Spontaneous Symmetry Breaking Preview

Spontaneous symmetry breaking is the situation where the equations or Hamiltonian have a symmetry, but the physically relevant state, phase, or long-distance description does not display the full symmetry.

The phrase is easy to misuse. In ordinary finite quantum mechanics, a state can fail to be invariant under a symmetry even when the Hamiltonian is symmetric. That alone is not the many-body phenomenon called spontaneous symmetry breaking.

The essential many-body idea is that symmetric dynamics can support stable symmetry-broken phases in an appropriate large-system limit.

We will use a simple two-branch model to see why the limits matter. For the many-body development, continue to spontaneous symmetry breaking, order parameters, and the thermodynamic limit.

Exact Symmetry Remains in the Equations

Start with a family of finite systems whose Hamiltonians obey

$$U(g)H_NU(g)^{-1} = H_N$$

for every symmetry element $g$ in a group $G$. The Hamiltonian symmetry is exact at every finite size $N$.

Spontaneous symmetry breaking is not the statement

$$U(g)H_NU(g)^{-1}\ne H_N.$$

That would be explicit symmetry breaking.

Instead, the symmetry remains in the Hamiltonian, while the relevant limiting states preserve only a subgroup $H\subset G$:

$$G \longrightarrow H.$$

The broken generators or broken group directions label transformations that move one symmetry-broken state into another physically distinct state in the same phase.

Why Finite Systems Are Subtle

For a finite system with a unique ground state, an exact unitary symmetry of the Hamiltonian implies that the ground-state ray is invariant. The symmetry must preserve the Hamiltonian domain, including boundary conditions. If

$$H\lvert\Omega\rangle = E_0\lvert\Omega\rangle$$

is nondegenerate and $[H,U(g)]=0$, then

$$U(g)\lvert\Omega\rangle$$

is another ground state with the same energy. Nondegeneracy forces

$$U(g)\lvert\Omega\rangle = e^{i\phi(g)} \lvert\Omega\rangle.$$

Thus a finite, nondegenerate ground state cannot literally choose one of several symmetry-related ordered states.

What changes in many-body systems is the large-$N$ or infinite-volume limit. Energy splittings between symmetry-related combinations can vanish with size, local measurements can become insensitive to the symmetric superposition, and tiny symmetry-breaking fields can select one branch.

Order Parameter

An order parameter is an observable or field whose expectation value diagnoses the broken symmetry. If $O$ transforms nontrivially under $G$, a symmetry-broken state may have

$$\langle O\rangle\ne0.$$

For a discrete spin-flip symmetry, a magnetization component can be an order parameter. If a spin-flip operation sends

$$M_z \mapsto -M_z,$$

then a phase with

$$\langle M_z\rangle \ne 0$$

does not preserve that spin-flip symmetry.

For a system with a continuous $U(1)$ symmetry, a complex order parameter can choose a phase:

$$\langle\Psi\rangle = |\Psi|e^{i\theta}.$$

Different values of $\theta$ are related by the original symmetry. A finite symmetric state may have $\langle\Psi\rangle=0$, while a symmetry-broken thermodynamic-limit state has a chosen phase after the correct limiting procedure.

Order of Limits

Use an intensive order parameter, such as magnetization per spin,

$$O_N=\frac1N\sum_{i=1}^N o_i.$$

A source $h$ conjugate to each local observable adds an extensive energy $-NhO_N$. This normalization lets the limiting order parameter stay finite. The source selects one branch:

$$H_{N,h} = H_N-NhO_N.$$

One first takes the large-system limit, then removes the source:

$$\lim_{h\to0^+} \lim_{N\to\infty} \langle O_N\rangle_{N,h}.$$

This can differ from taking $h\to0$ first at fixed finite $N$:

$$\lim_{N\to\infty} \lim_{h\to0^+} \langle O_N\rangle_{N,h}.$$

The noncommuting limits are not a technical nuisance. They are the mathematical signal that an infinitesimal perturbation can select a stable ordered phase only after the system is large enough.

