Remarks on Scattering Matrices for Schrödinger Operators with Critically Long-Range Perturbations

Abstract

We consider scattering matrix for Schrödinger-type operators on $\mathbb {R}^d$ with perturbation $V(x)=O(\langle x \rangle ^{-1})$ as $|x|\rightarrow \infty$. We show that the scattering matrix (with time-independent modifiers) is a pseudodifferential operator and analyze its spectrum. We present examples of which the spectrum of the scattering matrices has dense point spectrum, and absolutely continuous spectrum, respectively. These give a partial answer to an open question posed by Yafaev (Scattering theory: some old and new problems. Springer Lecture Notes in Mathematical, vol 1735, 2000).

Appendices

Appendix A: Functional Calculus of Unitary Pseudodifferential Operators

In Appendices A and B, we consider pseudodifferential operators on $\mathbb {R}^d$, but it can be generalized easily to pseudodifferential operators on manifolds. We restrict ourselves to the $\mathbb {R}^d$ case mostly to simplify notations related to Beal’s characterization of pseudodifferential operators.

Let $\delta \in [0,1)$, and we consider a unitary operator U on $L^2$ with the symbol $u\in \bigcap _{\delta >0}S^\delta _{1,0}$. We consider operators on $\mathbb {R}^d$, or in a local coordinate in a d-dimensional manifold. We show that f(U), the function of U, is a pseudodifferential operator and compute the principal symbol. At first, we note

Lemma A.1

Suppose $a\in S^1_{1,0}$, and the symbol is bounded. Then $\mathrm {Op}(a)$ is bounded in $L^2$.

Proof

The proof is essentially the same as the Gårding inequality. Without loss of generality, we may suppose a is real valued, and we write $A=\mathrm {Op}(a)$. Let $M>\sup |a|$. We set $b(x,\xi )= (M^2-a(x,\xi )^2)^{1/2}\in S^1_{1,0}$, and $B=\mathrm {Op}(b)$. Then by the symbol calculus, we learn

$$\begin{aligned} R=A^* A+ B^* B -M^2 \in \mathrm {Op}(S^0_{1,0}). \end{aligned}$$

Hence

$$\begin{aligned} \Vert A u \Vert ^2 \le \Vert Au \Vert ^2+\Vert Bu \Vert ^2 \le M^2 \Vert u \Vert ^2 + \Vert Ru \Vert \Vert u \Vert \le C\Vert u \Vert ^2 \end{aligned}$$

since R is bounded in $L^2$.

Lemma A.2

Suppose $U=\mathrm {Op}(u)$ is unitary with $u\in S^\delta _{1,0}$, $\delta \in [0,1)$. Then for any $s\in \mathbb {R}$,

$$\begin{aligned} \bigl \Vert U^k \bigr \Vert _{H^s\rightarrow H^s}\le C_s\langle k \rangle ^{|s|/(1-\delta )}, \quad k\in \mathbb {Z}. \end{aligned}$$

Proof

We let $\nu =1-\delta \in (0,1]$, $s=N\nu$ and show

$$\begin{aligned} \bigl \Vert U^k \bigr \Vert _{H^{N\nu }\rightarrow H^{N\nu }} \le C\langle k \rangle ^N, \quad k\in \mathbb {Z}. \end{aligned}$$

We first suppose $k>0$. We consider the commutator:

$$\begin{aligned}{}[\langle D_x \rangle ^\nu , U^k] = \sum _{j=1}^{d-1} U^j [\langle D_x \rangle ^\nu ,U] U^{k-1-j}. \end{aligned}$$

Since the symbol of the operator $[\langle D_x \rangle ^\nu ,U]$ is in $S^0_{1,0}$, it is bounded in $L^2$, and hence $\bigl \Vert [\langle D_x \rangle ^\nu ,U^k] \bigr \Vert \le C\langle k \rangle$. This implies $\Vert U^k \Vert _{H^\nu \rightarrow H^\nu }\le C\langle k \rangle$.

