Introduction to Elementary Particles Instructor s Solution Manual. 29th August 2008

Transcript

1 Introduction to Elementary Particles Instructor s Solution Manual 9th August 8

2 V Acknowledgments: I thank Robin Bjorkquist, who wrote and typeset many of the solutions in the first four chapters; Neelaksh Sadhoo, who typeset solutions from the first edition; and all those who sent me solutions or suggestions. I have tried to make every entry clear and accurate, but please: if you find an error, let me know griffith@reed.edu. I will post errata on my web page

3 VI Contents Historical Introduction to the Elementary Particles Elementary Particle Dynamics 9 3 Relativistic Kinematics 7 4 Symmetries 37 5 Bound States 57 6 The Feynman Calculus 79 7 Quantum Electrodynamics 97 8 Electrodynamics and Chromodynamics of Quarks 47 9 Weak Interactions 7 Gauge Theories 9

4 Contents VII Neutrino Oscillations 33 What s Next 37 A The Dirac Delta Function 47

5 Historical Introduction to the Elementary Particles Problem. For an undeflected charged particle, qe qvb v E B. With just a magnetic field, qvb m v R q m v BR E B R. Problem. r 5 m; h 6.58 MeV s; c 3. 8 m/s; so m h hc r c r c 98.7 MeV/c. Observed m π 38 MeV/c. Off by a factor of.4. Problem.3 r 5 m; h 6.58 MeV s; c 3. 8 m/s; m e.5 MeV/c. x p h E min so p min h r hc c r p min c + m e c MeV MeV/c. The energy of an electron emitted in the beta decay of tritium is < 7 kev.

6 Historical Introduction to the Elementary Particles Problem.4 m Λ 3 m N + m Ξ m Σ. m N 938.9; m Ξ 38.; m Σ 9.5. So m Λ MeV/c. Observed m Λ 5.7 MeV/c. Off by.7%. Problem.5 m K + m K m π 4m K 3 m π. m η 3 m K ; m π m η m η MeV/c. Actually m η MeV/c. Off by 3.5%. Problem.6 M M Σ M Σ M Ξ Average: 5. M Ω M Ξ MeV/c. Actually M Ω 67 MeV/c. Off by.7%. Problem.7 a n + π or Σ + K Σ + p + K ; Σ + + π ; Σ + + η; Σ + π + ; Λ + π + ; Ξ + K + Ξ Σ + K ; Σ + K ; Λ + K ; Ξ + π ; Ξ + π ; Ξ + η

7 3 b Kinematically allowed: n + π Σ + Σ + + π ; Σ + π + ; Λ + π + Ξ Ξ + π ; Ξ + π Problem.8 a With a strangeness of 3, the Ω would have to go to Ξ + K or Ξ + K to conserve S and Q. But the ΞK combination is too heavy at least 88 MeV/c, whereas the Ω is predicted see Problem.6 to have a mass of only 684. b About.5 cm ; t d/v 5 3 m/3 7 m/s s. Actually, t.8 s. Problem.9 Σ + Σ } p n + Ξ Ξ % difference. Problem. Roos lists a total of 3 meson types; in the first column is the particle name at the time, in the second column the quoted mass in MeV/c, and in the third its current status.

8 4 Historical Introduction to the Elementary Particles meson mass status exotic π 38 π K 496 K K 3 63 dead yes χ 34 f 37? κ 3 75 f 85? K 6 K 7? f 53 f 7? K5 5 dead yes χ 45 a 98? κ 4 dead κ φ ψ 5 99 dead yes K 888 K 89 ϕ η 985? ω 78 ω78 meson mass status exotic ρ 755 ρ77 ρ 78 f 6? ρ 7 f 6? ψ 4 76 dead yes K 73 dead δ 645 dead α 65 dead ψ dead yes ζ 556 dead η 549 η ϕ 5 dead ψ 44 dead yes ϕ 395 dead ψ 33 dead yes ω ABC 37 dead Problem. From the last column in Problem. I count 7 exotic species, all of them now dead. Of the surviving particles of course none is exotic. Problem. quark u: one uū; quarks u, d: four uū, ud, dū, dd; 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: thirty-six. The general formula for n flavors is n. Problem.3 quark u baryon uuu; quarks u, d 4 baryons uuu, uud, udd, ddd; 3 quarks u, d, s baryons baryon decuplet.

9 5 For n quarks, we can have all three quarks the same : two the same, one different : all three different : n ways nn ways nn n /6 ways. For the third type of combination, divide by six to cover the equivalent permutations uds usd dus dsu sud sdu. So the total is n + nn + nn n /6 n + n n + nn n /6 n 6n + n n 6 n 6n + n 3n + 6 n n + 3n + 6 nn + n +. 6 Thus for four quarks we have baryon types, for five quarks, 35, and for six quarks, 56. Problem.4 uuu uuc ucc ccc uud udc dcc has C 3 udd ddc scc ddd usc 3 have C uus dsc uds ssc dds 6 have C uss dss sss have C

10 6 Historical Introduction to the Elementary Particles Problem.5 uū cū u c c c ud c s d c dū cd s c dd 3 have C 3 have C u s sū d s sd s s have C including c c Problem.6 q q meson mass year uū π * ud π dd π * u s K d s K s s η * cū D cd D q q meson mass year c s D s c c η c S u b B d b B s b Bs c b B c b b ΥS All masses are in MeV/c ; * indicates that the particle is a combination of different quark states. qqq baryon mass year qqq baryon mass year uuu uud p udd n ddd 3 95 uus Σ uds Λ dds Σ uss Ξ dss Ξ sss Ω uuc Σ ++ c udc Λ + c ddc Σ c usc Ξ + c 467.9? dsc Ξ c 47.? ssc Ω c 697.5? ucc Ξ ++ cc dcc Ξ + cc scc Ω + cc ccc Ω ++ ccc

11 7 qqq baryon mass year uub Σ + b udb Λ b 564? ddb Σ qqq baryon mass year b usb Ξ ubb Ξ b bb dsb Ξ dbb Ξ b bb ssb Ω sbb Ω bb b ucb Ξ + cbb Ω ccb cb dcb Ξ bbb Ω bbb cb scb Ω cb ccb Ω + ccb Blank spaces indicate that the particle has not yet been found 8. Problem.7 N : ; actually 939; error:.7%. Σ : ; actually 93; error: 3.6%. Λ : ; actually 6; error: 3.%. Ξ : ; actually 38; error:.7%. Problem.8 Quarks and leptons: ccc q : e c c c q : e + ccn q 3 : ū c c n q 3 : u cnn q 3 : d c n n q 3 : d nnn q : ν e n n n q : ν e The neutrinos could be switched, but this seems the most natural assignment. Mediators: ccc n n n q : W nnn c c c q + : W + ccc c c c q : Z } or vice versa nnn n n n q : γ Gluons: We need matching triples of particles and antiparticles,

12 8 Historical Introduction to the Elementary Particles ccn 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 8 possibilities. Problem.9 There are at least four distinct answers, depending on the particle in question: Antiparticles such as the positron annihilate with the corresponding particle the electron, in this case, and since there are lots of electrons in the lab, positrons don t stick around long enough to have any role in ordinary chemical processes. But if you could work in a total vacuum you could make atoms and molecules of antimatter, and all of chemistry would proceed just the same as with ordinary matter. Most elementary particles such as muons, pions, and intermediate vector bosons are intrinsically unstable; they disintegrate spontaneously in a tiny fraction of a second not long enough to do any serious chemistry. You can make short-lived exotic atoms, with say muons in orbit around the nucleus instead of electrons. Some of these systems last long enough to do spectroscopy. Neutrinos interact so feebly with matter that they have no impact on chemistry, even though we are in fact bathed in them all the time. Quarks are the basic constituents of protons and neutrons, so in an indirect way they do play a fundamental role in chemistry. And gluons play a fundamental role in holding the nucleus together. But because of confinement, they do not occur as free particles, only in composite structures, so they don t act as individuals.

