Spontaneous Symmetry Breaking

Balance a pencil perfectly on its tip. The setup is rotationally symmetric — gravity and the pencil treat every horizontal direction alike — yet the pencil must fall some way, and once it does the symmetry is gone from the world even though it was never absent from the laws. Spontaneous symmetry breaking (SSB) is this scene made macroscopic and irreversible. A Hamiltonian carries a symmetry group $G$ ; the equilibrium state that actually forms carries less. The Ising ferromagnet is the textbook case: $H=-J\sum_{\langle ij\rangle}s_is_j$ is invariant under flipping every spin, but below the Curie temperature the magnetization points up or down and stays there. The paradox — a symmetric law with an asymmetric state — is resolved by two ideas working together: degeneracy and the thermodynamic limit. The broken-symmetry states come in a family related by $G$ (up and down magnetizations, the continuum of orientations of a Heisenberg magnet, the phase of a superconducting condensate). A finite system tunnels among them and restores the symmetry on average, so its order parameter is exactly zero. Only as the number of degrees of freedom $N\to\infty$ does the tunnelling amplitude vanish, the degenerate sectors decouple into separate worlds, and the system commit to one. SSB is therefore a genuinely emergent, infinite-system phenomenon — it does not exist in any finite box. The quantitative fingerprint is the order parameter, and its definition hides the whole subtlety in an order of limits: apply an infinitesimal field $h$ , send $N\to\infty$ first, then send $h\to0^+$ . Doing the limits in the other order gives zero. That non-commutativity is the mathematical heart of the effect. When the broken symmetry is continuous, the degenerate manifold is itself continuous, and gliding along it costs vanishingly little energy — these soft excitations are the Goldstone modes, and their violence in low dimensions (the Mermin–Wagner theorem) even decides where SSB can occur at all. This article owns the SSB mechanism and its singular order of limits, and hands the massless-mode count forward to the Goldstone theorem.

Key takeaways

Spontaneous symmetry breaking occurs when the equilibrium (or ground) state of a system fails to respect a symmetry that its Hamiltonian does respect; the symmetry survives only as a map among a degenerate family of states.
The signature is a nonzero order parameter — zero in the symmetric phase, finite below the transition — which selects one member of the degenerate broken states.
SSB requires the thermodynamic limit and a non-commuting order of limits: apply a field $h$ , take $N\to\infty$ , then $h\to0^+$ , giving $\lim_{h\to0^+}\lim_{N\to\infty}m\neq0$ ; the reverse order gives $0$ .
Breaking a continuous symmetry generates gapless Goldstone modes; the Mermin–Wagner theorem forbids SSB of a continuous symmetry with short-range forces in $d\le2$ at finite $T$ .
Discrete symmetries (Ising $\mathbb{Z}_2$ ) can break in $d\ge2$ because they have no Goldstone modes; the mean-field Ising transition sits at $k_BT_c=zJ$ with $\beta=\tfrac12$ .
Key idea: a symmetric Hamiltonian can have an asymmetric state because, in the thermodynamic limit, symmetry-related ground states decouple into separate superselection sectors — the realized state minimizes the free energy by choosing one, and the choice is detected by $\lim_{h\to0^+}\lim_{N\to\infty}\langle\phi\rangle_h\neq0$ .

The core insight

A symmetric Hamiltonian can support an asymmetric equilibrium state because, in the thermodynamic limit, the symmetry-related ground states decouple into distinct superselection sectors between which no local dynamics connects. The realized state minimizes the free energy by selecting one of a degenerate family; the symmetry is not destroyed but realized collectively, as a permutation of the degenerate states, and its breaking is diagnosed by a nonvanishing order parameter $\lim_{h\to0^+}\lim_{N\to\infty}\langle\phi\rangle_h\neq0$ .

