Stone’s Theorem

Stone’s theorem is the exact correspondence between continuous unitary evolution and self-adjoint generators. In the physical sign convention:

A family $\{U(t):t\in\mathbb R\}$ is a strongly continuous one-parameter unitary group if and only if there is a unique self-adjoint operator $H$ such that

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

for every real $t$.

The generator domain is not an auxiliary choice. It is recovered from the group itself:

$$D(H) = \left\{ \psi\in\mathcal H: \lim_{t\to0} \frac{U(t)\psi-\psi}{t} \text{ exists in }\mathcal H \right\},$$

and, on this domain,

$$H\psi = i\hbar \lim_{t\to0} \frac{U(t)\psi-\psi}{t}.$$

Strong continuity is essential. It is weak enough to allow unbounded Hamiltonians and strong enough to force a unique self-adjoint generator.

Required background. Self-Adjoint Operators, Strongly Continuous Unitary Groups, and the Unbounded Spectral Theorem supply all analytic prerequisites.

The two directions of the theorem

Stone’s theorem contains two logically distinct assertions.

Self-adjoint generator to unitary group

If $H$ is self-adjoint, the bounded Borel function

$$f_t(\lambda)=e^{-it\lambda/\hbar}$$

defines

$$U(t)=f_t(H) = \int_{\mathbb R}e^{-it\lambda/\hbar}\,dE_H(\lambda).$$

Because $|f_t|=1$, every $U(t)$ is unitary. Functional calculus gives $U(t+s)=U(t)U(s)$ and $U(0)=I$.

For any fixed $\psi$,

$$\|[U(t)-I]\psi\|^2 = \int_{\mathbb R} |e^{-it\lambda/\hbar}-1|^2 \,d\mu_\psi^H(\lambda).$$

The integrand tends pointwise to zero and is bounded by $4$. Since $\mu_\psi^H$ is finite, dominated convergence proves strong continuity. No finite energy moment is required for this continuity statement.

Unitary group to self-adjoint generator

Conversely, start with a strongly continuous unitary group. Define $H$ by the strong derivative formula on precisely those vectors for which the limit exists. The theorem proves that:

• $D(H)$ is dense;
• $H$ is self-adjoint, not merely symmetric;
• the original group equals $e^{-itH/\hbar}$;
• no other self-adjoint operator generates the same group.

These conclusions are not formal consequences of differentiating the group law. Density and self-adjointness are the core content of the theorem.

The generator domain and differentiable vectors

If $\psi\in D(H)$, then $U(t)\psi\in D(H)$ for every $t$, and

$$HU(t)\psi = U(t)H\psi.$$

Indeed, using the group law,

$$\frac{U(s)U(t)\psi-U(t)\psi}{s} = U(t) \frac{U(s)\psi-\psi}{s},$$

and the bounded operator $U(t)$ may be passed through the strong limit. Consequently the orbit is norm-differentiable at every time:

$$i\hbar\frac{d}{dt}U(t)\psi = HU(t)\psi = U(t)H\psi.$$

This is the strong Schrödinger equation for an autonomous Hamiltonian.

If $\psi\notin D(H)$, the vector $U(t)\psi$ still exists for every $t$, has constant norm, and depends continuously on $t$. It need not be differentiable as a Hilbert-space-valued function. Thus the unitary evolution law applies to more states than the strong differential equation.

The spectral domain formula makes the distinction quantitative:

$$\psi\in D(H) \quad\Longleftrightarrow\quad \int_{\mathbb R}\lambda^2 \,d\mu_\psi^H(\lambda)<\infty.$$

A normalized state can have a well-defined continuous evolution while its second spectral moment diverges.

Generator uniqueness

Uniqueness is immediate once the derivative-domain characterization has been established. The group determines the set of vectors for which the derivative exists, so it determines $D(H)$. On that domain it determines $H\psi$ by the strong limit. Two generators of the same group therefore have the same domain and action.

