Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 301 (Quantum Field Theory)

For an infinitesimal field variation $\delta \varphi$ such that $\delta \mathcal{L}={\partial }_{\mu }{K}^{\mu }$, integration by parts givesOn the Euler-Lagrange equations, Noether's theorem therefore givesFor transformations of spacetime, the variation at fixed coordinates and the change of the integration measure give the corresponding energy-momentum tensor terms.

Translations produce the conserved symmetric tensorThe conserved charges areandIndeed,when the field and its derivatives decay sufficiently rapidly.

Lorentz invariance gives the Lorentz currentAn infinitesimal rotation through angle $\theta$ around the $z$-axis hascorresponding, up to the stated index convention, to the only nonzero components ${\omega }^{12}=-{\omega }^{21}=\theta$. Its conserved angular momentum is

Scale invariance requires the potential to obeyThus, apart from the zero choice,for a dimensionless constant $g$; an additive constant is excluded because it introduces a scale in the action. The canonical dilatation current isUsing $\square \varphi +V^{\prime }(\varphi )=0$,Equivalently, after improving the stress tensor to make it traceless, the current is $x_{\nu }{T}_{\mathrm{imp}}^{\mu \nu }$.

The interaction-picture Hamiltonian density is ${\mathcal{H}}_I=\lambda \varphi {\Phi }^{†}\Phi$. The first-order Dyson series term cannot connect the four external particles, so the leading contribution isBy the Wick theorem, the nonzero terms annihilate the incoming complex particle using the annihilation part of $\Phi$, create the outgoing complex particle using the creation part of ${\Phi }^{†}$, annihilate the incoming real particle using the annihilation part of $\varphi$, and create the outgoing real particle using its creation part. The remaining $\Phi$ and ${\Phi }^{†}$ form an internal Feynman propagator. Interchanging which vertex absorbs the incoming real scalar gives the two contractions; this factor of two cancels the Dyson factor $1/2$.

There are therefore an $s$-channel internal momentum $p+k$ and a crossed channel internal momentum $p-k^{\prime }$. With covariantly normalized external states,whereEquivalently,with $s=(p+k{)}^2$ and $u=(p-k^{\prime }{)}^2$.

Define the spinor generatorsRepeated use of the Clifford algebra givesThe Spinor representation of the Lorentz group isThe adjoint relation for the Dirac matrices impliesExponentiating yields

For $x^{\prime }=\Lambda x$, a Dirac spinor transforms asHenceUsing the result of part i, the Dirac adjoint transforms as

The Dirac action isTo first order,The commutator found in part i givesTogether with $\det \Lambda =1$, the transformations in part ii therefore leave both the kinetic and mass terms invariant through first order. Thus $S_D^{\prime }=S_D+O({\Omega }^2)$.

Under parity, $\overline{\psi }(t,\mathbf{x})\mapsto \overline{\psi }(t,-\mathbf{x}){\gamma }^0$. Since
${\gamma }^0{\gamma }^0{\gamma }^0={\gamma }^0$ and
${\gamma }^0{\gamma }^i{\gamma }^0=-{\gamma }^i$,Moreover ${\gamma }^0{\gamma }^5{\gamma }^0=-{\gamma }^5$, so the axial current is a pseudovector:

In Feynman gauge, use the mode expansionwith $k^0=|\mathbf{k}|$ and the covariant polarization completeness relation. For $x^0>y^0$, time ordering retains the annihilation-creation contraction and gives the positive-frequency term; for $x^0<y^0$, it gives the negative-frequency term. Combining them by a $k^0$ contour integral givesThe Feynman i-epsilon prescription places the positive-energy pole $k^0=|\mathbf{k}|-iϵ$ below the real axis and the negative-energy pole $k^0=-|\mathbf{k}|+iϵ$ above it. Equivalently the denominator is $k^2+iϵ$ with the contour along the real axis.

The free complex field has the global unitary group $U(1)$ symmetryGauging it permits $\alpha =\alpha (x)$ and introduces a gauge field withThe scalar quantum electrodynamics Lagrangian isIts interaction terms are

With momentum flowing along the scalar arrow, the momentum-space Feynman rules are: