Advanced quantum field theory for particle physics course of the University of Oxford (MMathPhys/MSc paper)

a

Solution

The scalar field path integral integrates over real field histories with vacuum boundary conditions. With

S_{0} [ϕ] = \int d^{4} x L_{0}

, the partition function is

Z = N \int D ϕ exp {i \int d^{4} x [\frac{1}{2} \partial_{μ} ϕ \partial^{μ} ϕ - \frac{1}{2} (m^{2} - i 0) ϕ^{2} - \frac{λ}{4 !} ϕ^{4}]} .

The constant

N

fixes the desired vacuum normalization. A generating functional adds

\int d^{4} x J ϕ

to the action; the required partition function is its zero-source value. Equivalently, a Wick rotation gives

Z_{E} = N_{E} \int D ϕ e^{- S_{E} [ϕ]}

, with the positive-sign kinetic and potential terms in the Euclidean action when

m^{2} \geq 0

and

λ > 0

b

Solution

Vary the action, taking the field variation to vanish at the boundary. Integration by parts gives

δ S_{0} = \int d^{4} x [\partial^{μ} ϕ \partial_{μ} δ ϕ - m^{2} ϕ δ ϕ - \frac{λ}{6} ϕ^{3} δ ϕ] = - \int d^{4} x (□ ϕ + m^{2} ϕ + \frac{λ}{6} ϕ^{3}) δ ϕ .

Thus the Euler-Lagrange field equation is

□ ϕ + m^{2} ϕ + \frac{λ}{6} ϕ^{3} = 0.

For the field redefinition in part f, define the functional derivative

E (ϕ) = δ S_{0} / δ ϕ = - (□ ϕ + m^{2} ϕ + λ ϕ^{3} /6)

. The sign does not affect the equation of motion, but it fixes the sign of an added redundant operator.

c

Solution

The action is dimensionless, so in four dimensions the Lagrangian density has mass dimension four. Its kinetic term fixes

[ϕ] = 1

and

[\partial_{μ}] = 1

. An interaction

g_{D} O_{D}

with

[O_{D}] = D

has

[g_{D}] = 4 - D

. Therefore

D \leq 4 is renormalizable by power counting; D > 4 has a negative-dimensional coupling .

In an effective field theory with heavy scale

Λ

, a higher-dimensional local operator enters as

c_{D} O_{D} / Λ^{D - 4}

. For characteristic scales

E, m ≪ Λ

and coefficients compatible with this expansion, its effects are suppressed by powers of the small ratios. Omitting these terms defines the leading approximation and its truncation error in effective field theory. It is not a fundamental prohibition: such terms must be retained if that accuracy is required. A continuum renormalizable quantum field theory can instead be defined by the closed set of counterterms of dimension at most four.

Up to total derivatives, all Lorentz scalar terms of dimension at most four built from one real scalar field are

1, ϕ, ϕ^{2}, ϕ^{3}, ϕ^{4}, \partial_{μ} ϕ \partial^{μ} ϕ .

The term

ϕ □ ϕ

is equivalent to the kinetic term by integration by parts;

□ ϕ

and

□ (ϕ^{2})

contribute only at the boundary. The displayed Lagrangian density omits the linear and cubic interactions and a field-independent constant. The exact Z2 symmetry

ϕ \mapsto - ϕ

forbids both odd interactions, and radiative corrections preserving this symmetry cannot generate them. The constant shifts the vacuum energy and the normalization of

Z

and can be dropped when gravity and absolute vacuum energy are not being studied. A constant vacuum counterterm may still be used when renormalizing unnormalized vacuum diagrams. The kinetic coefficient has already been fixed by field normalization.

d

Solution

For dimension-six operators of a real scalar field, let

n

count fields and

r

count partial derivatives. Mass dimension gives

n + r = 6

. Lorentz invariance requires even

r

. Discard derivatives acting on the entire expression, but do not yet identify different expressions by integration by parts. A complete list of seven representatives is

O_{1} O_{2} O_{4} O_{6} = ϕ^{6}, = ϕ^{2} \partial_{μ} ϕ \partial^{μ} ϕ, = ϕ □^{2} ϕ, = (□ ϕ)^{2}, O_{3} O_{5} O_{7} = ϕ^{3} □ ϕ, = \partial_{μ} ϕ \partial^{μ} □ ϕ, = (\partial_{μ} \partial_{ν} ϕ) (\partial^{μ} \partial^{ν} ϕ) .

For

(n, r) = (6, 0)

there is one choice; for

(4, 2)

the derivative distribution is either

2

1 + 1

; for

(2, 4)

it is

4

3 + 1

, or

2 + 2

, the last allowing two distinct metric contractions. Commuting partial derivatives identifies alternative orderings. Contractions with a Levi-Civita symbol vanish because an antisymmetric index pair necessarily contracts symmetric derivatives. The remaining case

(0, 6)

gives no field-dependent expression. No independent odd-

r

Lorentz scalar exists.

The source's seven means this unreduced list. If “ignoring total derivative operators” were interpreted as already taking the full quotient by total derivatives, seven would be incorrect: the next part gives three representatives of that quotient. Dimensionful factors multiplying lower-dimensional terms do not count as new canonical-dimension-six field monomials.

e

Solution

Write

X ≃ Y

when the two densities differ by a total derivative. Under the boundary conditions used in the action, they have the same integral. The following identities establish the integration-by-parts reduction of scalar operators:

\partial_{μ} (ϕ^{3} \partial^{μ} ϕ) \partial_{μ} (ϕ \partial^{μ} □ ϕ) \partial_{μ} (\partial^{μ} ϕ □ ϕ) \partial_{μ} (\partial_{ν} ϕ \partial^{μ} \partial^{ν} ϕ) = 3 ϕ^{2} (\partial ϕ)^{2} + ϕ^{3} □ ϕ, = \partial_{μ} ϕ \partial^{μ} □ ϕ + ϕ □^{2} ϕ, = (□ ϕ)^{2} + \partial_{μ} ϕ \partial^{μ} □ ϕ, = (\partial_{μ} \partial_{ν} ϕ)^{2} + \partial_{ν} ϕ \partial^{ν} □ ϕ, O_{2} O_{4} O_{5} O_{7} ≃ - \frac{1}{3} O_{3}, ≃ - O_{5}, ≃ - O_{6}, ≃ - O_{5} .

Consequently

{ϕ^{6}, ϕ^{3} □ ϕ, (□ ϕ)^{2}}

spans the operator basis in effective field theory before equation-of-motion reduction of scalar operators. These four independent relations reduce the seven representatives to three; the three sectors have different field or derivative content, so integration by parts cannot identify them further.

f

Solution

Use the sign convention

E (ϕ) = δ S_{0} / δ ϕ

from part b, and let

S_{η} [ϕ] = S_{0} [ϕ] + η \int d^{4} x f (ϕ) E (ϕ) .

In the path integral, make the local field redefinition

ϕ_{old} = χ - η f (χ)

. The functional Taylor expansion of the action gives

S_{0} [χ - η f (χ)] η \int d^{4} x f (ϕ_{old}) E (ϕ_{old}) = S_{0} [χ] - η \int d^{4} x f (χ) \frac{δ S _{0}}{δχ ( x )} + O (η^{2}), = η \int d^{4} x f (χ) E (χ) + O (η^{2}) .

The two first-order terms cancel. Neglecting the functional Jacobian as permitted,

Z_{η} = N \int D χ exp {i S_{0} [χ] + O (η^{2})} .

Thus

f E

is a redundant operator at first order, by the equivalence theorem for field redefinitions. The same argument works for a local derivative-dependent

f

, which is needed to remove

(□ ϕ)^{2}

. Insertions and sources must be transformed too: arbitrary untransformed off shell correlation functions need not be equal.

Sign qualification. If instead

E_{+} = □ ϕ + m^{2} ϕ + λ ϕ^{3} /6

is used, then

E_{+} = - δ S_{0} / δ ϕ

. An added

+ η f E_{+}

is removed by

ϕ_{old} = χ + η f (χ)

. The minus shift requested in the source therefore fixes which sign is meant by its otherwise unspecified

E

g

Solution

Let

a = λ /6

. Within terms already suppressed by

Λ^{- 2}

, the leading equation of motion gives

□ ϕ = - m^{2} ϕ - a ϕ^{3}

. Equality here means equivalence under a perturbative field redefinition, together with integration by parts, rather than a pointwise identity inside arbitrary off shell correlation functions. Hence

ϕ^{3} □ ϕ (□ ϕ)^{2} ≐ - m^{2} ϕ^{4} - \frac{λ}{6} ϕ^{6}, ≐ m^{4} ϕ^{2} + \frac{m ^{2} λ}{3} ϕ^{4} + \frac{λ ^{2}}{36} ϕ^{6} .

For the second relation, factor the difference as

(□ ϕ - m^{2} ϕ - a ϕ^{3}) (□ ϕ + m^{2} ϕ + a ϕ^{3})

, which is explicitly a redundant operator. If

Δ L = \frac{1}{Λ ^{2}} [c_{1} ϕ^{6} + c_{2} ϕ^{3} □ ϕ + c_{3} (□ ϕ)^{2}],

the reduced correction is

Δ L ≐ \frac{1}{Λ ^{2}} [c_{3} m^{4} ϕ^{2} + (- c_{2} m^{2} + \frac{c _{3} m ^{2} λ}{3}) ϕ^{4} + (c_{1} - \frac{λ}{6} c_{2} + \frac{λ ^{2}}{36} c_{3}) ϕ^{6}] .

