Symmetric and Self-Adjoint Operators

Symmetric Operator

A densely defined operator on a Hilbert space is symmetric if for all , we have . That is, , and .

Self-Adjoint Operator

A densely defined operator on is self-adjoint if . That is, is symmetric, and .

Attention

Notice that when we say a symmetric (or self-adjoint) operator, we always mean a densely defined operator, unless otherwise specified.

Proposition

A symmetric operator is always closable. In fact, we have . Moreover, for closed symmetric operators, we have . For self-adjoint operators, we have .

Proof For a symmetric operator , is an extension of , and is closed, hence is closable, and by the theorem.

Essentially Self-adjoint Operator

A symmetric operator is essentially self-adjoint if its closure is self-adjoint.

Core

The core of an essentially self-adjoint operator refers to a dense subset of its domain on which the operator remains essentially self-adjoint when restricted to it.

Proposition

If is essentially self-adjoint, then it has a unique self-adjoint extension. Specifically,

e.g. Let be the momentum operator with domain , where denotes the set of all absolute continuous functions on . Then is symmetric, but not self-adjoint. Indeed, we have for all , so is symmetric. However, notice that we only require to make the above derivation valid, hence is not self-adjoint because . has uncountably many different self-adjoint extensions, each corresponding to a different choice of boundary conditions, such as for some unit length complex number . So it is not essentially self-adjoint.

Criterion for Self-Adjointness

Theorem

Let be a symmetric operator on a Hilbert space , then the followings are equivalent:

• is self-adjoint;
• is closed and ;
• .

Proof