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Consider two real scalar fields $\mathbf{\varphi }=({\varphi }_1,{\varphi }_2)$ with positive $\lambda ,v$, standard kinetic terms, and the displayed scalar potential. At $h=0$, the circle group rotates one field into the other: this is a global symmetry in field theory, with the same rotation at every point in space. The vacuum manifold is the circle ${\mathbf{\varphi }}^2=v^2$. Choose $(v,0)$ and differentiate the scalar potential twice:The scalar mass matrix assigns a restoring force to the radial displacement, whereas its tangent is one of the Goldstone directions in the scalar mass matrix. Moving around the circle costs no potential energy; moving between neighbouring points in space can still cost gradient energy. Flat angular potential energy therefore means a soft collective excitation, not that every spatial pattern is free.

A positive source $h$ explicitly selects the positive ${\varphi }_1$ direction. The minimum is $(w,0)$, where $w>v$ is the positive solution of $\lambda w(w^2-v^2)=h$. The two eigenvalues of the scalar mass matrix there areFor example, $\lambda =v=1$ and $h=0.15$ give $w\simeq 1.06792$, $m_{\mathrm{radial}}^2\simeq 2.42138$ and $m_{\mathrm{angular}}^2\simeq 0.140459$. The light angular excitation becomes a Goldstone boson as $h\downarrow 0$, while the radial mass stays finite. The source changes both the preferred direction and the radius; expanding about the old point $(v,0)$ at nonzero $h$ would miss that shift.

In a local quantum field theory obeying special relativity, with a positive physical state space, a conserved continuous internal global symmetry in field theory and a selected Lorentz-invariant ground state, a broken generator requires a massless Goldstone boson. For $G\to H$, there are $\dim G-\dim H$ symmetry-required Goldstone bosons in this setting; additional accidental massless excitations are possible. Jeffrey Goldstone's institutional biography identifies his work with Abdus Salam and Steven Weinberg on the zero-mass theorem.^[ref]

The Goldstone directions in the scalar mass matrix give a short classical proof at a smooth minimum. The quantum argument uses a conserved current $j^{\mu }$ and a scalar local operator $O$ whose expectation changes under the broken generator. The Ward identity for the current–local operator correlation function then has a nonzero contact term. After a Fourier transform, its divergence remains nonzero as four-momentum tends to zero. Lorentz invariance forces the vector–scalar correlator to have the form $p^{\mu }F(p^2)$, so $F$ must have a pole at $p^2=0$. The spectral representation associates that pole with a physical zero-mass excitation. A properly regulated spatial integral of the conserved current implements the broken generator; a well-defined global charge acting on the broken ground state cannot simply be assumed.

The hypotheses explain the boundaries of the result. A discrete symmetry has no infinitesimal generator and need not give a Goldstone boson. A gauge redundancy does not supply the physical global generator assumed here. In nonrelativistic matter the type-B Goldstone boson shows why one cannot universally count one independent mode per broken generator. Long-wavelength fluctuations can also prevent an assumed broken state from existing, as in the Mermin-Wagner theorem.

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.

Gauge invariance of the scalar potential implies the infinitesimal identityThe irrelevant common factor $i$ from the real representation's infinitesimal rotation has been divided out. Differentiate with respect to ${\varphi }_k$ and evaluate at the stationary vacuum $v$:The second term vanishes by stationarity, leavingWhenever $T^{a}v\ne 0$, this is a nonzero vector in the kernel of the scalar mass-squared matrix, so it is a zero-eigenvalue direction. It is tangent to the degenerate vacuum orbit. In a global theory it is a Goldstone direction; in the gauged theory the corresponding would-be Goldstone field supplies a longitudinal polarization of a massive vector.

For the specific potential, stationarity at a nonzero vacuum gives $v^2=m^2/(4\lambda )$, assuming $m^2>0$ and $\lambda >0$. Its Hessian isThe antisymmetry of the triplet generators gives $v_i(T^{a}v{)}_i=0$, verifying the claimed kernel directly.

The unbroken group is the stabilizer subgroup of the chosen vacuum expectation value:Its Lie algebra consists of generators $T$ with $T{\varphi }_0=0$. The vacuum manifold is the orbit $G{\varphi }_0\simeq G/H$ under the stated transitivity assumption. Its tangent space is the image of $T\mapsto T{\varphi }_0$, so its dimension is $r=\dim G-\dim H$.

Let $K_{ij}={\partial }_i{\partial }_{j}V({\varphi }_0)$ be the scalar mass matrix, with canonical scalar kinetic terms. Invariance of the scalar potential gives $({\partial }_{i}V)(T_a\varphi {)}_i=0$ for every $\varphi$. Differentiate with respect to ${\varphi }_j$ and evaluate at the minimum, where ${\partial }_{i}V=0$:Thus every independent tangent direction is a zero eigenvector of $K$. These are the Goldstone directions in the scalar mass matrix; for a global symmetry they are $r$ physical massless Goldstone bosons. For a generic nondegenerate minimum transverse to the orbit, they are precisely all the zero eigenvalues. Without that nondegeneracy assumption, additional massless scalars are possible.

