Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2025 (A15272H1)

Source PDF: mmathphys.physics.ox.ac.uk/sites/g/files/vzsiac2991/files/mmathphys/documents/media/a15272h1corrected.pdf

Solutions to all seventeen lettered parts. The source is the version publicly labelled corrected; no claim about the contents of an earlier version is needed. Use natural units, the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$, and $\square ={\partial }_{\mu }{\partial }^{\mu }$ for the d'Alembert operator. In question 2, every denominator has its Feynman i-epsilon prescription $+i0$, and ${\int }_k={\mu }_R^{2ϵ}\int d^{4-2ϵ}k/(2\pi {)}^{4-2ϵ}$. Here $ϵ$ is the dimensional regularization parameter; the small parameter in 1(f) is denoted $\eta$ to distinguish it. Signs follow the insertion Feynman rules printed on source page 4.

The scalar field path integral integrates over real field histories with vacuum boundary conditions. With $S_0[\varphi ]=\int d^{4}x{\mathcal{L}}_0$, the partition function isThe constant $\mathcal{N}$ fixes the desired vacuum normalization. A generating functional adds $\int d^{4}xJ\varphi$ to the action; the required partition function is its zero-source value. Equivalently, a Wick rotation gives $Z_E={\mathcal{N}}_E\int \mathcal{D}\varphi e^{-S_E[\varphi ]}$, with the positive-sign kinetic and potential terms in the Euclidean action when $m^2\ge 0$ and $\lambda >0$.

Vary the action, taking the field variation to vanish at the boundary. Integration by parts givesThus the Euler-Lagrange field equation isFor the field redefinition in part f, define the functional derivative $E(\varphi )=\delta S_0/\delta \varphi =-(\square \varphi +m^2\varphi +\lambda {\varphi }^3/6)$. The sign does not affect the equation of motion, but it fixes the sign of an added redundant operator.

The action is dimensionless, so in four dimensions the Lagrangian density has mass dimension four. Its kinetic term fixes $[\varphi ]=1$ and $[{\partial }_{\mu }]=1$. An interaction $g_D{\mathcal{O}}_D$ with $[{\mathcal{O}}_D]=D$ has $[g_D]=4-D$. ThereforeIn an effective field theory with heavy scale $\Lambda$, a higher-dimensional local operator enters as $c_D{\mathcal{O}}_D/{\Lambda }^{D-4}$. For characteristic scales $E,m\ll \Lambda$ and coefficients compatible with this expansion, its effects are suppressed by powers of the small ratios. Omitting these terms defines the leading approximation and its truncation error in effective field theory. It is not a fundamental prohibition: such terms must be retained if that accuracy is required. A continuum renormalizable quantum field theory can instead be defined by the closed set of counterterms of dimension at most four.

Up to total derivatives, all Lorentz scalar terms of dimension at most four built from one real scalar field areThe term $\varphi \square \varphi$ is equivalent to the kinetic term by integration by parts; $\square \varphi$ and $\square ({\varphi }^2)$ contribute only at the boundary. The displayed Lagrangian density omits the linear and cubic interactions and a field-independent constant. The exact Z2 symmetry $\varphi \mapsto -\varphi$ forbids both odd interactions, and radiative corrections preserving this symmetry cannot generate them. The constant shifts the vacuum energy and the normalization of $Z$ and can be dropped when gravity and absolute vacuum energy are not being studied. A constant vacuum counterterm may still be used when renormalizing unnormalized vacuum diagrams. The kinetic coefficient has already been fixed by field normalization.

For dimension-six operators of a real scalar field, let $n$ count fields and $r$ count partial derivatives. Mass dimension gives $n+r=6$. Lorentz invariance requires even $r$. Discard derivatives acting on the entire expression, but do not yet identify different expressions by integration by parts. A complete list of seven representatives isFor $(n,r)=(6,0)$ there is one choice; for $(4,2)$ the derivative distribution is either $2$ or $1+1$; for $(2,4)$ it is $4$, $3+1$, or $2+2$, the last allowing two distinct metric contractions. Commuting partial derivatives identifies alternative orderings. Contractions with a Levi-Civita symbol vanish because an antisymmetric index pair necessarily contracts symmetric derivatives. The remaining case $(0,6)$ gives no field-dependent expression. No independent odd-$r$ Lorentz scalar exists.

