Mathematical and theoretical physics course of the University of Oxford (MMathPhys)

Official course description. Fourth-year transfer route; shared teaching with the MSc in mathematical and theoretical physics course of the University of Oxford.

Fourth-year transfer route after the first three years of an eligible Oxford undergraduate degree. Teaching is shared with the MSc in mathematical and theoretical physics course of the University of Oxford. Examined subjects include Groups and representations course of the University of Oxford.

The held 2022 paper uses this title. Its source title and formal code are retained separately from later papers headed advanced quantum field theory for particle physics course of the University of Oxford.

Source: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/10/Physics%20Papers%202021.pdf?cookie_passthrough=1

Source bundle pages 111–115. The paper cover establishes 2022; the attachment filename uses 2021.

There are two one-loop Feynman diagrams: the two orders of the labelled photons around a top quark fermion loop, with one Yukawa interaction insertion for the outgoing Higgs boson. They are added, rather than identified by a Feynman-diagram symmetry factor. Each has the same color charge trace $N_c=3$.

In the first routing, traverse the fermion loop from the $p_2$ insertion to the $p_1$ insertion with momentum $l$, then to the Higgs boson insertion with momentum $l+p_1$, then back to the $p_2$ insertion with momentum $l-p_2$. Conservation at the Higgs boson insertion gives $p_h=p_1+p_2$. The second routing exchanges $(p_1,\mu )$ and $(p_2,\nu )$: its three internal momenta are $l$, $l+p_2$ and $l-p_1$. Reversing the orientation is another way to represent this second ordering. Two vector insertions give no Furry theorem cancellation.

Use Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$ and the convention $S_{fi}={\delta }_{fi}+i(2\pi {)}^4{\delta }^{(4)}(p_f-p_i){\mathcal{M}}_{fi}$. Put $m=m_t$, $y={\lambda }_t/\sqrt{2}$, $Q=e_t=2/3$ andThe Feynman i-epsilon prescription is understood in subsequent denominators. Contracting the spinor indices around the fermion loop gives the traceHere $/p={\gamma }^{\rho }{p}_{\rho }$ is Feynman slash notation. The closed fermion loop sign is $-1$, the three interaction vertices contribute $(-iy)(-ieQ{)}^2$, and the three Dirac propagators contribute $i^3$. Their product, including the color charge trace, is $-N_{c}y{e}^2{Q}^2$. Thus one ordering hasMultiplying by the incoming polarization vectors and adding the exchanged Feynman diagram givesThe displayed four-dimensional Feynman integral is formal until regularized; subsequent manipulations use dimensional regularization, which permits the required loop momentum shifts.

Set $s=(p_1+p_2{)}^2=m_h^2$, so $p_1^2=p_2^2=0$ and $p_1\cdot p_2=s/2$. The physical polarization vectors obey ${ϵ}_1\cdot p_1={ϵ}_2\cdot p_2=0$. Write $X_{\mu \nu }\simeq Y_{\mu \nu }$ when they agree after contraction with these polarization vectors; terms proportional to $p_{1,\mu }$ or $p_{2,\nu }$ may then be discarded.

The gamma matrix trace identities kill the terms with an odd number of gamma matrices. The remaining three two-momentum terms and the term with three factors of $m$ areIn the $g_{\mu \nu }$ coefficient the terms linear in $l\cdot p_i$ cancel, leaving $m^2-l^2-s/2$. Among the other terms, the surviving mixed products of $l$ and $p_i$ also cancel. ConsequentlyIn the second Feynman diagram, exchange the external labels and change the dummy loop momentum $l\mapsto -l$. Its projected numerator and denominator become identical to those of the first Feynman diagram. HenceWith the common factor $8m$ extracted, the primitive integer coefficients areEquivalently, extracting only $m$ gives $(A,B,C)=(32,8,8)$ and prefactor $i{N}_c{\lambda }_t{e}^2{e}_t^2{m}_t/\sqrt{2}$. A common rescaling of all three coefficients changes the prefactor inversely. The physical scattering amplitude is ${ϵ}_1^{\mu }{ϵ}_2^{\nu }{\mathcal{M}}_{\mu \nu }^{\prime }$; the projected integrand is not an identity for an arbitrary uncontracted off-shell tensor.

