Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 304 (Advanced Quantum Field Theory)

Insert a complete set of momentum eigenstates into the quantum-mechanical propagator:Completing the square and evaluating the resulting Gaussian integral, with the usual i-epsilon prescription, gives the free-particle propagatorThe square-root branch is fixed by requiring $K(x_f,T;x_i,0)\to \delta (x_f-x_i)$ as $T\downarrow 0$.

The Euler-Lagrange equation is $m\ddot{x}=0$. The unique path with the prescribed endpoints is thereforeIt is a minimum of the Euclidean action and a stationary point of the real-time action. Its classical action isThe principle of stationary action consequently fixes the position-dependent phase of the semiclassical propagator asBecause the action is quadratic, the stationary-phase evaluation of the path integral is exact. Composition of propagators, or the Van Vleck determinant, gives $C(T)=\sqrt{m/(2\pi i\hslash T)}$, reproducing part i.

The particle on a circle has the complete orthonormal basis of energy eigenstates $\{|n⟩:n\in \mathbb{Z}\}$. Its spectral representation givesThe integer $n$ is the quantized angular momentum in units of $\hslash$.

Choose a real lift $\Delta \theta ={\theta }_f-{\theta }_i$. Classical paths fall into winding number sectors $j\in \mathbb{Z}$ and areEach sector has the same fluctuation determinant, so the image-sum form of the propagator isSet $a=\hslash T/(2m{R}^2)$. Applying the Poisson summation formula to the Gaussian $e^{i{x}^2/(4a)}$ giveswhich is exactly the spectral propagator found in part i. Thus the angular-momentum sum is dual to a sum over homotopy classes of classical paths.

With a nonconstant potential energy, momentum no longer diagonalizes the Hamiltonian operator. The operator derivation must use the energy eigenfunctions of the full Hamiltonian, a Dyson series, or a time-sliced Trotter product formula. In the classical derivation, the straight paths are replaced by every solution of the nonlinear Euler-Lagrange equation with the specified endpoints. The semiclassical propagator becomes a sumwhere the prefactor is the Van Vleck determinant and ${\mu }_{\mathrm{cl}}$ is a Maslov index. Unlike a quadratic theory, the classical-path sum is generally only an asymptotic approximation: the exact path integral includes fluctuations of every order. On the circle, the sum must still include all winding number sectors.

For the Euclidean quartic scalar field theory, expansion of $e^{-S}$ gives the momentum-space rules

At one loop, the four-point one-particle-irreducible correlation function receives the three bubble diagrams in the $s$, $t$, and $u$ channels. If $P$ is the momentum through one channel, its contribution isUsing a Feynman parameter, shifting the loop momentum, and writing ${\Delta }_x=m^2+x(1-x)P^2$ gives, up to terms caused by shifting the boundary of a hard cutoff,Thus every channel has the logarithmic ultraviolet divergenceThe complete one-loop vertex isLet $I_{m,\mathrm{os}}$ denote the bracket evaluated at the chosen on-shell kinematic point. The on-shell renormalization scheme requires $V_a^{(4)}{|}_{\mathrm{os}}=-{\lambda }_{\mathrm{phys}}$, hence the perturbative solution isThe cutoff dependence of ${\lambda }_a$ is the coupling counterterm needed to hold the measured coupling fixed.

After integration by parts, the quadratic ghost operator is $c_1(-{\partial }^2)+c_2{M}^2$. Matching its propagator to the scalar propagator after $M\mapsto m$ requiresDifferentiating $c_3\lambda \bar{\eta }{\varphi }^2\eta$ with respect to its two identical scalar fields gives the ghost-scalar vertex $-2c_3\lambda$. Matching it to the four-scalar vertex therefore requiresThe additional rules are an oriented ghost propagator $(p^2+M^2{)}^{-1}$, a two-scalar two-ghost vertex $-\lambda$, and a minus sign for every closed loop of Grassmann-valued fields.

Besides the scalar bubbles from part a, each channel now has a closed heavy-ghost bubble. Its two directed internal lines cannot be interchanged, so it has no scalar bubble's factor $1/2$. ConsequentlyIf $I_{M,\mathrm{os}}$ denotes the corresponding sum of heavy integrals, on-shell matching givesThe root continuously connected to ${\lambda }_b={\lambda }_{\mathrm{phys}}$ is the perturbative, small positive root.

After ${\lambda }_a$ and ${\lambda }_b$ have been matched to the same measured coupling, the nonanalytic low-energy dependence from the light scalar loops is identical. For $|P|\ll M$, a heavy loop has the local expansionIts constant term is already absorbed into the matched quartic coupling, while the remaining terms are higher-dimensional local interactions suppressed by powers of $M$. Dependence on the finite ultraviolet cutoff is similarly suppressed by powers of $\Lambda$. This is heavy-field decoupling: low-energy scattering agrees up to $O(p^2/M^2,p^2/{\Lambda }^2)$ corrections after all relevant parameters are matched. At energies comparable to $M$, the two theories differ sharply because the second theory contains wrong-statistics heavy states.

