Compact scattering defects and the zero-energy obstruction

The real multiplier $\mathrm{\mid }p{\mathrm{\mid }}^2$ is self-adjoint on its maximal multiplication domain: truncation to $\mathrm{\mid }p\mathrm{\mid }\le n$ proves density, and testing the adjoint identity on those sets identifies the adjoint vector with $\mathrm{\mid }p{\mathrm{\mid }}^2$ times the original vector. Plancherel conjugation therefore makes $H_0$ self-adjoint on $H^2$. For $\mu >\mathrm{\parallel }V{\mathrm{\parallel }}_{\mathrm{\infty }}$, $R_0(\pm i\mu )={\mathcal{F}}_p^{\ast }{M}_{(\mathrm{\mid }p{\mathrm{\mid }}^2\mp i\mu {)}^{-1}}{\mathcal{F}}_p$ has norm at most $1\mathrm{/}\mu$. The factorization $H-z=(I+V{R}_0(z))(H_0-z)$ at $z=\pm i\mu$ has a norm-convergent Neumann inverse, so both $H\pm i\mu$ are onto. The symmetric densely defined $H$ is self-adjoint: if $y\in D(H^{\ast })$, solve $(H-i\mu )x=(H^{\ast }-i\mu )y$; then $y-x\in \ker (H^{\ast }-i\mu )=\mathrm{ran}(H+i\mu {)}^{\perp }=\{0\}$, giving equality of adjoint domains. Its quadratic form is bounded below by $-\mathrm{\parallel }V{\mathrm{\parallel }}_{\mathrm{\infty }}$. Thus no general Kato–Rellich theorem is required.

For each nonreal $z$, $m_z(p)=(\mathrm{\mid }p{\mathrm{\mid }}^2-z{)}^{-1}$ belongs to $L^2({\mathbb{R}}^3)$: its denominator is bounded away from zero on bounded $p$ sets, and its squared magnitude is $O(\mathrm{\mid }p{\mathrm{\mid }}^{-4})$ at infinity. Put $g_z=(2\pi {)}^{-3\mathrm{/}2}{\mathcal{F}}_p^{\ast }{m}_z$. The free resolvent has convolution kernel $g_z(x-y)$, since this normalization makes convolution have Fourier multiplier $m_z$. The kernel $k_z(x,y)=V(x)g_z(x-y)$ satisfies $\mathrm{\parallel }{k}_z{\mathrm{\parallel }}_2^2=\mathrm{\parallel }V{\mathrm{\parallel }}_2^2\mathrm{\parallel }{g}_z{\mathrm{\parallel }}_2^2<\mathrm{\infty }$ by Tonelli. For a finite rectangular kernel $k_m={\sum }_j{a}_j{\mathbf{1}}_{A_j}(x){\mathbf{1}}_{B_j}(y)$, its action is $T_{k_m}u={\sum }_j{a}_j{\mathbf{1}}_{A_j}{\int }_{B_j}u$, so its range is finite dimensional. Rational box density provides $k_m\to k_z$ in product $L^2$, and Cauchy–Schwarz gives $\mathrm{\parallel }{T}_{k_m}-T_{k_z}\mathrm{\parallel }\le \mathrm{\parallel }{k}_m-k_z{\mathrm{\parallel }}_2\to 0$. Thus $V{R}_0(z)$ has actual finite-rank norm approximants and is compact, proving (a) at $z=i$.

Apply CC spectral calculus on a Hilbert space with a fixed countable dense sequence to $H_0,H$ and the half-line self-adjoint operators used below, taking the fixed rational box-step dense sequences (and their half-line versions). It supplies their unique regular PVMs, exact square-integrability operator domains, bounded Borel multiplication calculus, strongly continuous unitary groups and exact generator domains under CC. In particular the Cayley unitary is $Z_A=(A-i)(A+i{)}^{-1}=I-2i{R}_A(-i)$ and its inverse-Cayley multiplier is $\lambda (z)=i(1+z)\mathrm{/}(1-z)$. We also record the elementary circle polynomial approximation: for a continuous circle function $F$, convolve it with $(2\pi N{)}^{-1}\mathrm{\mid }{\sum }_{j=0}^{N-1}{e}^{ijt}{\mathrm{\mid }}^2$. Finite expansion yields Laurent polynomials, the kernel is nonnegative with integral one, and for $\delta \le \mathrm{\mid }t\mathrm{\mid }\le \pi$ it is bounded by $[2\pi N{\sin }^2(\delta \mathrm{/}2){]}^{-1}$. Uniform continuity controls the part $\mathrm{\mid }t\mathrm{\mid }<\delta$ and this bound controls the complement, proving uniform convergence of the polynomials to $F$. Thus both the precise CC calculus and this needed norm approximation have proved local inputs.

