Continuous Symmetries and Conservation Laws

Continuous transformations connect a finite operation to an infinitesimal generator. When that transformation also preserves the dynamics, its generator is conserved. This chapter develops the distinction: momentum generates translations even in a trap, but the trap prevents momentum conservation. It also explains why a conserved observable can have a spread of outcomes and why explicitly time-dependent observables need a more general test.

Helpful background. Unitary symmetries give the finite transformation law; commutators give its infinitesimal action.

Enter this chapter

Question	Read next
Which continuous families have a fixed generator?	One-parameter unitary groups
How do we find and use the generator?	Generators and infinitesimal transformations
When does a symmetry give conservation?	Commutators and conservation laws and the quantum Noether principle
What if an observable depends explicitly on time?	Constants of motion
Which labels and matrix elements are restricted?	Good quantum numbers and selection rules
What survives a perturbation?	Broken symmetry

A group law is stronger than a smooth family

A strongly continuous one-parameter unitary group obeys $U(0)=I$ and $U(\alpha+\beta)=U(\alpha)U(\beta)$. Stone’s theorem gives a self-adjoint $G$ such that

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

Continuity means $U(\alpha)|\psi\rangle$ varies continuously in norm for every vector. Differentiability is stronger: the derivative at zero exists for vectors in the generator’s domain. On that domain,

$$U(\delta\alpha)|\psi\rangle =|\psi\rangle-\frac{i\delta\alpha}{\hbar}G|\psi\rangle +o(\delta\alpha).$$

A general smooth family of unitaries need not have one fixed generator or obey the additive group law. Time evolution under a varying Hamiltonian is an important example: its two-time propagators compose, but generally do not depend only on elapsed time.

The product $\alpha G$ has units of action. For translations, $\alpha$ is length and $G$ is momentum; for rotations, the angle is dimensionless and $G$ has angular-momentum units. See Stone’s theorem for the precise domain statement.

Commutators describe the first change

For finite matrices, or on vectors where the required unbounded products and derivatives exist,

$$\begin{aligned} UAU^\dagger&=A-\frac{i\delta\alpha}{\hbar}[G,A]+O(\delta\alpha^2),\\ U^\dagger A U&=A+\frac{i\delta\alpha}{\hbar}[G,A]+O(\delta\alpha^2). \end{aligned}$$

The opposite signs answer opposite transformation questions. The second expression computes a fixed observable’s expectation in the transformed state. For instance, with $T(a)=e^{-iaP/\hbar}$, $T^\dagger XT=X+aI$, so the wavepacket moves right by $a$. The first-order formula is not a substitute for the finite exponential when the parameter change is large.

Conservation includes explicit time dependence

For a possibly time-dependent Schrödinger-picture observable $A(t)$,

$$\frac{d}{dt}\langle A\rangle = \frac{i}{\hbar}\langle[H,A]\rangle + \left\langle \frac{\partial A}{\partial t} \right\rangle.$$

The corresponding operator criterion for a constant of motion is

$$\frac{\partial A}{\partial t} + \frac{i}{\hbar}[H,A] =0.$$

Thus $[H,A]=0$ is sufficient only when $A$ has no explicit time dependence. Conversely, an explicitly time-dependent $A(t)$ can still be conserved when its explicit derivative cancels its commutator term.

For a continuous unitary symmetry generated by $G$, the ordinary quantum Noether pattern is

$$\begin{aligned} U(\alpha)HU^\dagger(\alpha)&=H \\ &\Downarrow \\ [G,H]&=0 \\ &\Downarrow \\ \frac{dG_H}{dt}&=0, \end{aligned}$$

where the last implication assumes $\partial G/\partial t=0$.

For unbounded operators, these differentiations require suitable states and domains; strong commutation of spectral measures supplies the finite symmetry statement. The commutator argument alone is not the full field-theory theorem. Classical Noether theory begins with an action and treats boundary terms. Field theory produces a local current $j^\mu$ and a continuity equation $\partial_\mu j^\mu=0$ before a conserved charge is obtained. The ordinary quantum-mechanical chapter stays with Hilbert-space transformations, Hamiltonians, generators, and commutators. The field-theory bridge begins at From Quantum Generators to Noether Currents.

