Asymptotic Completeness

Existence of wave operators constructs interacting states from free data. Asymptotic completeness asserts that these data account for every state in the intended scattering sector. This page proves an inverse-limit criterion, gives an explicit incomplete comparison with a unitary scattering operator, and states how a short-range completeness theorem closes the gap.

Required background. Wave Operators proves that the ranges are closed reducing subspaces of $\mathcal H_{\rm ac}(H)$.

Completeness is a statement about the range

Let $W_\pm=W_\pm(H,H_0)$ exist on $\mathcal H_{\rm ac}(H_0)$. We call them asymptotically complete on the absolutely continuous sector when

$$\operatorname{Ran}W_\pm=\mathcal H_{\rm ac}(H).$$

Equivalently,

$$W_\pm W_\pm^*=P_{\rm ac}(H).$$

A complete wave operator is a unitary equivalence between the absolutely continuous restrictions of $H_0$ and $H$. It is generally only a partial isometry when extended by zero to the entire reference Hilbert space.

Terminology varies. Teschl distinguishes completeness onto $\mathcal H_{\rm ac}(H)$ from the stronger exhaustion of $\mathcal H_{\rm c}(H)$. The latter includes singular continuous states and implies their absence in this comparison. Our displayed convention follows the common absolutely continuous formulation; whenever absence of singular continuous spectrum is used, it is stated separately.

For $\Psi\in\mathcal H_{\rm ac}(H)$, completeness supplies unique incoming and outgoing vectors $\phi_\pm=W_\pm^*\Psi$. The uniqueness follows from isometry, not from a choice of stationary eigenfunction normalization.

Reverse wave operators characterize completeness

Suppose $W_\pm(H,H_0)$ exists. It is complete if and only if

$$\widetilde W_\pm =\operatorname*{s-lim}_{t\to\pm\infty} e^{itH_0/\hbar}e^{-itH/\hbar}P_{\rm ac}(H)$$

exists on all of $\mathcal H$.

First assume completeness. Every $\Psi$ in the interacting absolutely continuous sector is $W_\pm\phi$. The inverse-limit calculation on the wave-operator page shows that the reverse comparison converges to $\phi$. It vanishes on the complementary initial sector because of its projection. Thus the reverse limit exists and equals $W_\pm^*$.

Conversely, suppose both limits exist. Test the reverse limit against $\phi\in\mathcal H_{\rm ac}(H_0)$. Moving the finite-time unitaries across the inner product and taking the forward limit gives

$$\widetilde W_\pm\Psi=W_\pm^*\Psi, \qquad \Psi\in\mathcal H_{\rm ac}(H).$$

The reverse wave operator has its range in $\mathcal H_{\rm ac}(H_0)$ by spectral intertwining, so the tested identity determines it fully. Reverse isometry then gives

$$\|W_\pm^*\Psi\|=\|\Psi\|.$$

Since $W_\pm W_\pm^*$ is an orthogonal projection, this equality forces $\Psi$ into $\operatorname{Ran}W_\pm$. The desired range equality follows.

This criterion explains why a proof controlling only free trajectories may establish existence without completeness. The reverse problem controls the interacting trajectories.

A unitary scattering operator can miss a whole continuum

Let $\mathcal K=L^2(0,1)$ and let $A$ be multiplication by $x$. On $\mathcal H=\mathcal K\oplus\mathcal K$, choose

$$H_0=A\oplus0,\qquad H=A\oplus A.$$

The first summand of $H_0$ is absolutely continuous; the second is a zero-energy eigenspace. Hence

$$P_{\rm ac}(H_0)=I_{\mathcal K}\oplus0,\qquad P_{\rm ac}(H)=I_{\mathcal H}.$$

After applying the initial projection, the comparison operator is constant:

$$e^{itH}e^{-itH_0}P_{\rm ac}(H_0)=I_{\mathcal K}\oplus0.$$

Both wave operators exist with the same proper range, the first copy of $\mathcal K$. Their scattering operator is the identity on the reference absolutely continuous space. It is unitary, yet the entire second interacting continuum is missing.

This elementary example isolates a channel-selection problem. No norm calculation of $S^*S$ can recover a channel that was absent from the initial comparison space.

A short-range theorem supplies the missing exhaustion

For $H_0=-\Delta$ on $L^2(\mathbb R^d)$ and real bounded $V$ satisfying

$$|V(x)|\leq C\langle x\rangle^{-1-\epsilon}, \qquad\epsilon>0,$$

the standard short-range theorem gives existence and completeness of $W_\pm(H_0+V,H_0)$. The associated spectral result excludes singular continuous spectrum. A sufficient operator-tail version and the proof strategy are developed in Short-Range Potentials.

The proof requires propagation information beyond compactness of a resolvent difference. In an Enss approach, energy-localized interacting states are split into incoming and outgoing spatial-motion sectors; decay and compactness estimates identify their distant evolution with free motion. The resulting exhaustion is a theorem, not a definition of the word “scattering.”

For many-particle systems, a single completely separated channel is generally insufficient. Bound subsystems create additional asymptotic channels. A completeness statement must name the direct sum of all included cluster channels.

Exercises

1. Under completeness and absence of singular continuous spectrum, describe an arbitrary initial vector.

Solution

It decomposes orthogonally as $\Psi=\Psi_{\rm pp}+W_\pm\phi_\pm$. The pure point part evolves through its eigenvalue phases; the other part has the stated free asymptote in the chosen time direction. This need not be a finite sum of bound levels, and no uniform error rate follows from strong convergence alone.

2. In the two-copy example, can adding more accurate phase shifts to the first channel restore completeness?

Solution

No. Its established range remains in the first summand. One must enlarge or replace the comparison channels to describe the second continuum. Improving amplitudes within an incomplete channel space does not change its range.

References

• G. Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014, Chapter 12, author’s text.
• D. Yafaev, “Lectures on scattering theory,” 2004, arXiv:math/0403213.