Continue the loop momentum to $D=4-\epsilon$ dimensions, keeping $\mathrm{tr}1=4$ and the external physical polarization vectors four dimensional. Introduce Feynman parameters $x,y\ge 0$, $x+y\le 1$, denoting this integration triangle by $\mathcal{T}$. Completing the square givesOdd terms in $k$ integrate to zero. Lorentz invariance of dimensional regularization gives $k_{\mu }{k}_{\nu }\mapsto g_{\mu \nu }{k}^2/D$. The general projected numerator therefore becomes, under the Feynman integral,Only the $k^2$ term can give an ultraviolet divergence. Define, with the conventional scale factor implicit,The gamma function formula givesThus the pole of the numerator integral, including $2{\int }_{\mathcal{T}}dxdy=1$, isThe coefficients $(4,1,1)$ satisfy this condition. A nonzero local divergence would have to be removed by a counterterm. The Ward identity forbids a local $h{A}_{\mu }{A}^{\mu }$ term, and the first electromagnetic gauge-invariant coupling with one Higgs boson and two photons has the form $h{F}_{\mu \nu }{F}^{\mu \nu }$, of engineering dimension five. It is not an independent counterterm of the Standard Model, a renormalizable quantum field theory. The top quark contribution is separately electromagnetic gauge-invariant; no additional W boson loop Feynman diagram is needed to cancel this pole.

The finite remainder must still be retained. For $A=4C$,Putting $D=4$ before evaluating the divergent Feynman integral would incorrectly set this finite contribution to zero.

Use the Feynman parameters and $J_3$ defined in Solution. With $A=4C$, the finite term just found is $-C\Delta J_3$. Combining it with the remaining $g_{\mu \nu }$ coefficient givesThe other coefficient is $(B-4Cxy)J_3$. Thus, writing ${\mathcal{I}}_{\mu \nu }$ for the Feynman integral without the overall prefactor,For $T_{\mu \nu }=\frac{s}{2}{g}_{\mu \nu }-p_{2,\mu }{p}_{1,\nu }$, the two coefficients must have the corresponding ratio. The mismatch is simply $(B-C)J_3$, independent of $xy$. Therefore the required relation iswhich is satisfied. In the primitive normalization the result isand the full top quark scattering amplitude, in our stated $S$ convention, isIndeed $p_1^{\mu }{T}_{\mu \nu }=p_2^{\nu }{T}_{\mu \nu }=0$: replacing either polarization vector by its photon momentum makes the scattering amplitude vanish. This is the Ward identity, expressing electromagnetic gauge invariance. The result also is symmetric under exchange of the two photons. The finite term in Solution is essential for this Ward identity.

As an independent low-energy check, for $s/m^2\to 0$ the Feynman parameter integral isThe coefficient is consequently $-N_{c}y{e}^2{Q}^2/(6{\pi }^{2}m)$ at leading order. It has the inverse-mass dimension required for an effective $h{F}_{\mu \nu }{F}^{\mu \nu }$ coupling. If $y=m/v$, this coefficient stays finite in the large-$m$ limit at fixed $v$; that limit increases the Yukawa coupling, so it is not a fixed-coupling heavy-field decoupling limit.

At the stated order the renormalization-group beta function is separable:ConsequentlyFor a positive initial running coupling, this expression is used on the connected interval containing ${\mu }_i$ where its denominator is positive and perturbation theory remains valid.

If $b_0<0$, the running coupling decreases at high renormalization scale and tends to zero: this is asymptotic freedom. Decreasing the renormalization scale increases the running coupling, with a formal infrared pole at $\Lambda ={\mu }_i\exp [1/(b_0{\alpha }_i)]$. Near $\Lambda$, perturbation theory fails; the pole alone is not a proof of confinement. Both QCD and the SU(2) weak gauge coupling have this sign for the Standard Model particle content above the relevant thresholds.

