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Feynman Diagrams for Beginners
Krešimir Kumeriˇcki†
Department of Physics, Faculty of Science, University of Zagreb, Croatia
Abstract We give a short introduction to Feynman diagrams, with many exer- cises. Text is targeted at students who had little or no prior exposure to quantum field theory. We present condensed description of single-particle Dirac equation, free quantum fields and construction of Feynman amplitude using Feynman diagrams. As an example, we give a detailed calculation of cross-section for annihilation of electron and positron into a muon pair. We also show how such calculations are done with the aid of computer.
Contents
1 Natural units 2
2 Single-particle Dirac equation 4 2.1 The Dirac equation......................... 4 2.2 The adjoint Dirac equation and the Dirac current......... 6 2.3 Free-particle solutions of the Dirac equation............ 6
3 Free quantum fields 9 3.1 Spin 0: scalar field......................... 10 3.2 Spin 1/2: the Dirac field....................... 10 3.3 Spin 1: vector field......................... 10
4 Golden rules for decays and scatterings 11
5 Feynman diagrams 13 ∗Notes for the exercises at the Adriatic School on Particle Physics and Physics Informatics, 11
arXiv:1602.04182v1 [physics.ed-ph] 8 Feb 2016
2 1 Natural units
6 Example: e+e−^ → μ+μ−^ in QED 16 6.1 Summing over polarizations.................... 17 6.2 Casimir trick............................ 18 6.3 Traces and contraction identities of γ matrices........... 18 6.4 Kinematics in the center-of-mass frame.............. 20 6.5 Integration over two-particle phase space.............. 20 6.6 Summary of steps.......................... 22 6.7 Mandelstam variables........................ 22
Appendix: Doing Feynman diagrams on a computer 22
1 Natural units
To describe kinematics of some physical system or event we are free to choose units of measure of the three basic kinematical physical quantities: length (L), mass (M) and time (T). Equivalently, we may choose any three linearly indepen- dent combinations of these quantities. The choice of L, T and M is usually made (e.g. in SI system of units) because they are most convenient for description of our immediate experience. However, elementary particles experience a different world, one governed by the laws of relativistic quantum mechanics. Natural units in relativistic quantum mechanics are chosen in such a way that fundamental constants of this theory, c and ℏ, are both equal to one. [c] = LT −^1 , [ℏ] = M L−^2 T −^1 , and to completely fix our system of units we specify the unit of energy (M L^2 T −^2 ): 1 GeV = 1. 6 · 10 −^10 kg m^2 s−^2 ,
approximately equal to the mass of the proton. What we do in practice is:
- we ignore ℏ and c in formulae and only restore them at the end (if at all)
- we measure everything in GeV, GeV−^1 , GeV^2 ,...
Example: Thomson cross section
Total cross section for scattering of classical electromagnetic radiation by a free electron (Thomson scattering) is, in natural units,
σT =
8 πα^2 3 m^2 e
To restore ℏ and c we insert them in the above equation with general powers α and β, which we determine by requiring that cross section has the dimension of area
4 2 Single-particle Dirac equation
2 Single-particle Dirac equation
2.1 The Dirac equation
Turning the relativistic energy equation
E^2 = p^2 + m^2. (6)
into a differential equation using the usual substitutions
p → −i∇ , E → i
∂t
results in the Klein-Gordon equation:
( + m^2 )ψ(x) = 0 , (8)
which, interpreted as a single-particle wave equation, has problematic negative energy solutions. This is due to the negative root in E = ±
p^2 + m^2. Namely, in relativistic mechanics this negative root could be ignored, but in quantum physics one must keep all of the complete set of solutions to a differential equation. In order to overcome this problem Dirac tried the ansatz∗
(iβμ∂μ + m)(iγν^ ∂ν − m)ψ(x) = 0 (9)
with βμ^ and γν^ to be determined by requiring consistency with the Klein-Gordon equation. This requires γμ^ = βμ^ and
γμ∂μγν^ ∂ν = ∂μ∂μ , (10)
which in turn implies (γ^0 )^2 = 1 , (γi)^2 = − 1 , {γμ, γν^ } ≡ γμγν^ + γν^ γμ^ = 0 for μ 6 = ν.
