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Infunctional analysis, one is interested inextensions of symmetric operators acting on aHilbert space. Of particular importance is the existence, and sometimes explicit constructions, ofself-adjoint extensions. This problem arises, for example, when one needs to specify domains of self-adjointness for formal expressions ofobservables inquantum mechanics. Other applications of solutions to this problem can be seen in variousmoment 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, ofsymmetric operators is equivalent to finding unitary extensions of suitablepartial isometries.

Let${\displaystyle H}$ be a Hilbert space. Alinear operator${\displaystyle A}$ acting on${\displaystyle H}$ withdensedomain${\displaystyle \mathrm{dom}(A)}$ issymmetric if

${\displaystyle ⟨Ax,y⟩=⟨x,Ay⟩,\mathrm{\forall }x,y\in \mathrm{dom}(A).}$

If${\displaystyle \mathrm{dom}(A)=H}$, theHellinger-Toeplitz theorem says that${\displaystyle A}$ is abounded operator, in which case${\displaystyle A}$ isself-adjoint and the extension problem is trivial. In general, a symmetric operator is self-adjoint if the domain of its adjoint,${\displaystyle \mathrm{dom}(A^{\ast })}$, lies in${\displaystyle \mathrm{dom}(A)}$.

When dealing withunbounded operators, it is often desirable to be able to assume that the operator in question isclosed. In the present context, it is a convenient fact that every densely defined, symmetric operator${\displaystyle A}$ isclosable. That is,${\displaystyle A}$ has the smallest closed extension, called theclosure of${\displaystyle A}$. This canbe shown by invoking the symmetric assumption andRiesz representation theorem. Since${\displaystyle A}$ and its closure have the same closed extensions, it can always be assumed that the symmetric operator of interest is closed.

In the next section, a symmetric operator will be assumed to bedensely defined and closed.

If an operator${\displaystyle A}$ on the Hilbert space${\displaystyle H}$ is symmetric, when does it have self-adjoint extensions? An operator that has a unique self-adjoint extension is said to beessentially self-adjoint; equivalently, an operator is essentially self-adjoint if its closure (the operator whose graph is the closure of thegraph of${\displaystyle A}$) is self-adjoint. In general, a symmetric operator could have many self-adjoint extensions or none at all. Thus, we would like a classification of its self-adjoint extensions.

The first basic criterion for essential self-adjointness is the following:^[1]

Equivalently,${\displaystyle A}$ is essentially self-adjoint if and only if the operators${\displaystyle A^{\ast }\pm i}$ have trivialkernels.^[2] That is to say,${\displaystyle A}$fails to be self-adjoint if and only if${\displaystyle A^{\ast }}$ has an eigenvector withcomplex eigenvalues${\displaystyle \pm i}$.

Another way of looking at the issue is provided by theCayley transform of a self-adjoint operator and the deficiency indices.^[3]

${\displaystyle W(A)}$ isisometric on its domain. Moreover,${\displaystyle \mathrm{ran}(1-W(A))}$ is dense in${\displaystyle A}$.

Conversely, given any densely defined operator${\displaystyle U}$ which is isometric on its (not necessarily closed) domain and such that${\displaystyle 1-U}$ is dense, then there is a (unique) densely defined symmetric operator

${\displaystyle S(U):\mathrm{ran}(1-U)\to \mathrm{ran}(1+U)}$

such that

${\displaystyle S(U)(x-Ux)=i(x+Ux),x\in \mathrm{dom}(U).}$

The mappings${\displaystyle W}$ and${\displaystyle S}$ are inverses of each other, i.e.,${\displaystyle S(W(A))=A}$.

The mapping${\displaystyle A\mapsto W(A)}$ is called theCayley transform. It associates apartially defined isometry to any symmetric densely defined operator. Note that the mappings${\displaystyle W}$ and${\displaystyle S}$ aremonotone: This means that if${\displaystyle B}$ is a symmetric operator that extends the densely defined symmetric operator${\displaystyle A}$, then${\displaystyle W(B)}$ extends${\displaystyle W(A)}$, and similarly for${\displaystyle S}$.

This immediately gives us a necessary and sufficient condition for${\displaystyle A}$ to have a self-adjoint extension, as follows:

A partially defined isometric operator${\displaystyle V}$ on a Hilbert space${\displaystyle H}$ has a unique isometric extension to the norm closure of${\displaystyle \mathrm{dom}(V)}$. A partially defined isometric operator with closed domain is called apartial isometry.