More generally, we compute

$$\begin{aligned}{}[\langle D_x \rangle ^{N\nu }, U^k]&=\sum _{j=1}^{k-1} U^j[\langle D_x \rangle ^{N\nu },U] U^{k-1-j}\\&= \sum _{j=1}^{k-1} \sum _{\ell =0}^{N-1} U^j \langle D_x \rangle ^{\ell \nu } [\langle D_x \rangle ^\nu ,U]\langle D_x \rangle ^{(N-1-\ell )\nu } U^{k-1-j}. \end{aligned}$$

Now we use the induction in N. Suppose the claim holds for $N\le N_0$. Then we have

$$\begin{aligned}&[\langle D_x \rangle ^{N_0\nu },U^k] \langle D_x \rangle ^{-N_0\nu }\\&\quad = \sum _{j=1}^{k-1} \sum _{\ell =0}^{N_0-1} U^j \langle D_x \rangle ^{\ell \nu } [\langle D_x \rangle ^\nu ,U]\langle D_x \rangle ^{(N_0-1-\ell )\nu } U^{k-1-j}\langle D_x \rangle ^{-N_0\nu }\\&\quad = \sum _{j=1}^{k-1} \sum _{\ell =0}^{N_0-1} U^j (\langle D_x \rangle ^{\ell \nu }[\langle D_x \rangle ^\nu ,U]\langle D_x \rangle ^{-\ell \nu }) \\&\qquad \times (\langle D_x \rangle ^{(N_0-1)\nu } U^{k-1-j}\langle D_x \rangle ^{-(N_0-1)\nu }) \langle D_x \rangle ^{-1}. \end{aligned}$$

By the induction hypothesis and the fact $[\langle D_x \rangle ^\nu ,U]$ is bounded in $H^{\ell \nu }$, each term in the sum is bounded in $L^2$, and the norm is $O(\langle k \rangle ^{(N_0-1)\nu })$. By summing up these norms, we arrive at the claim with $N=N_0$. For $k<0$, we use the same argument for $U^{-1}=U^*$. Then the assertion for general $s\in \mathbb {R}$ follows by the interpolation and the duality argument.

Now we consider functional calculus of a unitary operator U. For $f\in C^\infty (S^1)$, we write the Fourier series expansion by $\hat{f}[k]$, i.e.,

$$\begin{aligned} \hat{f}[k]= \frac{1}{2\pi }\int _0^{2\pi } e^{-ik\theta } f(e^{i\theta })\mathrm{d}\theta , \quad k\in \mathbb {Z}, \end{aligned}$$

and hence

$$\begin{aligned} f(e^{i\theta }) =\sum _{k\in \mathbb {Z}} \hat{f}[k] e^{ik\theta }, \quad \theta \in [0,2\pi ). \end{aligned}$$

We recall $\hat{f}[n]$ is rapidly decreasing in n. Then we write

$$\begin{aligned} f(U)=\sum _{k\in \mathbb {Z}} \hat{f}[k] U^k \in \mathcal {B}(L^2). \end{aligned}$$

It is well-known that f(U) is the same function of U defined in terms of the spectral decomposition. We show f(U) is a pseudodifferential operator using the Beals characterization of pseudodifferential operators.

For an operator A, we write

$$\begin{aligned} K_jA= i[D_{x_j},A],\quad L_j A= -i [x_j,A], \quad j=1,\dots , d, \end{aligned}$$

and multiple commutators by $L^\alpha A$, $K^\beta A$, etc., for $\alpha ,\beta \in \mathbb {Z}_+^d$. We recall $A=\mathrm {Op}(a)$ with $a\in S^\delta _{1,0}$ if and only if $K^\alpha L^\beta A$ is bounded from $L^2$ to $H^{-\delta +|\beta |}$ for any $\alpha ,\beta \in \mathbb {Z}_+^d$ (cf. Dimassi-Sjöstrand [2], Zworski [16]). We compute

$$\begin{aligned} K^\alpha L^\beta (U^k)&= \sum _{\begin{array}{c} \alpha ^1+\cdots +\alpha ^N =\alpha , \\ \beta ^1+\cdots +\beta ^N=\beta ,\\ \alpha ^j+\beta ^j\ne 0, \\ k_1+\cdots +k_{N+1}=k \end{array}} U^{k_1} (K^{\alpha ^1}L^{\beta ^1} U) U^{k_2} (K^{\alpha ^2}L^{\beta ^2} U) \\&\quad \,\times \cdots \cdots \times U^{k_N} (K^{\alpha ^N}L^{\beta ^N} U) U^{k_{N+1}}. \end{aligned}$$