13 9 Elementary Particle Dynamics Problem. F g Gm r ; F e e 4πɛ r F g F e 4πɛ Gm e Problem. Problem.3 Eight are built on ; eight are built on ; one is special.

14 Elementary Particle Dynamics Problem.4 e e e e g g e e e e Energy and momentum are conserved at each vertex. Thus, for the virtual photon in the horizontal diagram E m e c and p m m e and v and in the vertical diagram, E and p m and v.

15 Problem.5 W d u p - W d u p - X - s s d u s d L X - s d W u u d d n a Ξ n + π would be favored kinematically, but since two s quarks have to be converted, it requires an extra W hence two extra weak vertices, and this makes it much less likely. Λ + π is favored. Experimentally, % go to Λ + π ; the branching ratio for n + π is less than.9 5. b D K + π + : Neither vertex crosses generations: W u d p+ c s D u u K - D π + π + : One vertex crosses generations: W u d p + c D d p- u u D π + K + : Both vertices cross generations:

16 Elementary Particle Dynamics W u s K + c D d p- u u Because weak vertices within a generation are favored, K + π + is most likely, K + + π least. Experimentally, the branching ratios are: 3.8% for K + π +,.4% for π + + π, and.4% for K + + π. c The b quark prefers to go to c V cb.4, whereas V ub.4, so B should go to D. Problem.6 e W + e W + e W + e W _ e Z W _ e g W _ Problem.7 a Impossible charge conservation b Possible, electromagnetic c Impossible energy conservation d Possible, weak e Possible, electromagnetic f Impossible muon number conservation

17 3 g Possible, strong h Possible, weak i Impossible baryon number conservation j Possible, strong k Impossible baryon and lepton number conservation l Possible, strong m Possible, strong n Impossible charge conservation o Possible, weak p Impossible charge conservation q Possible, electromagnetic r Possible, weak s Possible, weak t Possible, strong u Possible, electromagnetic v Possible, weak Problem.8 a K + µ + + ν µ + γ; weak and electromagnetic interactions are involved: e m W g n e e e n m

18 4 Elementary Particle Dynamics b Σ + p + γ; weak and electromagnetic interactions are involved: W W g S + s u u d u p u u Problem.9 The lifetime tells us this is an OZI-suppressed strong interaction. Evidently the B meson must weigh more than half the Υ just as the D weighs more than half the ψ. Thus the B meson should weigh more than 473 MeV/c and it does. Problem. d u p- y' c c Here s a typical contributing diagram. Yes, it is a strong interaction. Yes, it is OZI-suppressed. We should expect a lifetime around seconds. u d c c p+ y Problem. a Particle X has charge + and strangeness ; it was presumably a proton. b K + p K + K + + Ω strong; Ω Ξ + π weak; Ξ Λ + π weak, π γ + γ electromagnetic, and both photons undergo pair production γ e + + e electromagnetic; Λ π + p weak.

19 5 Problem. A pion seems most likely it requires only one weak vertex, with no generation crossing, and it s light hence kinematically favored: p u u d d u p+ u u d p W -

20

21 7 3 Relativistic Kinematics Problem 3. { x γ x vt x } /γ x vt t γ t v x vt /γ vt v x c c Adding these two equations: x + vt x v γ c x/γ x γ x + vt. x γ x vt v x c γ v x v t c c t γ t v x t /γ t v x c c Adding these two equations: t + v γ c x t v c Also, y y, and z z. This confirms Eq t/γ t γ t + v x. c Problem 3. a From the Lorentz transformations Eq. 3., t γ t v x. c t A t B γ t A t B v c x A x B. If simultaneous in S so that t A t B, then t A t B + γ v c x B x A. b γ ; x B x A 4 km 4 3 m.

22 8 3 Relativistic Kinematics t A t B c 4 3 m t B m c 3 8 m/s t B + 5 s. So B went on first ; A came on 5 seconds later. Problem 3.3 a The dimension along the direction of motion undergoes length contraction, but the other two dimensions do not. Thus, volumes transform according to V V /γ. b The number of moleculues N is invariant. Density is ρ N/V, so density transforms by ρ N/V γn/v γρ. Problem 3.4 a d vt m/s. 6 s 659 m. No. b γ d v γt γ ,4 m. Yes. c No. They only travel, /. 6 3 m. Problem 3.5 d 6 m; τ. 6 s. The half-life t / is related to the lifetime τ by t / ln τ. Thus, d vγt / v ln τ v.8 c v/c

23 9 Problem 3.6 a Velocity of bullet relative to ground c + 3 c 6 5 c c, but the getaway car goes 3 4 c 9 c slower, so bullet does reach target. b v c + 3 c c c 8 c, which is less than 3 4 c 8 c, so bullet doesn t reach target. Problem 3.7 γ γβ M γβ γ γ γβ γ γβ γ β ΛM γβ γ γβ γ γ β Problem 3.8 From Eq. 3.8: x x γ x βx γ x βx γ x βx x + β x x + βx x β x γ x β x β γ β Since x x and x 3 x 3, it follows that x x x x. x x x x 3 x x x x 3.

24 3 Relativistic Kinematics Problem 3.9 a µ 3, 4,, b µ 5,, 3, 4 a b a b a 3 a 9 -, so a µ is spacelike. b 5 b 5 5, so b µ is lightlike. a b 35 a b 5 4. Problem 3. a s, s, s, s 33, and s, s, s 3, s, s 3, s 3 b 6 the latter six the first four are zero in this case. c s νµ ΛκΛ ν µ σs κσ ΛκΛ ν µ σs σκ Λ µ σλκs ν σκ s µν, so s µν is symmetric. a νµ ΛκΛ ν µ σa κσ ΛκΛ ν µ σ a σκ Λ µ σλκa ν σκ a µν, so antisymmetry is also preserved by Lorentz transformations. d s νµ g νκ g µλ s κλ g νκ g µλ s λκ g µλ g νκ s λκ s µν, so s µν is symmetric. a νµ g νκ g µλ a κλ g νκ g µλ a λκ g µλ g νκ a λκ a µν, so a µν is antisymmetric.

