Solution manual for Introduction to Elementary Particles by Griffiths D.J.

By Griffiths D.J.

In advent to undemanding debris, moment, Revised version, writer David Griffiths moves a stability among quantitative rigor and intuitive realizing, utilizing a full of life, casual kind. the 1st bankruptcy presents an in depth historic creation to the topic, whereas next chapters supply a quantitative presentation of the traditional version. A simplified advent to the Feynman ideas, in line with a "toy" version, is helping readers study the calculational thoughts with no the issues of spin. it's via obtainable remedies of quantum electrodynamics, the powerful and vulnerable interactions, and gauge theories. New chapters tackle neutrino oscillations and customers for physics past the traditional version. The publication features a variety of labored examples and plenty of end-of-chapter difficulties. a whole resolution guide is out there for teachers. Revised variation of a well-established textual content on basic particle physics With a couple of labored examples and lots of end-of-chapter difficulties is helping the coed to grasp the Feynman principles answer guide on hand for teachers

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B) c2 s = p2A + 2p A · p B + p2B = p2A + 2 E A EB − p A · pB c c + p2B 33 In the CM frame, p A · pB = −p2A , so p2A = E2 E2A − p2A =⇒ p2A = 2A − p2A = −p A · pB 2 c c and p2B = =⇒ EB2 E2 − p2B = 2B − p2A 2 c c EB = c p2B + p2A = p2B + E2A − p2A . c2 So  E c2 s = p2A + 2  A c c2 s + p2A − p2B − 2  2 2 E E p2B + 2A − p2A + 2A − p2A  + p2B c c E2A E =2 A 2 c c p2B + E2A − p2A c2 Squaring both sides, (c2 s + p2A − p2B )2 − E2A = 4 4 E✓ E✓ 4 2 4 2 2 2 2 2 2 A A ✓ ✓ ( c s + p ( p − p − p ) E ) E + 4 + 4 = B A A A A c2 c2 B ✓ c4 ✓ c4 (c2 s + p2A − p2B )2 4s ECM A = (s + m2A − m2B )c2 √ .

Now, p2A = m2A c2 , But p2B = m2B c2 , p2C = m2C c2 , p A · pB = E A EB − p A · pB . c c p A = 0, and E A = m A c2 , so m2C c2 = m2A c2 + m2B c2 − 2m A EB , 2m A EB = −(m2C − m2A − m2B )c2 , EB = (−m2C + m2A + m2B ) 2 c ; 2m A EC = so or (m2A − m2B + m2C ) 2 c 2m A (b) EB2 − p2B c2 = m2B c4 =⇒ p2B = c 2m A c = 2m A |p B | = (m2A + m2B − m2C )2 c2 4m2A m2B c2 EB2 − m2B c2 = − 2 c 4m2A 4m2A m4A + m4B + m4C + 2m2A m2B − 2m2A m2C − 2m2B m2C − 4m2A m2B λ(m2A , m2B , m2C ) ∴ | p B | = | pC | = c 2m A λ(m2A , m2B , m2C ) (c) The decay is kinematically forbidden if m A < m B + mC (not enough energy to produce the final particles, in CM frame).

20 (a) [σi , σj ] = σi σj − σj σi = δij + i But δij = δji , jik =− ijk . ijk σk − δji − i So [σi , σj ] = 2i jik σk ijk σk . , (iθσz )2 (iθσz )3 + +··· 2 3! θ4 θ3 θ5 θ2 + − · · · + iσz θ − + +··· = 1− 2 4! 3! 5! eiθσz = 1 + iθσz + = cos θ + iσz sin θ In particular eiπσz /2 = cos π/2 + iσz sin π/2 = iσz . (b) Similarly (all we need is σy2 = 1) eiθσy = cos θ + iσy sin θ, so U = cos 90◦ − i sin 90◦ σy = −iσy = −i 1 0 0 −1 1 0 1 0 = 0 1 U (θ) = e−iθ·σ /2 = 1 + −i θ·σ 2 + U = , 0 −i i 0 0 −1 1 0 = which is spin down.

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