Conserved does not mean sharp

If a time-independent observable strongly commutes with the Hamiltonian, all its spectral probabilities are preserved by time evolution. A free wavepacket may therefore have a broad, unchanged momentum distribution. By contrast, an expectation value can happen to be stationary in one special state even when the operator is not conserved: the harmonic oscillator has $\langle P\rangle=0$ in an energy eigenstate, although $[H,P]\ne0$.

Good quantum numbers label compatible spectral sectors. In a discrete-spectrum problem, $[H,Q]=0$ with the appropriate operator assumptions lets energy states be chosen with a $Q$ label. A second proposed label must also be compatible with the first. Rotationally invariant dynamics conserves each $J_i$, but the noncommuting components cannot all label a basis sharply; the usual choice is $J^2,J_z$.

Selection rules constrain amplitudes

Let $Q$ generate a continuous symmetry, and suppose

$$Q|i\rangle=q_i|i\rangle, \qquad Q|f\rangle=q_f|f\rangle.$$

If a transition operator $T$ has a definite commutator with $Q$,

$$[Q,T]=q_T T,$$

then

$$\begin{aligned} (q_f-q_i)\langle f|T|i\rangle &= \langle f|[Q,T]|i\rangle \\ &= q_T\langle f|T|i\rangle. \end{aligned}$$

A nonzero matrix element therefore requires

$$q_f-q_i=q_T.$$

This is a symmetry constraint, not a dynamical prediction of the transition rate. Satisfying the condition is generally necessary, not sufficient: radial integrals, additional symmetries, kinematics, and accidental cancellations can still make the matrix element vanish.

If a perturbation mixes the old symmetry sectors, the old rule may become approximate. Small coupling is not enough by itself to make this mixing small: it must be compared with the relevant energy gaps, and near-degenerate sectors can mix strongly. See good quantum numbers for a two-level example.

Exercises

1. A particle has $H=P^2/(2m)+\tfrac12m\omega^2X^2$. Is spatial translation generated by $P$ a symmetry? Evaluate the relevant commutator.

Solution

Using $[X,P]=i\hbar$,

$$\begin{aligned} [P,X^2] &= [P,X]X +X[P,X] \\ &=-2i\hbar X. \end{aligned}$$

Hence

$$\begin{aligned} [P,H] &= \frac12m\omega^2[P,X^2] \\ &= -i\hbar m\omega^2X. \end{aligned}$$

For $\omega\ne0$ this is nonzero, so translations are not a symmetry and momentum is not conserved. The oscillator center selects a preferred position.

2. Show that the explicitly time-dependent operator $A(t)=X-tP/m$ is a constant of motion for a free particle with $H=P^2/(2m)$.

Solution

The explicit derivative is

$$\frac{\partial A}{\partial t} =-\frac{P}{m}.$$

Because $[P^2,X]=-2i\hbar P$,

$$\frac{i}{\hbar}[H,A] = \frac{i}{\hbar} \left[\frac{P^2}{2m},X\right] = \frac{P}{m}.$$

The two terms cancel:

$$\frac{\partial A}{\partial t} + \frac{i}{\hbar}[H,A] =0.$$

Thus $A_H$ is constant even though the Schrödinger-picture expression has explicit time dependence. This is why the commutator-only shortcut would be insufficient.

3. Suppose $Q|i\rangle=2\hbar|i\rangle$, $Q|f\rangle=5\hbar|f\rangle$, and $[Q,T]=3\hbar T$. Does the $Q$ selection rule force $\langle f|T|i\rangle$ to vanish?

Solution

The necessary condition for a nonzero matrix element is

$$q_f-q_i=q_T.$$

Here

$$q_f-q_i = 5\hbar-2\hbar =3\hbar =q_T.$$

The matrix element is symmetry allowed. The selection rule does not prove that it is nonzero; dynamics or another symmetry may still make it vanish.

References

• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics I: Functional Analysis, revised and enlarged ed., Academic Press, 1980.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• M. H. Stone, “On one-parameter unitary groups in Hilbert space,” Annals of Mathematics 33, 643–648 (1932).