25 e s µν a µν s νµ a νµ just by renaming the summation indices s µν a µν by symmetry/antisymmetry s µν a µν. But if x x, then x. f Let s µν tµν + t νµ and a µν tµν t νµ. Clearly s µν s νµ, a µν a νµ, and t µν s µν + a µν. Problem 3. γ η µ γc, v x, v y, v z 5 c, 35 4 c,, c 5, 3,, 4 Problem 3. p µ A + pµ B pµ C + pµ D Multiply by Λ ν µ to transform to S : Λ ν µp µ A + pµ B Λν µp µ C + pµ D p ν A + p ν B pc ν + p ν D Problem 3.3 p m c > Eq. 3.43, so it s timelike Eq. 3.5; if m, then p, so it s lightlike. The 4-momentum of a virtual particle could be anything. Problem 3.4 Say the potato weighs 4 kg at room temperature C, and we heat it to the boiling point C. If it were pure water specific heat c Cal/kg C

26 3 Relativistic Kinematics 4 J/kg C, the heat required would be Q mc T 4 J/kg C 4 kg 8 C 8 4 J. The increase in mass is m Q c 8 4 J 3 8 m/s kg. Not a substantial addition! Problem 3.5 p π p µ + p ν 4-vectors p µ p π p ν p µ p π + p ν p π p ν. p µ m µc, p π m πc, p ν ; p π p ν E π c E ν c p π p ν, but p π p ν, so m µc m πc E πe ν c. But E π γm π c and E ν p ν c. So m π m µc γm π p ν c p ν m π m µc γm π. Also, p π γm π v. So tan θ p ν p π m π m µc γm π γm π v m µ/m π βγ where β v c. Problem 3.6 Before the collision, in the lab frame, p µ TOT EA c + m B c, p A, so p TOT EA c + m Bc p A E A c + E Am B + m B c p A E A m B + m B + m A c I used E A p A c m A c4. After the collision, in the CM frame, at threshold: p µ TOT m + m m n c, Mc, ; p TOT Mc. But p µ TOT is conserved same before as after, in either frame and p TOT is invariant same value in both frames, so E A m B + m A + m B c M c, or E A M m A m B c m B.

27 3 Note that this generalizes the result in Example 3.4. There, m A m B m p ; M 4m p ; so E A 6 m pc m p 7m p c. Problem 3.7 a E A M m A m B c m B M m p + m π m A m B m p E MeV b M m p + m π + + m π m A m B m p E MeV c M m p + m n m A m π 39.57; m B m p E MeV d M m K + m Σ m A m π 39.57; m B m p E MeV

28 4 3 Relativistic Kinematics e M m p + m Σ + + m K m A m B m p E MeV Problem 3.8 For the second reaction K + p Ω + K + K +, M m Ω + m K + m K , m A m K and m B m p 938.7, so the threshold energy for the K is E K MeV. In the first reaction, at threshold, the K goes in the forward direction, and the other particles emerge as a group there is no point in wasting energy in transverse motion, or in internal motion of the p, p, K +. p p C K - before after We have, in effect, an outgoing particle C, of mass m C m p + m K MeV/c. Conservation of energy/momentum says p + p p C + p K, or p p K p C p. Squaring: Now p + p K p p K p C + p p C p, E m pc + m E K c K c c p p K m C c + m pc EC c m pc. E C E + m p c E K, p p K p p K, p c E m p c,

29 5 and EK m Kc 38.3 c c a c. p K c EE K + a E m p c m C c m K c m p c E + m p c E K, where b Squaring, a E m p c EE K m p c + b, m C c m K c + m p c E K m p c a E m p c E E K m p c + b + be E K m p c, E a E K m p c be E K m p c b + a m p c. Numerically, E K m p c 44., a E K m p c , be K m p c.59, b + a m p c So or E E , E E E ± ± But E has to be positive, so we need the upper sign: E 56.9 T MeV.

30 6 3 Relativistic Kinematics Problem 3.9 a p µ A pµ B + pµ C, or pµ C pµ A pµ B. Square both sides: p C p A + p B p A p B. Now, p A m A c, p B m B c, p C m C c, p A p B E A c E B c p A p B. But p A, and E A m A c, so m C c m A c + m B c m A E B, so m A E B m C m A m B c, or E B m C + m A + m B m A c ; E C m A m B + m C m A c b EB p B c m B c4 p B E B c m B c m A + m B m C c 4m 4m A m B c A 4m A p B c m 4 m A + m4 B + m4 C + m A m B m A m C m B m C 4m A m B A c λm m A, m B, m C A p B p C c λm m A, m B, m C A c The decay is kinematically forbidden if m A < m B + m C not enough energy to produce the final particles, in CM frame. Problem 3. a E µ 9.8 MeV; E ν 9.8 MeV b E γ 67.5 MeV c E π MeV; E π 45.6 MeV d E p MeV; E π 7. MeV e E Λ 35 MeV; E K MeV

31 7 Problem 3. The velocity of the muon is v m π m µ m π+m µ c Example 3.3. So γ m π m µ m π+m µ m π + m µ m π + m µ m π m µ m π + m µ m 4 π + m 4 µ + m πm µ m 4 π m 4 µ + m πm µ m π + m µ m π m µ. Lab lifetime is γτ, so distance is vγτ, or m π m µ d m π + m c m π + m µ τ m π m µ cτ. µ m π m µ m π m µ d m/s. 6 s 86 m Problem 3. a The minimum is clearly E min m B c, with B at rest. For maximum E B we want B going one way, C, D,... moving as a unit in the opposite direction there is no point in wasting energy in relative motion of the other particles. Thus the others act as a single particle, of mass M m C + m D +, and we can simply quote the result for two-body decays Problem 3.9a: E max m A + m B M m A c. In case that argument is not completely compelling, let s look more closely at the case of just three outgoing particles. For maximum E B we certainly want C and D to come off opposite to B, and the only question is how best to apportion the momentum between them. Let s say p C xp B, p D x p B so that p C + p D p B.

32 8 3 Relativistic Kinematics Here x ranges from C at rest to D at rest. From conservation of energy, m A c E B + E C + E D. But E C m C c4 + x p B c, E D m D c4 + x p B c, and p B c E B m B c4. So m A c E B + m C c4 + x EB m B c4 + m D c4 + x EB m B c4. Differentiate with respect to x: de B dx + + xe B m B c4 + x de E B B dx x EB m B c4 + x de E B B dx m C c4 + x EB m B c4 ; m D c4 + x EB m B c4. For the maximum we want de B /dx, so Therefore xe B m B c4 + x E B m B c4, x x +, x x x m D c4 + x E B m B c4 x m C c4 + x E B m B c4, x m D c4 x m C c4 xm D xm C ; xm C + m D m C, x m C m C + m D. What is the corresponding E B? m A c E B + +, but x x, so x, x x and hence m A c E + + x E + x E + x m C c4 + E m B c4. E E B,

33 9 m A c E m C + m D c 4 + E m B c4 m A c4 Em A c + E m C + m D c 4 + E m B c4 Em A m A + m B m C + m D c. m E max A + m B m C + m D c. m A Are we sure this is a maximum? Maybe it s a minimum, and the maximum occurs at the end of the interval x or x. We can test for this by calculating E at the end points: At x, m A c E + m C c + m D c4 + E m B c4. m A m C c E m D m B c4 + E m A m C c 4 Em A m C c + E m D m B c4 + E E m A m C + m B m D c m A m C For x, just interchange C and D. Now c E max E m A + m B m C + m D m A m m A + m C m B m D A m A m C m A m C m B m C + m D + m A m C m A m C m A m B m D. m A m A m C So m A m A m C c E max E m A m B m C m D m Cm D + m A m C m C m B + m C m B m C m D m Cm D m D m C ma m C m D + m A m B + m C + m D + m Cm D. m A m A m C m C c E max E m A + m C + m D m Am C m A m D + m C m D m B m A m C m D m B. But m A > m B + m C + m D m A m C m D > m B, so the right side is positive, and therefore E max > E. By the same token, E max > E at x, which differs only by C D. Conclusion: E max is a maximum, not a minimum.

