Essential Self-Adjointness

A symmetric operator is essentially self-adjoint when its graph closure is self-adjoint. Its initial test domain then determines a unique self-adjoint realization without additional extension parameters. The initial operator need not itself be closed, and calling it self-adjoint can conceal exactly the limiting argument that needs proof.

Helpful background. Closed and Closable Operators constructs the closure. Self-Adjoint Operators supplies the self-adjointness criterion; this page concentrates on proving uniqueness from a smaller domain.

Equivalent tests for a symmetric seed

Let $S$ be densely defined and symmetric. Fix a positive reference scale $\eta$ with the same units as $S$. The following conditions are equivalent:

• $\overline S$ is self-adjoint.
• $\ker(S^*-i\eta)=\ker(S^*+i\eta)=\{0\}$.
• Both $\operatorname{Ran}(S+i\eta)$ and $\operatorname{Ran}(S-i\eta)$ are dense.

The range and kernel conditions are equivalent because

$$\operatorname{Ran}(S-z)^\perp=\ker(S^*-\overline z).$$

To understand why they establish self-adjointness of the closure, first use symmetry to obtain $\|(S\pm i\eta)\psi\|\geq\eta\|\psi\|$. The same estimate holds for the closed operator $\overline S$, whose shifted ranges are closed. Density therefore upgrades those ranges to the whole Hilbert space. The two-sided range criterion then makes $\overline S$ self-adjoint.

Conversely, a self-adjoint operator has bounded resolvents at both nonreal shifts, so neither adjoint kernel can contain a nonzero vector. The argument requires both half-planes. Vanishing of only one deficiency space does not imply essential self-adjointness.

Why the self-adjoint realization is unique

Suppose $S$ is essentially self-adjoint and $T$ is any self-adjoint extension. Then

$$\overline S\subseteq T.$$

Taking adjoints reverses the inclusion:

$$T=T^*\subseteq(\overline S)^*=\overline S.$$

Hence $T=\overline S$. This proves uniqueness without needing an explicit classification of every possible boundary condition.

Conversely, the standard extension theorem says that a symmetric operator admitting exactly one self-adjoint extension is essentially self-adjoint. If its equal deficiency dimensions were nonzero, different unitary maps between the deficiency spaces would produce different extensions.

Unique realization is a statement about a specified Hilbert space, differential expression, and initial domain. Removing a point from the initial test domain can change the answer even when the same formal expression is retained.

Momentum on the line: a complete deficiency calculation

Set $\hbar=1$ and start with $S=-i\,d/dx$ on $C_c^\infty(\mathbb R)$. Its adjoint acts weakly as $-i\,d/dx$ on $H^1(\mathbb R)$.

The two equations are

$$(S^*-i)f=0\ \Longrightarrow\ f(x)=Ce^{-x}, \qquad (S^*+i)f=0\ \Longrightarrow\ f(x)=Ce^x.$$

Neither function belongs to $L^2(\mathbb R)$ unless $C=0$, since each grows at one end. Both deficiency spaces vanish. Thus the compactly supported seed is essentially self-adjoint, and its closure is momentum on $H^1(\mathbb R)$.

On a finite interval both exponential solutions are square-integrable, so the same differential expression has deficiency dimensions $(1,1)$ and needs an extension choice. On a half-line only one solution survives, giving unequal dimensions and no self-adjoint extension. Integrability at every endpoint is therefore part of the proof, not an optional afterthought.

Proving a core can be easier than solving deficiencies

Another strategy is to identify a self-adjoint candidate $A$ and show that the seed domain is a core for it. Then $\overline S=A$ by definition.

For real multiplication by $x$, truncation controls the weighted norm

$$\|\psi\|_Q^2=\int_{\mathbb R}(1+x^2)|\psi(x)|^2\,dx.$$

After truncation to a bounded interval, smooth approximation gives convergence in this weighted norm as well. Thus $C_c^\infty(\mathbb R)$ is a core for maximal position multiplication. The proof separates tail control from local smoothing.

For Schrödinger Hamiltonians, potential terms can obstruct such an argument. One must state hypotheses on the potential and show the needed graph estimates, or invoke a precise theorem such as Kato–Rellich. Merely recognizing the formal expression $-\Delta+V$ does not establish essential self-adjointness for an arbitrary singular $V$.

Exercises

Let $A e_n=a_n e_n$ on $\ell^2$, where all $a_n$ are real, and initially define $S$ only on finite sequences. Prove essential self-adjointness.

Solution

The maximal diagonal operator has domain $D(A)=\{c:\sum |a_n|^2|c_n|^2<\infty\}$ and is self-adjoint. Truncating $c$ to its first $N$ coordinates makes both the ordinary and weighted tails tend to zero. Finite sequences are therefore a core, so $\overline S=A$.

A proof finds $\ker(S^*-i)=\{0\}$ and concludes essential self-adjointness. Supply the missing condition and a counterexample to the incomplete argument.

Solution

It must also show $\ker(S^*+i)=\{0\}$. Take the negative of the half-line momentum on $H_0^1(0,\infty)$; its deficiency dimensions are $(0,1)$. The first kernel vanishes but the second does not, and no self-adjoint extension exists.

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 — the three momentum 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 — essential self-adjointness, cores, and deficiency criteria.
• G. Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., American Mathematical Society, 2014, author’s book page — self-adjointness and Schrödinger operators.