Adjoints of Unbounded Operators

The adjoint of a densely defined operator is determined by every vector for which its matrix-element pairing is continuous in the Hilbert norm. A formal integration by parts identifies a differential expression; the adjoint domain requires a separate maximality argument. This page develops that computation, using the definition and observable criterion in Self-Adjoint Operators.

Helpful background. Dense Domains explains uniqueness of the adjoint action. Closed and Closable Operators gives the graph language used below.

The adjoint domain is a continuity condition

Let $A:D(A)\subseteq\mathcal H\to\mathcal H$ have dense domain, and take the inner product linear in its second slot. A vector $\phi$ belongs to $D(A^*)$ precisely when there is $C_\phi<\infty$ such that

$$|\langle\phi,A\psi\rangle|\leq C_\phi\|\psi\|, \qquad \psi\in D(A).$$

The Riesz theorem then supplies a unique $\eta$ satisfying $\langle\phi,A\psi\rangle=\langle\eta,\psi\rangle$; define $A^*\phi=\eta$.

The bound is in $\|\psi\|$, not in $\|\psi\|_A$. The graph-norm estimate is automatic by Cauchy–Schwarz and would incorrectly admit every $\phi$. The Hilbert-norm bound is what turns the pairing into an actual output vector.

The adjoint is always closed. Indeed, if $\phi_n\to\phi$ and $A^*\phi_n\to\eta$, then for every $\psi\in D(A)$,

$$\langle\phi,A\psi\rangle =\lim_n\langle A^*\phi_n,\psi\rangle =\langle\eta,\psi\rangle.$$

This proves membership in the adjoint domain and identifies the limit action.

Computing a differential adjoint

Consider $A_0=-i\,d/dx$ on $C_c^\infty(0,1)$. Testing the adjoint relation against compactly supported functions shows, in the distributional sense, that $\phi$ must have a weak derivative in $L^2$. Therefore $\phi\in H^1(0,1)$ and $A_0^*\phi=-i\phi'$.

Conversely, every $H^1$ function satisfies this relation because the test functions vanish near the endpoints. Thus

$$D(A_0^*)=H^1(0,1).$$

This argument establishes necessity and sufficiency. Integration by parts on smooth functions alone would establish only part of the claim.

Now begin instead with the maximal derivative $A_{\max}$ on $H^1(0,1)$. Its boundary pairing is

$$\langle\phi,A_{\max}\psi\rangle -\langle-i\phi',\psi\rangle =-i[\overline{\phi(x)}\psi(x)]_0^1.$$

Since $\psi(0)$ and $\psi(1)$ can be varied independently, the boundary term vanishes for every $\psi$ exactly when $\phi(0)=\phi(1)=0$. Hence

$$D(A_{\max}^*)=H_0^1(0,1).$$

The minimal and maximal realizations exchange under the adjoint. A periodic domain instead reproduces its own periodic condition. The boundary condition belongs to the operator, not to its formal differential expression.

Restrictions reverse under adjoints

If $A\subseteq B$, with both densely defined, then

$$B^*\subseteq A^*.$$

A vector satisfying the adjoint relation for every test vector in the larger domain $D(B)$ automatically satisfies it on $D(A)$. Fewer input tests impose fewer restrictions on the adjoint domain.

This explains why shrinking a self-adjoint operator’s domain usually enlarges its adjoint domain and destroys self-adjointness. It also prevents a common mistaken repair: adding arbitrary boundary conditions to both the operator and its supposed adjoint without recomputing the maximal adjoint domain.

For densely defined $A$, graph orthogonality gives

$$G(A)^\perp=\{(-A^*\phi,\phi):\phi\in D(A^*)\}.$$

Taking orthogonal complements again shows that $A$ is closable exactly when $D(A^*)$ is dense. When it is, the closed graph obtained is $G(A^{**})$, so $\overline A=A^{**}$. In particular, a closed densely defined operator obeys $A=A^{**}$.

Adjoint algebra has domain qualifications

For bounded operators, $(AB)^*=B^*A^*$. For unbounded operators the generally safe statement is

$$B^*A^*\subseteq(AB)^*$$

when $A$, $B$, and $AB$ are densely defined. The relation on the left is checked only where both successive adjoint actions exist; the adjoint on the right may have a larger domain.

Likewise $A^*+B^*\subseteq(A+B)^*$ when the sum is densely defined. Equality needs additional hypotheses. A useful sufficient case is a bounded everywhere-defined perturbation $B$, for which

$$(A+B)^*=A^*+B^*, \qquad D((A+B)^*)=D(A^*).$$

Subtracting the bounded pairing isolates exactly the defining continuity condition for $A^*$. These domain-aware rules are more reliable than moving symbols mechanically through a dagger.

Exercises

For $D(A_\theta)=\{f\in H^1(0,1):f(1)=e^{i\theta}f(0)\}$ and $A_\theta f=-if'$, recover the adjoint boundary condition.

Solution

Testing first against compactly supported functions gives $\phi\in H^1$ and the same differential action. The remaining boundary term is $-i(\overline{\phi(1)}e^{i\theta}-\overline{\phi(0)})f(0)$. It vanishes for every endpoint value precisely when $\phi(1)=e^{i\theta}\phi(0)$. Thus the adjoint has exactly the same domain.

Why does imposing $\phi(0)=\phi(1)=0$ when computing $A_0^*$ above give an incorrect answer?

Solution

Those conditions are sufficient but unnecessary: compactly supported input tests already eliminate all boundary terms. The adjoint domain must include every admissible vector, including constants. Imposing extra conditions constructs a restriction of the adjoint and cannot establish the adjoint’s maximal domain.

References

• G. Bonneau, J. Faraut, and G. Valent, “Self-adjoint extensions of operators and the teaching of quantum mechanics,” American Journal of Physics 69, 322–331, 2001, doi:10.1119/1.1328351, author manuscript — boundary pairings and derivative domains.
• K. Schmüdgen, Unbounded Self-adjoint Operators on Hilbert Space, Graduate Texts in Mathematics 265, Springer, 2012, doi:10.1007/978-94-007-4753-1, Chapter 1 — adjoints, graph orthogonality, and closure.