Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 52 (The Standard Model)

Use metric signature $+---$ and the parity matrix $P^{\mu }{}_{\nu }=\mathrm{diag}(1,-1,-1,-1)$. If a Dirac field transforms as ${\psi }^{\prime }(x)={\eta }_P{S}_P\psi (x_P)$, the chain rule changes the spatial derivatives' signs. Covariance of its Dirac equation therefore requiresThe Clifford algebra gives precisely these identities for $S_P={\gamma }^0$. In an irreducible Dirac representation any other solution differs by a scalar phase, because its ratio with ${\gamma }^0$ commutes with all gamma matrices. With $|{\eta }_P|=1$, the Dirac mass bilinear and kinetic term consequently obeyThe factor ${\gamma }^0$ compensates the sign reversal of spatial derivatives. A transformation containing only the argument change would not preserve the massive Dirac equation. This establishes parity symmetry of the free or vector-coupled Dirac theory; it does not make a chiral weak interaction parity symmetric.

For charge conjugation, transpose the adjoint Dirac equation to obtain $(i{\gamma }^{\mu T}{\partial }_{\mu }+m){\bar{\psi }}^T=0$. Multiplying by $C$ gives the original Dirac equation for ${\psi }^c=C{\bar{\psi }}^T$ ifOne may choose $C$ unitary; in four dimensions a conventional choice in a standard Dirac or chiral basis is $C=i{\gamma }^2{\gamma }^0$, with $C^T=-C$. Its overall phase and ${\eta }_C$ do not affect the following bilinear transformation. Reordering the Grassmann fermion fields gives, for a color matrix $X$ commuting with the gamma matrices,Thus a vector color current changes sign and transposes its color generator under charge conjugation.

Write the Hermitian color connection as ${\mathcal{A}}_{\mu }=A_{\mu }^a{T}^a$. The interaction coming from $D_{\mu }={\partial }_{\mu }+ig{\mathcal{A}}_{\mu }$ is $-g\bar{\psi }{\gamma }^{\mu }{\mathcal{A}}_{\mu }\psi$. Its charge-conjugation symmetry requires ${\mathcal{A}}_{\mu }^C=-{\mathcal{A}}_{\mu }^T$. Parity acts on the Lorentz index as on a vector connection, so combining the two givesIn components, ${\mathcal{A}}_0\mapsto -{\mathcal{A}}_0^T(x_P)$ and ${\mathcal{A}}_i\mapsto +{\mathcal{A}}_i^T(x_P)$. The transpose is a color-space transpose, independent of the spinor matrix $C$. Intrinsic fermion phases cancel between a field and its adjoint.

Define the matrix gauge field strength unambiguously byThe derivative terms transform with the expected parity factors and a minus transpose. For the commutator, the crucial identity is $[X^T,Y^T]=-[X,Y{]}^T$, so the nonlinear term has the same transformation as the derivative terms. This is the CP transformation of non-Abelian field strength:For example, ${\mathcal{F}}_{0i}\mapsto +{\mathcal{F}}_{0i}^T(x_P)$ while ${\mathcal{F}}_{ij}\mapsto -{\mathcal{F}}_{ij}^T(x_P)$. With $\mathrm{tr}(T^a{T}^b)=\frac{1}{2}{\delta }^{ab}$, the color sum in the theta operator is twice a trace of two field strengths. The two charge-conjugation minus signs cancel, and transposition reverses the trace factors without changing their trace. The four parity matrices contracted with the Levi-Civita symbol contribute $\det P=-1$. HenceThe theta operator is CP odd; a generic fixed nonzero coefficient violates CP. Setting the coefficient to zero removes this term. Although the density is locally a total derivative, nontrivial gauge topology means that it need not be irrelevant to the quantum theory. If a properly normalized theta angle is identified periodically, invariance at special values equivalent to their negatives is a separate global question; it does not change the CP odd transformation of the local operator.

A stable symmetry-breaking minimum requires $\lambda >0$ and $m^2<0$. Put $v^2=-m^2/\lambda >0$. The common scalar potential can then be written, up to a constant, asIts minima form the sphere $\Phi \cdot \Phi =v^2$. Choose the vacuum direction ${\Phi }_0=(0,0,v{)}^T$. If instead $m^2>0$ with positive $\lambda$, the minimum is at the origin and the assumed nonzero vacuum expectation value is not a stable vacuum of these models.

In the ungauged theory, the internal symmetry is global $O(3)$; its connected part is $SO(3)$. Constant orthogonal transformations preserve both the kinetic term and the potential. The chosen vacuum is invariant under $O(2)$ acting on its first two components, so the continuous breaking is $SO(3)\to SO(2)$ with two broken generators. The vacuum orientations are genuinely different degenerate global vacua.

