Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 51 (Advanced Quantum Field Theory)

Use units $\hslash =1$. The fluctuations about a classical path satisfy $f(0)=f(T)=0$. Expanding the action functional and integrating the term $m{\dot{q}}_c\dot{f}$ by parts givesThe linear term vanishes by the Euler-Lagrange equation and the endpoint conditions. Integrating the quadratic kinetic term by parts yields the Dirichlet fluctuation determinant of a semiclassical propagator:A finite time-slicing first turns this into a product of ordinary Gaussian integrals, whose limit is proportional to $(\det {\mathcal{O}}_c{)}^{-1/2}$. A useful normalization relative to the free operator ${\mathcal{O}}_0=-m{d}^2/d{t}^2$ isThe oscillatory square root is fixed by the short-time limit and the Feynman i-epsilon prescription. Away from zero modes this gives the semiclassical propagator; if several classical paths contribute, their saddle contributions are summed with their corresponding determinant phases.

For a general potential the prefactor depends on the endpoints through $q_c$; writing only $D(T)$ suppresses that dependence. If $V(q)=a{q}^2/2+bq+c$, the Taylor expansion has no terms beyond quadratic and $V^{\prime \prime }=a$ is independent of $q_c$. The Gaussian functional integral is then exact, and its prefactor genuinely depends only on $T$ and the potential parameters. Singular focusing times are understood by continuation of the kernel, rather than by assigning a finite value to a zero-mode determinant.

For the free-particle propagator, $q_c(t)=q_0+(q_1-q_0)t/T$ and $S_c=m(q_1-q_0{)}^2/(2T)$. The free determinant supplies the factor $\sqrt{m/(2\pi iT)}$. Its normalization can also be checked by the momentum-space representationwhere completing the square evaluates the Gaussian integral and the short-time limit gives the position-space delta function.

For the gravitational potential, the original PDF contains $-mgq$; the TeX aid drops the final $q$. The Euler-Lagrange equation is ${\ddot{q}}_c=g$, and its endpoint solution isSubstitution into the kinetic and potential terms givesSince $V^{\prime \prime }=0$, the fluctuation operator is the free one. Thus the quantum-mechanical propagator in a constant force is exactly

The quantum-mechanical propagator is $⟨q_1|e^{-iHT}|q_0⟩$, with Hamiltonian operator $H=-{\partial }_q^2/(2m)+V(q)$. For imaginary time $-i\tau$, insert an orthonormal complete set of energy eigenstates to obtainHere ${\psi }_n(q)=⟨q|n⟩$. This discrete sum assumes a discrete complete spectrum; continuous spectral components require the corresponding spectral integral. The heat-semigroup expression is well defined for an appropriate self-adjoint Hamiltonian bounded below. It is this spectral condition, not an unrestricted formal rotation for every conceivable potential, that justifies the imaginary-time kernel.

A path on the circle lifts to a path on the real line ending at $q_1+2\pi w$, where the integer $w$ is its winding number. The periodic quantum theory sums all these winding sectors with equal weight, giving the real-time winding representation of a quantum rotor kernel. For imaginary time, put $x=q_1-q_0$ and $y=\tau /m$ in the supplied Gaussian summation identity:Comparison with the spectral expansion determinesThe normalization uses $0\le q<2\pi$. The ground state is nondegenerate; levels with $n\ne 0$ have the degeneracy $n\leftrightarrow -n$. Returning to real time replaces $e^{-\tau E_n}$ by $e^{-iT{E}_n}$, with the usual limiting prescription. No twisted boundary condition or flux phase is present in the stated image sum.

Write $\epsilon ={\epsilon }_a{t}_a$ and $A_{\mu }=A_{\mu a}{t}_a$. The proposed gauge-field transformation law becomes $\delta A_{\mu }=-{\partial }_{\mu }\epsilon +[\epsilon ,A_{\mu }]$. ConsequentlyBoth $\varphi$ and its gauge covariant derivative therefore transform as the same multiplet, with no derivative of the local parameter left over. The original algebraic invariance of the Lagrangian density now applies pointwise to $\mathcal{L}(\varphi ,D\varphi )$, so this constructs the local symmetry.

