Goldstone's Theorem

Derivation

Goldstone's Theorem

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Statement

Let a Lorentz-invariant theory of real scalar fields \(\phi_a\) (\(a=1,\dots,N\)) have a Lagrangian invariant under a continuous global symmetry group \(G\), acting infinitesimally as \(\delta\phi_a=\epsilon\,(T\phi)_a\). If the potential \(V(\phi)\) is minimised at a field configuration \(\phi_a=v_a\) that is not invariant under all of \(G\) — i.e. some generators satisfy \((Tv)_a\neq 0\) — then the tree-level mass matrix \(M^2_{ab}=\partial^2 V/\partial\phi_a\partial\phi_b\big|_{v}\) has exactly one zero eigenvalue for each such broken generator. Each zero eigenvalue is a massless spin-0 excitation, the Nambu–Goldstone boson, and the total number equals \(\dim G-\dim H\), where \(H\subseteq G\) is the subgroup that leaves \(v\) invariant.

Why it matters

Goldstone's theorem is the organising principle behind every phase with a spontaneously broken continuous symmetry: the phonon of a superfluid, the magnon of a ferromagnet or antiferromagnet, the pion triplet of low-energy QCD, and the "would-be" Goldstones eaten in the Higgs mechanism are all instances of the same counting rule. It converts a statement about the geometry of the vacuum — that there is a continuous family of degenerate ground states — into a hard prediction about the particle spectrum: gapless modes must exist.

Crucially, the massless mode is not an accident to be tuned away; its masslessness is protected by the symmetry to all orders. This makes Goldstone bosons the natural light degrees of freedom of an effective field theory, whose low-energy interactions are fixed by the broken-symmetry structure rather than by microscopic detail.

Assumptions

The symmetry is continuous, not discrete.The proof differentiates the invariance identity with respect to a continuous transformation parameter; a discrete symmetry (e.g. \(\phi\to-\phi\)) has no generator to differentiate and produces no massless mode — only degenerate isolated vacua.

The symmetry is a symmetry of the full Lagrangian, exactly (no explicit breaking).If \(V\) contains a small non-invariant piece, the flat direction tilts and the mode acquires a mass: it becomes a pseudo-Goldstone boson (the pion mass from quark masses is the canonical example).

The vacuum is not invariant: \((Tv)_a\neq 0\) for at least one generator.If every generator annihilates \(v\) the symmetry is unbroken (Wigner–Weyl mode); the identity \(M^2_{ab}(Tv)_b=0\) is satisfied trivially and predicts nothing — there are no Goldstone bosons.

The symmetry is global, not gauged.If \(G\) is a local gauge symmetry the Goldstone modes are removed from the physical spectrum (eaten) and reappear as the longitudinal polarisations of massive gauge bosons — the Higgs mechanism. The counting argument still runs, but the "massless scalar" conclusion is evaded.

Lorentz invariance and a positive-definite kinetic term.The identification "zero eigenvalue of \(V''\) = one massless particle" uses the canonical relativistic kinetic term \(\tfrac12\partial_\mu\phi_a\partial^\mu\phi_a\). In non-relativistic or finite-density systems the dispersion need not be \(\omega\propto|\mathbf k|\) and the count of Goldstone modes can fall below the number of broken generators (type-B Goldstones).

A positive-definite Hilbert space (unitarity).The all-orders quantum proof rests on a state \(|\pi\rangle\) with non-zero overlap \(\langle 0|j^0|\pi\rangle\) and positive norm. Indefinite-metric spaces (covariantly quantised gauge theories) supply the loophole through which the Higgs mechanism escapes the theorem.