Since $K^{\alpha ^j} L^{\beta ^j} U$ is bounded from $H^s$ to $H^{-\delta +|\beta ^j|}$, we have, using Lemma A.2,

$$\begin{aligned} \bigl \Vert K^\alpha L^\beta (U^k) \bigr \Vert _{L^2\rightarrow H^{-N_0\delta +|\beta |}} \le C \langle k \rangle ^{N_1}, \end{aligned}$$

where $N_0=|\alpha +\beta |$, $N_1= (N_0\delta +|\beta |)/(1-\delta ) + N_0$. Thus we learn

$$\begin{aligned} K^\alpha L^\beta (f(U))\in \mathcal {B}(L^2, H^{-|\alpha +\beta |\delta +|\beta |}), \end{aligned}$$

and we have the following lemma: We write

$$\begin{aligned} S^{+0}_{1,0} =\bigcap _{\delta >0} S^{\delta }_{1,0}. \end{aligned}$$

Lemma A.3

Suppose $U=\mathrm {Op}(u)$ is unitary with $u\in S^{+0}_{1,0}$. Then f(U) is a pseudodifferential operator with the symbol in $S^{+0}_{1,0}$.

We then compute the principal symbol of f(U). If $U=\mathrm {Op}(u)$ is unitary with $u\in S^\delta _{1,0}$, then the symbol of $1=U^*U$ is $1=|u(x,\xi )|^2$ modulo $S^{\delta -1}_{1,0}$. Thus, we may assume $u_0$, the principal symbol of U modulo $S^{\delta -1}_{1,0}$, has modulus 1. This implies, in particular, $u_0^j\in S_{1,\delta }^0$ for any $j\ge 0$. We show f(U) has the principal symbol $f\circ u_0$. We note

$$\begin{aligned} U^k -\mathrm {Op}(u_0^k)&= \sum _{j=0}^{k-1} (U^{j+1} \mathrm {Op}(u_0^{k-j-1}) -U^j \mathrm {Op}(u_0^{k-j}))\\&=\sum _{j=0}^{k-1} U^j (U-\mathrm {Op}(u_0))\mathrm {Op}(u_0^{k-j-1}) \\&\quad -\sum _{j=0}^{k-1} U^j (\mathrm {Op}(u_0^{k-j}-u_0\#(u_0^{k-j-1}))), \end{aligned}$$

where $a\#b$ denotes the operator composition: $\mathrm {Op}(a\# b)=\mathrm {Op}(a)\mathrm {Op}(b)$. By the symbol calculus, we learn $u_0^{k-j} -u_0\# (u_0^{k-j-1})\in S^{\delta -1}_{1,\delta }$, and each seminorm of it is bounded by $C\langle k \rangle ^M$ with some $M>0$. Thus, after direct computations, we learn that $U^k-\mathrm {Op}(u_0^k) \in S^{\delta -1}_{1,\delta }$ and its seminorm is bounded by $C\langle k \rangle ^M$ with some M. Hence we have the following claim: We note $\bigcap _{\delta>0}S^{\delta -1}_{1,0}=\bigcap _{\delta >0} S_{1,\delta }^{\delta -1}$.

Theorem A.4

Suppose $U=\mathrm {Op}(u)$ is unitary with $u\in S^{+0}_{1,0}$, and let $u_0$ be a principal symbol such that $|u_0(x,\xi )|=1$. Let $f\in C^\infty (S^1)$. Then f(U) is a pseudodifferential operator with its symbol in $S^{+0}_{1,0}$ and the principal symbol is given by $f\circ u_0$ modulo $S^{\delta -1}_{1,0}$ with any $\delta >0$.

