Physics 222 UCSD/225b UCSB. Lecture 3 Weak Interactions (continued) muon decay Pion decay

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1 Physics UCSD/5b UCSB Lecture 3 Weak Interactions (continued) muon decay Pion decay

2 Muon Decay Overview (1) Feynman diagram: µ! " u p ( ) Matrix Element: M = G u ( k )! µ ( 1"! 5 )u p [ ( )] u p'! " u ( k) e! " u p' ( )! "! k' ( ) [ ( )! ( µ 1"! 5 )v( k' )] G = effective coupling of a 4-fermion interaction Structure of 4-fermion interaction is (V-A)x(V-A) We clearly want to test this experimentally, e.g. compare against (V-A)x(V+A), S, P, T, etc.

3 Muon Decay Overview () Differential width: Phase space differential: d! = 1 E M dq dq = d 3 p' (! ) 3 E' d 3 k d 3 k' (! ) 3 " (! ) 3 "'! ( ) 4 # (4) p $ p'$k $ k' ( )

4 Calculational Challenges There s a spin averaged matrix element involved, requiring the use of some trace theorems. The phase space integral is not trivial. I ll provide you with an outline of how these are done, and leave it up to you to go through the details in H&M.

5 Phase Space Integration (1) We have: dq = d 3 p' (! ) 3 E' d 3 k (! ) 3 " d 3 k' (! ) 3 "' (! ) 4 # (4) ( p $ p'$k $ k' ) We can get rid of one of the three d 3 p/e by using: $ d 3 k! = $ d 4 k "(! )# k ( ) And eliminating the d 4 k integral against the 4d delta-function. This leads to: dq = 1 d 3 p' (! ) 5 E' d 3 k' "' (! ) 4 # E $ E'$"' (( ) ) ( )% p $ p'$k' This means E - E - ω > 0

6 Phase Space Integration () As was done for beta-decay, we replace: d 3 p'd 3 k' = 4!E' de'!"' d"'d cos# And evaluate delta-fct argument in muon restframe: ( ) =! m " me'"m#' +E'#' 1" cos$! ( p " p'"k' ) ( ( )) Recall: primed variables are from second W vertex.

7 Spin Average Matrix Element We neglect the mass of the electron and neutrinos. And use the trace theorem H&M (1.9) to arrive at: M = 1! M = 64G k " p' spins ( )( k'"p)

8 Muon Restframe To actually do the phase space integral, we go into the muon restframe where we find: kp =(k+p ) = (p-k ) = m - pk = m - m ω Mass of e and nu are ignored Mass nu is ignored k p = ω m 4-mom. cons. And as a result we get: Mu restframe M = 3G ( m! m"' )m"'

9 Putting the pieces together and doing the integration over cosθ, the opening angle of e and its anti-neutrino, we arrive at: d! = G de'd#'m# 3 ( m $ #' ) " 1 m $ E'% #'% 1 m 0 % E'% 1 m The inequality come from the requirement that cosθ is physical. And are easily understood from the allowed 3-body phasespace where one of the 3 is at rest.

10 Electron Energy Spectrum Integrate over electron antineutrino energy: d! de' = d! de' = G 0, E', 1 m 1 m % 1 m$e' ( ) " 3 d#'m# m $ #' G " 3 m E' ( 3$ 4E' ' m & ) + * The spectrum can be used to test V-A. This is discussed as measurement of Michel Parameters in the literature.

11 Michel Parameters It can be shown that any 4-fermion coupling will lead to an electron spectrum like the one we derived here, once we allow a Michel Parameter ρ, as follows: 1! ρ=0 for (V-A)x(V+A),S,P; ρ=1 for T ρ=0.75 for (V-A)x(V-A) x = E e m µ d! dx =1x * 1" x + 3 # $ & 4 % 3 x "1 '-, )/ + (. Measured Value:! µ = ± ! " = ± With polarized muon beams and measurement of electron polarization, other Michel Parameters come into play.

12 Total Decay Width of Muon Integrate over electron energy: 1! = 1 " = G m % # 3 m de' E' ' 3 $ 4E' ( + * & m ) 0! = G m 5 19# 3 µ! " e! # e # µ $! " e! # e # $ $! " µ! # µ # $ Note: Comparing muon and tau decays, as well as tau decays to electron and muon, allows for stringent tests of lepton universality to better than 1%.

13 Pion Decay u W - µ! " u p ( ) d V ud! "! k ( ) Leptonic vertex is identical to the leptonic current vertex in muon decay. Hadronic vertex needs to be parametrized as it can NOT be treated as a current composed of free quarks.