Derivation

\[ \mathcal{L}=\tfrac12\,\partial_\mu\phi_a\,\partial^\mu\phi_a-V(\phi),\qquad \delta\phi_a=\epsilon\,(T\phi)_a=\epsilon\,T_{ab}\phi_b \]

Set-up. \(T\) is a real antisymmetric generator of the continuous group \(G\); \(\epsilon\) is an infinitesimal constant parameter. Global symmetry means \(\mathcal{L}\) is unchanged, and since the kinetic term is manifestly invariant the content is that \(V\) is invariant. A

\[ V(\phi+\delta\phi)=V(\phi)\ \Rightarrow\ \frac{\partial V}{\partial\phi_a}\,T_{ab}\,\phi_b=0 \]

Expand to first order in \(\epsilon\): \(V(\phi+\epsilon T\phi)-V(\phi)=\epsilon\,(\partial_a V)(T_{ab}\phi_b)+O(\epsilon^2)\). Invariance for arbitrary \(\epsilon\) forces the linear coefficient to vanish. This is an identity in field space — it holds for every \(\phi\), not just the vacuum. (It is the potential-side statement whose spacetime counterpart is the conserved Noether current.) B

\[ \frac{\partial}{\partial\phi_c}\!\left(\frac{\partial V}{\partial\phi_a}\,T_{ab}\,\phi_b\right)=\frac{\partial^2 V}{\partial\phi_c\,\partial\phi_a}\,T_{ab}\,\phi_b+\frac{\partial V}{\partial\phi_a}\,T_{ac}=0 \]

Differentiate the identity of Step 2 with respect to \(\phi_c\). Because the identity holds throughout field space its derivative also vanishes identically. Product rule; \(\partial\phi_b/\partial\phi_c=\delta_{bc}\) and \(T_{ab}\) is a constant matrix. B

\[ \frac{\partial V}{\partial\phi_a}\bigg|_{v}=0\quad\Longrightarrow\quad \frac{\partial^2 V}{\partial\phi_c\,\partial\phi_a}\bigg|_{v}\,T_{ab}\,v_b=0 \]

Evaluate at the vacuum \(\phi_a=v_a\), a stationary point of \(V\), so the first-derivative factor in the second term vanishes and that whole term drops. Only the curvature term survives. This is the key move: it converts an off-shell identity into a statement about the second derivative at the minimum. B

\[ M^2_{ca}\equiv\frac{\partial^2 V}{\partial\phi_c\,\partial\phi_a}\bigg|_{v},\qquad M^2_{ca}\,(Tv)_a=0 \]

Define the symmetric mass matrix \(M^2\) as the Hessian of \(V\) at the minimum — it is exactly the coefficient of the quadratic term when \(V\) is expanded in fluctuations \(\phi_a=v_a+\eta_a\). The result of Step 4 reads: the vector \((Tv)_a\) lies in the kernel of \(M^2\). B

\[ (Tv)_a\neq 0\ \Rightarrow\ M^2\,(Tv)=0\ \text{with eigenvalue }0,\qquad (\Box+0)\,\eta_{\parallel}=0 \]

If the generator is broken, \((Tv)_a\) is a non-zero eigenvector of \(M^2\) with eigenvalue zero. Diagonalising \(M^2\), the fluctuation \(\eta_\parallel\) along this direction obeys a Klein–Gordon equation with mass\(^2=0\): it is a massless scalar. Unbroken generators give \(Tv=0\) and impose no constraint. Physically \((Tv)_a\) is tangent to the vacuum manifold — the flat direction along which \(V\) does not change. C

\[ \#\{\text{Goldstone bosons}\}=\#\{\text{broken generators}\}=\dim G-\dim H \]

The broken generators are precisely those not in the stabiliser \(H=\{g\in G: g\,v=v\}\). Their number is \(\dim G-\dim H\); the surviving \(v\) directions of the coset \(G/H\) are the independent flat directions, each supplying one massless mode. C

Result

\[ M^2_{ab}\,(T^{(\alpha)}v)_b=0\quad\text{for every generator }T^{(\alpha)};\qquad n_{\rm GB}=\dim G-\dim H \]

Reading. The curvature of the potential vanishes in every direction obtained by acting on the vacuum with a broken symmetry generator. Each such direction is a valley floor of the "Mexican-hat" landscape: rolling along it costs no energy, so the corresponding quantum has zero rest mass. The number of these massless Nambu–Goldstone bosons equals the number of symmetry generators that fail to leave the vacuum invariant — the dimension of the coset \(G/H\).