34 3 3 Relativistic Kinematics b E min m e c.5 MeV ; E max m µ + m e c 5.8MeV. m µ Problem 3.3 a The CM moves at speed v relative to lab. Classically, u v + v v, but relativistically the velocities add by Eq. 3.5: u v + v + vv v. Solve for v : c + v c u + u v c v u c v v + u v + ± 4 4 u c u c ± u u c c If we use the + sign, for small u we get v c u, which is wrong it should be u, so we need the sign: v c u u c. b In the CM frame, then: γ v c c u c + + c u u u c Let γ u c u c γ ; u γ γ γ γ γ γ + u + u c c the γ in the lab frame. Then u c u c c γ. Therefore u c. γ γ γ γ + γ γ +

35 3 c So T γ mc γ + mc. Hence γ + γ + T mc +. Therefore T T m c 4 + T mc + m c 4 + 4T mc +. T γ mc kinetic energy in lab, so T T m c 4 + 4T mc mc T + 4T 4T mc + T mc. Problem 3.4 Let p A be the 4-momentum of A before the collision, and let q A be the 4- momentum after the collision, in the CM frame; p A, q A are the corresponding quantities in the Breit fame. Now p A q A p A q A since the 4-dimensional dot product is invariant: E A c p A q A E A c p A q A. Note: The incoming A and outgoing A have the same energy in CM only the direction of A s momentum changes. Likewise, the incoming and outgoing A have same energy E A in Breit frame, since their momenta are opposite. Now p A q A p A cos θ, where So p A c E A m A c4, and p A q A p A c E A m A c4. E A c c E A m A c4 cos θ E A c + c E A m A c4 ; E A cos θ + m A c4 cos θ E A m A c4. cos θ + cos θ E A E A + m A c4 E θ A sin + m A c4 cos θ or dropping the subscripts E E sin θ + m c 4 cos θ.

36 3 3 Relativistic Kinematics y y' p A q A p' A q' A q/ q x x' q B p B q' B p' B v points along the line bisect- Velocity makes angle θ/ with incident A. ing p A and q A see diagram. q A x γ q Ax v E c c v c q Ax E v cos θ c c q A cos θ c E E cos θ ma c E E c m A c4 Problem 3.5 a s + t + u c p A + p B + p A p C + p A p D p c A + p A p B + p B + p A p A p C + p C + p A p A p D + p D m c A c + m B c + m C c + m D c + p A p A + p B p C p D Conservation of energy and momentum requires p A + p B p C + p D, so s + t + u m A + m B + m C + m D. b c s p A + p A p B + p B p A + EA c E B c p A p B + p B

37 33 In the CM frame, p A p B p A, so So p A E A c and E B c p A p A E A c p B E B c p B E B c p A c s p A + p A p A p B p B + p A p B + E A c p A. E A c c s + p A p B E A c Squaring both sides, p B + E A c E A c p A + E A c p B + E A c p A p A + p B c s + p A p B 4 c c s + p A p B E A + 4 E4 A c 4 4 c p B p A E A + 4 E4 A c 4 E A c s + p A p B 4s E CM A s + m A m B c. s c In the lab frame, p B and E B m B c, so c s p A + E A c E lab A s m A m B c m B. m B c + p B c m A c + m B E A + m B c d In the CM frame, p A + p B EA c + E B E TOT c c E CM TOT sc Problem 3.6 s c p A + p B EA + E B c p c A + p B

38 34 3 Relativistic Kinematics In the CM frame, p A + p B and E A E B p c + m c 4, so s 4p c + m c 4 c c 4p + m c c. t c p A p C EA E C c p c A p C But E A E C and p A p C p A + p C p A p C p cos θ, so t p cos θ c. u is the same as t, except that p A p D p A p D p cos θ: u p + cos θ c. Problem 3.7 E g before e g q f e after E', p g p e In this problem I ll use p e for the three-momentum of the charged particle, and p γ for the three-momentum of the photon. Conservation of momentum: or p e sin φ p γ sin θ sin φ E cp e sin θ E c p γ cos θ + p e cos φ E E c cos θ + p e p e c sin θ E E cos θ p e c E sin θ. p e c E EE cos θ + E cos θ + E sin θ p e c E EE cos θ + E

39 35 Conservation of energy: E + mc E + m c 4 + p e c E + m c 4 + E EE cos θ + E E E + mc E + E + m c 4 EE + Emc E mc mc E E EE cos θ; m c hc λ λ mc λ λ λλ m c 4 + E EE cos θ + E h c E hc λ, cos θ λλ E hc λ h λλ cos θ λ λ + h cos θ mc

40

41 37 4 Symmetries Problem 4. movement of corners symmetry operation A A, B B, C C I A A, B C, C B R a A B, B A, C C R c A B, B C, C A R A C, B B, C A R b A C, B A, C B R + Problem 4. I R + R R a R b R c I I R + R R a R b R c R + R + R I R b R c R a R R I R + R c R a R b R a R a R c R b I R R + R b R b R a R c R + I R R c R c R b R a R R + I The group is not Abelian; the multiplication table is not symmetrical across the main diagonal for example, R + R a R b, but R a R + R c.

42 38 4 Symmetries Problem 4.3 a DI ; DR ; DR a ; DR b ; DR c. b I, R + and R are represented by ; R a, R b, and R c are represented by. This representation is not faithful. Problem 4.4 d a c b The square has 8 symmetry operations: doing nothing I, clockwise rotation through 9 R +, rotation through 8 R π, counterclockwise rotation through 9 R, and flipping about the vertical axis a R a, the horizontal axis b R b, or the diagonal axes c R c or d R d. I R + R π R R a R b R c R d I I R + R π R R a R b R c R d R + R + R π R I R c R d R b R a R π R π R I R + R b R a R d R c R R I R + R π R d R c R a R b R a R a R d R b R c I R π R R + R b R b R c R a R d R π I R + R R c R c R a R d R b R + R I R π R d R d R b R c R a R R + R π I The group is not Abelian for example, R + R a R c, but R a R + R d.

43 39 Problem 4.5 a Let A and B be two n n unitary matrices. Then ÃB AB B Ã AB B B, so AB is also unitary the set of n n unitary matrices is closed under multiplication. The usual matrix identity is unitary. All unitary matrices have inverses A Ã, which are themselves unitary. Finally, matrix multiplication is associative, so Un a group. b Suppose that the unitary matrices A and B both have determinant. Then detab deta detb. We already know that AB is unitary, so the set of n n unitary matrices with determinant is closed. The matrix identity has determinant, and deta / deta, so the set contains the appropriate inverse and identity elements; SUn is a group. c On, the real subset of Un is closed by the same argument that Un is closed: ÃBAB BÃAB BB. The identity, inverse, and associativity requirements are still met, so On is a group. d The set of n n orthogonal matrices is closed, as is the set of n n matrices with determinant. Therefore, SOn is closed. If A SOn, so is A, and I SOn; SOn is a group. Problem 4.6 y' y a' y A a q f a' x a x y x' x a x a cos φ; a y a sin φ a length of A a x a cosφ θ acos φ cos θ + sin φ sin θ cos θa x + sin θa y a y a sinφ θ asin φ cos θ cos φ sin θ cos θa y sin θa x

44 4 4 Symmetries a x a y cos θ sin θ sin θ cos θ cos θ sin θ RR sin θ cos θ so R is orthogonal. ax a y cos θ sin θ sin θ cos θ cos θ sin θ. R. sin θ cos θ cos θ + sin θ cos θ sin θ sin θ cos θ sin θ cos θ cos θ sin θ sin θ + cos θ detr cos θ + sin θ Group is SO. Rθ Rθ cos θ sin θ cos θ sin θ sin θ cos θ sin θ cos θ cos θ cos θ sin θ sin θ cos θ sin θ + sin θ cos θ sin θ cos θ cos θ sin θ sin θ sin θ + cos θ cos θ cosθ + θ sinθ + θ Rθ sinθ + θ cosθ + θ + θ Rθ + θ Rθ Rθ. Yes, it is Abelian. Problem 4.7 MM Yes, it is in O. detm. No, it is not in SO. ax ax M, so a x a x ; a y a y. a y a y No, this is not a possible rotation of the xy plane. It s a reflection in the x axis. Problem 4.8 For a spinning solid sphere, I 5 mr and ω v/r, so v L Iω h 5 mr, v 5 h r 4 mr.