If $b_0>0$, the running coupling decreases towards low renormalization scale and grows towards high renormalization scale. The formal ultraviolet Landau pole is ${\mu }_L={\mu }_i\exp [1/(b_0{\alpha }_i)]$. The pole signals failure of this perturbative evolution, rather than a controlled physical infinity. The hypercharge gauge coupling and ordinary QED with charged matter have this sign.

If $b_0=0$, the running coupling is constant at this order; higher-order terms may still make it run. No Standard Model gauge factor has a vanishing one-loop order coefficient for the usual particle content.

For precise normalization, take ${\alpha }_i=g_i^2/(4\pi )$ and $Q=T_3+Y$. The Standard Model one-loop gauge coefficients then giveThese coefficients use $d/d\ln \mu$, not $d/d\ln {\mu }^2$. Threshold matching changes the active particle content, so the single constant-$b_0$ expression is applied piecewise across thresholds.

There is one tree-level Feynman diagram: annihilation of the up quark and its antiquark into an $s$-channel gluon, followed by production of the down quark and its antiquark. A gluon interaction vertex preserves quark flavor, so a $t$-channel line joining the incoming up quark to the outgoing down quark is impossible.

Let $i,j,k,l$ label the initial quark, initial antiquark, final quark, final antiquark color charges. Up to an irrelevant common sign, the scattering amplitude isThe Mandelstam variables are $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$, $u=(p_1-p_4{)}^2$, with $s+t+u=0$ in the zero-mass limit. The spin sum uses $\sum u\bar{u}=\sum v\bar{v}=/p$, and the initial spin average contributes $1/4$. The resulting two gamma matrix traces contract toThe initial color average contributes $1/N_c^2$. Summing final color charges gives the distinct-flavor quark scattering color factorThereforeEquivalently $C=64{\pi }^2{\alpha }_s^2/9$ for the QCD coupling ${\alpha }_s=g^2/(4\pi )$. There is no initial gluon spin average: the gluon is internal.

The two Bhabha scattering tree-level Feynman diagrams are the two panels of Solution with a photon mediator: $s$-channel annihilation with internal momentum $p_1+p_2$, and $t$-channel exchange with internal momentum $p_1-p_3$. Each interaction vertex preserves the electron species. There is no independent $u$-channel Feynman diagram, since the outgoing electron and Positron are distinguishable.

Removing the color charge factor from Solution and replacing $g$ by $e$ givesCrossing symmetry interchanges $s$ and $t$ for the exchange Feynman diagram, givingSquaring the sum of the two scattering amplitudes also produces the interference term $2\mathrm{Re}\overline{{\mathcal{M}}_s{\mathcal{M}}_t^{\ast }}$. The given third term must be this cross term: it has both propagator denominators $st$, whereas a squared individual Feynman diagram has $s^2$ or $t^2$. With the correct relative fermionic sign, its value is $4e^4{u}^2/(st)$. ThusThe coefficient of the last displayed fraction is positive, but $s>0$ and $t<0$ in the physical scattering region, so this interference is destructive. The relative sign originates in the two possible external fermion contractions; changing an external-state phase changes both scattering amplitudes together.^[ref]

Both $s$-channel annihilation and $t$-channel gluon exchange are now allowed. Their spin sum factors are the Bhabha scattering factors with $e\mapsto g$. Their color charge tensors, with the external indices of Solution, areEach squared term has $K_{ss}=K_{tt}=(N_c^2-1)/(4N_c^2)$. The cross term instead has the color averageTo evaluate it, use the SU(N) fundamental completeness relationThis follows by completing the orthogonal basis of traceless Hermitian matrices with the identity. Since $\mathrm{tr}{T}^b=0$, it gives ${\sum }_a{T}^a{T}^b{T}^a=-T^b/(2N_c)$, and henceMultiplying the two squared Bhabha scattering terms by $K_{ss}$ and its cross term by $K_{st}$ givesFor the SU(3) gauge group, this isThe negative color charge overlap reverses the sign of the QED interference. Since $st<0$, this cross contribution is positive in the physical region. No additional spinor calculation is needed; all the new information is in the SU(N) fundamental completeness relation.