Units check. In natural units (\(\hbar=c=1\), 4 spacetime dimensions) a scalar field has mass dimension 1, so \([V]=\text{mass}^4\) and \([M^2_{ab}]=[\partial^2 V/\partial\phi^2]=\text{mass}^2\). The eigenvalue equation \(M^2(Tv)=0\) equates a mass\(^2\) times a field (\(\text{mass}^3\)) to zero on both sides; the vanishing eigenvalue is literally \(m^2=0\), a genuine zero of the squared mass. The counting \(n_{\rm GB}=\dim G-\dim H\) is a pure integer, as it must be.

Limiting cases

Symmetric vacuum (\(Tv=0\) for all \(T\), i.e. \(H=G\)): no broken generators, \(n_{\rm GB}=0\); the whole spectrum is massive (Wigner–Weyl realisation).
Maximal breaking (\(H\) trivial): \(n_{\rm GB}=\dim G\) massless modes — e.g. \(N\) real fields with the vacuum breaking \(O(N)\to O(N-1)\) give \(N-1\) Goldstones.
Abelian \(U(1)\) (\(\dim G=1\), \(H\) trivial): exactly one Goldstone, the phase mode of a complex order parameter (superfluid phonon).
Weak explicit breaking: the exact zero shifts to a small \(m^2\propto\) (breaking parameter); in the limit of vanishing breaking the Goldstone masslessness is recovered continuously.
Large-\(|v|\) (deep well): the massive radial mode mass \(m_\sigma\sim\sqrt{2\lambda}\,v\) grows and decouples, leaving a pure non-linear sigma model of the Goldstones.

Breaks when

The symmetry is gauged (local). The Higgs mechanism removes the Goldstone from the spectrum: it becomes the longitudinal polarisation of a gauge boson that acquires mass \(m_A=g\,v\). No physical massless scalar remains. The evasion is legitimate because covariant gauge quantisation uses an indefinite-metric Hilbert space, violating the unitarity assumption of the quantum proof.
The symmetry is only approximate (explicit breaking present). With a non-invariant term the flat direction tilts; the mode is a light pseudo-Goldstone boson with \(m^2\neq 0\). Pions are pseudo-Goldstones of chiral \(SU(2)_L\times SU(2)_R\to SU(2)_V\), lifted by the up/down quark masses.
Lorentz invariance is absent (non-relativistic / finite density). Goldstone counting is modified: pairs of broken generators whose charges have a non-vanishing ground-state commutator combine into a single type-B mode with quadratic dispersion \(\omega\propto\mathbf k^2\). The ferromagnet magnon is one such mode from two broken spin generators, so \(n_{\rm GB}<\dim G-\dim H\).
Spacetime dimension \(d\le 2\) at finite temperature. The Coleman–Mermin–Wagner theorem forbids spontaneous breaking of a continuous symmetry: infrared fluctuations of the would-be Goldstone field diverge and restore the symmetry, so no long-range order and no Goldstone boson form in the first place.

Failure modes

Counting \(\dim G\) instead of \(\dim G-\dim H\). Forgetting that unbroken generators annihilate the vacuum (\(Tv=0\)) and contribute no massless mode; the correct count is the dimension of the coset, not of the whole group.
Calling the radial "Higgs" mode a Goldstone. The direction along \(v\) itself is the massive mode \(m_\sigma^2=2\lambda v^2\); Goldstones are the transverse (angular) directions orthogonal to \(v\).
Expecting Goldstones from a discrete symmetry. \(\mathbb{Z}_2\) breaking gives degenerate isolated vacua but no continuous flat direction and hence no massless boson.
Evaluating \(M^2\) at the symmetric point \(\phi=0\) rather than at the true minimum \(v\). At \(\phi=0\) the curvature is negative (\(-\mu^2\)); the theorem is a statement about the Hessian at the stable vacuum.
Assuming the mode is exactly massless even with explicit breaking. Any non-invariant term lifts it; the mass is small only if the breaking is small (pseudo-Goldstone).
Applying the relativistic one-to-one count in a magnet or superfluid at finite density. Ignoring type-B Goldstones over-counts the gapless modes.

