1 The muon decay in the Fermi theory

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1 Quantum Field Theory-I Problem Set n. 9 UZH and ETH, HS-015 Prof. G. Isidori Assistants: K. Ferreira, A. Greljo, D. Marzocca, A. Pattori, M. Soni Due: The muon decay in the Fermi theory Consider the Fermi Lagrangian for muon decay L F = G F [ ψνµ γ α (1 γ 5 )ψ µ [ ψe γ α (1 γ 5 )ψ νe + h.c., (1) where G F = GeV is the Fermi constant. This effective Lagrangian arises as a low energy description of the Standard Model interactions of fermions with the W boson: L SM W = g [ Wα J α + W αj α, J α = ψ l γ α (1 γ 5 )ψ νl, J α = () ψ νl γ α (1 γ 5 )ψ l. l=e,µ I. Derive the Feynman rules for the interactions above. In order to derive the Feynman rules for the interactions above we can proceed by analogy to the QED case, or explicitly compute the Wick contractions for some process. We are going to do this for the next question. However, a simple and effective way to derive the Feynman rules for interaction terms is to take functional derivatives of the interaction Lagrangian with respect to the fields, and taking then the Fourier transform (fixing all momenta incoming). The justification for this method arises in the functional integral formalism which will be studied in the QFT-II course. As an example let us consider the λφ 4 theory with L I = λ 4! φ4. The Feynman rule can be obtained by 4 il φ 4 I = iλ, which corresponds to the one derived in the lectures. The Feynman rule for the Fermi interaction in eq. (1) is given by (note that for fermions the order in which field derivatives are taken is relevant) by removing the delta function: l=e,µ ψµ(p a 1 ) ψ ν b µ (p ) ψν d e (p 3 ) ψ e(p c 4 ) il F = = d 4 ze iz(p 1+p +p 3 +p 4 ) ( i) G F [γ α (1 γ 5 ) ba [γ α (1 γ 5 ) cd = = (π) 4 4 (p 1 + p + p 3 + p 4 )( i) G F [γ α (1 γ 5 ) ba [γ α (1 γ 5 ) cd Similarly, the Feynman rules for the weak interactions are i G F [γ α (1 γ 5 ) ba [γ α (1 γ 5 ) cd. (4) (3) ψν b l ψ l a W µ il SM W i g [γµ (1 γ 5 ) ab, ψ b l ψ a ν l il SM W i g W µ [γµ (1 γ 5 ) ab. (5) 1

2 Figure 1: Feynman diagram for muon decay in the Fermi effective theory. II. Compute the Feynman amplitude M for µ (p, s 1 ) e (k, s ) (q 1, s 3 ) (q, s 4 ) in the two theories at leading order. We have to compute (π) 4 4 (p k q 1 q )im = = out e (k, s ) (q 1, s 3 ) (q, s 4 ) µ (p, s 1 ) in = e (k, s ) (q 1, s 3 ) (q, s 4 ) it µ (p, s 1 ) ( { [ } ) = 0 e (k, s ) (q 1, s 3 ) (q, s 4 ) T exp i d 4 zl int (z) µ (p, s 1 ) 0 connected, amputated In the Fermi theory, at O(G F ), this is given by the Feynman diagram of Fig. 1, i.e. { } 0 e (k, s ) (q 1, s 3 ) (q, s 4 ) T i d 4 zl F (z) µ (p, s 1 ) 0 = = d 4 z e iz(p k q 1 q ) i G F [γ α (1 γ 5 ) ba [γ α (1 γ 5 ) cd ū s 3 ν (q µ,b 1)u s 1 µ,a(p)ū s e,c(k)v s 4 ν (q e,d ) = that is = (π) 4 4 (p k q 1 q ) i G F (ū s 3 µ,a(p)) (ū s (q )), im F = i G F ū s 3 µ,a(p) ū s (q ). (8) Using the Feynman rule derived above, and those for the initial- and final-state spinors, we would have obtained the same result. Let us now compute the same amplitude in the Standard Model, as described by the interaction terms of eq. (). We use directly the Feynman rules (5), and the W boson propagator (in momentum space) ) D µν W (q) = i (g µν q m W + iɛ qµ q ν. (9) m W The Feynman diagram for the process is shown in Fig.. It is easy to convince ourselves that this is the only diagram contributing to this process at O(g ). The amplitude is given by (q = p q 1 = k + q ) g im W = i ūs 3 µ,a(p) = i g 8m W i q m W + iɛ m W. (6) (7) ) (g αβ qα q β g i ūs e (k)γ β (1 γ 5 )v s 4 (q ) m W gαβ q α q β q m W + iɛ ūs 3 µ,a(p) ū s e (k)γ β (1 γ 5 )v s 4 (q ). (10)

