Nelson's Commutator Theorem

These notes 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 $H$ be a self-adjoint and positive operator on a Hilbert space \newcommand{\H}{\mathcal{H}}\newcommand{\L}{\mathscr{L}}\newcommand{\R}{\mathbb{R}}(\H,\langle\cdot,\cdot\rangle). For any $k\in \mathbb{R}$, we define the Hilbert space (\H^{k},\langle\cdot,\cdot\rangle_{k}) as the completion of the domain $D(H^{k/2})$ ($H^{k/2}$ is the operator obtained from functional calculus) with respect to the inner product: $⟨f,g{⟩}_k:=⟨(1+H{)}^{k/2}f,(1+H{)}^{k/2}g⟩.$ These are called scales of \H associated with $H$. We also define \H^\infty := \bigcap_{k \in \R} \H^{k}. Moreover, for g\in\H^{-k} and f\in \H^{k}, we define $⟨g,f{⟩}_{-k,k}:=⟨(1+H{)}^{-k/2}g,(1+H{)}^{k/2}f⟩.$

Remark

Scales generalize Sobolev spaces. The powers $H^{k/2}$ define norms that distinguish different levels of regularity. Intuitively, higher $k$ corresponds to more “smooth” or “regular” elements (stronger norms), while lower (even negative) $k$ relate to dual or distribution-type spaces. Specifically, $⟨\cdot ,\cdot {⟩}_{-k,k}$ provides a duality between \H^{k} and \H^{-k}.

Throughout this note, we will assume that $H$ is a self-adjoint positive operator on \H.

Proposition

The following properties holds

1. For $k_1>k_2$, we have a continuous inclusion \H^{k_1} \subset \H^{k_2}.
2. The operator $(1+H{)}^t$ is a unitary map from \H^k \to \H^{k-2t}.

Proof

Proposition

\H^{\infty} is dense in \H^k for all $k\in \mathbb{R}$.

Proof By the spectral theorem, we can think of vectors in \H as a function $\psi$ and the operator $H$ as multiplication by an a.e. finite function $h$. Formally, there exists a measure space $(X,\mu )$ so that \psi\in\H^{k} if and only if $\int (1+h(x){)}^k|\psi (x){|}^2\text{dd}\mu$ is finite. Now suppose \psi\in\H^{k}. We can construct a sequence of approximating vectors, ${\psi }_n(x):=\{\begin{array}{ll}\psi (x),& \text{if }h(x)\le n,\\ 0,& \text{otherwise}.\end{array}$Then $\int (1+h(x){)}^l|{\psi }_n(x){|}^2\text{dd}\mu$ is finite for all $l\in \mathbb{R}$. So \psi_{n}\in\H^{l} for all $l\in \mathbb{R}$, which means \psi_{n}\in \H^{\infty} for all $n\in \mathbb{N}$. Note that $\Vert \psi -{\psi }_n{\Vert }_k^2={\int }_{h(\lambda )>n}(1+h(x){)}^k|\psi (x){|}^2\text{dd}\mu <\infty ,$because we started with the assumption that \psi\in\H^{k}, we know the total integral is finite. Then this integral approaches to zero as $n\to \infty$. We’ve just found an element in \H^{\infty} that can be arbitrarily close to $\psi$, this means \H^{\infty} is dense in \H^{k}. $\square$

Operator Spaces

For $-\infty \le k,l\le \infty$, we define two spaces of linear operators:

• \L(\H^k, \H^l) is the space of bounded linear operators from \H^k to \H^l, equipped with the operator norm denoted by $||\cdot |{|}_{k,l}$.
• \L_0(\H^k, \H^l) is the space of linear operators with domain \H^\infty which can be extended by continuity to an operator in \L(\H^k, \H^l).

As usual, if $k=l$, we simplify our notation to \L(\H^{k}) and \L_{0}(\H^{k}) respectively.

Remark

\L_{0}(\H^{k},\H^{l}) and \L(\H^{k},\H^{l}) are in fact isomorphic. Any T\in \L(\H^k, \H^l) can be restricted to \H^{\infty}.

Commutator Properties

We define the adjoint action of $H$ on an operator $A$ $(\mathrm{ad}H)A:=[H,A]=HA-AH$ as usual.

Lemma

If A \in \L_0(\H^k, \H^l), then its commutator with $H$, $(\text{ad}H)A$, belongs to \L_0(\H^{k+2}, \H^{l-2}).

Proof Let f \in \H^{k+2}. By definition, this means (1+H)f \in \H^k.

• Hf \in \H^k, so since A \in \L_0(\H^k, \H^l), we have AHf \in \H^l.
• f \in \H^{k+2} \subset \H^k, so Af \in \H^l. This implies HAf \in \H^{l-2}.

Combining these, (HA-AH)f \in \H^{l-2}, which shows that $(adH)A$ maps \H^{k+2} to \H^{l-2}. $\square$

Regularity Lemma

If (\ad H)^m A \in \L_0(\H^k, \H^l) for $m=0,1,\dots ,n$, then A \in \L_0(\H^{k+2n}, \H^{l+2n}). Consequently, if this holds for all $m\ge 0$, then $A$ is a smooth operator from \H^\infty to \H^\infty, i.e., A \in \L(\H^\infty).

Proof It suffices to prove the case for $n=1$. Assume A, [H,A] \in \L_0(\H^k, \H^l). We want to show A \in \L_0(\H^{k+2}, \H^{l+2}). This is equivalent to showing that $(1+H)A$ is a bounded operator from \H^{k+2} to \H^l. By the triangle inequality:$||(1+H)A|{|}_{k+2,l}\le ||(1+H)A-A(1+H)|{|}_{k+2,l}+||A(1+H)|{|}_{k+2,l}$The first term is $||[H,A]|{|}_{k+2,l}\le ||[H,A]|{|}_{k,l}<\infty$. The second term is $||A|{|}_{k,l}<\infty$ since $1+H$ is a unitary map from \H^{k+2} to \H^k. Thus, $||(1+H)A|{|}_{k+2,l}\le ||[H,A]|{|}_{k,l}+||A|{|}_{k,l}<\infty$. The general statement follows by induction. $\square$

Construction of Smooth Operators

We can construct operators with good regularity properties by integrating operator-valued functions.

Proposition

Let a\colon \R \to \L_0(\H^k, \H^l) be a Schwartz function, meaning for all u \in \H^k, v \in \H^{-l}, the map $t\mapsto ⟨v,a(t)u⟩$ is a Schwartz function on $\mathbb{R}$. Define the operator $A$ as: $A={\int }_{-\infty }^{\infty }{e}^{itH}a(t)e^{-itH}dt$ Then A \in \L(\H^\infty), i.e., it maps \H^\infty to \H^\infty.

Proof First, we show A \in \L_0(\H^k, \H^l). The operator $U(t)=e^{itH}$ is unitary on each space \H^j. [cite: 92, 100] Therefore, $||e^{itH}a(t)e^{-itH}|{|}_{k,l}=||a(t)|{|}_{k,l}$. [cite_start]Since $a(t)$ is a Schwartz function, its norm is absolutely integrable. [cite: 87, 105] [cite_start]$||A|{|}_{k,l}\le {\int }_{-\infty }^{\infty }||e^{itH}a(t)e^{-itH}|{|}_{k,l}dt={\int }_{-\infty }^{\infty }||a(t)|{|}_{k,l}dt<\infty$ [cite: 105] Next, we show that (adH)^m A \in \L_0(\H^k, \H^l) for all $m\ge 0$. Let’s compute the first commutator. Differentiating $e^{itH}a(t)e^{-itH}$ with respect to $t$ gives: $\frac{d}{dt}(e^{itH}a(t)e^{-itH})=i[H,e^{itH}a(t)e^{-itH}]+e^{itH}{a}^{\prime }(t)e^{-itH}$ [cite_start]Integrating from $-\infty$ to $\infty$, the left-hand side vanishes because $a(t)\to 0$ as $t\to \pm \infty$. [cite: 113, 115] $0={\int }_{-\infty }^{\infty }\left(i[H,e^{itH}a(t)e^{-itH}]+e^{itH}{a}^{\prime }(t)e^{-itH}\right)dt$ [cite_start]$0=i[H,A]+{\int }_{-\infty }^{\infty }{e}^{itH}{a}^{\prime }(t)e^{-itH}dt$ [cite: 117] This gives an expression for the commutator: [cite_start]$[H,A]=i{\int }_{-\infty }^{\infty }{e}^{itH}{a}^{\prime }(t)e^{-itH}dt$ [cite: 118] Since $a(t)$ is a Schwartz function, its derivative $a^{\prime }(t)$ is also a Schwartz function valued in \L_0(\H^k, \H^l). By the first part of the proof, this means [H,A] \in \L_0(\H^k, \H^l). By induction, we can show that for any integer $m\ge 0$: [cite_start]$(adH{)}^{m}A=i^m{\int }_{-\infty }^{\infty }{e}^{itH}{a}^{(m)}(t)e^{-itH}dt$ [cite: 128] [cite_start]Since all derivatives $a^{(m)}(t)$ are Schwartz functions, (adH)^m A \in \L_0(\H^k, \H^l) for all $m$. By the Regularity Lemma, we conclude that A \in \L(\H^\infty). $\square$

