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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.

With positive canonical kinetic terms and nonzero coupling $g$, expansion of the covariant scalar kinetic term gives the displayed gauge-boson mass matrix. It is a Gram matrix: for real $c_a$, $c^T{M}_A^{2}c=g^2|{\sum }_a{c}_a{T}_a{\varphi }_0{|}^2$. Its kernel is the Lie algebra of the stabilizer subgroup of the vacuum. Thus its rank is $\dim G-\dim H$, and exactly that many gauge-field combinations acquire nonzero squared masses. Removal of these gauge-orbit directions does not remove accidental zero eigenvalues of the scalar mass matrix on the transverse complement.

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 $m_h^2\to 2{\mu }^2$.

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.

Write $H={\varphi }_1$ and $P={\varphi }_2$, both Hermitian matrices. Through quadratic order around the chosen vacuum expectation value,The trace of a commutator vanishes. The needed matrix traces are thereforeUsing $m^2=-(\alpha +N\beta )C^2$, all linear terms cancel. The quadratic scalar potential reduces towith no $P^2$ term. The kinetic term issince its mixed term is the trace of a commutator. A Hermitian matrix has $N^2$ independent real components, so $P$ describes exactly $N^2$ massless scalar modes.

To determine the other masses, take $T_0=1/\sqrt{N}$ and traceless Hermitian $T_A$ with $\mathrm{tr}(T_A{T}_B)={\delta }_{AB}$, and expand $H=h_0{T}_0+{\sum }_{A=1}^{N^2-1}{h}_A{T}_A$. These are canonically normalized real scalar fields. The scalar mass matrix is diagonal in this basis:Thus $H$ contains one massive singlet and a degenerate multiplet of $N^2-1$ massive fields in the Adjoint representation of the unbroken diagonal $SU(N)$. Both squared masses are strictly positive for the specified parameters. The $P$ fields are the $N^2$ Goldstone bosons associated with the broken generators: $N^2-1$ traceless relative left-right rotations and one overall phase rotation.

Goldstone's theorem guarantees those massless directions. It does not alone forbid additional accidental massless modes; the positive scalar mass matrix computed here supplies the further conclusion that every $H$ component is massive. The fields describe a global symmetry, so these Goldstone bosons are physical rather than being absorbed into massive gauge bosons.

Put $r^2={\Phi }^T\Phi$. The potential has its minimum atChoose the vacuum expectation value ${\Phi }_0=(0,0,v{)}^T$. Since the weak hypercharge is zero, $Q=T^3$ on this real scalar triplet, and $T^3{\Phi }_0=0$. Neither $T^1$ nor $T^2$ annihilates ${\Phi }_0$. The stabilizer subgroup is consequentlyIn particular, spontaneous symmetry breaking has not reduced the two neutral gauge generators to a single one: $Y$ annihilates every component and remains unbroken independently of $T^3$.

The scalar mass matrix isWith canonical kinetic normalization $\frac{1}{2}(\partial h{)}^2$, the radial Higgs boson therefore hasThe two tangential zero eigenvalues are Goldstone modes, absorbed by the two massive gauge bosons through the Higgs mechanism.

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.

First assume global four-dimensional N=1 supersymmetry, canonical positive Kähler potential, and tree level. Write $W_i={\partial }_{i}W$ and $m_{ij}=W_{ij}$ at the stationary vacuum. The chiral-superfield fermion mass matrix is the symmetric matrix $m$, and the sum of squared Weyl spinor masses is $\mathrm{tr}(m^{†}m)$. From the F-term scalar potential $V={\sum }_k|W_k{|}^2$, its mixed and holomorphic Hessian matrix blocks areFor canonically normalized real and imaginary parts of the scalar fields, the real scalar mass matrix has trace twice the mixed trace. Its holomorphic blocks can split the two real scalar masses but do not change their sum. HenceA Weyl spinor has two spin states, so this proves the tree-level supertrace mass sum rule,The paper defines its supertrace with $(-1{)}^{2j+1}$, the negative of the conventional boson-minus-fermion supertrace used here. The asserted zero is the same in either convention. The result requires the stated canonical tree-level hypotheses; a noncanonical Kähler metric, supergravity, or radiative corrections can change the sum rule.

If vector multiplets are also present, the gauge contributions to the F-term supertrace cancel rather than being omitted. At $D_a=0$, define $C={\sum }_a{g}_a^2\Vert T_{a}v{\Vert }^2$ for the scalar expectation vector $v$. Differentiating $\frac{1}{2}{\sum }_a{D}_a^2$ adds $2C$ to the real-scalar trace. The symmetric gaugino-matter mass matrix has off-diagonal entries $\sqrt{2}{g}_a{T}_{a}v$, giving an extra $4C$ in its fermion squared-mass trace. The covariant scalar kinetic term gives vector squared-mass trace $2C$. Thus their contribution is $2C-2(4C)+3(2C)=0$. The gauge-theory result uses all spin states, including the vector weight three; it is not obtained by dropping massive gauge partners.

