physics 222 ucsd/225b ucsb
DESCRIPTION
Physics 222 UCSD/225b UCSB. Lecture 3 Weak Interactions (continued) muon decay Pion decay. Muon Decay Overview (1). Feynman diagram: Matrix Element:. G = effective coupling of a 4-fermion interaction. Structure of 4-fermion interaction is (V-A)x(V-A). - PowerPoint PPT PresentationTRANSCRIPT
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Physics 222 UCSD/225b UCSB
Lecture 3
• Weak Interactions (continued)• muon decay• Pion decay
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Muon Decay Overview (1)
• Feynman diagram:
• Matrix Element:
€
M =G
2u k( )γ μ 1− γ 5
( )u p( )[ ] u p'( )γ μ 1− γ 5( )v k'( )[ ]
€
ν ≡ν k'( )
€
e− ≡ u p'( )
€
ν ≡u k( )
€
μ−≡u p( )
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.
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Muon Decay Overview (2)
• Differential width:
• Phase space differential:
€
dΓ =1
2EM
2dQ
€
dQ =d3 p'
2π( )32E '
d3k
2π( )32ω
d3k'
2π( )32ω'
2π( )4δ (4 ) p − p'−k − k '( )
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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.
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Phase Space Integration (1)• We have:
• We can get rid of one of the three d3p/E by using:
• And eliminating the d4k integral against the 4d delta-function. This leads to:€
d3k
ω= d4k θ ω( )δ k 2
( )∫∫€
dQ =d3 p'
2π( )32E '
d3k
2π( )32ω
d3k'
2π( )32ω'
2π( )4δ (4 ) p − p'−k − k '( )
€
dQ =1
2π( )5
d3 p'
2E '
d3k'
2ω'2π( )
4θ E − E '−ω'( )δ p − p'−k '( )
2
( )
This means E - E’- ’ > 0
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Phase Space Integration (2)
• As was done for beta-decay, we replace:
• And evaluate delta-fct argument in muon restframe:€
d3 p'd3k '= 4πE '2 dE '2πω'2 dω'dcosθ
€
δ p − p'−k '( )2
( ) = δ m2 − 2mE '−2mω'+2E 'ω' 1− cosθ( )( )
Recall: primed variables are from second W vertex.
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Spin Average Matrix Element
• We neglect the mass of the electron and neutrinos.
• And use the trace theorem H&M (12.29) to arrive at:
€
M2
=1
2M
2
spins
∑ = 64G2 k ⋅ p'( ) k'⋅p( )
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Muon Restframe
• To actually do the phase space integral, we go into the muon restframe where we find:
• 2 kp’ =(k+p’)2 = (p-k’)2 = m2 - 2pk’ = m2 - 2m ’
• k’p = ’m
• And as a result we get:
Mass of e and nuare ignored
4-mom. cons.
Mass nu is ignored
€
M2
= 32G2 m2 − 2mω'( )mω'
Mu restframe
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Putting the pieces together and doing the integration over cos, the opening angle of e and
its anti-neutrino, we arrive at:
€
dΓ =G2
2π 3dE 'dω'mω m − 2ω'( )
1
2m − E '≤ ω'≤
1
2m
0 ≤ E '≤1
2m
The inequality come from the requirement that cos is physical. And are easily understood from the allowed 3-body phasespace where one of the 3 is at rest.
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Electron Energy Spectrum
• Integrate over electron antineutrino energy:
€
dΓ
dE '=
G2
2π 3dω'mω m − 2ω'( )
1
2m−E '
1
2m
∫
dΓ
dE '=
G2
2π 3m2E '2 3 −
4E '
m
⎛
⎝ ⎜
⎞
⎠ ⎟
0 ≤ E '≤1
2m
The spectrum can be used to test V-A.This is discussed as “measurement of Michel Parameters” in the literature.
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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” , as follows:
=0 for (V-A)x(V+A),S,P; =1 for T =0.75 for (V-A)x(V-A)• With polarized muon beams and measurement of electron
polarization, other “Michel Parameters” come into play.
€
x =2Ee
mμ
€
1
Γ
dΓ
dx=12x 2 1− x +
2
3ρ
4
3x −1
⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥
€
μ =0.7509 ± 0.0010
ρ τ = 0.745 ± 0.008
Measured Value:
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Total Decay Width of Muon
• Integrate over electron energy:
€
Γ=1
τ=
G2
2π 3m2 dE ' E '2 3−
4E '
m
⎛
⎝ ⎜
⎞
⎠ ⎟
0
1
2m
∫
Γ =G2m5
192π 3
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%. €
μ−→ e−ν eν μ
τ − → e−ν eν τ
τ − → μ−ν μν τ
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Pion Decay
€
ν ≡ν k( )
€
μ−≡u p( )
€
u
€
d
W-
Vud
• 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.