Discussion

The physical heart of the theorem is geometry. Spontaneous breaking means the vacuum is not a point but a manifold — the orbit \(G/H\) of degenerate minima, all related by the symmetry and all with identical energy. A Goldstone boson is a slow, long-wavelength wave that rotates the local order parameter along this manifold. At infinite wavelength (\(\mathbf k\to 0\)) such a rotation is a global symmetry transformation, which by definition costs no energy; hence the energy of the mode must vanish as \(\mathbf k\to 0\), which is exactly a gapless dispersion \(\omega=c|\mathbf k|\) and a massless particle. The tangent vectors to \(G/H\) are precisely the \((Tv)_a\) that appear in the derivation.

This is why Goldstone modes govern low-energy physics. Being the only gapless excitations, they dominate correlation functions, specific heats and response at long distances, and their self-interactions are dictated by the coset geometry through a non-linear sigma model rather than by microscopic parameters. The superfluid phonon, the magnon, the pion, and the phase mode of a superconductor's order parameter (before it is eaten by the photon) are the same object wearing different clothes. Goldstone's low-energy theorems — the vanishing of the amplitude to emit a soft Goldstone (Adler zero) — follow directly and are among the most model-independent predictions in physics.

Common misconceptions. Masslessness is not fine-tuning: it is enforced order by order because a Goldstone couples to the broken current \(j^\mu\) with \(\langle 0|j^\mu(x)|\pi(\mathbf k)\rangle\propto f\,k^\mu\), and current conservation \(\partial_\mu j^\mu=0\) then forces \(k^2=0\). Nor is the Higgs boson a Goldstone — it is the massive radial excitation; the three Goldstones of the electroweak \(SU(2)\times U(1)\) breaking are the ones eaten to give \(W^\pm\) and \(Z\) their masses.

The all-orders quantum proof dispenses with the tree potential entirely and rests on the spectral (Källén–Lehmann) representation of the current–order-parameter correlator \(\langle 0|\,[\,j^0(x),\phi(0)\,]\,|0\rangle\). Broken symmetry means this equal-time commutator has a non-zero vacuum expectation \(\langle 0|[Q,\phi]|0\rangle=\langle 0|\delta\phi|0\rangle\neq0\), the order parameter. Conservation of \(j^\mu\) together with this non-vanishing charge forces the spectral function to contain a \(\delta(k^2)\) pole: a massless one-particle state must couple to the current. The same reasoning exposes both loopholes — a Lorentz-non-invariant \(\langle 0|[Q,\phi]|0\rangle\) alters the pole structure and counting, while the indefinite metric of a gauge theory allows the massless pole to be a pure gauge artefact that cancels against a ghost, so no physical Goldstone survives.

Worked examples

Example 1 — \(U(1)\)/\(O(2)\) "Mexican hat": one Goldstone, one massive radial mode.

\[ V(\phi_1,\phi_2)=\frac{\lambda}{4}\left(\phi_1^2+\phi_2^2-v^2\right)^2,\qquad \rho^2\equiv\phi_1^2+\phi_2^2 \]

A real doublet with \(O(2)\cong U(1)\) symmetry; \(\dim G=1\). Minima lie on the circle \(\rho=v\). A

\[ \frac{\partial V}{\partial\phi_a}=\lambda(\rho^2-v^2)\phi_a,\qquad \frac{\partial^2 V}{\partial\phi_a\partial\phi_b}=\lambda\big[\,2\phi_a\phi_b+(\rho^2-v^2)\delta_{ab}\,\big] \]

Differentiate twice, using \(\partial\rho^2/\partial\phi_a=2\phi_a\). Symbols first. B

\[ v_a=(v,0):\quad M^2_{ab}=2\lambda\,\phi_a\phi_b\big|_v=\begin{pmatrix}2\lambda v^2 & 0\\[2pt]0 & 0\end{pmatrix} \]

Pick the vacuum on the \(\phi_1\) axis; the \((\rho^2-v^2)\) term vanishes at the minimum. Radial (\(\phi_1\)) direction is massive, angular (\(\phi_2\)) direction is flat. \((Tv)=(0,v)\neq0\) confirms one broken generator. B

\[ m_\sigma=\sqrt{2\lambda}\,v,\qquad m_{\rm GB}=0 \]