Also transform $J$ by $\delta J=\epsilon J$. Antisymmetry gives $(t_{a}J)\cdot \varphi +J\cdot (t_a\varphi )=0$, so the source coupling is invariant. Let $S_{A_{\mu a}}$ denote the functional derivative at a point. Vary the action functional and integrate the derivative of ${\epsilon }_a$ by parts, taking the local parameter to have compact support:Since every ${\epsilon }_a(x)$ is arbitrary, its coefficient vanishes. This proves the requested local identity, including the derivative sign and the order of the structure-constant indices.

For the generating functional, the integral of the first term is zero as a change-of-variable identity:The divergence of this vector field in field space is proportional to $\mathrm{tr}{t}_a=0$. Thus an invariant regulated functional measure, with no gauge anomaly, leaves no Jacobian term. Inserting the classical identity into the integral givesThe same first-order differential operator annihilates the connected generating functional $W$, because $Z=e^{iW}$. Its source term is $(t_{a}J)\cdot \phi$.

For the specified Legendre transform, vary $\Gamma =-W+\int J\phi$. Since $W_J=\phi$, the terms involving $\delta J$ cancel, leavingAntisymmetry also gives $(t_{a}J)\cdot \phi =-J\cdot (t_a\phi )$. Substitution in the identity for $W$, followed by multiplication by $-1$, proves the background gauge invariance of a scalar effective action:

The derivatives of the quantum effective action are proper, amputated one-particle-irreducible vertices. At each loop order they receive contributions from one-particle-irreducible Feynman diagrams with the indicated external scalar legs, including tree vertices and the required counterterms. Diagrams which disconnect after cutting one internal line are not additional proper-vertex contributions. Translation invariance supplies the overall momentum-conserving delta function factored out in the definition. The signs of these vertices must follow the chosen overall convention for $\Gamma$.

Differentiate the effective-action identity twice with respect to ${\phi }_i(y)$ and ${\phi }_j(z)$ and set $\phi =A=0$. In an invariant zero-field vacuum this gives the Ward identity contact termsThe term proportional to $A$ vanishes. With the printed Fourier factors $e^{i{p}_r{x}_r}$, integration by parts turns ${\partial }_{\mu }$ into $-i{p}_{1\mu }$. The first delta function combines $p_1$ with $p_2$, and the second combines $p_1$ with $p_3$. Using $(t_a{)}_{ki}=-(t_a{)}_{ik}$ and symmetry of the scalar two-point kernel then yieldsThis derivation fixes the relative signs independently of any vertex rule.

There is an overall-sign inconsistency in the final printed example. At tree level, stationarity of the source integral gives $W=S_0[\phi ,A]+\int J\phi$, hence the specified transform gives ${\Gamma }_{\mathrm{tree}}=-S_0$. For the free scalar multiplet this implies, after factoring out the delta function,The mixed result follows by expanding $-S_0$ to first order in $A$: its contribution is $\int A_{\mu a}({\partial }^{\mu }{\phi }_i)(t_a{)}_{ij}{\phi }_j$. The two scalar derivatives yield $i{p}_2^{\mu }(t_a{)}_{ij}+i{p}_3^{\mu }(t_a{)}_{ji}$. Momentum conservation givesso both sides of the Ward identity equal $i(p_3^2-p_2^2)(t_a{)}_{ij}$. Because the scalar theory is Gaussian for prescribed $A$, its fluctuation determinant depends on $A$ but not on $\phi$; it cannot add further two-scalar vertices. Thus this quadratic verification is sufficient, not merely the first term of an omitted scalar-loop correction.

