Feynman Diagrams and How to Use Them

Introducon to the famous Feynman diagrams and how to use them

Rosi Reed First Diagram

Diagrams introduced at the the Pocono Conference in early April, 1948 • An hours drive from here! • Discussion over the problems in QED First published in May, 1949

Rosi Reed – Feynman 2 Feynman Van

Rosi Reed – Feynman 3 Imagery and Understanding Feynman always thought about the representaon of the problems he studied • Invented his own trigonometry symbols in high-school • He had great creave power as a problem solver Who else would have thought of drawing pictures to calculate quanes?

Rosi Reed – Feynman 4 Quantum Electrodynamics Relavisc quantum ﬁeld theory of electrodynamics Inial descripon by Dirac à ensemble of harmonic oscillators w/creaon + annihilaon operators • Calculaons γ+charged parcles were reliable only to 1st order in perturbaon theory – At higher orders inﬁnies emerged – Fundamental incompability between special relavity and quantum mechanics

Rosi Reed – Feynman 5 Quantum Electrodynamics Shin'ichirō Tomonaga, Julian Schwinger, Richard Feynman à Fully covariant formulaons ﬁnite at any order • Nobel Prize in 1965 • Schwinger and Tomonaga’s approach was ﬁeld- theorec and operator-based • Feynman's approach was based on his diagrams • Freeman Dyson à 2 approaches are equivalent

Rosi Reed – Feynman 6 How are the Diagrams used?

Rosi Reed – Feynman 7 Fermi’s Golden Rule

Describes the transion rate from one energy eigenstate of a quantum system into other energy eigenstates in a connuum 2 Γ fi = 2π | Tfi | ρ(Ei )

Rosi Reed – Feynman 8 Fermi’s Golden Rule Describes the transion rate from one energy eigenstate of a quantum system into other energy eigenstates in a connuum

2 Γ fi = 2π | Tfi | ρ(Ei )

Density of States Natural Units (Kinemacs) c = ! =1 Transion Matrix (Diagrams)

Rosi Reed – Feynman 9 Fermi’s Golden Rule Describes the transion rate from one energy eigenstate of a quantum system into other energy eigenstates in a connuum 2 Γ fi = 2π | Tfi | ρ(Ei )

Transion Matrix Derived using Time Dependent (Diagrams) Perturbaon Theory

ˆ ˆ ˆ Hamiltonian is split into a me H = H0 + H ' independent (H0) and an interacon (H’) part

Rosi Reed – Feynman 10 Fermi’s Golden Rule Describes the transion rate from one energy eigenstate of a quantum system into other energy eigenstates in a connuum 2 Γ fi = 2π | Tfi | ρ(Ei )

ˆ ˆ ˆ T | Hˆ ' | H = H0 + H ' fi =< ψ f ψi >

1st order calculaon of the transion matrix from the inial state (ψi) to the ﬁnal state (ψf)

Rosi Reed – Feynman 11 Fermi’s Golden Rule Describes the transion rate from one energy eigenstate of a quantum system into other energy eigenstates in a connuum 2 Γ fi = 2π | Tfi | ρ(Ei )

T =< ψ | Hˆ ' |ψ > fi f i ρ(Ei ) = ∫ δ(Ei − E)dn dn = # of states with momentum p Combine together + some normalizaon from quanzaon (and insure Lorentz Invariance!)

Rosi Reed – Feynman 12 Lecture 22

Last time we discussed the scenario a 1 + 2, but since much particle physics is done in accelerators, we want to be! able to calculate things for the scenario a + b 1 + 2. First! we should deﬁne a few things: # of particles =ﬂux= Unit time Unit area Cross-Secons ⇥# of particles n =numberdensity= Unit Volume = cross-section = e↵ective area The transion probability is related to Γ fi →σ the cross-secon (σ)

σ depends on the momentum of the parcles and the Figure 1:parcular interacon force On the left are the two tubes (”beams”) of particles a and b, colliding together. On the right is the cross-section area, showing the overlap between the beam and particles b. Rosi Reed – Feynman 13 From Figure 1 we can see that the volume of the overlap region can be determined as: V = A(va + vb)t The total number of b particles in the overlap volume is: Nb = nbV = nbA(va + vb)t The probability that a particle a interacts is the area covered by the b particles divided by the total area of the beam, also shown in 1:

Nb nbA(va + vb)t P = = = n A(v + v )t A A b a b The interaction rate is the probability of interacting per unit time:

P r = = n (v + v ) a t b a b The reaction rate for the entire beam of a particles is:

Ra = Nara = naVra =(na(va + vb))(nbV ) = Nb

From here we can see that Rate = Flux # target particles cross-section. ⇥ ⇥

1 Cross-Secon (Center of Mass) 1 p* f | M |2 d * σ = 2 * ∫ fi Ω 64π s pi

p* Tfi → M fi f * p i

s à Total energy in the center of mass frame

Rosi Reed – Feynman 14 Cross-Secon (Center of Mass) 1 p* f | M |2 d * σ = 2 * ∫ fi Ω 64π s pi

Calculate using diagrams! 1 α ≈ g2 ≈ EM EM 137 QED Vertex Element Time

Rosi Reed – Feynman 15

Figure 1: Standard model vertices. Time runs from left to right. For the left- 2 1 most vertex (EM) the coupling constant is ↵EM gEM 137 . All charged particles can interact through the EM force, and particle⇡ flavor⇡ must be conserved in all interactions. The second vertex represents the strong (QCD) force. the cou- 2 pling constant is ↵s gs 1. Only quarks can interact through QCD, but like the EM force particle⇡ flavor⇡ must be conserved in all interactions. The next two vertices are for the weak force, with the W ± being the EM charged boson and Z 2 1 being the neutral boson. The weak coupling constant is ↵W—Z gW—Z 30 . All fermions can interact through the weak force, with the W the⇡ particle flavor⇡ is always changed and with the Z the flavor is never changed.

Figure 2: This is a LO diagram for the e + e e + e process in the t-channel conﬁguration. There is a second diagram! that also contributes, the u-channel, since the particles in the ﬁnal state are identical.

-hne,sneteprilsi h nlsaeaeidentical. are state ﬁnal the in particles the since u-channel,

-hne oﬁuain hr sascn iga htas otiue,the contributes, also that diagram second a is There conﬁguration. t-channel

rcs nthe in process e + e e + e the for diagram LO a is This 2: Figure

h ao snvrchanged. never is ﬂavor the Z the with and changed always is

l emoscnitrc hog h ekfre ihteWtepril ﬂavor particle the W the with force, weak the through interact can fermions All

⇡ ⇡

W—Z W—Z

. g ↵ is constant coupling weak The boson. neutral the being

2 1

± Z and boson charged EM the being W the with force, weak the for are vertices

h Mfrepril ao utb osre nalitrcin.Tenx two next The interactions. all in conserved be must ﬂavor particle force EM the

⇡ ⇡

s s .Ol urscnitrc hog C,btlike but QCD, through interact can quarks Only 1. g ↵ is constant pling 2

l neatos h eodvre ersnstesrn QD oc.tecou- the force. (QCD) strong the represents vertex second The QED Vertex interactions. all lscnitrc hog h Mfre n atceﬂvrms ecnevdin conserved be must ﬂavor particle and force, EM the through interact can cles

⇡ ⇡

137

EM EM l hre parti- charged All . g ↵ is constant coupling the (EM) vertex most

1 Fermions are parcles iue1 tnadmdlvrie.Tm usfo ett ih.Frteleft- the For right. to left from runs Time vertices. model Standard 1: traveling forward in me Figure

An-fermions are parcles traveling backwards in me • Stueckelberg interpretaon

Figure 1: Standard model vertices. Time runs from left to right. For the left- 2 1 most vertex (EM) the coupling constant is ↵EMTime g . All charged parti- + e+ EM 137 cles can interact throughe the EM force, and particle⇡ flavor⇡ must be conserved in all interactions. The second vertex represents the strong (QCD) force. the cou- Rosi Reed – Feynman 2 16 pling constant is ↵s gs 1. Only quarks can interact through QCD, but like the EM force particle⇡ flavor⇡ must be conserved in all interactions. The next two vertices are for the weak force, with the W ± being the EM charged boson and Z 2 1 being the neutral boson. The weak coupling constant is ↵W—Z gW—Z 30 . All fermions can interact through the weak force, with the W the⇡ particle flavor⇡ is always changed and with the Z the flavor is never changed.