Where this lives today

Spontaneous symmetry breaking is one of the few ideas that spans the whole of physics. In condensed matter it names the ordered phases: ferromagnets and antiferromagnets breaking spin-rotation symmetry, crystals breaking continuous translations (with phonons as the Goldstone modes), superfluids and superconductors breaking the $U(1)$ phase symmetry of the condensate, liquid crystals breaking rotations to a preferred axis. In a charged superconductor the would-be Goldstone mode is absorbed by the electromagnetic field, giving it an effective mass — the Anderson–Higgs mechanism, and the Meissner effect it explains. In particle physics the identical mathematics gives the electroweak vacuum its structure and, through the Higgs field, the masses of the $W$ and $Z$ bosons and the fermions. The renormalization group, the theory of critical phenomena, and the classification of matter into universality classes are all built on which symmetry breaks and in what dimension. SSB is the conceptual bridge between a laboratory magnet and the vacuum of the universe.

The physical picture

Picture the free energy as a landscape over order-parameter space. Above $T_c$ it is a single bowl centred at $m=0$ : symmetric and unbroken. Below $T_c$ the centre bulges up into a hill and the minima slide outward — into two wells for a discrete ( $\mathbb{Z}_2$ ) symmetry, or into a continuous ring for an $O(2)$ symmetry. The landscape keeps the symmetry of the Hamiltonian; a ball resting in it cannot. The state breaks what the landscape preserves. This is why "spontaneous" is the right word: nothing in the dynamics prefers a well, yet the system inhabits one.

Figure

The O(2) "Mexican hat" free energy: for r below zero the minimum is a degenerate circle |ψ| = ρ₀ of broken-symmetry vacua. Rolling around the valley (angular θ direction) costs no potential energy → the massless Goldstone mode; climbing the wall (radial σ direction) is gapped → the massive amplitude mode

Why does an infinitesimal field decide everything? In a finite system the two wells communicate by tunnelling, so the thermal average sits at the symmetric centre. A field tilts the landscape by an amount proportional to $hN$ — the total, extensive bias. In the thermodynamic limit even $h\to0^+$ produces an infinite tilt $hN\to\infty$ , pinning the system in one well. The control variable is the product $Nh$ , not $h$ alone, which is exactly why the order of the limits $N\to\infty$ then $h\to0$ is decisive and cannot be exchanged. Take $N$ to infinity first at any fixed $h>0$ and the field-favoured well wins completely; only then remove the field, and the selected value survives. The degenerate broken states are not connected by any local dynamics in the infinite system. Flipping a macroscopic magnetization means reorienting $\sim N$ spins, or nucleating and sweeping a domain wall whose free-energy cost $\sim\sigma L^{d-1}$ grows without bound. The tunnelling probability $P_{\text{flip}}\sim e^{-\beta\sigma L^{d-1}}\to0$ , so the degenerate sectors become genuinely separate worlds — superselection sectors — between which no observable connects. A prepared magnet stays magnetized for longer than the age of the universe. It could flip; it simply never does.

Figure

Punchline: the order parameter m(h) as a function of applied field, for a finite system versus the thermodynamic limit. Finite N (blue): m(h) passes smoothly through the origin, so m(0) = 0 — no SSB. As N → ∞ (red) m(h) develops a jump at h = 0, a finite right-limit m(0⁺) = m₀ ≠ 0 — the analytic fingerprint of SSB

When the broken symmetry is continuous, the degenerate vacuum manifold is a continuous surface, and long-wavelength distortions along it cost energy only through gradients, not through any potential — vanishingly little as the wavelength grows. These are the Goldstone modes: spin waves in magnets, phonons in crystals, second sound in a superfluid. Their softness has teeth. In low dimensions the accumulated Goldstone fluctuations are so violent that they destroy the ordered state entirely, which is the content of the Mermin–Wagner theorem — no continuous symmetry can break with short-range forces in $d\le2$ at finite temperature. Discrete symmetries, having no Goldstone modes, escape this fate and order even in $d=2$ .

