Tracking Domains

Tracking a domain means recording which vectors make every stage of an operator expression meaningful. For an unbounded operator, a formula for its action is only part of its definition. Products, cancellations, and limits can change the allowed inputs, so a formally correct calculation can establish equality on a test space without establishing equality of the full operators.

Required background. Operator Theory Minimum introduces operators with specified domains. The present page teaches how to follow those domains through a calculation.

Read an operator expression from the inside out

For a sum $A+B$, the natural domain is $D(A)\cap D(B)$. For a product,

$$D(AB)=\{\psi\in D(B):B\psi\in D(A)\}.$$

The second condition is not optional. Knowing that $\psi$ belongs to both initial domains does not show that the intermediate vector $B\psi$ belongs to $D(A)$.

When reading $ABC\psi$, first ask whether $C\psi$ exists, next whether it is in $D(B)$, and finally whether $BC\psi$ is in $D(A)$. This same procedure applies to differential expressions: a weak derivative may exist as a distribution while failing to be a Hilbert-space vector.

Record a claimed equality as either equality of operators, equality on a specified common domain, or equality of scalar forms. Equality of operators requires equal domains and equal actions there. A restriction $A|_{\mathcal D}$ can agree with $B|_{\mathcal D}$ without implying $A=B$.

A cancellation example that remembers its domain

Work in $\mathcal H=L^2(1,\infty)$, with dimensionless coordinate $x$. Define

$$(A\psi)(x)=x\psi(x), \qquad D(A)=\{\psi:x\psi\in L^2\},$$

and let $B$ be bounded multiplication by $1/x$. Because $x(B\psi)=\psi$ for every $\psi\in\mathcal H$, the product $AB$ has domain $\mathcal H$ and equals $I$.

For the reversed product, one must first apply $A$. Consequently,

$$AB=I\quad\hbox{on }\mathcal H, \qquad BA=I|_{D(A)}.$$

The second operator is a restriction of the identity, not the identity with its full domain. For a concrete witness, $\psi(x)=1/x$ lies in $L^2(1,\infty)$ but not in $D(A)$, since $x\psi=1$ is not square integrable. Then $AB\psi$ exists and equals $\psi$, while $BA\psi$ is undefined.

This does not invalidate the statement that $B=A^{-1}$: the inverse identities are understood on their proper spaces. It invalidates an unrestricted cancellation that silently enlarges a product’s domain.

The same issue occurs in $A+(-A)$. Its natural action is zero, but its natural domain remains $D(A)$. Its closure is the everywhere-defined zero operator because $D(A)$ is dense. Taking that closure is an additional operation, not a consequence of canceling symbols.

Keep boundary data attached to differential operators

For a differential operator, record the ambient space, regularity, and boundary conditions together. A derivative formula such as $-i\hbar\,d/dx$ on an interval can have several self-adjoint realizations, distinguished by endpoint conditions. Integration by parts identifies the boundary form that those domains must control.

After multiplying by a phase or another coordinate-dependent function, recheck the endpoint data. The transformed function may retain its Sobolev regularity but violate the original boundary condition. Domain Subtleties of the CCR gives the explicit interval calculation.

Taking an adjoint is another domain calculation. If $A$, $B$, and $AB$ are densely defined, one has the inclusion $B^*A^*\subseteq(AB)^*$, not an unconditional equality. The adjoint’s maximal domain must be determined from its defining bounded-functional condition; formal integration by parts supplies candidates, not the whole answer.

Identify the topology used to pass to a limit

Suppose $\psi_n\in D(A)$ and $\psi_n\to\psi$ in Hilbert norm. This alone does not imply $\psi\in D(A)$ or $A\psi_n\to A\psi$. If $A$ is closed, the valid inference requires both

$$\psi_n\longrightarrow\psi, \qquad A\psi_n\longrightarrow\eta$$

in norm; then $\psi\in D(A)$ and $A\psi=\eta$. A core supplies an approximation in this graph sense, which is stronger than ordinary density.

When a proof says “extend by density,” identify whether boundedness makes Hilbert-norm density sufficient or whether an unbounded action requires a core or form-core argument. Closed and Closable Operators and Dense Domains give the exact definitions.

Separate first actions from products and forms

The scalar $\langle A\psi,B\psi\rangle$ needs $\psi\in D(A)\cap D(B)$. Rewriting it as $\langle\psi,A^*B\psi\rangle$ requires $B\psi\in D(A^*)$. A scalar expression can therefore be meaningful even when the operator product suggested by moving a symbol across the bracket is not.

This is particularly useful for energy and uncertainty calculations. A form domain can admit trial states outside the operator domain, and a commutator form can exist outside both product domains. Use Commutation Relations on Domains for the canonical example rather than inferring domains from the appearance of a bracket.

Exercise: extend a restricted identity

In the multiplication example, is $BA$ closed? What is its closure, and what does this say about the distinction between an operator and its closure?

Solution

$BA$ is the identity restricted to the proper dense domain $D(A)$, so it is not closed. Truncating $\psi(x)=1/x$ to $1<x<n$ gives domain vectors converging to $\psi$, while their images under $BA$ converge to $\psi$ as well. The limiting graph point is missing from $BA$. Its graph closure is the graph of $I$ on all of $\mathcal H$. Thus agreeing after closure is a weaker claim than having the same domain initially.

References

• B. C. Hall, Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013 — domains, adjoints, and unbounded self-adjoint operators.
• G. Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014, Chapter 2 — closed operators, adjoints, and domain-sensitive identities.