-hne,sneteprilsi h nlsaeaeidentical. are state ﬁnal the in particles the since u-channel,

-hne oﬁuain hr sascn iga htas otiue,the contributes, also that diagram second a is There conﬁguration. t-channel

rcs nthe in process e + e e + e the for diagram LO a is This 2: Figure . Z 1 30 ⇡ 2 W—Z g process in the ⇡ e . All charged parti- W—Z + 1 ↵ 137

e h ao snvrchanged. never is ﬂavor the Z sawy hne n ihthe with and changed always is ⇡

l emoscnitrc hog h ekfre ihteWtepril ﬂavor particle the W the with force, weak the through interact can fermions All !

⇡

⇡ 2 EM

30 W—Z

W—Z g . g ↵

en h eta oo.Tewa opigcntn is constant coupling weak The boson. neutral the being

1 e

⇡

± Z and boson charged EM the being W etcsaefrtewa oc,wt the with force, weak the for are vertices +

being the EM charged boson and

h Mfrepril ao utb osre nalitrcin.Tenx two next The interactions. all in conserved be must ﬂavor particle force EM the EM

⇡

⇡ e

↵ ±

s .Ol urscnitrc hog C,btlike but QCD, through interact can quarks Only 1. g ↵ is constant pling 2 W -hne ofiuain hr sascn iga htas otiue,the identical. are contributes, state also final that the diagram the in second particles for the a diagram since is LO u-channel, There a configuration. is t-channel This 2: Figure flavor particle the W the the with with and force, is changed weak always constant the is through coupling interact weak can The fermions two All boson. next The neutral the interactions. with the all force, being weak in the conserved for be are must vertices cou- flavor the particle force force. EM (QCD) in the strong conserved the be is represents must left- constant flavor vertex pling the particle second For and The force, right. interactions. EM to all the left through interact from can is runs constant cles coupling Time the (EM) vertices. vertex model most Standard 1: Figure 2

l neatos h eodvre ersnstesrn QD oc.tecou- the force. (QCD) strong the represents vertex second The QED Vertex interactions. all

lscnitrc hog h Mfre n atceflvrms ecnevdin conserved be must flavor particle and force, EM the through interact can cles the flavor is never changed. ⇡ ⇡

137

EM l hre parti- charged All . g ↵

otvre E)teculn osatis constant coupling the (EM) vertex most Z

1 e+ iue1 tnadmdlvrie.Tm usfo ett ih.Frteleft- the For right. to left from runs Time vertices. model Standard 1: Figure 1. Only quarks can interact through QCD, but like ⇡ 2 e− s g ↵ s ⇡ ⇡ s ↵ g s 2 e+ ⇡ .Ol urscnitrc hog C,btlike but QCD, through interact can quarks Only 1.

Z e−

Figure changed. never is flavor the 1: Standard model vertices. Time runs from left to right. For the left- most vertex (EM) the coupling constant is ↵Time g2 1 . All charged parti- + + EM EM 137 2 e e Figure 1: Standardmost model vertex vertices. (EM) the Time couplingcles constant runs is can from interact through left theall to EM interactions. right. force, The and For second particle thepling vertex flavor constant left- must represents is be the conserved strongthe in (QCD) EM force. force particle the flavor cou- vertices must are be for conserved the in weakbeing force, all the with interactions. the neutral The next boson.All two fermions The can weak interact coupling throughis the constant always weak changed is force, and with with the the W the particle flavor Figure 2: Thist-channel is configuration. a Thereu-channel, LO is since diagram a the for particles second in the diagram the that final also state contributes, are identical. the cles can interactW through the EM force, and particle⇡ flavor⇡ must be conserved in ± ↵ e all interactions. The secondEM vertex represents the strong (QCD) force. the cou- en h Mcagdbsnand boson charged EM the being Rosi Reed – Feynman 2 17 + pling constant is ↵s gs 1. Only quarks can interact through QCD, but like ⇡ ⇡⇡ e the EM force particle flavor must be conserved in all interactions. The next two g EM 2

! vertices are for the weak force, with the W ± being the EM charged boson and Z 2 1 being the neutral boson. The⇡ weak coupling constant is ↵W—Z gW—Z . e ⇡ ⇡ 30 137 All fermions↵ can interact through the weak force, with the W the particle ﬂavor 1 + W—Z

is always changed and with parti- charged All . the Z the ﬂavor is never changed. e ⇡ rcs nthe in process g W—Z 2 ⇡ 30 1 Z .

