Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 305 / 3

The Goldstone theorem states that every spontaneously broken generator of a continuous internal global symmetry in a Lorentz-invariant quantum field theory produces a massless scalar particle.

For the classical proof, let $V(\varphi )$ be the invariant scalar potential and let $v_i$ be a vacuum. Infinitesimal invariance givesDifferentiate with respect to ${\varphi }_k$ and evaluate at the stationary point $v$, where ${\partial }_{i}V(v)=0$. The scalar mass matrix then obeysFor every broken generator, $(T^{a}v{)}_i\ne 0$, so $T^{a}v$ is a zero eigenvector of the hessian matrix. It is a massless fluctuation tangent to the vacuum manifold.

For the quantum proof, spontaneous breaking means that some local field hasWrite $Q_a=\int d^{3}x{j}_a^0(x)$ and insert a complete set of momentum eigenstates into the current-field correlation function. Current conservation and Lorentz covariance imply that a scalar intermediate state couples asThe nonzero equal-time commutator requires a pole at $p^2=0$; otherwise the conserved-current spectral integral vanishes at zero momentum. Thus a massless Goldstone boson exists for every independent broken direction.

For $U(x)\in SU(3)$ and $A_{\mu }=A_{\mu }^a{T}^a$, choose $D_{\mu }={\partial }_{\mu }-ig{A}_{\mu }$. The Yang-Mills gauge transformation iswhich ensures $D_{\mu }^{\prime }(U\Phi )=U{D}_{\mu }\Phi$. Sinceconjugating this commutator immediately givesThus the gauge field strength is gauge covariant.

For one complex scalar in the fundamental representation, the most general power-counting-renormalizable gauge-invariant Lagrangian isapart from a constant and the four-dimensional Yang-Mills theta term. Stability requires $\lambda >0$. A cubic candidate ${ϵ}_{ijk}{\Phi }_i{\Phi }_j{\Phi }_k$ vanishes because the scalar components commute.

For ${\mu }^2>0$, the minima satisfy ${\Phi }^{†}\Phi =v^2/2$ with $v^2={\mu }^2/\lambda$. A global $SU(3)$ rotation can chooseThe transformations acting as $SU(2)$ on the first two entries leave this vector fixed, so the symmetry-breaking pattern isThere are $8-3=5$ broken generators. If the symmetry were global, the Goldstone theorem would therefore give five massless scalar modes, exactly the five tangent directions of the vacuum manifold $SU(3)/SU(2)$.

Parameterize the scalar locally aswhere the five $X^{\alpha }$ are broken generators. A gauge transformation removes all ${\pi }^{\alpha }$ in unitary gauge. Their derivative terms combine with the corresponding gauge fields in $(D_{\mu }\Phi {)}^{†}{D}^{\mu }\Phi$, supplying the longitudinal polarizations of five massive vectors. This is the Higgs mechanism.

With $T^a={\lambda }^a/2$, the gauge bosons $A_{\mu }^{1,2,3}$ of the unbroken $SU(2)$ remain massless. The four bosons $A_{\mu }^{4,5,6,7}$ haveand $A_{\mu }^8$ hasThe five eaten Goldstone modes complete their third polarizations. The sixth real component of the complex triplet remains as a radial Higgs boson withThis realizes the degree-of-freedom count summarized by Fundamental-Higgs breaking of SU(3) to SU(2).