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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.

An isotropic spin-polarized metal has a degenerate magnetization direction and a long-wavelength transverse magnon with quadratic dispersion. Nonzero spin density pairs the two transverse broken generators into one type-B Goldstone boson. Exchange-split Fermi surfaces also support fermionic low-energy excitations; the magnon is distinct from longitudinal amplitude motion and the spin-flip continuum.

Use dimensionless spin angular momentum operators, so their commutators have no additional $\hslash$. A fully polarized state $|F⟩={⨂}_n|S,S{⟩}_n$ is an exact ground state of the Heisenberg ferromagnet, with $E_0=-NJ{S}^2$. On each bond the largest eigenvalue of ${\mathbf{S}}_n\cdot {\mathbf{S}}_{n+1}$ is $S^2$, obtained in the two-spin total-spin-$2S$ sector. Hence the bondwise lower bound on the Hamiltonian is attained by this state.

The ferromagnetic ground-state multiplet has total spin $NS$. Applying the global lowering operator $(S_{\mathrm{tot}}^{-}{)}^r$, $r=0,\dots ,2NS$, generates its $2NS+1$ magnetic states; their linear combinations are also ground states. Global spin rotations generate polarized ground states pointing in arbitrary directions and span the same multiplet. Uniform reorientation therefore costs no energy. Slowly varying orientations give gapless ferromagnetic magnons, with quadratic rather than linear long-wavelength dispersion. In the ordered thermodynamic ground state, the two broken spin generators are canonically paired because $⟨[S_{\mathrm{tot}}^x,S_{\mathrm{tot}}^y]⟩=i⟨S_{\mathrm{tot}}^z⟩\ne 0$, producing one type-B Goldstone boson. For a finite chain the zero-wavevector motion remains inside the exact ground multiplet; nonzero wavevectors have a finite-size spacing that vanishes as $N^{-2}$. This is a ground-state statement, not a claim of finite-temperature long-range order in one dimension.

Use $a_n=N^{-1/2}{\sum }_k{e}^{ikn}{a}_k$ with $k=2\pi j/N$, choosing $N$ distinct values modulo $2\pi$ in the Brillouin zone. The lattice spacing is one; for spacing $a$ the dimensionless variable here would be $ka$. Orthogonality of the discrete Fourier modes diagonalizes the bilinear Hamiltonian:This is the nearest-neighbour ferromagnetic magnon dispersion. The printed expression is missing a factor of four for the stated exchange Hamiltonian. If ${\omega }_k$ denotes a physical frequency, $\hslash {\omega }_k={\epsilon }_k$; setting $\hslash =1$ does not remove that factor. The printed coefficient would require an exchange constant $J/4$ in the original Hamiltonian.

An independent check uses one spin lowering at site $n$. The exact one-magnon Hamiltonian relative to $E_0$ has diagonal element $2JS$ and off-diagonal elements $-JS$ at $n\pm 1$, giving the same Fourier eigenvalue. At small $k$, ${\epsilon }_k\simeq JS{k}^2$, and the $k=0$ state belongs to the degenerate ground multiplet. The quadratic low-energy branch is the ferromagnetic type-B Goldstone boson.

The spin-independent term cancels in the signed sum. The linearized stationary equation becomesThe even powers vanish because the equation is odd under $M\mapsto -M$. Equivalently, the mean-field action has curvatureThus the unpolarized stationary point becomes unstable whenAt low temperature, the derivative of the Fermi distribution samples a narrow window around $\mu$, so ${\chi }_0\simeq \nu (\mu )$ for a smooth density of states. Measuring energy from the Fermi level gives $U_c\nu (0)\simeq 1$, the Stoner criterion in the spin-summed convention fixed above. This identifies local instability of the paramagnetic state; the detailed transition order also depends on higher terms in the action.

In the isotropic ordered phase of itinerant ferromagnetism, continuous spin-rotation symmetry leaves a degenerate direction of magnetization. Its transverse fluctuations produce one gapless, quadratically dispersing mean-field itinerant ferromagnetic Goldstone mode, $E(\mathbf{q})\propto q^2$. The two broken spin generators form a canonical pair, as for the type-B Goldstone boson of the localized ferromagnet. The system also remains metallic, with exchange-split Fermi surfaces and low-energy spin-conserving particle-hole excitations. Longitudinal-amplitude fluctuations and the spin-flip continuum are distinct from the long-wavelength magnon; their energy and damping scales depend on the band structure. Spin anisotropy, if added, could gap the transverse mode, but it is absent from the stated Hamiltonian.

The constant term in the spin sum cancels. Linearizing the saddle point equation givesThe paramagnetic auxiliary-field curvature isConsequently the Stoner criterion for the loss of local paramagnetic stability isThe printed $\nu (0)$ uses energy measured from the Fermi level. For a smooth density of states, $-n_F^{\prime }$ becomes a narrow unit-area peak at $\mu$; a singular density of states or appreciable temperature requires the integral form. The curvature establishes local instability, not the order or location of every possible thermodynamic transition.

