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.
Let
H
A
H
\operatorname{dom}(A)
\langleAx,y\rangle=\langlex,Ay\rangle, \forallx,y\in\operatorname{dom}(A).
If
\operatorname{dom}(A)=H
A
A
\operatorname{dom}(A*)
\operatorname{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
A
A
A
In the next section, a symmetric operator will be assumed to be densely defined and closed.
If an operator
A
H
A
The first basic criterion for essential self-adjointness is the following:[1] Equivalently,
A
A*\pmi
A
A*
\pmi
Another way of looking at the issue is provided by the Cayley transform of a self-adjoint operator and the deficiency indices.
W(A)
\operatorname{ran}(1-W(A))
A
Conversely, given any densely defined operator
U
1-U
S(U):\operatorname{ran}(1-U)\to\operatorname{ran}(1+U)
such that
S(U)(x-Ux)=i(x+Ux), x\in\operatorname{dom}(U).
The mappings
W
S
S(W(A))=A
The mapping
A\mapstoW(A)
W
S
B
A
W(B)
W(A)
S
This immediately gives us a necessary and sufficient condition for
A
A partially defined isometric operator
V
H
\operatorname{dom}(V)
Define the deficiency subspaces of A by
\begin{align} K+&=\operatorname{ran}(A+i)\perp\\ K-&=\operatorname{ran}(A-i)\perp\end{align}
A
K+
K-
The deficiency indices of a partial isometry
V
\begin{align} n+(V)&=\dim\operatorname{dom}(V)\perp\\ n-(V)&=\dim\operatorname{ran}(V)\perp \end{align}
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 be essentially self-adjoint. Symmetric operators which are not essentially self-adjoint may still have a canonical self-adjoint extension. Such is the case for non-negative symmetric operators (or more generally, operators which are bounded below). These operators always have a canonically defined Friedrichs 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 the Laplacian operator), so the issue of essential adjointness for these operators is less critical.
Suppose
A
A
A*
A\subseteqB
B
B\subseteqA*
\operatorname{dom}(A*)
Consider the Hilbert space
L2([0,1])
A
Af=i
d | |
dx |
f.
Integration by parts shows
A
A*
\operatorname{dom}(A*)
\operatorname{dom}(A)
\operatorname{dom}(A*)
Direct calculation shows that
K+
K-
\begin{align} K+&=\operatorname{span}\{\phi+=c ⋅ ex\}\\ K-&=\operatorname{span}\{\phi-=c ⋅ e-x\} \end{align}
where
c
A\alpha
A
T=\{\alpha\inC:|\alpha|=1\}
U\alpha:K-\toK+
U\alpha(\phi-)=\alpha\phi+
there corresponds an extension
A\alpha
\operatorname{dom}(A\alpha)=\{f+\beta(\alpha\phi--\phi+)|f\in\operatorname{dom}(A), \beta\inC\}.
If
f\in\operatorname{dom}(A\alpha)
f
\left| | f(0) |
f(1) |
\right|=\left|
e\alpha-1 | |
\alpha-e |
\right|=1.
Conversely, if
f
f(0)=\gammaf(1)
\gamma\inT
f
The self-adjoint operators
A\alpha
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.
A symmetric operator
A
\langleAx,x\rangle\ge0, \forallx\in\operatorname{dom}(A).
A
\operatorname{dim}K+=\operatorname{dim}K-
A
For two positive operators
A
B
A\leqB
(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 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
\Gamma
H
\begin{align}&D =(1-\Gamma*\Gamma
| ||||
) |
\\ &D | |
\Gamma* |
=(1-\Gamma\Gamma*)
| ||||
\end{align}
The defect spaces of
\Gamma
\begin{align}&l{D}\Gamma =\operatorname{ran}(D\Gamma
)\ &l{D} | |
\Gamma* |
=\operatorname{ran}(
D | |
\Gamma* |
) \end{align}
The defect operators indicate the non-unitarity of
\Gamma
\Gamma
\Gamma= \begin{bmatrix} \Gamma1&
D | |||||||
|
\Gamma2\\ \Gamma3
D | |
\Gamma1 |
&-\Gamma3
* | |
\Gamma | |
1 |
\Gamma2+
D | |||||||
|
\Gamma4
D | |
\Gamma2 |
\end{bmatrix}
where each
\Gammai
The Cayley transform for general symmetric operators can be adapted to this special case. For every non-negative number
a
\left| | a-1 |
a+1 |
\right|\le1.
This suggests we assign to every positive symmetric operator
A
CA:\operatorname{ran}(A+1) → \operatorname{ran}(A-1)\subsetH
defined by
CA(A+1)x=(A-1)x. i.e. CA=(A-1)(A+1)-1.
which have matrix representation
CA= \begin{bmatrix} \Gamma1\\ \Gamma3
D | |
\Gamma1 |
\end{bmatrix} :\operatorname{ran}(A+1) → \begin{matrix} \operatorname{ran}(A+1)\\ ⊕ \\ \operatorname{ran}(A+1)\perp\end{matrix}.
It is easily verified that the
\Gamma1
CA
\operatorname{ran}(A+1)=\operatorname{dom}(CA)
A
A=(1+CA)(1-
-1 | |
C | |
A) |
with
\operatorname{dom}(A)=\operatorname{ran}(CA-1)
\tilde{C}
CA
\tilde{A}=(1+\tilde{C})(1-\tilde{C})-1
defined on
\operatorname{ran}(1-\tilde{C})
A
A
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
CA
By the preceding subsection, all self-adjoint extensions of
CA
\tilde{C}(\Gamma4)= \begin{bmatrix} \Gamma1&
D | |
\Gamma1 |
* | |
\Gamma | |
3 |
\\ \Gamma3
D | |
\Gamma1 |
&-\Gamma3\Gamma1
* | |
\Gamma | |
3 |
+
D | |||||||
|
\Gamma4
D | |||||||
|
\end{bmatrix}.
So the self-adjoint positive extensions of
A
\Gamma4
l{D} | |||||||
|
\Gamma3
\tilde{C}(-1)
\tilde{C}(1)
A0
Ainfty
A
A0\leqB\leqAinfty
for any positive self-adjoint extension
B
A
Ainfty
A
A0
A
Similar results can be obtained for accretive operators.