Remark A.1

We can actually compute the asymptotic expansion of f(U) in terms of derivatives of $f\circ u$ and derivatives of u. Thus, in particular, the support of these terms is contained in the support of $f\circ u$, and hence the essential support of the symbol of f(U) is contained in the support of $f\circ u$.

Remark A.2

In our application, we consider the cace $u\in S(1,\tilde{g})$, i.e., for any $\alpha ,\beta \in \mathbb {Z}_+d$,

$$\begin{aligned} \bigl | \partial _x^\alpha \partial _\xi ^\beta u(x,\xi ) \bigr |\le C_{\alpha \beta } \langle \xi \rangle ^{-|\beta |}\langle \log \langle \xi \rangle \rangle ^{|\alpha |}. \end{aligned}$$

Then we can apply Theorem A.4 to learn f(U) is a pseudodifferential operator with the symbol in $S^{+0}_{1,0}$. Moreover, since the principal symbol is $f\circ u\in S(1,\tilde{g})$, and the remainder is in $S^{-1+\delta }_{1,0}$ for any $\delta >0$, we actually learn the symbol is in $S(1,\tilde{g})$.

Appendix B: Logarithm of Unitary Pseudodifferential Operators

For notational convenience, we write $\ell (\xi ) =\langle \log \langle \xi \rangle \rangle$ for $\xi \in \mathbb {R}^d$. We use the following metrics on $T^*\mathbb {R}^d$:

$$\begin{aligned} g= \mathrm {d}x^2 +\frac{\mathrm {d}\xi ^2}{\langle \xi \rangle ^2}, \quad \tilde{g} =\ell (\xi )^2\mathrm {d}x^2 +\frac{\mathrm {d}\xi ^2}{\langle \xi \rangle ^2}. \end{aligned}$$

We recall, $a\in S(m,g)$ if and only if, for any $\alpha ,\beta \in \mathbb {Z}^d$, $\exists C_{\alpha \beta }>0$ such that

$$\begin{aligned} \bigl | \partial _x^\alpha \partial _\xi ^\beta a(x,\xi ) \bigr | \le C_{\alpha \beta }m(x,\xi ) \langle \xi \rangle ^{-|\beta |}, \quad x,\xi \in \mathbb {R}^d, \end{aligned}$$

and $a\in S(m,\tilde{g})$ if and only if, for any $a\,\beta \in \mathbb {Z}^d$, $\exists C_{\alpha \beta }>0$ such that

$$\begin{aligned} \bigl | \partial _x^\alpha \partial _\xi ^\beta a(x,\xi ) \bigr | \le C_{\alpha \beta }m(x,\xi ) \ell (\xi )^{|\alpha |}\langle \xi \rangle ^{-|\beta |}, \quad x,\xi \in \mathbb {R}^d. \end{aligned}$$

Assumption E

Let $\psi _0\in S(\ell (\xi ),g)$, real-valued, and $\partial _\xi \psi _0\in S(\langle \xi \rangle ^{-1},g)$. Let U be a unitary pseudodifferential operator on $L^2(\mathbb {R}^d)$ such that the principal symbol is given by $e^{i\psi _0}$, i.e., $U\in \mathrm {Op}S(1,\tilde{g})$ and $U-\mathrm {Op}(e^{i\psi _0}) \in \mathrm {Op}S(\ell (\xi )/\langle \xi \rangle ,\tilde{g})$.

We note $e^{i\psi _0}\in S(1,\tilde{g})$, and natural remainder terms are in the symbol class $S(\ell (\xi )/\langle \xi \rangle ,\tilde{g})$.

Theorem B.1

Suppose $\psi _0$ and U as in Assumption E. Then there is $\psi \in S(\ell (\xi ),g)$ such that $U-\exp (i\mathrm {Op}(\psi ))\in \mathrm {Op}S(\langle \xi \rangle ^{-\infty },g)$, and $\psi -\psi _0\in S(\ell (\xi )/\langle \xi \rangle ,\tilde{g})$.