First let $g\in C_c^{\mathrm{\infty }}((0,\mathrm{\infty }))$ have support in $(a^2,b^2)$. The Duhamel integrand $F(t)=e^{itH}V{e}^{-it{H}_0}g(H_0)P_{+}$ factors as $e^{itH}KB(t)$, where $K=V{R}_0(i)$ and $B(t)=(H_0-i)e^{-it{H}_0}g(H_0)P_{+}$. The latter is bounded and norm-continuous, because the scalar multiplier $(E-i)e^{-itE}g(E)$ has bounded compact $E$ support and its difference at $t,s$ is bounded by a constant times $\mathrm{\mid }t-s\mathrm{\mid }$. The map $t\mapsto e^{itH}K$ is norm-continuous: approximate $K$ by finite-rank operators; for each finite-rank summand $u\mapsto ⟨u,v⟩w$, strong group continuity gives norm continuity of $u\mapsto ⟨u,v⟩e^{itH}w$, and the approximation error is unchanged by unitarity. Consequently $F(t)$ is norm-continuous and compact for every $t$. If $\mathrm{supp}V\subseteq B_R$, [F2] gives, for $t\ge \max (1,R\mathrm{/}(2a))$, $\mathrm{\parallel }F(t)\mathrm{\parallel }\le \mathrm{\parallel }V{\mathrm{\parallel }}_{\mathrm{\infty }}{C}_{N}t(1+t{)}^{-N}$; choose $N=3$. On the finite preceding interval its norm is bounded by $\mathrm{\parallel }V{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}$. Therefore $F$ is norm-integrable.

The resolvent identity gives $R_H(z)-R_0(z)=-R_H(z)V{R}_0(z)$, which has finite-rank norm approximants by step 1.2. Set $Z_A=(A-i)(A+i{)}^{-1}=I-2i{R}_A(-i)$ for $A=H,H_0$. Then $Z_H-Z_0$ is compact, as is its adjoint; telescoping products show the same for the difference of every Laurent polynomial in these unitaries. For $f\in C_0(\mathbb{R})$, the circle function $F(z)=f(i(1+z)\mathrm{/}(1-z))$ off $z=1$, with $F(1)=0$, is continuous. Its Fejér polynomials $p_n$ converge uniformly by the explicit kernel estimate in step 2.1. The Laurent polynomials $q_n=p_n-p_n(1)$ still converge uniformly to $F$ and vanish at $1$. Hence $q_n(Z_A)\to f(A)$ in norm by that step's multiplication calculus, and $f(H)-f(H_0)$ is a norm limit of the compact polynomial differences. This proves the functional-calculus assertion of (c) for all $C_0$, including nonzero values at zero, without Stone–Weierstrass or general Gelfand theory.

The range of $g(H_0)P_{+}$ lies in $D(H_0)=D(H)$ because $H_{0}g(H_0)$ is bounded. Differentiating $e^{itH}{e}^{-it{H}_0}g(H_0)P_{+}$ on each vector gives $iF(t)$, so Cook's limit yields $({\mathrm{\Omega }}_{+}-I)g(H_0)P_{+}=i{\int }_0^{\mathrm{\infty }}F(t)dt$. Step 3.1 makes the integral converge in operator norm. On each finite time interval its Riemann sums are compact, and the integrable tail tends to zero in norm; [F3] therefore makes the defect compact. For the incoming sign the formula is $-i{\int }_{-\mathrm{\infty }}^0{e}^{itH}V{e}^{-it{H}_0}g(H_0)P_{-}dt$; the incoming bound in [F2] and the same norm-continuity argument prove compactness. Existence and the isometry property of the wave operators themselves follow independently by Cook on Schwartz vectors: for $\mathrm{\mid }t\mathrm{\mid }\ge 1$, the explicit free Gaussian kernel gives $\mathrm{\parallel }V{e}^{-it{H}_0}\psi {\mathrm{\parallel }}_2\le \mathrm{\parallel }V{\mathrm{\parallel }}_2(4\pi \mathrm{\mid }t\mathrm{\mid }{)}^{-3\mathrm{/}2}\mathrm{\parallel }\psi {\mathrm{\parallel }}_1$, while for $\mathrm{\mid }t\mathrm{\mid }\le 1$ it is bounded by $\mathrm{\parallel }V{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }\psi {\mathrm{\parallel }}_2$. Schwartz density and unitary uniform boundedness extend both strong limits to all of $L^2$, with norm one. Thus this compactness proof assumes no completeness theorem.

