Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 53 (The Standard Model)

Write $t=\log \mu$ and use ${\alpha }_i=g_i^2/(4\pi )$. The chain rule turns the renormalization-group beta function intoIntegrating gives the one-loop inverse gauge coupling evolutionThis expression applies over a range with unchanged active field content and in the perturbative regime. It uses $\log \mu$, not $\log {\mu }^2$.

For gauge coupling unification, define the normalized inverse couplings $A_1=3/(5{\alpha }_1)$, $A_2=1/{\alpha }_2$, $A_3=1/{\alpha }_3$, and slopes $b_1=(3/5){\beta }_1$, $b_2={\beta }_2$, $b_3={\beta }_3$. At the common scale each inverse coupling is $A_U$. With $L=\log (M_{\mathrm{GUT}}/M_Z)$,Subtract the equations for factors 1 and 2 to solve $L=2\pi (A_1-A_2)/(b_1-b_2)$, then insert this into the difference of factors 3 and 2:This one-loop normalized hypercharge unification relation requires $b_1\ne b_2$. A scale above $M_Z$ also requires the inferred $L$ to be positive. If those two slopes coincide, unification instead requires $A_1=A_2$ and their difference cannot fix the scale.

In the convention $Q=T_3+Y$, one matter generation and one Higgs doublet have the following Standard Model representations. Spin is in units of $\hslash$; L/R describe chirality, not the spin projection. The $U(1{)}_Y$ entry is its hypercharge eigenvalue, defining the one-dimensional representation $e^{iY\omega }$.

There is no right-handed neutrino in the minimal field content being counted. Right-handed matter can alternatively be written as left-handed charge-conjugate Weyl fermions, in conjugate colour representations with opposite hypercharges. The quadratic indices and squared charges in the beta coefficients are unchanged.

For $SU(3)$, one generation supplies two colour fundamentals from $Q_L$ and one each from $u_R,d_R$. With three generations there are $n_f=3(2+1+1)=12$ Weyl fermions counted as fundamental copies, and no coloured scalar. ThusFor $SU(2)$, the three colours of $Q_L$ give three weak-doublet copies and $L_L$ gives one. There are $n_f=3(3+1)=12$ doublets across three generations and $n_s=1$ complex Higgs doublet, givingThese are spectator multiplicities in gauge beta functions: the colour index multiplies the weak contribution, while the weak index multiplies the colour contribution. Each two-component fermion is counted once; it must not be counted again as a separate antiparticle.

For $U(1{)}_Y$, sum squared hypercharges over all colour and weak components. One generation givesThe two complex Higgs components give ${\sum }_s{Y}_s^2=2(1/2{)}^2=1/2$. There is no Abelian gauge self-interaction contribution. ThereforeThe Higgs is counted only once for the whole model, whereas the fermionic generation is repeated three times. These are the Standard Model one-loop gauge coefficients in the unnormalized hypercharge convention. The grand-unified normalization instead has $b_1=41/10$. The ratio of slopes in the unification relation is $({\beta }_3-{\beta }_2)/(b_1-{\beta }_2)=-115/218$.

The local matrix $U(x)$ is a gauge transformation in the representation acting on the matter field. With Hermitian Lie algebra generators it can be written locally in spacetime, and pointwise for $SU(3)$, asThe real functions ${\omega }^a$ can instead include a factor of $g$ by redefining the transformation parameters. The gauge covariant derivative convention in this paper is $D_{\mu }={\partial }_{\mu }+ig{A}_{\mu }$. Inserting the stated transformed $A_{\mu }$ shows explicitly that the ${\partial }_{\mu }U$ term cancels, giving $D_{\mu }^{\prime }(U\psi )=U{D}_{\mu }\psi$.

In QCD, each quark field is in the colour fundamental representation $3$, with ${\psi }^i\mapsto U^i{}_j{\psi }^j$. Thus its colour generators are eight $3\times 3$ matrices, conventionally $T^a={\lambda }^a/2$ for the Gell-Mann matrices; their number is not their matrix dimension. The eight gluons themselves transform in the adjoint representation of SU(3).

