Von Neumann's theorem on self-adjoint extensions

Remarks

Every self-adjoint extension $B$ of $A$ satisfies $A \subseteq B = B^* \subseteq A^*$, so it is a restriction of $A^*$ to a subspace between $D(A)$ and $D(A^*)$. On $D(A^*)$, the form $\ip{A^* x}{y} - \ip{x}{A^* y}$ measures the failure of symmetry, and it vanishes on $D(A)$. Part 1 and a computation of this form on $K_+ \dotplus K_-$ show that the maximal subspaces on which it vanishes are exactly the graphs of the isometric isomorphisms $U \colon K_+ \to K_-$. In particular, $A$ is self-adjoint if and only if $n_+ = n_- = 0$, and a symmetric operator is essentially self-adjoint if and only if its closure has $n_+ = n_- = 0$. Example: for $A = -i \, d/dx$, closed from the smooth functions with compact support in $(0,1)$ in $L^2([0,1])$, the indices are $(1,1)$. The isometric isomorphisms from $K_+$ to $K_-$ form a circle, and the self-adjoint extensions are the operators $-i \, d/dx$ with boundary conditions $u(1) = e^{i \theta} u(0)$, $\theta \in [0, 2\pi)$. On $L^2([0, \infty))$, the same operator has indices $(1, 0)$, so it has no self-adjoint extension. A symmetric operator that commutes with a conjugation, such as a real differential operator, has $n_+ = n_-$ and so always has self-adjoint extensions.

Going deeper

For a student

In quantum mechanics, an observable must be self-adjoint, while a formula such as $-d^2/dx^2$ only gives a symmetric operator on nice functions. The theorem turns the choice of a self-adjoint realization into linear algebra on the deficiency subspaces, and the choices correspond to boundary conditions. For Sturm–Liouville operators, the Weyl limit point and limit circle alternative computes the deficiency indices from the behaviour of solutions at the endpoints. Other constructions, such as the Friedrichs extension, pick out a distinguished self-adjoint extension of a semibounded symmetric operator.