Lecture 11: ABC Theory and Feynman Diagrams
H. A. Tanaka Reminder
• Problem Set 2 due in Box 7 (basement) today by 1700. Today:
• Introduction into Feynman rules for a basic theory • We’ve already played around with Feynman diagrams • Start of the process to learn how to turn these into definite expressions for the amplitude of a process • This has been the missing ingredient in our study so far. • this will be the bulk of the rest of the class • Introduce the “ABC” theory • Realistic theory three species of scalar (spin 0) particles (A, B, C) with an interaction • but it doesn’t correspond to any interaction/particles that we know of • we study it because it is particularly simple and allows us to study the “recipe” without too much mathematical complication. • We’ll see a “real” theory (QED) in the next chapter. Basic Review of Particle Dynamics Basic Review of Particle Dynamics • Interactions between particles are effected by the exchange of particles • All interactions are effected by the exchange of particles • Electromagnetic interactions are the result of the exchange of photons •betweenElectromagnetic electrically forces charged are the particles: result of the exchange of photons between charged particles
µ 0 µ 0 qq1 p1 =(p1, p1) p =(p , p ) EE== 1 ˆrˆr 3 3 3 RR22 FF== qq22EE qµ =(q0, q) q1 q2 q1 q2
µ 0 µ 0 p4 =(p4, p4) p2 =(p2, p2)
Thursday, February 4, 2010 3 TimeTime ordering ordering of of vertices vertices
+
t • The Feynman diagram incorporates possible time orderings of particle exchange. • The “vertical” nature of the exchange is just to illustrate that the amplitude to • Feynman diagrams incorporate possible time orderings corresponding to the diagram is agnostic as to which “direction the exchange particle goes • The vertical exchange illustrates that the amplitude corresponding to the • In illustrating a process, we just “say” that this is the case. diagram is agnostic as to which “direction the exchange particle goes” • In the mathematical statement of the amplitude, the Feynman rules incorporate the possible orderings automatically. • The derivation of the Feynman rules through Quantum Field Theory includes• if we werethis into to derive consideration. them from QFT, we incorporate this in the derivation. Thursday, February 4, 2010 5 Components of a Feynman Diagram
• External Lines: • particles that come in and out in the initial and final state, respectively. • for spin-0, there is no factor. • Vertex factors: • each vertex (i.e. where A, B, C meet) has a factor. • determines “order” of diagram: order=number of vertices • Internal lines and Propagators: • Factors for internal particles exchanged between vertices • Only applies to particles “internal” to the diagram, not external lines • Momentum Conservation at vertex: • (4d) delta function at each vertex enforcing 4-momentum conservation • Integrals over internal momentum: • Internal lines have any momentum consistent with 4-momentum conservation B Basic Structure
TheoryExamples: of 3 “scalar” (spin 0) particles that are distinct with one interaction A, B, C are different types of particles A B B A Vertices C Vertex Internal line C External lines A B External lines
Thursday, February 4, 2010 8 MoreMore examples Examples
A A A C B B B A C C C B A A A B A C B B A C B A C A B
Thursday, February 4, 2010 9 Step The1 Feynman Calculus: Step I
• For a given process (e.g. A→B+C, A+C→A+C), specify the external lines: • For a given process (e.g. A!B+C, A+C!A+C), the particles specify the external lines:
B A C A ? ? A C C
Thursday, February 4, 2010 10 StepDiagrams 2: diagram and Order by “order”
• •IdentifyNow we all identifydiagrams all thatdiagrams contain that ≤N contain vertices, N wherevertices, N iswhere the order N is thewe orderwant towe calculatewant to thecalculate amplitude the amplitude to: to:
B A
C
• •CheckNote thatthat eachin order vertex to connect follows twothe vertexvertices, “rules” one has to “eliminate” one or two legs by connecting it to the other. • There are no second order diagrams for this process • There are no second order diagrams for this process
Thursday, February 4, 2010 11 Writing down the Amplitude (I)
• For each diagram, label the “flow” of energy momentum along each leg
p 3 3 StepWriting 3: Label down the2 the momentum Amplitude pflow, (I) vertex factorsp p 1 B p 1 • ForA each diagram, label the “flow” of energy momentum along each leg q q p1 3 3 p p 2 p p 2 pp 1 2 p p p p 3 CB 4 4 1
A q q p1 • Now we start writing down terms: 2 2 p p p p 3 C p • For each vertex, write down a factor of -ig 4 4 ig ( ig)2 ( ig)2 • Now we start writing down terms: • WeThursday, can February direct 4, 2010 the arrows in any way so long as we are consistent 13 • For each vertex, write down a factor of -ig • Introduce a factorig of (-ig) for each vertex.( (Feynmanig)2 rule (for igvertex))2
Thursday, February 4, 2010 13 Writing down the amplitude StepWriting 4: Propagators down the amplitude
• •ForFor each each internal internal line, line, write write a apropagation propagation factor: factor: i • Introduce a propagator term for each internal line i q2q2 mm2c22c2 3 3 p 3 3 2 p p p 2 p p p p 1 1 BB p p 1 1 AA q q q q p1p1 2 2 2 p p p p 2 p p p p p 3 C p 3 C 4 4 4 4 2 i i 2 i 2 2 i ( (igig) ) ( (igig) ) 2 ( ig) q2 2 m c22 2 ( ig) 2 2 22 q mB c q q2 mmBc c2 B B
• Note that we have to use the momentum we assigned to the line!
