Paper 301 (Quantum Field Theory)

At this order there is one annihilation tree-level Feynman diagram. The squared-amplitude diagram is its product with its complex conjugate, with the same external momenta and spin labels in both factors:

 e-(p,s) --->\          /---> mu-(p',s')
             o ~~~~~~ o
 e+(q,r) <---/          \<--- mu+(q',r')

                     times

 e-(p,s) --->\          /---> mu-(p',s')
             o ~~~~~~ o                       *
 e+(q,r) <---/          \<--- mu+(q',r')

A wavy line is the virtual photon; the arrows show fermion-number flow. The star means complex conjugation of the entire second graph, including its external spinors. This represents

∣ M ∣^{2} = M M^{*}

for fixed spins, without adding a new physical loop to the amplitude.

For the spin sum in part (iv), identify each external spin label between the two copies and sum it. The fermion spin sum sews each spinor chain into a gamma-matrix trace. The two incoming unpolarized spin states are then handled by a spin average, giving the factor

1/4

. In particular, the diagrammatic contraction becomes

\overline{∣ M ∣^{2}} = \frac{e ^{4}}{4 s ^{2}} tr (\neq q γ^{μ} \neq p γ^{ν}) tr (\neq p^{'} γ_{μ} \neq q^{'} γ_{ν}) .

(1)

iv

 0  0

Solution

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First derive the needed gamma-matrix traces directly from the Clifford algebra. The cyclic property of the trace gives

2 tr (γ^{μ} γ^{ν}) = tr {γ^{μ}, γ^{ν}} = 2 η^{μν} tr I_{4},

(1)

tr (γ^{μ} γ^{ν}) = 4 η^{μν}

. For

T = tr (γ^{μ} γ^{ν} γ^{ρ} γ^{σ})

, move

γ^{μ}

successively through the other three matrices and cycle the final term back to the front:

2 T = 2 η^{μν} tr (γ^{ρ} γ^{σ}) - 2 η^{μ ρ} tr (γ^{ν} γ^{σ}) + 2 η^{μ σ} tr (γ^{ν} γ^{ρ}) .

(2)

Therefore

tr (γ^{μ} γ^{ν} γ^{ρ} γ^{σ}) = 4 (η^{μν} η^{ρ σ} - η^{μ ρ} η^{ν σ} + η^{μ σ} η^{ν ρ}) .

(3)

These are all the trace identities required.

For massless external Dirac spinors, the fermion spin sums are

\sum_{s} u_{s} (p) \overset{u}{ˉ}_{s} (p) = \neq p

and

\sum_{r} v_{r} (q) \overset{v}{ˉ}_{r} (q) = \neq q

. With the initial spin average, part (iii) becomes a contraction of

L^{μν} = 4 (q^{μ} p^{ν} + q^{ν} p^{μ} - η^{μν} p \cdot q), H_{μν} = 4 (p_{μ}^{'} q_{ν}^{'} + p_{ν}^{'} q_{μ}^{'} - η_{μν} p^{'} \cdot q^{'}) .

(4)

Their contraction is

L^{μν} H_{μν} = 32 [(q \cdot p^{'}) (p \cdot q^{'}) + (q \cdot q^{'}) (p \cdot p^{'})] .

(5)

The massless Mandelstam variables give

p \cdot p^{'} = q \cdot q^{'} = - t /2

and

p \cdot q^{'} = q \cdot p^{'} = - u /2

, hence

\overline{∣ M ∣^{2}} = \frac{2 e ^{4}}{s ^{2}} (t^{2} + u^{2}) .

(6)

In the center of mass frame each particle has energy

s /2

, and

t = - \frac{s}{2} (1 - cos θ), u = - \frac{s}{2} (1 + cos θ) .

(7)

Substitution gives the concise angular answer:

\overline{∣ M ∣^{2}} = e^{4} (1 + cos^{2} θ) .

(8)

v

 0  0

Solution

 0  0

Substitute the preceding spin average into the supplied differential scattering cross-section. Since

e^{4} = 16 π^{2} α^{2}

\frac{d σ}{d Ω} = \frac{α ^{2}}{4 s} (1 + cos^{2} θ) .