With $\mathrm{tr}(T^a{T}^b)={\delta }^{ab}/2$, the pure Yang-Mills theory has Lagrangian ${\mathcal{L}}_{\mathrm{YM}}=-\frac{1}{4}{F}_{\mu \nu }^a{F}^{a\mu \nu }=-\frac{1}{2}\mathrm{tr}(F_{\mu \nu }{F}^{\mu \nu })$. Put $G_{\mu \nu }^a={\partial }_{\mu }{A}_{\nu }^a-{\partial }_{\nu }{A}_{\mu }^a$ and $C_{\mu \nu }^a=f_{abc}{A}_{\mu }^b{A}_{\nu }^c$. Since $F=G-gC$,The cross term has two derivative contributions. Interchange $\mu ,\nu$ and then $b,c$ in the second; antisymmetry of $f_{abc}$ makes it equal to the first. The cubic and quartic Yang-Mills self-interactions are thereforeThe positive cubic sign follows from the negative $gfAA$ term in the given Yang-Mills field strength; changing the covariant-derivative convention consistently changes intermediate signs.

For the left-handed quark doublet $Q_L=(u_L,d_L{)}^T$, the hypercharge is $Y=1/6$ and the weak generators are ${\sigma }^{\alpha }/2$. Its electroweak term isThe Pauli matrices give diagonal entries $\pm {\mathcal{A}}^3$ and off-diagonal entries ${\mathcal{A}}^1\mp i{\mathcal{A}}^2$. Expanding the term givesIn particular the off-diagonal weak charged current coupling is $-g(W_{\mu }{\bar{u}}_L{\gamma }^{\mu }{d}_L+W_{\mu }^{\ast }{\bar{d}}_L{\gamma }^{\mu }{u}_L)/\sqrt{2}$.

Write $s_W=\sin {\theta }_W$, $c_W=\cos {\theta }_W$. Invert the neutral-field rotation: ${\mathcal{A}}_{\mu }^3=c_W{Z}_{\mu }+s_W{A}_{\mu }$, $B_{\mu }=c_W{A}_{\mu }-s_W{Z}_{\mu }$. The original PDF gives $\tan {\theta }_W=g^{\prime }/g$, soThe photon coefficients inside the derivative are $g{s}_W/2+g^{\prime }{c}_W/6=2e/3$ for $u_L$, and $-g{s}_W/2+g^{\prime }{c}_W/6=-e/3$ for $d_L$. Multiplication by the $i$ in the kinetic Lagrangian gives the requested electromagnetic coupling of a weak multiplet:Equivalently it is $-e{A}_{\mu }{\overline{Q}}_L{\gamma }^{\mu }(T_3+Y)Q_L$, making the electric charges explicit. The remaining neutral terms are $-(g/c_W)Z_{\mu }\{(1/2-2s_W^2/3){\bar{u}}_L{\gamma }^{\mu }{u}_L+(-1/2+s_W^2/3){\bar{d}}_L{\gamma }^{\mu }{d}_L\}$. The TeX's missing prime in the mixing-angle ratio and truncated first Pauli matrix are conversion errors, not physical assumptions.

At momentum transfers small compared with $M_W$, the two weak charged current vertices each carry $g/(2\sqrt{2})$ when written with $(1-{\gamma }_5)$, and the W boson propagator reduces to a local factor proportional to $1/M_W^2$. Matching their product to the Fermi interaction gives, at tree level,The Fermi constant is thus determined by the electroweak coupling and symmetry-breaking scale; the four-fermion description is a low-energy effective interaction.