We prove the needed essential-spectrum comparison directly under CC. If every spectral neighborhood of $\lambda$ for a self-adjoint $A$ has infinite rank, recursively project a fixed dense sequence onto $E_A((\lambda -1\mathrm{/}n,\lambda +1\mathrm{/}n))\mathcal{H}$ intersected with the orthogonal complement of the preceding finite family, select its least nonzero member and normalize. This constructs an orthonormal sequence $u_n$ with $\mathrm{\parallel }(A-\lambda )u_n\mathrm{\parallel }\le 1\mathrm{/}n$; orthonormality makes it weakly null by Bessel's inequality. Conversely, a weakly-null unit sequence with $(A-\lambda )u_n\to 0$ excludes a finite-rank spectral $\epsilon$-neighborhood: its projection tends to zero, while its complementary component has norm at most $\mathrm{\parallel }(A-\lambda )u_n\mathrm{\parallel }\mathrm{/}\epsilon \to 0$, a contradiction. Suppose $K=R_B(z)-R_A(z)$ is compact and use such a sequence for $A$. Put $v_n=R_B(z)(A-z)u_n=u_n+K(A-z)u_n$. Since $(A-z)u_n=(\lambda -z)u_n+o(1)$ is bounded and weakly null, $v_n-u_n\to 0$. Also $(B-\lambda )v_n=(A-\lambda )u_n+(z-\lambda )(v_n-u_n)\to 0$. Normalizing $v_n$ transfers the criterion to $B$; reversing roles transfers it back. The free momentum spectral projections have infinite rank near each $\lambda \ge 0$, witnessed by normalized indicators of a countable sequence of disjoint rational subballs in the corresponding momentum shell, and vanish in a sufficiently small neighborhood of each $\lambda <0$. Therefore ${\sigma }_{\mathrm{e}\mathrm{s}\mathrm{s}}(H_0)=[0,\mathrm{\infty })$ and the compact resolvent difference in step 3.2 gives ${\sigma }_{\mathrm{e}\mathrm{s}\mathrm{s}}(H)=[0,\mathrm{\infty })$. This proves the rest of (c) without an abstract Weyl/Fredholm supplier.

If $f\in C_c(\mathbb{R})$ and $f(0)=0$, its restriction to $[0,\mathrm{\infty })$ is uniformly approximated by $g_n\in C_c^{\mathrm{\infty }}((0,\mathrm{\infty }))$. Explicitly, multiply by a fixed smooth cutoff removing $[0,1\mathrm{/}n]$ and equal to one beyond $2\mathrm{/}n$; the error tends uniformly to zero by continuity at zero. Extend the resulting positive-half-line function by zero to negative energies and convolve with a fixed compact smooth approximate identity of width less than $1\mathrm{/}(2n)$, taking a smaller width if required by its uniform modulus of continuity. This keeps support strictly positive and gives uniform approximation; only countably many choices are needed. Free multiplier calculus and $\mathrm{\parallel }{\mathrm{\Omega }}_{\pm }\mathrm{\parallel }=1$ give $\mathrm{\parallel }({\mathrm{\Omega }}_{\pm }-I)[f(H_0)-g_n(H_0)]P_{\pm }\mathrm{\parallel }\le 2{\sup }_{E\ge 0}\mathrm{\mid }f(E)-g_n(E)\mathrm{\mid }\to 0$. Thus (b) is the norm-limit consequence of step 4.1, with the whole stated continuous cutoff class.

