Broken Symmetry Preview

A symmetry can fail to appear in more than one way. Sometimes the Hamiltonian itself is not invariant. Sometimes the Hamiltonian is invariant, but a chosen state is not. In many-body physics, a symmetric Hamiltonian can even have stable phases whose local observables do not display the full symmetry after an appropriate large-system limit.

Those are different statements. This preview sorts them before the dedicated symmetry-breaking chapter.

For a continuous unitary symmetry,

$$U(\alpha) = \exp\!\left(-\frac{i\alpha G}{\hbar}\right),$$

the exact Hamiltonian test is

$$[H,G]=0.$$

If this commutator fails, the generator is not conserved by the full Hamiltonian. If it holds, the equations have the symmetry, but individual states may still transform nontrivially.

Hamiltonian, state, and phase symmetry

When someone says a symmetry is broken, ask three separate questions.

First, is the Hamiltonian or dynamical law invariant?

$$U(g)HU(g)^{-1} = H.$$

Second, is the state or phase invariant?

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

as a ray, or does the transformation move it to a physically different state?

Third, if a state is not invariant, is that non-invariance merely a choice of state in a finite Hilbert space, or is it a stable phase selected in a large-system limit?

These questions separate explicit symmetry breaking, ordinary nonsymmetric states, and spontaneous symmetry breaking.

Explicit Breaking

Explicit breaking means the symmetry is no longer a symmetry of the Hamiltonian. A standard model is

$$H = H_0+\lambda V,$$

with

$$[H_0,G]=0, \qquad [V,G]\ne0.$$

Then

$$[H,G] = \lambda[V,G],$$

so $G$ is not exactly conserved for $\lambda\ne0$. If $G$ has no explicit time dependence,

$$\frac{d}{dt}\langle G\rangle = \frac{i\lambda}{\hbar} \langle[V,G]\rangle.$$

The old quantum number may remain useful when $\lambda V$ is small, but it is not exact. Selection rules protected only by the broken symmetry can become approximate, and degeneracies can split. Perturbative state mixing additionally requires the coupling to be small compared with the relevant level spacings.

Examples include:

• a fixed magnetic field reducing rotational symmetry to rotations about the field axis;
• a static electric field breaking inversion symmetry;
• a crystal environment reducing continuous rotations to a finite point group;
• spin–orbit coupling reducing independent spin and orbital rotations to total rotations.

For further examples, see Explicit Symmetry Breaking.

Residual Symmetry

Broken does not usually mean no symmetry remains. If a full group $G$ is reduced to a subgroup $K$, then the exact labels associated with $K$ can survive.

For example, a fixed field in the $z$ direction can reduce

$$SO(3) \longrightarrow SO(2).$$

Full angular momentum multiplets need not remain degenerate, but the projection along the field can remain good:

$$[H,J_z]=0.$$

The practical rule is to test the full Hamiltonian after all fields, perturbations, boundaries, and approximations have been specified. The surviving symmetry, not the old ideal symmetry, controls exact labels.

Nonsymmetric States Are Not Automatically Broken Phases

A Hamiltonian can be symmetric even when a state is not. A wave packet localized on one side of an even potential is not a parity eigenstate, but that does not mean the Hamiltonian has lost parity symmetry.

If $[H,U]=0$ and

$$H\lvert\psi\rangle=E\lvert\psi\rangle,$$

then $U\lvert\psi\rangle$ is also an eigenstate with the same energy:

$$H(U\lvert\psi\rangle) = U(H\lvert\psi\rangle) = E(U\lvert\psi\rangle).$$

If the energy level is nondegenerate, this forces

$$U\lvert\psi\rangle = e^{i\phi}\lvert\psi\rangle.$$

A nondegenerate energy eigenstate is therefore invariant as a ray under each unitary symmetry of $H$. In degenerate subspaces, one can choose bases that do or do not make the symmetry manifest. That is a basis and representation issue, not by itself spontaneous symmetry breaking.

Finite Doublet Picture

A symmetric double well gives a useful warning. Let $\lvert L\rangle$ and $\lvert R\rangle$ be localized states in the left and right wells of a parity-symmetric potential. A two-state effective Hamiltonian often has the form

$$H_{\mathrm{eff}} = \begin{pmatrix} E_0 & -t \\ -t & E_0 \end{pmatrix}, \qquad t>0.$$

Taking the localized basis vectors to be orthonormal and exchanged by parity, the eigenstates within this two-state model are the even and odd combinations

$$\lvert+\rangle = \frac{1}{\sqrt2} (\lvert L\rangle+\lvert R\rangle), \qquad \lvert-\rangle = \frac{1}{\sqrt2} (\lvert L\rangle-\lvert R\rangle),$$

with energies

$$E_+=E_0-t, \qquad E_-=E_0+t.$$

The splitting is

$$\Delta E=2t.$$

A localized state changes little on times short compared with $\hbar/t$, but it is not an energy eigenstate for any nonzero $t$. The finite-model eigenstates still have definite parity. This is the finite-size version of a recurring theme: apparent symmetry-broken branches can be separated by exponentially small splittings, yet the exact finite problem may still have symmetric eigenstates.

Spontaneous Breaking

Spontaneous symmetry breaking is different from explicit breaking. The equations keep the symmetry:

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

for each finite system size $N$, but the physically relevant large-system states or phases preserve only a subgroup

$$K\subset G.$$

An order parameter $O$ transforms nontrivially under $G$. A symmetry-broken phase can have

$$\langle O\rangle\ne0,$$

even though the Hamiltonian is symmetric. The usual careful procedure adds a small source $h$, takes the large-system limit, and then removes the source:

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

This can differ from removing the source first at finite $N$. The noncommuting limits are the signal that an infinitesimal explicit bias can select a stable branch only after the system is large enough.