Only the cross term $J_L^{†}{J}_H$ can annihilate the pion and create the stated charged-lepton/antineutrino pair. Its leptonic matrix element is ${\bar{u}}_e(k){\gamma }^{\mu }(1-{\gamma }_5)v_{\nu }(q)$. Lorentz covariance permits a vacuum-to-spin-zero current matrix element only proportional to $p^{\mu }$. At rest this has only a temporal component. The vacuum is parity-even and the pion is parity-odd, while $V_H^0$ is parity-even, so $⟨0|V_H^0|\pi ⟩$ equals its own negative and vanishes. Lorentz covariance then makes the full vector matrix element vanish. The temporal axial current is parity-odd and is allowed: this is the vacuum-to-pseudoscalar current selection rule.

The minus sign in $J_H=V_H-A_H$ cancels the interaction's minus sign, up to an irrelevant common S-matrix phase. ThereforeThe pion decay constant convention here includes the given $\sqrt{2}$ in its current matrix element. Momentum conservation and the external Dirac equations give $p=k+q$, ${\bar{u}}_e/k=m_e{\bar{u}}_e$, $/q{v}_{\nu }=0$. Since $/q(1-{\gamma }_5)=(1+{\gamma }_5)/q$, the neutrino term vanishes andThis is helicity suppression: the amplitude vanishes in the zero charged-lepton-mass limit despite the larger available phase space.

For the spin sum, the Dirac adjoint of $1-{\gamma }_5$ is $1+{\gamma }_5$, and the completeness relations giveThe chiral projectors automatically eliminate the unused neutrino helicity when the full massless completeness relation is inserted. There is no initial spin average for a spin-zero pion.

In the pion rest frame, let $r=|\mathbf{k}|=|\mathbf{q}|$, $E=\sqrt{r^2+m_e^2}$. After integrating the spatial delta function and angles, the two-body Lorentz-invariant phase space becomesThe root is $r_{\ast }=(m_{\pi }^2-m_e^2)/(2m_{\pi })$ and $E_{\ast }=(m_{\pi }^2+m_e^2)/(2m_{\pi })$. The absolute derivative of the delta-function argument is $r_{\ast }/E_{\ast }+1=m_{\pi }/E_{\ast }$. Thus ${\Phi }_2=r_{\ast }/(4\pi m_{\pi })$. Multiplying by the spin sum and the initial factor $1/(2m_{\pi })$ yields the leptonic pseudoscalar decay widthIf the decay constant were instead defined by a matrix element $i{f}_{\pi }{p}^{\mu }$ without $\sqrt{2}$, then $f_{\pi }=\sqrt{2}{F}_{\pi }$ and the width coefficient would be $f_{\pi }^2/(8\pi )$; mixing these conventions would introduce a factor-of-two error.

The same derivation for a muon cancels the common hadronic and weak factors in the pion electron-to-muon decay ratio:The numerical value uses approximate masses $0.511,105.66,139.57$ MeV. Agreement with the corresponding Standard Model prediction tests lepton universality and the chiral, helicity-suppressing structure of the weak charged current, rather than the poorly calculable hadronic decay constant. A difference in the electron and muon weak couplings would multiply the ratio by their squared coupling ratio. Precision agreement must be assessed after radiative corrections, with matching photon-inclusive experimental definitions; the displayed tree-level ratio is not itself the radiatively corrected precision prediction.

An antilinear involution need not preserve norms, so the required physical assumption is that momentum reversal maps a complete normalized basis to a normalized basis. For a spinless sector take $\hat{T}|\mathbf{p}⟩=\eta (\mathbf{p})|-\mathbf{p}⟩$ with $|\eta |=1$ and relativistic normalization $⟨\mathbf{p}|\mathbf{q}⟩=(2\pi {)}^{3}2E_p{\delta }^3(\mathbf{p}-\mathbf{q})$. Any extra degeneracy labels must likewise be mapped by a norm-preserving permutation or unitary matrix. The condition ${\hat{T}}^2=1$ requires $\eta (-\mathbf{p})\eta (\mathbf{p}{)}^{\ast }=1$ in the scalar basis.