The source's seven means this unreduced list. If “ignoring total derivative operators” were interpreted as already taking the full quotient by total derivatives, seven would be incorrect: the next part gives three representatives of that quotient. Dimensionful factors multiplying lower-dimensional terms do not count as new canonical-dimension-six field monomials.

Write $X\simeq Y$ when the two densities differ by a total derivative. Under the boundary conditions used in the action, they have the same integral. The following identities establish the integration-by-parts reduction of scalar operators:Consequentlyspans the operator basis in effective field theory before equation-of-motion reduction of scalar operators. These four independent relations reduce the seven representatives to three; the three sectors have different field or derivative content, so integration by parts cannot identify them further.

Use the sign convention $E(\varphi )=\delta S_0/\delta \varphi$ from part b, and letIn the path integral, make the local field redefinition ${\varphi }_{\mathrm{old}}=\chi -\eta f(\chi )$. The functional Taylor expansion of the action givesThe two first-order terms cancel. Neglecting the functional Jacobian as permitted,Thus $fE$ is a redundant operator at first order, by the equivalence theorem for field redefinitions. The same argument works for a local derivative-dependent $f$, which is needed to remove $(\square \varphi {)}^2$. Insertions and sources must be transformed too: arbitrary untransformed off shell correlation functions need not be equal.

Sign qualification. If instead $E_{+}=\square \varphi +m^2\varphi +\lambda {\varphi }^3/6$ is used, then $E_{+}=-\delta S_0/\delta \varphi$. An added $+\eta f{E}_{+}$ is removed by ${\varphi }_{\mathrm{old}}=\chi +\eta f(\chi )$. The minus shift requested in the source therefore fixes which sign is meant by its otherwise unspecified $E$.

Let $a=\lambda /6$. Within terms already suppressed by ${\Lambda }^{-2}$, the leading equation of motion gives $\square \varphi =-m^2\varphi -a{\varphi }^3$. Equality here means equivalence under a perturbative field redefinition, together with integration by parts, rather than a pointwise identity inside arbitrary off shell correlation functions. HenceFor the second relation, factor the difference as $(\square \varphi -m^2\varphi -a{\varphi }^3)(\square \varphi +m^2\varphi +a{\varphi }^3)$, which is explicitly a redundant operator. Ifthe reduced correction isThe first two coefficients shift the already present mass and quartic coupling constant. After these shifts,This is the equation-of-motion reduction of scalar operators through $O({\Lambda }^{-2})$; terms of $O({\Lambda }^{-4})$ are outside that truncation. The lower-dimensional shifts should not be silently dropped when $m\ne 0$.

The first panel is the one-particle-irreducible Feynman diagram for photon vacuum polarization. The two internal Dirac propagators carry $k$ and $k+q$, and the external photons carry $q$. The other panels fix the routings used in parts c and d.

The closed fermion loop sign contributes a minus sign; there is no extra symmetry factor. Applying the given Feynman rules, with the external propagators amputated,Equivalently, the required amplitude isThe Feynman slash notation is $/k={\gamma }^{\rho }{k}_{\rho }$ and $\mathrm{tr}1=4$ in the chosen dimensional regularization convention.

The Ward identity implies $q^{\mu }{\Pi }_{\mu \nu }=0$. Lorentz invariance then gives ${\Pi }_{\mu \nu }=(q^2{g}_{\mu \nu }-q_{\mu }{q}_{\nu })\Pi$, so $g^{\mu \nu }{\Pi }_{\mu \nu }=(d-1)q^2\Pi$. The gamma matrices satisfy ${\gamma }_{\mu }/k{\gamma }^{\mu }=(2-d)/k$. Odd traces vanish, and thereforeSubstituting gives the requested scalar coefficient:To extract the ultraviolet divergence without introducing a Feynman parameter, set $A=(k+q{)}^2-m^2$, $B=k^2-m^2$ and useTranslation invariance of dimensional regularization gives ${\int }_k{A}^{-1}={\int }_k{B}^{-1}={\mathcal{I}}_1$. ConsequentlyIn the pole coefficient one may set $d=4$: the bracket becomesThus the ultraviolet pole of QED vacuum polarization isThe cancellation of the apparent $m^2/q^2$ divergence is essential: gauge invariance forbids a photon mass counterterm. The scalar result at $q^2=0$ is understood by continuation after this cancellation.