(1)

The angular integral is

2 π \int_{- 1}^{1} (1 + x^{2}) d x = 16 π /3

. Consequently the relativistic scattering cross-section for the massless electron-positron annihilation into a muon pair is

σ = \frac{4 π α ^{2}}{3 s} .

(2)

The outgoing particles are distinct, so there is no identical-particle factor. Comparing with the requested monomial form gives

B = \frac{4 π}{3}, j = 2, k = - 1, l = 0, m = 0.

(3)

The integrated answer depends only on the incoming invariant

s

; the angular variables

t, u

have been integrated out.

2

 0  0

i

 0  0

Solution

 0  0

For a time-independent Hamiltonian operator and a Schrödinger picture observable with no explicit time dependence, the Heisenberg picture operator is

O_{H} (t) = e^{i H t} O_{S} e^{- i H t} .

(1)

Differentiation gives

\dot{O}_{H} = i H O_{H} - i O_{H} H = i [H, O_{H}]

, so

[H, O_{H}] = - i \frac{d O _{H}}{d t} .

(2)

O_{S}

itself depends explicitly on time, one must add

e^{i H t} (\partial_{t} O_{S}) e^{- i H t}

\dot{O}_{H}

; the displayed commutator relation uses the usual convention of no explicit time dependence for the observable.

ii

 0  0

Solution

 0  0

For a free real scalar field, the Lagrangian density contains its quadratic kinetic term and mass term:

L_{1} = \frac{1}{2} \partial_{μ} ϕ_{1} \partial^{μ} ϕ_{1} - \frac{1}{2} m_{1}^{2} ϕ_{1}^{2} = \frac{1}{2} \dot{ϕ}_{1}^{2} - \frac{1}{2} ∣\nabla ϕ_{1} ∣^{2} - \frac{1}{2} m_{1}^{2} ϕ_{1}^{2} .

(1)

The field's conjugate momentum is

π_{1} = \partial L_{1} / \partial \dot{ϕ}_{1} = \dot{ϕ}_{1}

. The positive sign of the time-derivative term uses the metric convention fixed in Question 1.

iii

 0  0

Solution

 0  0

Apply the Legendre transform in mechanics pointwise to the field Lagrangian density. Using

π_{1} = \dot{ϕ}_{1}

, the Hamiltonian density is

H_{1} = π_{1} \dot{ϕ}_{1} - L_{1} = \frac{1}{2} π_{1}^{2} + \frac{1}{2} ∣\nabla ϕ_{1} ∣^{2} + \frac{1}{2} m_{1}^{2} ϕ_{1}^{2} .

(1)

The free Hamiltonian operator is its spatial integral:

H_{1} = \frac{1}{2} \int_{R^{3}} d^{3} x [π_{1}^{2} + (\nabla ϕ_{1})^{2} + m_{1}^{2} ϕ_{1}^{2}] .

(2)

A constant vacuum-energy subtraction, if made, does not affect the commutators or the ensuing field equation.

iv

 0  0

Solution

 0  0

The canonical quantization of a real scalar field promotes the classical equal-time brackets to the canonical commutation relations, in units

ℏ = 1

[ϕ_{1} (t, x), ϕ_{1} (t, y)] [π_{1} (t, x), π_{1} (t, y)] [ϕ_{1} (t, x), π_{1} (t, y)] = 0, = 0, = i δ^{(3)} (x - y) .

(1)

The Dirac delta distribution expresses locality of the conjugate variables. These are identities of operator-valued distributions, interpreted after spatial smearing where necessary.

v

 0  0

Solution

 0  0

Use the free Hamiltonian operator and the canonical commutation relations. The kinetic momentum term gives

[H_{1}, ϕ_{1} (t, x)] = \frac{1}{2} \int d^{3} y [π_{1} (t, y)^{2}, ϕ_{1} (t, x)] = - i π_{1} (t, x) .

(1)

The Heisenberg picture equation therefore yields

\dot{ϕ}_{1} = π_{1}

For the momentum commutator, the mass term gives

i m_{1}^{2} ϕ_{1} (x)

. The gradient term gives

\frac{1}{2} \int d^{3} y [(\nabla ϕ_{1} (y))^{2}, π_{1} (x)] = i \int d^{3} y \nabla ϕ_{1} (y) \cdot \nabla_{y} δ^{(3)} (y - x) = - i \nabla^{2} ϕ_{1} (x),

(2)

where integration by parts differentiates the field instead of the Dirac delta distribution. Thus

[H_{1}, π_{1}] = i (m_{1}^{2} - \nabla^{2}) ϕ_{1}, \overset{π}{˙}_{1} = (\nabla^{2} - m_{1}^{2}) ϕ_{1} .