45 4 An electron has mass m 9 3 kg and radius r < 8 m, giving v > 4 m/s, much faster than the speed of light. Evidently this classical model cannot be taken literally. Problem 4.9 a Before: 8 states.,,,,,,, After: s 3 is 4 states; s is states, and occurs twice, so we get b and 3, or. 3 and / 7/ 8 states 5/ 6 states and / 5/ 6 states 3/ 4 states and / 3/ 4 states / states states in all We get total angular momenta 7, 5 twice, 3 twice, and. We had states to begin with, and 3 at the end. Problem 4. Combining the spins of p and e, we can get s or s, but not s. So this process would violate conservation of angular momentum. If the p e system carries orbital angular momentum we could achieve s >, but still only integer values: remains inaccessible. would do, for combining s with s yields 3 and and combining s with s yields 3. But would also be OK, since combining s with s 3 yields 5, 3, and. If we allow for orbital angular momentum in the final state, any half-integer spin would work. But seems the most likely possibility.

46 4 4 Symmetries Problem 4. The proton has spin and the pion has spin, so the total spin in the final state is. This must combine with orbital angular momentum to give s 3. Thus l could be s + l 3 or or s + l 5 or 3. Problem 4. From the Clebsch Gordan table, I read }{{}. spin up Probability of spin up is /3. Problem 4.3 From the table, I read So you could get j 4, probability 8 35 ; j, probability 7 ; or j, probability 5. What this means, of course, is that the total angular momentum squared could come out 45 h h j 4 or 3 h 6 h j or j. Do the probabilities add to? Problem 4.4 From the 3 table, I read

47 43 So you could get m, probability 7/7; or, probability 3/35; or, probability 5/4; or, probability 6/35. That is, S z for the spin- particle could be h,, h, or h. Do the probabilities add to? Problem 4.5 Ŝ x χ + h / / h χ +; eigenvalue h. Ŝ x χ h / / h χ ; eigenvalue h. Problem 4.6 a + b α + β + α β α + β α + β + α β α β α + α β + β α + β + α α β β α + β α + β. Problem 4.7 a h i i α λ β α β i h/β i h/α i h β λα; i h α λβ i h β λ i h λβ λα λβ h 4 β λ β λ ± h

48 44 4 Symmetries Plus sign: i h β h α β iα χ + Minus sign: i h β h α β iα χ α β α. iα α β α. iα In both cases normalization α + β gives α /. Conclusion: Eigenvalues are ± h/. Eigenspinors are χ ±. ±i b α cχ + + dχ c + d c + d β i i i c d α c + d β i c d α iβ c c α iβ or α + iβ d d α + iβ You could get ± h, probability α iβ. Problem 4.8 a From Eq. 4.4, probability of h is α + β Probability of h is α β 5 5. b From Problem 4.7, probability of h is 5 i i + i Probability of h is 5 + i 5.

49 45 c Probability of h is α 5 ; probability of h is β 4 5. Problem 4.9 a b σx σy i i i i σz σ x σ y i i i i i iσ z i i σ y σ x iσ z i i i i σ y σ z i iσ x i i i i σ z σ y iσ x i i i σ z σ x i iσ y i σ x σ z iσ y Problem 4. a σ i, σ j σ i σ j σ j σ i δ ij + iɛ ijk σ k δ ji iɛ jik σ k But δ ij δ ji, ɛ jik ɛ ijk. So σ i, σ j iɛ ijk σ k.

50 46 4 Symmetries b {σ i, σ j } σ i σ j + σ j σ i δ ij + iɛ ijk σ k + δ ji + iɛ jik σ k δ ij c σ aσ b i,j i,j σ i a i σ j b j i,j a i b j σ i σ j a i b j δ ij + iɛ ijk σ k i,j a i b j δ ij + i ɛ ijk a i b j σ k a b + iσ a b i,j Problem 4. a Noting that σ z, σ 3 z σ z, etc., e iθσ z + iθσ z + iθσ z + iθσ z 3 + 3! θ + θ4 4! + iσ z θ θ3 3! + θ5 5! + cos θ + iσ z sin θ In particular e iπσ z/ cos π/ + iσ z sin π/ iσ z. b Similarly all we need is σ y e iθσ y cos θ + iσ y sin θ, so c U U cos 9 i sin 9 σ y iσ y i Uθ e iθ σ/ + i i, which is spin down. i θ σ + i θ σ + i θ σ 3 + 3! But θ σ θ θ + iσ θ σ θ. So θ σ Uθ i θ + θ iθ σ 3! 3 + θ + θ 4 i θ σ 4! θ cos θ i ˆθ σ sin θ. θ 3! θ 3 +

51 47 Problem 4. a U cos θ i ˆθ σ sin θ ; U cos θ + i ˆθ σ sin θ But σ σ the Pauli matrices are Hermitian, so U cos θ + i ˆθ σ sin θ. UU cos θ i ˆθ σ sin θ cos θ + i ˆθ σ sin θ cos θ i ˆθ σ sin θ cos θ + i ˆθ σ cos θ sin θ + ˆθ σ ˆθ σ sin θ cos θ + sin θ. Note: I used Problem 4. c to show that ˆθ σ ˆθ σ ˆθ ˆθ + iσ ˆθ ˆθ. b ˆθ σ ˆθ x i + ˆθ y i + ˆθ z ˆθ z ˆθ x i ˆθ y ˆθ x + i ˆθ y ˆθ z Uθ cos θ i sin θ ˆθ z ˆθ x i ˆθ y ˆθ x + i ˆθ y ˆθ z cos θ i ˆθ z sin θ i sin θ ˆθ x i ˆθ y i sin θ ˆθ x + i ˆθ y cos θ + i ˆθ z sin θ det U cos θ i ˆθ z sin θ cos θ + i ˆθ z sin θ + sin θ ˆθ x + i ˆθ y ˆθ x i ˆθ y cos θ + ˆθ z sin θ + sin θ ˆθ x + ˆθ y cos θ + sin θ ˆθ x + ˆθ y + ˆθ z cos θ + sin θ.

52 48 4 Symmetries Problem 4.3 a Call the three column vectors χ +, χ, χ. We have: b Ŝ z χ + hχ + ; Ŝ z χ ; Ŝ z χ hχ ; so Ŝ z h. Ŝ + ; Ŝ + h h ; Ŝ + h h. Ŝ + ; Ŝ + h ; h Ŝ + h. Ŝ + h. Ŝ h + h ; Ŝ h + h ; Ŝ. Ŝ h. h c Ŝ x Ŝ + + Ŝ h ; Ŝ y i Ŝ + Ŝ h i i i i d Represent the four spin 3 states m s + 3, +,, 3 by the column vectors,,,.