The printed mixed vertex instead has a minus sign. It satisfies the same Ward identity if one uses the opposite effective-action convention ${\Gamma }_{\mathrm{alt}}=W-\int J\phi =S_0$ at tree level, because its two-point vertex is then $-(p^2+m^2){\delta }_{ij}$ as well. Either overall convention works, but the printed Legendre transform and printed mixed vertex cannot both be used unchanged. For example, with $p_2^2\ne p_3^2$ and a nonzero generator, they give opposite sides of the claimed identity.

A renormalizable quantum field theory requires only finitely many independent field and parameter redefinitions to remove regulator dependence from its perturbative correlation functions, order by order. The allowed local counterterms must close within that finite set; it is not necessary that the unrenormalized integrals be finite.

In four spacetime dimensions the kinetic term fixes the mass dimension of $\varphi$ to one. A coefficient multiplying ${\varphi }^k$ therefore has dimension $4-k$. Power counting excludes couplings of negative mass dimension for perturbative renormalizability. For a single scalar with a conventional nonderivative polynomial potential, the allowed form isThe constant controls vacuum energy and the linear term a possible tadpole. A field-reflection symmetry can remove the odd terms but is not necessary for renormalizability. Terms of degree above four generally require an infinite tower of counterterms when treated as fundamental interactions; they can instead be used in an effective field theory. Stability of the potential is a separate requirement, for example a positive quartic leading term, and should not be confused with the power-counting criterion.

For a graph with $E$ external scalar lines, $I$ internal lines, $L$ loops and $V_k$ vertices of degree $k$, the superficial ultraviolet divergence degree is $\omega =4L-2I$. The identities $L=I-{\sum }_k{V}_k+1$ and $2I+E={\sum }_{k}k{V}_k$ giveWhen every $k\le 4$, only the finitely many low-point local divergences, including the kinetic term, can require subtraction. Subdivergences are removed recursively by the same local counterterms. A vertex with $k>4$ instead permits arbitrarily high divergence degrees as more such vertices are added. This explains the restriction rather than merely asserting it.

The bare Lagrangian density is written in terms of regulator-dependent bare quantities, before taking the regulator away:The coefficients in $V_B$ are related to the renormalized parameters by regulator-dependent redefinitions. The wave-function renormalization supplies a kinetic counterterm. In dimensional regularization $d=4-ϵ$, a quartic coupling has a bare relation of the form ${\lambda }_B={\mu }^{ϵ}[\lambda +\delta \lambda (\lambda ,ϵ)]$, with field factors incorporated according to the chosen definition. The bare parameters and bare correlation functions are held fixed when changing the renormalization scale.

Even without a physical mass, interacting loop subtractions require a reference scale $\mu$ to define the renormalized dimensionless coupling and field normalization. Finite logarithms then involve dimensionless combinations such as $\mu |x_r-x_s|$ or $p^2/{\mu }^2$. This scale is a subtraction convention, not a new physical mass parameter. Its explicit dependence is compensated by the running coupling and field normalization.

Let $G_n=⟨\varphi (x_1)\cdots \varphi (x_n)⟩$ at separated points. Its bare counterpart satisfies $G_{B,n}=Z_{\varphi }^{n/2}{G}_n$. DefineDifferentiating $G_{B,n}$ at fixed bare theory givesThis Callan-Symanzik equation expresses independence of the arbitrary subtraction scale. The renormalization-group beta function describes the compensating coupling change; the field-renormalization anomalous dimension describes the compensating rescaling of each insertion. Coincident composite insertions would require their own additional renormalization, rather than automatically obeying this separated-point equation.