Read off the eigenvalues. Now numbers: take \(\lambda=0.13\) and \(v=246\ \text{GeV}\). Then \(m_\sigma=\sqrt{2(0.13)}\,(246)=\sqrt{0.26}\times246=0.5099\times246\approx125.5\ \text{GeV}\). B

\[ m_\sigma\approx 125\ \text{GeV},\qquad m_{\rm GB}=0 \]

Reading. Breaking \(U(1)\) with one generator yields exactly one massless Goldstone (the phase mode) and one massive radial mode. With these numbers the radial mode sits near the physical Higgs mass — a deliberate reminder that the Higgs is the massive companion, not the Goldstone. Units check: \(\sqrt{\lambda}\) is dimensionless, \([v]=\text{GeV}\), so \([m_\sigma]=\text{GeV}\); the Goldstone eigenvalue is exactly zero.

Example 2 — \(O(3)\to O(2)\): two Goldstones, count matches broken generators.

\[ V(\boldsymbol\phi)=\frac{\lambda}{4}\left(\boldsymbol\phi\cdot\boldsymbol\phi-v^2\right)^2,\qquad \boldsymbol\phi=(\phi_1,\phi_2,\phi_3) \]

A real triplet with \(O(3)\) symmetry, \(\dim G=3\). The vacuum \(v_a=(0,0,v)\) is left invariant only by rotations about the 3-axis, so \(H=O(2)\), \(\dim H=1\). Broken generators: \(3-1=2\). A

\[ M^2_{ab}=2\lambda\,\phi_a\phi_b\big|_{v}=2\lambda v^2\,\delta_{a3}\delta_{b3}=\operatorname{diag}(0,\,0,\,2\lambda v^2) \]

Same Hessian as Example 1 generalised to three components; evaluated at \(v_a=(0,0,v)\) only the \(33\) entry survives. Directions 1 and 2 are flat. B

\[ m_\sigma=\sqrt{2\lambda}\,v\ (\text{along }\phi_3),\qquad m_{\rm GB}^{(1)}=m_{\rm GB}^{(2)}=0 \]

Two zero eigenvalues \(=\) two Goldstone bosons \(=\dim G-\dim H\). Numbers: \(\lambda=0.20\), \(v=100\ \text{GeV}\Rightarrow m_\sigma=\sqrt{0.40}\times100=0.6325\times100\approx63.2\ \text{GeV}\). B

\[ n_{\rm GB}=\dim O(3)-\dim O(2)=3-1=2,\qquad m_\sigma\approx63.2\ \text{GeV} \]

Reading. The two flat angular directions on the 2-sphere \(|\boldsymbol\phi|=v\) (the coset \(O(3)/O(2)\cong S^2\)) are the two Goldstone bosons; the one radial direction is massive. The count from the mass matrix agrees exactly with the geometric count of broken generators. Units check: \(n_{\rm GB}\) is a pure integer; \(m_\sigma\) carries GeV via \(v\), Goldstone masses are identically zero.