3 Figure : Feynman diagram for muon decay in the Standard Model. We see that the structure of this amplitude is very similar to the one in the Fermi theory, eq. (8). Moreover, the momentum q exchanged in the W propagator is necessarily of the order of the muon mass m µ GeV, while the W mass is m W 80.4 GeV. We can thus safely expand eq. (10) for q m W, in particular the term related to the W propagator becomes m W gαβ q α q β q m W + iɛ q m W g αβ (11) providing im W i g 8m W ū s 3 µ,a(p) ū s (q ). (1) Comparing with eq. (8) we can match the Fermi coupling to the one obtained in the Standard Model (at leading order): G F = g. (13) 8m W III. Compute the spin-avaraged amplitude squared from the Fermi theory. We have to compute 1 MM = G F 4 s 1,...s 4 s 1,...s 4 Tr [ u s 3 (q 1 )ū s 3 µ,a(p)ū s 1 µ,a(p)γ β (1 γ 5 ) Tr [ u s e (k)ū s (q ) v s 4 (q )γ β (1 γ 5 ) = =4G F Tr [/q 1 γ α P L (/p + m µ )γ β P L Tr [(/k + m e )γ α P L/q γ β P L = =4G F Tr [/q 1 γ α /pγ β P L Tr [/kγ α/q γ β P L, where in the first step we used s ua (p)ū s (p) = /p + m, assumed neutrinos as massless fermions, and defined P L,R = (1 γ 5 )/, and in the second step we cycled the chirality projectors to bring them together using γ µ P L = P R γ µ and viceversa. The terms proportional to the masses contain P L P R, which vanishes. In order to proceed we need the trace (14) Tr [ γ µ γ α γ ν γ β P L = ( g µα g νβ + g µβ g να g µν g αβ + iɛ µανβ), (15) 3

4 and we can notice that the only surviving term involving the Levi Civita tensor is the one where it is contracted with itself (since it is antisymmetric in αβ while the other part is symmetric). We thus need also ɛ αβµν ɛ αβρσ = ( µ ρ ν σ µ σ ν ρ). Using this we get M = 1 ) MM = 16G F (q α1 p β + q β1 p α (q 1 p)g αβ + iɛ µανβ q 1µ p ν s 1,...s 4 (k α q β + k β q α (kq )g αβ + iɛ ρασβ k ρ q σ ) = =16G F [(q 1 k)(pq ) + (q 1 q )(pk) + ( + 4 )(q 1 p)(kq ) +i ( 1) ɛ αβµν ɛ αβρσ q 1µ p ν k ρ q σ = [ =16G F (q1 k)(pq ) + (q 1 q )(pk) ( )( ρ µ σ ν σ µ ρ)q ν 1µ p ν k ρ q σ = =16G F [(q 1 k)(pq ) + (q 1 q )(pk) + (q 1 k)(pq ) (q 1 q )(pk) = =64G F (q 1 k)(pq ). (16) IV. Compute the differential decay width in the electron energy E e, dγ/de e, for m e 0. The fully differential decay width is given by dγ = 1 ( ) ( ) ( ) d M (π) q 1 d 3 q d 3 k (p k q 1 q ) = m µ (π) 3 E 1 (π) 3 E (π) 3 E e ( ) ( ) ( ) = 4G F d m µ (π) k µp 5 ν q µ 1 q ν 4 3 q 1 d 3 q d 3 k (p k q 1 q ). E 1 E E e (17) We have to integrate over the phase space. Let us start with the two neutrino momenta. The relevant integral is d I µν 3 q 1 d 3 q (q) = q µ 1 q ν 4 (q q 1 q ), (18) E 1 E where the result of the integration can only depend on q = p k. This integral is a rank two symmetric tensor (we can exchange 1 and the integral remains the same) so it can only be of the form I µν (q) = g µν A(q ) + q µ q ν B(q ), (19) from which follows g µν I µν = 4A(q ) + q B(q ), q µ q ν I µν = q A(q ) + (q ) B(q ). (0) Since we assume neutrinos to be massless q1 = q = 0, and q = (p k) = (q 1 + q ) = (q 1 q ). Let us now extract A(q ) and B(q ). We evaluate g µν I µν = q d 3 q 1 d 3 q 4 (q q 1 q ) = 4A(q ) + q B(q ), (1) E 1 E Since this is a scalar quantity we can compute it in any reference frame, we choose the c.o.m. frame of the two neutrinos, in which q 1 = q and the energy of the two neutrinos is ω = E 1 = q 1 = q = E. Therefore 4A(q ) + q B(q ) = q d 3 q 1 ω (q 0 ω) = q ω dωdω 1 ω (ω q 0/) = q π, () 4

5 [MeV -1 1 dγμ dee Γμ E e [MeV Figure 3: Normalized differential decay width of the muon with respect to the electron energy, in MeV. and similarly q µ q ν I µν = q4 4 ω dωdω ω 1 (ω q0 /) = q 4 π = q A(q ) + (q ) B(q ), (3) where in the first step we used (qq 1 ) = (qq ) = (q 1 q ) = q /. Solving these two equations for A and B we get finally I µν (q) = π ( g µν q + q µ q ν). (4) 6 Recalling the definition (18) and substituting this result in eq. (17) we get dγ = 4G F m µ (π) k π ( µp 5 ν g µν q + q µ q ν) d 3 k = π 6 E e 3 G ( F (kp)q + (kq)(pq) ) d 3 k. (5) m µ (π) 5 E e In order to integrate now over the electron momenta we go in the rest frame of the muon, in which p = (m µ, 0), k = q 1 q = q and q 0 = m µ E e : (kp) = m µ E e, (kq) = E e (m µ E e ) + k = E e m µ m e, (pq) = m µ m µ E e, q = (m µ E e ) k = (m µ E e ) (E e m e) = m µ + m e m µ E e, (6) where we used k = E e m e. Using also d 3 k = k d k dω = k E e de e dω and integrating over the solid angle Ω we get that is dγ = π 3 G F m µ (π) 5 ( (kp)q + (kq)(pq) ) 4π k de e, (7) dγ = G ( F E de e 1π 3 e m e Ee (m µ + m e m µ E e ) + (E e m µ m e)(m µ E e ) ). (8) In the limit m e = 0 this simplifies to (see Fig. 3) dγ de e = G F 4π 3 m µe e ( m µ 4 ) 3 E e. (9) 5

6 V. Setting m e = 0 compute the total width of the muon. Before doing the last integration in de e we need to fix the correct boundaries in the electron energy, which are given by energy and momentum conservation. Let us start by recognising that the muon neutrino energy is E 1 = q 1 = k + q = Ee + E + E e E cos θ [ E e E, E e + E. (30) Energy conservation, i.e. (m µ E e E 1 E ), fixes that is E e E E 1 = m µ E e E E e + E (31) E e E + E e + E m µ (E e + E ). (3) The right-hand side provides a lower bound E e m µ / E while the left-hand side gives an upper bound, in the limit E = 0 we get E e m µ /. Since the same argument can be repeated for E we get E m µ /, that is E e 0. We can thus integrate eq. (9) in de e from 0 to m µ /: Γ µ = 1 τ µ = G F m5 µ 19π 3. (33) The functional dependence of the total width could be also easily be argued on dimensional grounds. The coupling G F has dimensions E, the width depends on G F, that is a dimension E 4. However the width has the dimension of an energy. Since the muon mass is the only relevant scale to the process (neglecting the electron mass) then we necessarily need a factor m 5 µ in order to restore the correct mass dimensions. The measured muon lifetime is approximately τ µ s. Given the muon mass m µ GeV and the conversion factor 1GeV s 1 we can obtain the Fermi constant ( ) π 3 1/ G F = GeV. (34) τ µ m 5 µ 6