The MSSM tree-level sfermion mass constraint explains the phenomenological difficulty with direct visible-sector breaking. If F-term breaking is neutral under an unbroken electric and color gauge symmetry, and no D-term shifts are present, the same trace argument applies within each conserved-charge fermion block. Its corresponding scalar partners cannot all have squared masses above the mean of the light fermion squared masses. Thus canonical tree-level visible-sector supersymmetry breaking alone cannot make every squark and other scalar partner heavy while leaving the observed fermions light. The usual effective soft supersymmetry breaking terms arise after communicating breaking from a separate sector; integrating out that sector, noncanonical interactions and radiative effects evade the hypotheses. The global trace identity alone would not identify a particular light squark without the conserved-charge block argument.

For the specified O'Raifeartaigh model, phases may be chosen so that $\lambda ,\mu ,M$ are real and positive. Its superpotential derivatives areand its F-term scalar potential isFor fixed $x,z$, choose $y=-2\lambda xz/M$ to minimize the final square. Since $|z^2-{\mu }^2{|}^2\ge (|z{|}^2-{\mu }^2{)}^2$,Thus $M^2>2{\lambda }^2{\mu }^2$ makes $z=y=0$ a global minimum with arbitrary $x$. The hierarchy $M\gg \mu$ ensures this for fixed perturbative $\lambda$, rather than for arbitrarily large $\lambda$. The origin is one member of this classically flat family, with $V_0={\lambda }^2{\mu }^4>0$ and $F_X=\lambda {\mu }^2$. The complex field $x$ is a pseudomodulus.

At the origin the chiral-superfield fermion mass matrix, in the $({\psi }_X,{\psi }_Y,{\psi }_Z)$ basis, isIts physical fermion masses are $0,M,M$; the two massive Weyl spinors form one massive four-component fermion. The massless ${\psi }_X$ is the goldstino. More generally, stationarity gives $W_{ij}\overline{W_j}=0$, so the nonzero auxiliary field direction is a null vector of $m$, proving the goldstino zero mode from vacuum stationarity.

Writing $z=(z_R+i{z}_I)/\sqrt{2}$ and similarly for $x,y$, the quadratic scalar potential isThe mass spectrum of the quadratic-cubic O'Raifeartaigh model is thereforeThe scalar squared-mass sum is $4M^2$, while twice the fermion squared-mass sum is also $4M^2$, so the supertrace is zero, as required. The two massless real $x$ modes are tree-level pseudomoduli, which can be lifted by a quantum effective potential; their tree-level masslessness is not the exact symmetry protection enjoyed by the goldstino.

Let $V_i=\partial V/\partial {\varphi }_i$ and $M_{ij}^2={\partial }_i{\partial }_{j}V{|}_{{\varphi }_0}$ be the scalar mass matrix, which is the Hessian matrix of the scalar potential at the vacuum. Invariance under the infinitesimal Lie group action givesDifferentiate with respect to ${\varphi }_j$ and evaluate at the vacuum expectation value ${\varphi }_0$. Since $V_i({\varphi }_0)=0$ at a minimum,Thus every tangent vector $i{t}^a{\varphi }_0$ generated by a broken Lie algebra generator is a zero eigenvalue eigenvector of the scalar mass matrix. The unbroken generators are precisely those in the stabilizer Lie algebra $\mathfrak{h}$ and give the zero tangent vector; independent broken directions span $\mathfrak{g}/\mathfrak{h}$. Hence Goldstone theorem givesmassless scalar modes at the classical level.

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.

The mass of a canonically normalized scalar field is determined by its quadratic fluctuation potential at a stationary vacuum. With several real fields, physical squared masses are the eigenvalues of the scalar mass matrix. A negative squared mass signals an unstable direction rather than a stable massive particle.

For real scalar fields with canonically normalized quadratic kinetic terms, the tree-level scalar mass matrix is the Hessian matrix of the scalar potential at a stationary vacuum:Its eigenvalues are squared masses in units $c=\hslash =1$. A negative eigenvalue is a local instability of that candidate vacuum.

The kinetic normalization is essential. More generally, expand about ${\varphi }_0$ withwhere $H$ is the potential Hessian matrix and $K$ is a real symmetric positive-definite matrix. A mode $\eta =v{e}^{-ip\cdot x}$ obeys the generalized eigenvalue problemEquivalently, canonical field normalization gives the symmetric squared-mass matrix $K^{-1/2}H{K}^{-1/2}$. For one field with kinetic coefficient $Z>0$ and quadratic potential $m_0^2{\varphi }^2/2$, the physical squared mass is $m_0^2/Z$.

Broken continuous global-symmetry directions are zero vectors of the potential Hessian matrix, producing Goldstone bosons when the kinetic form is nonsingular. A Higgs mechanism changes which scalar modes remain independent physical particles; explicit symmetry breaking can lift a zero mass. Thus both the potential and kinetic structure are needed to read a physical spectrum.