This is stronger than saying that two Hamiltonians differ only by a constant. If

$$H'=H+cI,$$

then

$$e^{-itH'/\hbar} = e^{-itc/\hbar}e^{-itH/\hbar},$$

which is the same ray evolution but not the same unitary group on vectors unless the phase is identically one. Stone’s theorem gives uniqueness for the chosen Hilbert-space lift, not merely for its projective action.

Proof architecture from a unitary group

A complete converse proof is substantial. A standard architecture is:

1. Smooth vectors. For $f\in C_c^\infty(\mathbb R)$ and $\psi\in\mathcal H$, define the Bochner integral

   $$\psi_f = \int_{\mathbb R}f(t)U(t)\psi\,dt.$$

   Strong continuity makes the integral well defined, and differentiating $f$ rather than $U$ shows that $\psi_f$ is a differentiable vector.

2. Density. Choose an approximate identity $f_n$. Then $\psi_{f_n}\to\psi$, proving that differentiable vectors are dense.

3. Infinitesimal operator. Define $H$ by the derivative. The group law and unitarity show that $H$ is symmetric and closed.

4. Maximality. Laplace transforms of the positive- and negative-time orbits construct the resolvents at nonreal spectral parameters. Their range properties prove that the symmetric generator is self-adjoint.

5. Recovery. Solve the strong evolution equation on a dense invariant domain and extend by continuity to show $U(t)=e^{-itH/\hbar}$.

The imported ingredients are Bochner integration, approximate identities, closed-operator theory, and the self-adjoint range criterion. This outline identifies where they enter; it is not a substitute for the full functional- analytic proof.

Translations as the canonical example

On $L^2(\mathbb R)$, let

$$(T(a)\psi)(x)=\psi(x-a).$$

The family is a strongly continuous unitary group. Its derivative on the Sobolev domain $H^1(\mathbb R)$ is

$$\left. \frac{d}{da}T(a)\psi \right|_{a=0} = -\psi'.$$

Therefore the Stone generator in the convention $T(a)=e^{-iaP/\hbar}$ is

$$P\psi=-i\hbar\psi', \qquad D(P)=H^1(\mathbb R).$$

Vectors outside $H^1$ are translated continuously but are not differentiable at $a=0$ in the $L^2$ norm. The example also shows why operator-norm continuity would be too strong: $\|T(a)-I\|=2$ for every $a\ne0$.

Norm continuity and bounded generators

Stone’s theorem has a useful refinement:

$$t\mapsto U(t) \text{ is operator-norm continuous} \quad\Longleftrightarrow\quad H\text{ is bounded}.$$

If $H$ is bounded, the exponential series converges in operator norm and

$$\|e^{-itH/\hbar}-I\| \leq e^{|t|\|H\|/\hbar}-1.$$

Conversely, a norm-continuous unitary group has a bounded operator-norm derivative after using the group law and a local logarithm; its generator is bounded and defined on all of $\mathcal H$.

Most wave-mechanical Hamiltonians are unbounded, so their groups are strongly but not norm continuous. That is the generic regime, not a pathology.

Projective and nonautonomous boundaries

Stone’s theorem acts on a genuine unitary group on Hilbert-space vectors. A continuous symmetry may initially be given only as transformations of rays. One must first choose a coherent unitary lift and analyze any projective multiplier. Stone cannot be applied directly to an abstract projective action.

A generic time-dependent Hamiltonian produces a two-parameter propagator

$$U(t,s), \qquad U(t,r)U(r,s)=U(t,s),$$

rather than a translation-invariant group $U(t-s)$. Stone’s theorem does not by itself prove existence, uniqueness, or domain stability for such nonautonomous dynamics.

Contractive semigroups for irreversible evolution also lie outside the theorem: they may exist only for $t\geq0$, need not be unitary, and have a different generator theory.