The first two coefficients shift the already present mass and quartic coupling constant. After these shifts,

one genuinely new dimension-six interaction remains: \frac{c _{6, red}}{Λ ^{2}} ϕ^{6}, c_{6, red} = c_{1} - \frac{λ}{6} c_{2} + \frac{λ ^{2}}{36} c_{3} .

This is the equation-of-motion reduction of scalar operators through

O (Λ^{- 2})

; terms of

O (Λ^{- 4})

are outside that truncation. The lower-dimensional shifts should not be silently dropped when

m \neq = 0

2

a

Solution

The first panel is the one-particle-irreducible Feynman diagram for photon vacuum polarization. The two internal Dirac propagators carry

k

and

k + q

, and the external photons carry

q

. The other panels fix the routings used in parts c and d.

Figure 1.
Original labelled one-loop QED Feynman diagrams: photon vacuum polarization, fermion self-energy, and the electromagnetic interaction vertex
. **Details**.

The closed fermion loop sign contributes a minus sign; there is no extra symmetry factor. Applying the given Feynman rules, with the external propagators amputated,

i Π_{μν} (q) = (- 1) \int_{k} tr [(- i e γ_{μ}) \frac{i ( \neq k + m )}{k ^{2} - m ^{2} + i 0} (- i e γ_{ν}) \frac{i ( \neq k + \neq q + m )}{( k + q ) ^{2} - m ^{2} + i 0}] .

Equivalently, the required amplitude is

Π_{μν} (q) = i e^{2} \int_{k} \frac{tr [ γ _{μ} ( \neq k + m ) γ _{ν} ( \neq k + \neq q + m )]}{( k ^{2} - m ^{2} + i 0 ) (( k + q ) ^{2} - m ^{2} + i 0 )} .

The Feynman slash notation is

\neq k = γ^{ρ} k_{ρ}

and

tr 1 = 4

in the chosen dimensional regularization convention.

b

Solution

The Ward identity implies

q^{μ} Π_{μν} = 0

. Lorentz invariance then gives

Π_{μν} = (q^{2} g_{μν} - q_{μ} q_{ν}) Π

, so

g^{μν} Π_{μν} = (d - 1) q^{2} Π

. The gamma matrices satisfy

γ_{μ} \neq k γ^{μ} = (2 - d) \neq k

. Odd traces vanish, and therefore

g^{μν} tr [γ_{μ} (\neq k + m) γ_{ν} (\neq k + \neq q + m)] = 4 [- (d - 2) k \cdot (k + q) + d m^{2}] .

Substituting gives the requested scalar coefficient:

Π (q^{2}) = \frac{4 i e ^{2}}{( d - 1 ) q ^{2}} \int_{k} \frac{- ( d - 2 ) k \cdot ( k + q ) + d m ^{2}}{[( k + q ) ^{2} - m ^{2} + i 0 ] [ k ^{2} - m ^{2} + i 0 ]} .

To extract the ultraviolet divergence without introducing a Feynman parameter, set

A = (k + q)^{2} - m^{2}

B = k^{2} - m^{2}

and use

2 k \cdot (k + q) = A + B + 2 m^{2} - q^{2} .

Translation invariance of dimensional regularization gives

\int_{k} A^{- 1} = \int_{k} B^{- 1} = I_{1}

. Consequently

Π (q^{2}) = \frac{4 i e ^{2}}{( d - 1 ) q ^{2}} [- (d - 2) I_{1} + (2 m^{2} + \frac{d - 2}{2} q^{2}) I_{2} (q^{2})] .

In the pole coefficient one may set

d = 4

: the bracket becomes

\frac{i}{16 π ^{2} ϵ} [- 2 m^{2} + 2 m^{2} + q^{2}] = \frac{i q ^{2}}{16 π ^{2} ϵ} .

Thus the ultraviolet pole of QED vacuum polarization is

Π (q^{2})_{UV} = - \frac{e ^{2}}{12 π ^{2} ϵ}, Π_{μν}_{UV} = - \frac{e ^{2}}{12 π ^{2} ϵ} (q^{2} g_{μν} - q_{μ} q_{ν}) .

The cancellation of the apparent

m^{2} / q^{2}

divergence is essential: gauge invariance forbids a photon mass counterterm. The scalar result at

q^{2} = 0

is understood by continuation after this cancellation.

c

Solution

The middle panel in part a has an internal photon of momentum

k

joining two points on a fermion line; the internal fermion has momentum

p - k

. It is a one-particle-irreducible Feynman diagram with no closed fermion loop sign. In Feynman gauge,

- i Σ (p) = \int_{k} (- i e γ^{α}) \frac{i ( \neq p - \neq k + m )}{( p - k ) ^{2} - m ^{2} + i 0} (- i e γ^{β}) \frac{- i g _{α β}}{k ^{2} + i 0},

Σ (p) = - i e^{2} \int_{k} \frac{( 2 - d ) ( \neq p - \neq k ) + d m}{[ k ^{2} + i 0 ] [( p - k ) ^{2} - m ^{2} + i 0 ]} .

The decomposition in the source gives

tr (\neq p Σ) = 4 p^{2} S (p^{2})

. In particular,

S (p^{2}) = - \frac{i e ^{2} ( d - 2 )}{p ^{2}} \int_{k} \frac{p \cdot k - p ^{2}}{[ k ^{2} + i 0 ] [( p - k ) ^{2} - m ^{2} + i 0 ]} .

For a direct ultraviolet divergence extraction, use a Feynman parameter

x

multiplying the massive denominator, shift

r = k - x p

, and define

Δ = x m^{2} - x (1 - x) p^{2}

. Then

\neq p - \neq k = (1 - x) \neq p - \neq r

and

\int_{0}^{1} d x \int_{r} \frac{1}{( r ^{2} - Δ + i 0 ) ^{2}}_{UV} = \frac{i}{16 π ^{2} ϵ} \int_{0}^{1} d x .

Odd

r

terms integrate to zero. Taking the trace explicitly supplies the weight

(2 - d) (1 - x)

, whose integral tends to

- 1

. Hence

tr (\neq p Σ)_{UV} = - \frac{4 e ^{2} p ^{2}}{16 π ^{2} ϵ}, S (p^{2})_{UV} = - \frac{e ^{2}}{16 π ^{2} ϵ} .

As a check on the same fermion self-energy calculation,

Σ (p)_{UV} = \frac{e ^{2}}{16 π ^{2} ϵ} (- \neq p + 4 m) .

This is the ultraviolet pole of QED fermion self-energy. It is safe to extract it at generic external momentum; an on shell or massless limit must distinguish any infrared divergence from this ultraviolet divergence.

d

Solution

In the right panel of part a,

p

enters the fermion line,

p^{'} = p + q

leaves it, and a virtual photon of momentum

k

bridges the external photon vertex. Write

Γ^{μ} = γ^{μ} + Λ^{μ}

and

F_{1} = 1 + F_{1}^{(1)} + \dots

. The one-loop amplitude is

- i e Λ^{μ} = \int_{k} (- i e γ^{α}) \frac{i ( \neq p ^{'} - \neq k + m )}{( p ^{'} - k ) ^{2} - m ^{2} + i 0} (- i e γ^{μ}) \frac{i ( \neq p - \neq k + m )}{( p - k ) ^{2} - m ^{2} + i 0} (- i e γ^{β}) \frac{- i g _{α β}}{k ^{2} + i 0}, Λ^{μ} = - i e^{2} \int_{k} \frac{γ ^{α} ( \neq p ^{'} - \neq k + m ) γ ^{μ} ( \neq p - \neq k + m ) γ _{α}}{[ k ^{2} + i 0 ] [( p ^{'} - k ) ^{2} - m ^{2} + i 0 ] [( p - k ) ^{2} - m ^{2} + i 0 ]} .

This Feynman integral has a logarithmic superficial degree of divergence. Its ultraviolet pole is independent of the external momenta and

m

, because every such numerator insertion or denominator correction lowers the large-

k

degree. Preserve an auxiliary positive mass

M

while extracting the ultraviolet pole; setting everything to zero would create a scaleless integral and conceal the cancellation of ultraviolet and infrared divergences.

The leading numerator is

γ^{α} \neq k γ^{μ} \neq k γ_{α}

. Symmetric integration replaces

k_{ρ} k_{σ}

g_{ρ σ} k^{2} / d

. Two contractions of gamma matrices then give

γ^{α} \neq k γ^{μ} \neq k γ_{α} ⟼ \frac{( d - 2 ) ^{2}}{d} k^{2} γ^{μ}, tr [γ^{α} \neq k γ^{μ} \neq k γ_{α} γ_{μ}] = 4 (d - 2)^{2} k^{2} .

For the auxiliary-mass extraction of ultraviolet poles,

J_{M} = \int_{k} \frac{k ^{2}}{( k ^{2} - M ^{2} + i 0 ) ^{3}} = \int_{k} [\frac{1}{( k ^{2} - M ^{2} + i 0 ) ^{2}} + \frac{M ^{2}}{( k ^{2} - M ^{2} + i 0 ) ^{3}}], J_{M}_{UV} = \frac{i}{16 π ^{2} ϵ} .

The second term is ultraviolet convergent. Thus

\frac{1}{4 d} tr (Λ^{μ} γ_{μ})_{UV} = - i e^{2} \frac{( d - 2 ) ^{2}}{d} J_{M}_{UV}, F_{1}^{(1)} (q^{2})_{UV} = \frac{e ^{2}}{16 π ^{2} ϵ} .

The ultraviolet pole of the QED vertex is therefore

Λ^{μ} ∣_{UV} = e^{2} γ^{μ} / (16 π^{2} ϵ)

. The Pauli structure has zero trace against

γ_{μ}

and no ultraviolet pole at this order.

