Past exam of the quantum field theory course of the University of Oxford

Solution

Expanding

e^{i S_{int}}

gives the Feynman vertex

V_{ab} = - i g (γ^{5})_{ab} .

There is one scalar leg and a continuous oriented fermion line; no extra factor

i

is absorbed into

γ^{5}

. The Dirac propagator is

S_{F} (k) = i (\neq k + M) / (k^{2} - M^{2} + i 0)

, where

\neq k = γ^{μ} k_{μ}

. Each closed fermion loop supplies a minus sign.

Figure 1.
Pseudoscalar Yukawa vertex and the vanishing one-loop bosonic one-point and three-point diagrams
. **Details**.

The tadpole diagram is proportional to

- \int \frac{d ^{4} k}{( 2 π ) ^{4}} tr [(- i g γ^{5}) \frac{i ( \neq k + M )}{k ^{2} - M ^{2} + i 0}] = 0,

because both numerator traces vanish by a. This is zero before the loop momentum integration.

For either orientation of the triangle Feynman diagram, let

a, b, c

be its three internal four-momenta. Its numerator trace is

N = tr [γ^{5} (\neq a + M) γ^{5} (\neq b + M) γ^{5} (\neq c + M)] = tr [(\neq a - M) (\neq b + M) γ^{5} (\neq c + M)] = 0.

After using cyclic property of the trace, each expanded term is a trace of

γ^{5}

times at most three ordinary gamma matrices. Such traces vanish in four dimensions. For an odd number of ordinary gamma matrices, conjugation by

γ^{5}

changes the sign. For two ordinary gamma matrices, the contracted Clifford algebra identity

γ_{ρ} γ^{μ} γ^{ν} γ^{ρ} = 4 η^{μν} I

gives

4 tr (γ^{5} γ^{μ} γ^{ν}) = - 4 η^{μν} tr γ^{5} = 0.

The zero-matrix case is a. Hence both the one-point and three-point one-loop bosonic vertices vanish, independently of their external momenta. No cancellation between different loop orientations or delicate shift of a divergent integral is needed.

d

Solution

Writing

x_{P} = (t, - x)

, the standard parity transformation of a Dirac field is

ψ^{'} (x) = η_{P} γ^{0} ψ (x_{P}), \overset{ˉ}{ψ}^{'} (x) = η_{P}^{*} \overset{ˉ}{ψ} (x_{P}) γ^{0}, ∣ η_{P} ∣ = 1.

Since

γ^{0} γ^{5} γ^{0} = - γ^{5}

, the fermion bilinear

\overset{ˉ}{ψ} γ^{5} ψ

changes sign. Invariance of the pseudoscalar Yukawa interaction therefore requires

ϕ^{'} (x) = - ϕ (x_{P}) .

Thus

ϕ

is a pseudoscalar, despite having the usual scalar-form kinetic term and mass term. In this convention

(γ^{5})^{†} = - γ^{5}

, but

\overset{ˉ}{ψ} γ^{5} ψ

is Hermitian because of the additional

γ^{0}

in the Dirac adjoint. A real

g

is consequently consistent with a Hermitian Lagrangian density.

In perturbation theory about the parity-invariant vacuum state, renormalization can preserve this non-anomalous parity symmetry in quantum field theory. Every local counterterm must then be parity even, at every loop order. The allowed terms are precisely

L_{ct}^{(+)}

in b; all coefficients in

L_{ct}^{(-)}

are zero. In particular, no $ϕ$ or $ϕ^{3}$ counterterm is generated, whereas

ϕ^{4}

is allowed and required. The one-point and three-point pure-boson vertices also vanish to all orders: a parity-odd Lorentz scalar cannot be formed from their at most two independent external four-momenta. This last reasoning is specific to these low-point vertices; it does not assert that parity symmetry in quantum field theory is an independent

ϕ \mapsto - ϕ

internal symmetry with the fermions held fixed.

2

For the complex scalar fields, use the conventional particle names:

ϕ

annihilates a light

ϕ

particle and creates a light antiparticle

\overset{ˉ}{ϕ}

;

Φ

annihilates a heavy

Φ

particle and creates

\overset{ˉ}{Φ}

. These choices make the distinction between particle labels and field symbols explicit throughout ii and iii. The couplings

g, h

are real as required by the printed Hermitian Lagrangian density.

a

Solution

Independent global phase symmetries would rotate

ϕ \mapsto e^{- i α} ϕ

and

Φ \mapsto e^{- i β} Φ

. The cubic interaction vertex preserves only

2 α + β = 0

modulo

2 π

. The continuous internal symmetry of a classical field theory is therefore the circle group

U (1) : ϕ \mapsto e^{- i α} ϕ, Φ \mapsto e^{2 i α} Φ.

Normalize the Noether charge so that

q_{ϕ} = 1

. Then

q_{Φ} = - 2

q_{\overset{ˉ}{ϕ}} = - 1

and

q_{\overset{ˉ}{Φ}} = 2

. The Noether current is

j^{μ} = i ϕ^{†} \partial^{μ} ϕ - 2 i Φ^{†} \partial^{μ} Φ, \partial_{μ} j^{μ} = 0,

where

A \partial B = A \partial B - (\partial A) B

. For example, the Euler-Lagrange field equations give

\partial_{μ} (i ϕ^{†} \partial^{μ} ϕ) = ih (ϕ^{2} Φ - ϕ^{† 2} Φ^{†}), \partial_{μ} (i Φ^{†} \partial^{μ} Φ) = \frac{ih}{2} (ϕ^{2} Φ - ϕ^{† 2} Φ^{†}),

which cancel with the relative coefficient

- 2

. The quartic interaction vertex separately preserves both original phase rotations.

There is also a discrete charge conjugation symmetry

(ϕ, Φ) \mapsto (ϕ^{†}, Φ^{†})

for these real couplings. Including it, the internal symmetry is

U (1) ⋊ Z_{2} ≅ O (2)

, with charge conjugation reversing the Noether charge. The connected symmetry and its conserved current are the

U (1)

above.

b

Solution

The nonzero free scalar propagators join a field to its adjoint:

⟨ Tϕ ϕ^{†} ⟩ (p) = \frac{i}{p ^{2} - m ^{2} + i 0}, ⟨ T Φ Φ^{†} ⟩ (p) = \frac{i}{p ^{2} - M ^{2} + i 0} .

Contractions

⟨ Tϕϕ ⟩

⟨ T ΦΦ ⟩

and mixed-species contractions vanish. With all momenta incoming, the Feynman vertices are

V_{ϕ ϕ^{†} Φ Φ^{†}} = - i g, V_{ϕϕ Φ} = - ih, V_{ϕ^{†} ϕ^{†} Φ^{†}} = - ih .

The two identical light legs cancel the

1/2!

in each cubic interaction vertex; there is no remaining factor

1/2

in its Feynman rule. Multiply each Feynman vertex by

(2 π)^{4} δ^{(4)} (\sum p)

, and integrate an internal unconstrained four-momentum with

d^{4} p / (2 π)^{4}

Figure 1.
Charged scalar propagators and vertices, with arrows following positive conserved charge and heavy arrows carrying two units
. **Details**.

The arrows in the figure follow positive Noether charge: a light arrow carries one unit and a heavy arrow two. Consequently a heavy particle

Φ

travels opposite its charge arrow, whereas its antiparticle

\overset{ˉ}{Φ}

travels along it. At a

ϕ^{2} Φ

Feynman vertex, two unit-charge arrows enter and a two-unit arrow leaves; the conjugate Feynman vertex reverses every arrow. The quartic Feynman vertex has one incoming and one outgoing arrow of each weight. These rules prevent attaching a propagator between two unconjugated fields.

c

i

Solution

Let

p_{1}, p_{2}

be the incoming and

p_{3}, p_{4}

the outgoing four-momenta, and set

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

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

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

, the Mandelstam variables. The only connected tree-level Feynman diagram is the

s

-channel exchange of

\overset{ˉ}{Φ}

, with Noether charge

+ 2

. The two incoming light particles attach to

ϕ^{2} Φ

, and the two outgoing light particles to its conjugate. There is no four-light contact Feynman vertex, nor a charge-allowed

t

u

exchange with these external labels.

Figure 1.
Light-particle scattering through a heavy antiparticle in the s channel
. **Details**.

Define the invariant scattering amplitude by

⟨ f ∣ S - 1∣ i ⟩ = i (2 π)^{4} δ^{(4)} (P_{f} - P_{i}) M

. Then

i M = (- ih) \frac{i}{s - M ^{2} + i 0} (- ih), M (ϕϕ \to ϕϕ) = - \frac{h ^{2}}{s - M ^{2} + i 0} .

