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An integral over momentum space. For example, state counting in $d$ spatial dimensions uses a measure proportional to $d^{d}p/(2\pi \hslash {)}^d$, whereas a relativistic Feynman integral commonly uses $d^{d}p/(2\pi {)}^d$ in natural units. The convention and any regulator must be specified. Independent loop momenta supply separate integration variables in a loop Feynman diagram; ultraviolet divergences and infrared divergences can require different regularizations.

The rest mass of the muon, approximately $105.66\mathrm{MeV}$ in natural units. Keeping it in phase space is necessary in precision lepton universality tests.

For an especially simple neutral current example, scatter a muon neutrino elastically from an initially stationary unpolarized electron:Here $T_e$ is the final electron kinetic energy, $E_{\nu }$ is the incident neutrino energy, and ${\theta }_W$ is the Weinberg angle. These are electron chiral couplings in the $g/\cos {\theta }_W$ vertex convention of neutral-current vector and axial couplings. The chosen muon neutrino elastic channel has no additional W boson exchange amplitude; the electron-neutrino elastic channel would have one. Other charged-current final states can become kinematically accessible at sufficiently high incident energies and are not part of this elastic model.

Use natural units $\hslash =c=1$, retain only tree-level Feynman diagrams, and take the contact limit $|q^2|\ll m_Z^2$ with $E_{\nu }\gg m_e$. Neglect electron-mass terms relative to $E_{\nu }$, radiative corrections and detector response. Then$G_F$ is the Fermi constant. The approximate endpoint $y=1$ replaces the exact elastic endpoint $y_{\max }=2E_{\nu }/(m_e+2E_{\nu })$. The omitted leading mass-dependent interference term in $d{\sigma }_{\nu }/dy$ is $-K{g}_L{g}_R{m}_{e}y/E_{\nu }$. Thus the approximation is not appropriate for arbitrarily soft neutrinos.

The shape follows from the two chiral contributions to the massless scattering amplitude. The spin sum gives terms proportional to $s^2$ and $u^2$ in Mandelstam variables; in this limit $u/s=-(1-y)$, giving the square $(1-y{)}^2$. The two electron chiralities do not interfere when the relevant mass corrections are dropped. Replacing the neutrino by its antineutrino exchanges which coupling multiplies each kinematic term. This is the same chiral angular weighting that appears in neutrino scattering on an antiquark, now with a target whose couplings are particularly simple.

Integrating the two elementary terms givesFor an original teaching example choose $s_W^2=1/4$, so $g_L=-1/4$ and $g_R=1/4$. This is a stipulated parameter, not a measurement of the Weinberg angle. ThenThe normalized probability density function is $p(y)=3[1+(1-y{)}^2]/4$, not the unnormalized differential cross section. Its endpoints have ratio two, and the fraction with $y>1/2$ is $13/32$. The antineutrino spectrum happens to coincide at this chosen parameter because $g_L^2=g_R^2$; this equality is not a general prediction for arbitrary Weinberg angles.

Unlike a visible electron recoil, the scattered neutrino is unrecorded. A measured recoil spectrum must therefore be compared with incident neutrino flux, energy distribution, acceptance, detection efficiency and backgrounds. A shape or rate tests a specified theoretical model only after these experimental ingredients have been supplied.

An elastic weak interaction of a neutrino with an electron. For a stationary electron of mass $m$ and an effectively massless incident neutrino of energy $E$, the maximum recoil kinetic energy in natural units is $T_{\max }=2E^2/(m+2E)$. This kinematic limit does not specify the interaction probability.

After neutron capture on $(A,Z)$, the compound nucleus is $(A+1,Z)$. With negligible incoming kinetic energy, conservation of energy gives $E^{\ast }=M(A,Z)+m_n-M(A+1,Z)$ in natural units. This is the neutron separation energy of the compound nucleus. At finite incident energy, add the kinetic energy in the initial centre-of-momentum frame; in a nonrelativistic heavy-target approximation it is $A/(A+1)$ times the laboratory neutron kinetic energy. Nuclear pairing can significantly change capture excitation between neighboring even- and odd-mass targets.