For the full distinction, see Spontaneous Symmetry Breaking Preview.

Continuous Versus Discrete Breaking

Breaking a discrete symmetry can give multiple stable branches, near-degenerate finite-size doublets, and domain walls. It does not by itself imply a gapless mode.

Breaking a continuous global symmetry is more constrained. Under appropriate locality and stability assumptions, broken continuous global symmetries lead to low-energy collective modes. The physics is previewed in Goldstone Modes Preview and bridged toward field theory in From Symmetry Breaking to Goldstone Theorem.

Gauge redundancy requires separate care. A gauge choice that makes a field look nonsymmetric is not the same thing as breaking a physical global symmetry.

Level Splitting

Symmetry breaking often shows up spectroscopically as level splitting. Suppose $H_0$ has a degenerate subspace $\mathcal D$ with projector $P$. A perturbation $V$ gives the first-order effective operator

$$W=PVP.$$

If $W$ is not proportional to the identity on $\mathcal D$, the degeneracy splits at first order. If a residual symmetry remains, $W$ is block diagonal in the residual symmetry sectors, and some degeneracy may survive.

Thus a broken symmetry changes both labels and spectra:

$$\text{larger symmetry} \quad \longrightarrow \quad \text{smaller symmetry} \quad \longrightarrow \quad \text{finer spectral structure}.$$

The computational side belongs to degenerate perturbation theory. The symmetry interpretation is Degeneracy Lifting.

Selection Rules After Breaking

Selection rules depend on exact symmetry labels. If a perturbation breaks the protecting symmetry, matrix elements formerly forced to vanish may become nonzero.

In an approximate model one often finds

$$\langle f_{\mathrm{sym}}|T|i_{\mathrm{sym}}\rangle=0,$$

but the full states contain small admixtures:

$$\lvert i_{\mathrm{full}}\rangle = \lvert i_{\mathrm{sym}}\rangle +\epsilon\lvert i_1\rangle +O(\epsilon^2).$$

Then the full matrix element can be of order $\epsilon$. Such a transition is weakly allowed, not absolutely forbidden. This is why broken and approximate symmetries are inseparable from real spectroscopy.

Exercises

1. Explicit breaking and nonconservation.

Let $H=H_0+\lambda V$, with $[H_0,G]=0$ and $[V,G]\ne0$. Assuming $G$ has no explicit time dependence, derive $d\langle G\rangle/dt$.

Solution

The expectation-value equation is

$$\frac{d}{dt}\langle G\rangle = \frac{i}{\hbar}\langle[H,G]\rangle.$$

Since

$$[H,G] = [H_0,G]+\lambda[V,G] = \lambda[V,G],$$

we get

$$\frac{d}{dt}\langle G\rangle = \frac{i\lambda}{\hbar} \langle[V,G]\rangle.$$

The generator is not generally conserved by the full Hamiltonian.

2. Finite symmetric doublet.

Diagonalize

$$H_{\mathrm{eff}} = \begin{pmatrix} E_0 & -t \\ -t & E_0 \end{pmatrix}$$

in the $\lvert L\rangle,\lvert R\rangle$ basis.

Solution

The symmetric and antisymmetric combinations are

$$\lvert+\rangle = \frac{1}{\sqrt2} (\lvert L\rangle+\lvert R\rangle), \qquad \lvert-\rangle = \frac{1}{\sqrt2} (\lvert L\rangle-\lvert R\rangle).$$

Acting with the matrix gives

$$H_{\mathrm{eff}}\lvert+\rangle = (E_0-t)\lvert+\rangle, \qquad H_{\mathrm{eff}}\lvert-\rangle = (E_0+t)\lvert-\rangle.$$

Thus the splitting is $\Delta E=2t$. A state initially in $|L\rangle$ has right-well probability $\sin^2(t\tau/\hbar)$ after elapsed time $\tau$. Small tunneling delays transfer between wells; it does not make the splitting exactly zero. The even and odd probabilities remain one half each, so the evolving pure state never becomes a parity eigenstate, even when the two wells have equal populations.

3. Unique ground state and symmetry.

Assume $[H,U]=0$ and $H\lvert\Omega\rangle=E_0\lvert\Omega\rangle$, with $E_0$ nondegenerate. Show that $\lvert\Omega\rangle$ is invariant under $U$ up to phase.

Solution

Because $[H,U]=0$,

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

Thus $U\lvert\Omega\rangle$ is also a ground state. Nondegeneracy means it must be proportional to $\lvert\Omega\rangle$:

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

4. Why do old selection rules become weak rather than exact?

Suppose a symmetry of $H_0$ forces $\langle f_{\mathrm{sym}}|T|i_{\mathrm{sym}}\rangle=0$, but the full state is

$$\lvert i_{\mathrm{full}}\rangle = \lvert i_{\mathrm{sym}}\rangle +\epsilon\lvert i_1\rangle +O(\epsilon^2).$$

What is the leading possible order of the matrix element?

Solution

Substitute the full state:

$$\langle f_{\mathrm{sym}}|T|i_{\mathrm{full}}\rangle = \langle f_{\mathrm{sym}}|T|i_{\mathrm{sym}}\rangle +\epsilon \langle f_{\mathrm{sym}}|T|i_1\rangle +O(\epsilon^2).$$

The symmetry-forbidden leading term vanishes. The first possible contribution is therefore order $\epsilon$, assuming the mixed component has the needed symmetry character.

References

• A. Altland and B. Simons, Condensed Matter Field Theory, 2nd ed., Cambridge University Press, 2010.
• P. Coleman, Introduction to Many-Body Physics, Cambridge University Press, 2015.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020, chapters 4–5.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• M. Tinkham, Group Theory and Quantum Mechanics, Dover, 2003 reprint of the 1964 edition.