Past exam of the theoretical physics major option course of the University of Oxford / 2016 / 7

Impose the discrete Z2 symmetry $\boxed{\Phi \mapsto -\Phi }$. The quadratic and quartic trace invariants are unchanged, while $\mathrm{tr}{\Phi }^3$ changes sign. Invariance therefore forbids a nonzero coefficient of that cubic operator. This is an additional symmetry of the scalar potential, not a consequence of SU(5) conjugation alone; cubic traces are permitted by the latter.

The spectral theorem for a Hermitian matrix gives a unitary diagonalizer $V$ with $V\Phi V^{†}=\mathrm{diag}(x_1,\dots ,x_5)$ and real $x_i$. Similarity preserves the trace, so ${\sum }_i{x}_i=0$.

To ensure that the diagonalizer belongs to SU(5) rather than only the unitary group $U(5)$, write $\det V=e^{i\alpha }$ and put $U=e^{-i\alpha /5}V$. Its determinant is one, and the scalar phase cancels from its conjugation action. ThusThe same argument shows why no additional reality condition beyond Hermiticity is needed. Permuting the real eigenvalues merely changes their order in the diagonal form.

Write $R^2={\sum }_i{x}_i^2$ and impose tracelessness by a Lagrange multiplier $\zeta$. Differentiating $V-\zeta {\sum }_i{x}_i$ givesFor $b\ne 0$, each eigenvalue is a real root of the same nonzero cubic polynomial, so at most three distinct values occur at any stationary point, in particular at any existing minimum. If the multiplicities are $n_1,\dots ,n_k$, the unbroken subgroup is the block centralizerIts connected Lie algebra is ${⨁}_{\alpha }\mathfrak{su}(n_{\alpha })\oplus \mathfrak{u}(1{)}^{k-1}$. The partitions allowed by this upper bound are $5$, $4+1$, $3+2$, $3+1+1$, and $2+2+1$. These are candidate patterns, not a proof that every one is a stable minimum for given couplings. For three distinct roots, the absent quadratic coefficient also implies that their unweighted sum is zero; tracelessness imposes an additional multiplicity-weighted relation.

The unrestricted printed claim has an exception. If $b=0$, $a>0$ and $m>0$, every traceless matrix with $R^2=m^2/(2a)$ is a global minimum. For examplehas five distinct eigenvalues and attains that minimum. Its unbroken subgroup is the maximal torus $U(1{)}^4$. The cubic argument therefore requires a nonzero $b$, or another condition excluding the identically vanishing stationarity polynomial. Couplings for which the potential is unbounded below have no global minimum to classify.

Use a Hermitian gauge field $A_{\mu }=A_{\mu }^A{T}^A$ and $\mathrm{tr}(T^A{T}^B)={\delta }^{AB}/2$. With fundamental covariant derivative ${\partial }_{\mu }-ig{A}_{\mu }$, the adjoint derivative isIt transforms as $D_{\mu }\Phi \mapsto U(D_{\mu }\Phi )U^{†}$, provided $A_{\mu }^{\prime }=U{A}_{\mu }{U}^{†}-(i/g)({\partial }_{\mu }U)U^{†}$.

The infinitesimal parameter in $U=1+iϵ+O({ϵ}^2)$ must be Hermitian and traceless, from unitarity and unit determinant. It is not generally antisymmetric as the hint states. The valid generator $ϵ=\mathrm{diag}(1,-1,0,0,0)$ is a concrete counterexample to that assertion.

Choose canonical adjoint kinetic normalization ${\mathcal{L}}_{\Phi }=\mathrm{tr}(D_{\mu }\Phi D^{\mu }\Phi )$, which gives $\frac{1}{2}\partial {\Phi }^A\partial {\Phi }^A$. For the diagonal vacuum expectation value $D=v\mathrm{diag}(2,2,2,-3,-3)$, the matrix unit $E_{ij}$ obeys $[D,E_{ij}]=(d_i-d_j)E_{ij}$. Thus the mass of the gauge boson associated with either normalized Hermitian generator joining $i,j$ is $g|d_i-d_j|$.

Within the three-dimensional and two-dimensional blocks, or for diagonal generators, the commutator vanishes. There are therefore $8+3+1=12$ massless gauge bosons. The $3\times 2$ interblock pairs supply two real generators each, hence 12 equally massive gauge bosons:The absolute mass formula assumes the stated kinetic normalization, which the question leaves implicit. Writing instead $\frac{1}{2}\mathrm{tr}(D\Phi {)}^2$ without rescaling $\Phi$ would multiply these squared masses by $1/2$.

This is the gauge group of the Standard Model, up to its global quotient, arising from one adjoint Higgs mechanism in a simple unified group. The unbroken abelian generator can be chosen as $Y=\mathrm{diag}(-1/3,-1/3,-1/3,1/2,1/2)$, so $D=-6vY$. This is the characteristic first breaking step of an SU(5) grand unified theory, rather than electroweak breaking all the way to electromagnetism.