Past exam of the theoretical physics major option course of the University of Oxford / 2018 / 2

Goldstone's theorem states that a spontaneously broken continuous internal global symmetry in field theory of a local relativistic quantum field theory produces massless Goldstone bosons. For a Lorentz invariance vacuum breaking $G$ to its stabilizer $H$, there are $\dim G-\dim H$ independent symmetry-required massless directions. Additional accidentally massless modes are possible. A gauge redundancy does not obey this physical-state counting, and nonrelativistic systems need not have one independent mode per broken generator.

For the classical proof, consider real scalar fields ${\varphi }_i$ with a positive canonical kinetic term and an invariant smooth scalar potential $V$. An infinitesimal generator $T^a$ satisfiesLet $v$ be a stationary vacuum and differentiate this identity with respect to ${\varphi }_j$. At $v$, the term proportional to ${\partial }_{i}V$ vanishes, leavingThe Hessian matrix ${\mathsf{M}}^2$ is the scalar mass matrix. An unbroken generator has $T^{a}v=0$; broken generators span the tangent space of the vacuum manifold $G/H$. Each independent nonzero tangent vector is a zero eigenvector of ${\mathsf{M}}^2$. The quadratic Euler-Lagrange equations in these directions are $\square {\pi }^a=0$, giving the massless dispersion relation ${\omega }^2={\mathbf{p}}^2$. This proves the classical result and explains why the Goldstone bosons describe motion along the degenerate minima rather than against a restoring curvature.

Both the canonical kinetic term and $V({\varphi }^{\mathsf{T}}\varphi )$ are invariant under the orthogonal group $O(n)$. The stated nonzero vacuum expectation value has length $|v|$. Its stabilizer is the orthogonal group $O(n-1)$ acting on the perpendicular subspace. HenceThe two nonzero coordinates of this particular vacuum do not break more generators than an axis-aligned vacuum: the two choices are related by a common orthogonal transformation.

Explicitly, with $s={\varphi }^{\mathsf{T}}\varphi$,stationary point at nonzero $v$ requires $V^{\prime }(v^2)=0$. The $n-1$ transverse eigenvalues vanish, while the radial squared mass is $4v^2{V}^{\prime \prime }(v^2)$. If this curvature also vanishes, there is an additional accidental massless radial mode; the number of symmetry-protected Goldstone bosons remains $n-1$. If $v=0$, the symmetry is unbroken and none is required.

Assume ${\mu }^2,\lambda >0$. At zero source, the vacuum manifold is the sphere of radius $v_0=\mu /\sqrt{\lambda }$. The source tilts the scalar potential by $-ϵ{\varphi }_1$. Its force should be small compared with the radial restoring-force scale ${\mu }^2{v}_0$, so a useful quantitative condition isFor nonzero $ϵ$, the exact global symmetry in field theory of the Lagrangian is $O(n-1)$, rotating ${\varphi }_2,\dots ,{\varphi }_n$. The global vacuum aligns with the source:This vacuum preserves the entire remaining $O(n-1)$ symmetry. There is therefore no exact Goldstone boson for $ϵ\ne 0$. Expanding the scalar potential to quadratic order givesThus there are $n-1$ degenerate light pseudo-Goldstone bosons and one heavy radial scalar. The small-source condition makes their squared masses parametrically smaller than the radial squared mass. Sending $ϵ$ to zero restores the original massless modes.

The breaking of the internal global symmetry in field theory $SU(n)\times SU(n)$ to $SU(n)$ removes $2(n^2-1)-(n^2-1)$ generators. The relativistic Goldstone theorem therefore requiresOther states fall into multiplets of the unbroken special unitary group, but their masses are not fixed by this counting. For the usual diagonal embedding $SU(n{)}_V\subset SU(n{)}_L\times SU(n{)}_R$, the Goldstone bosons form its adjoint multiplet.

If small explicit symmetry-breaking terms are added, the broken-generator directions generally acquire small curvature, producing light pseudo-Goldstone bosons rather than exactly massless states. Their squared masses tend to zero with those breaking parameters. Small breaking can also split multiplet degeneracies. These claims concern a small perturbation of the broken phase; approximate symmetry alone does not establish which vacuum is selected.

The light states have the counting $3+4+1=8$. This suggests approximate chiral symmetry breakingwith a nonzero quark chiral condensate $⟨{\bar{q}}_R^j{q}_L^i⟩\propto {\delta }^{ij}$ for the three light flavours $u,d,s$. The three pions, four kaons and predominantly octet eta meson are the resulting pseudo-Goldstone bosons. Their small nonzero masses reflect explicit chiral symmetry breaking by the light-quark masses; for example, $m_{\pi }^2$ is proportional to $m_u+m_d$ at leading order. The larger strange-quark mass makes the kaons and octet eta meson heavier than the pions, while electromagnetic interactions help split charged and neutral partners.^[ref]

Although the massless strong-interaction terms admit independent rotations of all six flavours, the charm, bottom and top masses are too large to treat six-flavour chiral symmetry as a small explicit breaking at the light-hadron scale. Full approximate $SU(6{)}_L\times SU(6{)}_R$ breaking would suggest 35 light modes, rather than this octet. The classical axial $U(1)$ is also broken by the chiral anomaly, so it does not supply a ninth equally light Goldstone boson.

Terminology qualification: these mesons have spin zero but negative parity: they are pseudoscalar mesons, with $J^P=0^{-}$, rather than positive-parity scalar particles with $J^P=0^{+}$. Calling the listed states “scalar” is acceptable only if it is being used loosely to mean spin zero.