Form-factor qualification. The two electron electromagnetic vertex form factors in the source describe matrix elements between external on shell Dirac spinors. A general off shell vertex has more structures, so the unprojected trace is not a projector for the complete finite on-shell

F_{1}

. It does correctly isolate the divergent local

γ^{μ}

coefficient, which is all that is required here.

e

Solution

d = 4 - 2 ϵ

, the modified minimal subtraction scheme subtracts

Δ_{\overline{MS}} = 1/ ϵ - γ_{E} + lo g (4 π)

for each one-loop pole, with no further finite, kinematics-dependent subtraction. Equivalently, use a rescaled renormalization scale

\overline{μ}^{2} = 4 π e^{- γ_{E}} μ_{R}^{2}

and subtract only

1/ ϵ

in that convention. This differs from the minimal subtraction scheme by a fixed scale convention.

Let

C = e^{2} Δ_{\overline{MS}} / (16 π^{2})

. With the source's counterterm insertion signs, cancellation proceeds as follows. The vertex has

- i e γ^{μ} C

from the loop and

- i e γ^{μ} δ_{1}

from its counterterm, giving

δ_{1} = - C

. The kinetic part of the fermion self-energy insertion is

+ i \neq pC

and its counterterm is

+ i \neq p δ_{2}

, giving

δ_{2} = - C

. Finally the photon vacuum polarization insertion is

- i (q^{2} g_{μν} - q_{μ} q_{ν}) (4 C /3)

and its counterterm is

- i (q^{2} g_{μν} - q_{μ} q_{ν}) δ_{3}

, giving

δ_{1} = δ_{2} = - \frac{e ^{2}}{16 π ^{2}} Δ_{\overline{MS}}, δ_{3} = - \frac{e ^{2}}{12 π ^{2}} Δ_{\overline{MS}} .

Writing

Z_{i} = 1 + δ_{i}

therefore verifies the Ward identity for QED renormalization constants,

Z_{1} = Z_{2}

, at one loop. The proper-vertex Ward-Takahashi identity in QED independently gives

Λ^{μ} (p, p) = - \partial Σ (p) / \partial p_{μ}

, whose divergent part reproduces the opposite signs of

F_{1}^{(1)}

and

S

found above. The equality holds in a common gauge fixing and subtraction prescription; the individual constants

Z_{1}, Z_{2}

are gauge dependent.

3

a

Solution

Start with the formally gauge-redundant path integral

Z = vol (G)^{- 1} \int D A e^{i S [A]}

. Use

A_{μ}^{θ} = A_{μ} + \partial_{μ} θ

, and the generalized Lorenz gauge condition

F_{ω} [A] = \partial_{μ} A^{μ} - ω

. In a perturbative neighborhood where residual zero modes are removed by boundary conditions, insert the Faddeev-Popov gauge-orbit identity

1 = Δ_{FP} [A] \int D θ δ [F_{ω} (A^{θ})], \frac{δ F _{ω} ( A ^{θ} ) ( x )}{δ θ ( y )} = □_{x} δ^{(4)} (x - y) .

Thus

Δ_{FP} = det □

. Gauge invariance lets the

θ

integration factor out and cancel

vol (G)

. Average over

ω

with the Gaussian integral weight

exp [- i \int ω^{2} / (2 ξ)]

; the delta functional fixes

ω = \partial \cdot A

and gives

L_{gf} = - \frac{1}{2 ξ} (\partial \cdot A)^{2} .

Represent the functional determinant by the independent Grassmann fields

\overset{c}{ˉ}, c

. A field-independent determinant sign can be absorbed in normalization, so a convenient convention is

det □ \propto \int D \overset{c}{ˉ} D c exp [- i \int d^{4} x \overset{c}{ˉ} □ c] .

Therefore

Z = N \int D A D \overset{c}{ˉ} D c exp {i \int d^{4} x [- \frac{1}{4} F_{μν} F^{μν} - \frac{1}{2 ξ} (\partial \cdot A)^{2} - \overset{c}{ˉ} □ c]} .

The Faddeev-Popov ghosts are free and decouple because

□

is independent of

A

. For a Non-Abelian gauge theory, an infinitesimal gauge transformation gives

δ A_{μ}^{a} = D_{μ}^{ab} θ^{b}

, so the Faddeev-Popov operator is

\partial^{μ} D_{μ}^{ab} [A]

and

L_{gh} = - \overset{c}{ˉ}^{a} \partial^{μ} D_{μ}^{ab} [A] c^{b} .

It contains a ghost-gluon vertex: these Faddeev-Popov ghosts do not decouple in covariant gauge fixing of Lorenz gauge type.

b

Solution

Choose the gauge covariant derivative

D_{μ} = \partial_{μ} + i e A_{μ}

. Write

ρ = f + h

r = 1 + h / f

and

B_{μ} = \partial_{μ} π / f + e A_{μ}

. The polar parametrization of the Abelian Higgs field gives

D_{μ} ϕ = \frac{e ^{iπ / f}}{2} [\partial_{μ} h + i (f + h) B_{μ}], ϕ^{†} ϕ - \frac{f ^{2}}{2} = f h + \frac{h ^{2}}{2} .

Hence the full Lagrangian density is

L = - \frac{1}{4} F_{μν} F^{μν} + \frac{1}{2} (\partial h)^{2} + \frac{1}{2} (1 + \frac{h}{f})^{2} (\partial_{μ} π + e f A_{μ})^{2} - \frac{λ}{2} f^{2} h^{2} - \frac{λ}{2} f h^{3} - \frac{λ}{8} h^{4} .

Assume

f > 0

and

λ > 0

, so the minimum has

∣ ϕ ∣ = f / 2

. In this convention the local gauge symmetry acts as

ϕ \mapsto e^{i α} ϕ, A_{μ} \mapsto A_{μ} - \frac{1}{e} \partial_{μ} α, h \mapsto h, π \mapsto π + f α .

The combination

\partial_{μ} π + e f A_{μ}

, the radial field

h

, and the gauge field strength are each invariant, proving invariance of the displayed Lagrangian density.

This parametrization follows the circle of classical minima:

h

measures distance away from the vacuum manifold, whereas

π

moves along a gauge orbit. It separates the massive radial Higgs mode from the would-be Goldstone boson and makes its removal by the Higgs mechanism transparent. At quadratic order,

L_{2} = - \frac{1}{4} F^{2} + \frac{1}{2} (\partial h)^{2} - \frac{1}{2} λ f^{2} h^{2} + \frac{1}{2} (\partial π)^{2} + e f A \cdot \partial π + \frac{1}{2} e^{2} f^{2} A^{2}, m_{h}^{2} = λ f^{2}, m_{A}^{2} = e^{2} f^{2} .

The polar parametrization of the Abelian Higgs field is a local description near the chosen nonzero vacuum expectation value; it is singular at

ϕ = 0

and is not a global coordinate system for configurations with zeros or winding.

c

Solution

Unitary gauge fixes the scalar phase to zero. Locally choose

α = - π / f

in part b; then

π^{'} = 0

and

A_{μ}^{'} = A_{μ} + \partial_{μ} π / (e f)

. The remaining fields are the radial Higgs mode and a massive vector. Its quadratic Lagrangian density is

- F^{2} /4 + m_{A}^{2} A^{2} /2

and its gauge-boson propagator is

D_{μν}^{U} (k) = \frac{- i}{k ^{2} - m _{A}^{2} + i 0} (g_{μν} - \frac{k _{μ} k _{ν}}{m _{A}^{2}}) .

For

k^{2} \neq = 0

, introduce the transverse and longitudinal orthogonal projectors

P_{μν}^{T} = g_{μν} - k_{μ} k_{ν} / k^{2}

and

P_{μν}^{L} = k_{μ} k_{ν} / k^{2}

. Omitting the infinitesimal in this algebraic decomposition,

D^{U} = - \frac{i P ^{T}}{k ^{2} - m _{A}^{2}} + \frac{i P ^{L}}{m _{A}^{2}} .

At large momentum in Euclidean space, the transverse coefficient falls as

k^{- 2}

but the longitudinal coefficient approaches a constant of order

m_{A}^{- 2}

. Thus power counting applied directly to unitary gauge Feynman diagrams suggests increasingly severe ultraviolet divergences. This hides the renormalizability of physical scattering amplitudes; it does not disprove it. A finite-

ξ

R-xi gauge exposes the better propagator scaling while preserving the fields needed to track the cancellations.

d

Solution

Let

m_{A} = e f

r = 1 + h / f

, and

H = \partial \cdot A - ξ m_{A} r^{2} π

, so the given

G = H / ξ

contributes

L_{gf} = - H^{2} / (2 ξ)

. Its quadratic cross term is

+ m_{A} π \partial \cdot A

, which cancels

m_{A} A \cdot \partial π

from the scalar kinetic term after integration by parts. The free vector and would-be Goldstone boson propagators are therefore diagonal;

π

has gauge-dependent mass squared

ξ m_{A}^{2}

The field-dependent factor

r^{2}

cancels more than just the constant-coefficient mixing, but a residual derivative interaction remains. Indeed,

L_{mix} + L_{gf, cross} = m_{A} r^{2} (A \cdot \partial π + π \partial \cdot A) = \partial_{μ} (m_{A} r^{2} A^{μ} π) - m_{A} π A^{μ} \partial_{μ} r^{2} ≃ - 2 e (1 + \frac{h}{f}) π A^{μ} \partial_{μ} h .

Thus the nonlinear R-xi gauge in polar Higgs variables removes quadratic mixing; it does not remove every interaction containing

A

and

π

Use the gauge transformation in part b. Since

h

is invariant,

δ_{α} H = - \frac{1}{e} □ α - ξ m_{A} r^{2} f α = - \frac{1}{e} [□ + ξ m_{A}^{2} r^{2}] α .