(3)

Combining the two first-order operator equations gives the Klein-Gordon equation:

(\partial_{t}^{2} - \nabla^{2} + m_{1}^{2}) ϕ_{1} = (□ + m_{1}^{2}) ϕ_{1} = 0.

(4)

Suitable decay or spatial smearing removes the boundary term in the integration by parts.

vi

 0  0

Solution

 0  0

For the two free real scalar fields, let

L = L_{1} + L_{2}

. The infinitesimal internal rotation symmetry of two real scalar fields has

δ ϕ_{1} = θ ϕ_{2}

and

δ ϕ_{2} = - θ ϕ_{1}

, with constant

θ

. Its kinetic terms are invariant because this is an orthogonal rotation of the two-component field.

Under the assumed invariance of the full Lagrangian density, Noether's theorem gives the current per unit transformation parameter:

j^{μ} = \sum_{a = 1}^{2} \frac{\partial L}{\partial ( \partial _{μ} ϕ _{a} )} \frac{δ ϕ _{a}}{θ} = ϕ_{2} \partial^{μ} ϕ_{1} - ϕ_{1} \partial^{μ} ϕ_{2} .

(1)

The Noether current and Noether charge are therefore

j^{μ} = ϕ_{2} \partial^{μ} ϕ_{1} - ϕ_{1} \partial^{μ} ϕ_{2}, Q = \int d^{3} x (ϕ_{2} π_{1} - ϕ_{1} π_{2}) .

(2)

Independent species commute at equal time, so these products require no extra ordering sign. With vanishing boundary flux,

Q

is conserved. Part (vii) checks precisely what the symmetry assumption requires of the masses.

vii

 0  0

Solution

 0  0

Differentiate the Noether charge and use

\dot{ϕ}_{a} = π_{a}

and the Klein-Gordon equation for each species. The momentum products cancel because the two species commute:

\dot{Q} = \int d^{3} x [π_{2} π_{1} + ϕ_{2} (\nabla^{2} - m_{1}^{2}) ϕ_{1} - π_{1} π_{2} - ϕ_{1} (\nabla^{2} - m_{2}^{2}) ϕ_{2}] = \int d^{3} x [ϕ_{2} \nabla^{2} ϕ_{1} - ϕ_{1} \nabla^{2} ϕ_{2}] + (m_{2}^{2} - m_{1}^{2}) \int d^{3} x ϕ_{1} ϕ_{2} .

(1)

The first integral is the boundary flux of

ϕ_{2} \nabla ϕ_{1} - ϕ_{1} \nabla ϕ_{2}

, so it vanishes under the same boundary condition used for charge conservation. Hence

\frac{d Q}{d t} = (m_{2}^{2} - m_{1}^{2}) \int d^{3} x ϕ_{1} ϕ_{2} .

(2)

Conservation must hold for arbitrary configurations, rather than only a state in which this particular integral happens to vanish. Thus the internal rotation symmetry of two real scalar fields requires

m_{1}^{2} = m_{2}^{2}, and therefore m_{1} = m_{2} for nonnegative masses .

(3)

Equivalently, the local divergence is

\partial_{μ} j^{μ} = (m_{2}^{2} - m_{1}^{2}) ϕ_{1} ϕ_{2}

, exhibiting exactly how unequal masses break the symmetry.

3

 0  0

i

 0  0

Solution

 0  0

In phi cubed theory, each vertex has three legs. A connected four-point tree-level Feynman diagram therefore needs two cubic vertices and one internal propagator. There are exactly three ways to partition the labeled external legs between these vertices:

       s channel               t channel              u channel

 p \          / p'         p ---o--- p'            p ---o--- q'
    o--------o                  |                       |
 q /          \ q'         q ---o--- q'            q ---o--- p'

 internal: p+q             internal: p-p'          internal: p-q'

Solid lines denote the real scalar field propagator. The labels specify which external pair meets at each vertex, so the last two graphs are different even though their unlabeled shapes agree. This is the four-point tree amplitude in phi cubed theory:

three exchange diagrams, in the s, t, u channels; each is of order λ^{2} .

