IntroductiontoElementaryParticles Instructor’s Solution Manual

1 Introduction to Elementary ParticlesInstructor s Solution Manual29th August 2008 VAcknowledgments:I thank Robin Bjorkquist, who wrote and typeset many ofthe solutions in the first four chapters; Neelaksh Sadhoo, who typeset solu-tions from the first edition; and all those who sent me solutions or have tried to make every entry clear and accurate, but please: if you find anerror, let me know I will post errata on my web Set of Solutions for Introduction to Elementary Particles. David GriffithsCopyrightc 2005 WILEY-VCH Verlag GmbH & Co. KGaA, WeinheimISBN: 3-527-XXXXX-XVIC ontents1 Historical Introduction to the Elementary Particles12 Elementary Particle Dynamics93 Relativistic Kinematics174 Symmetries375 Bound States576 The Feynman Calculus797 Quantum Electrodynamics978 Electrodynamics and Chromodynamics of Quarks1479 Weak Interactions17110 Gauge Theories209 ContentsVII11 Neutrino Oscillations23312 What s Next237 AThe Dirac Delta Function247 Partial Set of Solutions for Introduction to Elementary Particles.

2 David GriffithsCopyrightc 2005 WILEY-VCH Verlag GmbH & Co. KGaA, WeinheimISBN: 3-527-XXXXX-X11 Historical Introduction to the Elementary ParticlesProblem an undeflected charged particle,qE=qvB= v= just a magnetic field,qvB=mv2R= qm=vBR= 15m; h= 10 22 MeV s;c= 108m/s;som= h2r0c=( hc2r0)1c2= =138 by a factor of 15m; h= 10 22 MeV s;c= 108m/s;me= MeV/c2. x p h2sopmin= h2r0=( hc2r0)1c= p2minc2+m2ec4= energy of an electron emitted in the beta decay of tritium is < 17 Historical Introduction to the Elementary ParticlesProblem =13[2(mN+m ) m ].mN= ;m = ;m = =13[2( ) ]= = by =13[2(m2K+m2K) m2 ]=13(4m2K m2 ).mK= ;m = =13[ 105]= 105 m = = by M =1232 1385= 153M M =1385 1533= : 151. M =M +151=1533+151=1684 =1672 by (a) n+ or +K0 + p+ K0; ++ 0; ++ ; 0+ +; + +; 0+K+ 0+K ; + K0; +K ; 0+ ; + 0; + 3(b)Kinematically allowed: n+ + ++ 0; 0+ +; + + 0+ ; + 0 Problem (a)With a strangeness of 3, the would have to go to( 0+K )or( + K0)to conserveSandQ.

3 But the Kcombination is too heavy(at least 1808 MeV/c2, whereas the is predicted see Problem to have a mass of only 1684).(b) cm;t=d/v=(5 10 3m)/(3 107m/s) =2 10 10s.(Actually,t= 10 10s.)Problem + = n+ 0 = + }3% lists a total of 30 meson types; in the first column is the particle name atthe time, in the second column the quoted mass (in MeV/c2), and in the thirdits current Historical Introduction to the Elementary Particlesmesonmassstatusexotic 138 K496KK31630deadyes 21340f0(1370)? 31275f1(1285)?K 1260K1(1270)?f1253f2(1270)?K 51150deadyes 11045a0(980)? 21040dead 11020 (1020) 5990deadyesK 1888K (892) 3885 (985)? 781 (782)mesonmassstatusexotic 755 (770) 2780f0(600)? 1720f0(600)? 4760deadyesK 1730dead 645dead 625dead 3597deadyes 556dead 549 2520dead 2440deadyes 1395dead 1330deadyes ABC317deadProblem the last column in Problem I count 7 exotic species, all of them nowdead.