The factor

- 1/ (e ξ)

δ G / δ α

is field independent and is absorbed into the partition function normalization. Exponentiating the remaining Faddeev-Popov determinant with the sign convention of part a gives

L_{gh} = - \overset{c}{ˉ} [□ + ξ e^{2} (f + h)^{2}] c = - \overset{c}{ˉ} (□ + ξ m_{A}^{2}) c - 2 ξ e^{2} f h \overset{c}{ˉ} c - ξ e^{2} h^{2} \overset{c}{ˉ} c .

The ghost coupling in the Abelian Higgs model therefore survives even though the gauge group is Abelian: these Faddeev-Popov ghosts interact with

h

and do not decouple for generic $ξ$ . The distinction from part a is the scalar dependence of the gauge fixing function. In the

ξ \to 0

Landau gauge limit these particular ghost interactions vanish.

e

Solution

After the quadratic mixing cancellation, the vector inverse quadratic operator is

K = - (k^{2} - m_{A}^{2}) P^{T} - (\frac{k ^{2}}{ξ} - m_{A}^{2}) P^{L} .

Inverting its two orthogonal projectors and multiplying by

i

gives

D_{μν}^{(ξ)} (k) = \frac{- i}{k ^{2} - m _{A}^{2} + i 0} [g_{μν} - (1 - ξ) \frac{k _{μ} k _{ν}}{k ^{2} - ξ m _{A}^{2} + i 0}] .

Equivalently

D^{(ξ)} = - i P^{T} / (k^{2} - m_{A}^{2} + i 0) - i ξ P^{L} / (k^{2} - ξ m_{A}^{2} + i 0)

. At fixed finite

ξ

, both sectors fall as

k^{- 2}

at large momentum in Euclidean space;

ξ = 1

gives

- i g_{μν} / (k^{2} - m_{A}^{2} + i 0)

, and taking

ξ \to \infty

at fixed

k

recovers unitary gauge. The latter limit is not uniform in ultraviolet momentum.

The expected power counting conclusion is that this

k^{- 2}

behaviour, together with polynomial interactions of mass dimension at most four in Cartesian scalar variables and the gauge symmetry identities, permits renormalization of spontaneously broken gauge theories using a finite set of physical parameters. The propagator alone is not sufficient to establish that claim in the polar variables of this question. In fact, part b contains

h (\partial π)^{2} / f

and

h^{2} (\partial π)^{2} / (2 f^{2})

, of dimensions five and six, and the nonlinear gauge fixing term contains

- ξ e^{2} h^{4} π^{2} / (2 f^{2})

, also of dimension six. Ordinary power counting is consequently not manifest in these variables.

For a representation where it is manifest, write

ϕ = (f + H + i χ) / 2

and choose the usual polynomial R-xi gauge function

\partial \cdot A - ξ e f χ

. It yields the same free vector propagator, cancels the quadratic mixing, and leaves only dimension-at-most-four interactions, including its Faddeev-Popov ghosts. This supplies the missing vertex condition in the source's proposed inference; the equivalence theorem for field redefinitions relates physical predictions near the vacuum. Good propagator falloff is the necessary ingredient illustrated by the requested formula, rather than a complete proof for arbitrary parametrizations or nonlinear choices of gauge fixing.

2024 (A15272W1)

Source PDF: mmathphys.physics.ox.ac.uk/sites/g/files/vzsiac2991/files/mmathphys/documents/media/aqft_for_particle_physics_2024.pdf

Solutions to all seventeen lettered parts. Use natural units, the Minkowski metric

g_{μν} = diag (1, - 1, - 1, - 1)

, Feynman slash notation

\neq p = γ^{μ} p_{μ}

, and the Feynman i-epsilon prescription in internal propagators. Define the scattering amplitude by

S_{f i} = δ_{f i} + i (2 π)^{4} δ^{(4)} (p_{f} - p_{i}) M_{f i}

. The gauge covariant derivative convention in question 3 is specified there to match the printed three-boson Feynman rule.

1

a

Solution

For a Dirac field, the path integral is a Berezin integral over independent Grassmann fields

ψ

and

\overset{ˉ}{ψ}

Z = N \int D \overset{ˉ}{ψ} D ψ exp [i \int d^{4} x \overset{ˉ}{ψ} (i \neq \partial - m) ψ] .

The vacuum boundary conditions and the Feynman i-epsilon prescription are understood;

N

fixes the vacuum normalization. This is the vacuum generating functional without sources, rather than a finite-temperature partition function.

The Dirac adjoint transforms as

\overset{ˉ}{ψ} \mapsto \overset{ˉ}{ψ} e^{- i α}

. For constant

α

\partial_{μ} (e^{i α} ψ) = e^{i α} \partial_{μ} ψ

. Both the kinetic term and the Dirac mass term therefore retain their values. This proves the global phase symmetry, whose group is the unitary group

U (1)

b

Solution

For the local variation

δ ψ = i α (x) ψ

δ \overset{ˉ}{ψ} = - i α (x) \overset{ˉ}{ψ}

, the terms proportional to

α

cancel and the differentiated phase gives

δ L = i (- i α \overset{ˉ}{ψ}) γ^{μ} \partial_{μ} ψ + i \overset{ˉ}{ψ} γ^{μ} \partial_{μ} (i α ψ) - m δ (\overset{ˉ}{ψ} ψ) = - \overset{ˉ}{ψ} γ^{μ} ψ \partial_{μ} α .

Thus, for a compactly supported

α

, integration by parts gives

j^{μ} = \overset{ˉ}{ψ} γ^{μ} ψ, δ S = - \int d^{4} x j^{μ} \partial_{μ} α = \int d^{4} x α \partial_{μ} j^{μ} .

This Dirac current is the Noether current in the convention

δ S = - \int j^{μ} \partial_{μ} α

. Using the convention with a plus sign reverses the Noether current and the conserved charge together. There is one independent global symmetry in field theory, with four spacetime components of its Noether current. The associated conserved charge is

Q = \int d^{3} x j^{0}

, with vacuum subtraction when required.

c

Solution

An operator equation in a path integral

O (x) = 0

means that insertion of

O (x)

into every time-ordered correlation function vanishes when

x

is separated from the other insertion points. At coincident points, Ward identity contact terms can remain; an operator equation does not assert that these distributions vanish.

Let

X

be a product of local insertions. Make the local phase change of integration variables from part b. The functional measure has unit functional Jacobian: the opposite phases of

ψ

and

\overset{ˉ}{ψ}

cancel in a vector-symmetry-preserving regularization in quantum field theory. There is no vector gauge anomaly. Invariance under a change of integration variables gives

0 = ⟨ δ_{α} X ⟩ + i \int d^{4} x α (x) ⟨ T {\partial_{μ} j^{μ} (x) X}⟩ .

X = \prod_{r} O_{r} (x_{r})

and

δ_{α} O_{r} = i q_{r} α (x_{r}) O_{r}

, comparison of the coefficients of the arbitrary

α (x)

yields the Ward-Takahashi identity

⟨ T {\partial_{μ} j^{μ} (x) X}⟩ = - r \sum q_{r} δ^{(4)} (x - x_{r}) ⟨ TX ⟩ .

Consequently

\partial_{μ} j^{μ} = 0

is an operator equation at separated points. This is stronger than just

⟨ \partial_{μ} j^{μ} ⟩ = 0

. For example,

X = ψ (y) \overset{ˉ}{ψ} (z)

has

q_{y} = 1

q_{z} = - 1

, so its two Ward identity contact terms are

[- δ^{(4)} (x - y) + δ^{(4)} (x - z)] ⟨ T ψ (y) \overset{ˉ}{ψ} (z)⟩

; dropping them would give a false identity. Equivalently, the Dirac equation and the adjoint Dirac equation give

\partial_{μ} (\overset{ˉ}{ψ} γ^{μ} ψ) = 0

, with the same coincident-point qualification in time-ordered correlation functions.

d

Solution

Couple a nondynamical background gauge field

B_{μ}

to the Dirac current, choosing a plus sign for the source term:

Z [B] = N \int D \overset{ˉ}{ψ} D ψ exp {i \int d^{4} x [\overset{ˉ}{ψ} (i \neq \partial - m) ψ + B_{μ} \overset{ˉ}{ψ} γ^{μ} ψ]} .

There is no integration over

B_{μ}

. The gauge covariant derivative is

\partial_{μ} - i B_{μ}

, so the simultaneous transformations are

ψ^{'} = e^{i α (x)} ψ, \overset{ˉ}{ψ}^{'} = \overset{ˉ}{ψ} e^{- i α (x)}, B_{μ}^{'} = B_{μ} + \partial_{μ} α .

The kinetic term changes by

- j^{μ} \partial_{μ} α

and the source term changes by

+ j^{μ} \partial_{μ} α

. The action is therefore invariant, exactly for finite

α

as well as infinitesimally. Changing integration variables and using the invariant functional measure from part c proves the background gauge invariance of a current generating functional:

Z [B + \partial α] = Z [B] .

The allowed transformations preserve the vacuum boundary conditions. Its infinitesimal version is

\partial_{μ} δ Z / δ B_{μ} = 0

. Because

B_{μ}

is only a source, this statement needs no gauge fixing or photon kinetic term. The absence of a vector gauge anomaly is essential to the change-of-variables argument.

e

Solution

Collect the two Dirac fields into

Ψ = (ψ_{1}, ψ_{2})^{T}

, with a common nonzero mass

m

. Then

L = i = 1 \sum 2 \overset{ˉ}{ψ}_{i} (i \neq \partial - m) ψ_{i} = \overset{ˉ}{Ψ} (i \neq \partial - m 1_{2}) Ψ.