Under the global U(1) gauge symmetry, the phases in $\bar{\psi }\psi$ cancel and $\alpha$ is constant, soBoth the fermion term and $F_{\mu \nu }{F}^{\mu \nu }$ are therefore invariant. To extract the Noether current, temporarily promote $\alpha$ to a function. The variation of the action isThe Euler-Lagrange equations then imply the current conservation law ${\partial }_{\mu }{j}^{\mu }=0$.

Perform the infinitesimal local change of integration variablesin the Euclidean path integral for $⟨j^{\mu }(x)\psi (x_1)\bar{\psi }(x_2)⟩$. The vector transformation has no quantum anomaly, so its functional measure is invariant. The action variation supplies $-i\int \alpha {\partial }_{\mu }{j}^{\mu }$, while varying the two charged insertions supplies contact terms at $x_1$ and $x_2$ with opposite signs. Since a change of integration variables cannot change the integral, the coefficient of the arbitrary function $\alpha (x)$ vanishes:This Schwinger-Dyson equation is the position-space Ward-Takahashi identity.

Fourier transformation sends ${\partial }_{\mu }$ to $i{q}_{\mu }$. Write the connected current three-point function as the two full Dirac propagators joined to the amputated vertex:The two contact terms remove one propagator at a time. Multiplying the transformed identity by $G^{-1}(p_1)$ on the left and $G^{-1}(p_2)$ on the right yieldsThis is the momentum-space Ward-Takahashi identity for the full vertex and full propagator.

Taking $q\to 0$ in the Ward-Takahashi identity givesup to the displayed Euclidean $i$ conventions. The ultraviolet divergence of the zero-momentum vertex is therefore exactly the derivative of the fermion self-energy divergence. In renormalization-constant notation this isso the vertex and fermion wave-function renormalization are not independent. The remaining charge renormalization is controlled by photon wave-function renormalization; in a common convention $e_0=Z_3^{-1/2}e$.

The derivation used only an exact change of variables, invariance of the action, and invariance of the measure. It therefore holds nonperturbatively whenever the regulator and definition of the theory preserve the vector symmetry. A symmetry-breaking regulator requires symmetry-restoring counterterms; a genuine quantum anomaly would obstruct the identity, but vector QED has no such anomaly.

Substituting $\delta A_{\mu }=g^{-1}{\partial }_{\mu }\alpha -i[A_{\mu },\alpha ]$ into the curvature and collecting terms gives the covariant transformationEquivalently, $\delta F_{\mu \nu }^a=f^{abc}{F}_{\mu \nu }^b{\alpha }^c$ in the stated conventions. Since $F_{\mu \nu }^a{F}^{\mu \nu ,a}$ is proportional to $\mathrm{tr}(F_{\mu \nu }{F}^{\mu \nu })$, cyclicity of the matrix trace givesThus the Yang-Mills action is gauge-invariant.

In components the BRST transformation isThe last two equations immediately give $s^2\bar{c}=s^{2}B=0$. Applying the graded Leibniz rule to $s^{2}c$ produces a sum of three ghost monomials whose coefficient is the Jacobi identity $f^{e[ab}{f}^{c]de}$, so $s^{2}c=0$. Finally,the derivative terms cancel by anticommutation of the Faddeev-Popov ghost fields, and the remaining terms again cancel by the Jacobi identity. Hence $s^2=0$ on every field: $s$ is a nilpotent Grassmann-odd differential.

The Yang-Mills action is BRST invariant because its field strength transforms covariantly, and the gauge-fixing contribution is BRST exact. Nilpotence therefore givesThe gauge-fixing functional itself is generally not closed: $sL[A]=L[Dc]$. Applying the graded Leibniz rule givesThe Nakanishi-Lautrup field $B^a$ is auxiliary. Its algebraic equation $B^a=L[A^a]/\xi$ turns the middle terms into $L[A{]}^2/(2\xi )$, while the final term is the Faddeev-Popov ghost field action.

For $L[A^a]={\partial }^{\mu }{A}_{\mu }^a$, eliminating the Nakanishi-Lautrup field gives the covariant gauge LagrangianThe first term supplies the gluon kinetic term, three-gluon vertex, and four-gluon vertex. The second makes the quadratic gauge-field operator invertible. The last supplies the ghost propagator and ghost-antighost-gluon vertex.

The nonzero one-loop contributions to the gluon propagator are a gluon bubble with two three-gluon vertices and a closed ghost bubble with two ghost-antighost-gluon vertices. A four-gluon tadpole is also present with a cutoff regulator; for massless fields it is a scaleless integral and vanishes in dimensional regularization.

For $L[A^a]=n^{\mu }{A}_{\mu }^a$, eliminating $B^a$ gives the axial gauge term $(n\cdot A^a{)}^2/(2\xi )$ and ghost operatorIn the strict $\xi \to 0$ gauge, $n\cdot A=0$. The gauge-field-dependent part of $n\cdot D$ then vanishes, the Faddeev-Popov determinant becomes field independent, and the ghosts decouple. The one-loop gluon propagator therefore receives the gluon bubble but no ghost bubble. As in the covariant gauge, the four-gluon tadpole can occur with a hard cutoff and vanishes as a scaleless integral in dimensional regularization. Gauge-invariant observables agree between the two gauges even though their individual propagators and diagrammatic decompositions differ.