Explicit Symmetry Breaking

Explicit symmetry breaking occurs when a Hamiltonian, Lagrangian, or experimental setup contains a term that is not invariant under a transformation that would otherwise be a symmetry.

The clean model is

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

where

$$SH_0S^{-1}=H_0, \qquad SVS^{-1}\ne V.$$

For $\lambda\ne0$, the full Hamiltonian generally fails to satisfy

$$SHS^{-1}=H.$$

The breaking is called explicit because it is visible in the equations themselves. A term has been written that names a direction, distinguishes two regions, singles out a spin component, or otherwise violates the original symmetry.

Reference Symmetry and Broken Term

Explicit breaking is always relative to a reference symmetry. One must state:

• the unbroken Hamiltonian $H_0$;
• the candidate symmetry $S$ or group $G$;
• the added term $V$;
• the residual symmetries, if any, of $H=H_0+\lambda V$.

For a group $G$ represented by operators $U(g)$, the exact symmetry group of the full Hamiltonian is

$$G_H = \{g\in G:U(g)HU(g)^{-1}=H\}.$$

Explicit breaking often means that

$$G_H \subsetneq G_{H_0}.$$

The full symmetry may be smaller, not absent. This is why one says a magnetic field breaks rotational symmetry down to rotations about the field axis.

Consequences for Conserved Quantities

If $G$ is the generator of a continuous symmetry of $H_0$, then

$$[H_0,G]=0.$$

After adding a breaking term,

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

If $[V,G]\ne0$, the generator is not conserved by the full dynamics. In expectation-value form,

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

when $G$ has no explicit time dependence.

For small $\lambda$, the nonconservation can be slow or perturbative, but it is not zero. This distinction matters: an approximately conserved quantity is useful, but it is not an exact quantum number.

Parity Broken by a Linear Perturbation

Consider a one-dimensional Hamiltonian with an even reference potential:

$$H_0 = \frac{P_x^2}{2m}+V_0(X), \qquad V_0(-X)=V_0(X).$$

Parity about the origin is exact for $H_0$:

$$\mathsf P H_0\mathsf P^{-1} = H_0.$$

Add a uniform-force perturbation

$$V_{\mathrm{break}} = -FX.$$

Since

$$\mathsf P X\mathsf P^{-1}=-X,$$

the perturbation transforms as

$$\mathsf P V_{\mathrm{break}}\mathsf P^{-1} = +FX \ne -FX.$$

Thus parity about the origin is explicitly broken when $F\ne0$.

For a harmonic oscillator, the full potential can be re-centered by completing the square. That does not mean the original origin-parity symmetry survived; it means the special quadratic problem has a different inversion symmetry about a displaced center. Always name the transformation being tested.

Magnetic Field as a Fixed Background

An isolated spin with Hamiltonian proportional to the identity has full spin-rotation symmetry:

$$H_0=aI.$$

Add a fixed static magnetic field along $z$:

$$H = aI-\gamma B S_z.$$

The full Hamiltonian commutes with $S_z$:

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

But for nonzero $B$ it does not commute with $S_x$ or $S_y$:

$$[H,S_x] = -\gamma B[S_z,S_x] = -i\gamma B\hbar S_y,$$

and similarly for $S_y$. The fixed field explicitly breaks full spin-rotation symmetry down to rotations about the $z$ axis.

The word “fixed” is doing real work. If one rotates the spin system and the external field together, the covariant relation between them is unchanged. If the field is held as a background in the laboratory, it selects an axis and reduces the system symmetry.

Electric Field and Inversion

A particle with an inversion-symmetric Hamiltonian can lose inversion symmetry in a static electric field. The perturbation is often of the form

$$V_E = -q\mathbf E\cdot\mathbf X.$$

Under inversion,

$$\mathbf X\mapsto-\mathbf X,$$

while a fixed external $\mathbf E$ is held fixed as a laboratory background. Therefore

$$V_E \mapsto +q\mathbf E\cdot\mathbf X,$$

so inversion is explicitly broken.

This is the symmetry reason that Stark-type perturbations can mix states of opposite parity. The quantitative energy shifts belong to perturbation-theory and atomic-physics pages; this page only tracks the symmetry content.

Degeneracy Splitting and Sector Mixing

Exact symmetry often lets one label states by quantum numbers. Explicit breaking can remove those labels.

Suppose $S$ is exact for $H_0$, and the eigenstates of $H_0$ can be chosen with labels $s$. If a perturbation $V$ does not commute with $S$, then $V$ can connect different $s$ sectors:

$$\langle s'|V|s\rangle \ne 0 \qquad \text{for} \qquad s'\ne s.$$

As a result:

• formerly exact selection rules can fail;
• degenerate multiplets can split;
• eigenstates can become mixtures of old symmetry sectors;
• only quantum numbers associated with residual symmetries remain exact.