Among complex-linear rotations of the Dirac fields, the internal symmetry of a classical field theory is

Ψ \mapsto U Ψ

U \in U (2)

, with

\overset{ˉ}{Ψ} \mapsto \overset{ˉ}{Ψ} U^{†}

. Unitarity preserves the kinetic term, and the mass matrix commutes with every

U

. This unitary flavour symmetry of degenerate Dirac fields has group

(S U (2) \times U (1)) / Z_{2}

, where the quotient avoids counting the common central element twice.

Choose the four Hermitian matrices

T^{0} = 1_{2} /2

T^{a} = σ^{a} /2

for

a = 1, 2, 3

, with

σ^{a}

the Pauli matrices. Promoting the four parameters to functions gives

δ S = - \int d^{4} x (\partial_{μ} α^{A}) \overset{ˉ}{Ψ} γ^{μ} T^{A} Ψ

. The four Noether currents are therefore

j_{A}^{μ} = \overset{ˉ}{Ψ} γ^{μ} T^{A} Ψ, A = 0, 1, 2, 3.

Their divergences vanish by the equal-mass Dirac equations. The singlet is half the total Dirac current; the other three currents rotate the two species.

Free-theory qualification. If “global internal symmetry” allows mixing the Dirac fields with charge-conjugate fields, the completely free theory written here has a larger enhanced internal symmetry of free Dirac fields. Write

ψ_{i} = (λ_{2 i - 1} + i λ_{2 i}) / 2

using four Majorana fermions. Up to a total derivative,

L = \frac{1}{2} \sum_{A = 1}^{4} \overset{ˉ}{λ}_{A} (i \neq \partial - m) λ_{A} .

The full orthogonal group

O (4)

acts on the four real species; its connected part is

SO (4)

. Its six Noether currents can be chosen as

J_{A B}^{μ} = i \overset{ˉ}{λ}_{A} γ^{μ} λ_{B}

for

A < B

, up to the common sign convention. The antisymmetry of this Majorana fermion vector bilinear makes these the six independent rotations. A concrete rotation outside the stated complex-linear

U (2)

λ_{2} \mapsto cos θ λ_{2} + sin θ λ_{3}

λ_{3} \mapsto - sin θ λ_{2} + cos θ λ_{3}

: it preserves the displayed Lagrangian density while mixing

ψ_{i}

with their charge-conjugate fields. The usual

U (2)

answer specifies the subgroup preserving the chosen common fermion number.

f

Solution

In the Weyl representation of the gamma matrices, write each Dirac field as

ψ_{i} = (ℓ_{i α}, r_{i}^{\overset{α}{˙}})^{T}

, where

ℓ_{i}

and

r_{i}

have opposite chirality. With the Weyl sigma matrices

σ^{μ} = (1, σ)

and

\overset{σ}{ˉ}^{μ} = (1, - σ)

L = i = 1 \sum 2 [i ℓ_{i}^{†} \overset{σ}{ˉ}^{μ} \partial_{μ} ℓ_{i} + i r_{i}^{†} σ^{μ} \partial_{μ} r_{i} - m (ℓ_{i}^{†} r_{i} + r_{i}^{†} ℓ_{i})] .

The Dirac mass term couples the two Weyl fields. Under

ℓ \mapsto U_{L} ℓ

r \mapsto U_{R} r

, its first term acquires

U_{L}^{†} U_{R}

, so a common nonzero mass requires

U_{L} = U_{R}

. At zero mass there is no such constraint:

G_{number preserving} = U (2)_{L} \times U (2)_{R} .

This is the chiral flavour symmetry of massless Dirac fields. Its eight Noether currents are

j_{L, A}^{μ} = ℓ^{†} \overset{σ}{ˉ}^{μ} T^{A} ℓ

and

j_{R, A}^{μ} = r^{†} σ^{μ} T^{A} r

. The sum

j_{R} + j_{L}

is the vector current, and

j_{R} - j_{L}

is the axial current, for each flavour generator. This statement concerns the free theory in flat spacetime; an interacting theory can have a chiral anomaly.

For completeness, allowing the extra transformations described in part e gives four left-handed Weyl fields,

q = (ℓ_{1}, ℓ_{2}, r_{1}^{c}, r_{2}^{c})^{T}

, where

r_{i}^{c}

is the left-handed charge-conjugate field of

r_{i}

. After integration by parts their kinetic term is

i q^{†} \overset{σ}{ˉ}^{μ} \partial_{μ} q

, whose maximal linear unitary group is

U (4)

. The symmetric mass matrix for equal Dirac masses is

M = m (0 1_{2} 1_{2} 0), L_{m} = - \frac{1}{2} (q^{T} Mq + Hermitian conjugate),

with the antisymmetric spinor contraction implicit. A unitary change of basis makes

M = m 1_{4}

; preserving both the kinetic term and

M

requires

V^{†} V = V^{T} V = 1_{4}

, hence real

V \in O (4)

. Thus the unrestricted free-theory enhancement is

U (4)

at zero mass and

O (4)

at equal nonzero mass; the familiar

U (2)_{L} \times U (2)_{R}

and diagonal

U (2)

are their common-fermion number-preserving subgroups.

2

a

Solution

The two one-loop Feynman diagrams are fermion loops with three photon interaction vertices. Fix vertex 1; the two cyclic orders are

(1, 2, 3)

and

(1, 3, 2)

. A cyclic rotation does not produce another Feynman diagram. Keeping all three external labels in the same positions, these two orders can be represented by the same triangle with opposite fermion flow, as in the first two panels.

Figure 1.
Original Feynman diagrams for the two orientations of the three-photon fermion loop and a representative four-photon box diagram
. **Details**.

The photons are labelled

(k_{1}, μ)

(k_{2}, ν)

(k_{3}, ρ)

, with all four-momenta incoming and

k_{1} + k_{2} + k_{3} = 0

. The amputation of external propagators removes the three external photon propagators, leaving their Lorentz indices open. There are no other connected one-loop Feynman diagrams with only three external photons in QED.

b

Solution

Put

S (p) = i (\neq p + m) / (p^{2} - m^{2} + i 0)

for the Dirac propagator, and

\int_{ℓ} = μ_{R}^{2 ϵ} \int d^{4 - 2 ϵ} ℓ / (2 π)^{4 - 2 ϵ}

for a translation-invariant dimensional regularization of the loop momentum integral. The two oriented scattering amplitudes are

i M_{A}^{μν ρ} i M_{B}^{μν ρ} = - \int_{ℓ} tr [(- i e γ^{μ}) S (ℓ + k_{1}) (- i e γ^{ν}) S (ℓ - k_{3}) (- i e γ^{ρ}) S (ℓ)], = - \int_{ℓ} tr [(- i e γ^{μ}) S (ℓ + k_{1}) (- i e γ^{ρ}) S (ℓ - k_{2}) (- i e γ^{ν}) S (ℓ)] .

The common minus sign is the closed fermion loop sign. There is no additional symmetry factor for either labelled orientation. Each expression is a trace over spinor indices; all external photon propagators are amputated. If

K (p) = (\neq p + m) / (p^{2} - m^{2} + i 0)

, the common factor is

- (- i e)^{3} i^{3} = - e^{3}

, so, for example,

i M_{A} = - e^{3} \int_{ℓ} tr [γ^{μ} K (ℓ + k_{1}) γ^{ν} K (ℓ - k_{3}) γ^{ρ} K (ℓ)]

c

Solution

Use the source's charge-conjugation matrix convention

C γ^{μ} C^{- 1} = - (γ^{μ})^{T}

. It implies

C^{- 1} (γ^{μ})^{T} C = - γ^{μ}, C^{- 1} K (p)^{T} C = K (- p) .

A trace equals the trace of its matrix transpose, which reverses the order of all factors. Inserting the displayed identities gives

T_{A} (ℓ) = tr [γ^{μ} K (ℓ + k_{1}) γ^{ν} K (ℓ - k_{3}) γ^{ρ} K (ℓ)] = - tr [K (- ℓ) γ^{ρ} K (- ℓ + k_{3}) γ^{ν} K (- ℓ - k_{1}) γ^{μ}] = - tr [γ^{μ} K (- ℓ) γ^{ρ} K (- ℓ + k_{3}) γ^{ν} K (- ℓ - k_{1})] .

There is one minus sign from each of the three photon interaction vertices, and none from the transformed Dirac propagators. In the last line, change loop momentum to

ℓ^{'} = - ℓ - k_{1}

. Since

k_{1} + k_{3} = - k_{2}

, its three Dirac propagators become

K (ℓ^{'} + k_{1})

K (ℓ^{'} - k_{2})

K (ℓ^{'})

, exactly those of the reversed orientation. Hence

M_{A}^{μν ρ} = - M_{B}^{μν ρ}, M_{one loop}^{μν ρ} = 0.

This is Furry's theorem. The regularization in quantum field theory matters: the individual Feynman integrals are superficially divergent, so an unregulated shift of a divergent integral is not a proof. A common dimensional regularization preserves charge conjugation and the vector Ward identity, making the change of variables legitimate. No chirality matrix occurs, so the axial triangle anomaly is not involved.

d

Solution

For an

n

-vertex fermion loop, reversing the fermion flow reverses the ordered product inside its trace. The same charge-conjugation matrix calculation transforms every Dirac propagator by

p \mapsto - p

and contributes

- 1

for each vector photon interaction vertex. Relabel the loop momentum so that the external labels are fixed. The reversed Feynman diagram therefore obeys

M_{reversed} = (- 1)^{n} M_{original} .

