Regularity Assumptions

Regularity is the continuity hypothesis that lets bounded canonical displacements possess self-adjoint infinitesimal generators. A Weyl representation is regular when every one-parameter subgroup $t\mapsto W(tz)$ is strongly continuous. The algebraic Weyl relations alone do not imply this property. An explicit nonregular representation below satisfies the relations and is irreducible, yet has no momentum generator for its translation subgroup.

Required background. The Weyl Form of the CCR fixes the phase convention. Stone’s Theorem states the exact link between strong continuity and self-adjoint generators.

Strong continuity concerns each fixed vector

For a one-parameter unitary group $U(t)$, strong continuity at the identity means

$$\|(U(t)-I)\psi\|\longrightarrow0 \qquad(t\to0)$$

for every fixed $\psi\in\mathcal H$. The group law transfers this to continuity at every parameter value. It is not a uniform estimate over the unit sphere.

For unitary operators there is a convenient matrix-element test:

$$\|(U(t)-I)\psi\|^2 =2\|\psi\|^2-2\operatorname{Re}\langle\psi,U(t)\psi\rangle.$$

Thus continuity of these diagonal matrix elements at zero implies strong continuity. It suffices to verify the norm limit on a dense subset: if $\phi$ approximates $\psi$, then

$$\|(U(t)-I)\psi\| \leq2\|\psi-\phi\|+\|(U(t)-I)\phi\|.$$

The fixed unitary norm bound controls the approximation error. These tests are often easier than differentiating a wavefunction and avoid assuming a generator before proving it exists.

For a finite-dimensional phase space, continuity along a symplectic coordinate basis gives joint strong continuity of $W(z)$ by a finite product decomposition and the continuous Weyl phase. For an infinite-dimensional phase space, continuity along each line is still called regularity, but continuity in an additional phase-space norm is a separate condition. One must specify that topology before asserting more.

Norm continuity would be too strong

In the Schrödinger representation, $M(b)$ multiplies by $e^{ibx/\hbar}$. For every nonzero $b$,

$$\|M(b)-I\| =\operatorname*{ess\,sup}_{x\in\mathbb R} |e^{ibx/\hbar}-1|=2.$$

It is therefore not norm continuous at zero. Nevertheless it is strongly continuous, by dominated convergence for each fixed square-integrable wavefunction. Fourier transformation gives the same operator-norm discontinuity for nonzero translations.

The distinction is physically useful: a small displacement acts mildly on each fixed regular wavepacket, while a state chosen to have an arbitrarily large conjugate scale can still detect it strongly. Requiring operator-norm continuity would exclude the standard unbounded canonical generators.

A nonregular representation with exact Weyl relations

Let $\mathcal H=\ell^2(\mathbb R)$ with counting measure, whose orthonormal vectors are labeled $e_x$ for every real $x$. This is not $L^2(\mathbb R,dx)$; it is a nonseparable Hilbert space. Define

$$T(a)e_x=e_{x+a}, \qquad M(b)e_x=e^{ibx/\hbar}e_x.$$

Both maps extend to unitaries, satisfy their additive group laws, and obey

$$T(a)M(b)=e^{-iab/\hbar}M(b)T(a).$$

Checking the phase on $e_x$ proves the bounded-operator identity by linearity and density of finite-support vectors. But for $a\ne0$,

$$\|T(a)e_x-e_x\|=\sqrt2.$$

The translation subgroup is not strongly continuous. Hence no self-adjoint $P$ can satisfy $T(a)=e^{-iaP/\hbar}$ for all $a$.

The modulation subgroup is strongly continuous. Every vector has countable support, and dominated convergence applies to the sum of its squared coordinates. Its self-adjoint generator is the diagonal position operator

$$Qe_x=xe_x, \qquad D(Q)=\left\{c:\sum_x x^2|c_x|^2<\infty\right\}.$$

This model therefore has sharp position eigenvectors and the algebraic translation unitaries, but lacks a canonical momentum observable generating those translations. One cannot transfer all Schrödinger-model predictions to it merely because the exponentiated algebra looks familiar.

Regularity does not follow from irreducibility

The same nonregular example is irreducible. If bounded $B$ commutes with every modulation, then

$$(e^{ibx/\hbar}-e^{iby/\hbar})\langle e_y,Be_x\rangle=0$$

for all $b$. For $x\ne y$ one can choose a phase distinguishing the labels, so all off-diagonal matrix elements vanish. Thus $B$ is diagonal. Commutation with every translation then forces its diagonal values to be constant, giving $B=cI$.

Irreducibility removes reducing subspaces; it does not restore continuity. Conversely, a direct sum of Schrödinger copies is regular and reducible. These two independent tests are both required by the finite-degree uniqueness theorem.

Regularity also does not make every vector differentiable. Once a subgroup is strongly continuous, Stone’s theorem defines its generator domain through the strong derivative. That domain can still be a proper dense subspace. Continuity of the full orbit and existence of an infinitesimal operator action remain distinct assertions.

Exercise: test continuity before differentiating

Compute $\langle e_0,T(a)e_0\rangle$ in the counting-measure representation. Use the matrix-element criterion to diagnose regularity without proposing a formal derivative.

Solution

The value is one at $a=0$ and zero at every $a\ne0$. It is discontinuous at the identity. The norm identity consequently gives squared displacement two for every nonzero $a$. A derivative calculation is unnecessary: the missing continuity already excludes a self-adjoint generator for this subgroup.

References

• J. Dereziński, “Introduction to Representations of the Canonical Commutation and Anticommutation Relations”, arXiv:math-ph/0511030, 2005 — regular Weyl representations and their generators.
• B. C. Hall, Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013 — strong continuity and the finite-degree canonical uniqueness theorem.