The Feynman vertices already contain the identical-leg combinatorics. No extra factor of two multiplies this invariant scattering amplitude; the distinct identical final-state symmetry factor enters a cross section integral.

ii

Solution

Keep the particle labels specified under 2. There are two connected tree-level Feynman diagrams: the quartic contact interaction vertex and an

s

-channel light antiparticle exchange. The initial Noether charge is

1 - 2 = - 1

, exactly the charge of

\overset{ˉ}{ϕ}

. At the first cubic Feynman vertex, the incoming

ϕ

and

Φ

leave the second

ϕ

leg internal; the conjugate Feynman vertex supplies both outgoing particles.

Figure 1.
Scattering of a light particle and a heavy particle, with contact and s-channel light-antiparticle exchange diagrams
. **Details**.

Thus

i M = - i g + (- ih) \frac{i}{s - m ^{2} + i 0} (- ih), M (ϕ Φ \to ϕ Φ) = - g - \frac{h ^{2}}{s - m ^{2} + i 0} .

Here

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

for incoming

ϕ (p_{1}), Φ (p_{2})

. Both terms have the same sign in the chosen S-matrix convention.

The distinction from antiparticles matters. If the paper's heavy “

Φ

particle” is instead intended to mean the charge-

+ 2

state

\overset{ˉ}{Φ}

, so that its printed decay into

ϕϕ

is allowed, the exchange diagram is crossed. For that alternative physical process,

M (ϕ \overset{ˉ}{Φ} \to ϕ \overset{ˉ}{Φ}) = - g - \frac{h ^{2}}{u - m ^{2} + i 0}, u = (p_{1} - p_{4})^{2} .

This is a consistent alternative naming convention only if it is also applied to the decay in iii; one cannot use the same heavy particle label for both charge assignments.

iii

Solution

With the conventional particle labels, the allowed cubic decay channels are

Φ \to \overset{ˉ}{ϕ} \overset{ˉ}{ϕ}, \overset{ˉ}{Φ} \to ϕϕ, M = - h for either channel.

Each has one tree-level Feynman diagram, consisting of one cubic Feynman vertex. The final identical-leg contractions supply

2!

and cancel the

1/2!

in the Lagrangian density.

Figure 1.
The two charge-conjugate heavy-scalar decays, with explicit antiparticle labels and conserved charge arrows
. **Details**.

Taken literally with these labels, the printed $Φ \to ϕϕ$ channel is forbidden: its initial Noether charge is

- 2

and its final Noether charge is

+ 2

. Its decay amplitude is zero, at every order preserving the global phase symmetry. If the source uses “

Φ

” to name the heavy antiparticle instead, its intended answer is the nonzero decay amplitude

- h

for

\overset{ˉ}{Φ} \to ϕϕ

. The following decay width calculation concerns either allowed channel, whose widths are equal by charge conjugation.

d

Solution

For an allowed decay channel in iii,

∣ M ∣^{2} = h^{2}

. Integrating the spatial Dirac delta distribution in the printed Lorentz-invariant phase-space measure sets

p_{2} = - p_{1}

. With

p = ∣ p_{1} ∣

and

E_{p} = p^{2} + m^{2}

, its remaining value is

I_{2} = \int d Φ_{2} = \frac{1}{4 ( 2 π ) ^{2}} \int \frac{d ^{3} p}{E _{p}^{2}} δ (2 E_{p} - M) = \frac{1}{4 π} \int_{0}^{\infty} \frac{p ^{2} d p}{E _{p}^{2}} δ (2 E_{p} - M) .

The positive root is

p_{*} = (M /2) β

, where

β = 1 - 4 m^{2} / M^{2}

, and the Dirac delta distribution contributes the Jacobian determinant

E_{*} / (2 p_{*})

. Hence

I_{2} = \frac{p _{*}}{8 π E _{*}} = \frac{β}{8 π}, E_{*} = \frac{M}{2} .

It follows that the literal printed integral evaluates to

Γ_{printed} = \frac{h ^{2}}{16 π M} 1 - \frac{4 m ^{2}}{M ^{2}} .

The printed integration ranges count each final configuration twice, once as

(p_{1}, p_{2})

and once in the opposite order. For two identical final bosons, the physical decay width therefore requires an additional identical final-state symmetry factor

1/2!

Γ_{phys} = \frac{h ^{2}}{32 π M} 1 - \frac{4 m ^{2}}{M ^{2}} ≃ \frac{h ^{2}}{32 π M} (M ≫ m) .

There is one allowed decay channel per heavy particle, not an extra factor of two from summing the two charge-conjugate heavy species. Both expressions vanish below the two-body decay threshold

M = 2 m

, and the limiting value at threshold is zero. For the literally forbidden labels

Φ \to ϕϕ

, the decay width is instead zero because its decay amplitude vanishes, as explained in iii.

e

Solution

For

g = 0

, the invariant scattering amplitude from the tree-level Feynman diagram in i has a pole at

s = M^{2}

. The intermediate heavy antiparticle can go on shell and live for a time of order its mean lifetime. Near this scattering resonance, inserting a single self-energy is no longer a uniformly small correction. Dyson resummation gives, to leading accuracy in a narrow decay width,

M (s) ≃ - \frac{h ^{2}}{s - M ^{2} + i M Γ _{phys}} .

Here

M

is understood as the renormalized resonance mass; the real self-energy shifts the original pole position. The imaginary term removes the unphysical divergence of the fixed-order invariant scattering amplitude.

For equal incoming and outgoing light masses, the two-body momentum ratio cancels in the cross section. The angle-independent invariant scattering amplitude, integrated over the full solid angle with the identical final-state symmetry factor

1/2!

, gives

σ = ∣ M ∣^{2} / (32 π s)

. Thus the Breit-Wigner resonance cross section is

σ (s) ≃ \frac{h ^{4}}{32 π s [ ( s - M ^{2} ) ^{2} + M ^{2} Γ _{phys}^{2} ]}, Γ_{phys} = \frac{h ^{2} β}{32 π M} .

s ≫ M^{2}

it reproduces

h^{4} / (32 π s^{3})

. At the resonance peak,

σ (M^{2}) ≃ \frac{32 π}{M ^{2} β ^{2}} ≃ \frac{32 π}{M ^{2}} (M ≫ m) .

This is also

8 π / k_{*}^{2}

for

k_{*} = Mβ /2

, the elastic identical-boson S wave partial-wave unitarity limit when this is the only open decay channel. The approximation assumes

Γ_{phys} ≪ M

and a resonance sufficiently far from threshold for the width to vary slowly across it.

If one mechanically uses the unsymmetrized

Γ_{printed} = 2 Γ_{phys}

from d, the same displayed Breit-Wigner resonance cross section would have that doubled width and predict a peak

8 π / (M^{2} β^{2})

. This is the literal-width substitution, rather than the consistent physical result: the given high-energy cross section already includes the identical-final-particle factor that the printed decay width integral omits.

3

a

Solution

For three odd fermionic operators, the identity

[A, BC] = {A, B} C - B {A, C}

converts the commutator with a quadratic Hamiltonian into the given canonical anticommutation relations. Writing a mode label as

i = (p, s)

, this yields

[a_{i}, a_{j}^{†} a_{j}] = δ_{ij} a_{j}, [b_{i}^{†}, b_{j}^{†} b_{j}] = - δ_{ij} b_{j}^{†} .

The opposite-species number terms commute with each fermionic creation operator or fermionic annihilation operator, because two anticommutations give a positive sign. Restoring the continuous normalization, the Dirac delta distribution in the mode brackets cancels

d^{3} q / (2 π)^{3}

. Consequently

[a_{p}^{s}, H] = E_{p} a_{p}^{s}, [b_{p}^{s †}, H] = - E_{p} b_{p}^{s †} .

Substitute these into the mode expansion of a Dirac field. The result is

[ψ (x), H] = \int \frac{d ^{3} p}{( 2 π ) ^{3} 2 E _{p}} \sum_{s} E_{p} [e^{- i p \cdot x} a_{p}^{s} u^{s} (p) - e^{i p \cdot x} b_{p}^{s †} v^{s} (p)] .

Since

i \partial_{t} e^{- i p \cdot x} = E_{p} e^{- i p \cdot x}

and

i \partial_{t} e^{i p \cdot x} = - E_{p} e^{i p \cdot x}

, this is exactly

i \partial_{t} ψ (t, x) = [ψ (t, x), H] .

The Heisenberg equation of motion uses an ordinary commutator with the even Hamiltonian, despite the odd fermionic operators obeying anticommutators among themselves.

b

Solution

The canonical anticommutation relations give

{a_{1}^{†}, a_{2}^{†}} = 0

, where each label includes both momentum and spin. Therefore

∣2, 1 ⟩ = a_{2}^{†} a_{1}^{†} ∣0 ⟩ = - a_{1}^{†} a_{2}^{†} ∣0 ⟩ = - ∣1, 2 ⟩ .