Formation of a deuteron from a neutron and a proton with photon emission. For an initial pair at rest, two-body energy and momentum conservation gives $E_{\gamma }=B-B^2/[2(m_p+m_n)]$ in natural units, where $B=m_p+m_n-m_d$. The missing binding energy becomes nuclear recoil.

The lowest initial centre-of-mass frame energy at which a specified final state can be made. For free massive final particles in natural units, the minimum is the sum of their masses, attained when their relative momenta vanish. The cross section at the exact threshold can vanish even though the channel becomes kinematically accessible there.

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/6/Physics%20Papers%202017.pdf?cookie_passthrough=1

The dated cover establishes Trinity Term 2018, despite the containing bundle's 2017 filename. Use natural units, the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$ and the Feynman i-epsilon prescription. In question 1(a), $D=4-ϵ$ and $e_D={\mu }^{ϵ/2}e$. The supplied transverse photon propagator selects Landau gauge within the Lorenz-gauge family. In question 2(a), distinguish the kernels in the stated generating functional from time-ordered Wick contractions, as explained below.

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/7/Physics%20Papers%202018.pdf?cookie_passthrough=1

The bundle filename says 2018, but the paper's cover establishes Trinity Term 2019, 24 April 2019, and formal code A15272W1. These are solutions to that 2019 paper. Use natural units, the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$, and the Feynman i-epsilon prescription. Write $/p={\gamma }^{\mu }{p}_{\mu }$. In question 1, $D=4-ϵ$, $g_D={\mu }^{ϵ/2}g$, and the amputated gluon self-energy insertion is $i{\Pi }^{\mu \nu ,ab}$. This fixes the overall sign and factor of $i$ in the results below; defining the insertion as $-i\Pi$ instead reverses all corresponding definitions of $\Pi$. The quadratic Casimir operator in the Adjoint representation is $C_A=N_c$ for $SU(N_c)$, hence $C_A=3$ in QCD.

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/8/Physics%20Papers%202019.pdf?cookie_passthrough=1

Official standalone paper: mmathphys.physics.ox.ac.uk/sites/default/files/mmathphys/documents/media/aqft_2020.pdf

The annual bundle's filename says 2019, but this paper's cover says Trinity Term 2020 and identifies A15272W1. These are solutions to that 2020 paper. Use natural units, the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$ and the Feynman i-epsilon prescription. Write $/p={\gamma }^{\mu }{p}_{\mu }$. In question 1, $D=4-ϵ$, $e_D={\mu }^{ϵ/2}e$, and $\mu$ is the renormalization scale; keeping the dimensionful coupling $e_D$ explicit prevents a factor-of-two ambiguity in the ultraviolet pole. In question 2(a), the factors of $i$ in the stated generating functional fix the normalization of its kernels, as explained in the solution.

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/9/Physics%20Papers%202020.pdf?cookie_passthrough=1

Official standalone paper: mmathphys.physics.ox.ac.uk/sites/default/files/mmathphys/documents/media/aqft_2021.pdf

Use natural units, the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$ and the Feynman i-epsilon prescription. In question 1 use dimensional regularization with $d=4-2ϵ$ and write $C_{ϵ}=1/(16{\pi }^2ϵ)$; in the alternative convention $d=4-ϵ$, replace $C_{ϵ}$ by $1/(8{\pi }^2ϵ)$. Only ultraviolet poles are retained there. Figure 1 is unreadable in the released source: the corresponding specific gaps are marked below. All other parts have complete derivations.

Source PDF: mmathphys.physics.ox.ac.uk/sites/g/files/vzsiac2991/files/mmathphys/documents/media/aqft_for_particle_physics_2024.pdf

Solutions to all seventeen lettered parts. Use natural units, the Minkowski metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, Feynman slash notation $/p={\gamma }^{\mu }{p}_{\mu }$, and the Feynman i-epsilon prescription in internal propagators. Define the scattering amplitude by $S_{fi}={\delta }_{fi}+i(2\pi {)}^4{\delta }^{(4)}(p_f-p_i){\mathcal{M}}_{fi}$. The gauge covariant derivative convention in question 3 is specified there to match the printed three-boson Feynman rule.

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.