4 Of the surviving particles (of course) none is quark (u):one(u u); 2 quarks (u,d):four(u u,u d,d u,d d);3 quarks (u,d,s):nine; 4 quarks (u,d,s,c):sixteen;5 quarks (u,d,s,c,b):twenty-five; 6 quarks (u,d,s,c,b,t) general formula fornflavors quark (u)= 1baryon (uuu);2 quarks (u,d)= 4baryons (uuu,uud,udd,ddd);3 quarks (u,d,s)= 10baryons (baryon decuplet).5 Fornquarks, we can haveall three quarks the same :nwaystwo the same, one different :n(n 1)waysall three different :n(n 1)(n 2)/6 ways.[For the third type of combination, divide by six to cover the equivalent per-mutations (uds=usd=dus=dsu=sud=sdu).]So the total isn+n(n 1)+n(n 1)(n 2)/6=n+n2 n+n(n 1)(n 2)/6=n6[6n+(n 1)(n 2)]=n6(6n+n2 3n+2)=n6(n2+3n+2)=n(n+1)(n+2) for four quarks we have20baryon types, for five quarks,35, and for sixquarks, hasC=3uddddcsccdddusc3 haveC=2uusdscudssscdds6 haveC=1ussdsssss10 haveC=061 Historical Introduction to the Elementary ParticlesProblem uc uu cc cu dc sd cd uc ds cd d3 haveC=1 3 haveC= 1u ss ud ss ds s10 haveC=0(includingc c)Problem qmesonmassyearu u 0(*) d + d 0(*) sK+ s (*) dD+ qmesonmassyearc sD+ c c(1S) bB+ bB+c62861998b b (1S) masses are in MeV/c2; (*) indicates that the particle is a combination ofdifferent quark ++ 12321951uus + ++ + + ++ccdcc + +ccccc ++ccc7qqqbaryonmassyearuub + 0b5624?

5 Ddb 0b57921995dsb bucb +cbdcb 0cbscb 0cbccb +ccbqqqbaryonmassyearubb 0bbdbb bbsbb bbcbb 0ccbbbb bbbBlank spaces indicate that the particle has not yet been found (2008).Problem : 3 336 62=946; actually 939; : 2 336+540 62=1150; actually 1193; error: : 2 336+540 62=1150; actually 1116; : 336+2 540 62=1354; actually 1318; and leptons:ccc(q= 1):e c c c(q=1):e+ccn(q= 23): u c c n(q=23):ucnn(q= 13):d c n n(q=13): dnnn(q=0): e n n n(q=0): e(The neutrinos could be switched, but this seems the most natural as-signment.)Mediators:ccc n n n(q= 1):W nnn c c c(q=+1):W+ccc c c c(q=0):Z0nnn n n n(q=0): }or vice versaGluons:We need matching triples of particles and antiparticles,81 Historical Introduction to the Elementary Particlesccn c c n(3 different orderings for each triple, so 9 possibilities);cnn c n n(3 different orderings for each triple, so 9 possibilities);leading to a total of 18 are at least four distinct answers, depending on the particle in question: Antiparticles (such as the positron)annihilatewith the correspondingparticle (the electron, in this case), and since there are lots of electronsin the lab, positrons don t stick around long enough to have any role inordinary chemical processes.

6 But if you could work in a total vacuumyou could make atoms and molecules of antimatter, and all of chemistrywould proceed just the same as with ordinary matter. Most elementary particles (such as muons, pions, and intermediate vec-tor bosons) are intrinsicallyunstable; they disintegrate spontaneouslyin a tiny fraction of a second not long enough to do any serious chem-istry. You can make short-lived exotic atoms , with (say) muons in orbitaround the nucleus instead of electrons. Some of these systems last longenough to do spectroscopy. Neutrinosinteractsofeeblywith matter that they have no impact onchemistry, even though we are in factbathedin them all the time. Quarks are the basic constituents of protons and neutrons, so in an in-direct way theydoplay a fundamental role in chemistry. (And gluonsplay a fundamental role in holding the nucleus together.)

