Exact Symmetry

An exact symmetry is a transformation that leaves the physical structure of a quantum system invariant without approximation. The chapter guide compares this baseline with explicit, approximate, spontaneous, hidden, dynamical, emergent, and protected structures. For a time-independent closed system, the most common statement is:

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

where $H$ is the Hamiltonian and $S$ is the unitary or antiunitary operator representing the transformation. If the symmetry is unitary, this is equivalently

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

For a continuous unitary symmetry generated by a self-adjoint operator $G$,

$$S(\alpha) = e^{-i\alpha G/\hbar},$$

exact invariance for all $\alpha$ implies

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

The word “exact” matters. It means the symmetry is a property of the model or physical setup as stated, not a rough simplification, a low-energy approximation, a perturbative limit, or a pattern visible only in one specially chosen state.

Hamiltonian, State, and Observable Symmetries

A symmetry statement must say what object is invariant. In quantum mechanics, different statements have different meanings:

• a Hamiltonian may be invariant, $SHS^{-1}=H$;
• a state may be invariant, $S\lvert\psi\rangle=e^{i\phi}\lvert\psi\rangle$;
• a set of states may be invariant as a subspace;
• an observable algebra may be invariant, $A\mapsto SAS^{-1}$;
• a measurement setup may or may not transform along with the system.

For a time-independent unitary symmetry, applying the transformation and then evolving gives the same result as evolving and then applying the transformation. Antiunitary symmetries require a different time relation, given below.

For unitary time evolution

$$U(t) = e^{-iHt/\hbar},$$

an exact unitary symmetry $S$ with $[S,H]=0$ also commutes with the time-evolution operator:

$$SU(t) = U(t)S.$$

Thus symmetry is not merely a label on eigenstates. It is a compatibility condition between transformations and dynamics.

Exact Symmetry of a Hamiltonian

For a unitary symmetry, the Hamiltonian condition can be written in three equivalent ways:

$$\begin{aligned} SHS^{-1}&=H,\\ SH&=HS,\\ [S,H]&=0. \end{aligned}$$

The first form emphasizes invariance under transformation. The second says that the order of symmetry operation and Hamiltonian action does not matter. The third is the commutator test.

For antiunitary symmetries, such as time reversal, the compact condition remains

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

but antilinearity sends $i$ to $-i$. Consequently,

$$SU(t)S^{-1}=U(-t),$$

not generally $U(t)$. This is the dynamical content of time reversal at an invariant Hamiltonian.

For unbounded operators, invariance includes preservation of the self-adjoint domain. The generator statement means strong commutation of the spectral measures; a formal differential commutator on a small common core is not by itself sufficient. The finite-matrix manipulations here are automatic in a finite-dimensional Hilbert space.

Continuous Symmetries and Conserved Quantities

Suppose

$$S(\alpha) = e^{-i\alpha G/\hbar}$$

is an exact symmetry for every value of $\alpha$. Then

$$S(\alpha)HS(\alpha)^\dagger=H.$$

Differentiate at $\alpha=0$:

$$\frac{d}{d\alpha} \left. S(\alpha)HS(\alpha)^\dagger \right|_{\alpha=0} = -\frac{i}{\hbar}GH + \frac{i}{\hbar}HG = 0.$$

Therefore

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

If $G$ has no explicit time dependence, it is conserved under closed-system evolution:

$$\frac{d}{dt}\langle G\rangle = \frac{i}{\hbar}\langle[H,G]\rangle = 0.$$

This is the quantum version of the symmetry-conservation link.

Discrete Symmetries

Discrete symmetries do not require infinitesimal generators. Parity in one dimension is the simplest example:

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

For

$$H = \frac{P_x^2}{2m}+V(X),$$

on the full line with a parity-invariant self-adjoint domain, parity is exact when

$$V(-X)=V(X).$$

Then

$$[\mathsf P,H]=0.$$

Energy eigenstates can be chosen with definite parity, and parity-odd perturbations have vanishing diagonal matrix elements in any definite-parity state for which the expectation exists. Nondegeneracy is sufficient to enforce a parity eigenstate, but is not required once that parity is specified.