Thus exchanging the complete labels of the two identical fermions changes the sign of their state. If the labels coincide,

(a_{1}^{†})^{2} = 0

; the state vanishes, which is the Pauli exclusion principle. A negative state vector describes the same isolated physical ray, but its relative sign matters when exchange alternatives interfere.

c

Solution

For

i < j

, interchange the entries in slots

i, j

while preserving the order of all intermediate fermionic creation operators. Move

a_{j}^{†}

left through

j - i

factors, then move

a_{i}^{†}

right through the remaining

j - i - 1

intermediate factors. The canonical anticommutation relations give the overall fermionic sign

(- 1)^{(j - i) + (j - i - 1)} = (- 1)^{2 (j - i) - 1} = - 1.

Hence

∣1, \dots, j, \dots, i, \dots, n ⟩ = - ∣1, \dots, i, \dots, j, \dots, n ⟩ .

More generally any permutation

π

gives

∣ π (1), \dots, π (n)⟩ = sgn (π) ∣1, \dots, n ⟩ .

The sign of a permutation is independent of its decomposition into adjacent interchanges. Thus the

n

-particle sector is an exterior power of the one-particle state space, the defining antisymmetry of fermionic Fock space. Repeated complete mode labels make the state zero, so they satisfy the same identity trivially.

d

Solution

Put

z = x - y

and

d Π_{p} = d^{3} p / [(2 π)^{3} 2 E_{p}]

. The Dirac adjoint of the printed mode expansion of a Dirac field is

\overset{ˉ}{ψ}_{a} (x) = \int \frac{d ^{3} p}{( 2 π ) ^{3} 2 E _{p}} \sum_{s} [e^{i p \cdot x} a_{p}^{s †} \overset{u}{ˉ}_{a}^{s} (p) + e^{- i p \cdot x} b_{p}^{s} \overset{v}{ˉ}_{a}^{s} (p)] .

The Fock vacuum is annihilated by

a, b

. The canonical anticommutation relations therefore give the two non-time-ordered vacuum expectation values

⟨ 0∣ \overset{ˉ}{ψ}_{a} (x) ψ_{b} (y) ∣0 ⟩ ⟨ 0∣ ψ_{b} (y) \overset{ˉ}{ψ}_{a} (x) ∣0 ⟩ = \int d Π_{p} e^{- i p \cdot z} s \sum \overset{v}{ˉ}_{a}^{s} (p) v_{b}^{s} (p), = \int d Π_{p} e^{i p \cdot z} s \sum \overset{u}{ˉ}_{a}^{s} (p) u_{b}^{s} (p) .

In the first line, the only surviving contraction is

⟨ b b^{†} ⟩

; in the second it is

⟨ a a^{†} ⟩

. Both of these unordered contractions have positive coefficients.

However, fermionic time ordering includes a minus sign when it interchanges two odd fields:

T \overset{ˉ}{ψ}_{a} (x) ψ_{b} (y) = θ (z^{0}) \overset{ˉ}{ψ}_{a} (x) ψ_{b} (y) - θ (- z^{0}) ψ_{b} (y) \overset{ˉ}{ψ}_{a} (x) .

Thus the result following from the stated canonical anticommutation relations and standard time ordering is

C_{ab} (x, y) \equiv ⟨ 0∣ T \overset{ˉ}{ψ}_{a} (x) ψ_{b} (y) ∣0 ⟩ = \int d Π_{p} [θ (z^{0}) e^{- i p \cdot z} s \sum \overset{v}{ˉ}_{a}^{s} v_{b}^{s} - θ (- z^{0}) e^{i p \cdot z} s \sum \overset{u}{ˉ}_{a}^{s} u_{b}^{s}] .

The second term printed with a plus sign is not correct for standard fermionic time ordering. Its plus-plus expression would be an ungraded chronological product. The missing minus cannot be absorbed by a phase choice for

u

v

, since each spin sum contains a spinor and its adjoint.

For comparison with the conventional Dirac propagator, define

S_{F} (w)_{ba} = ⟨ 0∣ T ψ_{b} (w) \overset{ˉ}{ψ}_{a} (0) ∣0 ⟩

and let

μ

denote this free Dirac field's mass. Then

C_{ab} (x, y) = - S_{F} (y - x)_{ba} = - \int \frac{d ^{4} p}{( 2 π ) ^{4}} \frac{i ( \neq p + μ ) _{ba}}{p ^{2} - μ ^{2} + i 0} e^{- i p \cdot (y - x)} .

The exchange of the spinor indices is essential:

\overset{v}{ˉ}_{a} v_{b}

is the

(b, a)

component of

v \overset{v}{ˉ}

, not its

(a, b)

component.

e

Solution

The corrected expression in d is a vacuum two-point function, or Wick contraction, of the free Dirac field. It equals the negative spinor transpose of the usual Dirac propagator with reversed spacetime arguments. It supplies the internal fermion lines in Feynman diagrams, rather than a position-space detection probability.

For

x^{0} > y^{0}

ψ (y)

creates an antiparticle and

\overset{ˉ}{ψ} (x)

annihilates it. The

v \overset{v}{ˉ}

term describes positive-energy antiparticle propagation from

y

x

. For

y^{0} > x^{0}

\overset{ˉ}{ψ} (x)

creates a particle and

ψ (y)

annihilates it. The

u \overset{u}{ˉ}

term describes positive-energy particle propagation from

x

y

, with the minus sign required by exchanging the odd operators in fermionic time ordering. Neither term describes a physical negative-energy excitation.

The Heaviside step functions select these chronological alternatives. The factors

e^{\mp i p \cdot z}

are on-shell plane waves with

p^{0} = E_{p} > 0

, and

d Π_{p}

is the positive-energy Lorentz-invariant phase-space measure. The sum over

s

includes the two spin states. With standard relativistic spinor normalization, the fermion spin sums are

\sum_{s} u^{s} (p) \overset{u}{ˉ}^{s} (p) = \neq p + μ, \sum_{s} v^{s} (p) \overset{v}{ˉ}^{s} (p) = \neq p - μ .

They encode the spinor structure and solve the corresponding on-shell Dirac equations. In particular, the corrected answer can also be written

C_{ab} (x, y) = \int \frac{d ^{4} p}{( 2 π ) ^{4}} \frac{i ( \neq p - μ ) _{ba}}{p ^{2} - μ ^{2} + i 0} e^{- i p \cdot (x - y)} .

The Feynman i-epsilon prescription specifies the temporal boundary conditions of the Green function; its positive- and negative-frequency residues reproduce the two terms and their relative minus sign. With

S_{F}

defined in d,

(i \neq \partial - μ) S_{F} (w) = i δ^{(4)} (w) I .

The contact term is fixed by the equal-time canonical anticommutation relations. Choosing a value for

θ (0)

does not change the unequal-time sign demonstrated above.

Finally, microcausality requires field anticommutators to vanish at spacelike separation; the time-ordered vacuum two-point function itself need not vanish there. These facts are compatible with relativistic causality. The plus-plus expression printed in the paper lacks the fermionic sign needed for this standard Dirac propagator interpretation.

2024 (A15088W1)

Source PDF: mmathphys.physics.ox.ac.uk/sites/g/files/vzsiac2991/files/mmathphys/documents/media/a15088w1.pdf

Use natural units

ℏ = c = 1

and the mostly-minus Minkowski metric. Question 3 has two spatial dimensions: its gamma matrices are the specified two-by-two matrices. The printed

σ^{μ}

in its kinetic term is interpreted as those

γ^{μ}

, rather than as three unmodified Pauli matrices. All spin sums in question 2 instead use four-component Dirac spinors in four spacetime dimensions.

1

a

Solution

Write

\int_{p} = \int d^{3} p / (2 π)^{3}

. The canonical commutation relations and the product rule for a commutator give

[a_{p}, a_{q}^{†} a_{q}] = (2 π)^{3} δ^{3} (p - q) a_{q}, [b_{p}^{†}, b_{q}^{†} b_{q}] = - (2 π)^{3} δ^{3} (p - q) b_{q}^{†} .

The other-species terms vanish. Integrating the Dirac delta distributions in the Hamiltonian therefore gives

[a_{p}, H] = E_{p} a_{p}, [b_{- p}^{†}, H] = - E_{p} b_{- p}^{†},

where

E_{- p} = E_{p}

. Consequently the complex scalar field satisfies

[ϕ (t, x), H] = \int_{p} \frac{E _{p}}{2 E _{p}} (- b_{- p}^{†} e^{i E_{p} t} + a_{p} e^{- i E_{p} t}) e^{i p \cdot x} .

Direct differentiation of the time factors gives the same expression, proving the Heisenberg equation of motion

i \partial_{t} ϕ (t, x) = [ϕ (t, x), H] .

b

Solution

Both a particle and its antiparticle carry the momentum of their mode label. Thus the normal-ordered total momentum operator is

P = \int_{p} p (a_{p}^{†} a_{p} + b_{p}^{†} b_{p}) .

The canonical commutation relations give

[a_{p}, P] = p a_{p}, [b_{- p}^{†}, P] = - (- p) b_{- p}^{†} = p b_{- p}^{†} .

