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The minima of the scalar potentialsatisfyThe SU(2) group acts transitively on this three-sphere of vacua, and the stabilizer of a nonzero fundamental doublet is trivial. A global transformation may therefore choose ${\varphi }_0=(0,v{)}^T/\sqrt{2}$. Because the symmetry is gauged, unitary gauge removes all three angular Goldstone bosons, leaving only the real radial Higgs mode $h$:Substitution into the gauge-covariant kinetic term giveswithThe interaction terms produce $h^3$, $h^4$, $hBB$, $hhBB$, and the cubic and quartic non-Abelian gauge-boson vertices shown below.

The three broken generators supply the three longitudinal polarizations of the equally massive gauge bosons. No physical massless Goldstone boson remains, and because the unbroken subgroup is trivial there is no massless gauge boson either. The remaining physical spectrum has $3\times 3+1=10$ degrees of freedom, equal to the original $3\times 2+4=10$.

At energy $E\gg M_X$, the longitudinal polarization behaves as ${ϵ}_L^{\mu }\sim p^{\mu }/M_X$. A Proca theory with arbitrary interactions can therefore produce longitudinal-vector scattering amplitudes growing like powers of $E/M_X$, eventually violating partial-wave unitarity and invalidating perturbation theory.

In a gauge-theory completion, the Higgs mechanism identifies the longitudinal mode with a would-be Goldstone boson. Gauge relations fix the vector, Goldstone, and Higgs couplings. Higgs-exchange diagrams then cancel the leading energy growth of the gauge diagrams, as in longitudinal $W$-boson scattering, leaving amplitudes compatible with perturbative unitarity up to the scale where the full renormalizable theory itself becomes strongly coupled.

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.

The longitudinal polarization of a massive vector behaves as ${ϵ}_L^{\mu }(p)\sim p^{\mu }/m$ at high energy. Generic self-interactions therefore make amplitudes grow as powers of $E/m$, eventually violating partial-wave unitarity and destroying perturbative predictivity. Simply adding a Proca mass also spoils the gauge cancellations that control ultraviolet behavior.

The Standard Model obtains the $W$ and $Z$ masses through the Higgs mechanism inside a renormalizable gauge theory. Goldstone bosons supply the longitudinal polarizations, and relations among gauge and Higgs couplings enforce cancellations of the growing terms. In particular, Higgs exchange cancels the remaining $E^2$ growth in longitudinal vector-boson scattering.

For an unbroken internal symmetry, $[H,Q^a]=0$ and $Q^a|0⟩=0$. If $H|n⟩=E_n|n⟩$, then $Q^a|n⟩$ is either zero or another state of the same energy, so particles fill degenerate symmetry multiplets.

In a spontaneously broken phase, applying the finite symmetry changes the vacuum and moves to a distinct infinite-volume superselection sector. The charge is not a well-defined finite-norm operator within one such sector, so the usual argument that $Q^a|n⟩$ is a physical partner state can fail; particle masses need not occur in multiplets of the broken group. Formally, however,Thus the broken directions are degenerate with the vacuum. In a local relativistic theory these zero-energy, long-wavelength excitations become the Goldstone bosons.

At zero temperature, the Matsubara sum becomes a frequency integral:The frequency integral is $1/(2\sqrt{v^2{k}^2+m^2})$, and radial momentum integration givesDefine the critical coupling by the massless equationTaking the cutoff to infinity in the difference givesand henceA positive mass solution exists only for $g>g_c$. For $g<g_c$, the symmetric saddle cannot enforce the constraint with $m^2>0$; instead the $O(N)$ symmetry is spontaneously broken to $O(N-1)$, the field acquires Néel order, and the ordered phase contains massless Goldstone bosons.

Stability requires $\lambda >0$. The classical potential $r\rho +\lambda {\rho }^2$ has a symmetry-breaking minimum when $r<0$, atWriting $\rho =\bar{\rho }+\delta \rho$ and discarding constants and total derivatives gives the quadratic LagrangianThe density fluctuation is a gapped amplitude mode, while the phase is the prospective Goldstone boson.

The quadratic path integral isIntroduce the positive spatial operatorCompleting the square in the Gaussian functional integral over $\delta \rho$ yieldsThe derivative expansion $A^{-1}=(2\lambda {)}^{-1}+O({\nabla }^2)$ therefore givesThis is the long-wavelength Goldstone boson action. Keeping the spatial derivative in $A$ gives ${\omega }^2=v^2{k}^2+k^4/(4m^2)$.

For ${\mu }^2<0$ and $g>0$, minimizing the $O(N)$-invariant potential givesThe free energy has the full $O(N)$ symmetry, but choosing one point on this sphere leaves only the rotations $O(N-1)$ fixing that point. This is spontaneous symmetry breaking, $O(N)\to O(N-1)$. The $N-1$ tangent directions along the sphere cost no potential energy and are the massless Goldstone modes predicted by the Goldstone theorem.

Writing $\Phi =({\varphi }_1,{\varphi }_2{)}^T$, the scalar sector has a global $SU(2)$ flavour symmetry in addition to the gauged common $U(1)$ phase. A vacuum can be chosen as ${\Phi }_0=(0,v{)}^T$. A diagonal combination of the gauge phase and the $SU(2)$ rotation generated by ${\sigma }_3$ leaves it invariant.