Thursday, February 4, 2010 14 Thursday, February 4, 2010 14 4-momentum conservation 4-momentum conservation Step 5: energy-momentum conservation • For each vertex, write a delta function to enforce 4-momentum conservation • For each vertex, write a delta function to enforce 4-momentum conservation • Introduce• + if “into” a delta the vertex,function - enforcingif “out of” 4-momentumthe vertex conservation at each vertex • + if “into” the vertex, - if “out of” the vertex p 3 3 2 p p 3 3 p 1 p 2 p p p B 1 p 1
B p A 1 q
A q q
p1 q p 1 p 2 p 2 2 p p p p 3 C 2 p p 4 4 p 3 C p 4 4 4 4 2 i 2 i ( ig) (2⇥) (p1 p2 p3) ( ig) i ( ig) ⇥ 4 4 q22 m2 c2 22 i2 2 ( ig) (2⇥) (p1 p2 p3) ( ig) B ( ig)q mBc 2 2 2 2 2 2 ⇥ 4 4 q mBc 4 4 q m c (2⇥) (p1 p 3 q) (2⇥) (p p Bq) ⇥ 4 4 4 4 1 4 (2⇥4) 4 (p1 p3 q) ⇥ (2⇥4) 4 (p1 p4 q) ⇥(2⇥) (p2 p4 + q) (2⇥) (p p + q) 4 4 ⇥ 4 4 2 3 ⇥ (2⇥) (p2 p4 + q) ⇥ (2⇥) (p p + q) ⇥ ⇥ 2 3
Thursday,• convention:February 4, 2010 momentum in/out of vertex is (+/-) 15 Thursday, February 4, 2010 15 Integration4-momentum over conservationInternal Momenta 1
Step 6: integrate over internal momenta:1 • For• For each each internal vertex, line, write write a delta down function a factor to of: enforce 4-momentumd4q conservation (2 )4 • Introduce• where• + if “into”q an is integralthe the 4-momentum vertex, over the- if “outmomentum of thatof” theparticle vertexof each internal line: • integral over each such 4-momenta 3 3 3 3 p p 2 p p p 2 p p 1 p B 1 p B p 1
1 A
A q q q p1 q p 1 2 2 2 p p p p C 2 p p p 3 p p p 4 4 3 C 4 4 4 4 ( ig) (2⇥) (p1 p2 p3) 4 4 2 i 2 i ( ig⇥) (2⇥) ( p1 p2 p3) ( ig) 1 i 2 2 2 4( ig) 22 2 2 ⇥ q mBc d q ( ig)q m c 1 4 2 i 4 2 B2 2 4 4 (2⇥) q mBc 4 d q( ig) 2 2 2 4 4 (2⇥) (2⇥)q (p1m cp3 q) (2⇥) 4 (4p1 p4 q) ⇥ B ⇥ (2⇥) (p1 p4 q) 4(24⇥)4 4(p p + q) ⇥ 4 4 (2⇥) (p1 p23 q4) (2⇥) 4 (4p2 p3 + q) ⇥ ⇥ ⇥ (2⇥) (p2 p3 + q) (2⇥)4 4(p p + q) ⇥ ⇥ 2 4 Thursday, February 4, 2010 15 Thursday, February 4, 2010 16 4-momentumIntegration over conservation Internal Momenta 2 Step 7: perform the integral •• ForIntegrate each vertex,over each write internal a delta 4-momenta function to enforce 4-momentum conservation • Perform• + if “into” the integrals,the vertex, leaving - if “out a δof” function the vertex for overall 4-p conservation
3 33 p 3 2 p p pp p 2 p1 p 1 BB pp 1 1 AA q q q p q p11 2 p 2 p 2 p p 2 p p3 C p p p 3 C p4 4 4 4 ( ig) (2⇥)4 4(p p p ) 4 4 1 2 3 2 i 2 i ( ig) ⇥(2⇥) (p1 p2 p3) ( ig) ( ig) ⇥ i q2 m2 c2 q2 im2 c2 ( ig)2 B ( ig)2 B 4 42 2 2 4 4 2 2 2 (p4(2⇥)p2 ) (p1 m pc3 q) (2⇥) (p(3p p2p) qm) c ⇥ B 1 4 B 4 4 4 4 ⇥ 4 4 4 (2⇥) (p1(2⇥)p3 (pp24 + pp42)+ q) (2⇥(2)⇥ ) ( p1(p p3p +p4q+) p2) ⇥ ⇥ ⇥ ⇥ 2 3
Thursday, February 4, 2010 15 Thursday,• FebruaryIn lowest 4, 2010 order calculations, each integral will meet with a delta function. 17 4-momentumCancel the overall conservation delta function Result: Result:• For each vertex, write a delta function to enforce 4-momentum conservation • Eliminate the overall delta function (along with the (2")4): • Eliminate• + if “into” the the δ function, vertex, - andif “out the of” Resulting the vertex expression is -iM • The result is equal to -iM 3 3 3 p 3 p 2 p p p p 2 p1 p B 1 p 3 B p1 3 p
p 1
2 p A p 1
A q
p q
B q q p1 1 p A 1 2 p q 2 2 p p p q p p C 2 p 3 p p p 4 1 3 C p4 4 4 p 2 p 2 p p p (4 ig4 ) 3 C i i 2i 2 i 4 ( ig) (2⇥) (p1 p2 p3) 2 ( ig) 4 2( ig) ( ig) 2 2 2 ( ig) 2 2 22 ⇥ 2q m2 B2c (p pq )2 mmc c2 (p4 p2) mBc 3 2 B B ( ig) 4 4 i 4 4 i 2 (2⇥) (p1 p3 q) ( ig(2)2⇥) (p1 p4 q) ( ig)⇥ 2 2 2 ⇥ 2 2 2 (p4 4 p42) m c (p434 p2) mBc (2⇥) (p2 p4B+ q) (2⇥) (p2 p 3 + q) ⇥ ⇥ = g g2 g2 M = = 2 2 2 2 2 2 Thursday, February 4, 2010 M (p4 p2) m c M (p3 p2) mBc 15 Thursday, February 4, 2010 B 18 These amplitudes must be added
Thursday, February 4, 2010 19 DecayDecayDecayDecay RateRate RateRate ofof AA !of→of B+CB+CAA!!B+CB+C
•• RecallRecall•• RecallourRecallour formulaformula ourour formulaformula forfor thethe for fortwo-bodytwo-body thethe two-bodytwo-body decaydecay decayofofdecay aa particle:particle: ofof aa particle:particle:
SS 2 S= 2 2 p = = 2 p p22 288 mm21cc |M|2 | | 8 m1c |M|1 |M|| | | |
•• NowNow• •thatthatNowNow wewe thatthat cancan wewe calculatecalculate cancan calculatecalculate thethe amplitude,amplitude, thethe amplitude,amplitude, wewe cancan wewe finishfinish cancan ourour finishfinish calculation:calculation: ourour calculation:calculation:
1 2 1= 1 2 g2 p = = g 2 pg pB/CB/C 288 mm2Acc B/C | | 8 mAc A | || |