For the threshold obstruction take $V=-{\alpha }^2{\mathbf{1}}_{\mathrm{\mid }x\mathrm{\mid }<1}$ with $\alpha =\pi \mathrm{/}2$; its support is contained in $B_2$. The unitary radial embedding is $Ju(x)=u(\mathrm{\mid }x\mathrm{\mid })\mathrm{/}(\sqrt{4\pi }\mathrm{\mid }x\mathrm{\mid })$. Radial reduction gives $h_0=-d^2\mathrm{/}d{r}^2$ and $h=h_0-{\alpha }^2{\mathbf{1}}_{(0,1)}$ on $H^2(0,\mathrm{\infty })\cap H_0^1(0,\mathrm{\infty })$. This follows by differentiating $r\psi (r)$ on the radial smooth core; the Dirichlet operator is the odd-extension Fourier multiplier, whose quadratic form domain is $H_0^1$, and the bounded potential leaves its operator domain unchanged by the Neumann argument of step 1.1. Rotations commute with both groups, so their radial wave limits equal the half-line waves. The reduced zero-energy function $u_0(r)=\sin (\alpha r)$ for $r<1$ and $u_0(r)=1$ for $r\ge 1$ has matching value and derivative at $1$; $J{u}_0$ is regular at the origin and behaves as $1\mathrm{/}r$ at infinity, so it is a locally square-integrable zero solution but not $L^2$. This permitted resonance motivates the following exact calculation.

For $k>0$ put $q(k)=\sqrt{k^2+{\alpha }^2}$, $b(k)=({\sin }^{2}q(k)+(q(k{)}^2\mathrm{/}{k}^2){\cos }^{2}q(k){)}^{1\mathrm{/}2}$ and $\zeta (k)=e^{-ik}((q(k)\mathrm{/}k)\cos q(k)+i\sin q(k))\mathrm{/}b(k)$. Since sine and cosine cannot vanish simultaneously, $b>0$ and $\mathrm{\mid }\zeta \mathrm{\mid }=1$; all these functions are smooth on compact positive $k$ intervals. Define $s_k(r)=b(k{)}^{-1}\sin (q(k)r)$ for $0<r<1$, and $s_k(r)=(\zeta (k)e^{ikr}-\overline{\zeta (k)}{e}^{-ikr})\mathrm{/}(2i)$ for $r\ge 1$. At $r=1$ its value and derivative on both sides are $b^{-1}\sin q$ and $b^{-1}q\cos q$, respectively; it has zero trace at $0$ and satisfies $h{s}_k=k^2{s}_k$ with no distributional boundary term. Expansion of $q=\alpha +k^2\mathrm{/}(2\alpha )+O(k^4)$ gives $\sin q\to 1$, $(q\mathrm{/}k)\cos q=-k\mathrm{/}2+O(k^3)$, $b\to 1$ and $\zeta \to i$ as $k\downarrow 0$, without selecting a phase branch.

The position dilation generator restricts under $J$ to $D=-i(r{\mathrm{\partial }}_r+1\mathrm{/}2)$. The radial log map $u\mapsto e^{\rho \mathrm{/}2}u(e^{\rho })$ followed by the Fourier convention $e^{-i\eta \rho }$ makes it multiplication by $\eta$; there is no zero atom, and complex conjugation exchanges $P_{+}$ and $P_{-}$. Set $g(r)=e^{-r}$ and $u=P_{+}g$. Since $g$ is real and $\mathrm{\parallel }g{\mathrm{\parallel }}^2=1\mathrm{/}2$, the equal positive/negative projection norms give $P_{+}u=u$, $\mathrm{\parallel }u{\mathrm{\parallel }}^2=1\mathrm{/}4$ and $⟨u,g⟩=1\mathrm{/}4$. In particular $u$ is nonzero; it need not belong to a Hamiltonian domain because the defect acts on all of $L^2$.