A Two-Branch Calculation

Represent two macroscopically ordered branches by $|+\rangle$ and $|-\rangle$, with $O_N=m_0\sigma_z$ and $m_0>0$. A minimal low-energy model is

$$H_{N,h}^{\mathrm{eff}} =-\delta_N\sigma_x-Nhm_0\sigma_z, \qquad \delta_N>0.$$

At $h=0$, the tunneling term gives a unique symmetric ground state and a splitting $2\delta_N$. A positive source favors $|+\rangle$. Diagonalizing this two-by-two matrix gives

$$E_-=-\sqrt{\delta_N^2+(Nhm_0)^2}, \qquad \langle O_N\rangle_{N,h} =m_0\frac{Nhm_0}{\sqrt{\delta_N^2+(Nhm_0)^2}}.$$

For each finite $N$, sending $h$ to zero gives zero. If the tunneling splitting vanishes with size, for example $\delta_N=\delta_0e^{-cN}$ with $c>0$, then any fixed $h>0$ dominates it as $N\to\infty$. Consequently,

$$\lim_{h\to0^+}\lim_{N\to\infty}\langle O_N\rangle_{N,h}=m_0, \qquad \lim_{N\to\infty}\lim_{h\to0^+}\langle O_N\rangle_{N,h}=0.$$

A negative source instead selects $-m_0$. This effective model displays the branch-selection mechanism; deriving its size-dependent tunneling scale and establishing an ordered phase requires a microscopic many-body model.

Discrete Versus Continuous Breaking

Breaking a discrete symmetry can produce multiple ordered phases, domain walls, and near-degenerate finite-size structures. It does not by itself imply a Goldstone mode.

Breaking a continuous global symmetry is different. Under suitable locality and stability assumptions, it leads to low-energy collective excitations. These are Nambu–Goldstone modes. The physical preview is Goldstone Modes Preview, and the compact theorem card is Goldstone Theorem Preview.

The word “global” is important. Gauge redundancy is not broken in the same direct sense as a physical global symmetry. Gauge theories, superconductors, and Higgs-like systems require separate treatment.

Examples and Nonexamples

Useful examples include:

• a three-dimensional Ising ferromagnet choosing positive or negative magnetization below its critical temperature;
• a three-dimensional Heisenberg ferromagnet choosing a magnetization direction at sufficiently low temperature;
• a three-dimensional neutral superfluid choosing a phase in a thermodynamic description with a global $U(1)$ order parameter;
• a crystal choosing a lattice of positions and breaking continuous translations to a discrete subgroup.

Useful nonexamples include:

• a single spin in the state $|+z\rangle$ while the Hamiltonian is zero;
• one localized wave packet in a symmetric finite double well;
• an explicitly applied magnetic field selecting an axis;
• a gauge choice that makes a field component look nonzero.

The nonexamples may be excellent approximations or useful pictures. They are not, by themselves, spontaneous symmetry breaking in the many-body sense.

Relation to Explicit Breaking

Explicit breaking and spontaneous breaking often appear together in practice. A tiny external field can select a branch of an ordered phase. The field is explicit breaking; the stable ordered response in the large-system limit is the spontaneous-breaking phenomenon.

If a continuous symmetry is explicitly but weakly broken, the would-be Goldstone mode can acquire a small gap. Such excitations are often called pseudo-Goldstone modes. The weak-breaking language is an approximate symmetry statement, while the source term itself is explicit breaking. The exact terminology and counting rules depend on the physical system.

Temperature, dimension, interaction range, and the kind of symmetry all matter. The examples above are not claims that every low-dimensional model orders. Quantum phase transitions develops zero-temperature gap scaling, and the transverse-field Ising model makes discrete branch selection concrete. Beekman, Rademaker, and van Wezel give a systematic account of the singular limits and their physical interpretation.