Nelson’s Commutator Theorem

Formal Adjoint

If A \in \L_0(\H^k, \H^l), we define its formal adjoint A^\dagger \in \L_0(\H^{-l}, \H^{-k}) as the operator satisfying: \langle A^\dagger v, u \rangle = \langle v, Au \rangle \quad \forall u, v \in \H^\infty.

We can explicitly give an expression of $A^{†}$.

Lemma

Let $H$ be a positive self-adjoint operator on a Hilbert space $\mathcal{H}$, and let $A$ and $[HA]$ be in \L_{0}(\H^{1},\H^{-1}). For any $\epsilon >0$, the commutator of $A$ with the semigroup $e^{-ϵH}$ satisfies the identity: $[A,e^{-ϵH}]=-{\int }_0^{ϵ}{e}^{-(ϵ-s)H}[A,H]e^{-sH}ds$ Furthermore, this commutator is a bounded operator on $\mathcal{H}$, with its norm satisfying: $\Vert [A,e^{-ϵH}]{\Vert }_{\mathcal{L}(\mathcal{H})}\le \frac{\pi }{2e}\Vert [A,H]{\Vert }_{\mathcal{L}({\mathcal{H}}^1,{\mathcal{H}}^{-1})}$

Proof Let’s first establish the identity. Define a one-parameter family of operators $F(s)$ for $s\in [0,ϵ]$: $F(s)=e^{-(ϵ-s)H}A{e}^{-sH}$ We differentiate $F(s)$ with respect to $s$ using the product rule: $F^{\prime }(s)=\left(\frac{d}{ds}{e}^{-(ϵ-s)H}\right)A{e}^{-sH}+e^{-(ϵ-s)H}A\left(\frac{d}{ds}{e}^{-sH}\right)$ By the functional calculus for the self-adjoint operator $H$, we have $\frac{d}{ds}{e}^{-sH}=-H{e}^{-sH}$ and $\frac{d}{ds}{e}^{-(ϵ-s)H}=H{e}^{-(ϵ-s)H}$. Substituting these in gives: $F^{\prime }(s)=H{e}^{-(ϵ-s)H}A{e}^{-sH}-e^{-(ϵ-s)H}AH{e}^{-sH}$ Since $H$ commutes with $e^{-(ϵ-s)H}$, we can factor it out: $F^{\prime }(s)=e^{-(ϵ-s)H}(HA-AH)e^{-sH}=-e^{-(ϵ-s)H}[A,H]e^{-sH}$ By the fundamental theorem of calculus for operator-valued functions, we integrate $F^{\prime }(s)$ from $0$ to $ϵ$: $F(ϵ)-F(0)={\int }_0^{ϵ}{F}^{\prime }(s)ds$ $A{e}^{-ϵH}-e^{-ϵH}A=-{\int }_0^{ϵ}{e}^{-(ϵ-s)H}[A,H]e^{-sH}ds$ This is precisely the identity for $[A,e^{-ϵH}]$.

Next, we bound the norm. Taking the operator norm on both sides: $\Vert [A,e^{-ϵH}]\Vert \le {\int }_0^{ϵ}\Vert e^{-(ϵ-s)H}{\Vert }_{\mathcal{L}({\mathcal{H}}^{-1},\mathcal{H})}\cdot \Vert [A,H]{\Vert }_{\mathcal{L}({\mathcal{H}}^1,{\mathcal{H}}^{-1})}\cdot \Vert e^{-sH}{\Vert }_{\mathcal{L}(\mathcal{H},{\mathcal{H}}^1)}ds$ We estimate the norms of the semigroup mapping between Sobolev spaces. $\Vert e^{-sH}{\Vert }_{\mathcal{L}(\mathcal{H},{\mathcal{H}}^1)}=\Vert (1+H{)}^{1/2}{e}^{-sH}{\Vert }_{\mathcal{L}(\mathcal{H})}\le {\sup }_{\lambda \in [0,\infty )}(1+\lambda {)}^{1/2}{e}^{-s\lambda }\approx (2se{)}^{-1/2}$ A similar calculation yields $\Vert e^{-(ϵ-s)H}{\Vert }_{\mathcal{L}({\mathcal{H}}^{-1},\mathcal{H})}\le (2(ϵ-s)e{)}^{-1/2}$. Plugging these bounds into the integral: $\Vert [A,e^{-ϵH}]\Vert \le \Vert [A,H]{\Vert }_{1,-1}{\int }_0^{ϵ}(2(ϵ-s)e{)}^{-1/2}(2se{)}^{-1/2}ds=\frac{\Vert [A,H]{\Vert }_{1,-1}}{2e}{\int }_0^{ϵ}\frac{1}{\sqrt{s(ϵ-s)}}ds$ The integral evaluates to $\pi$. Thus, we arrive at the final bound: $\Vert [A,e^{-ϵH}]\Vert \le \frac{\pi }{2e}\Vert [A,H]{\Vert }_{1,-1}$ This shows that $[A,e^{-ϵH}]$ is a bounded operator on $\mathcal{H}$ for any $ϵ>0$. $\square$