(1)

There is no four-point contact vertex in the given cubic interaction.

ii

 0  0

Solution

 0  0

For the interaction sign in this paper, the cubic Feynman rule is

+ iλ

, while the scalar Feynman propagator is

i / (k^{2} - m^{2} + i ϵ)

. Each exchange graph therefore contributes

- i λ^{2} / (k^{2} - m^{2} + i ϵ)

i M

. The three graphs give

M = - λ^{2} (\frac{1}{s - m ^{2} + i ϵ} + \frac{1}{t - m ^{2} + i ϵ} + \frac{1}{u - m ^{2} + i ϵ}) .

(1)

The equal-mass Mandelstam variables satisfy

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

, so

u - m^{2} = 3 m^{2} - s - t

. In the physical elastic region with

m > 0

, none of these denominators vanishes, and the tree amplitude is real up to its irrelevant overall convention. Thus

∣ M ∣^{2} = λ^{4} [\frac{1}{s - m ^{2}} + \frac{1}{t - m ^{2}} + \frac{1}{3 m ^{2} - s - t}]^{2} .

(2)

The square includes interference between all three channels; summing their individual squared contributions would omit physical cross terms. If a denominator approaches a pole, retain the Feynman i-epsilon prescription in the modulus before taking a limit.

iii

 0  0

Solution

 0  0

Use the Lorentz-invariant phase-space measure from the original PDF, including the

1/ (2 E_{p^{'}})

and

1/ (2 E_{q^{'}})

factors missing from the local TeX. In the center of mass frame put

E = s /2

and

k = s - 4 m^{2} /2

, assuming

s > 4 m^{2}

. The spatial Dirac delta distribution sets

q^{'} = - p^{'}

, leaving

d Φ_{2} = \frac{1}{16 π ^{2}} \frac{k ^{'2} d k ^{'} d Ω}{E _{k^{'}}^{2}} δ (s - 2 E_{k^{'}}) .

(1)

Since

d (2 E_{k^{'}}) / d k^{'} = 2 k^{'} / E_{k^{'}}

, the radial integral gives

d Φ_{2} = \frac{k}{16 π ^{2} s} d Ω = \frac{1 - 4 m ^{2} / s}{32 π ^{2}} d Ω.

(2)

The invariant flux factor is

2 κ (s, m^{2}, m^{2}) = 2 s (s - 4 m^{2})

, using the Källén function. Dividing the phase space by this flux yields

d σ / d Ω = ∣ M ∣^{2} / (64 π^{2} s)

, with the final-state labels retained as in the printed formula.

Let

x = cos θ

. Of the Mandelstam variables,

t

t = - 2 k^{2} (1 - x) = - \frac{s - 4 m ^{2}}{2} (1 - x), \frac{d t}{d x} = \frac{s - 4 m ^{2}}{2} .

(3)

The azimuthal integral contributes

2 π

, and the angular endpoints are therefore

t_{m i n} = 4 m^{2} - s, t_{m a x} = 0.

(4)

Changing variables gives the requested expression from the formula supplied in the paper:

σ = \frac{1}{16 π} \int_{4 m^{2} - s}^{0} \frac{∣ M ∣ ^{2}}{s ( s - 4 m ^{2} )} d t, X = \frac{1}{16 π}, Y = 4.

(5)

There is a normalization qualification: the displayed starting formula integrates over labeled final momenta and contains no

1/2!

. For the physical cross-section of two indistinguishable outgoing quanta of this real scalar field, the identical-particle factor in a final-state phase-space integral divides the full integral by two. With that convention

X = 1/ (32 π)

; equivalently, integrate only one representative of each exchanged pair. The boxed value

1/ (16 π)

follows the given formula exactly.

iv

 0  0

Solution

 0  0

In four spacetime dimensions with

ℏ = c = 1

, a relativistic scattering cross-section has mass dimension

- 2

, since it is an area. The action is dimensionless, so the Lagrangian density has dimension four. The scalar kinetic term then gives

[ϕ] = 1

, and the cubic term gives

[λ] = 1

Each term of the four-point tree amplitude in phi cubed theory has dimension

[λ^{2}] - [s - m^{2}] = 2 - 2 = 0

. Hence

[∣ M ∣^{2}] = 0

. In the final integral,

[d t] = 2, [s] = [s - 4 m^{2}] = 2,

(1)

while the numerical factors

X, Y

are dimensionless. Therefore

[\int d t \frac{∣ M ∣ ^{2}}{s ( s - 4 m ^{2} )}] = 2 + 0 - 4 = - 2 = [σ] .

