Nelson's Commutator Theorem

This note explore Nelson’s commutator theorem, which delivers sufficient conditions for a symmetric operator to be essentially self-adjoint.

Scales of Hilbert Spaces

Scales of Hilbert Spaces

Let be a self-adjoint and positive operator on a Hilbert space . For any , we define the Hilbert space as the completion of the domain ( is the operator obtained from functional calculus) with respect to the inner product: These are called scales of associated with . We also define Moreover, for and , we define

Remark

Scales generalize Sobolev spaces. The powers define norms that distinguish different levels of regularity. Intuitively, higher corresponds to more “smooth” or “regular” elements (stronger norms), while lower (even negative) relate to dual or distribution-type spaces. Specifically, provides a duality between and .

Throughout this note, we will assume that is a self-adjoint positive operator on . The following properties are immediate from the definition:

Proposition

The following properties hold:

1. For , we have a continuous inclusion .
2. The operator is a unitary map from .

Proposition

is dense in for all . In particular, is dense in .

Proof By the spectral theorem, we can think of vectors in as a function and the operator as multiplication by an a.e. finite function . Formally, there exists a measure space so that if and only if is finite. Now suppose . We can construct a sequence of approximating vectors, Then is finite for all . So for all , which means for all . Note that because we started with the assumption that , we know the total integral is finite. Then this integral approaches to zero as . We’ve just found an element in that can be arbitrarily close to , this means is dense in .

Operator Spaces

For , we define two spaces of linear operators:

• is the space of bounded linear operators from to , equipped with the operator norm denoted by .
• is the space of linear operators with domain which can be extended by continuity to an operator in .

As usual, if , we simplify our notation to and respectively.

Remark

and are in fact isomorphic. Any can be restricted to .

Commutator Properties

We define the adjoint action of on an operator as usual.

Lemma

If , then its commutator with , , belongs to .

Proof Let . Then , , and . So . Moreover, we have the bound: Thus, .

Regularity Lemma

If for , then . Consequently, if this holds for all , then is smooth, i.e., .

Proof It suffices to prove the case for as the general statement follows by induction. Assume . We want to show . Firstly, , so it maps any to , thus maps to . Secondly, we need to show that . Since is unitary, we have . Note that Therefore by triangle inequality, $$\begin{aligned}|(I+H)A|{k+2,l}&\leq|(I+H)A-A(I+H)|{k+2,l}+|A(I+H)|{k+2,l}\&\leq|[H,A]|{k,l} + |A|_{k,l} < \infty.\end{aligned}$$$\square$

Nelson’s Commutator Theorem

Formal Adjoint

If , we define its formal adjoint as the operator satisfying: We say that is formally Hermitian if .

We can explicitly give an expression for by observing the following: Therefore .

Lemma

Let be a positive self-adjoint operator on a Hilbert space , and let and be in . For any , the following identity holds:

Proof Define a one-parameter family of operators for : We differentiate with respect to using the product rule: By the functional calculus for the self-adjoint operator , commutes with , so we can factor it out: By the fundamental theorem of calculus for operator-valued functions, we integrate from to : it follows that This is precisely the desired identity for .

Nelson's Commutator Theorem

Let be a positive self-adjoint operator on a Hilbert space , and let and be in . Then the closure of its formal adjoint equals . In particular, if is formally Hermitian, it is essentially self-adjoint.

and are in , then so are and . By the lemma, and are in . Note that , so and are densely defined operators on , and is well-defined.

If

Proof It is clear that , because for all , since is self-adjoint in , we have It remains to show that the graph of is contained in the graph of , denoted . Pick some . By the above lemma, we have the identity Taking the operator norm on both sides: Note that the norm continuity property of functional calculus gives A similar calculation yields . Therefore, plugging these bounds into the integral, we obtain So is a bounded operator on for any . Let . For any , Since is dense in , and as , we conclude that in the strong operator topology. Now consider a vector , we want to show that is in . For any , we have So . As the adjoints of strongly continuous operators are weakly convergent, is in the weak closure of . Note that the graph of any linear operator is convex, so weak convergence implies strong convergence. Therefore, is in the strong closure of , which is .

References

Nelson (1972), Time-ordered operator products of sharp-time quadratic forms