Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 304 (Advanced Quantum Field Theory)

With Euclidean source convention $+\int d^{d}x{J}_a{\varphi }_a$, the generating functional isHere $J_a(x)$ is a nondynamical source. Its functional derivatives insert fields:This is why $Z$ “generates” correlation functions.

With the notation of the question, $W[J]=-\log Z[J]$ is the Euclidean connected generating functional; it is distinct from a Wilsonian effective action. Define the classical fieldThe quantum effective action is the Legendre transformwhere $J$ is eliminated in favor of $\Phi$. Its first derivative isso at zero source the quantum expectation value is a stationary point of $\Gamma$. Moreover,where $G_{ab}=⟨{\varphi }_a{\varphi }_b{⟩}_{J,\mathrm{conn}}$ is the exact connected correlation function. Thus the second derivative of $\Gamma$ is the inverse exact propagator.

Perturbatively, $Z[J]$ sums all Feynman diagrams, including disconnected ones. The exponential formula for combinatorial structures says that its logarithm selects connected Feynman diagrams, so $W$ sums connected diagrams with the sign dictated by the convention above. The Legendre transform removes diagrams that disconnect when one internal line is cut; consequently $\Gamma$ sums one-particle-irreducible Feynman diagrams. Equivalently, every connected diagram is a tree assembled from one-particle-irreducible vertices and full propagators, and the Legendre transform inverts that tree construction.

Now let $J_a^{\prime }=J_b{U}_{ba}$. ThenChanging variables to ${\varphi }_b^{\prime }=U_{ba}{\varphi }_a$ and using invariance of both the action and functional measure gives $Z[J^{\prime }]=Z[J]$, hence $W[J^{\prime }]=W[J]$. In the Legendre transform, the pairing obeysChanging the source variable and using the invariance of $W$ therefore givesThe symmetry of the classical action and measure is inherited by the full quantum effective action when it has no quantum anomaly.

For incoming momenta, the Euclidean momentum-space Feynman rules are

The one-loop one-particle-irreducible diagrams with at most three external legs can be classified by their external species. There is a one-point $\chi$ tadpole diagram made from a $\varphi$ loop. There are two two-point bubble diagrams: the $\varphi$ self-energy has one internal $\varphi$ and one internal $\chi$, while the $\chi$ self-energy has two internal $\varphi$ lines and symmetry factor $1/2$. For three external legs, three vertices make a triangle: one triangle corrects the $\varphi \varphi \chi$ vertex and contains two internal $\varphi$ lines and one internal $\chi$ line; another has three external $\chi$ legs and a $\varphi$ loop, generating a ${\chi }^3$ interaction. The symmetry $\varphi \mapsto -\varphi$ forbids amplitudes with an odd number of external $\varphi$ legs. The classical $\varphi \varphi \chi$ vertex is the corresponding tree-level three-point diagram.

Adopt the self-energy conventionThis follows by summing the geometric series of exact propagators separated by amputated one-particle-irreducible two-point insertions. At one loop, after writing ${\lambda }_0={\mu }^{ϵ/2}\lambda +O({\lambda }^3)$,Introduce a Feynman parameter and shift the loop momentum. Withdimensional regularization givesUsing $\Gamma (-1+ϵ/2)=-2/ϵ+\gamma -1+O(ϵ)$ produceswhereand one convenient integral form of the finite part isChanging the definition of the dimensional-regularization scale only moves a finite constant between $C$ and the counterterm.

To make the two-point function finite, write ${\varphi }_0=Z_{\varphi }^{1/2}\varphi$, express $m_0^2$ in terms of a renormalized mass and a mass counterterm, and choose the pole parts of $\delta Z_{\varphi }$ and $\delta m^2$ to cancel $B{p}^2/ϵ$ and $A/ϵ$. In the minimal subtraction scheme no additional finite pieces are removed. The physical mass is the pole mass, so with the self-energy convention above it obeysThe explicit $\mu$ dependence of the finite self-energy cancels the running of $m(\mu )$, leaving $m_{\mathrm{phys}}$ independent of the renormalization scale.

The superficial degree of divergence counts the ultraviolet power before subdivergences and symmetry cancellations are considered. At $d=6$, a connected graph made from cubic vertices haswhere $E$ is its number of external legs. Hence one-, two-, and three-point functions can have quartic, quadratic, and logarithmic superficial divergences, whereas graphs with more external legs are superficially convergent.

Full renormalization also requires the $\chi$ mass and wave-function counterterms from its two-point function, a linear $\chi$ counterterm cancelling the tadpole, a coupling counterterm from the divergent $\varphi \varphi \chi$ triangle, and a ${\chi }^3$ counterterm from the three-$\chi$ triangle. A vacuum-energy counterterm removes divergent vacuum diagrams. These are precisely the local operators allowed by power counting in quantum field theory and the exact $\varphi \mapsto -\varphi$ symmetry.