2 Figure 1: Standard model vertices. Time runs from left to right. For the left- 2 1 most vertex (EM) the coupling constant is ↵EM gEM 137 . All charged particles can interact through the EM force, and particle⇡ flavor⇡ must be conserved in all interactions. The second vertex represents the strong (QCD) force. the cou- 2 pling constant is ↵s gs 1. Only quarks can interact through QCD, but like the EM force particle⇡ flavor⇡ must be conserved in all interactions. The next two vertices are for the weak force, with the W ± being the EM charged boson and Z 2 1 being the neutral boson. The weak coupling constant is ↵W—Z gW—Z 30 . All fermions can interact through the weak force, with the W the⇡ particle flavor⇡ is always changed and with the Z the flavor is never changed.

e-e- à e-e- Lowest level diagram that conserves energy and momentum • Leading Order (LO)

At each vertex the following quanes are conserved: • Energy • Momentum (linear+angular) • Charge • Lepton # Time • Baryon #

Rosi Reed – Feynman 18

e-e- à e-e- At each vertex the following quanes are conserved: • Energy • Momentum (linear+angular)

Virtual Parcle • Does not have its “on shell” mass • Conservaon E+p! • Exists for only a short while • Uncertainty Principle Time

Rosi Reed – Feynman 19

2 e-e- à e-e-

Time

FigureWe can not know whether the top electron emits a photon 3: This is a LO diagram for the e + e e + e process in the t-channel conﬁguration indicating the two di↵erent time! ordered processes that are representedwhich is absorbed by the boom or vice versa by the ﬁnal diagram to the right. à Both are represented in this diagram!

The standard model interactionRosi Reed – Feynman vertices are shown in Figure 1. Each vertex20 represents the coupling of a particle with its force carrier. The fermions are shown with arrows indicating the direction in time that they are going. At the center of each vertex is a coupling constant, which varies depending on the force in question. Energy and momentum must be conserved at each vertex, which means for a Feynman diagram with any number of vertices, energy and momentum must be conserved.

Figure 4: On the left is a more accurate depiction of what happens with electron scattering. On the right is a Next to Leading Order (NLO) diagram of electron scattering.

The problem with conserving momentum and energy at each vertex, is that this is not possible with a gauge boson of zero mass (photons and gluons). (See HW 2 ) The way this problem is solved is that the gauge bosons are allowed to go ”o↵ mass shell”, which means that they can have whatever mass is necessary

3 Figure 1: Standard model vertices. Time runs from left to right. For the left- Figure 1: Standard model vertices. Time runs from left to right. For the left- most vertex (EM) the coupling constant is ↵ g2 1 . All charged parti- most vertex (EM)EM theEM coupling137 constant is ↵ g2 1 . All charged particles can interact through the EM force, and particle⇡ flavor⇡ must be conservedEM in EM 137 cles can interact through the EM force, and particle⇡ flavor⇡ must be conserved in all interactions. The second vertex represents the strong (QCD) force. the cou- all interactions. The second vertex represents the strong (QCD) force. the coupling constant is ↵ g2 1. Only quarks can interact through QCD, but like s s pling constant is ↵ g2 1. Only quarks can interact through QCD, but like the EM force particle⇡ flavor⇡ must be conserved ins all interactions.s The next two the EM force particle⇡ flavor⇡ must be conserved in all interactions. The next two vertices are for the weak force, with the W being the EM charged boson and Z vertices are± for the weak force, with the W being the EM charged boson and Z being the neutral boson. The weak coupling constant is ↵ g2 ± 1 . being the neutral boson. TheW—Z weak couplingW—Z constant30 is ↵ g2 1 . All fermions can interact through the weak force, with the W the⇡ particle flavor⇡ W—Z W—Z 30 All fermions can interact through the weak force, with the W the⇡ particle flavor⇡ is always changed and with the Z the flavor is never changed. is always changed and with the Z the flavor is never changed.

e-e- à e-e-

Time

Figure 2: This is a LO diagram for the e + e e + e process in the We can not know whether the top and boom electron Figure 2: This is a LO diagram for the e + e e + e process in the t-channel configuration. There is a second diagram! that also contributes, the t-channel configuration. There is a second diagram! thatà also contributes, the u-channel, since the particlesswitch places aer the interacon (idencal parcles) in the final state are identical. Both diagrams are neededu-channel, since the particles in the final state are identical.