Lemma B.2

Let $\varphi \in S(\ell (\xi ),g)$, real-valued, and $\partial _\xi \varphi \in S(\langle \xi \rangle ^{-1},g)$. Then $\mathrm {Op}(\varphi )$ is essentially self-adjoint and $\exp (it\mathrm {Op}(\varphi )) \in \mathrm {Op}S(1,\tilde{g})$, $t\in \mathbb {R}$. Moreover,

$$\begin{aligned} e^{it\mathrm {Op}(\varphi )}- \mathrm {Op}(e^{it\varphi }) \in \mathrm {Op}S(\ell (\xi )/\langle \xi \rangle ,\tilde{g}), \end{aligned}$$

and is uniformly bounded for $t\in [0,1]$.

Proof

The essential self-adjointness of $\mathrm {Op}(\varphi )$ follows by the commutator theorem with an auxiliary operator $N=\langle D_x \rangle$.

In order to show $e^{it\mathrm {Op}(\varphi )}\in \mathrm {Op}S(1,\tilde{g})$, we use Beal’s characterization. Let $K_j$ and $L_j$ ($j=1,\dots ,d$) as in “Appendix A”. We note, by a simple commutator argument as in Appendix A, we can show, for any $k,\ell \in \mathbb {Z}$, $T>0$,

$$\begin{aligned} \sup _{|t|\le T} \bigl \Vert \langle D_x \rangle ^k\ell (D_x)^\ell e^{it\mathrm {Op}(\varphi )} \ell (D_x)^{-\ell } \langle D_x \rangle ^{-k} \bigr \Vert _{L^2\rightarrow L^2} <\infty . \end{aligned}$$

We compute, for example,

$$\begin{aligned} L_j[e^{it\mathrm {Op}(\varphi )}] =i\int _0^t e^{is\mathrm {Op}(\varphi )} L_j[\mathrm {Op}(\varphi )] e^{i(t-s)\mathrm {Op}(\varphi )} \mathrm{d}s. \end{aligned}$$

Since $L_j[\mathrm {Op}(\varphi )]=\mathrm {Op}(\partial _{\xi _j}\varphi ) \in \mathrm {Op}S(\langle \xi \rangle ^{-1},g)$, we learn $\langle D_x \rangle L_j[e^{it\mathrm {Op}(\varphi )}]$ is bounded in $H^s$ with any $s\in \mathbb {R}$. Similarly, since $K_j[\mathrm {Op}(\varphi )]=\mathrm {Op}(\partial _{x_j}\varphi )\in \mathrm {Op}S(\ell (\xi ),g)$, we learn $\ell (D_x)^{-1} K_j[e^{it\mathrm {Op}(\varphi )}]$ is bounded in $H^s$, $\forall s\in \mathbb {R}$. Iterating this procedure, we learn, for any $\alpha ,\beta \in \mathbb {Z}_+^d$,

$$\begin{aligned} \ell (D_x)^{-|\alpha |} \langle D_x \rangle ^{|\beta |} (K^\alpha L^\beta [e^{it\mathrm {Op}(\varphi )}]):\ H^s\rightarrow H^s, \text { bounded}, \end{aligned}$$

with any $s\in \mathbb {R}$. By Beal’s characterization, this implies $e^{it\mathrm {Op}(\varphi )}\in \mathrm {Op}S(1,\tilde{g})$, and bounded locally uniformly in t.