The Goldstone theorem states that spontaneous breaking of a continuous internal global symmetry in field theory in a local relativistic quantum field theory produces a zero-mass Goldstone boson for each independent broken generator, under the usual positive-state-space and conserved-current hypotheses. For $G\to H$ there are $\dim G-\dim H$ symmetry-required Goldstone bosons. This count assumes a Lorentz-invariant vacuum state; nonrelativistic counting can differ.

At the classical level, invariance of the scalar potential gives ${\partial }_{i}V(T^a\varphi {)}_i=0$. Differentiating at a stationary vacuum expectation value ${\varphi }_0$ yieldsThese are the Goldstone directions in the scalar mass matrix. Broken generators have $T^a{\varphi }_0\ne 0$, so their independent directions have zero quadratic mass with a canonical kinetic term. The theorem guarantees these modes; it does not exclude further accidental zero-mass modes. A local gauge symmetry is a redundancy and does not meet the physical global-symmetry hypothesis: the Higgs mechanism can instead supply longitudinal particle polarizations of massive gauge bosons.

Put $r^2={\varphi }_i{\varphi }_i$. Completing the square in the scalar potential givesSince $\lambda >0$, its global minima form the vacuum manifoldHere $\mu$ denotes the positive square root of the parameter ${\mu }^2$. The origin is a local maximum, not a ground state. Choose $⟨\mathbf{\varphi }⟩=(v,0,0)$. Its stabilizer subgroup is the SO(2) subgroup of rotations in the $(2,3)$ plane, so the stated SO(3) global symmetry in field theory is spontaneously broken to SO(2).

Write ${\varphi }_1=v+h$, ${\varphi }_2={\pi }_2$, ${\varphi }_3={\pi }_3$. The kinetic terms stay canonically normalized, while the scalar mass matrix isEquivalently, the quadratic scalar potential is ${\mu }^2{h}^2$. Therefore the particle masses areThe two angular Goldstone bosons match $\dim SO(3)-\dim SO(2)=3-1=2$. The radial displacement changes $r$ and costs potential energy, giving the massive Higgs mode. These are tree-level masses; the Goldstone theorem protects the angular zero-mass modes in the exact global theory.

For any fixed $r\ge 0$, the source contribution obeys $c{\varphi }_1\ge cr$, because $c<0$ and ${\varphi }_1\le r$. Equality occurs only at $({\varphi }_1,{\varphi }_2,{\varphi }_3)=(r,0,0)$ when $r>0$. Thus the global ground state lies on the positive first axis. Minimize there:For $0\le r\le v=\mu /\sqrt{\lambda }$, $f^{\prime }(r)<0$. For $r>v$, $f^{\prime \prime }(r)=3\lambda r^2-{\mu }^2>0$, so $f^{\prime }$ increases strictly from $c$ to $+\infty$. There is exactly one root $\tilde{v}>v$, and it is the unique global minimum. ThereforeThis argument selects the true ground state even if the full cubic has additional negative stationary roots.

Expand ${\varphi }_1=\tilde{v}+h$, ${\varphi }_2={\pi }_2$, ${\varphi }_3={\pi }_3$. The source is linear, so it shifts the expansion point but has zero Hessian matrix. The scalar mass matrix is diagonal at the new vacuum expectation value:Eliminating $\lambda$ with the stationary equation givesAll three squared masses are positive. The masses themselves are the positive square roots of these expressions.

The source explicitly reduces the stated SO(3) symmetry to SO(2); the selected vacuum state preserves that remaining SO(2). There is consequently no spontaneously broken exact continuous global symmetry in field theory to which the Goldstone theorem applies. For small $|c|$, the two angular excitations are pseudo-Goldstone bosons. Their squared masses vanish as $c\to 0^{-}$, while $\tilde{v}\to \mu /\sqrt{\lambda }$ and $m_h^2\to 2{\mu }^2$.