Common Mistakes

• Saying “the state is not symmetric” and calling that spontaneous symmetry breaking in a finite system.
• Forgetting the thermodynamic-limit or infinite-volume assumption.
• Confusing a tiny explicit field that selects a branch with the spontaneous phenomenon itself.
• Applying Goldstone theorem to a broken discrete symmetry.
• Treating gauge redundancy as an ordinary global symmetry.
• Assuming a finite symmetric ground state and a symmetry-broken thermodynamic state are contradictory.
• Using an order parameter without stating how it transforms.

Exercises

1. Why does a unique finite-system ground-state ray preserve an exact unitary symmetry?

Assume $[H,U]=0$ and $H|\Omega\rangle=E_0|\Omega\rangle$, with $E_0$ nondegenerate.

Solution

Because $[H,U]=0$,

$$H(U|\Omega\rangle) = UH|\Omega\rangle = E_0(U|\Omega\rangle).$$

Thus $U|\Omega\rangle$ is also a ground state. If the ground state is nondegenerate, it must be proportional to $|\Omega\rangle$:

$$U|\Omega\rangle = e^{i\phi}|\Omega\rangle.$$

The state is symmetric as a ray. A literal choice among distinct symmetry-related ground states requires degeneracy or a limiting procedure.

2. Explain the order of limits in a source-selected order parameter.

Why can

$$\lim_{h\to0^+}\lim_{N\to\infty}\langle O_N\rangle_{N,h}$$

differ from

$$\lim_{N\to\infty}\lim_{h\to0^+}\langle O_N\rangle_{N,h}?$$

Solution

At finite $N$, removing the source first can restore the exact symmetry and force a symmetry-odd order parameter to vanish. Taking $N\to\infty$ first can make the tunneling splitting between symmetry-related combinations vanish. In the two-branch model, the extensive bias $Nhm_0$ then dominates $\delta_N$ for any fixed $h>0$, giving $\langle O_N\rangle\to m_0$. Removing the source afterwards leaves this nonzero limit; no literal infinite potential barrier is required.

3. Does breaking a discrete spin-flip symmetry imply a Goldstone mode?

Solution

No. Goldstone modes are tied to broken continuous global symmetries under appropriate assumptions. A discrete symmetry has separated alternatives, not a continuous family of nearby states. Discrete symmetry breaking can produce domain walls or nearly degenerate finite-size states, but not a Goldstone mode merely from discreteness.

4. Identify explicit versus spontaneous breaking.

A tiny magnetic field $h$ is applied to a ferromagnet and then sent to zero after the thermodynamic limit. Which part is explicit breaking, and which part signals spontaneous breaking?

Solution

The finite $h$ term in the Hamiltonian is explicit symmetry breaking: it directly selects a magnetization direction. If, after taking the thermodynamic limit, the magnetization remains nonzero as $h\to0^+$, that persistent ordered response is the spontaneous symmetry-breaking signal.

References

• A. Altland and B. Simons, Condensed Matter Field Theory, 2nd ed., Cambridge University Press, 2010.

• P. W. Anderson, “Plasmons, gauge invariance, and mass,” Physical Review 130, 439–442, 1963.

• A. J. Beekman, L. Rademaker, and J. van Wezel, “An introduction to spontaneous symmetry breaking”, SciPost Physics Lecture Notes 11, 2019, arXiv:1909.01820.

• P. Coleman, Introduction to Many-Body Physics, Cambridge University Press, 2015.

• J. Goldstone, “Field theories with superconductor solutions,” Il Nuovo Cimento 19, 154–164, 1961.

• J. Goldstone, A. Salam, and S. Weinberg, “Broken symmetries,” Physical Review 127, 965–970, 1962.

• Y. Nambu, “Quasi-particles and gauge invariance in the theory of superconductivity,” Physical Review 117, 648–663, 1960.

• S. Weinberg, The Quantum Theory of Fields, Volume II, Cambridge University Press, 1996.