Free and Interacting Dynamics

Free and interacting evolutions are individually unitary, yet their long-time comparison need not converge. The distinction is visible before solving any scattering problem: two self-adjoint Hamiltonians can differ by an arbitrarily small bounded operator and still have no ordinary wave operators for the chosen comparison. This page studies the comparison dynamics, its topology, and the role of reference channels.

Helpful background. Hamiltonians as Generators fixes the phase and domain conventions. Scattering Overview identifies the distinct existence and completeness questions.

The comparison operator carries interaction history

For self-adjoint $H,H_0$ on $\mathcal H$, set

$$F(t)=U(t)^*U_0(t) =e^{itH/\hbar}e^{-itH_0/\hbar}.$$

Every $F(t)$ is unitary. The existence of each factor follows from self-adjointness, even if $D(H)$ and $D(H_0)$ differ. This does not assert that $F(t)$ is differentiable on every vector.

If $H=H_0+V$ with bounded self-adjoint $V$, both operators have domain $D(H_0)$. Differentiation there, followed by the bounded integral extension, gives

$$F(t)\phi-\phi =\frac{i}{\hbar}\int_0^t e^{isH/\hbar}V e^{-isH_0/\hbar}\phi\,ds.$$

The integral is a strong Hilbert-space integral. For negative $t$ it is oriented in the usual way. This identity is the starting point of Cook’s Method, which develops the convergence criterion and its proof.

The comparison is generally not a one-parameter group: $F(t+s)$ need not equal $F(t)F(s)$. The two generators can fail to commute, and the comparison retains the order in which their motions are undone and applied.

Small perturbations need not have small accumulated effects

Take $H=H_0+cI$ with real $c\ne0$. Then

$$F(t)=e^{ict/\hbar}I.$$

For any nonzero vector, this has no limit as $t\to+\infty$ or $t\to-\infty$. A constant energy shift can be as small as desired in operator norm while producing an unboundedly accumulating relative phase.

Individual ray trajectories of $H$ and $H_0$ agree. Ordinary wave operators nevertheless compare vectors with a specified phase convention. This example is not a contradiction: scattering amplitudes need consistent asymptotic phases, and the comparison Hamiltonian should include a persistent common energy offset.

The same calculation explains why a finite-time estimate such as

$$\|U(t)-U_0(t)\|\leq\frac{|t|}{\hbar}\|V\|$$

does not prove a scattering limit. Its right-hand side grows with time. Scattering depends on the interaction experienced by escaping packets, not merely the size of $V$ as a bounded operator.

Strong limits are weaker than norm limits

A strong limit asks for convergence separately on every chosen packet. It need not be uniform over all normalized packets, including packets whose initial locations or momenta are chosen differently at every late time.

There is also an algebraic obstruction to norm convergence. A norm limit of unitaries is unitary: taking adjoints and multiplying preserves the identities $F(t)^*F(t)=F(t)F(t)^*=I$. A strong limit of unitaries need only be an isometry.

For scattering with purely absolutely continuous $H_0$, a complete wave operator maps $\mathcal H$ onto $\mathcal H_{\rm ac}(H)$. If $H$ has bound states, this is a proper subspace, so that wave operator cannot be an operator-norm limit of the unitaries $F(t)$. The use of a strong limit is therefore essential.

This argument does not prove that a norm limit exists when there are no bound states. It supplies a necessary condition, not a convergence theorem.

A reference channel can have its own Hilbert space

A useful generalization allows a comparison space $\mathcal H_0$ and an identification $J:\mathcal H_0\to\mathcal H$:

$$F_J(t)=e^{itH/\hbar}J e^{-itH_0/\hbar}.$$

In a many-particle problem, a channel Hamiltonian includes the internal energy of a bound cluster and the motion of separated clusters. “Free” then means free relative to that channel; it does not mean that every constituent has zero interaction.

For bounded $J$, the operators $F_J(t)$ are uniformly bounded. They are not automatically isometric. A sufficient asymptotic normalization condition is

$$\|J U_0(t)\phi\|\longrightarrow\|\phi\|$$

on the channel’s initial space. If a strong comparison limit exists there, this condition makes the limit isometric by continuity of the norm. The case $J=0$ shows why existence alone says nothing about preservation of probabilities.

Identifications, channel projections, and omitted channels must be specified before claiming completeness. This chapter’s basic wave-operator theorem uses the simpler same-space comparison $J=I$; the multichannel physics is developed in Multichannel Scattering.

Exercises

1. For bounded $V$, derive the finite-time norm estimate above without assuming $H_0$ is bounded.

Solution

Use the strong integral identity and unitarity to obtain $\|(F(t)-I)\phi\|\leq |t|\|V\|\|\phi\|/\hbar$. Taking the supremum over normalized $\phi$ bounds $\|F(t)-I\|$. Multiplication by $U(t)$ identifies this norm with $\|U_0(t)-U(t)\|$.

2. Suppose $F(t)$ converges strongly to an isometry $W$ with a proper closed range. Can $F(t)^*$ also converge strongly to $W^*$?

Solution

No. Strong convergence of both uniformly bounded factors would imply $F(t)F(t)^*\to WW^*$ strongly. The left side is $I$, whereas $WW^*$ is the proper range projection. On the range of $W$, the adjoints do converge to the inverse; the obstruction lies outside that 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.