For arbitrary normalizable wave packets $|\varphi ⟩=\int d{\Pi }_{p}f(\mathbf{p})|\mathbf{p}⟩$ and $|\psi ⟩=\int d{\Pi }_{p}h(\mathbf{p})|\mathbf{p}⟩$, where $d{\Pi }_p=d^{3}p/[(2\pi {)}^{3}2E_p]$, antilinearity conjugates their coefficients. Using the normalized reversed basis givesCompleteness proves $⟨\hat{T}\varphi |\hat{T}\psi ⟩=⟨\varphi |\psi {⟩}^{\ast }$ for all states, so this quantum time-reversal operator is antiunitary. The same proof with sums over extra labels uses the unitarity of their reversal matrix.

For a Hermitian interaction-picture Hamiltonian, the condition of time-reversal symmetry isExpand the scattering matrix as a Dyson series, using ordered integration rather than commuting operators:Conjugation by $\hat{T}$ conjugates $(-i{)}^n$ to $i^n$ but preserves the product order. Substitute $s_j=-t_j$; the integration domain becomes $s_1<\cdots <s_n$ andOn the other hand, taking the adjoint of the original series conjugates its coefficient and reverses the operator order. Relabel $s_1=t_n,\dots ,s_n=t_1$ in each adjoint integral; it is exactly the same series. ThusEquivalently, time ordering turns into anti-time ordering after reversing the time arguments. One can first use a finite symmetric time interval and the same time-symmetric switching prescription before taking the scattering limit; the argument never assumes $[V(t),V(s)]=0$.

Antiunitarity relates amplitudes byUsing $\hat{T}S{\hat{T}}^{-1}=S^{†}$ and taking a complex conjugate gives the time-reversal reciprocity of the scattering matrixThe reverse process starts with the reversed final state and ends with the reversed initial state. Intrinsic state phases can appear when these reversed states are rewritten in a separately chosen basis, but the corresponding squared amplitudes agree. Equal amplitudes do not mean identical numerical cross sections when the reversed process has different flux or phase-space factors.

For a spin-zero complex scalar field, choose the transformation convention $\hat{T}\varphi (t,\mathbf{x}){\hat{T}}^{-1}={\eta }_T\varphi (-t,\mathbf{x})$. Its mode expansion of a free field isApply $\hat{T}$ to the expansion. Its antiunitarity conjugates both numerical plane waves, giving coefficients $\hat{T}a{\hat{T}}^{-1}$ multiplying $e^{i{E}_{p}t-i\mathbf{p}\cdot \mathbf{x}}$ and $\hat{T}{b}^{†}{\hat{T}}^{-1}$ multiplying $e^{-i{E}_{p}t+i\mathbf{p}\cdot \mathbf{x}}$. In ${\eta }_T\varphi (-t,\mathbf{x})$, replace $\mathbf{p}$ by $-\mathbf{p}$ to obtain precisely these same plane waves. Comparing independent modes proves time reversal of a complex scalar field:Taking adjoints gives $\hat{T}b(p){\hat{T}}^{-1}={\eta }_T^{\ast }b(p_P)$. Positive energy is preserved, and an annihilation operator remains an annihilation operator. Time reversal does not exchange particles with antiparticles.

The coefficient ${\eta }_T$ is an intrinsic time-reversal phase. Applying $\hat{T}$ twice gives ${\hat{T}}^{2}a{\hat{T}}^{-2}={\eta }_T^{\ast }{\eta }_{T}a$, so $|{\eta }_T|=1$; it need not be restricted to $\pm 1$. Write ${\eta }_T=e^{i\vartheta }$ and redefine $a^{\prime }=e^{i\vartheta /2}a$. Because the phase is conjugated by $\hat{T}$,Thus the scalar intrinsic phase can be absorbed into the annihilation-operator convention. Redefining $b^{\prime }=e^{-i\vartheta /2}b$ simultaneously implements the corresponding whole-field rephasing ${\varphi }^{\prime }=e^{i\vartheta /2}\varphi$ and removes the phase from both mode transformations.