Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 301 (Quantum Field Theory)

The translation changes the Lagrangian density by $\delta {\mathcal{L}}_0={ϵ}^{\nu }{\partial }_{\nu }{\mathcal{L}}_0={\partial }_{\mu }({ϵ}^{\mu }{\mathcal{L}}_0)$. The Noether current is thereforeThis is the canonical stress-energy tensor. Directly,up to the analogous adjoint-field term, so the Euler-Lagrange field equations imply ${\partial }_{\mu }{T}^{\mu \nu }=0$ on shell. The Dirac equation also gives ${\mathcal{L}}_0=0$ on shell.

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After using the Clifford algebra, integrations by parts and the Dirac equation, every allowed Lorentz-covariant symmetric rank-two expression with one derivative reduces, modulo terms vanishing on shell and identically conserved improvements, to an overall multiple ofThus the general nontrivial tensor in the stated class isIt is manifestly symmetric. Differentiating it, commuting partial derivatives and usingmakes the terms cancel pairwise, proving ${\partial }_{\mu }{\hat{T}}^{\mu \nu }=0$ on shell. The normalization $C$ remains free at this stage.

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Choose $C=i/4$. Expandingwith the Clifford algebra, applying the product rule and then using the Dirac equation and its adjoint givesThis is the Belinfante-Rosenfeld stress-energy tensor written as an improvement of the canonical tensor.

The translation charges areFor spatial $\nu$, their difference is the integral of ${\partial }_i{S}^{i0\nu }-{\partial }^{\nu }(\bar{\psi }{\gamma }^0\psi )$. For $\nu =0$, use conservation of the Dirac current to replace ${\partial }^0(\bar{\psi }{\gamma }^0\psi )$ by a spatial divergence; also $S^{00\nu }=0$. Hence ${\hat{P}}^{\nu }-P^{\nu }$ is always a spatial boundary integral. Under the usual decay boundary condition it vanishes, so both currents generate the same four-momentum.

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Under spatial reflection, ${\varphi }^{\prime }(t,\mathbf{x})={\gamma }_p\varphi (t,-\mathbf{x})$, soBecause ${\gamma }_p^2=1$, the Lorentz scalar ${\partial }_{\mu }\varphi {\partial }^{\mu }\varphi$ and the mass term $m^2{\varphi }^2$ are unchanged after evaluating at the reflected point. The spatial change of variables $\mathbf{x}\mapsto -\mathbf{x}$ has unit absolute Jacobian, so the action of the free real scalar field is invariant.

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Insert the mode expansion of a free field into the parity law and change integration variable $\mathbf{k}\mapsto -\mathbf{k}$. Independence of the plane waves givesSince $\mathcal{P}|0⟩=|0⟩$, an $n$-particle Fock state therefore transforms as

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LetThe canonical commutation relations give $[N,a_{\mathbf{k}}]=-a_{\mathbf{k}}$ and $[K,a_{\mathbf{k}}]=-a_{-\mathbf{k}}$. Hence exponentiating the adjoint action givesWriting $R{a}_{\mathbf{k}}=a_{-\mathbf{k}}$, one has $R^2=1$ and ${\mathrm{ad}}_K=-R$ on annihilation operators. Therefore

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The number operator $N$ is self-adjoint. Alsoafter $\mathbf{q}\mapsto -\mathbf{q}$. Thus both exponents are anti-Hermitian and ${\mathcal{P}}_1,{\mathcal{P}}_2$ are unitary. Their product obeysand likewise for creation operators. Both generators annihilate the vacuum, so the product leaves it invariant. Substitution in the free-field mode expansion yieldsHence $\mathcal{P}={\mathcal{P}}_1{\mathcal{P}}_2$ implements parity with intrinsic parity ${\gamma }_p$.