This can be compactly written in form of the anticommutation relations
{γμ, γν^ } = 2gμν^ , gμν^ =
.^ (11)
These conditions are obviously impossible to satisfy with γ’s being equal to usual numbers, but we can satisfy them by taking γ’s equal to (at least) four-by-four matrices.
∗ (^) ansatz: guess, trial solution (from German Ansatz: start, beginning, onset, attack)
2 Single-particle Dirac equation 5
Now, to satisfy (9) it is enough that one of the two factors in that equation is zero, and by convention we require this from the second one. Thus we obtain the Dirac equation:
(iγμ∂μ − m)ψ(x) = 0. (12)
ψ(x) now has four components and is called the Dirac spinor. One of the most frequently used representations for γ matrices is the original Dirac representation
γ^0 =
γi^ =
0 σi −σi^0
where σi^ are the Pauli matrices:
σ^1 =
σ^2 =
0 −i i 0
σ^3 =
This representation is very convenient for the non-relativistic approximation, since then the dominant energy terms (iγ^0 ∂ 0 −... − m)ψ(0) turn out to be diagonal. Two other often used representations are
- the Weyl (or chiral) representation — convenient in the ultra-relativistic regime (where E m)
- the Majorana representation — makes the Dirac equation real; convenient for Majorana fermions for which antiparticles are equal to particles
(Question: Why can we choose at most one γ matrix to be diagonal?)
Properties of the Pauli matrices:
σi
† = σi^ (15) σi∗^ = (iσ^2 )σi(iσ^2 ) (16) [σi, σj^ ] = 2iijkσk^ (17) {σi, σj^ } = 2δij^ (18) σiσj^ = δij^ + iijkσk^ (19)
where ijk^ is the totally antisymmetric Levi-Civita tensor (^123 = ^231 = ^312 = 1, ^213 = ^321 = ^132 = − 1 , and all other components are zero).
Exercise 3 Prove that (σ · a)^2 = a^2 for any three-vector a.
2 Single-particle Dirac equation 7
which after inclusion in the Dirac equation gives the momentum space Dirac equa- tion (/p − m)u(p) = 0. (21)
This has two positive-energy solutions
u(p, σ) = N
χ(σ) σ · p E + m
χ(σ)
,^ σ^ = 1,^2 ,^ (22)
where
χ(1)^ =
, χ(2)^ =
and two negative-energy solutions which are then interpreted as positive-energy antiparticle solutions
v(p, σ) = −N
σ · p E + m
(iσ^2 )χ(σ)
(iσ^2 )χ(σ)
,^ σ^ = 1,^2 ,^ E >^0.^ (24)
N is the normalization constant to be determined later. Spinors above agree with those of [1]. The momentum-space Dirac equation for antiparticle solutions is
(/p + m)v(p, σ) = 0. (25)
It can be shown that the two solutions, one with σ = 1 and another with σ = 2, correspond to the two spin states of the spin-1/2 particle.
Exercise 8 Determine momentum-space Dirac equations for ¯u(p, σ) and ¯v(p, σ).
Normalization
In non-relativistic single-particle quantum mechanics normalization of a wave- function is straightforward. Probability that the particle is somewhere in space is equal to one, and this translates into the normalization condition
ψ∗ψ dV = 1. On the other hand, we will eventually use spinors (22) and (24) in many-particle quantum field theory so their normalization is not unique. We will choose nor- malization convention where we have 2 E particles in the unit volume: ∫
unit volume
ρ dV =
unit volume
ψ†ψ dV = 2E (26)
This choice is relativistically covariant because the Lorentz contraction of the vol- ume element is compensated by the energy change. There are other normalization conventions with other advantages.
8 2 Single-particle Dirac equation
Exercise 9 Determine the normalization constant N conforming to this choice.
Completeness
Exercise 10 Using the explicit expressions (22) and (24) show that
∑
σ=1, 2
u(p, σ)¯u(p, σ) = (^) /p + m , (27)
∑
σ=1, 2
v(p, σ)¯v(p, σ) = (^) /p − m. (28)
These relations are often needed in calculations of Feynman diagrams with unpo- larized fermions. See later sections.
Parity and bilinear covariants
The parity transformation:
- P : x → −x, t → t
- P : ψ → γ^0 ψ
Exercise 11 Check that the current jμ^ = ψγ¯ μψ transforms as a vector under par- ity i.e. that j^0 → j^0 and j → −j.