For odd

n \geq 3

, the

(n - 1)!

cyclic orders pair into opposite orientations and cancel. For

n = 1

, the single photon tadpole maps into itself with a minus sign and is zero. The common closed fermion loop sign is present on both sides and does not supply the factor

(- 1)^{n}

. This proves Furry's theorem for all the requested one-loop Feynman diagrams. The charge conjugation argument also gives vanishing odd-photon correlation functions to all orders in vacuum QED, provided no charge conjugation symmetry breaking background or charged external state is introduced.

e

Solution

The third panel of the figure in part a is a representative four-photon box diagram. Fix the external photon labelled 1 and orient the fermion loop. The six inequivalent cyclic orders of the labelled interaction vertices are

(1234), (1243), (1324), (1342), (1423), (1432) .

They are obtained by permuting labels

2, 3, 4

on the displayed box, keeping label 1 fixed. Equivalently there are three undirected boxes, each with two opposite fermion flows. The reversal pairs are

(1234) \leftrightarrow (1432), (1243) \leftrightarrow (1342), (1324) \leftrightarrow (1423) .

For four photon interaction vertices, the sign in part d is positive, so the members of each pair add. They can be represented by three box diagrams multiplied by two. There is no tree photon self-coupling in QED, and therefore no triangle or bubble with a photon self-vertex to add at this loop order. This is the one-loop light-by-light scattering contribution, of order

e^{4}

f

Solution

At large loop momentum, each Dirac propagator scales as

ℓ^{- 1}

. One four-dimensional loop and four internal Dirac propagators therefore give the superficial degree of divergence

D = 4 - 4 = 0.

Thus a possible ultraviolet divergence in the sum is logarithmic. By locality its divergent coefficient is a degree-zero polynomial in the external four-momenta, namely a constant Lorentz tensor,

M_{div}^{μν ρ σ} = \frac{1}{ϵ} (a g^{μν} g^{ρ σ} + b g^{μ ρ} g^{ν σ} + c g^{μ σ} g^{ν ρ}) .

The photon Ward identities apply to the complete sum for arbitrary external four-momenta, including off shell ones:

k_{1 μ} M^{μν ρ σ} = 0

, and likewise for the other legs. A constant tensor with

k_{1 μ} M_{div}^{μν ρ σ} = 0

for every

k_{1}

must be zero. Hence

a = b = c = 0

and

M_{one loop}^{μν ρ σ} is ultraviolet finite .

Equivalently, a logarithmic divergence would require a local dimension-four

A^{4}

counterterm, forbidden by Abelian gauge invariance. A gauge-invariant four-photon local operator starts with four gauge field strengths, for example

(F_{μν} F^{μν})^{2}

, of mass dimension eight; it cannot be this power counting divergence. Individual boxes need not be finite; their forbidden divergent parts cancel in the sum. There are no divergent proper loop subgraphs at this loop order. With the massive Dirac field in the source, there is also no soft or collinear infrared divergence at generic external kinematics. The argument establishes ultraviolet finiteness, and does not claim the absence of physical threshold singularities.

3

a

Solution

Choose

D_{μ} = \partial_{μ} - i g A_{μ}^{a} t^{a}

on the Dirac field,

t^{a} = σ^{a} /2

in the fundamental representation of SU(2), and

[t^{a}, t^{b}] = i ϵ^{ab c} t^{c}

. Define the gauge field strength and the Adjoint representation gauge covariant derivative by

F_{μν}^{a} = \partial_{μ} A_{ν}^{a} - \partial_{ν} A_{μ}^{a} + g ϵ^{ab c} A_{μ}^{b} A_{ν}^{c}, (D_{μ} c)^{a} = \partial_{μ} c^{a} + g ϵ^{ab c} A_{μ}^{b} c^{c} .

The required gauge-fixed Yang-Mills Lagrangian in a covariant gauge, including the matter term, is

L = - \frac{1}{4} F_{μν}^{a} F^{a μν} + \overset{ˉ}{ψ} (i γ^{μ} D_{μ} - m) ψ - \frac{1}{2 ξ} (\partial^{μ} A_{μ}^{a})^{2} - \overset{c}{ˉ}^{a} \partial^{μ} (D_{μ} c)^{a} .

Here

c^{a}

and

\overset{c}{ˉ}^{a}

are independent Grassmann fields in the Adjoint representation: the Faddeev-Popov ghost fields and antighost fields. Under an infinitesimal gauge transformation,

δ A_{μ}^{a} = (D_{μ} ω)^{a}

; hence the Lorenz gauge function

G^{a} = \partial^{μ} A_{μ}^{a}

varies by

\partial^{μ} D_{μ}^{ab} ω^{b}

. Its Faddeev-Popov determinant is represented by the ghost term, with an irrelevant field-independent determinant sign absorbed into normalization. Integration by parts gives

L_{gh} = (\partial^{μ} \overset{c}{ˉ}^{a}) (D_{μ} c)^{a}

, explicitly showing the ghost-gluon vertex.

There is no scalar vacuum expectation value here, so R-xi gauge reduces to the linear covariant gauge displayed above, with no Goldstone boson term. At

ξ = 1

the Feynman gauge gauge boson propagator is

- i δ^{ab} g_{μν} / (k^{2} + i 0)

. With Fourier convention

e^{- ik x}

and all three boson four-momenta incoming, this choice gives exactly the source's

+ g ϵ^{ab c}

three-boson Feynman rule, and the matter interaction vertex is

+ i g γ^{μ} t^{a}

. The sign of

D_{μ}

has been explicitly chosen here; the QED convention in question 2 instead has vertex

- i e γ^{μ}

b

Solution

There are three tree-level Feynman diagrams: two ordered fermion-exchange diagrams and one s-channel gauge boson exchange diagram. Write the outgoing labels as

(k_{1}, a, μ)

and

(k_{2}, b, ν)

Figure 1.
Original tree-level Feynman diagrams for fermion and antifermion annihilation into two non-Abelian gauge bosons: two ordered exchanges and the three-boson annihilation channel
. **Details**.

Following the fermion flow from

u (p_{1})

\overset{v}{ˉ} (p_{2})

, the first diagram emits boson 1 and then boson 2; its internal four-momentum is

r_{t} = p_{1} - k_{1}

and its ordered colour factor is

t^{b} t^{a}

. The second emits boson 2 and then boson 1, giving

r_{u} = p_{1} - k_{2}

and

t^{a} t^{b}

. The third annihilates into one virtual gauge boson of four-momentum

q = p_{1} + p_{2} = k_{1} + k_{2}

and uses the three-boson interaction vertex. There is no two-fermion two-boson contact interaction vertex for the minimally coupled Dirac field, and no tree ghost contribution with these external states.

c

Solution

Use the Feynman rules and Feynman gauge convention from part a. Put

s = q^{2}

t = r_{t}^{2}

u = r_{u}^{2}

, the Mandelstam variables, and define the three-boson tensor with incoming four-momenta

(q, - k_{1}, - k_{2})

V^{λ μν} = g^{λ μ} (q + k_{1})^{ν} + g^{μν} (k_{2} - k_{1})^{λ} - g^{ν λ} (q + k_{2})^{μ} .

The contributions, with external Dirac spinors included and both external gauge boson propagators amputated, are

M_{t}^{ab μν} M_{u}^{ab μν} M_{s}^{ab μν} M^{ab μν} = - g^{2} \overset{v}{ˉ} (p_{2}) t^{b} t^{a} γ^{ν} \frac{\neq r _{t} + m}{t - m ^{2} + i 0} γ^{μ} u (p_{1}), = - g^{2} \overset{v}{ˉ} (p_{2}) t^{a} t^{b} γ^{μ} \frac{\neq r _{u} + m}{u - m ^{2} + i 0} γ^{ν} u (p_{1}), = - i g^{2} ϵ^{ab c} \overset{v}{ˉ} (p_{2}) t^{c} γ_{λ} u (p_{1}) \frac{V ^{λ μν}}{s + i 0}, = M_{t}^{ab μν} + M_{u}^{ab μν} + M_{s}^{ab μν} .

For example,

i M_{t}

contains

(+ i g)^{2} i = - i g^{2}

. In the annihilation channel, the current vertex, propagator, and three-boson vertex give

(+ i g) (- i) g = g^{2}

, so division by the defining overall

i

gives

- i g^{2}

. These relative signs are essential for the cancellation in part d. The antisymmetric colour factor makes the last expression symmetric under simultaneously exchanging

(a, μ, k_{1})

and

(b, ν, k_{2})

, as required for identical outgoing gauge bosons. The on shell external Dirac spinors obey

(\neq p_{1} - m) u (p_{1}) = 0

\overset{v}{ˉ} (p_{2}) (\neq p_{2} + m) = 0

d

Solution

For the outgoing massless gauge bosons, take

k_{1}^{2} = k_{2}^{2} = 0

. Work away from internal poles, with the Feynman i-epsilon prescription understood. The Dirac equations and four-momentum conservation give

\overset{v}{ˉ} \neq k_{2} \neq k_{2} u = \overset{v}{ˉ} (\neq r_{t} - m), = - (\neq r_{u} - m) u .

Multiplying by the neighboring Dirac propagator numerator cancels its denominator. The two fermion-exchange diagrams therefore yield the matrix commutator

k_{2 ν} (M_{t} + M_{u})^{ab μν} = g^{2} \overset{v}{ˉ} γ^{μ} (t^{a} t^{b} - t^{b} t^{a}) u = i g^{2} ϵ^{ab c} J_{c}^{μ}, J_{c}^{λ} = \overset{v}{ˉ} (p_{2}) γ^{λ} t^{c} u (p_{1}) .