Problems

Counting for \(O(N)\). A real \(N\)-component scalar with an \(O(N)\)-symmetric Mexican-hat potential develops a vev. How many Goldstone bosons appear, and how many massive modes?
Solution
The vacuum \(v_a=(0,\dots,0,v)\) is invariant under rotations of the remaining \(N-1\) directions, so \(H=O(N-1)\). Using \(\dim O(N)=\tfrac12 N(N-1)\): \(n_{\rm GB}=\tfrac12 N(N-1)-\tfrac12(N-1)(N-2)=(N-1)\). There are \(N-1\) massless Goldstones (the angular directions on \(S^{N-1}\)) and exactly one massive radial mode, \(m_\sigma=\sqrt{2\lambda}\,v\). Check \(N=2\Rightarrow1\), \(N=3\Rightarrow2\), consistent with the worked examples.
Electroweak eating. The Standard Model Higgs field is a complex \(SU(2)\) doublet (4 real fields) whose vev breaks \(SU(2)_L\times U(1)_Y\to U(1)_{\rm EM}\). How many would-be Goldstone bosons are there, what happens to them, and how many physical scalars remain?
Solution
\(\dim[SU(2)\times U(1)]=3+1=4\); \(\dim U(1)_{\rm EM}=1\); broken generators \(=4-1=3\). Because the symmetry is gauged, these 3 Goldstones are eaten: they become the longitudinal polarisations of \(W^+,W^-,Z\), giving those bosons mass while the photon stays massless. Of the 4 real scalar degrees of freedom, 3 are eaten and \(4-3=1\) remains as the physical Higgs boson.
Mass of the radial mode. For \(V=-\mu^2|\phi|^2+\lambda|\phi|^4\) with a complex scalar, express \(v\) and the radial (Higgs) mass in terms of \(\mu\) and \(\lambda\), and evaluate for \(\mu=88\ \text{GeV}\), \(\lambda=0.13\).
Solution
Write \(|\phi|^2=\tfrac12\rho^2\). Minimising: \(dV/d\rho^2=-\mu^2+2\lambda|\phi|^2=0\Rightarrow |\phi|^2=\mu^2/2\lambda\), i.e. \(v=\langle\rho\rangle=\mu/\sqrt{\lambda}\). The radial curvature gives \(m_\sigma^2=2\mu^2\), so \(m_\sigma=\sqrt2\,\mu\). Numerically \(v=88/\sqrt{0.13}=88/0.3606=244\ \text{GeV}\) and \(m_\sigma=\sqrt2\times88=124.5\ \text{GeV}\). The phase mode remains exactly massless.
Pseudo-Goldstone. Add a small explicit-breaking term \(\Delta V=-c\,\phi_1\) (with \(c>0\), small) to the \(O(2)\) potential of Worked Example 1, tilting the hat. Estimate the mass acquired by the formerly-Goldstone mode.
Solution
The tilt aligns the vacuum along \(\phi_1\) and lifts the angular flat direction. Parametrise \(\phi_1=v\cos\theta,\ \phi_2=v\sin\theta\) on the near-circular valley; then \(\Delta V=-c\,v\cos\theta\approx \text{const}+\tfrac12 c v\,\theta^2\). The canonically normalised angular field is \(\pi=v\theta\), so \(\Delta V\approx\tfrac12\,(c/v)\,\pi^2\), giving \(m_\pi^2=c/v\), hence \(m_\pi=\sqrt{c/v}\). For \(c=(50\ \text{GeV})^3\) and \(v=246\ \text{GeV}\): \(m_\pi^2=1.25\times10^5/246\ \text{GeV}^2=508\ \text{GeV}^2\Rightarrow m_\pi\approx22.5\ \text{GeV}\). The mode is light but no longer massless — a pseudo-Goldstone, with \(m_\pi\to0\) as \(c\to0\).
Type-B counting. A Heisenberg ferromagnet spontaneously breaks spin-rotation symmetry \(O(3)\to O(2)\) by developing a net magnetisation \(\langle S_z\rangle\neq0\). Naive Goldstone counting predicts 2 gapless modes; experiment finds a single magnon with quadratic dispersion \(\omega\propto\mathbf k^2\). Reconcile this with the theorem.
Solution
The Lorentz-invariant one-to-one count (\(\dim G-\dim H=3-1=2\)) assumes each broken generator gives an independent linearly-dispersing mode. In a non-relativistic system the two broken generators \(S_x,S_y\) have a ground-state commutator \(\langle[S_x,S_y]\rangle=i\langle S_z\rangle\neq0\). Non-commuting broken charges pair up into a single type-B Goldstone whose two would-be modes become the canonically conjugate coordinate and momentum of one degree of freedom, with quadratic dispersion \(\omega\propto\mathbf k^2\). Hence \(n_{\rm GB}=1\), not 2: the count is \(n_A+2n_B=\dim G-\dim H\) with \(n_A=0,\ n_B=1\). The antiferromagnet, with \(\langle S_z\rangle=0\), instead has vanishing commutator and recovers two linear (type-A) magnons.