For the two-point function, set $z=p^2/{\mu }^2$ and $t=\frac{1}{2}\log z$. We consider large positive spacelike $p^2$, with continuation to other momenta as required. The dimensionless propagator factor obeysDefine $\bar{g}(0)=g$ and $d\bar{g}/dt=\beta (\bar{g})$. The renormalization-group characteristic solution for a two-point function isTo verify it, the running-coupling flow has $\beta (g){\partial }_g\bar{g}(t)=\beta (\bar{g}(t))$. Also, $\beta (g){\partial }_g{\int }_0^t\gamma (\bar{g}(s))ds=\gamma (\bar{g}(t))-\gamma (g)$. Applying ${\partial }_t-\beta (g){\partial }_g$ to the displayed expression therefore gives $2\gamma (g)d$, exactly the required differential equation. The boundary value $d(1,g)$ records the renormalization condition. This shows how the ultraviolet behaviour is determined by the coupling trajectory and its accumulated anomalous dimension: it can approach zero coupling, a nonzero fixed point, or a finite-scale singularity, depending on $\beta$.

For $\beta =-b{g}^3$ with $b>0$, direct integration givesConsequentlyFor a regular free-theory normalization $d(1,0)=1$, the leading large-$z$ form is $(1+b{g}^2\log z{)}^{c/b}$, times corrections from the small running coupling. This is asymptotic freedom with a logarithmic, rather than a fixed-power, propagator correction. The hypothetical beta and gamma functions are being used as given; this calculation does not assert that they are the actual beta and gamma functions of an arbitrary quartic scalar theory.

For the second beta function put $B=-b>0$, so $\beta (g)=B{g}^3-a{g}^5$. Its positive nonzero fixed point and linearized slope areFor $0<g<g_{\ast }$ the coupling increases with momentum, while for $g>g_{\ast }$ it decreases; each positive trajectory approaches $g_{\ast }$ in the ultraviolet. Negative initial couplings approach $-g_{\ast }$ by the odd symmetry of the beta function, with the same limiting squared coupling. The exactly zero trajectory remains zero. Thus the nonzero ultraviolet fixed point of a cubic-quintic beta function replaces asymptotic freedom by a finite-coupling ultraviolet limit, with $\bar{g}(t)-g_{\ast }=O(e^{-2B^{2}t/a})$. If the anomalous dimension is regular there, the propagator factor has fixed-point behaviour $d(z,g)\sim A(g)z^{\gamma (g_{\ast })}$ when its fixed-point boundary value is nonzero. If $\gamma (g)=c{g}^2$ is retained, the exponent is $\gamma (g_{\ast })=-cb/a$. Trusting this perturbative truncation as a statement about the underlying theory requires $g_{\ast }$ to be sufficiently small; the flow result itself follows from the specified beta function.

Without gauge fixing, gauge-equivalent configurations are integrated repeatedly, and the quadratic gauge-field operator has zero modes along infinitesimal gauge transformations. It therefore has no inverse with which to define perturbative propagators. A gauge condition removes these directions locally in the perturbative expansion. The associated Faddeev-Popov determinant can be represented by anticommuting Faddeev-Popov ghost fields and Faddeev-Popov antighost fields.

First make the normalization explicit. For the fields literally occurring in the action with overall $1/g^2$, putThe resulting kinetic terms are canonical, and the interaction terms carry powers of $g$. We use signature $(-,+,+,+)$ and incoming Fourier modes $e^{ipx}$, consistent with the displayed $p^2-i0$ denominators. The quadratic action in the canonical fields isColor indices are diagonal. Introduce the longitudinal and transverse projectors $P_{L,\mu \nu }=p_{\mu }{p}_{\nu }/p^2$ and $P_T=\eta -P_L$. The gauge kinetic matrix is $-p^2(P_T+{\xi }^{-1}{P}_L)$, with inverse $-(P_T+\xi P_L)/p^2$. The ghost kinetic operator is $-p^2$. The Gaussian functional integral and Grassmann integral therefore giveThe longitudinal formula is understood as the Feynman-prescribed inverse of the kinetic operator; away from its poles the projector inversion is ordinary algebra. The propagators for the original unrescaled fields are $g^2$ times these expressions. Using $P_{T}p=0$ and $P_{L}p=p$ giveswith either normalization. This is the longitudinal gauge propagator contraction.