Define thedeficiency subspaces ofA by

In this language, the description of the self-adjoint extension problem given by the theorem can be restated as follows: a symmetric operator${\displaystyle A}$ has self-adjoint extensions if and only if the deficiency subspaces${\displaystyle K_{+}}$ and${\displaystyle K_{-}}$ have the same dimension.^[4]

Thedeficiency indices of a partial isometry${\displaystyle V}$ are defined as the dimension of theorthogonal complements of the domain and range:

We see that there is a bijection between symmetric extensions of an operator and isometric extensions of its Cayley transform. The symmetric extension is self-adjoint if and only if the corresponding isometric extension is unitary.

A symmetric operator has a unique self-adjoint extension if and only if both its deficiency indices are zero. Such an operator is said to beessentially self-adjoint. Symmetric operators which are not essentially self-adjoint may still have acanonical self-adjoint extension. Such is the case fornon-negative symmetric operators (or more generally, operators which are bounded below). These operators always have a canonically definedFriedrichs extension and for these operators we can define a canonical functional calculus. Many operators that occur in analysis are bounded below (such as the negative of theLaplacian operator), so the issue of essential adjointness for these operators is less critical.

Suppose${\displaystyle A}$ is symmetric densely defined. Then any symmetric extension of${\displaystyle A}$ is a restriction of${\displaystyle A^{\ast }}$. Indeed,${\displaystyle A\subseteq B}$ and${\displaystyle B}$ symmetric yields${\displaystyle B\subseteq A^{\ast }}$ by applying the definition of${\displaystyle \mathrm{dom}(A^{\ast })}$. This notion leads to thevon Neumann formulae:^[5]

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Consider the Hilbert space${\displaystyle L^2([0,1])}$. On the subspace of absolutely continuous function that vanish on the boundary, define the operator${\displaystyle A}$ by

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

Integration by parts shows${\displaystyle A}$ is symmetric. Its adjoint${\displaystyle A^{\ast }}$ is the same operator with${\displaystyle \mathrm{dom}(A^{\ast })}$ being theabsolutely continuous functions with no boundary condition. We will see that extendingA amounts to modifying the boundary conditions, thereby enlarging${\displaystyle \mathrm{dom}(A)}$ and reducing${\displaystyle \mathrm{dom}(A^{\ast })}$, until the two coincide.

Direct calculation shows that${\displaystyle K_{+}}$ and${\displaystyle K_{-}}$ are one-dimensional subspaces given by

where${\displaystyle c}$ is a normalizing constant. The self-adjoint extensions${\displaystyle A_{\alpha }}$ of${\displaystyle A}$ are parametrized by thecircle group${\displaystyle \mathbb{T}=\{\alpha \in \mathbb{C}:|\alpha |=1\}}$. For eachunitary transformation${\displaystyle U_{\alpha }:K_{-}\to K_{+}}$ defined by

${\displaystyle U_{\alpha }({\varphi }_{-})=\alpha {\varphi }_{+}}$

there corresponds an extension${\displaystyle A_{\alpha }}$ with domain

${\displaystyle \mathrm{dom}(A_{\alpha })=\{f+\beta (\alpha {\varphi }_{-}-{\varphi }_{+})|f\in \mathrm{dom}(A),\beta \in \mathbb{C}\}.}$

If${\displaystyle f\in \mathrm{dom}(A_{\alpha })}$, then${\displaystyle f}$ is absolutely continuous and

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

Conversely, if${\displaystyle f}$ is absolutely continuous and${\displaystyle f(0)=\gamma f(1)}$ for some${\displaystyle \gamma \in \mathbb{T}}$, then${\displaystyle f}$ lies in the above domain.

The self-adjoint operators${\displaystyle A_{\alpha }}$ are instances of themomentum operator in quantum mechanics.

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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.

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A symmetric operator${\displaystyle A}$ is calledpositive if

${\displaystyle ⟨Ax,x⟩\ge 0,\mathrm{\forall }x\in \mathrm{dom}(A).}$

It is known that for every such${\displaystyle A}$, one has${\displaystyle \dim K_{+}=\dim K_{-}}$. Therefore, every positive symmetric operator has self-adjoint extensions. The more interesting question in this direction is whether${\displaystyle A}$ has positive self-adjoint extensions.

For two positive operators${\displaystyle A}$ and${\displaystyle B}$, we put${\displaystyle A\le B}$ if

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

in the sense of bounded operators.