Any fermion current will be of the form ψ¯Γψ, where Γ is some four-by-four matrix. For construction of interaction Lagrangian we want to use only those currents that have definite Lorentz transformation properties. To this end we first define two new matrices:
γ^5 ≡ iγ^0 γ^1 γ^2 γ^3 Dirac rep. =
, {γ^5 , γμ} = 0 , (29)
σμν^ ≡
i 2
[γμ, γν^ ] , σμν^ = −σνμ^. (30)
Now ψ¯Γψ will transform covariantly if Γ is one of the matrices given in the following table. Transformation properties of ψ¯Γψ, the number of different γ
10 3 Free quantum fields
and this is the reason that it is named a creation operator. Similarly, a is an anni- hilation operator a(p, σ)|p, σ〉 = | 0 〉 , (34)
and ac†^ and ac^ are creation and annihilation operators for antiparticle states (c in ac^ stands for “conjugated”). Processes in particle physics are mostly calculated in the framework of the theory of such fields — quantum field theory. This theory can be described at various levels of rigor but in any case is complicated enough to be beyond the scope of these notes. However, predictions of quantum field theory pertaining to the elementary particle interactions can often be calculated using a relatively simple “recipe” — Feynman diagrams. Before we turn to describing the method of Feynman diagrams, let us just specify other quantum fields that take part in the elementary particle physics inter- actions. All these are free fields, and interactions are treated as their perturbations. Each particle type (electron, photon, Higgs boson, ...) has its own quantum field.
3.1 Spin 0: scalar field
E.g. Higgs boson, pions, ...
φ(x) =
d^3 p √ (2π)^32 E
[
a(p)e−ipx^ + ac†(p)eipx
]
3.2 Spin 1/2: the Dirac field
E.g. quarks, leptons
We have already specified the Dirac spin-1/2 field. There are other types: Weyl and Majorana spin-1/2 fields but they are beyond our scope.
3.3 Spin 1: vector field
Either
- massive (e.g. W,Z weak bosons) or
- massless (e.g. photon)
Aμ(x) =
λ
d^3 p √ (2π)^32 E
[
μ(p, λ)a(p, λ)e−ipx^ + μ∗(p, λ)a†(p, λ)eipx
]
4 Golden rules for decays and scatterings 11
μ(p, λ) is a polarization vector. For massive particles it obeys
pμμ(p, λ) = 0 (37)
automatically, whereas in the massless case this condition can be imposed thanks to gauge invariance (Lorentz gauge condition). This means that there are only three independent polarizations of a massive vector particle: λ = 1, 2 , 3 or λ = +, −, 0. In massless case gauge symmetry can be further exploited to eliminate one more polarization state leaving us with only two: λ = 1, 2 or λ = +, −. Normalization of polarization vectors is such that
∗(p, λ) · (p, λ) = − 1. (38)
E.g. for a massive particle moving along the z-axis (p = (E, 0 , 0 , |p|)) we can take
(p, ±) = ∓
±i 0
,^ (p,^ 0) =
m
|p| 0 0 E
Exercise 12 Calculate (^) ∑
λ
μ∗(p, λ)ν^ (p, λ)
Hint: Write it in the most general form (Agμν^ + Bpμpν^ ) and then determine A and B.
The obtained result obviously cannot be simply extrapolated to the massless case via the limit m → 0. Gauge symmetry makes massless polarization sum somewhat more complicated but for the purpose of the simple Feynman diagram calculations it is permissible to use just the following relation
∑
λ
μ∗(p, λ)ν^ (p, λ) = −gμν^.
4 Golden rules for decays and scatterings
Principal experimental observables of particle physics are
- scattering cross section σ(1 + 2 → 1 ′^ + 2′^ + · · · + n′)
- decay width Γ(1 → 1 ′^ + 2′^ + · · · + n′)
5 Feynman diagrams 13
where uα is the relative velocity of particles 1 and 2:
uα =
(p 1 · p 2 )^2 − m^21 m^22 E 1 E 2
and |M|^2 is the Feynman invariant amplitude averaged over unmeasured particle spins (see Section 6.1). The dimension of M, in units of energy, is
- for decays [M] = 3 − n
- for scattering of two particles [M] = 2 − n
where n is the number of produced particles.
