Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 301 (Quantum Field Theory)

In four dimensions, $[\psi ]=3/2$, $[A_{\mu }]=1$, and $[F_{\mu \nu }]=2$. The minimal-coupling operator $\bar{\psi }{\gamma }^{\mu }{A}_{\mu }\psi$ has dimension four, so $[e]=0$ and $e$ is a marginal coupling. The Pauli term $\bar{\psi }[{\gamma }^{\mu },{\gamma }^{\nu }]F_{\mu \nu }\psi$ has dimension five, so $[\lambda ]=-1$ and $\lambda$ is an irrelevant coupling by power counting in quantum field theory.

For $x^{\prime }=\Lambda x$, a Dirac spinor and vector field transform aswhere $S^{-1}{\gamma }^{\mu }S={\Lambda }^{\mu }{}_{\nu }{\gamma }^{\nu }$. Hence the spinor terms and $F_{\mu \nu }{F}^{\mu \nu }$ are Lorentz scalars. The Maxwell term is real. Integration by parts and $({\gamma }^{\mu }{)}^{†}={\gamma }^0{\gamma }^{\mu }{\gamma }^0$ show that the adjoint of $i\bar{\psi }{\gamma }^{\mu }{D}_{\mu }\psi$ differs from it by $-i{\partial }_{\mu }(\bar{\psi }{\gamma }^{\mu }\psi )$. Finally, $i[{\gamma }^{\mu },{\gamma }^{\nu }]$ is Dirac-Hermitian, so the real coefficient $\lambda$ makes the Pauli bilinear real. Thus the action is real.

Varying $\bar{\psi }$ and $\psi$ independently givesand its adjointwhere $D_{\mu }\bar{\psi }={\partial }_{\mu }\bar{\psi }-ie{A}_{\mu }\bar{\psi }$. Varying $A_{\nu }$ and integrating the Pauli term by parts gives the modified Maxwell equations

The local U(1) gauge symmetry isThe global phase subgroup has Noether current and chargeMultiplying the fermion equation by $\bar{\psi }$, its adjoint by $\psi$, and subtracting shows ${\partial }_{\mu }{j}^{\mu }=0$ on shell. Therefore $\dot{Q}=-\int d^{3}x\nabla \cdot \mathbf{j}=0$ under vanishing boundary flux.

The source in Maxwell's equation isThe second term is an identically conserved magnetization current, because it is the divergence of an antisymmetric tensor. It changes the local source but contributes only a boundary term to the total charge.

For the chiral transformation $\psi \mapsto e^{i\alpha {\gamma }^5}\psi$, one also has $\bar{\psi }\mapsto \bar{\psi }{e}^{i\alpha {\gamma }^5}$. The massless kinetic and vector-current terms are invariant because $\{{\gamma }^5,{\gamma }^{\mu }\}=0$. Since ${\gamma }^5$ commutes with $[{\gamma }^{\mu },{\gamma }^{\nu }]$, the Pauli bilinear transforms with $e^{2i\alpha {\gamma }^5}$ and is not invariant. Thus the continuous classical axial symmetry requires $\lambda =0$.

A fermion mass also breaks the symmetry, so in the massive theory one additionally needs $m=0$, which contradicts a genuinely massive fermion. Making $\alpha$ local produces $({\partial }_{\mu }\alpha )\bar{\psi }{\gamma }^{\mu }{\gamma }^5\psi$. No choice of the vector coupling $e$ or Pauli coupling $\lambda$ cancels this term; gauging it requires an axial gauge field, and at the quantum level one must also address the chiral anomaly.

Integrating by parts givesbecause the $b$ and $c$ terms differ only by a total derivative. Hence

The differential operators are self-adjoint under integration by parts, so the Euler-Lagrange field equations are

Using the original first-derivative Lagrangian, its momentum current isTranslation invariance and Noether theorem give the canonical stress-energy tensoron shell. The Hamiltonian is

When $\tilde{b}=0$, integration by parts makes the kinetic term $-\tilde{a}{\partial }_{\mu }{V}_{\nu }{\partial }^{\mu }{V}^{\nu }$. The Lorentzian contraction over the field index gives the time component and the three spatial components opposite kinetic-energy signs. Reversing the sign of $\tilde{a}$ merely exchanges which component is a ghost field; no nonzero choice makes all four energies positive. At $\tilde{a}=0$ there is no healthy kinetic term. Thus the Hamiltonian is never positive definite.