Rotational Symmetry and Reduced Symmetry

For a spinless particle in a central potential,

$$H = \frac{\mathbf p^2}{2m} + V(r), \qquad r=\lvert\mathbf r\rvert,$$

the Hamiltonian is invariant under all spatial rotations. Equivalently,

$$[H,L_i]=0 \qquad i=x,y,z.$$

The exact symmetry group is rotational symmetry. Energy eigenstates can be organized by angular momentum quantum numbers.

If an external magnetic field chooses the $z$ direction, full rotational symmetry is generally reduced. For example,

$$H = H_0 - \gamma B S_z$$

is invariant under rotations about the $z$ axis if $H_0$ commutes with all three spin components, but it is not invariant under arbitrary rotations unless the field is transformed as part of the physical setup. With the field held fixed, the surviving exact continuous symmetry is axial:

$$[H,S_z]=0, \qquad [H,S_x]\ne0, \qquad [H,S_y]\ne0$$

for nonzero $\gamma B$ and nonzero spin. If $H_0$ includes spin–orbit coupling, full rotational invariance concerns $\mathbf J=\mathbf L+\mathbf S$, and axial conservation is of $J_z$, not necessarily $S_z$ separately.

This distinction is common in applications. A term can reduce a larger symmetry to a subgroup rather than destroy all symmetry.

Consequences of Exact Symmetry

Exact symmetry has several practical consequences.

First, it produces conserved quantities for continuous symmetries. Translational invariance gives momentum conservation; rotational invariance gives angular momentum conservation; time-translation invariance gives energy conservation.

Second, it organizes Hilbert space into invariant sectors. If

$$[S,H]=0,$$

then $H$ maps each eigenspace of $S$ into itself, assuming the relevant spectral decomposition is well-defined. This lets one block-diagonalize the Hamiltonian by symmetry labels.

Third, it constrains matrix elements. If states and operators transform incompatibly under a symmetry, the corresponding matrix element must vanish. This is the logic behind parity selection rules and the more systematic Wigner-Eckart theorem.

Fourth, it can protect degeneracies. If symmetry maps a state to a linearly independent state with the same energy, then degeneracy follows. However, not every exact symmetry forces degeneracy: one-dimensional representations can label nondegenerate states.

Symmetry of a State Is Different

A Hamiltonian can have an exact symmetry even if a particular state does not look symmetric. For example, an even potential has parity symmetry, but a wave packet localized on one side of the origin is not itself a parity eigenstate.

Conversely, a particular state can be invariant under a transformation that is not a symmetry of the Hamiltonian. A spin state aligned with $z$ is an eigenstate of $S_z$, but that alone does not imply the Hamiltonian is invariant under rotations about $z$.

The symmetry of the dynamics and the symmetry of a state answer different questions:

$$\begin{array}{ccl} SHS^{-1}=H &\text{asks}& \text{whether the dynamics has the symmetry},\\ S\lvert\psi\rangle=e^{i\phi}\lvert\psi\rangle &\text{asks}& \text{whether the state carries a definite symmetry label}. \end{array}$$

Keeping these separate prevents many false conclusions about symmetry breaking.

Exact Versus Approximate or Broken

Exact symmetry is the baseline against which other notions are defined.

An explicitly broken symmetry is not a symmetry of the Hamiltonian after all terms are included. For instance, adding a small odd potential to an even one breaks parity explicitly.

An approximate symmetry is not exact in the full Hamiltonian, but the symmetry-breaking terms are small in a useful regime. Approximate symmetries can explain near-degeneracies and weak selection-rule violations.

A hidden symmetry or accidental symmetry may explain degeneracy beyond an obvious geometric symmetry. The hydrogen atom’s Coulomb degeneracy is the standard example.