The middle panel in part a has an internal photon of momentum $k$ joining two points on a fermion line; the internal fermion has momentum $p-k$. It is a one-particle-irreducible Feynman diagram with no closed fermion loop sign. In Feynman gauge,orThe decomposition in the source gives $\mathrm{tr}(/p\Sigma )=4p^{2}S(p^2)$. In particular,For a direct ultraviolet divergence extraction, use a Feynman parameter $x$ multiplying the massive denominator, shift $r=k-xp$, and define $\Delta =x{m}^2-x(1-x)p^2$. Then $/p-/k=(1-x)/p-/r$ andOdd $r$ terms integrate to zero. Taking the trace explicitly supplies the weight $(2-d)(1-x)$, whose integral tends to $-1$. HenceAs a check on the same fermion self-energy calculation,This is the ultraviolet pole of QED fermion self-energy. It is safe to extract it at generic external momentum; an on shell or massless limit must distinguish any infrared divergence from this ultraviolet divergence.

In the right panel of part a, $p$ enters the fermion line, $p^{\prime }=p+q$ leaves it, and a virtual photon of momentum $k$ bridges the external photon vertex. Write ${\Gamma }^{\mu }={\gamma }^{\mu }+{\Lambda }^{\mu }$ and $F_1=1+F_1^{(1)}+\cdots$. The one-loop amplitude isThis Feynman integral has a logarithmic superficial degree of divergence. Its ultraviolet pole is independent of the external momenta and $m$, because every such numerator insertion or denominator correction lowers the large-$k$ degree. Preserve an auxiliary positive mass $M$ while extracting the ultraviolet pole; setting everything to zero would create a scaleless integral and conceal the cancellation of ultraviolet and infrared divergences.

The leading numerator is ${\gamma }^{\alpha }/k{\gamma }^{\mu }/k{\gamma }_{\alpha }$. Symmetric integration replaces $k_{\rho }{k}_{\sigma }$ by $g_{\rho \sigma }{k}^2/d$. Two contractions of gamma matrices then giveFor the auxiliary-mass extraction of ultraviolet poles,The second term is ultraviolet convergent. ThusThe ultraviolet pole of the QED vertex is therefore ${\Lambda }^{\mu }{|}_{\mathrm{UV}}=e^2{\gamma }^{\mu }/(16{\pi }^2ϵ)$. The Pauli structure has zero trace against ${\gamma }_{\mu }$ and no ultraviolet pole at this order.

Form-factor qualification. The two electron electromagnetic vertex form factors in the source describe matrix elements between external on shell Dirac spinors. A general off shell vertex has more structures, so the unprojected trace is not a projector for the complete finite on-shell $F_1$. It does correctly isolate the divergent local ${\gamma }^{\mu }$ coefficient, which is all that is required here.

In $d=4-2ϵ$, the modified minimal subtraction scheme subtracts ${\Delta }_{\overline{\mathrm{MS}}}=1/ϵ-{\gamma }_E+\log (4\pi )$ for each one-loop pole, with no further finite, kinematics-dependent subtraction. Equivalently, use a rescaled renormalization scale ${\overline{\mu }}^2=4\pi e^{-{\gamma }_E}{\mu }_R^2$ and subtract only $1/ϵ$ in that convention. This differs from the minimal subtraction scheme by a fixed scale convention.