Nelson's Commutator Theorem

Let $H$ be a positive self-adjoint operator on a Hilbert space $\mathcal{H}$, and let $A$ and $[H,A]$ be in \L_{0}(\H^{1},\H^{-1}). Then the closure of its formal adjoint $\overline{A^{†}}$ equals $A^{\ast }$, i.e., $A^{\ast }=\overline{A^{†}}$.

Proof By the previous lemma, we have the identity $[A,e^{-ϵH}]=-{\int }_0^{ϵ}{e}^{-(ϵ-s)H}[A,H]e^{-sH}\text{dd}s.$Taking the operator norm on both sides:

Let $R_{ϵ}=[A,e^{-ϵH}]$. As shown, $R_{ϵ}\in \mathcal{L}(\mathcal{H})$. For any $f\in \mathcal{H}$, $e^{-ϵH}f\to f$ as $ϵ\to 0$. This implies that $R_{ϵ}\to 0$ in the strong operator topology (SOT).

Consider a vector $w\in D(A^{\ast })$, the domain of the Hilbert space adjoint. We want to show that $(w,A^{\ast }w)$ is in the graph of $\overline{A^{†}}$, denoted $\Gamma (\overline{A^{†}})$. Define a sequence $w_{ϵ}=e^{-ϵH}w$. As $ϵ\to 0$, we have $w_{ϵ}\to w$. Since the semigroup $e^{-ϵH}$ maps into the space of smooth vectors, $w_{ϵ}\in H^{\infty }\subset D(A^{†})$. We can therefore apply $A^{†}$ to $w_{ϵ}$: $A^{†}{w}_{ϵ}=A^{†}{e}^{-ϵH}w$ From the identity $A{e}^{-ϵH}=e^{-ϵH}A+R_{ϵ}$, we take the adjoint of both sides: $(A{e}^{-ϵH}{)}^{\ast }=(e^{-ϵH}A{)}^{\ast }+R_{ϵ}^{\ast }$. This gives us $e^{-ϵH}{A}^{†}\supset A^{\ast }{e}^{-ϵH}+R_{ϵ}^{\ast }$. Applying this to $w\in D(A^{\ast })$: $A^{†}{e}^{-ϵH}w=A^{\ast }{e}^{-ϵH}w+R_{ϵ}^{\ast }w$ So we have a sequence of points $(w_{ϵ},A^{†}{w}_{ϵ})\in \Gamma (A^{†})$. We check their limits as $ϵ\to 0$:

1. $w_{ϵ}=e^{-ϵH}w\to w$
2. $A^{†}{w}_{ϵ}=A^{\ast }{e}^{-ϵH}w+R_{ϵ}^{\ast }w\to A^{\ast }w$

The second limit holds because $A^{\ast }$ is a closed operator, $w_{ϵ}\to w$, and $A^{\ast }{e}^{-ϵH}w\to A^{\ast }w$. Furthermore, $R_{ϵ}^{\ast }\to 0$ in SOT because $R_{ϵ}\to 0$ in SOT.

Since we have found a sequence $(w_{ϵ},A^{†}{w}_{ϵ})\in \Gamma (A^{†})$ that converges to $(w,A^{\ast }w)$, the point $(w,A^{\ast }w)$ must belong to the closure of the graph, $\Gamma (\overline{A^{†}})$. By definition, this means $w\in D(\overline{A^{†}})$ and $\overline{A^{†}}w=A^{\ast }w$. As this holds for any $w\in D(A^{\ast })$, we conclude that $A^{\ast }\subset \overline{A^{†}}$. $\square$

References

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