•• from from• •PS PSfromfrom 2 2 PSPS 22
c cc 4 4 4 2 2 2 2 2 2 ppB/C == 4 mm44A++mm44B ++mm4C2 222mm2Amm2B2 222mm2Amm2C2 222mm2Bmm2C pB/C |=B/C | 2mmAA+ mBA + mCB 2mCAm B A2mBAm C A2mCBm C B C | || 2m|A 2mA
Thursday, February 4, 2010 20 Thursday, FebruaryThursday, 4, 2010 February 4, 2010 20 20 ScatteringScatteringScatteringScattering rate raterate rate of ofof ofA+A A+AA+A A+A!!!C+CC+CC+CC+C in inin CMin CMCM CM Frame FrameFrame Frame ScatteringScattering rate rate of of A+A A+A!!→C+CC+C in in CM CM Frame Frame
• From the textbook (6.47): • From•• •FromFromFrom the thethe thetextbook textbooktextbook textbook (6.47): (6.47):(6.47): (6.47), assume M=MA=MC, MB = 0 • From the textbook (6.47):2 2 22 2 22 p d⇥dd⇥⇥ c cc S SS 2 pfppff d⇥= = c 2 |M|S |M| 2| ||| f || d⇥== c S|M||M|2 22| p|f d dd =8 88 (E(1(EE+11 E++2EE) 22))2pi pp| ii | d 8⇥ ⇥⇥ (E1 +|M|E2) 2|pi | d 8 ⇥ (E1 + E2|) ||| p||i ⇥p3 | | p3pp33 2 1 11 1 11 = =g=2 gg22 1 + ++ 1 p3 = g 2 2 221 2 22 22 + 2 221 2 22 22 p1pp11 p2pp22MMM = g(p4((pp44 p2)pp22))2 m mmc2Bcc2 (p+3((pp33 p2)pp22))2 m mmc2Bcc2 p1 ### p2 M (p4 p2) 2 BmBB2c 2 (p3 p2) 2 BmBB2c⇥ 2⇥ p1 # p2 M ( p4 p 2) m c ( p3 p 2) m c⇥⇥ # B B ⇥ p4ppp444 p4 2 22 2 22 22 2 222 22 2 22 22 2 22 22 2 22 (p4((pp44p2)pp22=))2 m== mmc2C+cc2 m++mmc2Acc2 2p422pp4p42 pp=22 m== mmc2C+cc2 m++mmc2Acc2 2((2((2((E/cE/cE/c) ))2 p ppp4 2)pp22)) (p 4 p2) 2= CmCCc2 +2 AmAAc2 2 2p·4 ··p2 = CmCCc2 +2 AmAAc2 2 2((E/c) 2 4p·44 ··p2) (p 4 p2) = mC c + mAc 22p·4 22 p22 =2m22C c22 + mAc 2 222((E/c) p4· p2) = m== mmc2C· +cc2 m++mmc2Acc2 2((2((2((E/cE/cE/c) ))2 pf ppff pi p pcosii coscos )· )) = CmCCc2 +2 AmAAc2 2 2((E/c) 2| ||p|·|f |·||·||pi ||cos ) = mC c + mAc 2((E/c) | p|·|f p|i cos ) 2 22 2 22 22 2 222 22 2 22 22 2 22 22 |2 22|·| | (p3((pp33p2)pp22=))2 m== mmc2C+cc2 m++mmc2Acc2 2p322pp3p32 pp=22 m== mmc2C+cc2 m++mmc2Acc2 2((2((2((E/cE/cE/c) ))2 p ppp3 2)pp22)) (p 3 p2) 2= CmCCc2 +2 AmAAc2 2 2p·3 ··p2 = CmCCc2 +2 AmAAc2 2 2((E/c) 2 3p·33 ··p2) (p 3 p2) = mC c + mAc 22p·3 22 p22 =2m22C c22 + mAc 2 222((E/c) p3· p2) = m== mmc2C· +cc2 m++mmc2Acc2 2((2((2((E/cE/cE/c) +))2 ++pf ppff pi p pcosii coscos )· )) = CmCCc2 +2 AmAAc2 2 2((E/c) 2+| ||p|·|f |·||·||pi ||cos ) m mm=A m== mm=C m,== m,m, m m m=0B =0=0 = mC c + mAc 2((E/c) +| p|·|f p|i cos ) AmAA = CmCC = m,B mBB =0 | |·| | mA = mC = m, m2B 22=02 222 22 2 22 2 22 2 22 (p ((pp44p )pp22=2))2 =2=2m mmc2 cc2 2((2((2((E/cE/cE/c) ))2 p ppcos2 coscos )= )=)=2 p22 pp(12 (1(1coscoscos ) )) 4(p4 2p2) 2=2m c2 2 2((E/c) 2 p cos2 )= 2 p (12 cos ) (p 4 p2) =2m c 2((E/c ) | |||| |p|| cos )= | |||2| |p|| (1 cos ) 2 22 2 222 22 2 22 2 22 2 22 (p ((pp33p )pp22=2))2 =2=2m mmc2 cc2 2(( 2((2((E/cE/cE/c) +))2 ++p ppcos|2 |coscos )= )=)=2 p 22 pp(1|2 +(1|(1 cos + + cos cos ) )) 3(p3 2p2) 2=2m c2 2 2((E/c) +2 p cos2 )= 2 p 2(1 + cos ) (p 3 p2) =2m c 2((E/c)| +|||| |p|| cos )= | ||2|| p||| (1 + cos ) Thursday,Thursday, FebruaryFebruary 4,4, 20102010 2121 Thursday,Thursday, February February 4, 2010 4, 2010 | | | | 21 21 Thursday, February 4, 2010 21 ToTo thethe end:end: ToTo To thethe the end:end: end: Endgame
22 1 11 1 11 2 2 ==gg 1 ++1 = g2= g 1 22 + + 1 22 = g2 MM 2 2 22pp (1(1+ coscos ))2 2 22pp (1(1 + + cos cos )) M=Mg 2 p2 p(1 (1 cos|| |cos| )+ ) 2 p2 p(1 +(1 cos| +| | cos| ) ) ⇥⇥ M 2 p | 2|(1| | cos ) 2 p | 2|(1| | + cos ) ⇥ ⇥ | | 22 | | 22 ⇥ 2 22 22pp (1(1 + + cos cos2 ))2 22pp (1(1 coscos )) 22 p=2 pg(1 +(1 cos + cos ) )2 p2 p(1 (1coscos ) ) 2 2 p 2=(1g + cos|| ||) 2 p424(1 cos||22 ||) = g2= g2 p | |(1| | + cos )4 424p p|p |(1(1|(12| coscoscos ) )) = g | | 44p (1 | cos|2 ) 4 p42 (1 cos|| 2|| ) 2 4 p| gg|(1|2 | cos ) 2 g | | g2 == 2 = = g 2 22 2 = 2 2 2 pp sinsin = p2 psin2sin || | | p| 2||sin| | | 2 2 2 22 p dd⇥⇥2 cc 2 2SS p pff d⇥ d⇥ c 2c= S S 2 p f || || d⇥ = c =S 2 p|M||M|f| | 22 SS=1=1//22,E,E11==EE22,, pp == ppii d⇥ = cdd S88|M| |M|((EE|1f2++|EE2)) ppSi =1/2,E1 = E2, p = pffi =d 8 |M|(E1 + E2|2)1 | pi 2 S =1i /2,E1 = E2, pf =f pi|| || || || d 8 (E 1 +⇥E⇥2)2 pi S =1|| |/|2,E1 = E2, p| f ||= |p| i | | d 8 ⇥ (⇥E1 + E2) p| i | | | | | | ⇥ | | | | | | 22 44 dd⇤⇤ 2112 cc 4 gg d⇤ 1 2c 4 g d⇤ 1 c== g4 2 d⇤ =1= d c 2 8⇥ g 222 22 2 22 =d 2d 8⇥ 2 82⇥ 22 (2(22E2E p2p sin2sin )) d = 2 8⇥ (2 E(22 Ep⇥⇥2 psin2sin |)2| | |) d 2 8⇥ ⇥ (2⇥E2 p| 2||sin| )2 ⇥ | |
Thursday,Thursday, February February 4, 4, 2010 2010 2222 Thursday, February 4, 2010 22 Thursday, February 4, 2010 22 Thursday, February 4, 2010 22