The annihilation vector current obeys

q_{λ} J_{c}^{λ} = \overset{v}{ˉ} (\neq p_{1} + \neq p_{2}) t^{c} u = 0

. Direct contraction of the three-boson tensor gives

k_{2 ν} V^{λ μν} = s g^{λ μ} - q^{λ} k_{2}^{μ} - k_{2}^{λ} k_{1}^{μ} .

Thus

k_{2 ν} M_{s}^{ab μν} = - i g^{2} ϵ^{ab c} (J_{c}^{μ} - \frac{J _{c} \cdot k _{2}}{s} k_{1}^{μ}) .

The terms proportional to

J_{c}^{μ}

cancel, leaving the non-Abelian Ward identity for fermion annihilation

k_{2 ν} M^{ab μν} = i g^{2} ϵ^{ab c} \frac{J _{c} \cdot k _{2}}{s} k_{1}^{μ} .

The open-index tensor is generally nonzero: neither

ϵ^{ab c}

nor

J_{c} \cdot k_{2}

is constrained to vanish. For a concrete nonzero configuration, take centre-of-momentum frame kinematics

p_{1} = (E, 0, 0, p)

p_{2} = (E, 0, 0, - p)

p = E^{2} - m^{2}

k_{1} = (E, E, 0, 0)

k_{2} = (E, - E, 0, 0)

. Standard Dirac spinors with two-component spin labels

χ = (1, 0)^{T}

for

u

and

η = (0, 1)^{T}

for

v

give the spinor current

(0, 2 E, 2 i E, 0)

and its contraction with

k_{2}

2 E^{2}

. Choosing both colour spinors

(1, 0)^{T}

and

(a, b, c) = (1, 2, 3)

multiplies this by

1/2

, still nonzero. This explicitly rules out an unconditional tensor identity

k_{2 ν} M^{ab μν} = 0

For a physical polarization vector

ε_{1 μ}^{*}

, however,

k_{1} \cdot ε_{1}^{*} = 0

, and therefore

ε_{1 μ}^{*} k_{2 ν} M^{ab μν} = 0.

This is the required decoupling of a pure-gauge polarization vector on leg 2 when the other leg is physical. Interchanging the legs gives the corresponding identity for leg 1. The role of the three-boson diagram is explicit: it cancels the noncommuting colour-current term left by the two fermion exchanges.

e

Solution

The non-Abelian counterpart of the photon Ward identity is a Slavnov-Taylor identity, derived from BRST symmetry. A scattering amplitude with all other external states physical vanishes when one external massless gauge boson polarization vector is replaced by its four-momentum. An amputated tensor with the remaining indices open need not be transverse: part d gives its explicit nonzero longitudinal remainder.

Consequently, amplitudes involving several unphysical covariant polarization vectors need not vanish. For example, contracting the result of part d with a timelike vector

n_{μ}

gives

i g^{2} ϵ^{ab c} (J_{c} \cdot k_{2}) (n \cdot k_{1}) / s

, nonzero in the explicit configuration there for

n = (1, 0, 0, 0)

. Even with other states physical, the pure-gauge vector

k

is the relevant null combination; separate timelike and spatial longitudinal basis vectors are not themselves pure gauge, and their separate amplitudes need not be zero.

These auxiliary components do not define separately observable positive probabilities. Physical states are classes in BRST cohomology, and the S-matrix preserves that physical state space in this anomaly-free vectorlike gauge spectrum. There is no physical transition from a physical state to a nonphysical state. The incorrect stronger statement would be that every covariant component called “unphysical” has a zero matrix element.

For loop Feynman diagrams or unitarity cuts, summing covariant gauge indices with

- g_{μν}

propagates those auxiliary components. In a Non-Abelian gauge theory, their cancellation requires the interacting Faddeev-Popov ghost fields as well as the gauge boson diagrams. Ghost contributions carry their Grassmann field loop signs and restore the Slavnov-Taylor identities and physical unitarity. One may instead use a formulation with only physical propagating polarization vectors, but one cannot simply discard the ghost diagrams while retaining the Feynman gauge rules from part a.

2021 (A15272W1)

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/9/Physics%20Papers%202020.pdf?cookie_passthrough=1

Official standalone paper: mmathphys.physics.ox.ac.uk/sites/default/files/mmathphys/documents/media/aqft_2021.pdf

Use natural units, the Minkowski metric

g = diag (1, - 1, - 1, - 1)

and the Feynman i-epsilon prescription. In question 1 use dimensional regularization with

d = 4 - 2 ϵ

and write

C_{ϵ} = 1/ (16 π^{2} ϵ)

; in the alternative convention

d = 4 - ϵ

, replace

C_{ϵ}

1/ (8 π^{2} ϵ)

. Only ultraviolet poles are retained there. Figure 1 is unreadable in the released source: the corresponding specific gaps are marked below. All other parts have complete derivations.

1

i

Solution

The superficial degree of divergence of this scalar electrodynamics three-point vertex is one. Its ultraviolet pole is consequently a local expression at most linear in the external four-momenta. The allowed counterterm Lagrangian, with the charge-conjugation and Hermiticity properties of scalar electrodynamics, contains a single one-derivative scalar electrodynamics three-point vertex:

V_{ct}^{μ} (k, k^{'}) = - i e δ_{1} (k + k^{'})^{μ} .

Thus the coefficient

δ_{1}

is independent of the chosen external four-momenta. Setting

k^{'} = 0

leaves

- i e δ_{1} k^{μ}

, so a nonzero generic

k

still determines that coefficient. It does not determine the full finite scattering amplitude at other four-momenta.

Keep

m \neq = 0

and generic off shell

k

while extracting the ultraviolet pole. These choices avoid spurious infrared divergences; setting all scales to zero would instead produce scaleless integrals that can conceal the ultraviolet pole.

ii

Solution

TODO (source gap): Figure 1(a), including its internal connections and four-momentum labels, is absent from the recoverable source. The local PDF renderer reports missing image objects, and the matching official PDF's figure could not be retrieved for visual inspection. Therefore the individual Feynman diagram called (a) cannot be assigned an integrand and its separate independence of

ξ

cannot be verified from the available diagram. A guessed replacement drawing would not resolve this gap.

An independent determination of the complete ultraviolet pole, given in part (iv), remains possible from the scalar QED Ward identity. Together with the textually supplied integral in part (iv), it fixes the sum of the contributions outside that integral to

3 i e^{3} C_{ϵ} k^{μ}

in the stated scalar electrodynamics three-point vertex convention. That sum alone does not identify the contents of panel (a).

iii

Solution

TODO (source gap): Figure 1(c) and its four-momentum labels are unreadable. The assertion about this individual Feynman diagram therefore cannot be checked against its actual integrand. In particular, one must not identify the missing panel with a convenient vanishing seagull vertex graph by inference.

A useful criterion, once the drawing is recovered, is that dimensional regularization annihilates a vector integral of the form

\int \frac{d ^{d} l}{( 2 π ) ^{d}} l^{μ} f (l^{2}) = 0

by the change of variables

l \mapsto - l

. This would prove vanishing for every

ξ

if all remaining factors were functions of

l^{2}

; a denominator involving

l + k

must first be checked rather than discarded. Whether this criterion applies to panel (c) is precisely the unavailable information.

iv

Solution

The integral printed in the text can be evaluated without reconstructing Figure 1. Put

A = l^{2} - m^{2} + i 0

and

B = (l + k)^{2} - m^{2} + i 0

, and let

I^{μ}

denote that integral without its factor

ξ

. The numerator contraction responsible for its covariant gauge dependence is

l_{α} (g^{α β} - (1 - ξ) \frac{l ^{α} l ^{β}}{l ^{2}}) (l + 2 k)_{β} = ξ (l^{2} + 2 l \cdot k) .

With the scalar electrodynamics three-point vertex

- i e (p + p^{'})^{μ}

, three such vertices, two scalar propagators and one photon propagator have overall factor

(- i e)^{3} i^{2} (- i) = - e^{3}

. Thus the supplied integrand gives

V_{supplied}^{μ} = - e^{3} ξ I^{μ}

First isolate the numerator term

l^{2}

. Its contribution is

I_{0}^{μ} = \int \frac{d ^{d} l}{( 2 π ) ^{d}} \frac{( 2 l + k ) ^{μ}}{A B} .

Under

l \mapsto - l - k

the two denominators exchange and the numerator changes sign. A shift-invariant dimensional regularization therefore gives

I_{0}^{μ} = 0

, including the finite part.

For the remaining numerator

2 (l \cdot k) (2 l + k)^{μ}

, only its term

4 l^{μ} (l \cdot k)

contributes to the ultraviolet pole. Corrections from expanding the denominators, and the numerator

2 k^{μ} (l \cdot k)

, are ultraviolet finite by power counting. Use auxiliary-mass extraction of ultraviolet poles with any

M > 0

rather than literally integrating a massless scaleless integral. Lorentz covariance then gives

I^{μ}_{UV} V_{supplied}^{μ}_{UV} = \frac{4}{d} k_{ν} g^{μν} \int \frac{d ^{d} l}{( 2 π ) ^{d}} \frac{1}{( l ^{2} - M ^{2} + i 0 ) ^{2}}_{UV} = i C_{ϵ} k^{μ}, = - i e^{3} ξ C_{ϵ} k^{μ} .

TODO (source gap): The algebra of the supplied expression is complete, but its graphical identification with Figure 1(b), including the routing at each internal vertex, cannot be visually verified from the unreadable panel.

