Cook’s Method

Cook’s method proves existence of a wave operator by integrating the interaction along a free trajectory. It is a sufficient criterion with an explicit error bound. Its strength is that only free evolution needs to be estimated; its limitation is that the same calculation generally does not establish completeness.

Required background. Wave Operators explains strong limits and extension from a dense set. Helpful background. Domains of Hamiltonians supplies conditions making a perturbation self-adjoint on the free domain.

An integrable interaction makes the comparison Cauchy

Let $H_0,H$ be self-adjoint and suppose $D(H_0)\subseteq D(H)$. Write $V=H-H_0$ on $D(H_0)$. For a vector $\phi\in D(H_0)$, assume

$$\int_0^\infty\|V e^{-itH_0/\hbar}\phi\|\,dt<\infty.$$

Then the outgoing comparison limit exists on $\phi$, and for $T\geq0$,

$$\left\|W_+\phi-e^{iTH/\hbar}e^{-iTH_0/\hbar}\phi\right\| \leq\frac1\hbar\int_T^\infty \|V e^{-itH_0/\hbar}\phi\|\,dt.$$

Here $W_+\phi$ denotes the limit on this vector; if the criterion holds on a dense subset of $\mathcal H_{\rm ac}(H_0)$, the wave operator exists on that entire initial space. The incoming version integrates over $(-\infty,0]$.

To prove the claim, put $F(t)=e^{itH/\hbar}e^{-itH_0/\hbar}$. On the specified vector,

$$\frac{d}{dt}F(t)\phi =\frac{i}{\hbar}e^{itH/\hbar}V e^{-itH_0/\hbar}\phi.$$

The domain inclusion is important. It makes $H$ a bounded map from $D(H_0)$ with its graph norm into $\mathcal H$, by the closed graph theorem. Free evolution is continuous in that graph norm, so the derivative and its integral are well-defined.

For $t_2>t_1$,

$$\|F(t_2)\phi-F(t_1)\phi\| \leq\frac1\hbar\int_{t_1}^{t_2} \|V e^{-itH_0/\hbar}\phi\|\,dt.$$

The integrable tail makes $F(t)\phi$ Cauchy. Letting $t_2\to\infty$ gives the displayed error bound. Uniform boundedness of $F(t)$ then extends convergence from a dense set, as proved on the wave-operator page.

A complete existence proof in three dimensions

Choose units $\hbar=2m=1$, so $H_0=-\Delta$ on $H^2(\mathbb R^3)$. Let $V$ be real with $V\in L^\infty(\mathbb R^3)\cap L^2(\mathbb R^3)$. Bounded perturbation theory makes $H=H_0+V$ self-adjoint on $H^2$.

For $\phi$ in the Schwartz space, the free dispersive estimate gives

$$\|e^{-itH_0}\phi\|_\infty \leq (4\pi|t|)^{-3/2}\|\phi\|_1, \qquad t\ne0.$$

The free kernel underlying this estimate is derived in Free-Particle Propagator. Hölder’s inequality now yields

$$\|V e^{-itH_0}\phi\|_2 \leq (4\pi|t|)^{-3/2}\|V\|_2\|\phi\|_1.$$

This bound is integrable for $|t|\geq1$. Near zero, use the different estimate $\|V e^{-itH_0}\phi\|_2\leq\|V\|_\infty\|\phi\|_2$, avoiding the spurious short-time singularity of the dispersive bound.

Schwartz functions are dense in $L^2$, and the free Hamiltonian is purely absolutely continuous. Both wave operators therefore exist on all of $L^2(\mathbb R^3)$. For $T\geq1$, the explicit outgoing error is bounded by

$$2(4\pi)^{-3/2}\|V\|_2\|\phi\|_1\,T^{-1/2}.$$

The rate applies to this class of packets; it is not an operator-norm convergence estimate. The $L^1$ norm of a normalized packet is not uniformly bounded over the $L^2$ unit sphere.

What the estimate does not establish

The hypotheses above are convenient sufficient assumptions. They are not the optimal class of scattering potentials. For example, a bounded tail $\langle x\rangle^{-\rho}$ with $1<\rho\leq3/2$ is short range in three dimensions but is not in $L^2$. Short-Range Potentials uses velocity localization or stationary estimates to reach that class.

Failure of the Cook integral to converge is not a proof that a wave operator fails to exist. Absolute integrability is sufficient for convergence, while oscillatory cancellation can make some limits exist without an integrable derivative norm.

Nor does forward Cook control prove that every interacting absolutely continuous state is in the range. A reverse criterion would require information about $V e^{-itH}\Psi$, which is harder because the interacting evolution is the unknown object. Asymptotic Completeness explains this distinction.

Exercises

1. Why does the same $L^2$-potential dispersive argument not directly prove existence in dimensions one and two?

Solution

The free $L^1$-to-$L^\infty$ estimate decays like $|t|^{-d/2}$. Its tail is integrable only for $d>2$. The failure is in this estimate, not necessarily in scattering. Restricting free momentum away from zero and using spatial localization yields stronger estimates for suitable short-range potentials.

2. Suppose $\|V U_0(t)\phi\|\leq C(1+t)^{-1-\epsilon}$ for $t\geq0$. What error does Cook’s argument give?

Solution

Integrating the tail gives $C[\hbar\epsilon]^{-1}(1+T)^{-\epsilon}$. The estimate controls the norm difference between the limiting interacting packet and its finite-time comparison. If the constant depends on $\phi$, it does not establish a uniform operator-norm rate.

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.