Spontaneous symmetry breaking is a more subtle notion: the equations or Hamiltonian have a symmetry, while physically relevant states or phases fail to display it in an appropriate many-degree-of-freedom limit. It is not simply the observation that one finite-system state lacks a symmetry.

The word exact always applies to the stated Hamiltonian, domain, and fixed backgrounds. It does not assert that the ideal model includes every interaction of a real experiment.

Common Mistakes

• Calling a transformation a symmetry because it is mathematically available, without checking $SHS^{-1}=H$.
• Confusing symmetry of the Hamiltonian with symmetry of one state.
• Forgetting that external fields must either transform as part of the setup or be treated as fixed backgrounds.
• Assuming every exact symmetry creates degeneracy.
• Treating a weakly broken symmetry as exact when deriving selection rules.
• Using a continuous-symmetry generator after the corresponding symmetry-breaking perturbation has been added.
• Handling antiunitary symmetries as if they were ordinary unitary matrices.

Exercises

1. Parity of a one-dimensional Hamiltonian.

Let

$$H = \frac{P_x^2}{2m} + V(X),$$

with $\mathsf P X\mathsf P^{-1}=-X$ and $\mathsf P P_x\mathsf P^{-1}=-P_x$. Show that parity is an exact symmetry precisely when $V(-X)=V(X)$.

Solution

Transform the Hamiltonian:

$$\mathsf P H\mathsf P^{-1} = \frac{(\mathsf P P_x\mathsf P^{-1})^2}{2m} + V(\mathsf P X\mathsf P^{-1}) = \frac{P_x^2}{2m} + V(-X).$$

Thus $\mathsf P H\mathsf P^{-1}=H$ exactly when

$$V(-X)=V(X).$$

This also assumes the domain and boundary conditions preserve reflection; an even differential expression on an asymmetric domain is not enough.

2. Continuous symmetry and conservation.

Suppose $S(\alpha)=e^{-i\alpha G/\hbar}$ is an exact unitary symmetry of $H$ for all $\alpha$. Show that $G$ is conserved if it has no explicit time dependence.

Solution

Exact invariance gives

$$S(\alpha)HS(\alpha)^\dagger=H.$$

Differentiating at $\alpha=0$ yields

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

The expectation-value equation for an observable with no explicit time dependence is

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

Since $[H,G]=0$, the expectation value is constant in time.

3. Axial symmetry of a spin Hamiltonian.

Let

$$H = aI+bS_z.$$

Which spin rotations are exact symmetries?

Solution

Rotations about the $z$ axis are generated by $S_z$:

$$U_z(\theta)=e^{-i\theta S_z/\hbar}.$$

Since $H$ is a function of $S_z$,

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

so rotations about $z$ are exact symmetries. For nonzero $b$ and a nonzero spin representation, the Hamiltonian does not commute with $S_x$ or $S_y$, because

$$[S_z,S_x]=i\hbar S_y, \qquad [S_z,S_y]=-i\hbar S_x.$$

Thus full rotational symmetry is reduced to axial symmetry.

4. State symmetry versus Hamiltonian symmetry.

Give an example where the Hamiltonian has parity symmetry but a state does not.

Solution

Take an even potential, such as a harmonic oscillator potential $V(X)=m\omega^2X^2/2$. The Hamiltonian commutes with parity.

A localized wave packet centered at $x_0\ne0$ is not generally a parity eigenstate, because parity maps it to a packet centered near $-x_0$. The Hamiltonian has parity symmetry, but that particular state does not.

References

• H. F. Jones, Groups, Representations and Physics, 2nd ed., CRC Press, 1998.
• 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.
• M. Tinkham, Group Theory and Quantum Mechanics, Dover, 2003.
• E. P. Wigner, Group Theory and Its Application to the Quantum Mechanics of Atomic Spectra, Academic Press, 1959.