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Varying the displayed gauge-fixed Lagrangian density gives the kinetic operatorIts quantum field theory propagator should satisfyand inversion into transverse and longitudinal projectors gives the numerator

The paper instead prints $-{\eta }_{\mu \nu }+(\alpha -1)k_{\mu }{k}_{\nu }/k^2$. Except at $\alpha =1$, that is not the inverse of the displayed Lagrangian's kinetic operator. Taken literally, for $\alpha \ne 2$ it is the Green function ofand at $\alpha =2$ its longitudinal part is noninvertible. Thus the longitudinal sign in the printed propagator is a typographical error; the two forms coincide in Feynman gauge.

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Let $G_{\mu \nu }(x,y)=⟨\Omega |T{A}_{\mu }(x)A_{\nu }(y)|\Omega ⟩$. Invariance of the normalized path integral under $A\mapsto A+\delta A$ gives the Schwinger-Dyson equationtogether with $D^{\mu \rho }⟨A_{\rho }{⟩}_j=e{j}^{\mu }$. This assumes the functional measure is translation invariant, boundary terms in field space vanish, and vacuum bubbles are removed by normalization. The current is a fixed c-number and conserved; conservation removes dependence on the longitudinal, gauge-parameter part of the photon propagator. An $iϵ$ prescription and adiabatic switching select the interacting vacuum.

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Since $D\Delta =i\delta$, the source-induced one-point function isThe action is quadratic and the source is linear, so completing the square gives the exact full two-point functionDiagrammatically these are a free line joining $x$ to $y$ and a disconnected pair of lines, each joining one external insertion to one current cross. The connected two-point function remains exactly ${\Delta }_{\mu \nu }$. The expansion truncates at $e^2$ because a Gaussian integral has no interaction vertices and its mean is linear in $e$.

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For quantum electrodynamics, $j^{\mu }=\bar{\psi }{\gamma }^{\mu }\psi$ is an operator built from a dynamical Dirac field. Integrating over $\psi ,\bar{\psi }$ generates arbitrarily high powers of $e$, so the Gaussian-source truncation fails.

For the connected photon two-point function, the required diagrams through $e^4$ are: the bare photon line at $e^0$; one fermion-loop photon vacuum polarization insertion at $e^2$; and at $e^4$, a photon line with two successive one-loop polarization insertions together with the two-loop one-particle-irreducible fermion loop whose loop contains one internal photon chord. The latter includes the cyclic placements conventionally interpreted as self-energy and vertex corrections. Counterterm insertions must be added in a renormalized calculation. Disconnected vacuum bubbles cancel against the vacuum normalization.

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Expanding the covariant derivatives of scalar quantum electrodynamics givesandIn four dimensions, $[A_{\mu }]=[\varphi ]=1$ and $[{\partial }_{\mu }]=1$, so $[e]=0$. Both interactions have dimension four and $e$ is a marginal coupling by power counting in quantum field theory.

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With all momenta directed into each vertex, the momentum-space Feynman rules are:

Each vertex also carries its momentum-conserving delta function; integrate every independent loop momentum and attach the appropriate external wavefunctions and photon polarization vectors.

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Three connected tree-level Feynman diagrams contribute at order $e^2$: a scalar-exchange diagram in which photon $k_1$ is emitted before $k_2$ along the charged scalar line; the crossed scalar-exchange diagram with $k_1$ and $k_2$ interchanged; and the local seagull vertex joining both incoming scalars and both outgoing photons. The exchange denominators are $(p_1-k_1{)}^2-m^2$ and $(p_1-k_2{)}^2-m^2$.

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Let $p_1,p_2$ be the incoming scalar and antiscalar momenta and $k_1,k_2$ the outgoing photon momenta, so $p_1+p_2=k_1+k_2$. Up to the common overall phase fixed by the $S$-matrix convention, the connected leading amplitude iswhereOn shell, the denominators are $-2p_1\cdot k_1=-2p_2\cdot k_2$ and $-2p_1\cdot k_2=-2p_2\cdot k_1$. Contracting with $k_{1\mu }$ therefore givesusing momentum conservation and $k_2\cdot {ϵ}_2=0$. The same calculation with the photons interchanged gives $k_{2\nu }{\mathcal{M}}^{\mu \nu }{ϵ}_{1\mu }^{\ast }=0$. The two exchange diagrams and the seagull term are all required for this Ward identity.

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