Let $C=e^2{\Delta }_{\overline{\mathrm{MS}}}/(16{\pi }^2)$. With the source's counterterm insertion signs, cancellation proceeds as follows. The vertex has $-ie{\gamma }^{\mu }C$ from the loop and $-ie{\gamma }^{\mu }{\delta }_1$ from its counterterm, giving ${\delta }_1=-C$. The kinetic part of the fermion self-energy insertion is $+i/pC$ and its counterterm is $+i/p{\delta }_2$, giving ${\delta }_2=-C$. Finally the photon vacuum polarization insertion is $-i(q^2{g}_{\mu \nu }-q_{\mu }{q}_{\nu })(4C/3)$ and its counterterm is $-i(q^2{g}_{\mu \nu }-q_{\mu }{q}_{\nu }){\delta }_3$, givingWriting $Z_i=1+{\delta }_i$ therefore verifies the Ward identity for QED renormalization constants, $Z_1=Z_2$, at one loop. The proper-vertex Ward-Takahashi identity in QED independently gives ${\Lambda }^{\mu }(p,p)=-\partial \Sigma (p)/\partial p_{\mu }$, whose divergent part reproduces the opposite signs of $F_1^{(1)}$ and $S$ found above. The equality holds in a common gauge fixing and subtraction prescription; the individual constants $Z_1,Z_2$ are gauge dependent.

Start with the formally gauge-redundant path integral $Z=\mathrm{vol}(\mathcal{G}{)}^{-1}\int \mathcal{D}A{e}^{iS[A]}$. Use $A_{\mu }^{\theta }=A_{\mu }+{\partial }_{\mu }\theta$, and the generalized Lorenz gauge condition $F_{\omega }[A]={\partial }_{\mu }{A}^{\mu }-\omega$. In a perturbative neighborhood where residual zero modes are removed by boundary conditions, insert the Faddeev-Popov gauge-orbit identityThus ${\Delta }_{\mathrm{FP}}=\det \square$. Gauge invariance lets the $\theta$ integration factor out and cancel $\mathrm{vol}(\mathcal{G})$. Average over $\omega$ with the Gaussian integral weight $\exp [-i\int {\omega }^2/(2\xi )]$; the delta functional fixes $\omega =\partial \cdot A$ and givesRepresent the functional determinant by the independent Grassmann fields $\bar{c},c$. A field-independent determinant sign can be absorbed in normalization, so a convenient convention isThereforeThe Faddeev-Popov ghosts are free and decouple because $\square$ is independent of $A$. For a Non-Abelian gauge theory, an infinitesimal gauge transformation gives $\delta A_{\mu }^a=D_{\mu }^{ab}{\theta }^b$, so the Faddeev-Popov operator is ${\partial }^{\mu }{D}_{\mu }^{ab}[A]$ andIt contains a ghost-gluon vertex: these Faddeev-Popov ghosts do not decouple in covariant gauge fixing of Lorenz gauge type.

Choose the gauge covariant derivative $D_{\mu }={\partial }_{\mu }+ie{A}_{\mu }$. Write $\rho =f+h$, $r=1+h/f$ and $B_{\mu }={\partial }_{\mu }\pi /f+e{A}_{\mu }$. The polar parametrization of the Abelian Higgs field givesHence the full Lagrangian density isAssume $f>0$ and $\lambda >0$, so the minimum has $|\varphi |=f/\sqrt{2}$. In this convention the local gauge symmetry acts asThe combination ${\partial }_{\mu }\pi +ef{A}_{\mu }$, the radial field $h$, and the gauge field strength are each invariant, proving invariance of the displayed Lagrangian density.

This parametrization follows the circle of classical minima: $h$ measures distance away from the vacuum manifold, whereas $\pi$ moves along a gauge orbit. It separates the massive radial Higgs mode from the would-be Goldstone boson and makes its removal by the Higgs mechanism transparent. At quadratic order,The polar parametrization of the Abelian Higgs field is a local description near the chosen nonzero vacuum expectation value; it is singular at $\varphi =0$ and is not a global coordinate system for configurations with zeros or winding.