The minus sign in the antiparticle mode label is essential: both terms in the printed mode expansion of a free field contain

e^{i p \cdot x}

. Hence

[ϕ, P] = \int_{p} \frac{p}{2 E _{p}} (b_{- p}^{†} e^{i E_{p} t} + a_{p} e^{- i E_{p} t}) e^{i p \cdot x} = - i \nabla ϕ .

c

Solution

For a small real time displacement

ϵ

, the unitary time evolution operator is

T_{t} (ϵ) = 1 - i ϵH + O (ϵ^{2})

. The corresponding Heisenberg picture field transformation is conjugation by its inverse:

T_{t} (ϵ)^{†} ϕ (t, x) T_{t} (ϵ) = ϕ + i ϵ [H, ϕ] + O (ϵ^{2}) = ϕ + ϵ \partial_{t} ϕ + O (ϵ^{2}) = ϕ (t + ϵ, x) + O (ϵ^{2}) .

Here a gives

[H, ϕ] = - i \partial_{t} ϕ

Likewise let

T_{s} (a) = 1 + i a \cdot P + O (∣ a ∣^{2})

, with the paper's sign for its spatial translation operator. From b,

T_{s} (a)^{†} ϕ (t, x) T_{s} (a) = ϕ + i a \cdot [ϕ, P] + O (∣ a ∣^{2}) = ϕ + a \cdot \nabla ϕ + O (∣ a ∣^{2}) = ϕ (t, x + a) + O (∣ a ∣^{2}) .

These are precisely the first-order Taylor expansions. Since

H

and

P

commute in this translation-invariant theory, the finite result is

ϕ (x) = e^{i P \cdot x} ϕ (0) e^{- i P \cdot x}, P \cdot x = H t - P \cdot x .

d

Solution

Choose the zero of energy so that the translation-invariant vacuum state has

P^{μ} ∣Ω ⟩ = 0

. A one-particle state of rest mass

m_{ψ}

obeys the relativistic energy-momentum relation

p^{μ} = (E_{p}, p), E_{p} = m_{ψ}^{2} + ∣ p ∣^{2}, p^{2} = m_{ψ}^{2} .

The translation covariance of a quantum field in c remains valid for the full interacting four-momentum operator. Acting its exponential functions on the two momentum eigenstates gives

⟨ Ω∣ ϕ (x) ∣ ψ, p ⟩ = ⟨ Ω∣ e^{i P \cdot x} ϕ (0) e^{- i P \cdot x} ∣ ψ, p ⟩ = e^{- i p \cdot x} ⟨ Ω∣ ϕ (0) ∣ ψ, p ⟩ .

Here

p \cdot x = E_{p} t - p \cdot x

. A different additive vacuum energy must be subtracted consistently from both energy eigenvalues before identifying the physical

p^{0}

e

Solution

Assume

m_{ψ} > 0

, as required to have a rest frame. Choose the Lorentz boost

Λ

for which

Λ p = (m_{ψ}, 0)

, and write its unitary operator as

U

. The stipulated one-particle state normalization gives

U ∣ ψ, p ⟩ = ∣ ψ, 0 ⟩

without a momentum-dependent prefactor. A Lorentz scalar at the origin satisfies

U ϕ (0) U^{†} = ϕ (0)

, and the vacuum state satisfies

U ∣Ω ⟩ = ∣Ω ⟩

. Inserting

U^{†} U

therefore proves

⟨ Ω∣ ϕ (0) ∣ ψ, p ⟩ = ⟨ Ω∣ ϕ (0) ∣ ψ, 0 ⟩ .

With a noncovariant normalization of momentum eigenstates, the assumed boost law would instead contain a square-root energy factor, so this equality is specifically tied to the paper's assumption.

Let

q^{μ} = (E_{q}^{χ}, q)

denote the four-momentum of the other energy eigenstate. Translation covariance first gives

⟨ χ, q ∣ ϕ (x) ∣ ψ, p ⟩ = e^{i (q - p) \cdot x} ⟨ χ, q ∣ ϕ (0) ∣ ψ, p ⟩ = e^{i (q - p) \cdot x} ⟨ U χ, q ∣ ϕ (0) ∣ ψ, 0 ⟩ .

Thus a suitable ket, including the phase whose complex conjugate appears in its bra, is

∣Ψ ⟩ = e^{- i (q - p) \cdot x} U ∣ χ, q ⟩ .

Here

⟨ U χ, q ∣

means

(U ∣ χ, q ⟩)^{†}

. The overall phase depends on the fixed insertion point

x

; it does not change the energy eigenvalue. It cannot simply be dropped if the equality of complex matrix elements is to hold exactly.

The transformed state has four-momentum

q^{'} = Λ q

. Lorentz invariance of the Minkowski inner product gives

p \cdot q = (Λ p) \cdot (Λ q) = m_{ψ} q^{'0}, E^{Ψ} = \frac{p \cdot q}{m _{ψ}} = \frac{E _{p} E _{q}^{χ} - p \cdot q}{m _{ψ}} .

There is no boost to a rest frame for a nonzero massless four-momentum, so the rest-state statements require the positive-mass assumption.

2

a

Solution

Using Feynman slash notation, the symmetric tensor

p_{μ} p_{ν}

kills the antisymmetric commutator of the gamma matrices. The Clifford algebra gives

\neq p \neq p = \frac{1}{2} p_{μ} p_{ν} {γ^{μ}, γ^{ν}} = p^{2} I .

Consequently

(\neq p)^{2 r} = (p^{2})^{r} I

and

(\neq p)^{2 r + 1} = (p^{2})^{r} \neq p

. For a scalar-coefficient polynomial

F (z) = \sum_{n} c_{n} z^{n}

, split its even and odd powers:

F (\neq p) = A (p^{2}) \neq p + B (p^{2}) I, A (z) = r \sum c_{2 r + 1} z^{r}, B (z) = r \sum c_{2 r} z^{r} .

This reduction assumes scalar coefficients; arbitrary noncommuting matrix coefficients would not give the stated scalar functions

A, B

b

Solution

Every positive-energy Dirac spinor obeys

(\neq p - m) u^{s} (p) = 0

. Therefore the fermion spin sum satisfies

(\neq p - m) P = \sum_{s} [(\neq p - m) u^{s}] \overset{u}{ˉ}^{s} = 0.

Since

\overset{u}{ˉ}^{s} = u^{s †} γ^{0}

and

(γ^{0})^{2} = I

, the trace of each outer product is its contracted scalar. The prescribed normalization and the two spin states yield

tr (P γ^{0}) = s \sum u^{s †} u^{s} = 2 (2 E_{p}) = 4 E_{p} .

Now make exactly the stated polynomial assumption and apply polynomial reduction of a slashed vector. By a, on the mass shell

p^{2} = m^{2}

write

P = A \neq p + B I

. Then

0 = (\neq p - m) P = (A m^{2} - m B) I + (B - m A) \neq p .

The matrices

I

and

\neq p

are linearly independent for a physical nonzero

p

, so

B = m A

. The gamma matrix trace identities,

tr γ^{0} = 0

and

tr (γ^{μ} γ^{0}) = 4 η^{μ 0}

, give

4 E_{p} = tr [A (\neq p + m) γ^{0}] = 4 A E_{p} .

Thus

A = 1

and

P (p) = \neq p + m .

No explicit representation of the Dirac spinors has been used. The same reasoning covers

m = 0

provided

p^{0} > 0

c

Solution

Interpret the repeated species label in the Yukawa interaction as summed over

i = 1, 2

: each Feynman vertex is diagonal in species and equals

- i g

. Define the invariant scattering amplitude by

S - 1 = i (2 π)^{4} δ^{(4)} (P_{f} - P_{i}) M

. There is one tree-level Feynman diagram, an

s

-channel scalar exchange. In the left panel,

j = 2

for this part; the right panel only occurs in d. Arrows show fermion-number flow, so an antifermion arrow opposes its physical momentum.

Figure 1.
Labelled scalar-exchange annihilation and elastic diagrams, with arrows following fermion-number flow
. **Details**.

Writing

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

and

s = k^{2}

, the momentum-space Feynman rules give

i M = [\overset{v}{ˉ}^{s_{2}} (p_{2}) (- i g) u^{s_{1}} (p_{1})] \frac{i}{s - M ^{2} + i 0} [\overset{u}{ˉ}^{s_{3}} (p_{3}) (- i g) v^{s_{4}} (p_{4})], M = - \frac{g ^{2} [ v ˉ _{2} u _{1} ] [ u ˉ _{3} v _{4} ]}{s - M ^{2} + i 0} .

The scalar carries no spin label. Average over the

2 \times 2

incoming spin states and sum the four outgoing combinations. For the real scalar Yukawa interaction, the conjugate of

\overset{v}{ˉ}_{2} u_{1}

\overset{u}{ˉ}_{1} v_{2}

. The fermion spin sums then yield

\overline{∣ M ∣^{2}} = \frac{g ^{4}}{4 ( s - M ^{2} ) ^{2}} tr [(\neq p_{2} - m) (\neq p_{1} + m)] tr [(\neq p_{3} + m) (\neq p_{4} - m)] .