Use natural units and the Minkowski metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$. Varying the Proca action with respect to $A_{\nu }$ gives $\partial \mathcal{L}/\partial A_{\nu }=m^2{A}^{\nu }$ and $\partial \mathcal{L}/\partial ({\partial }_{\mu }{A}_{\nu })=-F^{\mu \nu }$. The Euler-Lagrange equation is thereforeTaking a divergence annihilates the antisymmetric field strength, since commuting derivatives are symmetric in their indices. For $m\ne 0$ this gives the Lorenz constraint in Proca theory ${\partial }_{\nu }{A}^{\nu }=0$, and hence $(\square +m^2)A^{\nu }=0$. The constraint follows from dynamics; it is not an additional gauge fixing condition for this massive theory.

For four-momentum $p^{\mu }=(E,0,0,p)$, longitudinal polarization of a massive vector boson lies in the time–$z$ plane, has $p\cdot \epsilon =0$ and is normalized by $\epsilon \cdot \epsilon =-1$. Takingsatisfies both requirements because $E^2-p^2=m^2$. It can also be obtained by boosting the rest-frame unit spatial polarization along the direction of motion. Together with two transverse polarizations it gives the three physical spin states of a massive spin one particle.

In the Higgs boson rest frame, choose $p^{\mu }=(E,0,0,p^{\ast })$ and $p^{\prime \mu }=(E,0,0,-p^{\ast })$, with $E=m_H/2$ and $p^{\ast 2}=m_H^2/4-m_W^2$. Consistent normalized longitudinal polarization vectors areTheir Lorentz contraction is $(p^{\ast 2}+E^2)/m_W^2=(2p^{\ast 2}+m_W^2)/m_W^2$. The Higgs boson coupling to W bosons consequently givesUsing the two-body Lorentz-invariant phase space, $p^{\ast }=m_H\sqrt{1-4x}/2$ and $v^{-2}=\sqrt{2}{G}_F$ yields the longitudinal partial decay widthThe charged W bosons are distinguishable, so there is no identical-particle factor of $1/2$. This expression includes the entire angular phase space but only the specified longitudinal final polarizations. The Higgs decay to two W bosons summed over all polarizations instead has the factor $1-4x+12x^2$; the additional transverse contribution is $8x^2$. At $m_H\gg m_W$ the longitudinal channel dominates and the leading width grows as $G_F{m}_H^3$.

Interactions produce Feynman diagrams with internal lines whose independent loop momenta are integrated without an upper bound. Their large-momentum contributions can have an ultraviolet divergence. Two explicit examples in a scalar quartic theory are a one-loop tadpole diagram contributing to the two-point function and a one-loop four-point bubble diagram. After a Wick rotation, a spherical cutoff $\Lambda$ gives the representative integralsCombinatorial factors and couplings multiply these integrals in a particular diagram. The first is quadratically divergent by cutoff power counting; the second is logarithmically divergent. Other schemes package the divergent pieces differently.

In renormalization, introduce a regularization, separate bare parameters into renormalized parameters and counterterms, and fix the latter by stated normalization conditions. Local divergent pieces of the scalar two- and four-point functions are absorbed into mass, field normalization and quartic-coupling counterterms, with all other operators allowed by the chosen symmetries included as needed. A perturbatively renormalizable quantum field theory requires this finite counterterm family to remain closed at all orders. It does not require every individual bare diagram to be finite. For gauge theories, identities expressing gauge symmetry and absence of a gauge anomaly are essential in addition to superficial counting.

The action is dimensionless in natural units, so in four spacetime dimensions $[\mathcal{L}]=4$ and $[{\partial }_{\mu }]=1$. The engineering dimensions of relativistic fields follow from matching the free kinetic terms:and henceFor a local operator $\mathcal{O}$ of mass dimension $d_{\mathcal{O}}$, its coupling has dimension $4-d_{\mathcal{O}}$. Operators with dimension at most four are candidates for ordinary perturbative power-counting renormalizability. This test presupposes consistent kinetic terms, symmetries and the full set of induced counterterms; it is not a proof of physical consistency for an arbitrary vector interaction.

Original public paper: Physical PDF pages 813–815. All three questions and all three alternatives in question 2 are answered. We use natural units and Minkowski metric signature $+---$.

