Unbounded Operators

Unbounded Operators

An unbounded operator is a linear mapping , where the linear subspace is called the domain of .

e.g. The position operator in quantum mechanics is unbounded. Specifically, consider and let be the set of all functions such that . For , the position operator acts on such that Then is clearly unbounded because we can consider satisfying Then , but , which diverges to as .

Closed Operator

An operator between two Hilbert spaces is called closed if for any sequence , and implies and . Equivalently, is closed if its graph is closed in .

e.g. Clearly, any bounded linear operator is closed.

Extension

Let be operators between two Hilbert spaces. We say that is an extension of if their graphs satisfy and we write .

Proposition

if and only if and for all .

Closable & Closure

An operator is closable or preclosed if it has a closed extension. The closure of a closable operator , denoted , is the smallest closed extension of .

Proposition

If is closable, then its closure is the operator whose graph is the closure of the graph of .

e.g. The derivative operator on is not closed but is closable; its closure is the operator on .

Adjoints of Densely Defined Operators

Adjoint of Densely Defined Operator

Let be a densely defined (i.e. is dense in ) operator between Hilbert spaces. Define the adjoint operator on the domain such that for all and .

The following proposition guarantees the unique existence of adjoints for unbounded operators under certain conditions:

Proposition

The adjoint of a densely defined operator is unique if exits.

Proof Fix some , suppose and satisfy and for all . Then, we have for all . Since is dense in , there exists a sequence converges to , which implies that hence .

Proposition

Suppose and are operators such that , then .

Proof If , then , and . For all , we have for all . So it also holds in , hence for all . This shows that , so . Therefore, .

Theorem

Let be a densely defined operator, then

• is closed;
• is closable if and only if is dense, in which case ;
• If is closable, then .

Proof For the first statement, suppose such that and , then for all , we have so and . This shows that is closed. Now we prove the second statement. Suppose is dense, then

Resolvent Set & Resolvent Operator

Let be a closed operator on a Hilbert space . Then the resolvent set of is defined as For any , the resolvent of at is defined as

Remark

Note that