Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2019 / 3

After integration by parts, the quadratic Proca action with an external source isSince $M>0$, this kernel is invertible with the Feynman i-epsilon prescription; the massless gauge redundancy is absent in this unitary gauge quadratic action. Complete the square by $A\mapsto A+K_F^{-1}J$ in the Gaussian functional integral. Its source-independent determinant cancels in the normalized generating functional:In Fourier space, the kernel isFor $k^2\ne 0$, use the transverse projector of a vector field $P_T^{\mu \nu }=g^{\mu \nu }-k^{\mu }{k}^{\nu }/k^2$ and the longitudinal projector of a vector field $P_L^{\mu \nu }=k^{\mu }{k}^{\nu }/k^2$. Their mixed-index products obey $P_T{P}_L=0$, soThe last expression also extends to $k^2=0$; that point only obstructs the intermediate projector decomposition. Two source functional derivatives yield the gauge-boson propagator $D_F=i{K}_F^{-1}$:The pole residue is the physical massive-vector massive vector polarization sum. The longitudinal term is essential for inverting the quadratic kernel, and makes the naive $M\to 0$ limit of this unitary gauge Proca propagator singular.

Expand the matrix trace explicitly:Write ${\Phi }_{ij}=(x_{ij}+i{y}_{ij})/\sqrt{2}$ and collect the $2N^2$ real fields into a vector $X$. ThenBoth quadratic forms are preserved by every constant orthogonal matrix $R$, acting as $X\mapsto RX$. Thus the full linear internal symmetry contains $O(2N^2)$, with connected special orthogonal group $SO(2N^2)$ as requested. The enlargement beyond unitary transformations is possible because the real and imaginary components have identical kinetic and mass terms and no interactions distinguishing them.

For a Hermitian matrix $\Phi$, there are $N$ real diagonal entries and $N(N-1)/2$ independent complex off-diagonal entries, hence $N^2$ real fields. Choose an orthonormal basis $T_A$ of Hermitian matrices for the trace inner product, $\mathrm{tr}(T_A{T}_B)={\delta }_{AB}$, and put $\Phi ={\sum }_A{X}_A{T}_A/\sqrt{2}$. The same calculation giveswhere these are the connected rotation groups; their full orthogonal extensions are $O(2N^2)$ and $O(N^2)$. A purely quadratic theory with $m^2<0$ would have an unstable scalar potential, but this algebraic symmetry statement still holds.

Let $X={\Phi }^{†}\Phi$. Under the global transformation of the chiral matrix scalar model,The phase and $U_L^{†}{U}_L$ cancel, and $U_R^{†}{U}_R=1$. The cyclic property of the trace impliesTherefore each of the three terms in the scalar potential is invariant separately. Since $\theta ,U_L,U_R$ are constant, ${\partial }_{\mu }{\Phi }^{\prime }=e^{i\theta }{U}_L({\partial }_{\mu }\Phi )U_R^{†}$, giving invariance of the kinetic matrix trace by the same argument. Henceis a global symmetry group of the Lagrangian density. Its action has a finite central kernel, which does not affect its Lie algebra or the broken-generator counting below. For special parameter choices the symmetry can be larger; the displayed group is a symmetry for the stated potential without such extra assumptions.

It is useful first to establish the global minimum, rather than only a stationary point. The positive semidefinite matrix $X={\Phi }^{†}\Phi$ has nonnegative eigenvalues $x_1,\dots ,x_N$. For fixed $r={\sum }_i{x}_i$, the Cauchy-Schwarz inequality gives ${\sum }_i{x}_i^2\ge r^2/N$, with equality exactly when all $x_i$ are equal. Since $\alpha >0$,For $m^2<0$ and $\beta >0$, this lower bound has its minimum atThe bound is attained, so every minimum has ${\Phi }^{†}\Phi =C^{2}1$, or $\Phi =CW$ with $W$ a unitary matrix. A global transformation removes $W$ by decomposing it into an overall phase and an $SU(N)$ matrix. Thus ${\Phi }_0=C1$ is a valid representative of the vacuum manifold.

Preservation of this chosen vacuum expectation value requiresInfinitesimally, with traceless Hermitian generators $L,R$, this reads $\theta 1+L-R=0$. Taking the trace forces $\theta =0$, and then $L=R$. The surviving continuous transformations are consequentlyAll other independent continuous directions change the chosen vacuum expectation value. Thus the number of broken generators isGlobally, taking the determinant of the stabilizer equation permits the discrete phases $e^{iN\theta }=1$, paired with $U_L=e^{-i\theta }{U}_R$. They induce the same conjugation action on $\Phi$. The faithful unbroken conjugation group is $SU(N)/{\mathbb{Z}}_N$; for example, at $N=2$, $U=-1$ acts exactly like $U=1$. Thus the question's description by $SU(N)$ is correct as a parametrization and at the Lie algebra level, with this central redundancy understood. It leaves the number of broken generators unchanged.

Write $H={\varphi }_1$ and $P={\varphi }_2$, both Hermitian matrices. Through quadratic order around the chosen vacuum expectation value,The trace of a commutator vanishes. The needed matrix traces are thereforeUsing $m^2=-(\alpha +N\beta )C^2$, all linear terms cancel. The quadratic scalar potential reduces towith no $P^2$ term. The kinetic term issince its mixed term is the trace of a commutator. A Hermitian matrix has $N^2$ independent real components, so $P$ describes exactly $N^2$ massless scalar modes.

To determine the other masses, take $T_0=1/\sqrt{N}$ and traceless Hermitian $T_A$ with $\mathrm{tr}(T_A{T}_B)={\delta }_{AB}$, and expand $H=h_0{T}_0+{\sum }_{A=1}^{N^2-1}{h}_A{T}_A$. These are canonically normalized real scalar fields. The scalar mass matrix is diagonal in this basis:Thus $H$ contains one massive singlet and a degenerate multiplet of $N^2-1$ massive fields in the Adjoint representation of the unbroken diagonal $SU(N)$. Both squared masses are strictly positive for the specified parameters. The $P$ fields are the $N^2$ Goldstone bosons associated with the broken generators: $N^2-1$ traceless relative left-right rotations and one overall phase rotation.

Goldstone's theorem guarantees those massless directions. It does not alone forbid additional accidental massless modes; the positive scalar mass matrix computed here supplies the further conclusion that every $H$ component is massive. The fields describe a global symmetry, so these Goldstone bosons are physical rather than being absorbed into massive gauge bosons.