The bar denotes the initial spin average, not a Dirac adjoint. The gamma matrix trace identities give

tr [(\neq p_{2} - m) (\neq p_{1} + m)] = 4 (p_{1} \cdot p_{2} - m^{2}) = 2 (s - 4 m^{2}),

and the second trace has the same value because

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

and all external masses are

m

. Therefore

\overline{∣ M ∣^{2}} = g^{4} \frac{( s - 4 m ^{2} ) ^{2}}{( s - M ^{2} ) ^{2}} = g^{4} (\frac{4 m ^{2} - s}{M ^{2} - s})^{2} .

This is the fixed-order expression away from

s = M^{2}

. If that pole is accessible, a physical scattering resonance requires a scalar self-energy and decay width; the tree formula alone is not finite there.

d

Solution

For a single species both panels of the figure in c contribute, now with

j = 1

. They are the

s

-channel annihilation and

t

-channel scalar exchange, where the Mandelstam variables are

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

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

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

and

s + t + u = 4 m^{2}

. Fix the external-state order consistently with c. The exchange of identical external fermionic operators gives a relative minus sign:

M = - g^{2} (\frac{( v ˉ _{2} u _{1} ) ( u ˉ _{3} v _{4} )}{s - M ^{2} + i 0} - \frac{( u ˉ _{3} u _{1} ) ( v ˉ _{2} v _{4} )}{t - M ^{2} + i 0}) .

Changing the convention for ordering all external states can change the overall sign, but not this relative fermionic sign.

The exchange

p_{1} \leftrightarrow - p_{4}

must include crossing symmetry the associated external spinors, as well as continuing the four-momenta. It interchanges

s

and

t

, leaves

u

fixed, and interchanges the two spinor pairings. Hence the analytically continued invariant scattering amplitude is antisymmetric:

M ⟼ - M, \overline{∣ M ∣^{2}} (s, t) = \overline{∣ M ∣^{2}} (t, s) .

The first relation is understood with consistently chosen spinor crossing phases. It is not a substitution of negative-energy arguments into the original positive-energy

u

spinors without changing external states.

Without repeating the spin sum calculation, c fixes the square of the first diagram. Crossing symmetry fixes the second by

s \leftrightarrow t

. Consequently the strongest immediate inference is

\overline{∣ M ∣^{2}} = g^{4} [\frac{( s - 4 m ^{2} ) ^{2}}{( s - M ^{2} ) ^{2}} + \frac{( t - 4 m ^{2} ) ^{2}}{( t - M ^{2} ) ^{2}} + \frac{C ( s , t )}{( s - M ^{2} ) ( t - M ^{2} )}], C (s, t) = C (t, s) .

The interference function is not determined by the separate diagram squares. In particular adding the two squares and dropping interference is unjustified. The Cauchy-Schwarz inequality applied to the arrays of spin amplitudes also gives

∣ C (s, t) ∣ \leq 2∣ (s - 4 m^{2}) (t - 4 m^{2}) ∣

on the physical scattering region.

For completeness, its value can be checked with one contracted trace, without another diagram calculation. The cross product is

K = \frac{1}{4} tr [(\neq p_{2} - m) (\neq p_{1} + m) (\neq p_{3} + m) (\neq p_{4} - m)] = - \frac{1}{2} s t - 2 m^{2} (s + t) + 8 m^{4} .

The relative minus in the invariant scattering amplitude gives

C = - 2 K

. Thus the fully evaluated result, beyond the symmetry inference alone, is

\overline{∣ M ∣^{2}} = g^{4} [\frac{( s - 4 m ^{2} ) ^{2}}{( s - M ^{2} ) ^{2}} + \frac{( t - 4 m ^{2} ) ^{2}}{( t - M ^{2} ) ^{2}} + \frac{s t + 4 m ^{2} ( s + t ) - 16 m ^{4}}{( s - M ^{2} ) ( t - M ^{2} )}] .

Every term respects the inferred

s, t

symmetry. In the massless limit the interference numerator reduces to

s t

, which is nonzero at a generic scattering angle.

3

a

Solution

2 + 1

spacetime dimensions, the dimensionless action gives

[L] = 3

and

[\partial_{μ}] = 1

. The scalar kinetic term requires

2 [ϕ] + 2 = 3

, whereas the fermion kinetic term requires

2 [ψ] + 1 = 3

. Thus the engineering dimensions of relativistic fields are

[ϕ] = \frac{1}{2}, [ψ] = [\overset{ˉ}{ψ}] = 1.

Matching each interaction to mass dimension three gives

[g_{1}] = 3 - (2 [ϕ] + 2 [ψ]) = 0, [g_{2}] = 3 - 4 [ϕ] = 1, [g_{3}] = 3 - 6 [ϕ] = 0.

Here the brackets mean exponents of mass in natural units. The coefficients

g_{1}, g_{3}

are marginal couplings by classical power counting, and

g_{2}

is a relevant coupling.

b

Solution

Let

I_{b}, I_{f}

count internal scalar and fermion lines and

L

count independent loops in a connected amputated Feynman diagram. At large loop momentum, the scalar propagator falls as

k^{- 2}

and the Dirac propagator falls as

k^{- 1}

, including its numerator. Each loop integration is

d^{3} k

. The nonderivative interaction vertices add no powers. Therefore

ω = 3 L - 2 I_{b} - I_{f} .

Count endpoints of each kind of line and use the connected-graph loop order identity:

2 I_{b} + b = 2 n_{1} + 4 n_{2} + 6 n_{3}, 2 I_{f} + f = 2 n_{1}, L = I_{b} + I_{f} - (n_{1} + n_{2} + n_{3}) + 1.

Eliminating

L, I_{b}, I_{f}

gives

ω = I_{b} + 2 I_{f} - 3 (n_{1} + n_{2} + n_{3}) + 3 = (n_{1} + 2 n_{2} + 3 n_{3} - \frac{b}{2}) + 2 (n_{1} - \frac{f}{2}) - 3 (n_{1} + n_{2} + n_{3}) + 3 = 3 - \frac{b}{2} - f - n_{2} .

The cancellation of

n_{1}, n_{3}

reflects their dimensionless couplings; every insertion of the positive-dimensional

g_{2}

improves the superficial degree of divergence by one.

The printed exact homogeneity is shorthand for ultraviolet power counting, not an identity for a massive integrated correlation function. For example, even the free amputated scalar two-point vertex

Γ^{(2)} (p) = p^{2} - M^{2}

obeys

Γ^{(2)} (λ p) = λ^{2} p^{2} - M^{2}

, which differs from

λ^{2} Γ^{(2)} (p)

when

M \neq = 0

. Logarithmic divergences, actual numerator cancellations and ultraviolet subdivergences also require separate treatment.

c

Solution

Use one-particle-irreducible correlation functions to identify independent local counterterms. The exact internal symmetry

ϕ \mapsto - ϕ

excludes odd

b

, and Dirac fermion number conservation requires equal numbers of

ψ

and

\overset{ˉ}{ψ}

, so

f

is even. From b, the complete list of potentially overall divergent vertices with external legs is:

$Γ_{ϕϕ}$ : $(b, f) = (2, 0)$ , $ω = 2 - n_{2}$ , potentially divergent for $n_{2} \leq 2$ .
$Γ_{ϕ^{4}}$ : $(b, f) = (4, 0)$ , $ω = 1 - n_{2}$ , potentially divergent for $n_{2} \leq 1$ .
$Γ_{ϕ^{6}}$ : $(b, f) = (6, 0)$ , $ω = - n_{2}$ , potentially divergent only for $n_{2} = 0$ .
$Γ_{\overset{ˉ}{ψ} ψ}$ : $(b, f) = (0, 2)$ , $ω = 1 - n_{2}$ , potentially divergent for $n_{2} \leq 1$ .
$Γ_{ϕ^{2} \overset{ˉ}{ψ} ψ}$ : $(b, f) = (2, 2)$ , $ω = - n_{2}$ , potentially divergent only for $n_{2} = 0$ .

The zero-leg vacuum diagrams, with

ω = 3 - n_{2}

, may also require a vacuum energy density subtraction. Four-fermion vertices have

ω \leq - 1

and do not require a new four-fermion counterterm. These are superficial bounds: a particular allowed diagram can vanish or be convergent.

Locality and Lorentz invariance restrict the divergent parts to a finite basis. The scalar two-point vertex can contain a constant and

p^{2}

; there is no Lorentz scalar linear in a single momentum. The fermion two-point vertex can contain

I

and

\neq p

. The remaining listed vertices have no allowed derivative divergence. A complete counterterm Lagrangian, up to total derivatives, is therefore

L_{ct} = - δ Λ + \frac{δ Z _{ϕ}}{2} (\partial_{μ} ϕ) (\partial^{μ} ϕ) - \frac{δ M ^{2}}{2} ϕ^{2} + δ Z_{ψ} \overset{ˉ}{ψ} i γ^{μ} \partial_{μ} ψ - δ m \overset{ˉ}{ψ} ψ - \frac{δ g _{1}}{2} ϕ^{2} \overset{ˉ}{ψ} ψ - \frac{δ g _{2}}{4 !} ϕ^{4} - \frac{δ g _{3}}{6 !} ϕ^{6} .

