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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.

Symmetric operators

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

<Ax, y> = <x, Ay>, 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 a densely defined and closed.

Extensions of symmetric operators

Problem Given a densely defined closed symmetric operator A, does it always have extension?

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 U_A is in fact an isometry between closed subspaces that takes (A + i)x to (A - i)x for x in Dom(A). ^[1]

The map

A\mapsto U_{A}

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

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

defined on Dom(A) = Ran(U - 1).^[2]

Now if

{\tilde {U}}

is an isometric extension of U_A, 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. ^[3]

^[4]

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

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 U_A is an unitary operator on H. ^[5]

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 U_A.

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 U_A 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^[6] . 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_{+}=span\{\phi _{+}=a\cdot e^{x}\}

and

K_{-}=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 U_α : K_- → K₊, defined by U_α(φ_-) = αφ₊), there corresponds an extension A_α with domain

Dom(A_{\alpha })=\{f+\beta (\alpha \phi _{-}-\phi _{+})|f\in Dom(A)\}.

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

|{\frac {f(0)}{f(1)}}|=|{\frac {e\alpha -1}{\alpha -e^{2}}}|=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 { A_α } 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 an 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 <Ax, x> ≥ 0 for all x in Dom(A). It is known that for every such A, one has dim(K₊) = dim(K_-)^[7]. Therefore every positive symmetric operator has self-adjoint extensions. The more interesting question in this direction is whether A has positive self-adjoint extensions.

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-D^{*}D)^{\frac {1}{2}}\quad {\mbox{and}}\quad D_{\Gamma ^{*}}=(1-DD^{*})^{\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 parametrizations. 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 Γ_i 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,

|{\frac {a-1}{a+1}}|\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 V_{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, C_A projected onto Ran(A + 1) = Dom(C_A), is self-adjoint.^[8]

The operator A can be written as

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

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

{\tilde {C}}

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

{\tilde {A}}=(1+{\tilde {C}}+1)(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) is self-adjoint.

^[9]

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

Theorem A symmetric positive operator B 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.

^[10]

Therefore finding self-adjoint extension for a positive symmetric operator becomes a matrix "completion problem". Specificly, we need to embed the column contraction 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 C_A 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 extension B of A. The operator A₀ is the Friedrichs extension of A and A_∞ is the von Neumann-Krein extension of A.

Similar results can be obtained for accretive operators.

Notes

^ The operator A + i is bounded below, the expression (A + i)^-1 makes sense on Ran(A + i). Closedness of A implies both Ran(A - i) and Ran(A + i) are closed subspaces.
^ We observe
$(A+i)^{-1}={\frac {i}{2}}(U-1)\quad {\mbox{and}}\quad {\frac {1}{2}}(U+1)=A(A+i)^{-1}$
are well defined on Ran(A + i). So
$Dom(A)=Ran(A+i)^{-1}=Ran(-{\frac {i}{2}}(U-1))=Ran(U-1).$
^ This is because
$Ran(-{\frac {i}{2}}({\tilde {U}}-1))\supset Ran(U-1),$
and that ${\tilde {A}}$ is symmetric follows from ${\tilde {V}}$ being an isometry.
^ Sidenote: it is see here that an isometry V between closed subspaces of H is the Cayley transform of a symmetric operator if and only if V - 1 is injective, i.e. 1 is not an eigenvalue of V.
^ To show this, we note that Ran(A ± i) ≠ H if and only if there exists y ∈ H such that <A ± x, y> for all x ∈ Dom(A). In turn, this is true if and only if y ∈ Dom(A*) and A*y = ± iy, which implies A is not self-adjoint. For the converse, if Ran(A ± i) = H, then for all y in Dom(A*), (A* + i)y = (A + i)x for some x in Dom(A). By A being symmetric, (A* + i)(x - y) = 0, i.e. x - y lies in Ker(A* + i). Since Ker(A* + i) = Ran(A - i)^⊥, we have x - y = 0.
^ Reed and Simon, v1
^ Dunford and Schwarz, v2
^ Because <C_A(A + 1)x, (A + 1)y> = <(A + 1)x, C_A(A + 1)y> for all x, y in Dom(A).
^ Note The connection between positive operators and contractions is a natural analog of that between self-adjoint operators and unitaries. See generalized semigroup of contractions, where a Stone-von Neumann type result is discussed.
^ The argument for this is as follows. If Ran(A + 1) = H, then for all y in Dom(A*), (A* + 1)y = (A + 1)x. By A being symmetric, (A* + 1)(x - y) = 0. But Ker(A* + 1) = Ran(A + 1)^⊥. So x = y and y is in Dom(A). Conversely, if Ran(A + 1) ≠ H, then there exists y in Dom(A*) such that A*y = -y. Therefore A cannot be self-adjoint.

References

Gr. Arsene and A. Gheondea, Completing matrix contractions, J. Operator Theory, 7, 1982, 179-189.

A. Alfonso and B. Simon, The Birman-Krein-Vishik theory of self-adjoint extensions of semibounded operators. J. Operator Theory, 4, 1980, 251-270.

N. Dunford and J.T. Schwartz, Linear Operators, Part II, Interscience, 1958.

M. Reed and B. Simon, Methods of Modern Mathematical Physics, vol. I and II, Academic Press, 1975.

[1] The operator A + i is bounded below, the expression (A + i)^-1 makes sense on Ran(A + i). Closedness of A implies both Ran(A - i) and Ran(A + i) are closed subspaces.

[2] We observe
$(A+i)^{-1}={\frac {i}{2}}(U-1)\quad {\mbox{and}}\quad {\frac {1}{2}}(U+1)=A(A+i)^{-1}$
are well defined on Ran(A + i). So
$Dom(A)=Ran(A+i)^{-1}=Ran(-{\frac {i}{2}}(U-1))=Ran(U-1).$

[3] This is because
$Ran(-{\frac {i}{2}}({\tilde {U}}-1))\supset Ran(U-1),$
and that ${\tilde {A}}$ is symmetric follows from ${\tilde {V}}$ being an isometry.

[4] Sidenote: it is see here that an isometry V between closed subspaces of H is the Cayley transform of a symmetric operator if and only if V - 1 is injective, i.e. 1 is not an eigenvalue of V.

[5] To show this, we note that Ran(A ± i) ≠ H if and only if there exists y ∈ H such that <A ± x, y> for all x ∈ Dom(A). In turn, this is true if and only if y ∈ Dom(A*) and A*y = ± iy, which implies A is not self-adjoint. For the converse, if Ran(A ± i) = H, then for all y in Dom(A*), (A* + i)y = (A + i)x for some x in Dom(A). By A being symmetric, (A* + i)(x - y) = 0, i.e. x - y lies in Ker(A* + i). Since Ker(A* + i) = Ran(A - i)^⊥, we have x - y = 0.

[6] Reed and Simon, v1

[7] Dunford and Schwarz, v2

[8] Because <C_A(A + 1)x, (A + 1)y> = <(A + 1)x, C_A(A + 1)y> for all x, y in Dom(A).

[9] Note The connection between positive operators and contractions is a natural analog of that between self-adjoint operators and unitaries. See generalized semigroup of contractions, where a Stone-von Neumann type result is discussed.

[10] The argument for this is as follows. If Ran(A + 1) = H, then for all y in Dom(A*), (A* + 1)y = (A + 1)x. By A being symmetric, (A* + 1)(x - y) = 0. But Ker(A* + 1) = Ran(A + 1)^⊥. So x = y and y is in Dom(A). Conversely, if Ran(A + 1) ≠ H, then there exists y in Dom(A*) such that A*y = -y. Therefore A cannot be self-adjoint.

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