Rosi Reed – Feynman 21

2 2 Figure 3: This is a LO diagram for the e + e e + e process in the t-channel conﬁguration indicating the two di↵erent time! ordered processes that are represented by the ﬁnal diagram to the right.

The standard model interaction vertices are shown in Figure 1. Each vertex represents the coupling of a particle with its force carrier. The fermions are shown with arrows indicating the direction in time that they are going. At the center of each vertex is a coupling constant, which varies depending on the force in question. Energy and momentum must be conserved at each vertex, which means for a Feynman diagram with any number of vertices, energy and momentum must be conserved. e-e- à e-e- 1 α ≈ g2 ≈ EM EM 137

For QCD this becomes

a problem as αs ~ 1!! • A problem for another day Time

All higher order diagrams also contribute, but fortunately Figure 4: On the left is a more accurate depiction of what happens with electron each set of verces contributes a factor of less scattering. On the right is a Next to Leading Order (NLO) diagram of electron αEM scattering. • NLO ~1% of LO contribuon Rosi Reed – Feynman 22 The problem with conserving momentum and energy at each vertex, is that this is not possible with a gauge boson of zero mass (photons and gluons). (See HW 2 ) The way this problem is solved is that the gauge bosons are allowed to go ”o↵ mass shell”, which means that they can have whatever mass is necessary

3 Figure 3: This is a LO diagram for the e + e e + e process in the t-channel conﬁguration indicating the two di↵erent time! ordered processes that are represented by the ﬁnal diagram to the right.

Figure 3: This is a LO diagram for the e + e e + e process in the ! The standard model interaction vertices are shown in Figure 1. Eacht-channel vertex conﬁguration indicating the two di↵erent time ordered processes that are represented by the ﬁnal diagram to the right. represents the coupling of a particle with its force carrier. The fermions are shown with arrows indicating the direction in time that they are going. At the center of each vertex is a coupling constant, which varies dependingThe on standard the model interaction vertices are shown in Figure 1. Each vertex force in question. Energy and momentum must be conserved at eachrepresents vertex, the coupling of a particle with its force carrier. The fermions are shown with arrows indicating the direction in time that they are going. At which means for a Feynman diagram with any number of vertices, energythe center and of each vertex is a coupling constant, which varies depending on the momentum must be conserved. force in question. Energy and momentum must be conserved at each vertex, - which- means- for- a Feynman diagram with any number of vertices, energy and e emomentum à e muste be conserved.

Mij

Time

All we measure are the kinemacs of two electrons in the Figure 4: On the left is a more accurate depiction of what happens with electron Figure 4: On the left is a more accurate depiction of what happens with electron inial and two electrons in the ﬁnal state! scattering. On the right is a Next to Leading Order (NLO) diagram of electron scattering. On the right is a Next to Leading Order (NLO) diagramscattering. of electron scattering. The problem with conserving momentum and energy at each vertex, is that Rosi Reed – Feynman this is not possible with a gauge boson of zero23 mass (photons and gluons). (See The problem with conserving momentum and energy at each vertex,HW 2 is ) that The way this problem is solved is that the gauge bosons are allowed to this is not possible with a gauge boson of zero mass (photons and gluons).go ”o↵ mass (See shell”, which means that they can have whatever mass is necessary HW 2 ) The way this problem is solved is that the gauge bosons are allowed to go ”o↵ mass shell”, which means that they can have whatever mass is necessary 3