The core equations

The definition of SSB is a statement about the mismatch between the symmetry of the law and the symmetry of the realized state. It is compact but carries the whole idea:

H[g\,s]=H[s]\ \ \text{for all }g\in G,\qquad\text{yet}\qquad g\,|\Omega\rangle\neq|\Omega\rangle

where: $H$ is the Hamiltonian (or free-energy functional), $G$ is the symmetry group leaving it invariant, $g$ is a group element acting on the degrees of freedom $s$ , and $|\Omega\rangle$ is the realized ground or equilibrium state. Read it as: the dynamics is $G$ -symmetric, but the state you find is not — the missing symmetry reappears only as a map $g$ carrying $|\Omega\rangle$ to a different, degenerate ground state. The diagnostic that makes this operational is the order parameter, and its definition is where the subtlety lives. The two limits must be taken in a fixed order:

m\equiv\lim_{h\to0^+}\ \lim_{N\to\infty}\ \frac{1}{N}\Big\langle\sum_i\phi_i\Big\rangle_h

where: $m$ is the order parameter (magnetization per site), $\phi_i$ is the local field conjugate to the symmetry-breaking source, $h$ is the infinitesimal external field that explicitly breaks $G$ , $N$ is the number of degrees of freedom, and $\langle\cdot\rangle_h$ is the thermal average with the field on. The physical reading: pin the system with a whisper of a field, make it macroscopic, then remove the whisper — and check whether any order survives. Landau theory packages the mechanism into a free energy whose minima migrate away from the symmetric point below $T_c$ :

f(m)=f_0+\tfrac12\,a\,(T-T_c)\,m^2+\tfrac14\,b\,m^4-h\,m,\qquad b>0

where: $f(m)$ is the free-energy density, $a,b>0$ are phenomenological coefficients, $T_c$ is the temperature at which the $m^2$ coefficient changes sign, and the $-hm$ term explicitly breaks $m\to-m$ . Minimizing at $h=0$ gives $m=0$ for $T>T_c$ and the degenerate pair $m_\pm=\pm\sqrt{a(T_c-T)/b}$ for $T<T_c$ : the state must pick one, and that choice is the symmetry breaking. For a continuous symmetry the broken manifold is continuous, and the general count of soft modes is fixed by how much symmetry is lost:

\omega(\mathbf{k})\to0\ \text{ as }\mathbf{k}\to0,\qquad N_{\text{Goldstone}}=\dim G-\dim H

where: $\omega(\mathbf{k})$ is the dispersion of a Goldstone mode (gapless at zero wavevector), $\dim G$ is the dimension of the full symmetry group, and $\dim H$ is the dimension of the unbroken subgroup $H\subset G$ . Reading it: each broken generator supplies exactly one massless mode — $O(3)\to O(2)$ breaks two generators and gives two magnons; $O(2)\to\{e\}$ breaks one and gives one phase mode.

Derivation AWhy the two limits cannot be exchanged

The claim at the center of SSB is that $\lim_{h\to0^+}\lim_{N\to\infty}m\neq\lim_{N\to\infty}\lim_{h\to0^+}m$ , and that only the first order gives a broken state. The goal of this derivation is to prove that non-commutativity explicitly, using a double-well free energy treated as a single collective coordinate with an extensive barrier. The idea is that a field biases the two wells by an amount proportional to $Nh$ , so the outcome depends entirely on whether $Nh$ is sent to infinity or to zero — and those two roads correspond exactly to the two orders of limits.

Step 1. Set up the collective average. Treat $m$ as one coordinate with free-energy density $f(m,h)$ , so the thermal average is $\langle m\rangle_{N,h}=\dfrac{\int \mathrm{d}m\;m\,e^{-\beta N f(m,h)}}{\int \mathrm{d}m\;e^{-\beta N f(m,h)}}$ . The extensivity — the factor $N$ multiplying $f$ — is what makes the barrier macroscopic.

Step 2. Two degenerate wells at $h=0$ . Write $f(m,0)=\tfrac14 b(m^2-m_0^2)^2$ , with exactly degenerate minima at $\pm m_0$ . The integrand is even in $m$ , so $\langle m\rangle=0$ for any finite $N$ : no symmetry breaking in a finite system, at any temperature. Why: the measure weights $+m_0$ and $-m_0$ configurations identically, and every odd observable cancels pairwise.