Let $S$ and $C$ be the half-line sine and cosine transforms with kernels $\sqrt{2\mathrm{/}\pi }\sin (kr)$ and $\sqrt{2\mathrm{/}\pi }\cos (kr)$. Odd and even Fourier extensions make them unitary self-adjoint involutions, with $S{h}_{0}S=M_{k^2}$. On each compact interval $I⋐(0,\mathrm{\infty })$ define $Tv(r)=\sqrt{2\mathrm{/}\pi }{\int }_I{s}_k(r)v(k)dk$. It is bounded from $L^2(I)$ into $L^2(dr)$: for $r\ge 1$ it is a sum of restricted Fourier transforms of the bounded coefficients $\zeta v$ and $\bar{\zeta }v$; for $r<1$ its kernel is bounded on the finite rectangle and Cauchy–Schwarz applies. Twice differentiating gives $Tv\in D(h)$ and $hTv=T(k^{2}v)$, with the matching and Dirichlet conditions from step 6.1; outside the well the derivative estimates are Plancherel bounds for bounded powers of $k$, and inside they are bounds on a finite rectangle. Thus uniqueness of strong Schrödinger evolution gives $e^{-ith}Tv=T(e^{-it{k}^2}v)$ on these packets. No interacting generalized eigenfunction expansion is assumed.

For $v\in C_c^{\mathrm{\infty }}(I)$ compare $S{e}^{-it{k}^2}v$ with $T\bar{\zeta }{e}^{-it{k}^2}v$. On $r\ge 1$ their $e^{ikr}$ coefficients agree, and their difference, up to an irrelevant overall sign, is $\sqrt{2\mathrm{/}\pi }(2i{)}^{-1}{\int }_I{e}^{-ikr-it{k}^2}(1-\bar{\zeta }(k{)}^2)v(k)dk$. For $t\to +\mathrm{\infty }$ its phase derivative is $-r-2tk$, whose magnitude is at least a positive constant times $r+t$ on $I$. Repeated integration by parts, with no endpoint terms, bounds it by $C_N(r+t{)}^{-N}$; integration of the squared bound over $r\ge 1$ tends to zero for $N\ge 1$. On $0<r<1$ both packet integrals are $O(t^{-N})$ uniformly in $r$, by integrating against $-t{k}^2$; their amplitude derivatives are bounded uniformly on the finite rectangle. Thus $\mathrm{\parallel }{e}^{-it{h}_0}Sv-e^{-ith}T\bar{\zeta }v\mathrm{\parallel }\to 0$. Multiplying by $e^{ith}$ identifies the actual Cook wave as ${\mathrm{\Omega }}_{+}Sv=T\bar{\zeta }v$ on this dense packet class. This identification has used neither completeness nor a stationary/dynamical correspondence theorem.

For $L>0$ let $B_{L}u(r)=L^{-1\mathrm{/}2}u(r\mathrm{/}L)$; then $S{B}_L=B_{1\mathrm{/}L}S$. For a compact positive-energy packet, the last step gives $(B_L^{\ast }{\mathrm{\Omega }}_{+}{B}_{L}Sv)(r)=\sqrt{2\mathrm{/}\pi }\int s_{k\mathrm{/}L}(Lr)\overline{\zeta (k\mathrm{/}L)}v(k)dk$. On $r\ge 1\mathrm{/}L$ this equals $\sqrt{2\mathrm{/}\pi }(2i{)}^{-1}\int (e^{ikr}-{\overline{\zeta (k\mathrm{/}L)}}^2{e}^{-ikr})v(k)dk$, which converges in $L^2$ to $-iCv$: $\zeta (k\mathrm{/}L)\to i$ uniformly on the support of $v$, and Plancherel bounds the difference by the $L^2$ norm of its Fourier coefficient. On $0<r<1\mathrm{/}L$, $b(k\mathrm{/}L{)}^{-1}$ is uniformly bounded for large $L$, $\mathrm{\mid }\sin (q(k\mathrm{/}L)Lr)\mathrm{\mid }\le 1$ and $v\in L^1$, so this omitted piece is uniformly bounded with norm $O(L^{-1\mathrm{/}2})$; the cosine packet has the same bound there. Density and $\mathrm{\parallel }{\mathrm{\Omega }}_{+}\mathrm{\parallel }=1$ therefore give $B_L^{\ast }{\mathrm{\Omega }}_{+}{B}_L\to Q=-iCS$ strongly on all $L^2(0,\mathrm{\infty })$.

