Scattering Unitarity and Energy-Shell Operators

The scattering operator acts on free asymptotic data. Its familiar unitarity statement hides two precise questions: whether the incoming and outgoing wave operators have the same range, and how an operator on packets becomes an energy-dependent channel matrix. This page proves the range-defect criterion and explains the almost-everywhere meaning of the spectral decomposition. The general physical construction and the usual completeness-based unitarity proof belong to S-Matrix.

Required background. Wave Operators supplies isometries, reducing ranges, and spectral intertwining.

Range defects of the scattering operator

Let $W_-=W_{\rm in}$ and $W_+=W_{\rm out}$ be existing wave operators with common initial space $\mathcal H_0^{\rm ac}$. In this section they are regarded as isometries from that space, so $W_\pm^*W_\pm=I_0$. Define

$$S=W_+^*W_-,\qquad Q_\pm=W_\pm W_\pm^*.$$

The $Q_\pm$ are the orthogonal projections onto the respective interacting ranges $\mathcal R_\pm$. No completeness assumption has yet been made. Direct multiplication gives the positive defect identities

$$\begin{aligned} I_0-S^*S&=W_-^*(I-Q_+)W_-,\\ I_0-SS^*&=W_+^*(I-Q_-)W_+. \end{aligned}$$

For example,

$$\|\phi\|^2-\|S\phi\|^2 =\|(I-Q_+)W_-\phi\|^2.$$

Thus $S$ is always a contraction. It is isometric exactly when $\mathcal R_-\subseteq\mathcal R_+$: vanishing of the first defect means every incoming packet lies in the established outgoing range. Similarly, $SS^*=I_0$ exactly when $\mathcal R_+\subseteq\mathcal R_-$.

Consequently,

$$S\text{ is unitary on }\mathcal H_0^{\rm ac} \quad\Longleftrightarrow\quad \mathcal R_-=\mathcal R_+.$$

Equality of ranges is sufficient even when that common range is a proper subspace of $\mathcal H_{\rm ac}(H)$. Unitarity of $S$ alone therefore does not prove asymptotic completeness. Completeness supplies equality with a particular larger space, as explained in Asymptotic Completeness.

These defects concern missing dynamical ranges. They should not be confused with the deficit of one elastic channel inside a complete unitary multichannel matrix.

Strong energy conservation gives decomposability

Spectral intertwining gives, for every Borel set $B$,

$$E_{H_0}(B)S=SE_{H_0}(B)$$

on $\mathcal H_0^{\rm ac}$. This bounded projection identity is stronger and better specified than an unqualified formal commutator $[H_0,S]=0$.

Assume the Hilbert space is separable. In a spectral representation of the free absolutely continuous part,

$$\mathcal F_0:\mathcal H_0^{\rm ac} \longrightarrow\int_{\mathbb R}^{\oplus}\mathfrak h_E\,dE,$$

$H_0$ acts by multiplication by $E$. A bounded operator commuting with all its spectral projections is decomposable:

$$(\mathcal F_0S\mathcal F_0^{-1}a)(E)=S(E)a(E).$$

The measurable fiber space $\mathfrak h_E$ accounts for angular variables and open-channel multiplicity. A zero-dimensional fiber may be used outside the spectral support. The operator field $S(E)$ is defined only up to changes on an energy set of Lebesgue measure zero.

If $S$ is unitary, then $S(E)$ is unitary for almost every $E$. One way to see the qualification is to test $S^*S=I$ against measurable packet fields supported in arbitrary energy sets. These integrated identities force the fiber identities almost everywhere, not at each individually chosen energy.

Continuity of a selected representative of $S(E)$, and its value at a threshold, require additional stationary estimates. They do not follow from abstract decomposability.

A packet does not determine a null-set value

For a measurable family of outgoing event projections $P(E)$, a normalized incoming packet $a$ has event probability

$$\int_{\mathbb R} \langle S(E)a(E),P(E)S(E)a(E)\rangle_{\mathfrak h_E}\,dE.$$

Changing $S(E)$ at one energy leaves every such packet probability unchanged. A displayed exact-energy matrix is therefore a representative of an operator field, or a limit justified by extra regularity.

The free three-dimensional Hamiltonian $H_0=-\Delta$ illustrates the multiplicity. With unitary Fourier transform, an energy representation is

$$(\mathcal F_0f)(E,\omega) =\frac{E^{1/4}}{\sqrt2}\widehat f(\sqrt E\,\omega), \qquad E>0,\quad\omega\in S^2.$$

The factor follows from $d^3k=\frac12\sqrt E\,dE\,d\omega$. Its fiber is $L^2(S^2)$, not a single complex number. The trace interpretation of this formula for weighted vectors is developed in Stationary Scattering Theory.

Changing channel conventions

A measurable unitary basis change $B(E)$ used for both incoming and outgoing data gives

$$S_{\rm new}(E)=B(E)S(E)B(E)^*.$$

Different incoming and outgoing conventions instead give $B_{\rm out}(E)S(E)B_{\rm in}(E)^*$. Measurable unitarity and packet probabilities transform consistently when states and measurement projections are transformed as well.

Channel rephasing changes individual matrix elements. It does not change the underlying scattering map. Amplitude normalization, partial-wave eigenvalues, and flux factors are developed in the physical treatment of the S-matrix.

Exercises

1. If $\|S\phi\|=\|\phi\|$ for one incoming packet, what follows?

Solution

The first defect identity shows $W_-\phi\in\mathcal R_+$. That packet has an outgoing representative $S\phi$. Nothing follows about other packets or about completeness of either range.

2. Can an abstract unitary scattering operator have a representative with $S(E_*)=0$?

Solution

Yes. Redefine any measurable representative at the single energy $E_*$. This does not change its direct-integral operator. The new field is still unitary almost everywhere. If a stationary theorem selects a continuous representative near $E_*$, this arbitrary change destroys that continuity and is no longer the selected representative.

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.