Keep the mostly-plus Minkowski metric and the Fourier transform

e^{i p \cdot x}

. Write

\neq p = γ^{a} p_{a}

, with

{γ^{a}, γ^{b}} = 2 η^{ab}

. To fix the otherwise unspecified phase of

γ^{5}

and the Dirac adjoint, take

γ^{a} = - i Γ^{a}, \overset{ˉ}{ψ} = - i ψ^{†} γ^{0}, γ^{5} = γ^{0} γ^{1} γ^{2} γ^{3},

(1)

where

Γ^{a}

are ordinary mostly-minus gamma matrices. Then

(γ^{5})^{2} = - 1

{γ^{5}, γ^{a}} = 0

, and the fermionic time-derivative term is

i ψ^{†} \partial_{t} ψ

. In these conventions the interaction with real

g

is Hermitian: in mostly-minus notation it is

i g \overset{ˉ}{ψ}_{std} Γ^{5} ψ ϕ

. If one instead calls

i γ^{5}

the square-one chirality matrix, its coefficient must be

- i g

to represent the same interaction. These phase choices leave physical relativistic scattering cross-sections unchanged.

Expanding

e^{i S_{int}}

yields the following Feynman rules with relativistically normalized external states:

A real scalar field line carrying four-momentum $r$ contributes $- i / (r^{2} + m^{2} - i 0)$ .
An oriented Dirac field line contributes the Dirac propagator
$S_{F} (r) = \frac{i ( M - i \neq r )}{r ^{2} + M ^{2} - i 0} .$
(2)
This follows from $(i \neq r + M) (M - i \neq r) = (r^{2} + M^{2}) I$ .
Each pseudoscalar Yukawa interaction vertex has one scalar leg, one incoming fermion arrow and one outgoing fermion arrow, and contributes $i g γ^{5}$ . It also contributes $(2 π)^{4} δ^{(4)} (\sum r)$ with all vertex momenta counted incoming.
External incoming particles contribute $u (p, s)$ and outgoing particles $\overset{u}{ˉ} (p, s)$ . External incoming antiparticles contribute $\overset{v}{ˉ} (p, s)$ and outgoing antiparticles $v (p, s)$ , with the corresponding fermion arrows. External scalar factors are one. Choose $u^{†} u = v^{†} v = 2 E$ , $(i \neq p + M) u = 0$ , and $(- i \neq p + M) v = 0$ ; all external momenta are on the appropriate mass shell.
Integrate each independent loop four-momentum with $d^{4} r / (2 π)^{4}$ . Keep matrix factors in their order along a fermion line, take a trace around a closed fermion loop, and include a minus sign for each closed fermion loop. Permuting external identical fermions contributes the corresponding fermionic sign; the Wick theorem determines the Feynman-diagram symmetry factors.

Strip the overall four-momentum conservation delta function when defining the scattering amplitude. There are no further bare interaction vertices, no gauge fixing and no Faddeev-Popov ghost fields in this theory. Renormalized higher-order calculations add the required counterterms; these are additional to the rules of the displayed classical Lagrangian density.

The two terms add with the same relative sign: neither diagram exchanges identical external fermions. Their interference must be retained when squaring the complete scattering amplitude. In the phase convention above,

γ^{5} (M - i \neq r) γ^{5} = - (M + i \neq r)

, which gives an equivalent simplified numerator. An overall phase depends on the S-matrix convention and does not change a relativistic scattering cross-section.

iii

 0  0

Solution

 0  0

For an unpolarized initial fermion, average over its two spin states and sum over the unobserved final spin:

\overline{∣ T ∣^{2}} = \frac{1}{2} \sum_{s, s^{'}} ∣ T_{s^{'} s} ∣^{2} .

(1)

The initial scalar has only one spin state. One can evaluate this sum directly from normalized Dirac spinors, or use consistent fermion spin sums to express it as a trace. It is the squared sum of both tree scattering amplitudes, not the sum of their separate squares.

The relativistic scattering cross-section is obtained by integrating the Lorentz-invariant phase-space measure and dividing by the invariant flux factor:

σ = \frac{1}{4 ( p \cdot k ) ^{2} - M ^{2} m ^{2}} \int \frac{d ^{3} q}{( 2 π ) ^{3} 2 E _{q}} \frac{d ^{3} l}{( 2 π ) ^{3} 2 E _{l}} (2 π)^{4} δ^{(4)} (p + k - q - l) \overline{∣ T ∣^{2}} .

(2)

There is no identical-final-particle factor, because the outgoing scalar and fermion are distinct. All energies are positive, with

E_{q}^{2} = ∣ q ∣^{2} + M^{2}

and

E_{l}^{2} = ∣ l ∣^{2} + m^{2}

Equivalently, in the centre-of-momentum frame set

s = - (p + k)^{2}

. Integrating the energy delta function in the relativistic two-body phase space gives

\frac{d σ}{d Ω} = \frac{1}{64 π ^{2} s} \frac{k _{f}}{k _{i}} \overline{∣ T ∣^{2}} .

(3)

For this elastic process

k_{f} = k_{i}

, so integrate

\overline{∣ T ∣^{2}} / (64 π^{2} s)

over the full solid angle. This supplies the requested prescription without evaluating the angular integral.

iv

 0  0

Solution

 0  0

The Dirac field has the global symmetry

ψ \mapsto e^{i α} ψ

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

, while the real scalar is unchanged. Both the free Dirac action and the pseudoscalar Yukawa interaction preserve this symmetry. Therefore Dirac fermion number conservation holds: its charge counts particles minus antiparticles.

The initial state has charge

+ 1

, whereas a final antiparticle and a neutral scalar have charge

- 1

. Since the S-matrix commutes with that charge,

(Q_{f} - Q_{i}) ⟨ f ∣ S ∣ i ⟩ = 0 ⟹ T (f ϕ \to \overset{ˉ}{f} ϕ) = 0, σ = 0 .

(1)

This holds at every order, not just tree level. In the Feynman rules, a continuous fermion arrow cannot connect these specified external states. Replacing only the outgoing

\overset{u}{ˉ}

by a

v

in the previous expression would not give a physical amplitude. Changing both external fermions to antiparticles would instead produce an allowed process with reversed fermion flow and the appropriate

v

spinors; moving a leg between initial and final states is a different operation governed by crossing symmetry.

Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 40 / 2

i

Solution

ii

Solution

iii

Solution

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Solution

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Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 40 / 2

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