# Extensions of symmetric operators

In functional analysis, one is interested in extensions of symmetric operators acting on a Hilbert space. Of particular importance is the existence, and sometimes explicit constructions, of self-adjoint extensions. This problem arises, for example, when one needs to specify domains of self-adjointness for formal expressions of observables in quantum mechanics. Other applications of solutions to this problem can be seen in various moment problems.

This article discusses a few related problems of this type. The unifying theme is that each problem has an operator-theoretic characterization which gives a corresponding parametrization of solutions. More specifically, finding self-adjoint extensions, with various requirements, of symmetric operators is equivalent to finding unitary extensions of suitable partial isometries.

## Symmetric operators

Let H be a Hilbert space. A linear operator A acting on H with dense domain Dom(A) is symmetric if

$\langle Ax,y\rangle =\langle x,Ay\rangle$  for all x, y in Dom(A).

If Dom(A) = H, the Hellinger-Toeplitz theorem says that A is a bounded operator, in which case A is self-adjoint and the extension problem is trivial. In general, a symmetric operator is self-adjoint if the domain of its adjoint, Dom(A*), lies in Dom(A).

When dealing with unbounded operators, it is often desirable to be able to assume that the operator in question is closed. In the present context, it is a convenient fact that every symmetric operator A is closable. That is, A has a smallest closed extension, called the closure of A. This can be shown by invoking the symmetric assumption and Riesz representation theorem. Since A and its closure have the same closed extensions, it can always be assumed that the symmetric operator of interest is closed.

In the sequel, a symmetric operator will be assumed to be densely defined and closed.

Problem Given a densely defined closed symmetric operator A, find its self-adjoint extensions.

This question can be translated to an operator-theoretic one. As a heuristic motivation, notice that the Cayley transform on the complex plane, defined by

$z\mapsto {\frac {z-i}{z+i}}$

maps the real line to the unit circle. This suggests one define, for a symmetric operator A,

$U_{A}=(A-i)(A+i)^{-1}\,$

on Ran(A + i), the range of A + i. The operator UA is in fact an isometry between closed subspaces that takes (A + i)x to (A - i)x for x in Dom(A). The map

$A\mapsto U_{A}$

is also called the Cayley transform of the symmetric operator A. Given UA, A can be recovered by

$A=-i(U+1)(U-1)^{-1},\,$

defined on Dom(A) = Ran(U - 1). Now if

${\tilde {U}}$

is an isometric extension of UA, the operator

${\tilde {A}}=-i({\tilde {U}}+1)({\tilde {U}}-1)^{-1}$

acting on

$Ran(-{\frac {i}{2}}({\tilde {U}}-1))=Ran({\tilde {U}}-1)$

is a symmetric extension of A.

Theorem The symmetric extensions of a closed symmetric operator A is in one-to-one correspondence with the isometric extensions of its Cayley transform UA.

Of more interest is the existence of self-adjoint extensions. The following is true.

Theorem A closed symmetric operator A is self-adjoint if and only if Ran (A ± i) = H, i.e. when its Cayley transform UA is a unitary operator on H.

Corollary The self-adjoint extensions of a closed symmetric operator A is in one-to-one correspondence with the unitary extensions of its Cayley transform UA.

Define the deficiency subspaces of A by

$K_{+}=Ran(A+i)^{\perp }$

and

$K_{-}=Ran(A-i)^{\perp }.$

In this language, the description of the self-adjoint extension problem given by the corollary can be restated as follows: a symmetric operator A has self-adjoint extensions if and only if its Cayley transform UA has unitary extensions to H, i.e. the deficiency subspaces K+ and K have the same dimension.

### An example

Consider the Hilbert space L2[0,1]. On the subspace of absolutely continuous function that vanish on the boundary, define the operator A by

$Af=i{\frac {d}{dx}}f.$

Integration by parts shows A is symmetric. Its adjoint A* is the same operator with Dom(A*) being the absolutely continuous functions with no boundary condition. We will see that extending A amounts to modifying the boundary conditions, thereby enlarging Dom(A) and reducing Dom(A*), until the two coincide.

Direct calculation shows that K+ and K are one-dimensional subspaces given by

$K_{+}=\operatorname {span} \{\phi _{+}=a\cdot e^{x}\}$

and

$K_{-}=\operatorname {span} \{\phi _{-}=a\cdot e^{-x}\}$

where a is a normalizing constant. So the self-adjoint extensions of A are parametrized by the unit circle in the complex plane, {|α| = 1}. For each unitary Uα : KK+, defined by Uα(φ) = αφ+, there corresponds an extension Aα with domain

$\operatorname {Dom} (A_{\alpha })=\{f+\beta (\alpha \phi _{-}-\phi _{+})|f\in \operatorname {Dom} (A),\;\beta \in \mathbb {C} \}.$

If f ∈ Dom(Aα), then f is absolutely continuous and

$\left|{\frac {f(0)}{f(1)}}\right|=\left|{\frac {e\alpha -1}{\alpha -e}}\right|=1.$

Conversely, if f is absolutely continuous and f(0) = γf(1) for some complex γ with |γ| = 1, then f lies in the above domain.

The self-adjoint operators { Aα } are instances of the momentum operator in quantum mechanics.

## Self-adjoint extension on a larger space

Every partial isometry can be extended, on a possibly larger space, to a unitary operator. Consequently, every symmetric operator has a self-adjoint extension, on a possibly larger space.

## Positive symmetric operators

A symmetric operator A is called positive if $\langle Ax,x\rangle \geq 0$  for all x in Dom(A). It is known that for every such A, one has dim(K+) = dim(K). Therefore, every positive symmetric operator has self-adjoint extensions. The more interesting question in this direction is whether A has positive self-adjoint extensions.

