the be interaction if that Model helicity show Standard neutrino, same the the the and of have lepton form to the minimal have Using would the massless. products In a were decay of neutrino decay neutrino. the the corresponding for and that its lepton Show and massless. are positron neutrinos dominates a 1 Model, Figure into Standard in decay shown process analogous decay the the why reason over basic the explain to is problem The f π π stecagdpo ea osatand constant decay pion charged the is + noacagdlepton charged a into J µ ± hre urnspriiaei hre indcy nteeetv hoy for theory, effective the In decay. pion charged in participate currents charged ¡ u d ¯ iue1 indcyit muon a into decay Pion 1: Figure f L π eff L = eff ` J J n t orsodn neutrino corresponding its and − 2 = µ µ − + G = = √ F W c − 2 √ 1 G 1 f 1 2 2 π F 2 ψ c ∂ ¯ ( 1 ν µ J f ` π γ π + + ∂ µ µ 1+ (1 ψ µ J ¯ π ν µ − c ` + 1 γ + µ sacntn htnest eobtained be to needs that constant a is γ 1+ (1 J 5 ) Z µ ψ J ` Zµ µ ν γ 5 µ + ) ) (p ψ , (p ` 4 3 . ) ) ν ` hs are these , π + trs,techarged the rest, at W µ + h charged the , (5) (4) (6) One obtains momentum-space Feynman rules for a quantum spinor field ψ by expanding in † P R d3p ip·x terms of bs(p), ds(p) annihilation and creation operators, ψ(x) = s=± 2(2π)3p0 [bs(p)us(p)e + † −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 m`u¯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 |T | = −(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 |T | = (GF c1fπm`) (−8p3 · p4) spins 2 2 2 = 4(GF c1fπm`) (mπ − m` ) (12)

The decay rate Γ is given by 1 |p3| 2 Γπ = = 2 h|T | 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 .