Step 3. Tilt with the field. Add $-hm$ : the $+m_0$ well is lowered relative to $-m_0$ by $\Delta f\approx2hm_0$ . The probability ratio of the two wells is then $\dfrac{P(+m_0)}{P(-m_0)}\approx e^{+2\beta N h m_0}$ . Why $Nh$ is the control knob: the exponent carries the product $Nh$ , not $h$ alone — the bias is extensive.

Step 4. $N\to\infty$ first. Hold $h>0$ fixed and send $N\to\infty$ : then $Nh\to\infty$ , the ratio diverges, and the $+m_0$ well dominates completely, giving $\lim_{N\to\infty}\langle m\rangle_{N,h}=m_0\,\mathrm{sgn}(h)$ — a sharp, field-aligned order parameter.

Step 5. Then remove the field. Now send $h\to0^+$ : the selected value persists, $\lim_{h\to0^+}\lim_{N\to\infty}\langle m\rangle=m_0\neq0$ . This is SSB. Why the reverse order fails: taking $h\to0$ first sends $Nh\to0$ at finite $N$ , the two wells re-equalize, and $\langle m\rangle=0$ is recovered. The order parameter $m(h)$ therefore has a jump at $h=0$ in the infinite system — a first-order-like discontinuity in the field — which is the analytic fingerprint of the broken phase. Verified symbolically in verify_spontaneous-symmetry-breaking.py (well-weight ratio $e^{+2\beta Nhm_0}$ , the $Nh$ scaling, and the two limit orders).

Go deeper · Derivation B · 1 minMean-field Ising: locating the broken phase and the exponent

Derivation BMean-field Ising: locating the broken phase and the exponent

To see SSB in a concrete microscopic model, take the Ising ferromagnet and solve it in mean-field theory, where each spin feels the average field of its neighbours. The goal is to locate the symmetry-breaking transition ( $k_BT_c=zJ$ ) and extract the order-parameter exponent ( $\beta=\tfrac12$ ) from the self-consistency equation. The method replaces the exact neighbour interaction by an effective field $h_{\text{eff}}=zJm+h$ and closes the loop by demanding that the resulting single-spin average equal $m$ .

Step 1. Mean-field decoupling. Approximate $H\approx-\sum_i(zJm+h)s_i+\tfrac12 NzJm^2$ , where $z$ is the coordination number, $m=\langle s\rangle$ , and the constant compensates double counting. Each spin now sees only the effective field $h_{\text{eff}}=zJm+h$ .

Step 2. Single-spin average. A two-state spin $s=\pm1$ in field $h_{\text{eff}}$ has thermal magnetization $\langle s\rangle=\tanh(\beta h_{\text{eff}})$ . Imposing self-consistency $\langle s\rangle=m$ gives $m=\tanh[\beta(zJm+h)]$ . At $h=0$ , $m=0$ is always a solution.

Step 3. Stability and $T_c$ . Expand for small $m$ at $h=0$ : $m\approx\beta zJ\,m-\tfrac13(\beta zJ\,m)^3$ . The origin is stable while the slope $\beta zJ<1$ and unstable once $\beta zJ>1$ . The marginal condition $\beta_c zJ=1$ fixes $k_BT_c=zJ$ . Why this is SSB: below $T_c$ the symmetric solution $m=0$ becomes a maximum and two new solutions $\pm m$ appear, so the state must break the $m\to-m$ symmetry.

Step 4. Order-parameter exponent. Solving the cubic near $T_c$ , $m^2\approx3(\beta zJ-1)/(\beta zJ)^3$ . With $\beta zJ=T_c/T$ , the factor $\beta zJ-1=(T_c-T)/T\propto(T_c-T)$ , so $m\propto(T_c-T)^{1/2}$ : the mean-field exponent $\beta=\tfrac12$ , exact for $d\ge4$ . Why only the qualitative picture is right in 3D: mean field ignores the correlated fluctuations that give the true $\beta\approx0.326$ . *Verified symbolically in verify_spontaneous-symmetry-breaking.py (nonzero $m^2=3(\beta zJ-1)/(\beta zJ)^3$ , the marginal-slope $T_c$ , and $m^2\propto(T_c-T)$ ).*

Limiting cases

case	condition	outcome
Discrete symmetry, $d\ge2$	$\mathbb{Z}_2:\ s\to-s$	order survives fluctuations; SSB at finite $T_c$ (destroyed only in $d=1$ )
Continuous symmetry, $d>2$	$O(n)$ , $n\ge2$	SSB allowed; gapless Goldstone modes; lower critical dimension $d_\ell=2$
Continuous symmetry, $d\le2$	Mermin–Wagner	$\langle\phi\rangle=0$ ; Goldstone fluctuations destroy long-range order at finite $T$
Symmetric phase	$\lim_{h\to0^+}\lim_{N\to\infty}m=0$	above $T_c$ the unique equilibrium respects all of $G$
Above upper critical dimension	$d\ge d_u=4$	fluctuations negligible; mean-field exponents exact, $\beta=\tfrac12$
Finite system, any $T>0$	$N<\infty$	$\langle\phi\rangle=0$ exactly — no SSB in a finite box

Common trapThere is strictly no symmetry breaking in a finite system

It is tempting to say a small magnet at low temperature "has" a broken symmetry. Strictly, it does not: for any finite $N$ at $T>0$ , the partition sum weights symmetry-related configurations equally, so $\langle\phi\rangle=0$ exactly. The nonzero order parameter appears only after the singular thermodynamic limit, and only if $N\to\infty$ is taken before $h\to0$ . Reverse the order and you recover zero. A second, related trap: naive mean-field or Landau treatments predict order for a continuous symmetry in $d\le2$ , but they ignore Goldstone fluctuations. Mermin–Wagner forbids it — a 2D Heisenberg magnet has no spontaneous magnetization at any finite temperature, and the $\int \mathrm{d}^dk/k^2$ infrared divergence is the proof.

Worked examples

Example 1 — How small a field is "infinitesimal"? For a macroscopic sample with $N=6.0\times10^{23}$ spins and $m_0=1$ , estimate the free-energy bias between the two wells at a field $h=10^{-9}$ (in units where the bias is $2Nhm_0$ ), and interpret.

Step 1. Form the extensive bias. The relevant exponent is $2Nhm_0$ .

Step 2. Substitute. $2\,(6.0\times10^{23})(10^{-9})(1)$ .

Step 3. Evaluate. The bias is $\mathbf{1.2\times10^{15}}$ — astronomically large despite $h$ being a billionth. The well ratio $e^{1.2\times10^{15}}$ locks the system into the favoured well utterly. This is why " $N\to\infty$ then $h\to0$ " is not a mathematical nicety: for any macroscopic $N$ , a field far too weak to measure already fixes the direction of the order parameter.

Example 2 — Mean-field magnetization just below $T_c$ . Using $m\simeq\sqrt{3}\,(1-T/T_c)^{1/2}$ from the cubic expansion, find the magnetization at $T/T_c=0.99$ .

Step 1. Substitute. $m\simeq\sqrt{3}\,(1-0.99)^{1/2}=\sqrt{3}\,(0.01)^{1/2}$ .

Step 2. Evaluate. $m\simeq1.732\times0.1=\mathbf{0.173}$ . The order parameter is already 17% of saturation just 1% below $T_c$ , the steep square-root onset of $\beta=\tfrac12$ .

Example 3 — Counting Goldstone modes. A Heisenberg ferromagnet has an $O(3)$ symmetry that breaks to $O(2)$ (rotations about the magnetization axis remain). How many Goldstone modes appear, and what are they?

Step 1. Use the count. $N_{\text{Goldstone}}=\dim O(3)-\dim O(2)$ .

