Rigorous Form of the Lippmann–Schwinger Equation

The Lippmann–Schwinger equation is not an ordinary inverse of $E-H_0$ on $L^2$ when $E$ lies in the continuum. A rigorous construction must specify the boundary value, the function space, and the uniqueness condition. For a real bounded compactly supported potential in three dimensions, these requirements reduce to an invertible Fredholm equation on the interaction region.

Required background. Stationary Scattering Theory explains weighted resolvent boundary values. The physical amplitude derivation remains with Lippmann–Schwinger Equation.

A local Fredholm equation for the scattering solution

Use $\hbar=2m=1$, $H_0=-\Delta$, and real $V\in L^\infty(\mathbb R^3)$ supported in a ball $B_R$. Fix $k>0$ and an incident direction $\omega\in S^2$. Write

$$\phi(x)=e^{ik\omega\cdot x},\qquad E=k^2.$$

Let $R_0(E+i0)$ denote the outgoing free resolvent in the convention $(H_0-z)^{-1}$. Its kernel is $e^{ik|x-y|}/(4\pi|x-y|)$, with derivation and physical sign conventions in Green Function for Scattering.

On $X=L^2(B_R)$ define

$$(K_k f)(x)=\int_{B_R} \frac{e^{ik|x-y|}}{4\pi|x-y|}V(y)f(y)\,dy, \qquad x\in B_R.$$

Solve

$$(I+K_k)u_R=\phi|_{B_R}.$$

Once $u_R$ is known, define the global distributional solution by

$$u=\phi-R_0(E+i0)(Vu_R),$$

where $Vu_R$ is extended by zero outside the ball. Restriction to $B_R$ reproduces $u_R$. Applying $H_0-E$ distributionally gives $(H-E)u=0$, and the scattered term has the outgoing radiation condition.

This formulation does not require the incident plane wave to belong to $L^2$. It is square-integrable on the bounded interaction region, which is the space where the equation is solved.

Compactness follows from the kernel singularity

The kernel of $K_k$ is square-integrable on $B_R\times B_R$. Near the diagonal its squared magnitude behaves like $|x-y|^{-2}$, while the three-dimensional volume element supplies $r^2dr$. Thus the local singularity is integrable, and $K_k$ is Hilbert–Schmidt and therefore compact.

Consequently $I+K_k$ is Fredholm of index zero. To prove invertibility it remains to show that its kernel is zero. Compactness alone does not accomplish this: it reduces the infinite-dimensional problem to a uniqueness question.

Outgoing uniqueness removes the Fredholm obstruction

Suppose $(I+K_k)f=0$. Define $w=-R_0(E+i0)Vf$. Then $w=f$ in the ball, $(H-E)w=0$, and $w$ is outgoing outside the potential support.

For a large sphere, Green’s identity and reality of $V$ imply

$$\operatorname{Im}\int_{|x|=r}\overline w\,\partial_r w\,dS=0.$$

The outgoing expansion has the form $w(r\theta)=r^{-1}e^{ikr}a(\theta)+O(r^{-2})$. Taking the limit in the flux identity gives $k\|a\|_{L^2(S^2)}^2=0$. Since $k>0$, its far-field amplitude vanishes.

Rellich’s uniqueness theorem for the exterior Helmholtz equation then gives $w=0$ outside the ball. Unique continuation for the Schrödinger equation with bounded potential gives $w=0$ everywhere. Hence $f=0$.

These last two PDE uniqueness results are substantive inputs; the argument identifies exactly where they are needed. The Fredholm alternative now proves existence and uniqueness of $u_R$ for every $k>0$. The argument does not include $k=0$, complex absorbing potentials, or long-range tails.

The solution space and the approximation space differ

The scattered field belongs to $L^2_{-s}$ for every $s>1/2$. The full solution includes the plane wave and belongs to $L^2_{-s}$ for $s>3/2$. Confusing these two weight conditions obscures what the limiting absorption principle actually estimates.

A Neumann expansion

$$u_R=\sum_{n=0}^\infty(-K_k)^n\phi|_{B_R}$$

is justified if $\|K_k\|<1$. The Fredholm proof gives invertibility even when this norm condition fails; solvability is therefore more general than convergence of a Born series.

For example, the Hilbert–Schmidt estimate gives

$$\|K_k\|\leq\sqrt{\frac23}\,\|V\|_\infty R^2.$$

Indeed, for each $y\in B_R$, enlarge the translated integration region to a ball of radius $2R$ when integrating $|x-y|^{-2}$. This produces the displayed sufficient bound. It is not an optimal coupling threshold and has no high-energy improvement because the absolute-value estimate discards oscillation.

A residual gives a meaningful numerical error

For an approximate local solution $u_{\rm app}$ define $r=(I+K_k)u_{\rm app}-\phi|_{B_R}$. Then

$$\|u_{\rm app}-u_R\|_X \leq\|(I+K_k)^{-1}\|\,\|r\|_X.$$

A small residual alone is insufficient without control of the inverse. A large inverse norm near a narrow resonance can amplify discretization or quadrature errors, even though every positive real energy remains uniquely solvable in this model.

Exercises

1. Why does the flux proof not exclude a zero-energy resonance?

Solution

The positive factor $k$ vanishes at threshold, and the oscillatory radiation asymptotics change. The identity no longer forces the coefficient of the threshold tail to vanish. Zero-energy uniqueness needs its own hypotheses.

2. Under $\|K_k\|=q<1$, bound the error after retaining terms $0$ through $N$ of the Neumann series.

Solution

The omitted norm is at most $q^{N+1}\|\phi|_{B_R}\|/(1-q)$ by the geometric-series bound. This is a controlled local error in $L^2(B_R)$; further mapping estimates translate it into an error in the scattered field or far-field amplitude.

References

• D. Yafaev, “Lectures on scattering theory,” 2004, arXiv:math/0403213.
• M. Zworski, “Mathematical study of scattering resonances,” Bulletin of Mathematical Sciences 7, 1–85, 2017, doi:10.1007/s13373-017-0099-4, author text.