The global triplet scalar symmetry breaking is visible directly in the fluctuation masses. Write $\Phi =({\pi }_1,{\pi }_2,v+h{)}^T$. The potential Hessian isThe quadratic Lagrangian is thereforeThere is one radial scalar of mass $m_h^2=2\lambda v^2=-2m^2$ and two massless Goldstone bosons. The two Goldstone modes describe motion tangential to the vacuum sphere. They remain physical, since a global rotation has only constant parameters and cannot remove arbitrary spacetime-dependent fluctuations. All three original real-scalar degrees of freedom remain. Interactions persist: the shifted potential is $V-V_{\min }=\frac{\lambda }{4}(2vh+h^2+{\pi }_1^2+{\pi }_2^2{)}^2$, containing cubic and quartic terms. Masslessness of the angular modes is protected by the exact continuous symmetry through Goldstone's theorem.

In the gauged theory, the three-component real field is the vector representation of local $SO(3)$, equivalently the Adjoint representation of $SU(2)$. The center of $SU(2)$ acts trivially on this field, so both descriptions have the same local field content and perturbative spectrum. Take $(t^a{)}_{bc}=-i{ϵ}^{abc}$, for which $[t^a,t^b]=i{ϵ}^{abc}{t}^c$. The covariant kinetic term and the gauge kinetic term are invariant under local transformations with the associated connection transformation. The scalar inversion $\Phi \mapsto -\Phi$, with the gauge field unchanged, is also a discrete invariance of the displayed action.

The vacuum leaves rotations about its third axis unbroken. The residual gauge group is $SO(2)$, or the corresponding $U(1)$ subgroup in the $SU(2)$ description. Expanding the covariant kinetic term about ${\Phi }_0$ givesThus the gauge-boson mass matrix is $g^2{v}^2\mathrm{diag}(1,1,0)$. The third generator annihilates the vacuum, while the first two do not. The masses are $m_{B^1}=m_{B^2}=gv$, $m_{B^3}=0$, and $m_h=\sqrt{2\lambda }v$.

The quadratic derivative mixing displays what happens to the Goldstone fields. With the generator convention just chosen,Local rotations can set the two angular fields to zero in unitary gauge. Then $\Phi =(0,0,v+h{)}^T$ andThe two Goldstone modes supply the longitudinal polarizations of the two massive vector bosons. They do not survive as additional physical massless scalars. The radial scalar is still physical, and $B^3$ retains only two transverse polarizations. This adjoint triplet Higgs spectrum realizes the Higgs mechanism; the gauge group is only partially broken, so one massless gauge boson remains.

The degree-of-freedom check isIn contrast, the global theory has just the three scalar modes, including its two physical Goldstone bosons. A vacuum expectation value in the local theory is a choice of gauge-fixed description: the direction can be changed by gauge transformations and is not itself an observable. Gauge-independent masses and the physical degree count express the actual effect. Both theories retain Lorentz symmetry in the constant scalar vacuum.

Put $s_W=\sin {\theta }_W$ and $c_W=\cos {\theta }_W$. The left-handed doublet haswhile the right-handed charged-lepton singlet has $T_R^3=0$ and $Q_R=-1$. There is no weakly coupled right-handed neutrino in the stated field content. For a fermion $f$, write $g_L^f=T_{3L}^f-Q_f{s}_W^2$ and $g_R^f=T_{3R}^f-Q_f{s}_W^2$, with $g_R^{\nu }=0$ here. Since ${\bar{f}}_L{\gamma }^{\mu }{f}_L=\bar{f}{\gamma }^{\mu }{P}_{L}f$ and similarly for $R$,Consequently the neutral-current vector and axial couplings in the specified overall normalization areThese coefficients accompany $g/c_W$. A convention with $g/(2c_W)$ uses twice these vector and axial coefficients; mixing the two conventions would make the widths wrong by a factor of four. For the neutrino, $c_V-c_A{\gamma }^5=\frac{1}{2}{P}_L$, so the chiral projection is already built into the vertex.

For either massless final pair with momenta $k_1,k_2$, $p=k_1+k_2$, the invariant amplitude is, up to an irrelevant overall phase,The spin-summed fermion tensor isThe symmetric part, using the four-gamma trace, isThe vector-axial interference is antisymmetric in $\mu ,\nu$, so it drops out of the symmetric $Z$ polarization sum. Also $p_{\mu }{T}_f^{\mu \nu }=0$ for massless external fermions, by their Dirac equations; hence the $p_{\mu }{p}_{\nu }/m_Z^2$ part contributes zero. Using $k_1\cdot k_2=m_Z^2/2$ givesThe massive-vector spin average supplies the factor $1/3$ for the three spin states of the massive $Z$. The massless two-body Lorentz-invariant phase space is $d{\Phi }_2=d\Omega /(32{\pi }^2)$, and so $\int d{\Phi }_2=1/(8\pi )$. The spin-averaged amplitude above is angle-independent in the rest frame, givingThe massless leptonic Z decay widths for one charged-lepton flavor and one active neutrino flavor respectively areThere is no color factor for leptons and no extra final-state identical-particle factor for a particle-antiparticle pair. The neutrino projectors already exclude sterile helicity states, so no additional factor of two is needed. These are tree-level massless widths; electroweak radiative corrections and charged-lepton masses have not been included. If desired, using $G_F=g^2/(4\sqrt{2}{m}_W^2)$ and $m_W=c_W{m}_Z$ converts the common prefactor to $G_F{m}_Z^3/(12\sqrt{2}\pi )$.