Throughout this question use natural units with $c=1$. For zero potential and nonzero kinetic energy, the homogeneous canonical scalar field has $P=\rho ={\dot{\phi }}^2/2$, so $\boxed{w=1}$. Its equation of motion becomes $d(a^3\dot{\phi })/dt=0$, giving $\dot{\phi }\propto a^{-3}$ and thereforeThe same scaling follows from cosmological energy conservation with constant equation-of-state parameter $w=1$. If the field is exactly constant, both $P$ and $\rho$ vanish and their ratio is undefined; that is the separate empty solution.

Use natural units and the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. Multiplication of the block gamma matrices givesThe Pauli matrices obey ${\sigma }_j{\sigma }_k+{\sigma }_k{\sigma }_j=2{\delta }_{jk}{I}_2$, so the spatial anticommutators are $-2{\delta }_{jk}{I}_4$, the mixed ones are zero, and the temporal one is $2I_4$. ThusBecause partial derivatives commute, only the symmetric part of ${\gamma }^{\mu }{\gamma }^{\nu }$ survives in their contraction. Multiplying the Dirac equation on the left by its opposite-mass factor therefore gives the flat-space Dirac factorizationHence every component satisfies $\boxed{(\square +m^2)\psi =0}$, the Klein-Gordon equation. This implication does not remove the first-order spinor constraints imposed by the Dirac equation.

For Lorentz covariance, transform coordinates by $x^{\prime }=Lx$ and the Dirac spinor by ${\psi }^{\prime }(x^{\prime })=S(L)\psi (x)$. The chain rule gives ${\partial }_{\mu }^{\prime }=(L^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$. ConsequentlyThe cancellation uses $L^{\mu }{}_{\rho }(L^{-1}{)}^{\nu }{}_{\mu }={\delta }^{\nu }{}_{\rho }$. Thus the Lorentz transformation maps every solution to another, establishing the Lorentz covariance of the Dirac operator.

The infinitesimal identity follows directly from the Clifford algebra. Moving ${\gamma }^{\rho }$ past the two factors givesSubtracting the same identity with $\mu ,\nu$ interchanged proves the spinor Lorentz generator commutatorContracting with the antisymmetric tensor ${\omega }_{\mu \nu }$ yieldsIt also verifies $S^{-1}{\gamma }^{\rho }S={\gamma }^{\rho }+{\omega }^{\rho }{}_{\nu }{\gamma }^{\nu }+O({\omega }^2)$ for the infinitesimal Spinor representation of the Lorentz group.

For a positive-energy four-momentum, put $\psi =u(p)e^{-ip\cdot x}$, $E=\sqrt{{\mathbf{p}}^2+m^2}>0$. The Dirac equation becomes $(/p-m)u=0$, where the Feynman slash notation uses $/p={\gamma }^{\mu }{p}_{\mu }=E{\gamma }^0-\mathbf{\gamma }\cdot \mathbf{p}$. Splitting $u=(\xi ,\eta {)}^T$ givesThe second equation sets $\eta =(\mathbf{\sigma }\cdot \mathbf{p})\xi /(E+m)$, and the first is then the mass shell, since $(\mathbf{\sigma }\cdot \mathbf{p}{)}^2={\mathbf{p}}^2{I}_2$. Choosing two orthonormal two-spinors ${\chi }_s$ gives the Dirac plane waves in the standard representation:The negative-frequency solutions are independentlywith orthonormal ${\eta }_s$. They are the antiparticle modes after field quantization. The normalization gives $u_r^{†}{u}_s=v_r^{†}{v}_s=2E{\delta }_{rs}$ and ${\overline{u}}_r{u}_s=2m{\delta }_{rs}$, ${\overline{v}}_r{v}_s=-2m{\delta }_{rs}$, using the Dirac adjoint $\overline{u}=u^{†}{\gamma }^0$.

For a spatial rotation, ${\sigma }^{ij}={ϵ}_{ijk}\mathrm{diag}({\sigma }_k,{\sigma }_k)$. Hence the intrinsic rotation generators areFor a massive particle in its rest frame, the two independent spinors therefore carry the two-dimensional spin one-half representation, with $J_3$ eigenvalues $\pm 1/2$ and Casimir $s(s+1)=3/4$. A $2\pi$ rotation acts as $-I$, the spinor sign under a full spatial rotation. Thus the particle has spin $\boxed{s=1/2}$; four spinor components encode particle and antiparticle sectors, not four spin states of one particle. In the massless limit the corresponding physical labels are helicities $\pm 1/2$ rather than a rest-frame spin basis.