The coefficients supply wavefunction renormalization, mass counterterms and renormalizations of the three interactions already present. Recursive subtraction of ultraviolet subdivergences then leaves only these local overall divergences. This finite counterterm closure is why the complete theory is expected to have perturbative renormalizability. It does not mean every listed coefficient first becomes nonzero at one loop, or that setting an allowed coupling to zero is stable under renormalization.

d

Solution

Set the renormalized

g_{3}

to zero. For a six-scalar vertex, b gives

ω = - n_{2}

, so a new overall logarithmic divergence can only occur in graphs with

n_{2} = 0

. With

b = 6, f = 0

, the line counts become

I_{f} = n_{1}

and

I_{b} = n_{1} - 3

, hence

L = n_{1} - 2

At one loop this forces

n_{1} = 3

and

I_{b} = 0

: the only candidate is a closed triangle Feynman diagram with two external scalar legs at each vertex. Its numerator at large loop momentum is proportional to

tr [(\neq k)^{3}] = k^{2} tr (\neq k) = 0.

This vanishes for the specified two-component gamma matrices, each of which is traceless. Replacing one leading

\neq k

by the fermion mass or an external momentum lowers the superficial degree of divergence from zero to at most

- 1

. Thus the apparent logarithm cancels. Graphs with a

g_{2}

vertex already have negative overall degree; the pure-scalar one-loop triangle, for instance, has

ω = - 3

. There is no one-loop ultraviolet subdivergence inside these proper one-loop vertices. Therefore

δ g_{3}^{one loop} = 0.

The six-point correlation function need not vanish: finite terms are allowed. Nor is it correct to say that all odd gamma-matrix traces vanish in

2 + 1

dimensions. In this representation,

γ^{0} γ^{1} γ^{2} = - i I_{2}, tr (γ^{0} γ^{1} γ^{2}) = - 2 i \neq = 0.

The cancellation above follows from repeating the same leading loop momentum, not from a false four-dimensional trace rule.

At two loops,

n_{1} = 4

I_{f} = 4

and

I_{b} = 1

are possible. A closed fermion box with a scalar line joining two of its vertices leaves six external scalar legs. It has

L = 2

ω = 0

, and four leading gamma matrices; the one-loop odd-numerator argument no longer applies.

A background-field check demonstrates that its logarithm is actually nonzero. Put

ϕ = v + η

with constant

v

. Define the background fermion mass

μ = m + g_{1} v^{2} /2

, the background scalar squared mass

B^{2} = M^{2} + g_{2} v^{2} /2

, and the induced three-leg Yukawa coupling

y = g_{1} v

. The two-loop quantum effective potential contains the sunset diagram with two fermion lines and one scalar line. After a Wick rotation, its relevant integral, apart from a nonzero overall combinatorial factor, is

y^{2} \int \frac{d ^{3} k d ^{3} l}{( 2 π ) ^{6}} \frac{2 ( - k \cdot l + μ ^{2} )}{( k ^{2} + μ ^{2} ) ( l ^{2} + μ ^{2} ) (( k - l ) ^{2} + B ^{2} )} .

Here the dots and gamma-matrix traces are Euclidean. With

A = k^{2} + μ^{2}

B = l^{2} + μ^{2}

C = (k - l)^{2} + B^{2}

, its numerator is exactly

2 (- k \cdot l + μ^{2}) = - A - B + C + 4 μ^{2} - B^{2} .

The first three terms reduce to products of one-loop tadpole integrals, which have no logarithmic pole near three dimensions in dimensional regularization. The remaining term is

y^{2} (4 μ^{2} - B^{2}) H, H = \int \frac{d ^{3} k d ^{3} l}{( 2 π ) ^{6}} \frac{1}{A BC} .

(6)

For positive regulator masses, the large-momentum region of

H

has a positive, nonzero logarithmic divergence. One can see this directly by restricting the relative angles and the ratio

∣ k ∣/∣ l ∣

to compact intervals away from zero: the remaining common radial integral is proportional to

\int^{Λ} d r / r

. Its proper one-loop subgraphs are ultraviolet convergent.

Finally, the coefficient multiplying

H

contains

y^{2} (4 μ^{2} - B^{2}) = g_{1}^{2} v^{2} [4 m^{2} - M^{2} + (4 m g_{1} - \frac{g _{2}}{2}) v^{2} + g_{1}^{2} v^{4}] .

(7)

The two-loop sextic counterterm in quartic Yukawa theory follows because its

v^{6}

coefficient is

g_{1}^{4} \neq = 0

for a nonzero coupling. Other two-loop vacuum topologies reduce to products of finite-in-dimensional-regularization tadpoles and do not cancel this logarithmic pole. Thus

δ g_{3}^{two loops} is generically nonzero and proportional to g_{1}^{4} .

(8)

A sextic counterterm is required even if the renormalized sextic coupling is set to zero at a chosen renormalization scale. A theory constrained to

g_{3} = 0

at every scale lacks counterterm closure when

g_{1} \neq = 0

. The decoupled exception

g_{1} = 0

is the three-dimensional scalar quartic interaction, which needs no sextic counterterm by power counting.

2023 (A15088W1)

Source PDF: mmathphys.web.ox.ac.uk/sitefiles/quantum-field-theory.pdf

1

a

Solution

Use

ℏ = c = 1

, the Minkowski metric

η = diag (1, - 1)

, and

p^{2} = (p^{0})^{2} - (p^{1})^{2}

. The momentum-space Feynman rules for the real scalar field are

internal line: \frac{i}{p ^{2} - m ^{2} + i ϵ}, four-leg vertex: - i g_{4} .

At each interaction vertex impose two-dimensional momentum conservation; before removing the overall conservation factor this is

(2 π)^{2} δ^{(2)} (\sum_{j} p_{j})

. Integrate each independent loop momentum with

d^{2} k / (2 π)^{2}

, and divide a Feynman diagram by its symmetry factor. External lines carry the same scalar propagator in an unamputated two-point correlation function; they are removed in an amputated scattering amplitude.

The quadratic action is inverted with the Feynman i-epsilon prescription to obtain the scalar propagator. Expanding the Dyson series gives

- i g_{4} /4!

per insertion; the

4!

attachments in the Wick theorem cancel the factorial. Fourier integration over the insertion point supplies the momentum conservation factor. Automorphisms that leave the labelled external attachments fixed account for the remaining symmetry factor. In two spacetime dimensions the mass dimensions are

[ϕ] = 0

and

[g_{4}] = 2

b

Solution

The one-loop one-particle irreducible correction is a tadpole diagram: two external lines meet one quartic interaction and the other two legs are joined. Interchanging the two ends of its loop gives the symmetry factor

2

Figure 1.
Original scalar diagrams: the two-dimensional tadpole, charge-conserving light-scalar exchange, and heavy-scalar box amplitudes. Dashed lines carry charge two; solid lines carry charge one.
**Details**.

Define the self-energy by an insertion

- i Σ (p^{2})

and a full scalar propagator

i / [p^{2} - m^{2} - Σ (p^{2}) + i ϵ]

before counterterms. The supplied cutoff regularization integral gives

T_{Λ} := \int_{∣ k_{E} ∣ < Λ} \frac{d ^{2} k _{E}}{( 2 π ) ^{2}} \frac{1}{k _{E}^{2} + m ^{2}} = \frac{1}{4 π} lo g (1 + \frac{Λ ^{2}}{m ^{2}}), - i Σ_{1} = \frac{- i g _{4}}{2} \int \frac{d ^{2} k}{( 2 π ) ^{2}} \frac{i}{k ^{2} - m ^{2} + i ϵ} .

Thus

Σ_{1} (p^{2}) = \frac{g _{4}}{2} T_{Λ} = \frac{g _{4}}{8 π} lo g (1 + \frac{Λ ^{2}}{m ^{2}}) .

It has no external momentum dependence. The correction to the unamputated two-point correlation function in momentum space is

Δ G (p) = \frac{i}{p ^{2} - m ^{2} + i ϵ} [- i Σ_{1}] \frac{i}{p ^{2} - m ^{2} + i ϵ} = \frac{i Σ _{1}}{( p ^{2} - m ^{2} + i ϵ ) ^{2}} .

The positive sign of

Σ_{1}

raises the unrenormalized squared pole mass for

g_{4} > 0

c

Solution

Write

ϕ_{B} = Z^{1/2} ϕ

Z = 1 + δ_{Z}

, and define the mass counterterm

δ_{m}

as a squared-mass coefficient:

L_{ct} = \frac{1}{2} δ_{Z} (\partial ϕ)^{2} - \frac{1}{2} δ_{m} ϕ^{2}, δ_{m} = Z m_{B}^{2} - m^{2} .

