PHY 396 K. Solutions for problem set #11. Problem 1: At the tree level, the σ ππ decay proceeds via the Feynman diagram

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1 PHY 396 K. Solutions for problem set #. Problem : At the tree level, the σ ππ decay proceeds via the Feynman diagram π i σ / \ πj which gives im(σ π i + π j iλvδ ij. The two pions must have same flavor index i j,,., (N, but the amplitude and hence decay rate is the same for all flavors. Thus, Γ(σ any ππ (N Γ(σ π π (N d 3 p d 3 p M σ (π 6 (π 4 δ (4 (p + p p σ λv E E 4(N λ v 4π p d p M σ (π (E δ(e M σ EEp p (N λ v (N λ M σ. 4πM σ 8π (S. Problem : As discussed in class, scattering of fermions in the Yukawa theory proceeds by exchange of virtual scalar quanta between the fermions. For the problem at hand, the two fermions involved are of distinct types, so there is just one tree-level Feynman diagram, f (p, s f (p, s... which results in the scattering amplitude f (p, s f (p, s M(f + f f + f g g q M s ū(p, s u(p, s ū(p, s u(p, s (S. where q p p p p.

2 The un-polarized -particle scattering cross-section is given by ( dσ c.m. 64π Ec.m. s s ( g g 64π Ec.m. q Ms M s s s,s ū(p, s u(p, s where the spin sums on the second line evaluate according to eq. (: ū(p, s u(p, s (m + p p s,s ū(p, s u(p, s (S.3 s,s (m + E p cos θ 4m + p ( cos θ in the center-of-mass frame, ū(p, s u(p, s (m + p p (S.4 s,s (m + E p cos θ 4m + p ( cos θ in the center-of-mass frame. Also, in the center of mass frame, p p, q 0 0 and q ( cos θp. Substituting all these formulae into eq. (S.3, we arrive at the partial cross section ( dσ c.m. g g 64π E c.m. (4m + q (4m + q (M s + q. (S.5 Next, we calculate the total cross-section by integrating over scattered particles directions. To that end, we notice that in the center-of-mass frame, π d( cos θ π p dq (S.6 and therefore, σ tot g g 64π E c.m. g g 6πE c.m. π p 4p 0 dq (4m + q (4m + q (M s + q [ + (4m M s (4m M s M s (M s + 4p + m + m M s p log M s + 4p ] Ms (S.7

3 where is the solution of the kinematical relation p 4 E c.m. (m + m + (m m 4E c.m. E c.m. E + E m + p + m + p. (S.8 It remains to prove eq. ( we have used above to derive eqs. (S.4. We begin by evaluating a simpler spin sum, for an arbitrary constant spinor w: wu(p, s ( wu(p, s ( (wu(p, w α u α (p, s s ū(p, sw ū β (p, sw β s s α β ( w α u α (p, sū β (p, s ( p + m αβ w β α,β s w( p + mw. Next, we substitute w u(p, s and sum over the spin s : ū(p, s u(p, s ( ū(p, s ( p + mu(p, s ū α (p, s ( p + m αβ u β (p, s s α,β ( ( p + m αβ u β (p, s ū α (p, s ( p + m βα (S.0 α,β s tr ( ( p + m( p + m. This proves the first equality in eq. (; to prove the second equality, we need to evaluate the trace. There is a whole technology for evaluating various Dirac traces and we shall study it in January, but the trace we need is simple enough to calculate by inspection of explicit Dirac matrices. In the Weyl basis, ( p + m ( m p µ σ µ p µ σ, ( p + m µ m (S.9 ( m p ν σ ν p ν, (S. σ ν m hence tr ( ( p + m( p + m ( m + p µ p ν σ µ σ ν tr m + p µ p ν σ µ σ ν m tr( + p µ p ν tr(σ µ σ ν + σ µ σ ν. (S. For the σ matrices we have σ 0 σ 0 while σ i σ i are Pauli matrices, which are traceless 3