What the theorem does not establish

• It does not make every symmetric operator a Hamiltonian; the generator is self-adjoint.
• It does not make every state differentiable; the strong equation holds on $D(H)$.
• It does not apply directly to a time-dependent two-parameter propagator.
• It does not turn a projective ray action into a unitary representation.
• It does not say the group is operator-norm continuous unless the generator is bounded.
• It does not choose a self-adjoint extension of a formal differential expression; that operator must already be specified.

Common pitfalls

Dropping strong continuity. Algebraic group laws alone can admit discontinuous representations and do not force a self-adjoint generator.

Defining the generator on all vectors. The strong derivative domain is generally proper. Continuity of an orbit does not imply differentiability.

Losing the sign or $\hbar$. With $U(t)=e^{-itH/\hbar}$, the recovery formula is $H\psi=i\hbar\lim_{t\to0}[U(t)\psi-\psi]/t$.

Confusing vector and ray uniqueness. Adding a scalar to $H$ changes the unitary group by a time-dependent phase even though the ray motion is unchanged.

Applying Stone to $H(t)$. A time-dependent Hamiltonian normally produces a two-parameter propagator, not one autonomous group.

Exercises

1. Strong continuity from spectral calculus

Fill in the dominated-convergence proof that a self-adjoint $H$ makes $e^{-itH/\hbar}$ strongly continuous on every vector.

Solution

For fixed $\psi$,

$$\|[U(t)-I]\psi\|^2 = \int|e^{-it\lambda/\hbar}-1|^2 \,d\mu_\psi^H(\lambda).$$

The integrand converges pointwise to zero and is bounded by $4$. The measure has total mass $\|\psi\|^2<\infty$, so dominated convergence sends the integral to zero.

2. Invariance of the generator domain

Use the group law to prove $U(t)D(H)=D(H)$ and $HU(t)\psi=U(t)H\psi$.

Solution

For $\psi\in D(H)$,

$$\frac{U(s)U(t)\psi-U(t)\psi}{s} = U(t)\frac{U(s)\psi-\psi}{s}.$$

The right side has a norm limit because $U(t)$ is bounded. This places $U(t)\psi$ in $D(H)$ and gives commutation with $H$. Replacing $t$ by $-t$ proves equality rather than only inclusion of the domains.

3. Translation generator

For $\psi\in C_c^\infty(\mathbb R)$, compute the generator of $T(a)\psi(x)=\psi(x-a)$ and reconcile it with the Fourier multiplier $\hbar k$.

Solution

The derivative at zero is $-\psi'$, so

$$P\psi = i\hbar(-\psi') = -i\hbar\psi'.$$

Under the Fourier convention with kernel $e^{-ikx}$, $\widehat{\psi'}(k)=ik\widehat\psi(k)$, hence $\widehat{P\psi}(k)=\hbar k\widehat\psi(k)$.

4. Bounded generator estimate

Assume $H$ is bounded. Prove the displayed operator-norm continuity estimate from the exponential series.

Solution

Subtract the zeroth term and use submultiplicativity:

$$\begin{aligned} \|e^{-itH/\hbar}-I\| &\leq \sum_{n=1}^{\infty} \frac{|t|^n\|H\|^n}{\hbar^n n!}\\ &= e^{|t|\|H\|/\hbar}-1. \end{aligned}$$

5. Same ray motion, different unitary groups

Let $H'=H+cI$ with real $c$. Show that $H$ and $H'$ induce the same ray motion but different vector groups when $c\ne0$.

Solution

Because $I$ commutes with $H$,

$$e^{-itH'/\hbar} = e^{-itc/\hbar}e^{-itH/\hbar}.$$

The prefactor is one global phase, so both vectors determine the same ray at each time. The operators differ for generic $t$ when $c\ne0$, so the unitary groups and their Stone generators are distinct.

References

• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Volume I: Functional Analysis, revised and enlarged ed., Academic Press, 1980.
• M. H. Stone, “On one-parameter unitary groups in Hilbert space,” Annals of Mathematics 33, 643–648, 1932, doi:10.2307/1968538.
• G. Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., American Mathematical Society, 2014.