With the self-energy convention in b, the inverse scalar propagator is

p^{2} - m^{2} + δ_{Z} p^{2} - δ_{m} - Σ (p^{2})

. Requiring the pole mass to equal

m

therefore gives the on-shell renormalization scheme condition

δ_{m} = m^{2} δ_{Z} - Σ (m^{2}) .

At one loop order,

Σ_{1}^{'} = 0

; hence a unit pole residue also gives

δ_{Z}^{(1)} = 0

, and

δ_{m}^{(1)} = - \frac{g _{4}}{8 π} lo g (1 + \frac{Λ ^{2}}{m ^{2}}), m_{B}^{2} = m^{2} + δ_{m}^{(1)} + O (g_{4}^{2}) .

The mass counterterm Feynman rule is

- i δ_{m}

, cancelling the tadpole diagram. A convention defining a positive mass subtraction by

m_{B}^{2} = m^{2} - δ_{m}^{sub}

instead has

δ_{m}^{sub} = - δ_{m}

; specifying this convention avoids a spurious sign disagreement.

There is no one-loop wave-function renormalization because the tadpole diagram is independent of momentum. Each one-loop four-point bubble diagram has two scalar propagators, so its large Euclidean loop momentum integral behaves as

\int d^{2} k_{E} / k_{E}^{4}

and is ultraviolet finite. Thus no divergent quartic interaction counterterm is required. Finite coupling redefinitions remain possible. At higher loop orders, finite wave-function renormalization may be needed if unit pole residue is imposed; power counting only rules out its ultraviolet divergence.

d

Solution

Let

I, L, V

count internal lines, independent loops, and interaction vertices of a connected Feynman diagram. The graph identity

L = I - V + 1

and the two inverse powers from each scalar propagator give its superficial degree of divergence:

ω = 2 L - 2 I = 2 (1 - V) .

The same formula applies to every connected subgraph. Graphs with

V \geq 2

have negative superficial degree of divergence, but this alone does not remove a possible ultraviolet subdivergence. A one-vertex subgraph with external legs in the quartic scalar field theory can diverge only by a tadpole diagram, whose subtraction was fixed in c. Subtract every such tadpole diagram; every remaining loop subgraph then has

V \geq 2

and negative superficial degree of divergence. Its ultraviolet divergence is absent. Equivalently, normal ordering the quartic interaction removes all self-contractions at a single interaction vertex. Thus no new divergent field-dependent counterterm structure is needed, and in a normal-ordering subtraction scheme in two dimensions the one-loop mass subtraction exhausts these divergent coefficients.

The literal assertion that there are no further counterterm contributions in the on-shell renormalization scheme is stronger and generally false. A concrete counterexample is the two-loop sunset diagram. After analytic continuation below its three-particle threshold, the magnitude of its self-energy at

p^{2} = m^{2}

contains the strictly positive finite integral

Σ_{sunset} (m^{2}) = \frac{g _{4}^{2}}{6 ( 4 π ) ^{2} m ^{2}} \int_{x, y, z \geq 0} d x d y d z \frac{δ ( 1 - x - y - z )}{x y + x z + yz - x yz} > 0 (g_{4} \neq = 0) .

The integrable corners do not spoil convergence. Its nonzero value shifts the pole mass, requiring an additional finite mass counterterm if

m

is to remain the exact pole mass; its derivative can likewise change the pole residue. These are compatible with the power counting result. If vacuum diagrams are retained rather than divided out in normalized correlation functions, a separate constant vacuum energy counterterm is also required: the one-vertex double-loop vacuum diagram is already divergent. The claim about the mass subtraction concerns field-dependent counterterms and normalized correlation functions.

e

Solution

The additional six-leg interaction vertex has Feynman rule

- i g_{6}

. Since both interactions lack derivatives, the graph identity still gives

ω = 2 (1 - V)

, with

V = V_{4} + V_{6}

. The difference is that a one-vertex contraction of the phi-six theory can leave four external legs: choosing the contracted pair gives

(2 6) = 15

. Its quartic interaction correction is

\frac{g _{6}}{6 !} 15 T_{Λ} ϕ^{4} = \frac{1}{4 !} \frac{g _{6} T _{Λ}}{2} ϕ^{4} .

Consequently a one-loop quartic interaction counterterm is now necessary:

δ g_{4}^{(1)} = - \frac{g _{6}}{2} T_{Λ} = - \frac{g _{6}}{8 π} lo g (1 + \frac{Λ ^{2}}{m ^{2}}) .

The one-loop mass counterterm from c is unchanged. Contracting four legs of a single six-leg interaction vertex gives

(2 6) 3 = 45

double contractions, hence a two-loop raw self-energy

g_{6} T_{Λ}^{2} /8

. Its subtraction must include the quartic interaction counterterm insertion, rather than subtracting this raw expression in isolation.

All divergent field-dependent coefficients are most simply organized by the normal-ordering subtraction scheme in two dimensions. With

T = T_{Λ}

, express a potential normal ordered at mass

m

in ordinary fields:

V_{B} (ϕ) = \frac{g _{4}}{24} (ϕ^{4} - 6 T ϕ^{2} + 3 T^{2}) + \frac{g _{6}}{720} (ϕ^{6} - 15 T ϕ^{4} + 45 T^{2} ϕ^{2} - 15 T^{3}) .

Thus its divergent subtraction coefficients, in this scheme, are

δ g_{4} = - \frac{g _{6} T}{2}, δ_{m} = - \frac{g _{4} T}{2} + \frac{g _{6} T ^{2}}{8}, δ g_{6} = δ_{Z} = 0.

Indeed the two-loop self-energy terms are

g_{6} T^{2} /8 + δ g_{4} T /2 = - g_{6} T^{2} /8

, which the displayed mass counterterm cancels. The constant terms only affect vacuum energy. After removing all one-vertex self-contractions, every remaining subgraph is ultraviolet finite. Additional finite on-shell renormalization scheme adjustments are subject to the same qualification as d.

2

a

Solution

For nonzero real

g

, the continuous internal symmetry of a classical field theory is the global U(1) symmetry

ϕ ⟼ e^{- i α} ϕ, Φ ⟼ e^{- 2 i α} Φ, α constant .

The kinetic terms and mass terms are invariant separately, while the phases in

ϕ^{2} Φ^{†}

cancel. Independent phase rotations would instead require

2 α_{ϕ} - α_{Φ} = 0

modulo

2 π

, leaving one connected circle. Simultaneous complex conjugation is also a discrete symmetry, reversing the continuous charge; including it gives

U (1) ⋊ Z_{2} ≅ O (2)

for generic parameters. At

g = 0

the continuous symmetry enlarges to

U (1) \times U (1)

The Noether current with particles assigned positive charges is

j^{μ} = i (ϕ^{†} \partial^{μ} ϕ - (\partial^{μ} ϕ^{†}) ϕ) + 2 i (Φ^{†} \partial^{μ} Φ - (\partial^{μ} Φ^{†}) Φ) .

For example, put

A = ϕ^{2} Φ^{†}

and

B = ϕ^{† 2} Φ

. The Euler-Lagrange equations give

\partial_{μ} j_{ϕ}^{μ} = i g (A - B)

and

\partial_{μ} j_{Φ}^{μ} = i g (B - A)

, with the factor

2

already included in

j_{Φ}^{μ}

. Their sum vanishes. The Noether charge

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

is conserved and, after normal ordering, counts

Q = N_{ϕ} - N_{\overset{ˉ}{ϕ}} + 2 (N_{Φ} - N_{\overset{ˉ}{Φ}}) .

Thus a

Φ

particle carries two units of the same conserved charge carried by a

ϕ

particle; their antiparticles carry the negatives. The interaction permits conversion of one

Φ

particle into two

ϕ

particles when allowed by conservation of energy, while preserving this weighted number.

b

Solution

The two complex scalar fields have scalar propagators

⟨ Tϕ ϕ^{†} ⟩ (p) = \frac{i}{p ^{2} - m ^{2} + i ϵ}, ⟨ T Φ Φ^{†} ⟩ (p) = \frac{i}{p ^{2} - M ^{2} + i ϵ};

contractions of two unconjugated fields or two conjugated fields vanish. Draw an arrow on each line in the direction of positive conserved charge flow: a solid line carries one unit and a dashed line two units. A particle arrow follows its physical momentum; an antiparticle arrow opposes it.

There are two conjugate three-leg interaction vertices, both with Feynman rule

- i g

: one has a charge-two arrow entering and two charge-one arrows leaving, and the other reverses every arrow. These describe the two conjugate terms

ϕ^{2} Φ^{†}

and

ϕ^{† 2} Φ

. The two identical

ϕ

attachments cancel the

1/2

in the Lagrangian density. All crossed assignments are obtained by moving external lines between initial and final states while maintaining the arrows. There is no interaction vertex with three equal-sign charges.