The flavor content is $K^0=d\bar{s}$ and ${\bar{K}}^0=s\bar{d}$. A charged-current box diagram changes strangeness by two units. The two internal quark lines can contain any up-type flavors $u_i,u_j\in \{u,c,t\}$, connected by two charged $W$ propagators. One allowed topology is shown below; its crossed counterpart also contributes to the full mixing amplitude.

Each charged-current vertex contains the appropriate CKM matrix element. The flavor sum has combinations ${\lambda }_i{\lambda }_j$, with ${\lambda }_i=V_{is}^{\ast }{V}_{id}$, multiplying mass-dependent loop functions. CKM matrix unitarity gives ${\sum }_i{\lambda }_i=0$, illustrating the GIM cancellation of flavor-independent loop terms. The diagram is at order $g^4$ and second order in the weak interaction.

Let $\Theta$ be the antiunitary CPT operator. It interchanges the neutral-kaon flavor states up to phases. For a Hermitian Hamiltonian invariant under CPT,since the diagonal expectation is real. Therefore CPT gives $R_{11}=R_{22}$. More generally an effective decay Hamiltonian has $R=M-i\Gamma /2$, where both $M$ and $\Gamma$ are Hermitian. CPT gives $M_{11}=M_{22}$ and ${\Gamma }_{11}={\Gamma }_{22}$, hence the same equality of the complex diagonal entries. It does not require $R_{12}=R_{21}$; that is a CP condition. Nor is $R_{21}=R_{12}^{\ast }$ true for the full decay matrix in general.

Choose the CP convention $CP|K^0⟩=-|{\bar{K}}^0⟩$ and $CP|{\bar{K}}^0⟩=-|K^0⟩$. Then CP acts as $-{\sigma }_x$ on the flavor basis. Invariance of $H^{\prime }$ means $R={\sigma }_{x}R{\sigma }_x$, and thereforeWith a different flavor-state phase convention this relation carries the corresponding phase factors; equality is the relation in the convention adopted here. For a Hermitian mass matrix it makes the off-diagonal element real.

For the requested production-time mass eigenstates, discard the absorptive part. From now on $R$ denotes the Hermitian part $(R_{\mathrm{original}}+R_{\mathrm{original}}^{†})/2$, so its entries have the formIf the original matrix already was Hermitian no replacement is needed. The neutral-kaon mass matrix diagonalization gives eigenvalues $r_0\pm |z|$, since $(r_0-\lambda {)}^2-|z{|}^2=0$. Write $z=r{e}^{i\phi }$, $r=|z|>0$. Use the kaon mixing square-root branch conventioncontinuously from the CP-conserving convention in which $z>0$. Independent unrelated root signs would interchange the eigenvalue labels. Direct multiplication gives $R(a,-b{)}^T=(r_0-r)(a,-b{)}^T$ and $R(a,b{)}^T=(r_0+r)(a,b{)}^T$. Assigning the larger mass to $K_L^0$ giveswith mass shifts $r_0-r$ and $r_0+r$ respectively. A common strong-interaction mass may simply be added to both. Their norms are one and their inner product is zero. A Hermitian mass-only calculation determines the lighter and heavier combinations; it does not determine their different lifetimes.

In the CP-conserving limit, the chosen lighter and heavier combinations areThey are CP eigenstates with eigenvalues $+1$ and $-1$ in the convention above. The sign of the real off-diagonal entry and the flavor-state convention are chosen together so that the stated $K_1^0=K_S^0$ limit is the lower-mass state.

Changing to this CP basis givesConsequently, provided $a+b\ne 0$, defineAbsorbing the common phase of $a+b$ into the definitions of the mass eigenstates and normalizing now yieldsFor this Hermitian approximation, $ϵ=i\tan (\phi /2)$ is purely imaginary, so $⟨K_S^0|K_L^0⟩=(ϵ+{ϵ}^{\ast })/(1+|ϵ{|}^2)=0$. The phase convention matters for the kaon CP mixing parameter. Physical kaon decay eigenstates instead diagonalize the generally non-Hermitian matrix $M-i\Gamma /2$, for which the mixing parameter can have a real part and the eigenstates need not be orthogonal. The formula here is the requested mass-only result, not a calculation of that full decay dynamics.