53 49 Ŝ z h 3 3 ; Ŝ z 3 h 3 ; Ŝ z 3 h 3 ; Ŝ z h Ŝ z h. 3 Ŝ ; Ŝ + 3 h h 3 3 ; Ŝ + 3 h h 3 ; Ŝ h h 3 ; 3 h Ŝ + h 3 h. Ŝ 3 3 h h 3 ; Ŝ 3 h h 3 ; Ŝ 3 h h 3 3 ; Ŝ 3 3 Ŝ 3 h h. 3 h Ŝ x Ŝ + + Ŝ h ; 3 Ŝ y i Ŝ + Ŝ h 3i 3i i i 3i. 3i

54 5 4 Symmetries Problem 4.4 Ω ; Σ + ; Ξ ; ρ+ ; η ; K. Problem 4.5 a b u : Q d : Q s : Q ū ; Q + 3 d ; Q s ; Q Problem 4.6 a Q I 3 + A + S + C + B + T b I 3 U + D ; A U + C + T D S B 3

55 5 c Q U + D + U + C + T D S B + S + C + B + T 3 U + B + T + D + S + B 3 Problem 4.7 I tot I + I I + I + I I. Here I I I tot in singlet state, so I I, or I I I tot + in triplet, so I I, or I I 4. Problem 4.8 a M a M f M 3 ; b c M b Me 3 M M ; M c M d 3 M M ; M g M h M i M j /3M 3 /3M. σ a : σ b : σ c : σ d : σ e : σ f : σ g : σ h : σ i : σ j 9 M 3 : M 3 + M : M 3 + M : M 3 + M : M 3 + M : 9 M 3 : M 3 M : M 3 M : M 3 M : M 3 M σ a : σ b : σ c : σ d : σ e : σ f : σ g : σ h : σ i : σ j 9 : 4 : : : 4 : 9 : : : :

56 5 4 Symmetries Problem 4.9 Using the method in Section 4.3: { π + p : π + + p : 3 3 K + Σ : K + + Σ : K + + Σ + : 3 3 M c M 3 ; M a 3 M 3 3 M ; M b 3 M M σ a : σ b : σ c 9 M 3 M : 9 M 3 + M : M 3. If the I 3/ channel dominates, M 3 M, so σ a : σ b : σ c : : 9. If the I / channel dominates, M M 3, so σ a : σ b : σ c : 4 : 9. Problem 4.3 { K + p : K + p : Σ + π : 3 3 Σ + + π : Σ + + π : + Σ + π + : a M a b M b 3 M ; I tot M + c M c M ; I tot d M d M ; I tot 3 M ; I tot or

57 σ a : σ b : σ c : σ d 6 M : 3 M + If I dominates, σ a : σ b : σ c : σ d : : : If I dominates, σ a : σ b : σ c : σ d : : : M : M : M. 53 Problem 4.3 If M M 3, Eq. 4.5 becomes σ a : σ c : σ j : :, so σ tot π + + p, σ tot π + p 3, and σ tot π + + p/σ tot π + p. There is no sign of a resonance in σ tot π + + p at 55, 688, or 9, so these must be I /. The ratios are not zero, of course evidently there is a lot of resonant background. There is a clear resonance in σ tot π + + p at 9; this must be I 3/. The ratio is σ tot π + + p/σ tot π + p 43/36., which is not very close to 3 presumably this is again due to nonresonant background. The nomenclature should be N55, N688, 9, N9. The Particle Physics Booklet lists N5, N68, 9, N9 and there are others with roughly the same mass. Problem 4.3 { Σ : a Σ + + π : b Σ + π : 3 3 c Σ + π + : M a M, M b, M c M. σ a : σ b : σ c : : I d expect to see about 5 decays each to Σ + + π and Σ + π +, but none to Σ + π. Problem 4.33 a Isospin must be zero. b The deuterons carry I, so the isospin on the left is zero. The α has I, and π has I, so the isospin on the right is one. This process does not conserve isospin, and hence is not a possible strong interaction.

58 54 4 Symmetries c There are five possible 4-nucleon states: nnnn, nnnp 4 H, nnpp 4 He, nppp 4 Li, pppp 4 Be In principle they could form an I multiplet, but since 4 H and 4 Li do not exist, this is out. No: 4 Be and nnnn should not exist. 4 H, 4 He, and 4 Li could make an I multiplet, but, again, 4 H and 4 Li do not exist, so this too is out. Evidently four nucleons bind only in the I combination, making 4 He. Problem 4.34 a Suppose f is an eigenfunction of P: P f λ f. Then P f PP f Pλ f λp f λλ f λ f. But P, so λ, and hence λ ±. b Let f ± x, y, z f x, y, z ± f x, y, z. Then f f + + f. Note that P f ± f x, y, z ± f x, y, z ± f ±, so f ± is an eigenfunction of P, with eigenvalue ±. Problem 4.35 a No. P would transform a left-handed neutrino into a right-handed neutrino, which doesn t even exist in the massless limit. b The K has intrinsic parity, so the two-pion decay violates conservation of parity.

59 55 Problem 4.36 a Equation 4.6 says G I C, so π : I, C G ρ : I, C G ω : I, C G η : I, C G η : I, C G φ : I, C G f : I, C G b????? Problem 4.37 a Since the η and the π s have spin zero, the final state would need to have orbital angular momentum zero: l by conservation of angular momentum. But this means that the parity of the final state is P π P π l. However, the parity of the η is, so this decay is forbidden by conservation of parity. b The G-parity of the final state is 3 Eq But the G-parity of the η is + Eq. 4.6 and Table 4.6. So η 3π violates conservation of G-parity, and hence is a forbidden strong interaction. Problem 4.38 Same baryon number as antiparticle meson; same charge as antiparticle neutral. We re not interested in q q, since these are already their own antiparitlces; we need q q, with Q Q Q: Q /3 : d s s d, d b b d, s b b s. Q /3 : u c cū only, since t forms no bound states. So the candidate mesons are K K, B B, B s B s, D D.

60 56 4 Symmetries Neutron/antineutron oscillation would violate conservation of baryon number. The vector mesons decay by the strong interaction long before they would have a chance to interconvert. Problem 4.39 First establish the convention for positive charge, as in Section 4.4.3: it is the charge carried by the lepton preferentially produced in the decay of the longlived neutral K meson. Then define right-handedness: it is the helicity of the charged lepton produced in the decay of a positively charged pion. Finally, define up as the direction away from the earth, and front as the side our eyes are on; cross up with front, using the right-hand rule, and the result is the side our hearts are on. Identifying the particle types is easy just make a list in order of increasing mass. Problem 4.4 Penguin: W - B b d t d u u d p - p + Tree: W d u p - - B b d u d p +

61 57 5 Bound States Problem 5. a m d m p m n , so the binding energy is. MeV, which is only.% of the total. No, it s not relativistic. b m π m d m u , so the binding energy is 536 MeV, which is 3.8 times the total. Yes, it is relativistic. Problem 5. Here n, l, m l, so 3 / Ψ e r/a L a r/ay θ, φ e iet/ h. But L, and Y / 4π so Ψ a 3/ e r/a π e ie t/ h πa 3 e r/a e ie t/ h, where E me 4 / h. Now f r /r d/drr d f /dr, so Ψ πa 3 e ie t/ h d r r e r/a dr a πa 3 e ie t/ h re r/a a r a r e r/a a Ψ ar. The left side of the Schrödinger equation is lhs h m a Ψ ar e r Ψ.