Use natural units with $c=\hslash =k_B=1$ and the unreduced Planck mass $M_{\mathrm{Pl}}=G^{-1/2}$. In a radiation-dominated universe, the effective number of relativistic energy degrees of freedom $g_{\ast }=\mathcal{N}$ givesThe ratio of weak conversion rate to expansion rate is proportional to $T^3$, so it decreases as the universe cools. Cosmological weak freeze-out occurs approximately at $\Gamma =H$, givingFor $g_{\ast }=10.75$, $G_F=10^{-5}{\mathrm{GeV}}^{-2}$ and $M_{\mathrm{Pl}}=1.22\times 10^{19}\mathrm{GeV}$ this estimate is $1.65\mathrm{MeV}$. Equivalently $g_{\ast }^{1/6}\simeq 1.5$ is only a mild multiplicative correction. The rate itself was given only to order of magnitude, so this is a decoupling scale, not a precise prediction for the weak freeze-out abundance.

Both neutrons and protons are nonrelativistic at this temperature. The nonrelativistic Maxwell--Boltzmann number density follows by the Gaussian momentum integral:Their equal spin degeneracies are $g_n=g_p=2$. Chemical equilibrium of the weak reaction gives ${\mu }_n+{\mu }_{{\nu }_e}={\mu }_p+{\mu }_e$; with negligible lepton asymmetry, ${\mu }_n\simeq {\mu }_p$. Taking the ratio and setting the nearly equal nucleon masses equal only in the prefactor gives the neutron-proton chemical equilibrium resultThe common baryon normalization cancels. Keeping the exact mass prefactor changes this by only order $Q/m_p$. Thus the printed simple exponential is the usual nonrelativistic equal-mass-prefactor approximation. Using $Q\simeq 1.29\mathrm{MeV}$ with the rough decoupling scale above gives a ratio of order one half; the rate's omitted numerical factors and gradual freeze-out must be included before using a more accurate neutron abundance.

Use natural units and conformal time $dt=ad\tau$. The conformal Hubble parameter is $\mathcal{H}=a^{\prime }/a=aH$. For the barotropic equation of state $P=(\gamma -1)\rho$, the Friedmann acceleration equation givesThe Friedmann equation gives $8\pi G{a}^2\rho /3={\mathcal{H}}^2+k$. Hence the constant-equation-of-state conformal Riccati equation isFor $\alpha \ne 0$, separation and integration of this Riccati equation yield, after choosing the origin of conformal time,The open positive-energy density branch uses the hyperbolic cotangent, rather than the cotangent. Integrating $a^{\prime }/a=\mathcal{H}$ giveswhere the positive constants are fixed by the energy density normalization. These formulas apply on branches where the bases are positive. For $\alpha >0$, take the Big Bang at $\tau =0$. The cosmic time is thenThe integrals also give the general answer after selecting a suitable reference time when a finite Big Bang endpoint does not exist. The exceptional coasting fluid $\gamma =2/3$ has $\alpha =0$, so $\mathcal{H}=h$ is constant: $a=A{e}^{h\tau }$ and $t=(A/h)e^{h\tau }+t_{\ast }$. If $h=0$, instead $a=A$ and $t=A\tau +t_{\ast }$. A flat fluid with $\gamma =0$ is the constant-density de Sitter spacetime case, for which the flat time integral is logarithmic.

During matter domination, $\gamma =1$ and $\alpha =1/2$. Absorbing numerical factors into separate positive constants gives the closed matter-dominated Friedmann solution, the flat Einstein-de Sitter universe and the open matter-dominated Friedmann solution:To compare the age of a dust universe without a cosmological constant, specify the normalization. At fixed present Hubble parameter $H_0$, with $x=a/a_0$ and present matter cosmological density parameter ${\Omega }_m>0$, the Friedmann equation givesThe integrand decreases strictly with ${\Omega }_m$ for $0<x<1$. Therefore age ordering of dust universes at fixed Hubble parameter gives, on the expanding branches,The final bound is approached by the empty Milne universe. This compares models with the same $H_0$; fixing instead the present scale factor and energy density gives a different comparison.