Symmetric Operators

A densely defined operator is symmetric when moving it between the two slots of an inner product does not change the pairing on its domain. Symmetry ensures real quadratic expectations, orthogonality of eigenvectors with different eigenvalues, and useful estimates. It does not establish the domain equality required of a self-adjoint observable.

Helpful background. Adjoints of Unbounded Operators explains adjoint domains. The canonical comparison with observables and dynamics is Self-Adjoint Operators.

Symmetry and polarization

Take the inner product linear in its second argument. A densely defined operator $S$ is symmetric if

$$\langle\phi,S\psi\rangle=\langle S\phi,\psi\rangle \qquad(\phi,\psi\in D(S)),$$

or equivalently $S\subseteq S^*$. Its quadratic form $q(\psi)=\langle\psi,S\psi\rangle$ is real.

The converse is also true on a complex linear domain. To prove it, define

$$b(\phi,\psi) =\langle\phi,S\psi\rangle-\langle S\phi,\psi\rangle.$$

Reality of $q$ gives $b(\psi,\psi)=0$. Expanding this equality for $\phi+\psi$ gives $b(\phi,\psi)+b(\psi,\phi)=0$. Expanding for $\phi+i\psi$ gives $b(\phi,\psi)-b(\psi,\phi)=0$ after dividing by $i$. Hence $b=0$.

Thus real expectations characterize symmetry, not self-adjointness. The proof tests vectors already in $D(S)$ and never tests whether $D(S^*)$ contains additional vectors.

Every densely defined symmetric operator is closable: $D(S^*)$ contains the dense set $D(S)$. Its closure remains symmetric. This allows a formal symmetric seed to become a closed symmetric operator, after which an extension question still may remain.

Eigenvalues and orthogonality

If $S\psi=\lambda\psi$ for nonzero $\psi\in D(S)$, then

$$\lambda\|\psi\|^2=\langle\psi,S\psi\rangle\in\mathbb R,$$

so $\lambda$ is real. If $S\phi=\mu\phi$ with $\mu\neq\lambda$, symmetry yields $(\lambda-\mu)\langle\phi,\psi\rangle=0$, giving orthogonality.

These facts resemble the finite-dimensional spectral theorem but are much weaker. They do not show that eigenvectors span the Hilbert space, that a spectral measure exists, or even that the whole spectrum is real. A closed symmetric operator that is not self-adjoint can have nonreal residual spectrum.

The distinction is observable in the half-line momentum example below. Its real eigenvalue constraint is compatible with the failure of a self-adjoint momentum operator on that Hilbert space.

Nonreal shifts: an estimate without surjectivity

Let $z=a+ib$ with $b\neq0$. Symmetry cancels the cross terms and gives

$$\|(S-z)\psi\|^2 =\|(S-a)\psi\|^2+b^2\|\psi\|^2.$$

Therefore $S-z$ is injective and its inverse on its range is bounded by $|b|^{-1}$. If $S$ is closed, this range is also closed: a convergent image sequence is Cauchy, the estimate makes its preimages Cauchy, and closedness supplies the limit.

Surjectivity is a different issue. The orthogonal complement is

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

Nonzero vectors on the right are missing directions in the range. These deficiency directions are precisely what self-adjointness eliminates. The estimate alone cannot prove existence of a resolvent on the whole Hilbert space.

Half-line momentum as a nonexample

Work in units $\hbar=1$. Define

$$Sf=-if', \qquad D(S)=H_0^1(0,\infty).$$

Here the trace $f(0)$ vanishes. The operator is closed and symmetric, and its adjoint has the same weak differential expression on $H^1(0,\infty)$ without the endpoint condition.

The equation $(S^*-i)f=0$ has solution $f(x)=Ce^{-x}$, which lies in $L^2(0,\infty)$. The equation $(S^*+i)f=0$ has solution $Ce^x$, which does not. The deficiency dimensions are therefore $(1,0)$ in this sign convention.

Consequently this symmetric momentum has no self-adjoint extension on the same Hilbert space. The missing translation direction is consistent with the geometry: unrestricted translations cannot preserve a half-line in both time directions. This interpretation supplements, rather than replaces, the deficiency calculation.

The half-line Hamiltonian $-d^2/dx^2$ behaves differently and admits self-adjoint boundary conditions. Squaring a differential expression changes the domain problem; failure for momentum does not rule out a well-defined kinetic Hamiltonian.

Exercises

Let $S$ be a bounded symmetric operator with full domain. Explain why it is self-adjoint and why the full-domain hypothesis matters.

Solution

The bounded adjoint is defined on all of $\mathcal H$. Symmetry gives equality of its action with $S$ on all vectors, and the domains already agree. A restriction of this operator to a proper dense subspace remains symmetric and bounded on that subspace, but its adjoint still has full domain, so the restriction is not self-adjoint.

For the half-line momentum above, show directly that $\operatorname{Ran}(S+i)$ is not dense.

Solution

The vector $e^{-x}$ belongs to $\ker(S^*-i)$ and is nonzero. The range-kernel identity gives $\operatorname{Ran}(S+i)^\perp=\ker(S^*-i)$. Thus every vector in that range is orthogonal to $e^{-x}$, so the range cannot be dense. The nonreal norm estimate still holds; it guarantees injectivity, not density.

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 — momentum on the line, interval, and half-line.
• K. Schmüdgen, Unbounded Self-adjoint Operators on Hilbert Space, Graduate Texts in Mathematics 265, Springer, 2012, doi:10.1007/978-94-007-4753-1 — symmetric operators and range criteria.