So calculation of some observable quantity consists of two stages:
- Determination of |M|^2. For this we use the method of Feynman diagrams to be introduced in the next section.
- Integration over the Lorentz invariant phase space
dLips = (2π)^4 δ^4 (p 1 + p 2 − p′ 1 − · · · − p′ n)
∏^ n
i=
d^3 p′ i (2π)^3 2 E′ i
5 Feynman diagrams
Example: φ^4 -theory
L =
∂μφ∂μφ −
m^2 φ^2 −
g 4!
φ^4
- Free (kinetic) Lagrangian (terms with exactly two fields) determines parti- cles of the theory and their propagators. Here we have just one scalar field: φ
- Interaction Lagrangian (terms with three or more fields) determines possible vertices. Here, again, there is just one vertex: φ
φ
φ
φ
14 5 Feynman diagrams
We construct all possible diagrams with fixed outer particles. E.g. for scatter- ing of two scalar particles in this theory we would have
M(1 + 2 → 3 + 4) = +^ +^ +...
1
2
3
4 t In these diagrams time flows from left to right. Some people draw Feynman diagrams with time flowing up, which is more in accordance with the way we usually draw space-time in relativity physics. Since each vertex corresponds to one interaction Lagrangian (Hamiltonian) term in (42), diagrams with loops correspond to higher orders of perturbation theory. Here we will work only to the lowest order, so we will use tree diagrams only. To actually write down the Feynman amplitude M, we have a set of Feynman rules that associate factors with elements of the Feynman diagram. In particular, to get −iM we construct the Feynman rules in the following way:
- the vertex factor is just the i times the interaction term in the (momentum space) Lagrangian with all fields removed:
iLI = −i
g 4!
φ^4 removing fields ⇒
φ
φ
φ
φ
= −i
g 4!
- the propagator is i times the inverse of the kinetic operator (defined by the free equation of motion) in the momentum space:
Lfree
Euler-Lagrange eq. −→ (∂μ∂μ^ + m^2 )φ = 0 (Klein-Gordon eq.) (48)
Going to the momentum space using the substitution ∂μ^ → −ipμ^ and then taking the inverse gives:
(p^2 − m^2 )φ = 0 ⇒ φ =
i p^2 − m^2
(Actually, the correct Feynman propagator is i/(p^2 − m^2 + i), but for our purposes we can ignore the infinitesimal i term.)
16 6 Example: e+e−^ → μ+μ−^ in QED
This is in principle almost all we need to know to be able to calculate the Feynman amplitude of any given process. Note that propagators and external-line polarization vectors are determined only by the particle type (its spin and mass) so that the corresponding rules above are not restricted only to the φ^4 theory and QED, but will apply to all theories of scalars, spin-1 vector bosons and Dirac fermions (such as the standard model). The only additional information we need are the vertex factors. “Almost” in the preceding paragraph alludes to the fact that in general Feyn- man diagram calculation there are several additional subtleties:
- In loop diagrams some internal momenta are undetermined and we have to integrate over those. Also, there is an additional factor (-1) for each closed fermion loop. Since we will consider tree-level diagrams only, we can ignore this.
- There are some combinatoric numerical factors when identical fields come into a single vertex.
- Sometimes there is a relative (-) sign between diagrams.
- There is a symmetry factor if there are identical particles in the final state.
For explanation of these, reader is advised to look in some quantum field the- ory textbook.
6 Example: e+e−^ → μ+μ−^ in QED
There is only one contributing tree-level diagram:
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6 Example: e+e−^ → μ+μ−^ in QED 17
fermion lines backwards. Order of lines themselves is unimportant.
−iM = [¯u(p 3 , σ 3 )(ieγν^ )v(p 4 , σ 4 )]
−igμν (p 1 + p 2 )^2
[¯v(p 2 , σ 2 )(ieγμ)u(p 1 , σ 1 )] , (55) or, introducing abbreviation u 1 ≡ u(p 1 , σ 1 ),
M =
e^2 (p 1 + p 2 )^2
[¯u 3 γμv 4 ][¯v 2 γμu 1 ]. (56)
Exercise 14 Draw Feynman diagram(s) and write down the amplitude for Comp- ton scattering γe−^ → γe−.