Unitary gauge fixes the scalar phase to zero. Locally choose $\alpha =-\pi /f$ in part b; then ${\pi }^{\prime }=0$ and $A_{\mu }^{\prime }=A_{\mu }+{\partial }_{\mu }\pi /(ef)$. The remaining fields are the radial Higgs mode and a massive vector. Its quadratic Lagrangian density is $-F^2/4+m_A^2{A}^2/2$ and its gauge-boson propagator isFor $k^2\ne 0$, introduce the transverse and longitudinal orthogonal projectors $P_{\mu \nu }^T=g_{\mu \nu }-k_{\mu }{k}_{\nu }/k^2$ and $P_{\mu \nu }^L=k_{\mu }{k}_{\nu }/k^2$. Omitting the infinitesimal in this algebraic decomposition,At large momentum in Euclidean space, the transverse coefficient falls as $k^{-2}$ but the longitudinal coefficient approaches a constant of order $m_A^{-2}$. Thus power counting applied directly to unitary gauge Feynman diagrams suggests increasingly severe ultraviolet divergences. This hides the renormalizability of physical scattering amplitudes; it does not disprove it. A finite-$\xi$ R-xi gauge exposes the better propagator scaling while preserving the fields needed to track the cancellations.

Let $m_A=ef$, $r=1+h/f$, and $H=\partial \cdot A-\xi m_A{r}^2\pi$, so the given $G=H/\sqrt{\xi }$ contributes ${\mathcal{L}}_{\mathrm{gf}}=-H^2/(2\xi )$. Its quadratic cross term is $+m_A\pi \partial \cdot A$, which cancels $m_{A}A\cdot \partial \pi$ from the scalar kinetic term after integration by parts. The free vector and would-be Goldstone boson propagators are therefore diagonal; $\pi$ has gauge-dependent mass squared $\xi m_A^2$.

The field-dependent factor $r^2$ cancels more than just the constant-coefficient mixing, but a residual derivative interaction remains. Indeed,Thus the nonlinear R-xi gauge in polar Higgs variables removes quadratic mixing; it does not remove every interaction containing $A$ and $\pi$.

Use the gauge transformation in part b. Since $h$ is invariant,The factor $-1/(e\sqrt{\xi })$ in $\delta G/\delta \alpha$ is field independent and is absorbed into the partition function normalization. Exponentiating the remaining Faddeev-Popov determinant with the sign convention of part a givesThe ghost coupling in the Abelian Higgs model therefore survives even though the gauge group is Abelian: these Faddeev-Popov ghosts interact with $h$ and do not decouple for generic $\xi$. The distinction from part a is the scalar dependence of the gauge fixing function. In the $\xi \to 0$ Landau gauge limit these particular ghost interactions vanish.

After the quadratic mixing cancellation, the vector inverse quadratic operator isInverting its two orthogonal projectors and multiplying by $i$ givesEquivalently $D^{(\xi )}=-i{P}^T/(k^2-m_A^2+i0)-i\xi P^L/(k^2-\xi m_A^2+i0)$. At fixed finite $\xi$, both sectors fall as $k^{-2}$ at large momentum in Euclidean space; $\xi =1$ gives $-i{g}_{\mu \nu }/(k^2-m_A^2+i0)$, and taking $\xi \to \infty$ at fixed $k$ recovers unitary gauge. The latter limit is not uniform in ultraviolet momentum.

The expected power counting conclusion is that this $k^{-2}$ behaviour, together with polynomial interactions of mass dimension at most four in Cartesian scalar variables and the gauge symmetry identities, permits renormalization of spontaneously broken gauge theories using a finite set of physical parameters. The propagator alone is not sufficient to establish that claim in the polar variables of this question. In fact, part b contains $h(\partial \pi {)}^2/f$ and $h^2(\partial \pi {)}^2/(2f^2)$, of dimensions five and six, and the nonlinear gauge fixing term contains $-\xi e^2{h}^4{\pi }^2/(2f^2)$, also of dimension six. Ordinary power counting is consequently not manifest in these variables.

For a representation where it is manifest, write $\varphi =(f+H+i\chi )/\sqrt{2}$ and choose the usual polynomial R-xi gauge function $\partial \cdot A-\xi ef\chi$. It yields the same free vector propagator, cancels the quadratic mixing, and leaves only dimension-at-most-four interactions, including its Faddeev-Popov ghosts. This supplies the missing vertex condition in the source's proposed inference; the equivalence theorem for field redefinitions relates physical predictions near the vacuum. Good propagator falloff is the necessary ingredient illustrated by the requested formula, rather than a complete proof for arbitrary parametrizations or nonlinear choices of gauge fixing.