For two positive operators A and B, we put AB if

$(A+1)^{-1}\geq (B+1)^{-1}$

in the sense of bounded operators.

### Structure of 2 × 2 matrix contractions

While the extension problem for general symmetric operators is essentially that of extending partial isometries to unitaries, for positive symmetric operators the question becomes one of extending contractions: by "filling out" certain unknown entries of a 2 × 2 self-adjoint contraction, we obtain the positive self-adjoint extensions of a positive symmetric operator.

Before stating the relevant result, we first fix some terminology. For a contraction Γ, acting on H, we define its defect operators by

$D_{\Gamma }=(1-\Gamma ^{*}\Gamma )^{\frac {1}{2}}\quad {\mbox{and}}\quad D_{\Gamma ^{*}}=(1-\Gamma \Gamma ^{*})^{\frac {1}{2}}.$

The defect spaces of Γ are

${\mathcal {D}}_{\Gamma }=Ran(D_{\Gamma })\quad {\mbox{and}}\quad {\mathcal {D}}_{\Gamma ^{*}}=Ran(D_{\Gamma ^{*}}).$

The defect operators indicate the non-unitarity of Γ, while the defect spaces ensure uniqueness in some parameterizations. Using this machinery, one can explicitly describe the structure of general matrix contractions. We will only need the 2 × 2 case. Every 2 × 2 contraction Γ can be uniquely expressed as

$\Gamma ={\begin{bmatrix}\Gamma _{1}&D_{\Gamma _{1}^{*}}\Gamma _{2}\\\Gamma _{3}D_{\Gamma _{1}}&-\Gamma _{3}\Gamma _{1}^{*}\Gamma _{2}+D_{\Gamma _{3}^{*}}\Gamma _{4}D_{\Gamma _{2}}\end{bmatrix}}$

where each Γi is a contraction.

### Extensions of Positive symmetric operators

The Cayley transform for general symmetric operators can be adapted to this special case. For every non-negative number a,

$\left|{\frac {a-1}{a+1}}\right|\leq 1.$

This suggests we assign to every positive symmetric operator A a contraction

$C_{A}:Ran(A+1)\rightarrow Ran(A-1)\subset {\mathcal {H}}$

defined by

$C_{A}(A+1)x=(A-1)x.\quad {\mbox{i.e.}}\quad C_{A}=(A-1)(A+1)^{-1}.\,$

which have matrix representation

$C_{A}={\begin{bmatrix}\Gamma _{1}\\\Gamma _{3}D_{\Gamma _{1}}\end{bmatrix}}:Ran(A+1)\rightarrow {\begin{matrix}Ran(A+1)\\\oplus \\Ran(A+1)^{\perp }\end{matrix}}.$

It is easily verified that the Γ1 entry, CA projected onto Ran(A + 1) = Dom(CA), is self-adjoint. The operator A can be written as

$A=(1+C_{A})(1-C_{A})^{-1}\,$

with Dom(A) = Ran(CA - 1). If

${\tilde {C}}$

is a contraction that extends CA and its projection onto its domain is self-adjoint, then it is clear that its inverse Cayley transform

${\tilde {A}}=(1+{\tilde {C}})(1-{\tilde {C}})^{-1}$

defined on

$Ran(1-{\tilde {C}})$

is a positive symmetric extension of A. The symmetric property follows from its projection onto its own domain being self-adjoint and positivity follows from contractivity. The converse is also true: given a positive symmetric extension of A, its Cayley transform is a contraction satisfying the stated "partial" self-adjoint property.

Theorem The positive symmetric extensions of A are in one-to-one correspondence with the extensions of its Cayley transform where if C is such an extension, we require C projected onto Dom(C) be self-adjoint.

The unitarity criterion of the Cayley transform is replaced by self-adjointness for positive operators.

Theorem A symmetric positive operator A is self-adjoint if and only if its Cayley transform is a self-adjoint contraction defined on all of H, i.e. when Ran(A + 1) = H.

Therefore, finding self-adjoint extension for a positive symmetric operator becomes a "matrix completion problem". Specifically, we need to embed the column contraction CA into a 2 × 2 self-adjoint contraction. This can always be done and the structure of such contractions gives a parametrization of all possible extensions.

By the preceding subsection, all self-adjoint extensions of CA takes the form

${\tilde {C}}(\Gamma _{4})={\begin{bmatrix}\Gamma _{1}&D_{\Gamma _{1}}\Gamma _{3}^{*}\\\Gamma _{3}D_{\Gamma _{1}}&-\Gamma _{3}\Gamma _{1}\Gamma _{3}^{*}+D_{\Gamma _{3}^{*}}\Gamma _{4}D_{\Gamma _{3}^{*}}\end{bmatrix}}.$

So the self-adjoint positive extensions of A are in bijective correspondence with the self-adjoint contractions Γ4 on the defect space

${\mathcal {D}}_{\Gamma _{3}^{*}}$

of Γ3. The contractions

${\tilde {C}}(-1)\quad {\mbox{and}}\quad {\tilde {C}}(1)$

give rise to positive extensions

$A_{0}\quad {\mbox{and}}\quad A_{\infty }$

respectively. These are the smallest and largest positive extensions of A in the sense that

$A_{0}\leq B\leq A_{\infty }$

for any positive self-adjoint extension B of A. The operator A is the Friedrichs extension of A and A0 is the von Neumann-Krein extension of A.

Similar results can be obtained for accretive operators.