The canonical ghost interaction comes directly from the covariant derivative:For incoming antighost momentum $p$, the derivative supplies $i{p}^{\mu }$; the expansion of $e^{iS}$ supplies another $i$. Thus the ghost-gluon vertex is $g{p}^{\mu }{f}_{abc}$, as requested. For unrescaled fields its coefficient is instead $p^{\mu }{f}_{abc}/g^2$. The printed order-$g$ rule presupposes canonical normalization; keeping the overall $1/g^2$ and an order-$g$ vertex for the same unrescaled variables would mix conventions. Choosing Fourier modes $e^{-ipx}$ instead reverses the vertex sign and must be done consistently for all momentum assignments.

In Feynman gauge, $\xi =1$, the one-loop ghost two-point graph has one internal ghost line and one internal gauge line joined by two such vertices. The external ghost line is open, so it does not carry a closed ghost-loop minus sign. With an orthonormal adjoint basis, the structure constants are totally antisymmetric, and the color product along this line is $f_{adc}{f}_{cdb}=-C{\delta }_{ab}$. The two internal propagators supply $(-i{)}^2=-1$, leaving a positive color-and-vertex factor for the amputated insertion. DefineThe amputated insertion is $i{\delta }_{ab}{g}^2{I}_d(p)C$. Restoring the two external canonical ghost propagators gives the one-loop contributionThe literal unrescaled correlator in the question is $g^2{G}^{(1)}$, hence of order $g^4$, whereas its tree term is $-i{\delta }_{ab}{g}^2/(p^2-i0)$. In $d\ne 4$ the dimensionless renormalized coupling convention adds ${\mu }^{4-d}$ multiplying $g^2$ in the canonical loop term.

To evaluate the Feynman integral, combine the denominators with a Feynman parameter:Shift $\ell =k-(1-\alpha )p$. The denominator becomes $[{\ell }^2+\alpha (1-\alpha )p^2-i0{]}^2$, and the numerator is $p\cdot \ell +(1-\alpha )p^2$. The odd term vanishes in translation-invariant dimensional regularization. For spacelike $p$, a Wick rotation converts the remaining integral into a Euclidean one and cancels the prefactor $1/i$. The needed radial integral isIndeed, spherical coordinates and $u={\ell }_E^2/M^2$ leave a beta integral ${\int }_0^{\infty }{u}^{d/2-1}(1+u{)}^{-2}du=\Gamma (d/2)\Gamma (2-d/2)$; the angular factor cancels $\Gamma (d/2)$. Establish this first in a convergent range and then analytically continue in $d$. Substituting $M^2=\alpha (1-\alpha )p^2$ yieldsThis is the printed result, including the factor $1/i$ in its Minkowski-space definition. Equivalently the parameter integral is $B(d/2-1,d/2)$. Symmetry under $k\leftrightarrow p-k$ also gives $I_d(p)$ as $p^2/2$ times the massless scalar bubble, independently checking the numerator factor.

Write $ϵ=4-d$. Since $\Gamma (ϵ/2)=2/ϵ+O(1)$ and the parameter integral tends to ${\int }_0^1(1-\alpha )d\alpha =1/2$, the pole isHere $p$ is off shell and nonzero, separating the ultraviolet pole from massless infrared issues. The unrescaled amplitude has an extra factor $g^2$. The divergence is proportional to the existing ghost kinetic operator, so the one-loop ghost kinetic counterterm in Feynman gauge is local:It inserts $-i\delta Z_c{p}^2$ into the inverse propagator, cancelling the divergent insertion $+i{g}^{2}C{I}_d(p)$. With the two external propagators attached, its contribution is $+i{\delta }_{ab}\delta Z_c/p^2$, cancelling the displayed connected amplitude pole. In the original variables the same action term is $-\delta Z_c\partial \bar{c}\cdot \partial c/g^2$. If instead one defines dimensional regularization by $d=4-2ϵ$, the identical counterterm is written $\delta Z_c=g^{2}C/(32{\pi }^2ϵ)$; the factor of two is purely the regulator convention.