For a local gauge group, use the orthogonal unitary-gauge slice near a scalar vacuum. Since ${\varphi }_0^T{T}_a{\varphi }_0=0$ by antisymmetry, the proposed condition is equivalent toIt sets the gauge-orbit components of the scalar fluctuation to zero. This is an admissible local unitary gauge: on a basis of independent broken tangent vectors $v_a=T_a{\varphi }_0$, the variation of these conditions under the broken gauge parameters is $v_a^T{v}_b$. This Gram matrix is invertible on that basis, so the implicit function theorem selects the required gauge transformation near the vacuum. Unbroken transformations remain as a residual $H$ gauge symmetry; no global gauge slice through zeros of the scalar field is being asserted.

Write $\varphi ={\varphi }_0+h$. To quadratic order, the covariant kinetic term containsThe middle term vanishes in this unitary gauge. The gauge-boson mass rank from a real scalar vacuum is therefore determined byThese are squared masses: the physical gauge boson masses are square roots of the matrix's nonzero eigenvalues after diagonalization. For real $c_a$,Its kernel consists exactly of the unbroken generators. With $g\ne 0$, the rank is $r$, so exactly $\dim G-\dim H$ gauge-field combinations become massive. Each removed scalar Goldstone mode supplies the extra longitudinal polarization of a massive gauge boson.

The remaining physical scalar fluctuations lie in the orthogonal complement of the orbit tangent space. They have no massless modes if $K$ is positive definite on that complement. The unconditional absence of massless scalars does not follow from the printed assumptions. For a concrete counterexample, take a real doublet, $G=SO(2)$ andIts entire minimum set is one circle orbit, as required, but its Hessian matrix vanishes at the minimum. Gauging the rotation gives one vector of squared mass $g^2{v}^2$ and removes the angular scalar. The radial scalar remains massless at quadratic order, since its potential starts at fourth order in the radial fluctuation. This is an accidental massless radial mode beyond Goldstone modes. Thus the intended no-massless-scalar conclusion needs nonzero transverse curvature.

In the Standard Model, the Higgs doublet contains four real components and breaks $SU(2{)}_L\times U(1{)}_Y$ to $U(1{)}_{\mathrm{em}}$. Three Goldstone bosons become the longitudinal components of $W^{\pm }$ and $Z$, while the photon remains massless. The remaining radial Higgs boson is massive for the ordinary quartic potential with positive curvature. With $⟨H⟩=(0,v_{\mathrm{EW}}/\sqrt{2}{)}^T$,The colour gauge group is unbroken. This is the Higgs mechanism, with the generic curvature condition realized by the standard Higgs potential.

Work with canonically normalized real components of the scalar field; a complex representation can first be written in terms of its real and imaginary components. At a vacuum expectation value ${\phi }_0$, expansion of the scalar potential givesThe linear term vanishes by stationarity. Thus the scalar mass matrix is the Hessian matrix of the scalar potential. Invariance under each Lie algebra generator gives the identity ${\partial }_{r}V(\phi )(i{T}_a\phi {)}_r=0$. Differentiating this identity with respect to ${\phi }_s$ and evaluating at the stationary point removes the term containing ${\partial }_{r}V$, leavingThe kernel of the map $T\mapsto iT{\phi }_0$ is precisely the Lie algebra of the stabilizer subgroup H. Consequently its restriction to a complementary space of broken Lie algebra generators is injective: a linear combination that annihilates ${\phi }_0$ would belong both to the stabilizer algebra and its complement, and hence would vanish. There are therefore $\dim G-\dim H$ independent Goldstone directions in the scalar mass matrix, tangent to the vacuum manifold. There are $\dim G-\dim H$ symmetry-protected massless scalars.

This Goldstone theorem guarantees at least that many zero quadratic masses. The claim that these are all the massless scalars additionally requires nonzero curvature transverse to the vacuum orbit. For example, $V=\kappa ({\phi }^T\phi -v^2{)}^4$, with $\kappa >0$, has an ordinary spherical vacuum manifold but its entire Hessian matrix vanishes there. These accidental massless radial modes beyond Goldstone modes are not excluded by the hypotheses. For the indicated orthogonal group breaking, the protected number is

For the Higgs doublet, put $r=\sqrt{2{\varphi }^{†}\varphi }=v+H$ and $z=\sqrt{2}\varphi /r$ wherever $r\ne 0$. If $z=(z_1,z_2{)}^T$, thenThis explicitly exhibits the three angular coordinates as an SU(2) transformation of a real radial field. In particular, the polar coordinates for the electroweak Higgs doublet can locally be chosen in the formAt the vacuum the three variations generated by $-i{\sigma }_1$, $-i{\sigma }_2$, and $i(I-{\sigma }_3)$ acting on $(0,1{)}^T$ are independent over the reals. They span the three broken directions; the remaining generator $Q=({\sigma }_3+I)/2$ annihilates the vacuum. This proves that the exponential supplies local angular coordinates. An inverse local gauge transformation, applied also to the gauge fields, removes these three fields in unitary gauge. They supply the longitudinal polarizations of the massive W bosons and Z boson through the Higgs mechanism. The parameterization is local about a nonzero vacuum, rather than a coordinate system through every possible zero of the Higgs doublet.

