Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 301 (Quantum Field Theory)

Under $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\alpha$, the field strength is unchanged, but the mass term in the Proca action changes bywhich is not generally a total derivative. The mass therefore breaks gauge invariance.

The Euler-Lagrange field equation isTaking its divergence and using the antisymmetry of $F^{\mu \nu }$ gives $m^2{\partial }_{\nu }{A}^{\nu }=0$. Since $m>0$, the Lorenz constraint in Proca theory follows. Substitution back into the field equation then gives

The symmetric stress-energy tensor obtained by metric variation isIt is conserved on shell. With signature $(+---)$ its energy density isThe canonical stress-energy tensor differs from this symmetric tensor by the divergence of $F^{\mu \rho }{A}^{\nu }$ plus terms proportional to the field equation. Thus its integrated energy agrees after discarding the corresponding total spatial derivative, and positivity holds on shell.

For the plane wave $A^{\mu }(x)={ϵ}^{\mu }(k)e^{-ik\cdot x}$, the equations from part a becomeIn the rest frame $k^{\mu }=(m,0)$, transversality sets ${ϵ}^0=0$ while the three spatial components remain independent. Hence a Proca field has three polarization states, as required for a massive spin-one particle. A covariant orthonormal choice obeys

The quadratic momentum-space operator is inverted by the Proca propagator. With the Feynman propagator boundary prescription,Multiplication by the Fourier-space Proca operator gives $i{\delta }_{\mu }{}^{\nu }$, with the overall sign determined by the convention for the Green function.

In four spacetime dimensions the action is dimensionless, so the mass dimension of the Lagrangian density is $[\mathcal{L}]=4$. The kinetic term gives $[\varphi ]=1$, and $g{\varphi }^3$ then givesThus $g$ has mass dimension one and is a relevant coupling under power counting in quantum field theory.

The momentum-space Feynman rules for Phi cubed theory areEach vertex carries $(2\pi {)}^4{\delta }^{(4)}$ of momentum conservation, each independent loop momentum is integrated with $d^4\ell /(2\pi {)}^4$, and a graph is divided by its Feynman-diagram symmetry factor. External propagators are omitted from an amputated scattering amplitude.

Let $G_5$ be the connected time-ordered five-point function. The LSZ reduction formula gives, up to one factor $Z^{-1/2}$ per external leg,Translation invariance factors this as

A connected Feynman diagram that is a cubic tree with five external legs has three vertices and two internal lines. Its unique unlabeled topology is a chain: two external legs meet at each end vertex and one external leg attaches to the middle vertex. The five-point tree amplitude in phi cubed theory therefore has fifteen labeled tree-level Feynman diagrams: choose the middle external leg in five ways, then partition the remaining four legs into two unordered pairs in three ways.

Take all external momenta incoming, $q_1=p_1$, $q_2=p_2$, and $q_{2+r}=-k_r$. For a diagram whose middle leg is $a$ and whose two end pairs are $\{b,c\}$ and $\{d,e\}$, the Feynman rules giveEquivalently,The full connected tree amplitude is the sum over the fifteen choices described in part d, multiplied by the overall momentum-conserving delta function from part c.

Use $D_{\mu }={\partial }_{\mu }+ie{A}_{\mu }$. The two Lagrangians areandBoth are invariant under the U(1) gauge symmetryFor the constant phase subgroup, normalized to unit matter charge, the Noether currents areTheir electric currents are $e{j}^{\mu }$.

Under charge conjugation, $\varphi \leftrightarrow {\varphi }^{\ast }$ must be accompanied byThen $D_{\mu }\varphi$ is exchanged with $(D_{\mu }\varphi {)}^{\ast }$, while $F_{\mu \nu }\mapsto -F_{\mu \nu }$. The scalar kinetic and mass terms are therefore exchanged with themselves and $F_{\mu \nu }{F}^{\mu \nu }$ is unchanged, proving invariance of scalar quantum electrodynamics.

Write ${\psi }^C=\mathcal{C}{\bar{\psi }}^T$ and ${\bar{\psi }}^C={\psi }^T\mathcal{C}$. Anticommuting the spinor fields and using ${\mathcal{C}}^{-1}({\gamma }^{\mu }{)}^T\mathcal{C}=-{\gamma }^{\mu }$ givesThe free Dirac terms are invariant up to the total derivative used to move the derivative between the two anticommuting fields. The QED interaction $-e\bar{\psi }{\gamma }^{\mu }\psi A_{\mu }$ is also invariant because both the Dirac current and $A_{\mu }$ are odd. Together with $F_{\mu \nu }{F}^{\mu \nu }$ invariance, this proves charge-conjugation invariance of quantum electrodynamics.

Let $\hat{C}|\Omega ⟩=|\Omega ⟩$. Since charge conjugation sends both $A_{\mu }$ and either Noether current to its negative and commutes with time ordering,For odd $n$ the correlator equals its negative and hence vanishes. This operator argument is exact and does not use a perturbation expansion. It is Furry's theorem: scattering amplitudes with an odd number of external photons and no charged external particles vanish to every order, whereas even-photon scattering is allowed.

In the Weyl representation of the gamma matrices,A Lorentz transformation acts as ${\psi }^{\prime }(x^{\prime })=S(\Lambda )\psi (x)$ with $S(\Lambda )=\exp [-\frac{i}{4}{\omega }_{\mu \nu }{\sigma }^{\mu \nu }]$. In this basis a boost is block diagonal and gives $u_L\mapsto e^{\mathbf{\chi }\cdot \mathbf{\sigma }/2}{u}_L$ and $u_R\mapsto e^{-\mathbf{\chi }\cdot \mathbf{\sigma }/2}{u}_R$.

For boosts, the two exponentials cancel in $u_L^{†}{u}_R$; for rotations, the unitary rotation and its inverse cancel. Hence this bilinear is a Lorentz scalar. The Pauli matrices obey the identityshows that $i{\sigma }^2{u}_L^{\ast }$ acquires the right-handed boost matrix and the usual rotation matrix, so it is right-handed.

The identities $S_L^{†}{\bar{\sigma }}^{\mu }{S}_L={\Lambda }^{\mu }{}_{\nu }{\bar{\sigma }}^{\nu }$ and $S_L^T{\sigma }^2{S}_L={\sigma }^2$ show respectively that the kinetic term and both mass bilinears are Lorentz invariant. Infinitesimally, the first identity makes $u_L^{†}{\bar{\sigma }}^{\mu }{u}_L$ transform by the vector law stated in the question, while the second makes $u_L^T{\sigma }^2{u}_L$ a scalar.

Varying the anticommuting components of $u_L^{†}$ givesThe factor two from varying the antisymmetric quadratic form cancels the $1/2$ in the mass term. This is a Majorana mass term: under $u_L\mapsto e^{iq\alpha }{u}_L$ it carries charge $2q$, so a nonzero mass is compatible with an unbroken electric charge only for $q=0$. A charged massive fermion instead needs an independent Weyl field of opposite chirality to form a Dirac mass.

The Dirac field Lagrangian isChoose the Majorana spinorSubstitution shows that its two chiral kinetic terms agree after integration by parts and thatThus ${\mathcal{L}}_D[{\psi }_M]$ equals twice the displayed one-Weyl-field Lagrangian, up to a total derivative.

The lower chiral component of the Dirac equation isPutting $u_R=i{\sigma }^2{u}_L^{\ast }$ gives exactly $i{\bar{\sigma }}^{\mu }{\partial }_{\mu }{u}_L-im{\sigma }^2{u}_L^{\ast }=0$; the upper component is its complex conjugate.