4 and satisfy tr(σ i σ j δ ij. Consequently, tr(σ µ σ ν tr( σ µ σ ν g µν (S.3 and hence the last line in eq. (S. evaluates to 4m + 4pp. In other words, tr ( ( p + m( p + m 4(m + pp 4(m + EE p p, (S.4 which proves the second equality in eq. (, Q.E.D. Problem 3(a: To lowest order in ĤI, out i ˆT ( in out T-exp i dt ĤI(t in i dt out ĤI(t in, (S.5 which for the problem at hand means e (p, s i ˆT e (p, s ie d 4 x A µ (x ( e +ip xū(p, s γ µ( e ipx u(p, s [ ie ū(p, s γ µ u(p, s ] d 4 x A µ (x e ip x ipx Ãµ(p p. (S.6 Note opposite signs between this formula and its analogue in the textbook. The difference is due to different sign conventions for the e: The textbook uses e < 0 while I (and most other people use e > 0. Problem 3(b: Strictly speaking, the electron scatters off the potential s source, which is basically a heavy particle e.g., an atomic nucleus or a system of particles. The static-source approximation arises when the source S is so heavy that its velocity in some frame is negligible both before and after the scattering event and its recoil un-observable. In this static-source frame, E S E S M S regardless of the p S and p S, thus conservation of the total energy of the electron plus the source implies the electron s energy conservation, E e E e. 4

5 Now consider eq. (4.79 of the textbook for the scattering cross-section; in the static-source frame, we have dσ (E e (M S v e d 3 p e (π 3 (E e d 3 p S (π 3 (M S M (π 4 δ (4 (p e + p S p e p S (E e v e d 3 p e (π 3 (E e M(e + S e + S (M S (πδ(e e E e. (S.7 Note that the amplitude M(e + S e + S here is normalized to e + S ˆT e + S M(e + S e + S (π 4 δ (4 (p e + p S p e p S, (S.8 but in terms of an electron scattering off a static potential rather than S it is more convenient to use M(e e normalized to e ˆT e M(e e (πδ(e e E e (S.9 (this is the normalization used in the problem. The relation between the two amplitudes follows from the relativistic normalization of the source particle s states, S S (E M S (π 3 δ (3 (p S p S. (S.0 Putting eqs. (S.8, (S.9 and (S.0 together, we obtain M(e + S e + S (E M S M(e e (S. and hence dσ (E e v e d 3 p e (π 3 (E e M(e e (πδ(e e E e. (S. Q.E.D. For the purpose of an actual calculation, we integrate eq. (S. over the magnitude p e of the final electron. This removes the remaining δ function and simplifies the rest of the kinematic 5

6 factors, the net result being ( dσ e 6π M(e e. (S.3 Problem 3(c: For the Coulomb source, we have Ã(q 0, Ã 0 (q Ze/q and hence M ( e(p, s e(p, s Ze q ū(p, s γ 0 u(p, s. (S.4 For non-relativistic electrons, ū(p, s γ 0 u(p, s m e ξ ξ, the scattering is spin-preserving and spin-independent, and ( dσ m e 4π ( Ze q α Z 4m ev 4 e sin 4 (θ/ where the second equality follows from q p e( cos θ (m e v e sin(θ/. Problem 3(d: (S.5 For the relativistic electrons, the Coulomb scattering is no longer spin-blind. For an un-polarized stream of initial electrons and a detector blind to the final electrons spins, we should sum M(e(p, s e(p, s over the final spin states s and average over the initial spin states s. Thus, According to eq. (3, ( dσ e M(e(p, 3π s e(p, s 3π ( Ze ū(p, s γ 0 u(p, s. q (S.6 ū(p, s γ 0 u(p, s 4(m e + E e E e + p e p e 8m e + 4p e( + cos θ, (S.7 which gives us the Mott s formula for relativistic Coulomb scattering, ( dσ m e + p e cos (θ/ 4π ( Ze q α Z 4m ev 4 e sin 4 (θ/ β e sin (θ/ γ e (S.8 where β e v e /c and γ e / β e. 6

7 It remains to prove eq. (3 for the spin sum. Proceeding exactly as in eqs. (S.9 and (S.0 of problem, we derive ū(p, s γ 0 u(p, s tr ( ( p + mγ 0 ( p + mγ 0. (3. Next, we observe that γ 0 ( p + mγ 0 γ 0 (Eγ 0 p γ mγ 0 Eγ 0 + p γ + m p + m (S.9 where p µ (+E, p. Consequently, tr ( ( p + mγ 0 ( p + mγ 0 tr ( ( p + m( p + m using eq. (S.4 4m + 4p p (3. 4m + 4E E + 4p p. Q.E.D. 7

Lecture 8 Feynman diagramms. SS2011: Introduction to Nuclear and Particle Physics, Part 2 2

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