6.1 Summing over polarizations
If we knew momenta and polarizations of all external particles, we could calculate M explicitly. However, experiments are often done with unpolarized particles so we have to sum over the polarizations (spins) of the final particles and average over the polarizations (spins) of the initial ones:
|M|^2 → |M|^2 =
σ 1 σ 2 ︸ ︷︷ ︸ avg. over initial pol.
sum over final pol. ︷︸︸︷∑
σ 3 σ 4
|M|^2. (57)
Factors 1 / 2 are due to the fact that each initial fermion has two polarization (spin) states.
(Question: Why we sum probabilities and not amplitudes?)
In the calculation of |M|^2 = M∗M, the following identity is needed
[¯uγμv]∗^ = [u†γ^0 γμv]†^ = v†γμ†γ^0 u = [¯vγμu]. (58)
Thus,
|M|^2 =
e^4 4(p 1 + p 2 )^4
σ 1 , 2 , 3 , 4
[¯v 4 γμu 3 ][¯u 1 γμv 2 ][¯u 3 γν v 4 ][¯v 2 γν^ u 1 ]. (59)
6 Example: e+e−^ → μ+μ−^ in QED 19
Tr 1 = 4
Trγμγν^ = Tr(2gμν^ − γν^ γμ) (2.) = 8gμν^ − Trγν^ γμ^ = 8gμν^ − Trγμγν ⇒ 2 Trγμγν^ = 8gμν^ ⇒ Trγμγν^ = 4gμν This also implies: Tr/a/b = 4a · b
- Exercise 15 Calculate Tr(γμγν^ γργσ). Hint: Move γσ^ all the way to the left, using the anticommutation relations. Then use 3. Homework: Prove that Tr(γμ^1 γμ^2 · · · γμ^2 n^ ) has (2n − 1)!! terms.
- Tr(γ^5 γμ^1 γμ^2 · · · γμ^2 n+1^ ) = 0. This follows from 1. and from the fact that γ^5 consists of even number of γ’s.
- Trγ^5 = Tr(γ^0 γ^0 γ^5 ) = −Tr(γ^0 γ^5 γ^0 ) = −Trγ^5 = 0
- Tr(γ^5 γμγν^ ) = 0. (Same trick as above, with γα^6 = μ, ν instead of γ^0 .)
- Tr(γ^5 γμγν^ γργσ) = − 4 iμνρσ, with ^0123 = 1. Careful: convention with ^0123 = − 1 is also in use.
Contraction identities
γμγμ =
gμν (γμγν^ + γν^ γμ) ︸ ︷︷ ︸ 2 gμν
= gμν gμν^ = 4
γμ^ γαγμ ︸ ︷︷ ︸ −γμγα+2gαμ
= − 4 γα^ + 2γα^ = − 2 γα
- Exercise 16 Contract γμγαγβ^ γμ.
- γμγαγβ^ γγ^ γμ = − 2 γγ^ γβ^ γα
Exercise 17 Calculate traces in |M|^2 :
Tr[(/p 1 + m 1 )γμ(/p 2 − m 2 )γν^ ] =? Tr[(/p 4 − m 4 )γμ(/p 3 + m 3 )γν ] =?
Exercise 18 Calculate |M|^2
20 6 Example: e+e−^ → μ+μ−^ in QED
6.4 Kinematics in the center-of-mass frame
In e+e−^ coliders often pi me, mμ, i = 1,... , 4 , so we can take
mi → 0 “high-energy” or “extreme relativistic” limit
Then
|M|^2 =
8 e^4 (p 1 + p 2 )^4
[(p 1 · p 3 )(p 2 · p 4 ) + (p 1 · p 4 )(p 2 · p 3 )] (63)
To calculate scattering cross-section σ we have to specialize to some particular frame (σ is not frame-independent). For e+e−^ colliders the most relevant is the center-of-mass (CM) frame:
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Exercise 19 Express |M|^2 in terms of E and θ.
6.5 Integration over two-particle phase space
Now we can use the “golden rule” (45) for the 1+2 → 3+4 differential scattering cross-section
dσ =
uα
|M|^2 dLips 2 (64)
where two-particle phase space to be integrated over is
dLips 2 = (2π)^4 δ^4 (p 1 + p 2 − p 3 − p 4 )
d^3 p 3 (2π)^3 2 E 3
d^3 p 4 (2π)^3 2 E 4
