One-Parameter Unitary Groups

A one-parameter unitary group is a continuous family of unitary operators $U(\alpha)$ labeled by one real parameter and satisfying the same composition law as the parameter addition:

$$U(\alpha+\beta) = U(\alpha)U(\beta), \qquad U(0)=I.$$

It is the mathematical object behind statements such as “momentum generates translations,” “angular momentum generates rotations,” and “the Hamiltonian generates time evolution.” The next page, Generators, focuses on the observable $G$ that appears infinitesimally. This page focuses on the unitary family itself.

For strong versus norm continuity and the domains of differentiable vectors, see Strongly Continuous Unitary Groups.

Parameter addition and the group law

A discrete symmetry, such as parity, can constrain spectra and matrix elements, but it usually cannot be differentiated. A continuous one-parameter symmetry can be studied near the identity. That local information is what produces a generator.

The parameter may be time, a displacement, an angle, or an internal phase. The essential feature is that two transformations compose by adding the parameters. If translating by $a$ and then by $b$ is the same as translating by $a+b$, the corresponding unitaries should obey

$$T(a+b) = T(a)T(b).$$

This is more than a convenient notation. It is the group law expressed on Hilbert space.

Definition

A one-parameter unitary group on a Hilbert space $\mathcal H$ is a map

$$\alpha\in\mathbb R \longmapsto U(\alpha)$$

such that:

• $U(\alpha)$ is unitary for every real $\alpha$;
• $U(0)=I$;
• $U(\alpha+\beta)=U(\alpha)U(\beta)$ for all real $\alpha,\beta$;
• $U(\alpha)$ depends continuously on $\alpha$ in the appropriate operator-theoretic sense.

The first three conditions are algebraic. The last condition is analytic: it rules out pathological group representations and allows an infinitesimal generator to be defined.

For finite-dimensional Hilbert spaces, ordinary matrix continuity is enough. For infinite-dimensional Hilbert spaces, the standard hypothesis is strong continuity:

$$\lim_{\alpha\to\alpha_0} \left\| U(\alpha)\psi-U(\alpha_0)\psi \right\| =0$$

for every $\psi\in\mathcal H$.

Exponential Form

With the physics sign convention, a strongly continuous one-parameter unitary group has the exact representation

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

where $G$ is self-adjoint and $\alpha G/\hbar$ is dimensionless. In finite dimensions this is the ordinary matrix exponential. In infinite dimensions it is defined by the spectral theorem for the self-adjoint operator $G$.

For a vector $\psi$ in the domain $D(G)$, the derivative exists in Hilbert-space norm:

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

The remainder is $O(\alpha^2)$ on $D(G^2)$. An operator-norm Taylor expansion is justified for bounded $G$, but generally fails for an unbounded generator.

Thus the same object has two faces:

• the finite transformation $U(\alpha)$;
• the infinitesimal generator $G$.

The finite family is often easier to identify physically, while the generator is often easier to use in commutator calculations.

Stone Theorem

The rigorous result behind the exponential form is Stone theorem. In units with no explicit $\hbar$, it says that a strongly continuous one-parameter unitary group has a unique self-adjoint generator $A$ such that

$$U(t)=e^{-itA}.$$

In the common quantum-mechanical convention,

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

The distinction between symmetric and self-adjoint operators matters here. A formal differential operator that is merely symmetric may fail to generate a unitary group until a self-adjoint domain or extension has been specified.

Standard Examples

The same pattern appears in several basic quantum transformations.

Transformation	Parameter	Unitary family	Generator
Time translation	$t$	$U(t)=e^{-itH/\hbar}$	Hamiltonian $H$
Space translation in one direction	$a$	$T(a)=e^{-iaP/\hbar}$	momentum $P$
Rotation about $\hat{\mathbf n}$	$\theta$	$R(\theta)=e^{-i\theta\hat{\mathbf n}\cdot\mathbf J/\hbar}$	$\hat{\mathbf n}\cdot\mathbf J$
Internal phase rotation	$\theta$	$U(\theta)=e^{-i\theta N}$	number operator $N$

The last example uses a dimensionless generator $N$, so no explicit $\hbar$ appears. If one uses a charge $Q$ with physical dimensions instead, the exponential must be normalized by the appropriate unit so that the exponent is dimensionless.

Time Evolution as a Special Case

For a time-independent closed-system Hamiltonian,

$$U(t+s) = U(t)U(s), \qquad U(t)=e^{-itH/\hbar}.$$

This is a one-parameter unitary group. The Hamiltonian is the generator of time translations, as discussed in Hamiltonians as Generators.

For a time-dependent Hamiltonian $H(t)$, the evolution operator is usually a two-parameter propagator $U(t,t_0)$, not a one-parameter group:

$$U(t_2,t_0) = U(t_2,t_1)U(t_1,t_0).$$

This composition law is useful, but it is not the same as $U(t+s)=U(t)U(s)$ unless the dynamics is time-translation invariant in the relevant sense.

Translations and Rotations

Spatial translations on the line form the additive group $(\mathbb R,+)$. In the active convention used here,

$$(T(a)\psi)(x)=\psi(x-a), \qquad T(a)=e^{-iaP/\hbar}.$$

The group law $T(a+b)=T(a)T(b)$ reflects ordinary addition of displacements. Its generator is the momentum operator, treated in Momentum Operator as Generator.