Then we show the principal symbol of $e^{it\mathrm {Op}(\varphi )}$ is $e^{it\varphi }$. We have

$$\begin{aligned}&e^{it\mathrm {Op}(\varphi )}-\mathrm {Op}(e^{it\varphi }) = \int _0^t \frac{\mathrm{d}}{\mathrm{d}s}\bigl (e^{is\mathrm {Op}(\varphi )} \mathrm {Op}(e^{i(t-s)\varphi })\bigr )\mathrm{d}s \\&\quad = i\int _0^t e^{is\mathrm {Op}(\varphi )}\bigl (\mathrm {Op}(\varphi )\mathrm {Op}(e^{i(t-s)\varphi })-\mathrm {Op}(\varphi e^{i(t-s)\varphi })\bigr ) \mathrm{d}s\\&\quad \in \mathrm {Op}S(\ell (\xi )/\langle \xi \rangle ,\tilde{g}) \end{aligned}$$

by the asymptotic expansion.

In particular, we have

$$\begin{aligned} U e^{-i\mathrm {Op}(\psi _0)}-1 \in \mathrm {Op}S(\ell (\xi )/\langle \xi \rangle ,\tilde{g}), \end{aligned}$$

and hence there is a real-valued symbol $\psi _1\in S(\ell (\xi )/\langle \xi \rangle ,\tilde{g})$ such that

$$\begin{aligned} U e^{-i\mathrm {Op}(\psi _0)} - \mathrm {Op}(e^{i\psi _1})\in \mathrm {Op}S(\ell (\xi )^2/\langle \xi \rangle ^2,\tilde{g}). \end{aligned}$$

This implies,

$$\begin{aligned} U e^{-i\mathrm {Op}(\psi _0)} e^{-i\mathrm {Op}(\psi _1)} -1 \in \mathrm {Op}S(\ell (\xi )^2/\langle \xi \rangle ^2,\tilde{g}). \end{aligned}$$

(4.1)

We use the next lemma to rewrite $e^{-i\mathrm {Op}(\psi _0)} e^{-i\mathrm {Op}(\psi _1)}$.

Lemma B.3

Let $\varphi \in S(\ell (\xi ),g)$, real-valued, and $\partial _\xi \varphi \in S(\langle \xi \rangle ^{-1},g)$. Let $\eta \in S(\ell (\xi )^{k}/\langle \xi \rangle ^k,\tilde{g})$, real-valued, with $k\ge 1$. Then

$$\begin{aligned} e^{i\mathrm {Op}(\eta )} e^{i\mathrm {Op}(\varphi )}- e^{i\mathrm {Op}(\varphi +\eta )} \in \mathrm {Op}S(\ell (\xi )^{k+1}/\langle \xi \rangle ^{k+1},\tilde{g}). \end{aligned}$$

Proof

We have, for any self-adjoint operators A and B, at least formally,

$$\begin{aligned}&e^{i(A+B)} e^{-iA} e^{-iB} -1 =\int _0^1 \frac{\mathrm{d}}{\mathrm{d}t} \Bigl (e^{it(A+B)} e^{-itA} e^{-itB}\Bigr ) \mathrm{d}t \\&\quad = i \int _0^1 \Bigl ( e^{it(A+B)} (A+B-A) e^{-itA} e^{-itB} - e^{-t(A+B)} e^{-itA} B e^{-itB}\Bigr ) \mathrm{d}t \\&\quad = i \int _0^1 e^{it(A+B)} \bigl [B, e^{-itA}\bigr ] e^{-itB} \mathrm{d}t \\&\quad = -\int _0^1\biggl (\int _0^t e^{it(A+B)} e^{i(t-s)A} [A,B] e^{-isA} e^{-itB} \mathrm{d}s\biggr ) \mathrm{d}t. \end{aligned}$$

This computation is easily justified when $A=\mathrm {Op}(\varphi )$ and $B=\mathrm {Op}(\eta )$, and since $[\mathrm {Op}(\varphi ),\mathrm {Op}(\eta )]\in \mathrm {Op}S(\ell (\xi )^{k+1}/\langle \xi \rangle ^{k+1},\tilde{g})$, $e^{it\mathrm {Op}(\varphi )} \in \mathrm {Op}S(1,\tilde{g})$, etc., we have

$$\begin{aligned} e^{i\mathrm {Op}(\varphi +\eta )}e^{-i\mathrm {Op}(\varphi )}e^{-i\mathrm {Op}(\eta )} -1 \in \mathrm {Op}S(\ell (\xi )^{k+1}/\langle \xi \rangle ^{k+1},\tilde{g}), \end{aligned}$$

and this implies the assertion.

