Domain Subtleties of Canonical Commutation Relations

The differential identity $[x,-i\hbar\partial_x]=i\hbar$ can hold on a dense invariant test space while the associated self-adjoint operators fail the Weyl relations. Momentum on a finite interval provides an explicit example. Its endpoint conditions obstruct the domain invariance required by the tempting exponentiation argument. The lesson is to check the operator domains before treating a local differential calculation as a global canonical representation.

Required background. Self-Adjoint Operators supplies interval momentum domains and boundary forms. The Weyl Form of the CCR specifies the bounded identities being tested.

A self-adjoint pair with a local canonical identity

Work in $L^2(0,L)$ with $L>0$. Let $Q$ be multiplication by $x$, bounded and self-adjoint on the whole space. For fixed $\theta\in[0,2\pi)$, take

$$P_\theta=-i\hbar\frac{d}{dx}, \qquad D(P_\theta)=\{\psi\in H^1(0,L): \psi(L)=e^{i\theta}\psi(0)\}.$$

The self-adjoint interval theorem proves this using the endpoint boundary form. Both operators preserve $\mathcal D=C_c^\infty(0,L)$, and the ordinary product rule gives

$$(QP_\theta-P_\theta Q)\psi=i\hbar\psi \qquad(\psi\in\mathcal D).$$

Thus this example has a dense common invariant domain, not merely a collection of isolated test vectors. Nevertheless $\mathcal D$ is not an operator core for $P_\theta$: its closure in the momentum graph norm is $H_0^1(0,L)$, with both endpoint traces zero. That closure is a proper subspace of $D(P_\theta)$.

Multiplication changes the boundary condition

For $\psi\in D(P_\theta)$, the vector $Q\psi$ has endpoint values

$$(Q\psi)(0)=0, \qquad (Q\psi)(L)=L\psi(L).$$

It belongs to $D(P_\theta)$ only if $\psi(L)=0$, hence also $\psi(0)=0$. The natural product intersection is therefore

$$D(QP_\theta)\cap D(P_\theta Q)=H_0^1(0,L).$$

To check the second inclusion, first note that $D(QP_\theta)=D(P_\theta)$ because $Q$ is bounded. On that domain, $x\psi$ is in $H^1$; only its endpoint condition remains to be tested. The canonical product commutator holds on this intersection, but not as an expression on every momentum-domain vector.

In particular, the normalized momentum eigenfunctions

$$\psi_n(x)=L^{-1/2}e^{i(2\pi n+\theta)x/L}$$

belong to $D(P_\theta)$ and do not belong to $D(P_\theta Q)$. A formal commutator expectation in these eigenstates is therefore not a legitimate product calculation.

Why the global Weyl identity is impossible

Define $M(b)=e^{ibQ/\hbar}$ and $T(a)=e^{-iaP_\theta/\hbar}$ by spectral calculus. Both are strongly continuous unitary groups individually. But

$$M(b)D(P_\theta) =D(P_{\theta+bL/\hbar}),$$

with the phase label understood modulo $2\pi$. A generic modulation changes the endpoint condition. It preserves the chosen momentum domain only when $bL/\hbar$ is an integer multiple of $2\pi$.

There is also a representation-independent spectral obstruction. If the full Weyl identity held for all $a,b$, uniqueness of self-adjoint generators would imply

$$T(a)QT(a)^*=Q-aI.$$

Unitary conjugation preserves spectrum, but $\sigma(Q)=[0,L]$ whereas $\sigma(Q-aI)=[-a,L-a]$. These are unequal for $a\ne0$. Thus the global Weyl relation cannot hold for this pair. The interval unitary generated by $P_\theta$ translates with a boundary identification; it is not unrestricted translation of the line coordinate.

This counterexample does not contradict Stone–von Neumann uniqueness. Its hypotheses include the global Weyl relations, which have failed. Nor does self-adjointness of each operator separately prove that their commutator exponentiates.

The boundary term repairs the uncertainty calculation

For a normalized $\psi\in D(P_\theta)$, the two inner products $\langle Q\psi,P_\theta\psi\rangle$ and $\langle P_\theta\psi,Q\psi\rangle$ exist even when the product operators do not. Integration by parts gives the quadratic-form identity

$$\langle Q\psi,P_\theta\psi\rangle -\langle P_\theta\psi,Q\psi\rangle =i\hbar\bigl(1-L|\psi(L)|^2\bigr).$$

This has a state-dependent boundary contribution. For a normalized momentum eigenfunction $L|\psi_n(L)|^2=1$, so the right side is zero, consistent with zero momentum variance. Inserting $i\hbar$ by ignoring the endpoint would yield a false positive uncertainty lower bound.

The correct general uncertainty estimate uses the quadratic-form imaginary part on the domain where both first operator actions exist. Replacing that form by an operator commutator requires the additional product domains.

Exercise: identify the missing invariance

For periodic momentum ($\theta=0$), determine which modulation parameters preserve its domain. Explain why agreement for those parameters cannot establish a regular canonical pair on the full real parameter plane.

Solution

The multiplier has endpoint ratio $e^{ibL/\hbar}$, so the periodic condition is preserved exactly when $b=2\pi k\hbar/L$ for $k\in\mathbb Z$. This is a discrete set, not the full real modulation group. The standard Weyl relation requires every real $b$ and $a$ with the specified phase. A restricted discrete compatibility cannot supply the missing continuous canonical representation.

References

• B. C. Hall, Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013 — domains, canonical commutators, and the hypotheses of Stone–von Neumann uniqueness.
• G. Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014 — self-adjoint extensions, boundary forms, and abstract commutation relations.