The radial scalar has $m_h^2=2\lambda v^2$, the gauge boson has $m_A^2=2e^2{v}^2$, and two real scalars remain massless. The four real scalar directions consist of one radial mode and three angular modes; one angular mode is eaten, leaving the two Goldstone bosons expected from the two broken physical global generators. The vacuum set before quotienting is $S^3$, and the manifold of gauge-inequivalent vacua is the complex projective line

Use unitary gauge and write $\varphi =v+h/\sqrt{2}$. The quadratic Lagrangian containsThus the radial Higgs excitation is a spin-zero particle with $m_h^2=2\lambda v^2$, while the gauge boson is a massive spin-one particle with $m_A^2=2e^2{v}^2$. The phase of $\varphi$ is the Goldstone boson eaten by the Higgs mechanism to supply the longitudinal polarization of $A_{\mu }$.

Put ${\rho }_i=|{\psi }_i{|}^2$. The uniform potential isIf both masses are positive, ${\rho }_1={\rho }_2=0$. If ${\mu }_1^2<\min (0,{\mu }_2^2)$, only ${\psi }_1$ condenses, with ${\rho }_1=-{\mu }_1^2/(2g)$; symmetrically, only ${\psi }_2$ condenses when ${\mu }_2^2<\min (0,{\mu }_1^2)$. The positive coordinate axes are continuous-transition lines. On ${\mu }_1^2={\mu }_2^2={\mu }^2<0$, every pair withis a minimum; crossing this diagonal exchanges the two condensates, so it is a coexistence line ending at the origin.

Away from the diagonal the symmetry is $U(1{)}_1\times U(1{)}_2$. It is unbroken in the normal phase. In either one-condensate phase one factor is broken and the other remains, giving one Goldstone boson. On the diagonal the symmetry is enhanced to $U(2)$. For positive equal mass it is unbroken; for negative equal mass it breaks as $U(2)\to U(1)$ and gives three Goldstone bosons. At the origin $U(2)$ is unbroken although both quadratic modes are critical.

Let $x_a\ge 0$ be the eigenvalues of the positive semidefinite matrix $M^{†}M$. The potential isFor $k<0$ and $\lambda >0$, each summand is minimized atso $M_0^{†}{M}_0=v^{2}1$. A symmetry transformation preserves the representative $M_0=v1$ precisely when $L=R$, givingThe number of broken generators is $2N^2-N^2=N^2$, so the Goldstone theorem predicts $N^2$ modes, each a Goldstone boson.

A nonzero vacuum value of the scalar quark density breaks the massless nonsinglet chiral symmetry to its diagonal vector subgroup. Indeed, the nonsinglet axial charge algebra of quark densities gives $⟨[Q_{5,a},P_b]⟩=-i{\delta }_{ab}⟨\bar{q}q⟩$, whereas invariance of the vacuum under every conserved axial charge would make these expectations vanish. The three broken generators produce three massless Goldstone bosons in the exact chiral limit. Small equal quark masses explicitly break that symmetry and turn the pions into pseudo-Goldstone bosons, with squared masses linear in the quark mass at leading order.

For $V=m^2{\varphi }^2/2+\lambda ({\varphi }^2{)}^2/4$, $m^2<0$, the vacuum manifold is a sphere of radius $v=\sqrt{-m^2/\lambda }$. At a chosen vacuum the Hessian matrix is $2\lambda v^{2}n{n}^T$, where $n$ is its unit direction. The radial scalar has mass squared $2\lambda v^2$; the $n-1$ angular modes are physical Goldstone bosons when the symmetry is global.

For the Abelian Higgs mechanism with scalar phase $\chi$ and vector mass $m$, take ${\mathcal{L}}_{\mathrm{gf}}=-(\partial \cdot A+\xi m\chi {)}^2/(2\xi )$ when the scalar kinetic mixing is $-m{A}_{\mu }{\partial }^{\mu }\chi$. The cross term cancels this mixing after integration by parts. The resulting vector propagator isAt fixed finite $\xi$ it decays as $1/k^2$. Would-be Goldstone bosons and the required Faddeev-Popov ghosts remain in intermediate calculations; BRST symmetry identifies the physical states and ensures cancellation of unphysical modes. The limit $\xi \to \infty$ gives the unitary gauge propagator and is not uniform in ultraviolet momentum.

A real scalar field with the displayed potential is invariant under the Z2 symmetry $\varphi \mapsto -\varphi$, but a chosen broken-phase pure vacuum has nonzero vacuum expectation value of one sign. Expanding about $\varphi =v$ gives $V=\lambda v^2{h}^2+\lambda v{h}^3+\lambda h^4/4$. The excitation is massive, and no continuous broken generator requires a Goldstone boson. In a finite-volume theory with symmetry-preserving conditions, tunneling can produce a symmetric ground state. A broken pure vacuum is selected by taking the infinite-volume limit before removing an infinitesimal source, or by specifying appropriate boundary conditions. The two signs are related by the symmetry, although the individual pure vacua are not invariant under it.

A superfluid exhibits superfluidity, with phase stiffness and a phase-gradient current. In an appropriate ordered or quasi-ordered regime it supports low-speed flow without viscous dissipation, and its long-wavelength Goldstone boson has a sound-like dispersion. True thermodynamic phase selection and Bose-Einstein condensation need separate infrared conditions.

At a vacuum obeying F-flatness and D-flatness, charged scalar field expectation values can break an internal gauge symmetry while preserving supersymmetry. For each broken generator, a massless vector multiplet combines with a chiral superfield into a massive N=1 vector multiplet. The vector absorbs one real Goldstone boson; a real scalar remains, and the gaugino mixes with a higgsino. All four bosonic and four fermionic on-shell states have a common mass.