62 58 5 Bound States But a h /me, so the second term cancels the third, leaving Meanwhile, the right side is i h Ψ i h t Normalization requires lhs h ma Ψ me4 h Ψ. ie h Ψ E Ψ me4 h Ψ. Ψ r sin θ dr dθ dφ. Here Ψ πa 3 e r/a, and sin θ dθ dφ 4π, so Ψ r sin θ dr dθ dφ πa 3 4π e r/a r dr 4 a 3 a 3. Problem 5.3 There are four states: Ψ Ψ Ψ ± 3 / a 4 3 e r/a L r/a Y e ie t/ h 4a πa 3 a 46 3 r e r/a e ie t/ h a / r e r/a L 3 a r/a Y e ie t/ h 4a πa re r/a cos θe ie t/ h 3 a 46 3 / r e r/a L 3 r/a Y± a e ie t/ h 8a πa re r/a sin θe ±iφ e ie t/ h where E me 4 /8 h, and I used Y 3 3, Y cos θ, Y± 4π 4π 8π sin θ e±iφ,

63 59 and L x x + 4, L3 6. Problem 5.4 Binomial expansion: / E c p + m c mc + p m c. So + ɛ / + ɛ 8 ɛ + E mc + p m c p 8m 4 c 4 mc + p m p4 8m 3 c. T E mc p m p4 8m 3 c. Problem 5.5 From Eq. 5.9, E 3/ E / By comparison, E α 4 mc 3 α4 mc E α 4 mc ev. 37 E E E E ev 4 so the fine structure is smaller by a factor of nearly a million. 4 3 Problem 5.6 From Eq. 5., using k, 3. From Eq. 5., E P α 5 mc 4 3 E S α 5 mc 3 k, α5 mc k, π3/ 3 α5 mc 3π

64 6 5 Bound States I dropped k, <.5 in comparison with /3π.. So the Lamb shift is E E S E P 3 + 3π 3 α5 mc Not bad! ev. E hν ν E h π Hz. Problem 5.7 In the absence of any splitting there are 6 completely degenerate states four orbital states, and for each of these two electron spin states and two proton spin states. Fine structure splits these into two distinct energy levels, one for j / and one for j 3/ the S / and P / states remain degenerate, P 3/ is slightly higher. The Lamb shift lifts the S / /P / degeneracy, so there are now three distinct energy levels. Hyperfine splitting further separates each of these into two, according to the value of f or, if j /; or, if j 3/. So there are 6 energy levels in all: n, l, j /, f one state; n, l, j /, f three states; n, l, j /, f one state; n, l, j /, f three states; n, l, j 3/, f three states; n, l, j 3/, f five states still 6 states in all. Using Eq. 5.3 with n, j /, f lower sign, and l, : m E S m p m E P m p α 4 mc γ p 6 α 4 mc γ p 6 // m α 4 mc γ p, 4 m p /3/ m α 4 mc γ p. m p E E S E P m α 4 mc γ p 6 m p ev This is about a tenth the Lamb shift.

65 6 Problem 5.8 For n 3 we have l S, l P, and l D. For l we can have either the singlet or the triplet spin configuration since l, j s automatically. For l we can have the singlet with j automatically or the tripet for which j can be,, or. For l we have the singlet with j automatic or the triplet for which j can be,, or 3. So there are levels in all. But it turns out that 3 P and 3 D are degenerate, so in fact there are just 9 distinct energy levels. The unperturbed energy of all these states is Eq. 5.7 E 3 α mc ev. 8 The fine/hyperfine correction is Eq. 5.9 E fs α 4 mc ɛ/ l + + ɛ/ l + with ɛ given by Eq In addition there is an annihilation correction, which applies only to the triplet S state Eq. 5.3 E ann α 4 mc ev. In units of µev 6 ev, the corrections are: S : l, s, j, ɛ ; E S : l, s, j, ɛ 4 3 ; E P : l, s, j, ɛ ; E P : l, s, j, ɛ 7 ; 3 E P : l, s, j, ɛ ; 5 E P : l, s, j, ɛ ; E D : l, s, j, ɛ ; E D 3 : l, s, j 3, ɛ ; 6 E D : l, s, j, ɛ 6 ; 3 E D : l, s, j, ɛ 5 6 ; 7 E

66 6 5 Bound States 3S - P 3 P 3 P D 3 D 3 3 D 3 D 3 P - S -3 S state 3 S state 3 3 P state P states D state D states Problem 5.9 The φ is quasi-bound, because the OZI-allowed decay into a pair of K s is kinematically permissible m φ 9 > m K 99. Problem 5. The parameters we have to work with in solving the Schrödinger equation are m units of kg J s /m, h units of J s, F units of N J/m. From these we must construct an energy: Evidently J m α h β F γ J s m α J s β J γ J α+β+γ s α+β m α γ. m α + β β α; α + γ γ α; α + β + γ 3α, so α /3, β γ /3, and hence E m /3 h /3 F /3 a where a is some numerical factor. hf /3 a, m

67 63 Problem 5. For the ψ s, M m c + E n 5 + E n : F M M M 3 M Expt For the Υ s, M m b + E n 9 + E n : F M M M 3 M Expt Level spacings: F M M M 3 M M 4 M Expt ψ Expt Υ Evidently F is between 5 and ; of the three choices, 5 MeV/fm is best though looks better for the masses themselves. The results are not exact, of course, because a the potential Eq is just an approximation, b the parameters used in creating Table 5. are very rough in particular, the quark mass is about right for the c, but way off for the b, and the strong coupling is only taken to one significant digit. All things considered, the agreement is surprisingly good. Problem 5. For the pseudoscalar mesons, S S 3 4 h, so M m + m + 4m u h h m m m + m 477 m u m m.

68 64 5 Bound States π : m m m u M K : m m u, m m s M uū : same as π : 39, dd : same as π : s s : m m m s M η : M η : M ?! 3 For the vector mesons, S S 4 h, so M m + m + 4m u h 59 4 h m m m + m + 59 m u. m m ρ : m m m u M K : m m u, m m s M φ : m m m s M ω : m m m u M Problem 5.3 a For the pseudoscalar mesons see Problem 5.: M m + m 477 m u m m. 38 η c : m m m c M D : m m c, m m u M D s + 38 : m m c, m m s M The observed masses are 98, 865, and 968, respectively.

69 65 For the vector mesons Problem 5.: M m + m + 59 m u m m. 38 ψ : m m m c M D : m m c, m m u M Ds + 38 : m m c, m m s M The observed masses are 397, 7, and, respectively. b For the pseudoscalar bottom mesons, we have u b : m m u, m m b M s b : m m s, m m b M c b : m m c, m m b M b b : m m m b M The observed masses are 579, 5368, 686,???, respectively. For the vector bottom mesons: u b : m m u, m m b M s b : m m s, m m b M c b : m m c, m m b M b b : m m m b M The observed masses are???,???,???, and 946, respectively. Evidently the effective masses of the heavy quarks are somewhat larger than their bare masses as listed in the Particle Physics Booklet hardly a surprise, since the same is true of the light quarks.

