Commutation Relations on Domains

A commutator can mean an operator difference or a quadratic form, and those meanings have different domains. For self-adjoint $A,B$, the scalar difference $\langle A\psi,B\psi\rangle-\langle B\psi,A\psi\rangle$ requires only the first operator actions. Writing it as $\langle\psi,[A,B]\psi\rangle$ additionally requires both products. This distinction allows rigorous uncertainty estimates for states on which the product commutator is not an operator expression at all.

Required background. Self-Adjoint Operators supplies domain-sensitive equality. The Schrödinger Representation gives position and weak-derivative momentum on the line.

Product commutators and invariant test spaces

The operator commutator has natural domain

$$D([A,B])=D(AB)\cap D(BA), \qquad D(AB)=\{\psi\in D(B):B\psi\in D(A)\}.$$

Repeated algebraic operations become easier on a linear subspace $\mathcal D$ preserved by both $A$ and $B$. Every finite polynomial in the two operators then acts on $\mathcal D$, and algebraic identities such as

$$[A,BC]\psi=[A,B]C\psi+B[A,C]\psi$$

hold when all operators preserve the chosen domain. The equality is a statement there; it does not assert that the closures of the displayed products have equal maximal domains.

Schwartz space is the standard invariant domain for polynomial position–momentum calculations on the line. Hilbert-norm density alone, however, cannot justify passing unbounded actions through a limit. One needs the relevant graph-norm estimates or a theorem establishing the closure. Even an invariant dense domain can fail to be a core for a selected self-adjoint realization, as the interval example demonstrates.

A commutator form needs fewer operator actions

For $\psi\in D(A)\cap D(B)$ define

$$c_\psi(A,B) =\langle A\psi,B\psi\rangle-\langle B\psi,A\psi\rangle =2i\operatorname{Im}\langle A\psi,B\psi\rangle.$$

Both inner products exist without requiring $AB\psi$ or $BA\psi$. If the product domains also hold, self-adjointness moves the first operator across the inner product and recovers $c_\psi(A,B)=\langle\psi,(AB-BA)\psi\rangle$.

For a normalized vector, let $a=\langle\psi,A\psi\rangle$, $b=\langle\psi,B\psi\rangle$, and define standard deviations by the norms of $(A-a)\psi$ and $(B-b)\psi$. Cauchy–Schwarz gives

$$\Delta_\psi A\,\Delta_\psi B \geq\left|\operatorname{Im} \langle(A-a)\psi,(B-b)\psi\rangle\right| =\frac12|c_\psi(A,B)|.$$

This form of the inequality is valid on the joint first-operator domain. It does not demand $\psi\in D(A^2)$ either: the variance is the squared norm, or equivalently the spectral second moment, rather than necessarily a literal $A^2\psi$ inner product.

The canonical form extends beyond product domains

For position $Q$ and momentum $P$ on $L^2(\mathbb R)$,

$$c_\psi(Q,P)=i\hbar\|\psi\|^2 \qquad\text{for }\psi\in D(Q)\cap D(P).$$

First establish it for compactly supported smooth functions by integration by parts, where there are no endpoint terms. Then extend by approximation in the joint graph norm

$$\|\psi\|_{Q,P}^2 =\|\psi\|^2+\ell^{-2}\|Q\psi\|^2 +\ell^2\hbar^{-2}\|P\psi\|^2,$$

with an arbitrary fixed length $\ell>0$.

The required density has a concrete proof. Smooth cutoffs equal to one on increasingly large intervals remove the position and derivative tails; their derivatives are uniformly of order the inverse cutoff length. Mollifying the resulting compactly supported $H^1$ functions then approximates both the function and its weak derivative, while the position weight is bounded on the common compact support. Thus $C_c^\infty$ is dense in the joint norm. The form is continuous in that norm, so the identity passes to the limit.

The resulting bound $\Delta Q\Delta P\geq\hbar/2$ holds whenever both second moments are finite. It is not a claim that the operator product commutator exists for every such state. Nor does the same no-boundary formula automatically hold on an interval with different endpoint data.

A state with a form commutator but no product commutator

Set $y=x/\ell$ and consider

$$\psi(x)=\sqrt{\frac{2}{\pi\ell}} \frac{e^{iy^2/2}}{1+y^2}.$$

It is normalized because $\int_{\mathbb R}(1+y^2)^{-2}dy=\pi/2$. Its amplitude decays as $|x|^{-2}$, so $Q\psi\in L^2$. Differentiation of its quadratic phase gives a leading derivative of order $|x|^{-1}$, so $P\psi\in L^2$ as well.

But multiplying that derivative by $x$ produces a nondecaying leading term. Thus $QP\psi$ is not in $L^2$; the leading derivative of $x\psi$ similarly makes $PQ\psi$ unavailable. The canonical form and the uncertainty bound remain valid, while the notation $\langle\psi,(QP-PQ)\psi\rangle$ would conceal nonexistent Hilbert-space products.

This example also shows why formal cancellation is not an adequate domain argument. Two individually non-square-integrable differential expressions may cancel to a square-integrable expression, but that cancellation does not define the difference of their Hilbert-space operators on the vector.

Exercise: identify the safe statement

Suppose $\psi\in D(A)\cap D(B)$ but $B\psi\notin D(A)$. Which of $\langle A\psi,B\psi\rangle$, $AB\psi$, and $c_\psi(A,B)$ is defined?

Solution

The two first actions are Hilbert-space vectors, so the inner product and the commutator form are defined. The product $AB\psi$ is not. The uncertainty estimate using the form remains available, but rewriting it as an expectation of the operator commutator requires additional domain information.

References

• B. C. Hall, Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013, Chapter 12 — uncertainty estimates and their domain qualifications.
• G. Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014, Chapters 7–8 — Sobolev domains, canonical operators, and commutation relations.