3 Feynman Example

Covers Griths 7.3 Feynman Rules for QED 1) For each external line, label the terms p1,p2,...,pn and s1,s2,....sn. For each internal line, label the line with q1,q2...., qn. Then give each line an arrow indicating the direction of ﬂow. For electrons, the arrows should point in the positive time directly (left to right), for positrons they should point in the opposite direction. The ﬂow for internal lines is ambiguous, but each vertex should have at least one arrow entering and leaving. 2) For each external line add the following terms:

e going into the diagram - u • e going out of the diagram -u ¯ • e+ going into the diagram -v ¯ • Feynman Example e+ going out of the diagram - v • Covers Griths 7.3 going into the diagram - ✏µ • Feynman Rules for QED µ 1) For each external line, label the terms p1,p2,...,pn and s1,s2,....sn. For going out of the diagram - ✏ ⇤ • each internal line, label the line with q1,q2...., qn. Then give each line an arrow µ indicating the direction of flow. For electrons, the arrows should point in the 3) For each vertex add igem . (Note,positive I time will directlydrop the (left em to subscript right), for and positrons just they should point in the Feynmanuse g since we Example have so many subscriptsopposite since direction. we will not The be flow doing for internal weak or lines QCD is ambiguous, but each vertex calculations.) should have at least one arrow entering and leaving. Covers4) Gri Forths each 7.3 internal line add the following2) For each propagators: external line add the following terms: Feynman Rules for QED µ e going into the diagram - u 1) For each externali( q line,µ + labelme) the terms p ,p,...,p and s ,s ,....s . For For e : (Note, I will•1 also2 dropn the1 e2 subscript)n each internal line,± label2 the line2 with q ,q ...., q . Then give each line an arrow • q me 1 2 ne going out of the diagram -u ¯ indicating the direction of flow. For electrons,• the arrows should point in the e+ going into the diagram -v ¯ positive time directlyig (leftµFeynman Rules ⌫ to right), for positrons they should point in the For : • opposite• direction. Theq2 flow for internal lines ise+ ambiguous,going out of but the each diagramµ vertex - v shouldFor each line place an arrow have at least one arrow enteringà and leaving.For each vertex • igγ 2)direcon of momentum For each external line add the following terms:4going4 into the diagram - ✏µ 5) For each vertex, add the term (2•⇡) (k1 + k2 + k3) where each term is positive for flow into the vertex and negative for flow out of the vertex.µ e going into the diagram - u + for flow into the going out of the diagram - ✏ ⇤ • • d4 µ e6)going For each out of internal the diagram line -addu ¯ the termvertex, − for flow out of 3) For each4 and vertex integrate. add igem . (Note, I will drop the em subscript and just • use g since(2⇡ we) have so many subscripts since we will not be doing weak or QCD e+ going into the diagram4 4 -v ¯ the vertex • 7) Remove a (2⇡) (momentumcalculations.) conservation) from the term. +8) The remainder is equal to iMFor each internal line 4)with For each one internal caveat. line Antisymmetrization add the following propagators: e going out of the diagram - v • µ needs to be applied. If two diagrams only di↵er byi( theqµ interchange+ me) of 2 fermions, going into the diagram - ✏µ For e : (Note, I will also drop the e subscript) then the diagrams need to di↵er by a minus sign± (So2 essentially2 M = M1 M2 • • q me µ instead going of outM of= theM1 diagram+ M2) as- ✏ the⇤ system has to follow Fermi statistics. It does • ig not matter which one you pick as negative since theµ⌫ relevant quantities is the µ For : 2 matrix3)Each symbol has a For each element vertex squared. add igem . (Note, I will• drop the emq subscript and just use g since we have so many subscripts since we willd 4 not be doing weak or QCD Matrix: The gamma matrix comes from the Dirac Equation (which4 4 is a mathemacal prescripon 5) For each4 vertex,Integrate add the term (2⇡) (k1 + k2 + k3) where each term is calculations.) ∫ (2π ) EOM4) For for each spin internal 1/2 line particles) add the that following wepositive will propagators: cover for flow the into origin the vertex of shortly. and negative However, for flow out of the vertex. 4 we can simply defineRosi Reed – Feynman it here as: 24 d µ 6) For each internal line add the term and integrate. i( qµ + me) (2⇡)4 For e± : (Note, I will also drop the e subscript) • q2 m2 4 4 e 7) Remove a (2⇡) (momentum conservation) from the term. 8)1 The remainder is equal to iM with one caveat. Antisymmetrization igµ⌫ needs to be applied. If two diagrams only di↵er by the interchange of 2 fermions, For : 2 • q then the diagrams need to di↵er by a minus sign (So essentially M = M M 1 2 4 instead4 of M = M1 + M2) as the system has to follow Fermi statistics. It does 5) For each vertex, add the term (2⇡) (k1 + k2 + k3) where each term is positive for flow into the vertex and negativenot for matter flow out which of theone vertex. you pick as negative since the relevant quantities is the matrixd4 element squared. 6) For each internal line add the term 4 Matrixand integrate.: The gamma matrix comes from the Dirac Equation (which is a (2EOM⇡) for spin 1/2 particles) that we will cover the origin of shortly. However, 4 7) Remove a (2⇡) 4(momentum conservation)we can simply from the define term. it here as: 8) The remainder is equal to iM with one caveat. Antisymmetrization needs to be applied. If two diagrams only di↵er by the interchange of 2 fermions, 1 then the diagrams need to di↵er by a minus sign (So essentially M = M M 1 2 instead of M = M1 + M2) as the system has to follow Fermi statistics. It does not matter which one you pick as negative since the relevant quantities is the matrix element squared. Matrix: The gamma matrix comes from the Dirac Equation (which is a EOM for spin 1/2 particles) that we will cover the origin of shortly. However, we can simply define it here as:

1 Let’s Calculate Something!

Rosi Reed – Feynman 25 Let’s Calculate Something! e-µ- à e-µ- At LO this only has a single diagram

Rosi Reed – Feynman 26 e-µ- à e-µ-

Choose direcon for photon

e− (and µ−) à forward in me

e+ (and µ+) à backward in me

Where do we start?

Rosi Reed – Feynman 27

Figure 1: Tree level Feynman diagram for the interaction e + µ e + µ on the right. On the left is the Feynman pattern used to write down! the matrix element.

Example 1: e + µ e + µ shown in Figure 1. Note: we can treat muons exactly as electrons,! the only di↵erence is that they are identiﬁably di↵erent.... With the gamma matrices and metric included in the Feynman diagram, the order that things are written down matters. The scheme is that you should work your way backwards in time along a fermion line, writing down the elements as you come to them. The pattern is also shown in Figure 1. We will start with particle 1, work our way to the vertex and then ﬁnish with particle 2, which gives us: s3 µ s1 u¯ (p3)ig u (p1) Next we will go down the propagator (line 3 in the pattern graphic) which gives us: igµ⌫ s3 µ s1 u¯ (p3)ig u (p1) 2 q1 Then we return to particle 4 and trace our way backwards through the vertex and to problem 2. Note that the gamma matrix in the second vertex has a di↵erent label. Each vertex must have a di↵erent label, as we will need to sum over all of the spatial and time dimensions to calculate anything. Following this gives us: igµ⌫ s3 µ s1 s4 ⌫ s2 u¯ (p3)ig u (p1) 2 u¯ (p4)ig u (p2) q1 I’ve written out everything explicitly to show that the spinors depend both on momentum and the spin, but in general we only need to keep track of which

2 e-µ- à e-µ-

Rosi Reed – Feynman 28

Figure 1: Tree level Feynman diagram for the interaction e + µ e + µ on the right. On the left is the Feynman pattern used to write down! the matrix element.

2 e-µ- à e-µ-

e- going out of the diagram

This is a plane-wave Dirac spinor

Figure 1: Tree level Feynman diagram for the interaction e + µ e + µ on the right. On the left isRosi Reed – Feynman the Feynman pattern used to write down! the matrix 29 element.

2 e-µ- à e-µ-

e- going out of the diagram Vertex

µ u3igγ

Figure 1: Tree level Feynman diagram for the interaction e + µ e + µ on the right. On the left isRosi Reed – Feynman the Feynman pattern used to write down! the matrix 30 element.

2 e-µ- à e-µ-

e- going out of the diagram Vertex e- going into the diagram µ u3igγ u1

Figure 1: Tree level Feynman diagram for the interaction e + µ e + µ on the right. On the left isRosi Reed – Feynman the Feynman pattern used to write down! the matrix 31 element.