14 Parametrization of Hadronic Current Matrix element is Lorentz invariant scalar. Hadronic current must be vector or axial vector Pion is spinless Q is the only vector to construct a current from. The current at the hadronic vertex thus must be of the form: q µ f! (q ) = q µ f! (m! ) = q µ f! However, as q = m π = constant, we refer to f π simply as the pion decay constant. All other purely leptonic decays of weakly decaying mesons can be calculated in the same way. There are thus decay constants for B 0, B s0, D +, D s+, K +,etc.

15 Aside: This sort of parametrization is reused also when extrapolating from semileptonic to hadronic decays at fixed q E.g. Using B -> D lnu to predict B -> D X where X is some hadron. This is crude, but works reasonably well in some cases.

16 Matrix Element for Pion Decay M = GV ud [ ( )] ( p µ + k µ ) f! u ( p)" ( µ 1# " 5 )v k => => Now use the Dirac Equation for muon and neutrino: u ( p) ( p µ! µ " m ) µ = 0 u ( p) p µ! ( µ 1"! 5 )v( k) = u ( p)m ( µ 1"! 5 )v ( k) k µ! µ v( k) = 0 u ( p)k µ! ( µ 1"! 5 )v( k) = 0 Note: this works same way for any av+ba. [ ( )v( k) ] => M = G m µ f! u ( p) 1" # 5

17 Trace and Spin averaging The spin average matrix element squared is then given by: M G = V ud f! m µ Tr p µ " µ + m µ M = 4G V ud f! m µ [( )( 1# " 5 )k µ " ( µ 1 + " 5 )] p $ k ( ) You can convince yourself that this trace is correct by going back to H&M (6.19), (6.0). The only difference is the + sign. This comes from pulling a gamma matrix past gamma5.

18 Going into the pion restframe We get: p! k = E" # pk = "( E + ") Where we used that muon and neutrino are back to back in the pion restframe.

19 Pion leptonic decay width Putting it all together, we then get: d! = 1 M d 3 p m " " ( ) 3 E d 3 k ( " ) 3 # " ( ) 4 $ ( q % p % k)! = G V ud " f " m µ ( ) m " d 3 p d 3 & k E # $ ( m % E %# )$ 3 " ( ) ( k + p)# E + # ( ) Energy conservation 3-momentum conservation Use this to kill int over d 3 p

20 Pion leptonic width I ll spare you the details of the integrations. They are discussed in H&M p.65f The final result is:! = G V ud 8" $ f " m µ m " 1# m ' µ & ) % m " ( Helicity Suppression

21 Helicity Suppression The pion has spin=0. Angular momentum is conserved. Electron and anti-neutrino have same helicity. However, weak current does not couple to J=0 electron & antineutrino pair. Rate is suppressed by a factor:! " = G V ud 8" $ f " m 3 " 1# m ' µ & ) % m " ( * m µ m " m µ m!! µ = G 19" m 5 3 µ Helicity suppression

22 Experimentally As the pion decay constant is not known, it is much more powerful to form the ratio of partial widths: ( ) ( ) = m e m µ! " # $ e # % e! " # $ µ # % µ & ( ' m " # m e m " # m µ ) + * =1.33,10 #4 Experimentally, we find: ( ) x 10-4 Aside: Theory number here includes radiative corrections!!! I.e., this is not just the mass ratio as indicated!!!

23 Experimental Relevance We ve encountered this a few times already, and now we have actually shown the size of the helicity suppression, and where it comes from. ( )! " # $ e # % e ( ) = m e! " # $ µ # % µ m µ & ( ' m " # m e m " # m µ Accordingly, pion decay produces a rather pure muon neutrino beam, with the charge of the pion determining neutrino or anti-neutrino in the beam. ) + * = 1.3, 10 #4

24 Origin of Helicity Suppression Recap The muon mass entered because of the vector nature of the leptonic current. Either V or A or some combination of av+ba will all lead to helicity suppression. In particular, a charged weak current with S,P, or T instead of V,A is NOT consistent with experiment. In addition, we used: Neutrinos are massless Electron-muon universality

25 Window for New Physics via leptonic decays Example B + decay u d V ud W - µ! " u p ( )! "! k ( )! = G V ub 8" f B m µ m B 1# m $ ' µ & % m ) B ( Helicity Suppression The smallness of V ub and muon mass allows for propagators other than W to compete, especially if they do not suffer from helicity suppression => e.g. charged Higgs

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