70 66 5 Bound States Problem 5.4 The proton is composed of u, u, and d. To make it antisymmetric in, with appropriate normalization, we want p ud duu. The other five corners are constructed in the same way: n ud dud, Σ + us suu, Σ ds sdd, Ξ us sus, Ξ ds sds. Actually, the overall phase of p is arbitrary, but having chosen it, the others are not entirely arbitrary. In terms of isospin, p ; applying the isospin lowering operator see Problem 4.3a for the angular momentum analog: For three particles, I p I 3 n. I I + I + I 3 I I + I + I 3 where the subscript indicates the particle acted upon. Meanwhile, for the quarks I u d and I d, so I p I + I + I 3 udu duu I udu + ui du + udi u I duu di uu dui u ddu + + udd ddu dud udd dud n. Evidently these phases are consistent. Now Σ +, and Eq I Σ + I Σ. So noting that s is an isosinglet, so I s : I Σ + I + I + I 3 usu suu I usu + ui su + usi u I suu si uu sui u dsu + + usd sdu sud dsu + usd sdu sud Σ Σ us sud + ds sdu

71 67 Finally, Λ is constructed from u, d, and s; the most general linear combination of the various permutations that is antisymmetric in is Λ αud dus + βus sud + γds sdu. Orthogonalize with respect to Σ : us sud + ds sdu αud dus + βus sud + γds sdu α + β + γ γ β. Orthogonalize with respect to ψ A : 6 uds usd + dsu dus + sud sdu αud dus + βus sud + γds sdu 6 α + β + γ α β γ β. So Λ β ud dus + us sud ds sdu. Normalize: β ud dus + us sud ds sdu ud dus + us sud ds sdu β β β. Λ ud dus + us sud ds sdu. Problem 5.5 We want a color singlet made from q and q see Eq This is the color analog to the flavor singlet Eq. 5.4: 3 r r + b b + gḡ

72 68 5 Bound States Problem 5.6 Equation 5.6 says ψ 3 ψ sψ f + ψ 3 sψ 3 f + ψ 3 sψ 3 f, so ψ ψ ψ 9 s ψ s ψ f ψ f + ψ s ψ 3 s ψ f ψ 3 f + ψ s ψ 3 s ψ f ψ 3 f + ψ 3 s ψ s ψ 3 f ψ f + ψ 3 s ψ 3 s ψ 3 f ψ 3 f + ψ 3 s ψ 3 s ψ 3 f ψ 3 f + ψ 3 s ψ s ψ 3 f ψ f + ψ 3 s ψ 3 s ψ 3 f ψ 3 f + ψ 3 s ψ 3 s ψ 3 f ψ 3 f. These states are normalized ψ s ψ s, etc., but they are not orthogonal. From Eqs. 5.5, 5.5, and 5.53, using the typical state : ψ s ψ 3 s ψ 3s ψ s ψ s ψ 3 s ψ 3s ψ s ψ 3 s ψ 3 s ψ 3s ψ 3 s. From the figures on pp , using the typical state in the upper right corner: ψ f ψ 3 f udu duu uud udu ψ 3 f ψ f ψ f ψ 3 f udu duu uud duu ψ 3 f ψ f ψ 3 f ψ 3 f uud duu uud udu ψ 3 f ψ 3 f. ψ ψ

73 69 Problem 5.7 From Eq. 5.6, using Eqs and the diagrams on pp , I find: Σ + : usu suu 3 + uus usu + uus suu 3 usu + suu + uus 3 u s u u s u u s u + s u u s u u s u u + u u s u u s u u s. Λ : uds dus + usd sud dsu + sdu 3 + sud sdu + dus dsu uds + usd + usd dsu + uds sdu dus + sud 6 3 uds dus usd sud dsu sdu {uds + dus + usd + sud + dsu + sdu }

74 7 5 Bound States u d s u d s + d u s d u s 3 + u s d u s d + s u d s u d + d s u d s u + s d u s d u. Problem 5.8 For notational simplicity, use ψ ij ψ ji for the spin-/ functions Eqs. 5.5, 5.5, and 5.53 and φ ij φ ji for the octet flavor states pp The following structure is antisymmetric, as you can easily check: ψ A ψ φ 3 + φ 3 + ψ 3 φ + φ 3 + ψ 3 s φ 3 + φ Problem 5.9 a First construct the wave functions, using Eq. 5.6 with 5.5, 5.5, and 5.53 for spin and the figures on pp for flavor: p : udu duu 3 + uud udu + uud duu 3 udu + perms 3 u d u u d u u d u + perms. n : udd dud 3 + dud ddu + udd ddu 3 udd perms 3 u d d + u d d + u d d + perms.

75 7 Σ + : usu suu 3 + uus usu + uus suu 3 usu + perms 3 u s u u s u u s u + perms. Σ : dsd sdd 3 + dds dsd + dds sdd 3 dsd + perms 3 d s d d s d d s d + perms. Ξ : uss sus 3 + sus ssu + uss ssu 3 uss perms 3 u s s + u s s + u s s + perms. Ξ : dss sds 3 + sds ssd + dss ssd 3 dss perms 3 d s s + d s s + d s s + perms. In each of these cases perm stands for three permutations two in addition to the one listed, in which the odd quark occupies the first, second, or third

76 7 5 Bound States position; Σ and Λ are a little trickier, because they involve all three quarks, and there are six permutations: Σ : 3 usd sud + dsu sdu + dus dsu + uds usd + uds sdu + dus sud usd + 6 perms 6 u s d u s d u s d + 6 perms. 6 Λ : 3 uds dus + usd sud dsu + sdu 6 + sud sdu + dus dsu uds + usd 6 + usd dsu + uds sdu dus + sud 6 6 3uds + 6 perms 3 u d s u d s + 6 perms. 3 Now use Eq to determine the magnetic moment of the proton, as in Example 5.3:

77 73 a 3 u d u b c µ i S iz a 3 µ u h µ a h a µ i S iz a 9 µ u µ d ; 3 u d u µ u h µ i S iz b 3 µ b h b µ i S iz b 3 u d u µ i S iz c 3 µ u h µ h d + µ h u a h 3 µ u µ d a + µ h d + µ h u b h 3 µ d b 8 µ d; µ c h c µ i S iz c 8 µ d. + µ h d µ h u c h 3 µ d c The total is the sum times three for the three permutations: µ 3µ a + µ b + µ c 3 9 µ u µ d + 8 µ d 3 4µ u µ d. Comparing the wave functions, we see that no new calculation is required for n, Σ +, Σ, Ξ, and Ξ for some of them there is an overall minus sign in the wave function, but this squares out; sometimes the permutation listed is different, but this doesn t affect the answer. n : same as p, only with u d : µ 3 4µ d µ u. Σ + : same as p, only with d s : µ 3 4µ u µ s. Σ : same as n, only with u s : µ 3 4µ d µ s. Ξ : same as n, only with d s : µ 3 4µ s µ u. Ξ : same as Ξ, only with u d : µ 3 4µ s µ d.

78 74 5 Bound States As for Σ, a 3 u s d µ i S iz a h µ u 3 µ h s + µ h d a h 3 µ u µ s + µ d a µ a h a µ i S iz a 9 µ u µ s + µ d ; b 6 u s d µ i S iz b h µ u 6 + µ h s + µ h d µ b h b µ i S iz b 36 µ u + µ s + µ d ; c 6 u s d µ u h µ i S iz c 6 + µ h s µ h d µ c h c µ i S iz c 36 µ u + µ s µ d. The total is the sum times six for the six permutations: b h 6 µ u + µ s + µ d b c h 6 µ u + µ s µ d c µ 6µ a + µ b + µ c 6 9 µ u µ s + µ d + 36 µ u + µ s + µ d + 36 µ u + µ s µ d 3 µ u + µ d µ s. Finally, for Λ, a 3 u d s b µ u h µ i S iz a 3 µ a h a µ i S iz a 3 u d s µ i S iz b 3 µ u h µ h d + µ h s a h 3 µ u µ s + µ s a µ u µ d + µ s ; µ b h b µ i S iz b µ u + µ d + µ s. + µ h d + µ h s b h 3 µ u + µ d + µ s b