When degeneracy is present, the correct first-order calculation is not to assign shifts to arbitrary old basis states. One diagonalizes the perturbation inside the degenerate subspace. The symmetry interpretation is degeneracy lifting; the calculation method is covered in degenerate perturbation theory.

Residual Symmetry

Explicit breaking often preserves a subgroup. For example:

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

when a fixed vector such as $\mathbf B=B\hat{\mathbf z}$ selects an axis. The original angular momentum components $J_x,J_y,J_z$ are no longer all conserved, but $J_z$ may remain conserved.

In a one-dimensional parity problem, adding a generic asymmetric potential can remove parity entirely. But adding a displaced symmetric potential may preserve inversion about a different point. The residual symmetry is found by testing the full Hamiltonian, not by remembering the labels of the unperturbed problem.

Explicit Versus Spontaneous

Explicit breaking is a statement about the equations:

$$SHS^{-1}\ne H.$$

Spontaneous symmetry breaking is different. There the equations may still have the symmetry, while physically relevant thermodynamic states fail to display it after a source or boundary condition selects a branch. The canonical many-body page develops the required limit and finite-size spectrum.

In finite-dimensional or ordinary finite-particle quantum mechanics, most simple examples called “symmetry breaking” are explicit breaking: a term has been added to the Hamiltonian or a background field has been fixed.

Common Mistakes

• Saying a symmetry is broken without naming the original symmetry and the full Hamiltonian.
• Treating a fixed external field as if it transformed along with the system when the laboratory problem holds it fixed.
• Forgetting residual symmetries after a perturbation is added.
• Applying old selection rules after the perturbation has broken their assumptions.
• Calling a small breaking term harmless because it is perturbative; small is not exact.
• Confusing explicit breaking with spontaneous symmetry breaking.
• Assuming degeneracy always disappears; some residual or antiunitary symmetry may still protect part of it.

Cross-Links

• Exact Symmetry
• Broken Symmetry Preview
• Spontaneous Symmetry Breaking
• Spontaneous Symmetry Breaking Preview
• Approximate Symmetry
• Degeneracy Lifting
• Symmetry Constraints on Hamiltonians
• Commutators and Conservation Laws
• Parity
• Spin in Magnetic Fields
• Central Potentials
• Selection Rules
• Nondegenerate Perturbation Theory
• Degenerate Perturbation Theory
• Commutator Table

References

• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Butterworth-Heinemann, 1977.
• H. F. Jones, Groups, Representations and Physics, 2nd ed., CRC Press, 1998.
• M. Tinkham, Group Theory and Quantum Mechanics, Dover, 2003.

Exercises

1. Show that a linear perturbation breaks origin parity.

Let

$$H = \frac{P_x^2}{2m} + \frac12m\omega^2X^2 - FX.$$

Use $\mathsf P X\mathsf P^{-1}=-X$ and $\mathsf P P_x\mathsf P^{-1}=-P_x$ to test parity about the origin.

Solution

The kinetic and quadratic potential terms are invariant:

$$\mathsf P P_x^2\mathsf P^{-1}=P_x^2, \qquad \mathsf P X^2\mathsf P^{-1}=X^2.$$

The linear term changes sign:

$$\mathsf P(-FX)\mathsf P^{-1} = +FX.$$

Therefore $\mathsf P H\mathsf P^{-1}\ne H$ when $F\ne0$. Origin parity is explicitly broken.

2. Determine the residual symmetry of a fixed-field spin Hamiltonian.

For

$$H=-\gamma B S_z,$$

which spin-rotation generator remains conserved?

Solution

The Hamiltonian is proportional to $S_z$, so

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

However,

$$[H,S_x] = -\gamma B[S_z,S_x] = -i\gamma B\hbar S_y,$$

which is nonzero for a generic state and nonzero $B$. Similarly $S_y$ is not conserved. The residual continuous symmetry is rotation about the $z$ axis.

3. Broken generator and nonconservation.

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

Solution

The expectation-value equation gives

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

Since

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

we get

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

Thus $G$ is not generally conserved by the full Hamiltonian.

4. Why do old selection rules fail?

Suppose parity is exact for $H_0$, but a perturbation $V$ is odd under parity. Why should parity selection rules for exact eigenstates of $H_0$ not be applied unchanged to eigenstates of $H_0+\lambda V$?

Solution

The selection rule assumes the Hamiltonian eigenstates can be chosen with definite parity. If the full Hamiltonian includes an odd perturbation, it no longer commutes with parity:

$$[H_0+\lambda V,\mathsf P]\ne0.$$

The new eigenstates are generally mixtures of the old parity sectors. Parity labels are no longer exact, so selection rules derived from exact parity symmetry no longer apply without approximation.