Positivity selectsfor which the kinetic terms combine, up to normalization and a total derivative, into the Proca field form $-F_{\mu \nu }{F}^{\mu \nu }/4$. Taking the divergence of the equation of motion givesFor the selected relation and $m\ne 0$, this implies the transverse vector field condition ${\partial }_{\mu }{V}^{\mu }=0$. The remaining three massive polarizations have positive on-shell energy when $\tilde{a}>0$.

Put $\tilde{c}=\tilde{a}+\tilde{b}$. Substituting the transverse-longitudinal decomposition $V_{\mu }=A_{\mu }+{\partial }_{\mu }\pi$ and integrating cross terms by parts givesThe equations aretogether with ${\partial }_{\mu }{A}^{\mu }=0$.

The scalar kinetic operator is proportional in momentum space to $k^2(\tilde{c}{k}^2-m^2)$. Its inverse has the partial-fraction decompositionup to the overall convention-dependent factor allowed in the question. Thus $\tilde{c}=\tilde{a}+\tilde{b}$. Under the physical relation $\tilde{b}=-\tilde{a}$, one has $\tilde{c}=0$, so the extra massive pole and its ghost residue disappear.

The classical equations areFor any time-ordered functional $\mathcal{O}$, the three Schwinger-Dyson equations state that insertion of each left-hand side equals the corresponding contact terms obtained by $-i$ times the functional derivative of $\mathcal{O}$ with respect to $phi$, $\bar{\psi }$, or $\psi$, with the usual reversed order and Grassmann signs for fermionic derivatives.

Define asymptotic states $|\varphi (k)⟩=a_k^{†}|0⟩$ and $|\psi (p,s)\bar{\psi }(q,s^{\prime })⟩=b_{p,s}^{†}{c}_{q,s^{\prime }}^{†}|0⟩$. The LSZ reduction formula reduces the decay matrix element from the connected three-point functionIn momentum space, amputate the scalar leg with $i(k^2-M^2)$ and the fermion legs with the inverse Dirac propagators, contract them with ${\bar{u}}^s(p)$ and $v^{s^{\prime }}(q)$, and take $k^2=M^2$, $p^2=q^2=m^2$. The spacetime integrations produce $(2\pi {)}^4{\delta }^{(4)}(k-p-q)$.

Iterating the Schwinger-Dyson equations once, the leading connected correlator is one scalar propagator and two Dirac propagators meeting at one Yukawa interaction vertex. Amputation givesThus the invariant amplitude, in the convention $S_{fi}=i(2\pi {)}^4{\delta }^{(4)}{\mathcal{M}}_{fi}$, is ${\mathcal{M}}_{s,s^{\prime }}=-\lambda {\bar{u}}^s(p)v^{s^{\prime }}(q)$.

The spin sum becomes a gamma-matrix trace:Since $(p+q{)}^2=M^2$, $p\cdot q=(M^2-2m^2)/2$. Therefore the normalization requested in the paper gives

The pseudoscalar vertex replaces $-i\lambda$ by $g{\gamma }^5$, soUsing ${\gamma }^5/q{\gamma }^5=-/q$ with the conjugation sign givesThe interactions are experimentally distinguishable through the decay rate and its threshold behavior: scalar decay is proportional to $(1-4m^2/M^2{)}^{3/2}$, whereas pseudoscalar decay is proportional to $(1-4m^2/M^2{)}^{1/2}$. Spin and parity correlations provide further discrimination.

Substitution of $\Phi =\phi +\lambda {\phi }^2$ givesand

With all momenta incoming, the Feynman rules areandEach vertex includes its momentum-conserving delta function. The combinatorial factors follow by assigning the identical fields to the differentiated and undifferentiated slots in every possible way.

At order ${\lambda }^2$, four connected tree-level Feynman diagrams contribute: one four-point contact diagram and three diagrams with two cubic vertices joined by a scalar propagator in the $s$, $t$, and $u$ channels.

On shell, the contact vertex is $4i{\lambda }^2{m}^2$. At a cubic vertex with two external on-shell legs and internal momentum $k$, the rule reduces to $2i\lambda (k^2-m^2)$. The propagator cancels one such factor, so the three exchange diagrams sum toFor equal-mass two-to-two scattering, $s+t+u=4m^2$, and the exchange sum is $-4i{\lambda }^2{m}^2$. It cancels the contact diagram exactly. Thus the connected on-shell amplitude vanishes, as required by the equivalence theorem for field redefinitions: an invertible local field redefinition cannot turn a free theory into a physically interacting one.