(2)

Including the identical-particle factor changes only a dimensionless constant and preserves this check.

4

 0  0

i

 0  0

Solution

 0  0

Under a gauge transformation, the electromagnetic field tensor changes by

δ F_{μν} = \partial_{μ} \partial_{ν} ξ - \partial_{ν} \partial_{μ} ξ = 0.

(1)

The partial derivatives commute for a smooth gauge parameter, or in the distributional formulation. Thus the Maxwell Lagrangian depends only on a gauge-invariant tensor:

F_{μν}^{'} = F_{μν}, L^{'} = - \frac{1}{4} F_{μν}^{'} F^{' μν} = L .

(2)

Strictly, the pointwise calculation requires the second derivatives to exist; smooth gauge parameters are the usual convention. Mere first differentiability in the wording can instead be understood through commuting distributional derivatives.

ii

 0  0

Solution

 0  0

A Lorentz transformation preserves the quadratic form of the Minkowski metric for every four-vector. Substituting

x^{' σ} = Λ^{σ}_{μ} x^{μ}

gives

η_{σ τ} Λ^{σ}_{μ} Λ^{τ}_{ν} x^{μ} x^{ν} = η_{μν} x^{μ} x^{ν} .

(1)

The coefficient difference is a symmetric matrix, and a symmetric quadratic form that vanishes on every vector is zero: evaluating it on basis vectors and their sums recovers each coefficient. Consequently

Λ^{T} η Λ = η, η_{μν} = η_{σ τ} Λ^{σ}_{μ} Λ^{τ}_{ν} .

(2)

This is the defining metric-preservation relation of the Lorentz group.

iii

 0  0

Solution

 0  0

Insert

Λ^{μ}_{ν} = δ^{μ}_{ν} + α ω^{μ}_{ν}

into the Lorentz group condition and keep the coefficient of

α

. It gives

η_{σ ν} ω^{σ}_{μ} + η_{μτ} ω^{τ}_{ν} = 0.

(1)

Define

ω_{μν} = η_{μ ρ} ω^{ρ}_{ν}

. The displayed equation is

ω_{νμ} + ω_{μν} = 0

. Raising both indices with the symmetric Minkowski metric preserves this antisymmetry:

ω_{μν} = - ω_{νμ}, ω^{μν} = - ω^{νμ} .

(2)

These are the generators of the Lie algebra of the Lorentz group. In particular

ω^{μ}_{μ} = 0

. The mixed-index matrix need not be antisymmetric in the ordinary Euclidean transpose; it is antisymmetric after lowering an index with

η

iv

 0  0

Solution

 0  0

For the active Lorentz transformation of a vector field, coordinates are held fixed and

A^{' μ} (x) = Λ^{μ}_{ν} A^{ν} (Λ^{- 1} x)

. Its field strength transforms as a two-index Lorentz tensor. The index contractions in the Maxwell Lagrangian are invariant by

Λ^{T} η Λ = η

, so the density transforms as a Lorentz scalar:

L^{'} (x) = L (Λ^{- 1} x) .

(1)

For

Λ = I + α ω

, its inverse is

I - α ω + O (α^{2})

. Expanding the coordinate argument therefore gives

δ L (x) = - α ω^{μ}_{ν} x^{ν} \partial_{μ} L (x) .

(2)

Set

v^{μ} (x) = α ω^{μ}_{ν} x^{ν}

. Part (iii) implies

\partial_{μ} v^{μ} = α ω^{μ}_{μ} = 0

. The product rule then converts the variation into a total derivative:

δ L = - \partial_{μ} (α ω^{μ}_{ν} x^{ν} L) .