Proof of Theorem B.1

Combining (4.1) with lemma B.3, we have

$$\begin{aligned} U e^{-i\mathrm {Op}(\psi _0+\psi _1)} -1 \in \mathrm {Op}S(\ell (\xi )^2/\langle \xi \rangle ^2,\tilde{g}). \end{aligned}$$

We note $\psi _0+\psi _1\in S(\ell (\xi ),g)+S(\ell (\xi )^2/\langle \xi \rangle ,\tilde{g}) \subset S(1,g)$. Iterating this procedure, we construct $\psi _k\in S(\ell (\xi )^{k}/\langle \xi \rangle ^k,\tilde{g})$, real-valued, such that

$$\begin{aligned} U e^{-i\mathrm {Op}(\psi _0+\cdots +\psi _k)} -1 \in \mathrm {Op}S(\ell (\xi )^{k+1}/\langle \xi \rangle ^{k+1},\tilde{g}). \end{aligned}$$

for $k=2,3,\dots$. Then we choose an asymptotic sum: $\psi \sim \sum _{k=0}^\infty \psi _k$, i.e., $\psi \in S(\ell (\xi ),g)$ and

$$\begin{aligned} \psi -\sum _{k=0}^N\psi _k \in S(\ell (\xi )^{N+1}/\langle \xi \rangle ^{N+1},\tilde{g}) \end{aligned}$$

for any $N>0$. Then we have

$$\begin{aligned} U e^{-i\mathrm {Op}(\psi )}-1\in \mathrm {Op}S(\langle \xi \rangle ^{-\infty },\tilde{g})=\mathrm {Op}S(\langle \xi \rangle ^{-\infty },g), \end{aligned}$$

and we complete the proof of Theorem B.1.

Appendix C: Trace Class Scattering for Unitary Operators

The next theorem, the unitary version of the Kuroda-Birman theorem, seems well-known, but the author could not find an appropriate reference. Here we give a proof for the completeness.

Theorem C.1

Let $U_1$ and $U_2$ be unitary operators on a separable Hilbert space, and suppose $U_1-U_2$ is a trace class operator. Then $\sigma _{\mathrm {ac}}(U_1)=\sigma _{\mathrm {ac}}(U_2)$.

Proof

Since the eigenvalues of $U_1$ and $U_2$ are at most countable, we can find $\theta \in \mathbb {R}$ such that $e^{-i\theta }$ is not an eigenvalue of both $U_1$ and $U_2$. Then, by replacing $U_1$ and $U_2$ by $e^{i\theta }U_1$ and $e^{i\theta }U_2$, respectively, we may suppose 1 is not an eigenvalue of both $U_1$ and $U_2$. Then we can define the Cayley transform of $U_1$ and $U_2$ by

$$\begin{aligned} H_j =i(U_j+1)(U_j-1)^{-1}, \quad j=1,2. \end{aligned}$$

By the definition, we have

$$\begin{aligned} U_j =(H_j+i)(H_j-i)^{-1}= 1+2i(H_j-i)^{-1}, \quad j=1,2, \end{aligned}$$

and hence

$$\begin{aligned} (H_1+i)^{-1}-(H_2+i)^{-1} = \frac{1}{2i}(U_1-U_2), \end{aligned}$$

is in the trace class. Thus we can apply the Kuroda-Birman theorem ([11], Theorem XI.9) to learn $\sigma _{\mathrm {ac}}(H_1)= \sigma _{\mathrm {ac}}(H_2)$. This implies the assertion since

$$\begin{aligned} \sigma _{\mathrm {ac}}(U_j)= \bigl \{(s-i)(s+i)^{-1}\bigm |s\in \sigma _{\mathrm {ac}}(H_j)\bigr \}, \quad j=1,2, \end{aligned}$$

by the spectral decomposition theorem.