With exact spin-rotation symmetry, itinerant ferromagnetism selects a magnetization direction. The two broken transverse generators form one quadratic type-B Goldstone boson, a long-wave ferromagnetic magnon with $E_q\sim D{q}^2$. Longitudinal amplitude motion is not symmetry-protected to be gapless. In a metallic ordered state, particle-hole excitations remain near each surviving exchange-split Fermi surface; these are distinct from the collective magnon. The itinerant spin-flip particle-hole continuum starts at exchange scale $2|M|$ at $q=0$ in this simple mean-field spectrum, while its finite-$q$ boundaries differ. Spin anisotropy, spin-orbit coupling or an external field can gap the transverse mode; none is present in the stated model.

The Holstein–Primakoff transformation converts a spin algebra into bosonic operators while retaining the finite physical spin Hilbert space. With $[a_n,a_m^{†}]={\delta }_{nm}$, writeThe Holstein–Primakoff occupation constraint is $0\le n_n\le 2S$. Acting on $|n⟩$, the raising matrix element is $\sqrt{n(2S-n+1)}$ and the lowering element is $\sqrt{(n+1)(2S-n)}$. Their squared difference is $2(S-n)$, reproducing $[S^{+},S^{-}]=2S^z$. Operator ordering inside the square roots is essential. For low boson occupation relative to $S$, expand them to obtain $S^{+}\simeq \sqrt{2S}a$ and $S^{-}\simeq \sqrt{2S}{a}^{†}$. Keeping quadratic terms gives the linear spin-wave approximation; higher terms describe magnon interactions. Large $S$ helps control the expansion, but a low-dimensional infrared divergence can invalidate it even at large finite $S$.

For the ferromagnet choose $H_F=-J{\sum }_n{\mathbf{S}}_n\cdot {\mathbf{S}}_{n+1}$, $J>0$, with unit lattice spacing and $\hslash =1$. Expand around the all-up state. A longitudinal bond contributes $-J{S}^2+JS(n_n+n_{n+1})$, and the transverse exchange contributes $-JS(a_n^{†}{a}_{n+1}+a_{n+1}^{†}{a}_n)$. After a Fourier transform,These are ferromagnetic magnons. The polarized vacuum is an exact ground state of the full ferromagnetic Hamiltonian: every bond is in its maximum total-spin sector, minimizing that bond energy. It therefore has no quantum zero-point reduction of its magnetization. The one-magnon dispersion is also exact, although interacting many-magnon states need not be free bosons. Rotating the polarization generates the maximum-total-spin ground-state multiplet. Selecting a definite direction in a thermodynamic symmetry-breaking description breaks $SU(2)$ to $U(1)$; the two transverse spin components form a canonical pair, producing a single quadratic type-B Goldstone boson. A small aligning field gaps it by the Zeeman energy.

At nonzero temperature the one-dimensional ferromagnet is different: its thermal magnon number contains $\int dk[e^{\beta JS{k}^2}-1{]}^{-1}$, with infrared behaviour $\int dkT/k^2$. The divergent magnetization reduction signals the absence of finite-temperature long-range order in an infinite short-range isotropic chain, consistent with the Mermin-Wagner theorem. It does not disallow the polarized zero-temperature ground states just constructed.^[ref]

For the antiferromagnet use $H_{AF}=+J{\sum }_n{\mathbf{S}}_n\cdot {\mathbf{S}}_{n+1}$, still $J>0$, on an even bipartite chain. The classical reference state alternates up and down. Rotate one sublattice through $\pi$ so that both local quantization axes point along their reference spins. In unrotated physical components, to leading order $S_A^z=S-a^{†}a$, $S_B^z=-S+b^{†}b$, $S_A^{+}\simeq \sqrt{2S}a$ and $S_B^{+}\simeq \sqrt{2S}{b}^{†}$. Choosing Fourier phases using the actual site positions givesThe reduced Brillouin zone has $M=N/2$ momenta and length $\pi$. The pair creation and annihilation terms express the fact that an antiparallel reference product state is not an eigenstate. A bosonic Bogoliubov transformation with $u_k=\cosh {\theta }_k$, $v_k=\sinh {\theta }_k$, $\tanh 2{\theta }_k=\cos k$ diagonalizes each two-mode block. With the sign convention $a_k=u_k{\alpha }_k-v_k{\beta }_{-k}^{†}$ and $b_{-k}=u_k{\beta }_{-k}-v_k{\alpha }_k^{†}$, the result isThus the harmonic antiferromagnetic modes have linear, rather than quadratic, long-wave dispersion. In an unfolded Brillouin-zone description the same result is represented by zeros at both zero and the antiferromagnetic wavevector; these are not extra independent degrees of freedom beyond the two-sublattice accounting.

The Bogoliubov vacuum has no $\alpha$ or $\beta$ excitations but contains original spin bosons, with $v_k^2=\frac{1}{2}(|\sin k{|}^{-1}-1)$. Its staggered magnetization is $m_s=S-M^{-1}{\sum }_{k\in \mathrm{RBZ}}{v}_k^2$. This quantum depletion of Néel order diverges logarithmically at small $k$ in one dimension. Linear spin-wave theory therefore does not establish a stable ordered antiferromagnetic ground state for this chain. Its nominal gapless dispersion is a result of expanding around an invalid ordered reference, not a general proof of the exact excitation spectrum. For the nearest-neighbour isotropic chain, integer-spin systems have a Haldane gap, whereas half-integer-spin systems are gapless; the spin-one-half chain in particular has fractional spin excitations rather than a simple dilute gas of ordered-state magnons. This distinction requires physics beyond the harmonic expansion, including the topological term in the continuum theory.^[ref]