Rotations about a fixed axis also give a one-parameter family:

$$R_{\hat{\mathbf n}}(\theta) = \exp\left( -\frac{i}{\hbar}\theta\, \hat{\mathbf n}\cdot\mathbf J \right).$$

For a fixed axis, angle addition gives the one-parameter subgroup law

$$R_{\hat{\mathbf n}}(\theta+\phi) = R_{\hat{\mathbf n}}(\theta)R_{\hat{\mathbf n}}(\phi).$$

Rotations about different axes do not generally commute. The full rotation group is not one-dimensional, but every fixed-axis rotation is a one-parameter subgroup of it.

Periodic Parameters

Not every one-parameter family is globally the real line in disguise. Angle variables are periodic. For an ordinary representation of a $2\pi$-periodic phase or rotation family, one expects

$$U(\theta+2\pi) = U(\theta).$$

If

$$U(\theta) = e^{-i\theta N},$$

then $U(2\pi)=I$ requires eigenvalues of $N$ to be integers in an ordinary single-valued representation. This is the elementary spectral content behind many quantization conditions for compact one-parameter groups.

Spin introduces an important caveat. Spinor states are naturally represented by $SU(2)$, the double cover of $SO(3)$. A $2\pi$ spatial rotation may act as $-I$ on a spinor representative even though physical rays are unchanged. For that reason, global questions about periodicity often require the language of projective representations and covering groups.

Symmetry of a Hamiltonian

A one-parameter unitary family is a symmetry of a Hamiltonian $H$ when

$$U(\alpha)HU^\dagger(\alpha) = H$$

for every $\alpha$. This is stronger than saying that $U(\alpha)$ is unitary. A unitary family can represent a possible transformation without being a symmetry of a particular Hamiltonian.

If

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

then differentiating the invariance condition at $\alpha=0$ gives

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

When $G$ has no explicit time dependence, this commutator condition is the route from continuous symmetry to conservation law. The systematic conservation-law discussion lives in Commutators and Conservation Laws.

Infinitesimal View

The infinitesimal transformation at the identity recovers the generator:

$$G = i\hbar \left. \frac{dU}{d\alpha} \right|_{\alpha=0}$$

when this derivative is understood on the appropriate domain. Acting on a state,

$$\delta\lvert\psi\rangle = -\frac{i}{\hbar}\alpha G\lvert\psi\rangle.$$

For an active transformation of an operator along with the system,

$$A\mapsto U(\alpha)AU^\dagger(\alpha),$$

and to first order,

$$\delta A = -\frac{i}{\hbar}\alpha[G,A].$$

The passive or Heisenberg-style convention uses $A\mapsto U^\dagger A U$ and reverses this sign. See Active and Passive Transformations for examples of these two conventions.

A curve need not be a group

The continuous unitary curve $V(\alpha)=e^{-i\alpha^2 G/\hbar}$ starts at $I$, but generally satisfies neither $V(\alpha+\beta)=V(\alpha)V(\beta)$ nor $V(-\alpha)=V(\alpha)^{-1}$. Its derivative at zero vanishes even when the curve is nonconstant. Continuity alone therefore does not justify recovering a whole family by exponentiating its derivative at zero.

For unbounded generators, also keep the domain fixed when testing a Hamiltonian symmetry. Commuting spectral projections of $G$ and $H$ guarantee the full group symmetry; a formal commutator on a small common domain does not.

Exercises

1. Let $U(\alpha)=e^{-i\alpha G/\hbar}$ with $G$ self-adjoint. Show that $U(\alpha+\beta)=U(\alpha)U(\beta)$.

Solution

Because $G$ commutes with itself,

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

Thus

$$U(\alpha+\beta) = U(\alpha)U(\beta).$$

The same argument is immediate for matrices and follows from the functional calculus for a self-adjoint operator in the infinite-dimensional case.

2. Suppose $U(\theta)=e^{-i\theta N}$ and $U(2\pi)=I$. What does this imply about eigenvalues of $N$?

Solution

If $N\lvert n\rangle=n\lvert n\rangle$, then

$$U(2\pi)\lvert n\rangle = e^{-i2\pi n}\lvert n\rangle.$$

The condition $U(2\pi)=I$ requires

$$e^{-i2\pi n}=1$$

for every eigenvalue in the representation. Hence every eigenvalue is an integer. More generally, the spectral theorem places all spectral support on the integers; this is not weakened by a domain qualification.

3. A time-dependent Hamiltonian produces a propagator $U(t,t_0)$. Why is this generally not a one-parameter group?

Solution

The propagator satisfies the composition rule

$$U(t_2,t_0) = U(t_2,t_1)U(t_1,t_0),$$

which depends on two endpoint times. A one-parameter group would instead have a single parameter and obey

$$U(t+s) = U(t)U(s).$$

Generic time-dependent dynamics depends on the absolute times at which the Hamiltonian is sampled, not only on the elapsed time. Therefore it does not usually define a one-parameter group.

4. In finite dimensions, let $U(\alpha)$ be a symmetry of a time-independent Hamiltonian $H$ and let $U(\alpha)=e^{-i\alpha G/\hbar}$. Differentiate the invariance condition to obtain the commutator relation.

Solution

The invariance condition is

$$U(\alpha)HU^\dagger(\alpha)=H.$$

Using

$$U(\alpha) = I-\frac{i\alpha G}{\hbar}+O(\alpha^2), \qquad U^\dagger(\alpha) = I+\frac{i\alpha G}{\hbar}+O(\alpha^2),$$

one finds

$$U(\alpha)HU^\dagger(\alpha) = H-\frac{i\alpha}{\hbar}GH +\frac{i\alpha}{\hbar}HG +O(\alpha^2).$$

The first-order term must vanish, so

$$GH-HG=0.$$

Thus $[G,H]=0$.

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).