Long-Range Potentials

A potential can tend to zero while leaving an infinite accumulated phase along an escaping trajectory. Ordinary free comparison then becomes the wrong asymptotic reference. Long-range scattering replaces it by a modified dynamics whose phase includes the persistent part of the interaction. The mechanism is proved here in an exactly solvable transport model and then related, with explicit limitations, to Coulomb scattering.

Helpful background. Free and Interacting Dynamics distinguishes individual unitary evolution from convergence of its comparison. Short-Range Potentials explains the integrable-tail criterion.

A vanishing force can leave a divergent phase

Along a nonzero-velocity ray $x(t)\simeq vt$, a tail $V(x)\sim c|x|^{-\rho}$ suggests

$$\int^t V(vs)\,ds \sim \begin{cases} \text{finite limit},&\rho>1,\\ (c/|v|)\log t,&\rho=1,\\ c|v|^{-\rho}t^{1-\rho}/(1-\rho),&0<\rho<1. \end{cases}$$

This estimate motivates the short/long-range boundary. It is not by itself a theorem about quantum wave operators: one must control packet spreading, slow velocities, and the residual interaction after modification.

The gradient of a Coulomb tail decays faster than its potential. Hence the momentum change along a distant trajectory can be integrable while its phase correction diverges. Small late-time acceleration does not establish ordinary free asymptotics.

An exact model with a logarithmic obstruction

Set $\hbar=1$ and take $H_0=P=-i\,d/dx$ on $H^1(\mathbb R)$. Let

$$H=P+v(Q),\qquad v(x)=\frac{\gamma}{\sqrt{1+x^2}}, \qquad \gamma\in\mathbb R\setminus\{0\}.$$

The perturbation is bounded and real, so $H$ is self-adjoint on $H^1$. Its evolution, obtained by solving the first-order transport equation along characteristics, is

$$(U(t)\psi)(x) =\exp\!\left[-i\int_{x-t}^{x}v(y)\,dy\right]\psi(x-t).$$

Substitution in $i\partial_t\psi=H\psi$ checks both the sign and the initial value. Undoing this evolution after free translation gives

$$(U(t)^*U_0(t)\psi)(x) =e^{\,i\gamma[\operatorname{arsinh}(x+t)-\operatorname{arsinh}(x)]}\psi(x).$$

As $t\to+\infty$, $\operatorname{arsinh}(x+t)=\log t+\log2+o(1)$ for fixed $x$. Thus the comparison contains the nonconvergent scalar factor $e^{i\gamma\log t}$.

This really prevents strong convergence. Compare times $t$ and $e^{\pi/|\gamma|}t$. Their phase ratio tends pointwise to $-1$, so dominated convergence gives a norm difference tending to $2\|\psi\|$ for every nonzero $\psi$. The comparison family is not Cauchy.

Removing the persistent phase restores a limit

For $|t|>1$, define the modified reference evolution

$$U_D(t)=U_0(t)e^{-i\gamma\,\operatorname{sgn}(t)\log|t|}.$$

It need not be a one-parameter group. For positive time, the exact calculation above gives

$$\operatorname*{s-lim}_{t\to+\infty}U(t)^*U_D(t) =M_{\exp[i\gamma(\log2-\operatorname{arsinh}x)]}.$$

For negative time the limit is multiplication by $\exp[-i\gamma(\log2+\operatorname{arsinh}x)]$. Both are unitary because their multipliers have modulus one. Dominated convergence proves each strong limit on all of $L^2$.

The model demonstrates three separate facts: the Hamiltonian is perfectly self-adjoint, ordinary wave operators fail, and an appropriate phase modification restores complete comparison. The failure was in the asymptotic reference, not in unitary time evolution.

This is a first-order transport model, not a proof of the Schrödinger Coulomb theorem. Its exactly known trajectories make the mechanism transparent.

Coulomb comparison uses a momentum-dependent phase

For $H_0=P^2$ in three dimensions, still in units $\hbar=2m=1$, the free velocity is $2P$. A Coulomb tail $g/|x|$ therefore suggests, on momenta away from zero,

$$U_D(t)=e^{-itP^2} \exp\!\left[-i\frac{g}{2|P|} \operatorname{sgn}(t)\log|t|\right], \qquad |t|>1.$$

The phase is defined by measurable functional calculus; its value at zero momentum is irrelevant to $L^2$ vectors, although estimates near zero remain delicate. The logarithm uses a fixed reference time in physical units. Changing that reference changes the asymptotic phase convention.

Dollard’s construction makes this Coulomb comparison rigorous. More general slowly decaying potentials require tailored eikonal or time-dependent modifiers and hypotheses controlling derivatives of the tail. A divergent Cook estimate alone does not identify the correct modifier or prove ordinary nonexistence.

The physical Coulomb amplitude and its forward singularity belong to Coulomb Scattering. Long-range modifications also affect stationary boundary conditions: a free outgoing spherical wave is no longer the complete far-field phase.

Exercises

1. In the transport model, compute the modified scattering operator $S_D=W_{D,+}^*W_{D,-}$.

Solution

Multiply the conjugate outgoing multiplier by the incoming multiplier. The $x$-dependent terms cancel, giving $S_D=e^{-2i\gamma\log2}I$. Its scalar phase depends on the chosen reference phase convention; its unitarity follows directly from the exact limits.

2. Why is convergence of $\int^\infty|\nabla V(vt)|dt$ insufficient for ordinary free wave operators?

Solution

It controls a classical momentum correction, while quantum comparison is sensitive to accumulated phase. For a Coulomb tail the gradient integral converges but $\int^\infty V(vt)dt$ diverges logarithmically. The exact transport model realizes this phase obstruction without any approximation.

References

• J. D. Dollard, “Asymptotic convergence and the Coulomb interaction,” Journal of Mathematical Physics 5, 729–738, 1964, doi:10.1063/1.1704171 — modified Coulomb asymptotics.
• D. Yafaev, “Lectures on scattering theory,” 2004, arXiv:math/0403213.