(3)

Its integral changes the action only by a boundary term, which is the condition needed for Noether's theorem.

v

 0  0

Solution

 0  0

For a translation with the sign used in the paper, the field variation at fixed coordinates is

δ A_{ρ} = β ϵ^{ν} \partial_{ν} A_{ρ}

, and

δ L = \partial_{μ} (β ϵ^{μ} L)

. Differentiating the Maxwell Lagrangian with respect to the field gradient gives

\frac{\partial L}{\partial ( \partial _{μ} A _{ρ} )} = - F^{μ ρ} .

(1)

Thus Noether's theorem yields

j^{μ} = ϵ_{ν} T^{μν}

, where the canonical stress-energy tensor of the Maxwell field is

T^{μν} = - F^{μ ρ} \partial^{ν} A_{ρ} - η^{μν} L = - F^{μ ρ} \partial^{ν} A_{ρ} + \frac{1}{4} η^{μν} F_{ρ σ} F^{ρ σ} .

(2)

It is conserved on the source-free Maxwell equations, but it is not generally symmetric. Indeed,

T^{μν} - T^{νμ} = - F^{μ ρ} \partial^{ν} A_{ρ} + F^{ν ρ} \partial^{μ} A_{ρ}

(3)

need not vanish. For example, take

A_{0} = a x^{1}

A_{2} = b x^{0}

, and the other components zero, with nonzero constants

a, b

. This produces constant field strength and satisfies the source-free equations, but

T^{12} = 0

and

T^{21} = ab

Nor is it gauge-invariant: while

F

and

L

are unchanged, a gauge transformation gives

δ_{ξ} T^{μν} = - F^{μ ρ} \partial^{ν} \partial_{ρ} ξ,

(4)

which is generally nonzero. These shortcomings motivate the improvement in part (vi); they do not contradict conservation of the canonical stress-energy tensor.

vi

 0  0

Solution

 0  0

Antisymmetry gives

- F^{ρ μ} \partial_{ρ} A^{ν} = F^{μ ρ} \partial_{ρ} A^{ν}

. Combining this with the canonical stress-energy tensor removes the potential in favour of the electromagnetic field tensor:

Θ^{μν} = - F^{μ ρ} (\partial^{ν} A_{ρ} - \partial_{ρ} A^{ν}) + \frac{1}{4} η^{μν} F_{ρ σ} F^{ρ σ} = - F^{μ ρ} F^{ν}_{ρ} + \frac{1}{4} η^{μν} F_{ρ σ} F^{ρ σ} .

(1)

This is the electromagnetic stress-energy tensor in the

(+, -, -, -)

convention. The product

F^{μ ρ} F^{ν}_{ρ}

is symmetric in

μ, ν

, since the contracted metric is symmetric. Its expression only in

F

makes it gauge-invariant. These two properties hold without using the equations of motion.

For conservation, the source-free Maxwell equations are

\partial_{ρ} F^{ρ μ} = 0

. Hence the given correction equals

- F^{ρ μ} \partial_{ρ} A^{ν} = - \partial_{ρ} (F^{ρ μ} A^{ν})

(2)

on shell. This is an antisymmetric superpotential improvement:

B^{ρ μν} = - F^{ρ μ} A^{ν}

is antisymmetric in

ρ, μ

, so

\partial_{μ} \partial_{ρ} B^{ρ μν} = 0

. Conservation of the canonical stress-energy tensor therefore implies

\partial_{μ} Θ^{μν} = 0

. Equivalently, direct differentiation gives

\partial_{μ} Θ^{μν} = - F^{ν}_{ρ} \partial_{μ} F^{μ ρ}

after the homogeneous Maxwell equations cancel the field-strength derivative terms.

There are four conserved currents, one for each fixed

ν

, with charges

P^{ν} = \int d^{3} x Θ^{0 ν}

. The Belinfante-Rosenfeld stress-energy tensor differs from the canonical tensor only by boundary contributions to these charges when the fields decay suitably.

Finally, in four spacetime dimensions,

Θ^{μ}_{μ} = - F^{μ ρ} F_{μ ρ} + \frac{4}{4} F_{ρ σ} F^{ρ σ} = 0.

(3)

Thus all requested properties are

\partial_{μ} Θ^{μν} = 0 on shell, Θ^{μν} = Θ^{νμ}, δ_{ξ} Θ^{μν} = 0, Θ^{μ}_{μ} = 0.

(4)

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 301 (Quantum Field Theory)

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