Past exam of the advanced quantum field theory course of the University of Oxford / 2022 / 3 / a

The Goldstone theorem states that spontaneous breaking of a continuous internal global symmetry in field theory in a local relativistic quantum field theory produces a zero-mass Goldstone boson for each independent broken generator, under the usual positive-state-space and conserved-current hypotheses. For $G\to H$ there are $\dim G-\dim H$ symmetry-required Goldstone bosons. This count assumes a Lorentz-invariant vacuum state; nonrelativistic counting can differ.

At the classical level, invariance of the scalar potential gives ${\partial }_{i}V(T^a\varphi {)}_i=0$. Differentiating at a stationary vacuum expectation value ${\varphi }_0$ yieldsThese are the Goldstone directions in the scalar mass matrix. Broken generators have $T^a{\varphi }_0\ne 0$, so their independent directions have zero quadratic mass with a canonical kinetic term. The theorem guarantees these modes; it does not exclude further accidental zero-mass modes. A local gauge symmetry is a redundancy and does not meet the physical global-symmetry hypothesis: the Higgs mechanism can instead supply longitudinal particle polarizations of massive gauge bosons.

Put $r^2={\varphi }_i{\varphi }_i$. Completing the square in the scalar potential givesSince $\lambda >0$, its global minima form the vacuum manifoldHere $\mu$ denotes the positive square root of the parameter ${\mu }^2$. The origin is a local maximum, not a ground state. Choose $⟨\mathbf{\varphi }⟩=(v,0,0)$. Its stabilizer subgroup is the SO(2) subgroup of rotations in the $(2,3)$ plane, so the stated SO(3) global symmetry in field theory is spontaneously broken to SO(2).

Write ${\varphi }_1=v+h$, ${\varphi }_2={\pi }_2$, ${\varphi }_3={\pi }_3$. The kinetic terms stay canonically normalized, while the scalar mass matrix isEquivalently, the quadratic scalar potential is ${\mu }^2{h}^2$. Therefore the particle masses areThe two angular Goldstone bosons match $\dim SO(3)-\dim SO(2)=3-1=2$. The radial displacement changes $r$ and costs potential energy, giving the massive Higgs mode. These are tree-level masses; the Goldstone theorem protects the angular zero-mass modes in the exact global theory.

For any fixed $r\ge 0$, the source contribution obeys $c{\varphi }_1\ge cr$, because $c<0$ and ${\varphi }_1\le r$. Equality occurs only at $({\varphi }_1,{\varphi }_2,{\varphi }_3)=(r,0,0)$ when $r>0$. Thus the global ground state lies on the positive first axis. Minimize there:For $0\le r\le v=\mu /\sqrt{\lambda }$, $f^{\prime }(r)<0$. For $r>v$, $f^{\prime \prime }(r)=3\lambda r^2-{\mu }^2>0$, so $f^{\prime }$ increases strictly from $c$ to $+\infty$. There is exactly one root $\tilde{v}>v$, and it is the unique global minimum. ThereforeThis argument selects the true ground state even if the full cubic has additional negative stationary roots.

Expand ${\varphi }_1=\tilde{v}+h$, ${\varphi }_2={\pi }_2$, ${\varphi }_3={\pi }_3$. The source is linear, so it shifts the expansion point but has zero Hessian matrix. The scalar mass matrix is diagonal at the new vacuum expectation value:Eliminating $\lambda$ with the stationary equation givesAll three squared masses are positive. The masses themselves are the positive square roots of these expressions.

The source explicitly reduces the stated SO(3) symmetry to SO(2); the selected vacuum state preserves that remaining SO(2). There is consequently no spontaneously broken exact continuous global symmetry in field theory to which the Goldstone theorem applies. For small $|c|$, the two angular excitations are pseudo-Goldstone bosons. Their squared masses vanish as $c\to 0^{-}$, while $\tilde{v}\to \mu /\sqrt{\lambda }$ and $m_h^2\to 2{\mu }^2$.