Each interaction vertex conserves four-momentum, supplying

(2 π)^{4} δ^{4} (\sum p_{j})

with all momenta entering. Integrate each independent loop momentum with

d^{4} k / (2 π)^{4}

and divide by the Feynman-diagram symmetry factor. No minus sign is assigned to a closed scalar loop. These rules are for

i M

after amputating external lines and removing the overall four-momentum conservation factor.

c

i

Solution

Label the incoming four-momenta

p_{1}, p_{2}

and the outgoing ones

p_{3}, p_{4}

. Define the Mandelstam variables

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

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

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

. Two incoming

ϕ

particles have conserved charge

+ 2

, so the only tree-level Feynman diagram is an

s

-channel

Φ

exchange, shown in panel (b) of b's figure. Each end converts two charge-one lines into one charge-two line. There are no

t

- or

u

-channel exchanges: an incoming and an outgoing particle at such an end would transfer zero charge, for which this theory has no cubic interaction vertex.

i M_{ϕϕ} = (- i g)^{2} \frac{i}{s - M ^{2} + i ϵ}, M_{ϕϕ} = - \frac{g ^{2}}{s - M ^{2} + i ϵ} .

The single Feynman diagram is already symmetric under exchanging the two outgoing particles. This does not add another copy of the scattering amplitude; the identical final-state symmetry factor enters the integrated scattering cross-section instead.

ii

Solution

Take

ϕ (p_{1}) + \overset{ˉ}{ϕ} (p_{2}) \to ϕ (p_{3}) + \overset{ˉ}{ϕ} (p_{4})

with the Mandelstam variables of i. The allowed tree-level Feynman diagram is the crossed

u

-channel

Φ

exchange in panel (c) of b's figure. At one end the incoming particle and outgoing antiparticle transfer charge

+ 2

; the other end absorbs that charge. The zero-charge

s

- and

t

-channel assignments are forbidden by the interaction vertices. Thus

M_{ϕ \overset{ˉ}{ϕ}} = - \frac{g ^{2}}{u - M ^{2} + i ϵ} .

In the center-of-mass frame with

∣ p ∣ ≪ m

s = 4 m^{2} + O (∣ p ∣^{2})

and

u = O (∣ p ∣^{2})

. For

M ≫ m

, both scattering amplitudes approach

g^{2} / M^{2}

. The elastic two-body Lorentz-invariant phase space and incident flux agree for the two experiments. Using

\frac{d σ}{d Ω} = \frac{∣ M ∣ ^{2}}{64 π ^{2} s}

for distinguishable final particles, and an extra factor

1/2!

when integrating the identical-particle final state over the whole sphere, gives

σ_{ϕϕ} ≃ \frac{g ^{4}}{32 π s M ^{4}}, σ_{ϕ \overset{ˉ}{ϕ}} ≃ \frac{g ^{4}}{16 π s M ^{4}}, \frac{σ _{ϕϕ}}{σ _{ϕ \overset{ˉ}{ϕ}}} ⟶ \frac{1}{2} .

There is no additional division for identical incoming particles in the definition of a beam scattering cross-section. Before taking

M / m \to \infty

, the zero-momentum ratio is

\frac{1}{2} [M^{2} / (M^{2} - 4 m^{2})]^{2}

, provided the threshold is away from the heavy pole.

Near

s = M^{2}

, the bare

s

-channel scalar propagator makes fixed-order perturbation theory fail: repeated self-energy insertions contain powers of

Σ_{Φ} (s) / (s - M^{2})

. The

Φ \to ϕϕ

channel is open for

M > 2 m

, giving the leading decay width

Γ_{Φ} = \frac{g ^{2}}{32 π M} 1 - \frac{4 m ^{2}}{M ^{2}} .

Resumming the self-energy gives a finite Breit-Wigner resonance cross section, with denominator

s - M_{pole}^{2} + i M_{pole} Γ_{Φ}

near the pole, rather than an infinite scattering cross-section. Also,

s ≃ M^{2}

with

M ≫ m

lies outside the assumed nonrelativistic incoming kinematics; the ratio

1/2

is not a resonance prediction.

d

i

Solution

Four external

Φ

lines require four cubic interaction vertices, since each such interaction vertex contains only one heavy field. Their other eight endpoints must form four light internal lines. A connected Feynman diagram therefore has

V = I = 4

and

L = 1

: the leading contribution is a one-loop box diagram, of order

g^{4}

, with four

ϕ

scalar propagators. Charge-two incoming and outgoing heavy lines attach alternately around the box, as shown in panel (d) of b's figure. Each incoming heavy line supplies the two outgoing light-charge arrows at its corner, and each outgoing heavy line absorbs them. With external labels fixed there is one such box and its symmetry factor is one.

For

∣ p ∣ ≪ M ≪ m

, the external energies are also small compared with

m

, so expand the loop integrand in all external four-momenta. Its leading term is

i M_{ΦΦ} = g^{4} \int \frac{d ^{4} k}{( 2 π ) ^{4}} \frac{1}{( k ^{2} - m ^{2} + i ϵ ) ^{4}} + O (\frac{i g ^{4} M ^{2}}{m ^{6}}) .

A Wick rotation and the Euclidean massive loop integral give

\int \frac{d ^{4} k}{( 2 π ) ^{4}} \frac{1}{( k ^{2} - m ^{2} + i ϵ ) ^{4}} = i \int \frac{d ^{4} k _{E}}{( 2 π ) ^{4}} \frac{1}{( k _{E}^{2} + m ^{2} ) ^{4}} = \frac{i}{96 π ^{2} m ^{4}} .

Hence, more precisely than just its requested scaling,

M_{ΦΦ} = \frac{g ^{4}}{96 π ^{2} m ^{4}} [1 + O (\frac{M ^{2}}{m ^{2}}, \frac{∣ p ∣ ^{2}}{m ^{2}})] .

The mass dimension

[g] = 1

makes

g^{4} / m^{4}

dimensionless, as a four-dimensional scattering amplitude should be. The leading amplitude has no extra identical-particle multiplicity.

ii

Solution

Cross the second incoming heavy particle and one outgoing heavy particle in i to obtain

Φ \overset{ˉ}{Φ} \to Φ \overset{ˉ}{Φ}

. Panel (e) of b's figure shows the resulting box diagram. Its four light-charge arrows again alternate source and sink corners; it is the only connected order-

g^{4}

topology. There is no tree exchange because every cubic interaction vertex has only one heavy leg. With

∣ p ∣ ≪ M ≪ m

, crossing does not affect the zero-external-momentum loop integral, so

M_{Φ \overset{ˉ}{Φ}} = \frac{g ^{4}}{96 π ^{2} m ^{4}} [1 + O (\frac{M ^{2}}{m ^{2}}, \frac{∣ p ∣ ^{2}}{m ^{2}})] .

An independent check is to integrate out the complex scalar field

ϕ

at constant

Φ

. Its two real-component squared masses are

m^{2} \pm g ∣Φ∣

. The one-loop quantum effective potential is therefore

V_{1} (Φ) = \frac{1}{2} \int \frac{d ^{4} k _{E}}{( 2 π ) ^{4}} [lo g (k_{E}^{2} + m^{2} + g ∣Φ∣) + lo g (k_{E}^{2} + m^{2} - g ∣Φ∣)] .

Expanding the logarithms to fourth order in

Φ

gives

V_{1} \supset - \frac{g ^{4} ∣Φ ∣ ^{4}}{4} \int \frac{d ^{4} k _{E}}{( 2 π ) ^{4}} \frac{1}{( k _{E}^{2} + m ^{2} ) ^{4}} = - \frac{g ^{4}}{384 π ^{2} m ^{4}} (Φ^{†} Φ)^{2} .

Thus the effective Lagrangian has the positive local coefficient

g^{4} / (384 π^{2} m^{4})

. Its

2! 2!

external attachments give the common scattering amplitude

g^{4} / (96 π^{2} m^{4})

in both channels. This also checks the box multiplicity and sign.

3

a

Solution

In terms of the Pauli matrices,

γ^{0} = σ_{3}

γ^{1} = i σ_{1}

γ^{2} = i σ_{2}

. The Pauli matrix multiplication law implies

σ_{j}^{2} = I_{2}

and

σ_{j} σ_{k} = - σ_{k} σ_{j}

for

j \neq = k

. Consequently

(γ^{0})^{2} = I_{2}, (γ^{1})^{2} = (γ^{2})^{2} = - I_{2}, γ^{μ} γ^{ν} = - γ^{ν} γ^{μ} (μ \neq = ν) .

These diagonal and off-diagonal cases are exactly the Clifford algebra relation

{γ^{μ}, γ^{ν}} = 2 η^{μν} I_{2} .

The identity matrix is implicit on the right-hand side in the source. Useful conventions for the remaining parts are

(γ^{0})^{†} = γ^{0}

(γ^{j})^{†} = - γ^{j}

for

j = 1, 2

, and

γ^{0 *} = γ^{0}

γ^{1 *} = - γ^{1}

γ^{2 *} = γ^{2}

b

Solution