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 extendingcontractions: 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${\displaystyle \mathrm{\Gamma }}$, acting on${\displaystyle H}$, we define itsdefect operators by

Thedefect spaces of${\displaystyle \mathrm{\Gamma }}$ are

The defect operators indicate the non-unitarity of${\displaystyle \mathrm{\Gamma }}$, 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${\displaystyle \mathrm{\Gamma }}$ can be uniquely expressed as

where each${\displaystyle {\mathrm{\Gamma }}_i}$ is a contraction.

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

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

This suggests we assign to every positive symmetric operator${\displaystyle A}$ a contraction

${\displaystyle C_A:\mathrm{ran}(A+1)\to \mathrm{ran}(A-1)\subset H}$

defined by

${\displaystyle C_A(A+1)x=(A-1)x.\text{i.e.}{C}_A=(A-1)(A+1{)}^{-1}.}$

which have matrix representation^{[clarification needed]}

${\displaystyle C_A=\left[\begin{array}{c}{\mathrm{\Gamma }}_1\\ {\mathrm{\Gamma }}_3{D}_{{\mathrm{\Gamma }}_1}\end{array}\right]:\mathrm{ran}(A+1)\to \begin{array}{c}\mathrm{ran}(A+1)\\ \oplus \\ \mathrm{ran}(A+1{)}^{\perp }\end{array}.}$

It is easily verified that the${\displaystyle {\mathrm{\Gamma }}_1}$ entry,${\displaystyle C_A}$ projected onto${\displaystyle \mathrm{ran}(A+1)=\mathrm{dom}(C_A)}$, is self-adjoint. The operator${\displaystyle A}$ can be written as

${\displaystyle A=(1+C_A)(1-C_A{)}^{-1}}$

with${\displaystyle \mathrm{dom}(A)=\mathrm{ran}(C_A-1)}$. If${\displaystyle \tilde{C}}$ is a contraction that extends${\displaystyle C_A}$ and its projection onto its domain is self-adjoint, then it is clear that its inverse Cayley transform

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

defined on${\displaystyle \mathrm{ran}(1-\tilde{C})}$ is a positive symmetric extension of${\displaystyle A}$. 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${\displaystyle A}$, its Cayley transform is a contraction satisfying the stated "partial" self-adjoint property.

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

Therefore, finding self-adjoint extension for a positive symmetric operator becomes a "matrix completion problem". Specifically, we need to embed the column contraction${\displaystyle C_A}$ 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${\displaystyle C_A}$ takes the form

So the self-adjoint positive extensions of${\displaystyle A}$ are in bijective correspondence with the self-adjoint contractions${\displaystyle {\mathrm{\Gamma }}_4}$ on the defect space${\displaystyle {\mathcal{D}}_{{\mathrm{\Gamma }}_3^{\ast }}}$ of${\displaystyle {\mathrm{\Gamma }}_3}$. The contractions${\displaystyle \tilde{C}(-1)}$ and${\displaystyle \tilde{C}(1)}$ give rise to positive extensions${\displaystyle A_0}$ and${\displaystyle A_{\mathrm{\infty }}}$ respectively. These are thesmallest andlargest positive extensions of${\displaystyle A}$ in the sense that

${\displaystyle A_0\le B\le A_{\mathrm{\infty }}}$

for any positive self-adjoint extension${\displaystyle B}$ of${\displaystyle A}$. The operator${\displaystyle A_{\mathrm{\infty }}}$ is theFriedrichs extension of${\displaystyle A}$ and${\displaystyle A_0}$ is thevon Neumann-Krein extension of${\displaystyle A}$.

Similar results can be obtained foraccretive operators.

• Akhiezer, Naum Ilʹich (1981).Theory of Linear Operators in Hilbert Space. Boston: Pitman.ISBN 0-273-08496-8.
• A. Alonso and B. Simon, The Birman-Krein-Vishik theory of self-adjoint extensions of semibounded operators.J. Operator Theory4 (1980), 251-270.
• Gr. Arsene and A. Gheondea, Completing matrix contractions,J. Operator Theory7 (1982), 179-189.
• N. Dunford and J.T. Schwartz,Linear Operators, Part II, Interscience, 1958.
• Hall, B. C. (2013),Quantum Theory for Mathematicians, Graduate Texts in Mathematics, vol. 267, Springer,Bibcode:2013qtm..book.....H,ISBN 978-1461471158
• Jørgensen, Palle E. T.; Kornelson, Keri A.; Shuman, Karen L. (2011).Iterated Function Systems, Moments, and Transformations of Infinite Matrices. Providence, RI: American Mathematical Soc.ISBN 978-0-8218-5248-4.
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