Step 2. Substitute. $\dim O(n)=n(n-1)/2$ , so $\dim O(3)=3$ and $\dim O(2)=1$ ; $N_{\text{Goldstone}}=3-1=\mathbf{2}$ .

Step 3. Interpret. The two modes are the transverse spin waves (magnons) — the two independent directions the magnetization can tilt at no potential-energy cost. A $U(1)$ superfluid, by contrast, breaks one generator and has exactly one Goldstone mode, the phase (second-sound) mode.

Origins · 1 minYoichiro Nambu and how this was discovered

Yoichiro Nambu

1921–2015; Japanese-American theoretical physicist; imported spontaneous symmetry breaking from condensed matter into particle physics (1960–61); Nobel Prize in Physics 2008 · SSB in superconductivity · the Nambu–Jona-Lasinio model · the origin of mass from broken symmetry

Yoichiro Nambu carried spontaneous symmetry breaking across the border between condensed-matter and particle physics. Studying the theory of superconductivity in 1960, he recognized that the BCS ground state broke a continuous $U(1)$ symmetry, and that this breaking required an accompanying massless collective mode — the seed of what became the Goldstone theorem. In 1961, with Giovanni Jona-Lasinio, he built a model (now the Nambu–Jona-Lasinio model) in which the mass of the nucleon arises spontaneously from a broken chiral symmetry of otherwise massless constituents, a mechanism that anticipated the modern understanding of how particles acquire mass. His insight that "the vacuum is a medium" — that empty space can order like a magnet or a superconductor — reshaped high-energy physics and underlies the Higgs mechanism. He shared the 2008 Nobel Prize in Physics "for the discovery of the mechanism of spontaneous broken symmetry in subatomic physics." Nambu's career is the clearest illustration that SSB is not a specialty of magnets: it is a single idea that unifies ferromagnetism, superfluidity, and the structure of the electroweak vacuum.

History

1928

— Heisenberg's exchange theory shows ferromagnetic order arises from a symmetric Hamiltonian, framing the puzzle SSB answers

1937

— Landau's order-parameter theory gives the free-energy mechanism by which symmetric laws yield asymmetric states

1960

— Nambu identifies spontaneous symmetry breaking in the BCS theory of superconductivity, with a required massless mode

1961

— Goldstone proves that breaking a continuous symmetry generates gapless bosons; Nambu and Jona-Lasinio model dynamical mass generation

1966

— Mermin and Wagner prove no continuous-symmetry SSB with short-range forces in two dimensions at finite temperature

2008

— Nambu awarded the Nobel Prize for the mechanism of spontaneous broken symmetry in subatomic physics

Going further

For the statistical-mechanics development see Pathria & Beale, Statistical Mechanics (3rd ed.), Ch. 11–13, and Kardar's Statistical Physics of Fields (Ch. 2–3), which carries the $O(n)$ model and the Goldstone expansion carefully. Sethna's Statistical Mechanics: Entropy, Order Parameters, and Complexity (Ch. 9) is the best conceptual treatment of order parameters and the order-of-limits argument; Landau & Lifshitz, Statistical Physics Part 1, §142–147 gives the classic Landau-theory foundation; and Goldenfeld's Lectures on Phase Transitions and the Renormalization Group (Ch. 2–5) contains the Mermin–Wagner fluctuation integral in full. Provenance: all equations cross-checked against Pathria & Beale Ch. 11–13 (mean-field Ising $k_BT_c=zJ$ , $\beta=\tfrac12$ ), Kardar SPF Ch. 2–3 (Goldstone/O(n)), Sethna Ch. 9 (limit-order argument), Landau & Lifshitz §142–147, and Goldenfeld Ch. 2–5 (Mermin–Wagner). Verified numerically in verify_spontaneous-symmetry-breaking.py. Prerequisites: order parameter · Landau theory of phase transitions · continuous phase transitions. Leads to: Goldstone modes · universality and universality classes · critical exponents · correlation length and its divergence.

What is the difference between spontaneous and explicit symmetry breaking?