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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:
• However, as q2 = m2 = constant, we refer to f 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 B0, Bs
0, D+, Ds+, K+,etc.
€
qμ fπ (q2) = qμ fπ (mπ2 ) = qμ fπ
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Aside:
• This sort of parametrization is “reused” also when extrapolating from semileptonic to hadronic decays at fixed q2
– 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.
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Matrix Element for Pion Decay
€
M =GVud
2pμ + k μ
( ) fπ u p( )γ μ 1− γ 5( )v k( )[ ]
Now use the Dirac Equation for muon and neutrino:
€
u p( ) pμγ μ − mμ( ) = 0
u p( ) pμγ μ 1− γ 5( )v k( ) = u p( )mμ 1− γ 5
( )v k( )
k μγ μv k( ) = 0
u p( )k μγ μ 1− γ 5( )v k( ) = 0
=>
=>
=>
€
M =G
2mμ fπ u p( ) 1− γ 5
( )v k( )[ ]
Note: this works same way for any aV+bA .
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Trace and Spin averaging• The spin average matrix element squared is
then given by:
• You can convince yourself that this trace is correct by going back to H&M (6.19), (6.20). The only difference is the “+” sign. This comes from “pulling” a gamma matrix past gamma5.
€
M2
= Vud
2 G2
2fπ
2mμ2Tr pμγ μ + mμ( ) 1− γ 5
( )k μγ μ 1+ γ 5( )[ ]
M2
= 4G2 Vud
2fπ
2mμ2 p ⋅k( )
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Going into the pion restframe
• We get:
• Where we used that muon and neutrino are back to back in the pion restframe.
€
p ⋅k = Eω − pk = ω E + ω( )
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Pion leptonic decay width
• Putting it all together, we then get:
€
dΓ =1
2mπ
M2 d3 p
2π( )32E
d3k
2π( )32ω
2π( )4δ q − p − k( )
Γ =G2 Vud
2fπ
2mμ2
2π( )22mπ
d3 p
E
d3k
ω∫ δ mπ − E −ω( )δ 3( ) k + p( )ω E + ω( )
Energy conservation
3-momentum conservationUse this to kill int over d3p
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Pion leptonic width
• I’ll spare you the details of the integrations. They are discussed in H&M p.265f
• The final result is:
€
Γ=G2 Vud
2
8πfπ
2mμ2mπ 1−
mμ2
mπ2
⎛
⎝ ⎜
⎞
⎠ ⎟
2
Helicity Suppression
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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:
€
Γ =G2 Vud
2
8πfπ
2mπ3 1−
mμ2
mπ2
⎛
⎝ ⎜
⎞
⎠ ⎟
2
×mμ
2
mπ2
Γμ =G2
192π 3mμ
5
€
mμ2
mπ2
Helicity suppression
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Experimentally
• As the pion decay constant is not known, it is much more powerful to form the ratio of partial widths:
€
Γ −→ e−ν e( )
Γ π − → μ−ν μ( )=
me2
mμ2
mπ2 − me
2
mπ2 − mμ
2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
2
=1.233×10−4
Experimentally, we find: (1.230 +- 0.004) x 10-4
Aside: Theory number here includes radiative corrections !!!I.e., this is not just the mass ratio as indicated !!!
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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.
Accordingly, pion decay produces a rather pure muon neutrino beam, with the charge of the pion determining neutrino or anti-neutrino in the beam.
Γ −→ e−ν e( )
Γ π − → μ −ν μ( )=
me2
mμ2
mπ2 − me
2
mπ2 − mμ
2
⎛
⎝⎜
⎞
⎠⎟
2
= 1.23 ×10−4
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Origin of Helicity SuppressionRecap
• 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
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Window for New Physics via leptonic decays
Γ =G2 Vub
2
8πfB
2mμ2mB 1−
mμ2
mB2
⎛
⎝⎜⎞
⎠⎟
2
Helicity Suppression
€
ν ≡ν k( )
€
μ−≡u p( )
€
u
€
d
W-
Vud
ExampleB+ decay
The smallness of Vub
and muon mass allows for propagators other than W to compete, especially if theydo not suffer from helicity suppression => e.g. charged Higgs
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