In unitary gauge the quadratic action is that of a Proca field. Use Minkowski metric $(+---)$ and add a source $J_{\mu }{A}^{\mu }$. Integration by parts, dropping the boundary term, givesWith Fourier transform convention $A(x)=\int d^{4}k{e}^{-ik\cdot x}A(k)/(2\pi {)}^4$, its kernel isFor $k^2\ne 0$, introduce the transverse and longitudinal momentum projectorsThey obey $P_T{P}_L=0$ and $P_T+P_L=1$ as mixed-index linear operators. HenceThe final expression is regular at $k^2=0$; the apparent singularities of the individual projectors have cancelled.

The normalized Gaussian functional integral is evaluated by completing the square with $A\mapsto A+K^{-1}J$:The source-independent functional determinant cancels in this ratio. Two source derivatives insert two fields, so the covariant Feynman propagator is $i{K}^{-1}$, with the Feynman i-epsilon prescription at the physical pole:This is the Proca propagator. Its pole residue has the three physical particle polarizations of a massive spin-one gauge boson. No additional gauge-fixing term is needed to invert this already gauge-fixed massive quadratic action. In a full interacting Higgs mechanism calculation, any field-dependent unitary gauge measure belongs to the interacting theory; the free quadratic gauge boson propagator above is unaffected.

Source PDF: mmathphys.physics.ox.ac.uk/sites/g/files/vzsiac2991/files/mmathphys/documents/media/a15271h1.pdf

Solutions to both questions and all nine parts. The two-sphere calculation uses the leading free-draining approximation, with negligible connector drag; the limitations of that approximation are made explicit.

To first order in fluid velocity divided by the speed of light, the Lorentz transformation of electromagnetic fields gives ${\mathbf{E}}^{\prime }=\mathbf{E}+\mathbf{u}\times \mathbf{B}$. Applying Ohm's law in the local fluid frame gives $\mathbf{E}+\mathbf{u}\times \mathbf{B}=\eta \mathbf{J}$. Neglecting the displacement current in Ampère's law, $\mathbf{J}=\nabla \times \mathbf{B}/{\mu }_0$. Substitution in Faraday's law givesHere ${\eta }_m$ is the magnetic diffusivity; the paper's $\eta$ is electrical resistivity, rather than magnetic diffusivity. Expanding the curl with the solenoidal magnetic-field constraint gives the resistive induction equation in its compressible form:The left side is the material derivative of the magnetic field. The three right-hand terms respectively describe field-line stretching and reorientation, amplification or dilution by compression or expansion, and magnetic diffusion. In particular, compression has $\nabla \cdot \mathbf{u}<0$ and amplifies a field component perpendicular to the compression.

The assumptions are nonrelativistic bulk motion, the low-frequency approximation permitting omission of the displacement current, a conducting-fluid continuum, isotropic local Ohm's law with spatially constant scalar electrical resistivity, constant magnetic permeability ${\mu }_0$, and $\nabla \cdot \mathbf{B}=0$. Hall, electron-pressure and other non-Ohmic corrections to the generalized Ohm's law are excluded. Incompressible flow is not required: imposing it would remove the compression term.

Let $E_B={\int }_V{B}^2/(2{\mu }_0)dV$ be the total magnetic energy. Multiply the curl form of the resistive induction equation by $\mathbf{B}/{\mu }_0$ and integrate over the fixed volume. The identityfollows from the divergence theorem. The stated zero boundary fields remove its surface term. Applying it to the motional and resistive contributions yields the magnetic energy balance in resistive magnetohydrodynamics:The first term is the negative of the mechanical power delivered to the fluid by the Lorentz force density. It transfers energy between fluid motion and the magnetic field, with either sign. The second is the nonpositive magnetic contribution to Joule heating. A nonzero positive electrical resistivity therefore dissipates magnetic energy at rate $\eta {\int }_V{J}^{2}dV$; it does not force the total magnetic energy to decrease when fluid motion supplies a larger positive power.