2 e-µ- à e-µ-

e- going out of the diagram Vertex e- going into the diagram Internal γ line

µ −igµν u3igγ u1 2 q1

gµν is the space-me metricà geometric and causal structure of spaceme

Figure 1: Tree level Feynman diagram for the interaction e + µ e + µ on the right. On the left isRosi Reed – Feynman the Feynman pattern used to write down! the matrix 32 element.

2 e-µ- à e-µ-

e- going out of the diagram Vertex e- going into the diagram Internal γ line µ- going out of the diagram

µ −igµν u3igγ u1 2 u4 q1

Figure 1: Tree level Feynman diagram for the interaction e + µ e + µ on the right. On the left isRosi Reed – Feynman the Feynman pattern used to write down! the matrix 33 element.

2 e-µ- à e-µ-

e- going out of the diagram Vertex e- going into the diagram Internal γ line µ- going out of the diagram Vertex

µ −igµν ν u3igγ u1 2 u4igγ q1 γµ and γν are 4x4 matrices

Figure 1: Tree level Feynman diagram for the interaction e + µ e + µ on the right. On the left isRosi Reed – Feynman the Feynman pattern used to write down! the matrix 34 element.

2 e-µ- à e-µ-

e- going out of the diagram Vertex e- going into the diagram Internal γ line µ- going out of the diagram Vertex µ- going into the diagram

µ −igµν ν u3igγ u1 2 u4igγ u2 q1

Then we need to place this into the integral to ﬁnd the matrix Figure 1: Tree level Feynman diagram for the interaction e + µ e + µ on the right. On the left isRosi Reed – Feynman the Feynman pattern used to write down! the matrix 35 element.

2 e-µ- à e-µ-

IN OUT Only 1 internal line à 1 OUT integraon IN

IN OUT

−ig (2π )4 u igγ µu µν u igγ ν u δ 4 (p − p − q )δ 4 (p − p + q )d 4q ∫ 3 1 q2 4 2 1 3 1 2 4 1 1 Figure 1: Tree level Feynman1 diagram for the interaction e + µ e + µ on the right. On the left is the Feynman pattern used to write down! the matrix element. Rosi Reed – Feynman 36

2 Dirac Delta Funcon

4 (x)d 4 x 4 à This is a 4 dimensional delta ∫ δ funcon, integrate over t,x,y,z

Integrang over the enre range “picks” a value

∞ ∫ f (x)δ(x)dx = f (0) −∞ This has a lot of ∞ convenient ∫ f (x − a)δ(x)dx = f (a) mathemacal −∞ properes

Rosi Reed – Feynman 37 e-µ- à e-µ-

−ig (2 )4 u ig µu µν u ig ν u 4 (p p q ) 4 (p p q )d 4q π ∫ 3 γ 1 2 4 γ 2δ 1 − 3 − 1 δ 2 − 4 + 1 1 q1 The metric will contract with one of gamma matrices: ν gµνγ = γν

Integrate and pick one delta funcon q1 = p1 – p3

2 4 µ 1 4 ig (2π ) u3γ u1 2 u4γν u2δ (p2 − p4 + p1 − p3 ) (p1 − p3 ) Remaining delta funcon is simply overall conservaon of energy and momentum and can be dropped (and quanzaon factor) Rosi Reed – Feynman 38 e-µ- à e-µ-

2 µ 1 ig u3γ u1 2 u4γν u2 (p1 − p3 )

The remainder is equal to –iMﬁ!

2 −g µ µ M fi = 2 [u3γ u1][u4γ u2 ] (p1 − p3 )

Rosi Reed – Feynman 39 e-µ- à e-µ-

2 −g µ µ M fi = 2 [u3γ u1][u4γ u2 ] (p1 − p3 )

1 p* f | M |2 d * σ = 2 * ∫ fi Ω 64π s pi

“Simply” square the matrix element (this can have an angular dependence) and calculate the cross secon

Rosi Reed – Feynman 40 Conclusions • Feynman Diagrams are pictorial representaons of mathemacal expressions describing the behavior of subatomic parcles • They have revoluonized nearly every aspect of theorecal physics – Extend past QFT into other ﬁelds – Describe many processes

Rosi Reed – Feynman 41