Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 301 (Quantum Field Theory)

For a Schrödinger-picture state satisfyingdefine the interaction picture byDifferentiation givesThe formal solution from $t_0$ is the Dyson seriesThrough quadratic order,Differentiating giveswhich checks the equation to the requested order.

At the endpoints, $|i{⟩}_I=e^{-i{H}_{0}T/2}|i{⟩}_S$ and $|f{⟩}_I=e^{i{H}_{0}T/2}|f{⟩}_S$. ThereforewhereFor ${\varphi }_1(p_1)\to {\varphi }_2(p_1^{\prime }){\varphi }_2(p_2^{\prime }){\varphi }_2(p_3^{\prime })$, the leading term isThere are $3!$ contractions of the identical outgoing fields. With covariantly normalized states,Equivalently, the interaction vertex from $-\lambda {\varphi }_1{\varphi }_2^3$ is $-i3!\lambda$.

Suppose an infinitesimal continuous transformation has variations $\delta \psi ,\delta \bar{\psi }$ and changes the Lagrangian density by $\delta \mathcal{L}={\partial }_{\mu }{K}^{\mu }$. Expanding by the chain rule and integrating derivatives by parts givesOn solutions of the Euler-Lagrange equations, the first two brackets vanish. Hence Noether's theorem gives the conserved currentFor spacetime transformations, the coordinate variation supplies the corresponding energy-momentum term.

The free massless Dirac field has actionUnder a finite dilation,The measure, derivative, and fields contribute ${\lambda }^4$, ${\lambda }^{-1}$, and ${\lambda }^{-2\Delta }$, so invariance requiresFor $\lambda =e^{\alpha }$, the infinitesimal active variations areThe associated dilatation current isup to an improvement term. Its divergence is the on-shell trace $T^{\mu }{}_{\mu }=0$.

The matrix ${\gamma }^5=-i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$ anticommutes with every Dirac matrix. Moving ${\gamma }^{\mu }$ through each power in the exponential series givesUnder the chiral transformationHermiticity of ${\gamma }^5$ and its anticommutation with ${\gamma }^0$ implyConsequentlyso the action is invariant. The corresponding classical axial current is

For metric signature $(+---)$, the momentum-space rules for the Yukawa interaction are:

At leading order the process has one $s$-channel scalar propagator. With $s=(p+q{)}^2$,Thus, up to an irrelevant overall sign,The fermion spin sums and $\mathrm{tr}(/a/b)=4a\cdot b$ giveand the analogous final sum is $4(p^{\prime }\cdot q^{\prime }-m_{\chi }^2)$. ThereforeUsing $p\cdot q=(s-2m_{\psi }^2)/2$ and its final-state analogue yields

In Feynman gauge, the Lagrangian differs by a total divergence fromIts Euler-Lagrange equation is $\square A^{\mu }=0$. Directly from the original gauge-fixed Lagrangian, the canonical momenta areequivalently ${\pi }^{\mu }=-{\partial }^0{A}^{\mu }$ after using the total-divergence form.

Using the mode expansions and oscillator commutator, the two nonzero cross terms combine towhere polarization completeness was used. The other equal-time field commutators vanish.

Substitution into the Hamiltonian and normal ordering giveEvery oscillator excitation carries positive energy $|\mathbf{p}|$, but the covariant state space has an indefinite inner product: time-like excitations have negative norm, and time-like and longitudinal polarizations are unphysical. Gupta-Bleuler quantization imposesand quotients by null states. The physical state space contains only the two transverse photon polarizations with positive norm, as required by gauge invariance.