Use the metric $(+---)$ and define the charged W bosons and neutral gauge fields bywhere $s_W=\sin {\theta }_W$, $c_W=\cos {\theta }_W$, and $\tan {\theta }_W=g^{\prime }/g$. Direct substitution into the gauge-covariant kinetic term givesand henceThe electroweak doublet gauge-boson mass matrix in the $(A^3,B)$ basis isIts normalized zero direction is $(s_W,c_W)$ and its massive direction is $(c_W,-s_W)$. Thus the scalar-vector terms in physical fields arewith the tree-level electroweak gauge-boson massesThese are also charge eigenstates. The unbroken U(1) gauge symmetry has generator $Q=T_3+Y$; on the Higgs doublet, $Y=I/2$. Since $[Q,T_{\pm }]=\pm T_{\pm }$, the combinations $W^{\pm }$ have charges $\pm 1$, while Z, the photon, and H are neutral. The photon disappears from the displayed scalar-vector terms because $g{A}^3-g^{\prime }B=\sqrt{g^2+g^{\prime 2}}Z$. The photon is massless and has no direct tree-level coupling to the remaining radial scalar in this sector. This does not remove its interactions with charged W bosons, or exclude a scalar-to-photon interaction generated by loops.

Finally, retain the printed normalization of the Higgs field potential:The kinetic term is $\frac{1}{2}(\partial H{)}^2$, so the Higgs radial mass for a half-normalized quartic potential is $\boxed{m_H^2=\lambda v^2}$. The extra factor of two familiar from a potential with prefactor $\lambda$, rather than $\lambda /2$, does not apply here.

Take real field coordinates ${\varphi }^I$; for complex scalar fields, split them into real and imaginary parts. Assume a smooth scalar potential and positive, canonically normalized kinetic terms in a relativistic theory. At a classical vacuum, $V_I({\varphi }_0)=0$. The Taylor expansion isThus the scalar mass matrix is the Hessian matrix, and its eigenvalues give squared masses of the linearized scalar excitations.

Put $X_a(\varphi )=i{t}^a\varphi$. Infinitesimal invariance of the scalar potential gives the identity $V_I{X}_a^I=0$ at every field value. Differentiate with respect to ${\varphi }^J$ and evaluate at the classical vacuum:Every nonzero infinitesimal symmetry direction therefore lies in the kernel of the scalar mass matrix. This is the classical Goldstone theorem expressed through Goldstone directions in the scalar mass matrix.

Let $H_0$ be the full stabilizer subgroup of ${\varphi }_0$. The linear map from the Lie algebra of $G$ to field space, $T\mapsto iT{\varphi }_0$, has kernel equal to the Lie algebra of $H_0$. The rank-nullity theorem givesThere are consequently at least $\dim G-\dim H_0$ massless scalar directions, namely the tangent directions to the symmetry orbit through the classical vacuum. If the intended $H$ is $H_0$, these are the $\dim G-\dim H$ symmetry-required Goldstone bosons. Exactly that many massless modes occur if the scalar mass matrix is positive definite on a complement of those tangent directions. A nonsingular positive field-space kinetic metric changes normalization, but not the number of zero masses.

Two qualifications are needed for the literal assumptions. A subgroup fixing the classical vacuum need not be the full stabilizer subgroup. For example, take $G=SO(3)$ acting on a real triplet and $V=\lambda (|\varphi {|}^2-v^2{)}^2/8$ with $\lambda ,v>0$. At ${\varphi }_0=v{e}_3$, the scalar mass matrix is $\lambda v^2\mathrm{diag}(0,0,1)$: there are two massless modes. Choosing $H=\{1\}$ satisfies the printed invariance condition but would incorrectly predict three. The full stabilizer subgroup is $SO(2)$.

Even with the full stabilizer subgroup, symmetry does not exclude an accidental massless scalar. Take $G=SO(2)$ rotating $(x,y)$, with an invariant singlet $z$, andAt $(v,0,0)$ the full continuous stabilizer is trivial, but the scalar mass matrix is $\mathrm{diag}(2\lambda v^2,0,0)$. One zero direction is the Goldstone boson; the other is an accidental massless scalar at quadratic order. The proof establishes the symmetry-required count, not unconditional equality with the total number of massless fields. The statement concerns global internal symmetry; gauging it changes the physical interpretation through the Higgs mechanism.