Explicit breaking means a term in the Hamiltonian itself violates the symmetry — like the $-hm$ field term, which biases the order parameter for all temperatures, even above $T_c$ . Spontaneous breaking occurs when the Hamiltonian (or free energy) is fully symmetric, with no such term, yet its minimum lies away from the symmetric point, so the realized state breaks a symmetry the dynamics respects. The tell is degeneracy: under spontaneous breaking the broken states form a family related by the symmetry group, whereas explicit breaking picks out one state uniquely from the start.

If the finite system average is exactly zero, in what sense is a real magnet "broken"?

A real magnet is finite, so its true equilibrium average magnetization is indeed zero — it would, given unimaginably long times, tunnel through all orientations. But the tunnelling time scales as $e^{+\beta\sigma L^{d-1}}$ , which for a macroscopic sample exceeds the age of the universe by enormous factors. The magnet is dynamically trapped in one broken state. So SSB in a finite system is a statement about time scales: the symmetry is restored only on a time no experiment (or cosmos) can wait for, and for every practical purpose the state is broken.

Why does breaking a continuous symmetry produce massless modes?

Because the broken states form a continuous, degenerate manifold — a ring or sphere of equivalent vacua. Sliding the order parameter uniformly along this manifold costs no energy, since all points are equivalent ground states. A long-wavelength distortion that varies the direction slowly in space therefore costs energy only through the gradient term, which vanishes as the wavelength grows: $\omega(\mathbf{k})\to0$ as $\mathbf{k}\to0$ . That gapless excitation is the Goldstone mode. A discrete symmetry has no continuous manifold to slide along, so it produces no Goldstone modes — which is exactly why discrete order survives in $d=2$ but continuous order does not.

Is spontaneous symmetry breaking the same thing as a phase transition?

They overlap heavily but are not identical. Most familiar continuous phase transitions are symmetry-breaking transitions — the paramagnet-to-ferromagnet, the normal-to-superfluid. But some transitions break no symmetry (the liquid–gas transition above its critical point is a smooth crossover; topological transitions like Kosterlitz–Thouless have no local order parameter), and some symmetry breaking is not associated with a thermal transition at all. The cleanest statement: SSB is the mechanism by which an ordered phase differs from a disordered one whenever the two are distinguished by a symmetry, and the order parameter is the quantity that measures it.

How does SSB reconcile a symmetric law with an asymmetric world?

The symmetry is never lost — it is realized collectively rather than on individual states. The full symmetry group maps each broken ground state to another, equally valid broken ground state; the set of all of them respects the symmetry perfectly. What breaks is only the realized state's individual invariance, because in the thermodynamic limit those symmetry-related states decouple into separate superselection sectors and the system inhabits just one. So the law's symmetry survives as a relationship among the possible worlds, even though the single world we observe does not display it.

Why is crystallization first order while the ferromagnetic transition is continuous?

Because of which symmetry breaks and what invariants that permits. A crystal breaks continuous translational (and rotational) symmetry, and the density-wave order parameters $\rho_{\mathbf{G}}$ admit cubic invariants in the free energy — three reciprocal-lattice vectors can sum to zero. A cubic term forces a discontinuous jump (first order), which is why freezing has a latent heat. The Ising ferromagnet, by contrast, has an exact $m\to-m$ symmetry that forbids odd terms, so its transition is generically continuous. The order of the transition follows from the symmetry-allowed terms, exactly as in Landau theory.

Does the Higgs mechanism count as spontaneous symmetry breaking?

It is SSB with a twist. When the broken continuous symmetry is a gauged (local) symmetry — as in a superconductor or the electroweak theory — the would-be Goldstone mode does not appear as an independent massless particle. Instead it is absorbed by the gauge field, which acquires a mass (the photon in a superconductor becomes massive, giving the Meissner effect; the $W$ and $Z$ bosons get their masses in the electroweak theory). This is the Anderson–Higgs mechanism. The underlying symmetry breaking is the same statistical-mechanics phenomenon; the gauge coupling just changes where the Goldstone degree of freedom ends up.