Renormalization introduces an arbitrary renormalization scale $\mu$ even though the classical massless theory has no dimensionful parameter. Independence of the bare correlation function from that auxiliary scale gives the Callan-Symanzik equationHereis the beta function and $\gamma (g)$ is the field anomalous dimension, with its sign fixed by the displayed equation. The beta function determines the running coupling. Its zeros are renormalization-group fixed points, where the theory can become scale invariant. A positive beta function makes a positive coupling increase toward larger $\mu$, while a negative one makes it decrease.

For the propagator coefficient $C$, follow a characteristic with $t=\log (\mu /{\mu }_0)$ and $dg/dt=\beta (g)$. The $n=2$ equation becomesIntegration giveswith

For $\beta (g)=-b{g}^3$ with $b>0$, the only real fixed point is $g^{\ast }=0$. It is ultraviolet-attractive: the theory is asymptotically free. Integrating the running equation givesorfor the branch with the same sign as $g({\mu }_0)$. The perturbative expression has an infrared Landau pole atand therefore

Finally set ${\mu }_0=p$ and use $\gamma (g)=c{g}^2$. The characteristic factor becomesIn terms of the dynamically generated scale,

Write $U(x)=e^{i{\alpha }^a(x)T^a}$ and $A_{\mu }=A_{\mu }^a{T}^a$. With the conventionthe gauge field must transform asThen $D_{\mu }^{\prime }{\psi }^{\prime }=U{D}_{\mu }\psi$. The conjugate field has $D_{\mu }\bar{\psi }={\partial }_{\mu }\bar{\psi }+ig\bar{\psi }{A}_{\mu }$. For the adjoint scalar,or $(D_{\mu }\varphi {)}^a={\partial }_{\mu }{\varphi }^a+g{ϵ}^{abc}{A}_{\mu }^b{\varphi }^c$. It transforms as $D_{\mu }^{\prime }{\varphi }^{\prime }=U(D_{\mu }\varphi )U^{-1}$. The gauge field strengthsimilarly transforms by conjugation.

Up to Euclidean sign conventions, the parity-even renormalizable Lagrangian isThe Yang-Mills term is invariant because $F_{\mu \nu }$ transforms by conjugation and the matrix trace is cyclic. Covariance of $D_{\mu }\psi$ makes $\bar{\psi }{\gamma }_{\mu }{D}_{\mu }\psi$ invariant, and the fermion mass is invariant because the factors $U^{-1}U$ cancel. Likewise $D_{\mu }\varphi$ transforms by conjugation, so its trace norm, ${\varphi }^a{\varphi }^a=2\mathrm{tr}{\varphi }^2$, and every power of that norm are invariant. Finally,which proves gauge invariance of the Yukawa interaction.

Gauge invariance and power counting also permit the parity-odd Yukawa interaction $i{y}_5\bar{\psi }{\gamma }_5\varphi \psi$, a pseudoscalar fermion mass $i{m}_5\bar{\psi }{\gamma }_5\psi$, and the Yang-Mills theta term. They are absent if parity and CP are imposed. There is no nonzero cubic scalar invariant: $\mathrm{tr}{\varphi }^3$ vanishes for $SU(2)$, equivalently ${ϵ}^{abc}{\varphi }^a{\varphi }^b{\varphi }^c=0$ for commuting scalar components.

The quadratic Yang-Mills operator has zero directions along gauge orbits, so it has no propagator until one chooses a gauge fixing. The Faddeev-Popov determinant generated by this choice is represented by anticommuting Faddeev-Popov ghost fields, which cancel unphysical gauge-field contributions in loop calculations. A Nakanishi-Lautrup field imposes the gauge condition algebraically and lets the gauge-fixing plus ghost action be written as a BRST-exact term with off-shell nilpotency.

Remove the common Grassmann parameter and write the BRST transformation as the odd derivation $s$:The last two equations immediately give $s^2{\bar{c}}^a=s^2{B}^a=0$. Applying $s$ to the ghost and using the graded product rule givesThe cancellation is the Jacobi identity for the $SU(2)$ structure constants together with anticommutation of the ghost fields. For the gauge field, the component calculation isThe graded product rule givesso the two terms cancel. Thus $s^2$ vanishes on every elementary field.

For two independent Grassmann parameters, ${\delta }_1={\eta }_{1}s$ and ${\delta }_2={\eta }_{2}s$ satisfy ${\delta }_1{\delta }_{2}O={\eta }_1{\eta }_2{s}^{2}O=0$. Since $s$ obeys the graded Leibniz rule, induction extends $s^{2}O=0$ from the generators $A,c,\bar{c},B$ to every polynomial $O(A,c,\bar{c},B)$. Hence the BRST transformations are nilpotent.