7 But becauseofconfinement, they do not occur as free particles, only in compositestructures, so they don t act as individuals .Partial Set of Solutions for Introduction to Elementary Particles. David GriffithsCopyrightc 2005 WILEY-VCH Verlag GmbH & Co. KGaA, WeinheimISBN: 3-527-XXXXX-X92 Elementary Particle DynamicsProblem ;Fe=e24 e0r2 FgFe=4 e0Gm2e2= 10 are built on; eight are built on; one is Elementary Particle DynamicsProblem and momentum are conserved at each vertex. Thus, for the virtualphoton in the horizontal diagramE=2mec2andp=0= m=2meandv=0and in the vertical diagram,E=0 andp=0= m=0 andv= (a) n+ would be favoredkinematically, but sincetwo squarkshave to be converted, it requires an extraW (hence two extra weakvertices), and this makes it much less likely. + is , go to + ; the branching ratio forn+ isless than 10 5.

8 (b)D0 K + +:Neithervertex crosses generations:sduuup+WcD0K-D0 + +:Onevertex crosses generations:duuup+WcD0-dpD0 +K+:Bothvertices cross generations:122 Elementary Particle Dynamicsuuu+WcD0-dpsKBecause weak vertices within a generation are favored,K + +ismostlikely,K++ , the branching ratios are: forK + +, for ++ , and forK++ .(c)Thebquark prefers to go toc(Vcb= , whereasVub= ), soBshould go +W_ZeeW_W+eeW_W+Problem (a)Impossible (charge conservation)(b)Possible,electromagnetic (c)Impossible (energy conservation)(d)Possible,weak(e)Possible ,electromagnetic(f)Impossible (muon number conservation)13(g)Possible,strong(h)Poss ible,weak(i)Impossible (baryon number conservation)(j)Possible,strong(k)Imposs ible (baryon and lepton number conservation)(l)Possible,strong(m)Possib le,strong(n)Impossible (charge conservation)(o)Possible,weak(p)Impossib le (charge conservation)(q)Possible,electromagnetic (r)Possible,weak(s)Possible,weak(t)Possi ble,strong(u)Possible,electromagnetic(v) Possible,weakProblem (a)K+ ++ + ; weak and electromagnetic interactions are involved:eWmnmnegee142 Elementary Particle Dynamics(b) + p+.

9 Weak and electromagnetic interactions are involved:suuugS+WuudpWProblem lifetime tells us this is an OZI-suppressed strong interaction. EvidentlytheBmeson must weighmore than halfthe (just as theDweighs more thanhalf the ). Thus theBmeson should weighmore than 4730 MeV/c2(and itdoes).Problem 'p+Here s a typical contributing , it is a strong , it isOZI-suppressed. We should expect a lifetimearound10 (a)ParticleXhas charge +1 and strangeness 0; it was presumably a proton.(b)K +p K0+K++ (strong); 0+ (weak); 0 0+ 0(weak), 0 + (electromagnetic), and both photons undergo pair produc-tion e++e (electromagnetic); 0 +p(weak).15 Problem pion seems most likely it requires only one weak vertex, with no genera-tion crossing, and it s light (hence kinematically favored):ddduup+uuuppW-Partial Set of Solutions for Introduction to Elementary Particles.

10 David GriffithsCopyrightc 2005 WILEY-VCH Verlag GmbH & Co. KGaA, WeinheimISBN: 3-527-XXXXX-X173 Relativistic KinematicsProblem {x = (x vt)= x / =x vtt = (t vc2x)= vt / =vt v2c2x}Adding these two equations:1 (x +vt )=x(1 v2c2)=x/ 2 x= (x +vt ). x = (x vt)= vc2x =vc2x v2c2tt = (t vc2x)= t / =t vc2x Adding these two equations:1 (t +vc2x )=t(1 v2c2)=t/ 2 t= (t +vc2x ).Also,y=y , andz=z . This confirms Eq. (a)From the Lorentz transformations (Eq. ),t = (t vc2x). tA tB = [tA tB vc2(xA xB)].If simultaneous inS(so thattA=tB), thentA =tB + vc2(xB xA).(b) =1 1 925=54;xB xA=4 km=4 Relativistic KinematicstA =tB + 5 4 3 5 cc 2( 4 103m)=tB +3 103m3 108m/s=tB +10 went on first; A came on10 5seconds (a)The dimension along the direction of motion undergoes length contrac-tion, but the other two dimensions do not. Thus, volumes transformaccording toV=V /.