The complete renormalization constant can nevertheless be obtained independently. Here is a short scalar QED wave-function pole calculation, to avoid assuming either missing panel's value. For an external four-momentum

p

, the exchange contribution to the scalar self-energy has numerator

N = (2 p - l)^{2} - (1 - ξ) \frac{[( 2 p - l ) \cdot l ] ^{2}}{l ^{2}} = 4 p^{2} - 4 ξ p \cdot l + ξ l^{2} - 4 (1 - ξ) \frac{( p \cdot l ) ^{2}}{l ^{2}} .

Its denominator is

l^{2} [(p - l)^{2} - m^{2} + i 0]

. The necessary ultraviolet poles, with the measure

d^{d} l / (2 π)^{d}

, are respectively

\int \frac{1}{l ^{2} D}_{UV} = i C_{ϵ}, \int \frac{l ^{μ}}{l ^{2} D}_{UV} = \frac{p ^{μ}}{2} i C_{ϵ}, \int \frac{1}{D}_{UV} = m^{2} i C_{ϵ}, \int \frac{l ^{μ} l ^{ν}}{( l ^{2} ) ^{2} D}_{UV} = \frac{g ^{μν}}{4} i C_{ϵ}, D = (p - l)^{2} - m^{2} + i 0.

The first two follow by a Feynman parameter and a shift of

l

; the third is a tadpole integral; the last follows by the same large-

l

extraction used above. The seagull vertex contributes only a photon tadpole integral, which is a scaleless integral, and in any case cannot affect the coefficient of

p^{2}

. Multiplication by the exchange graph's factor

- e^{2}

gives the scalar self-energy insertion

- i e^{2} C_{ϵ} [(3 - ξ) p^{2} + ξ m^{2}] .

The kinetic counterterm

i δ_{2} p^{2}

therefore requires

δ_{2} = e^{2} (3 - ξ) C_{ϵ}

. The scalar QED Ward identity sets

δ_{1} = δ_{2}

, so

Z_{1} = 1 + \frac{e ^{2} ( 3 - ξ )}{16 π ^{2} ϵ} + O (e^{4}) .

(7)

Indeed the full unrenormalized scalar electrodynamics three-point vertex has ultraviolet pole

i e^{3} (3 - ξ) C_{ϵ} k^{μ}

, which is cancelled by

- i e δ_{1} k^{μ}

. Subtracting the supplied contribution gives the gauge-independent sum stated in part (ii), without assigning either unreadable panel an integrand.

v

Solution

The scalar QED Ward identity equates the vertex renormalization and wave-function renormalization constants. Consequently, using part (iv),

Z_{2} = Z_{1} = 1 + \frac{e ^{2} ( 3 - ξ )}{16 π ^{2} ϵ} + O (e^{4}) .

The renormalization constants describe gauge-dependent fields and off shell vertices, so their dependence on

ξ

is permitted. A physical scattering amplitude includes all Feynman diagrams, the counterterms, and the external-state factors from LSZ reduction. The Ward identities ensure that the dependence on

ξ

cancels for physical external states. Infrared-sensitive probabilities also require the usual inclusive treatment of unobserved soft photons. Thus no dependence on

ξ

is implied for consistently calculated physical observables.

s = (p_{1} + p_{2})^{2}

t = (p_{1} - p_{3})^{2}

and

u = (p_{1} - p_{4})^{2}

. The external on shell relations are

p_{1}^{2} = p_{2}^{2} = m_{e}^{2}

and

p_{3}^{2} = p_{4}^{2} = m_{t}^{2}

. Expanding the three Mandelstam variables and using four-momentum conservation,

p_{3} + p_{4} = p_{1} + p_{2}

, gives

s + t + u = 3 p_{1}^{2} + p_{2}^{2} + p_{3}^{2} + p_{4}^{2} + 2 p_{1} \cdot (p_{2} - p_{3} - p_{4}) = p_{1}^{2} + p_{2}^{2} + p_{3}^{2} + p_{4}^{2} = 2 (m_{e}^{2} + m_{t}^{2}) .

ii

Solution

Let

Σ = m_{e}^{2} + m_{t}^{2}

Q_{e} = - 1

Q_{t} = 2/3

and

N_{c} = 3

. The single tree-level Feynman diagram is annihilation through a virtual photon. Up to an irrelevant common phase, its scattering amplitude is

M_{γ} = \frac{e ^{2} Q _{e} Q _{t}}{s} [\overset{v}{ˉ} (p_{2}) γ^{μ} u (p_{1})] [\overset{u}{ˉ} (p_{3}) γ_{μ} v (p_{4})] δ_{c d} .

There are two initial spin angular momentum states for each particle, so the spin average is

1/4

. The final color charge sum gives

\sum_{c, d} ∣ δ_{c d} ∣^{2} = N_{c}

. There is no additional photon polarization average for a virtual photon and no separate polarization label beyond the Dirac spinor spin sums.

The spin sums and the gamma-matrix trace identities give

L^{μν} H_{μν} = tr [(\neq p_{2} - m_{e}) γ^{μ} (\neq p_{1} + m_{e}) γ^{ν}] = 4 [p_{2}^{μ} p_{1}^{ν} + p_{2}^{ν} p_{1}^{μ} - \frac{s}{2} g^{μν}], = tr [(\neq p_{3} + m_{t}) γ_{μ} (\neq p_{4} - m_{t}) γ_{ν}] = 4 [p_{3 μ} p_{4 ν} + p_{3 ν} p_{4 μ} - \frac{s}{2} g_{μν}] .

Here

p_{1} \cdot p_{2} + m_{e}^{2} = p_{3} \cdot p_{4} + m_{t}^{2} = s /2

. Contracting gives

\frac{1}{4} L^{μν} H_{μν} = 8 [(p_{1} \cdot p_{3}) (p_{2} \cdot p_{4}) + (p_{1} \cdot p_{4}) (p_{2} \cdot p_{3})] + 4 s Σ.

Since

2 p_{1} \cdot p_{3} = 2 p_{2} \cdot p_{4} = Σ - t

and

2 p_{1} \cdot p_{4} = 2 p_{2} \cdot p_{3} = Σ - u

, this is the massive vector-current annihilation kernel

2 [(Σ - t)^{2} + (Σ - u)^{2} + 2 s Σ] = 2 [t^{2} + u^{2} + 4 s Σ - 2 Σ^{2}] .

Thus the absolutely normalized result is

\overline{∣ M_{γ} ∣^{2}} = \frac{2 N _{c} e ^{4} Q _{e}^{2} Q _{t}^{2}}{s ^{2}} [t^{2} + u^{2} + 4 s Σ - 2 Σ^{2}] = \frac{8 e ^{4}}{3 s ^{2}} [t^{2} + u^{2} + 4 s Σ - 2 Σ^{2}] .

After taking out the displayed prefactor,

A = 1, B = 4, C = - 2

. Because the source uses a proportionality sign, only their ratio is fixed until that prefactor is specified. Equivalently, after taking out just

e^{4} / s^{2}

, their values are

8/3, 32/3, - 16/3

. No mass has been neglected. At

m_{e} = 0

and

s = 4 m_{t}^{2}

, the bracket yields the finite isotropic threshold value of the scattering amplitude; the vanishing threshold scattering cross-section comes from relativistic two-body phase space.

iii

Solution

The exchanged particle is now a gluon, and the incoming up quark mass replaces

m_{e}

. Define

Σ_{u} = m_{u}^{2} + m_{t}^{2}, F_{u} = t^{2} + u^{2} + 4 s Σ_{u} - 2 Σ_{u}^{2} .

The Dirac trace and spin average are unchanged. Normalize the fundamental gauge generators by

tr (T^{A} T^{B}) = δ^{A B} /2

. Averaging over the

N_{c}^{2}

initial color charge choices and summing the final color charge choices gives

C_{g} = \frac{1}{N _{c}^{2}} \sum_{A, B} tr (T^{A} T^{B}) tr (T^{A} T^{B}) = \frac{N _{c}^{2} - 1}{4 N _{c}^{2}} = \frac{2}{9} .

Replace the factor

N_{c} e^{4} Q_{e}^{2} Q_{t}^{2}

from part (ii) by

g_{s}^{4} C_{g}

. Therefore

\overline{∣ M_{g} ∣^{2}} = \frac{4 g _{s}^{4}}{9 s ^{2}} F_{u} .

Keeping

m_{u}

is allowed; if the usual negligible up quark mass approximation is wanted, simply set

m_{u} = 0

. The factor

1/ N_{c}^{2}

is essential: initial quark color charge states are averaged, whereas final ones are summed.

iv

Solution

There are two tree-level Feynman diagrams, with intermediate photon and gluon respectively. Their color charge tensors are

δ_{ji} δ_{k l}

and

\sum_{A} T_{ji}^{A} T_{k l}^{A}

. The pure photon factor is

C_{γ} = \frac{1}{N _{c}^{2}} \sum_{i, j, k, l} ∣ δ_{ji} δ_{k l} ∣^{2} = 1.

Their interference vanishes after the color charge sum because of photon-gluon colour orthogonality:

C_{γ g} = \frac{1}{N _{c}^{2}} \sum_{A} tr (T^{A}) tr (T^{A}) = 0.

Thus there is no term proportional to

e^{2} g_{s}^{2}

, and with the

F_{u}

of part (iii),

\overline{∣ M_{γ} + M_{g} ∣^{2}} = \frac{2 F _{u}}{s ^{2}} [e^{4} Q_{u}^{2} Q_{t}^{2} + \frac{2}{9} g_{s}^{4}] = \frac{F _{u}}{s ^{2}} [\frac{32}{81} e^{4} + \frac{4}{9} g_{s}^{4}] .