Define the Carleman operator $Au(r)={\int }_0^{\mathrm{\infty }}u(s)\mathrm{/}(r+s)ds$. Its norm is at most $\pi$: the weighted Cauchy–Schwarz/Schur estimate uses $w(r)=r^{-1\mathrm{/}2}$ and ${\int }_0^{\mathrm{\infty }}{s}^{-1\mathrm{/}2}\mathrm{/}(r+s)ds=\pi r^{-1\mathrm{/}2}$. For $M_{\epsilon }v(k)=e^{-\epsilon k}v(k)$, direct damped integration gives $(C{M}_{\epsilon }S+S{M}_{\epsilon }C)(r,s)=(2\mathrm{/}\pi )(r+s)\mathrm{/}((r+s{)}^2+{\epsilon }^2)$. The left side tends strongly to $CS+SC$. Testing compact functions supported away from zero, dominated convergence on the right and the Schur bound prove the bounded-operator identity $(CS+SC)\mathrm{/}2=A\mathrm{/}\pi$. Moreover $⟨Au,u⟩={\int }_0^{\mathrm{\infty }}\mathrm{\mid }{\int }_0^{\mathrm{\infty }}{e}^{-tr}u(r)dr{\mathrm{\mid }}^{2}dt$: for compact test functions use absolute Fubini and $1\mathrm{/}(r+s)={\int }_0^{\mathrm{\infty }}{e}^{-t(r+s)}dt$; the Schur bound extends this bounded Laplace map and its identity to all $L^2$, and each fixed-$t$ value agrees with the integral since $e^{-tr}\in L^2$. For the $u$ in step 6.2 its Laplace value at $t=1$ is $1\mathrm{/}4$, and it is continuous for $t>0$ because $t\mapsto e^{-tr}$ is continuous in $L^2$. The quadratic form is consequently strictly positive. Thus $\mathrm{Re}⟨CSu,u⟩=⟨Au,u⟩\mathrm{/}\pi >0$. If $Qu=u$, then $CSu=iu$ would make this real part zero. Therefore $(Q-I)u\mathrm{\ne }0$.

Put $L_n=2^n$ and $u_n=B_{L_n}u$. This bounded sequence is weakly null: for compact-annulus test functions the dilated and fixed supports eventually separate, and density and unitarity extend that conclusion to arbitrary $L^2$ pairs. Also $B_L$ commutes with $P_{+}$ and $B_L^{\ast }{h}_0{B}_L=L^{-2}{h}_0$. For any $f\in C_c(\mathbb{R})$, scalar dominated convergence gives $f(L^{-2}{h}_0)\to f(0)I$ strongly. Consequently, for $K=({\mathrm{\Omega }}_{+}-I)f(h_0)P_{+}$, $B_{L_n}^{\ast }K{B}_{L_n}u=(B_{L_n}^{\ast }{\mathrm{\Omega }}_{+}{B}_{L_n}-I)f(L_n^{-2}{h}_0)u\to f(0)(Q-I)u$. If $f(0)\mathrm{\ne }0$, step 10.1 makes its limit nonzero, so $\mathrm{\parallel }K{u}_n\mathrm{\parallel }\to \mathrm{\mid }f(0)\mathrm{\mid }\mathrm{\parallel }(Q-I)u\mathrm{\parallel }>0$. A compact operator sends bounded weakly-null sequences to norm-null sequences, contradicting this limit. Thus the radial defect is not compact; if the full three-dimensional defect were compact, its bounded radial compression $J^{\ast }{C}_{+}J=K$ would be compact. The displayed triangular $f$ has $f(0)=1$ and proves the explicit obstruction in the statement.

Step 1.2 proves (a), steps 3.1, 4.1 and 5.1 prove the full continuous-cutoff claim (b), and steps 3.2 and 4.2 prove (c) for every generic real bounded compact potential. Steps 5.2, 6.1, 6.2, 7.1, 8.1, 9.1, 10.1 and 11.1 prove the nonzero-threshold obstruction within that same potential class. All sequences were explicit or chosen from countable approximation families; no arbitrary-index choice, AC/DC Fredholm theorem, abstract Gelfand construction, Hilbert–Schmidt spectral theorem or completeness assumption was used.