orEcm ∼
√
2 EMc^2 for high energy. The linear energy dependence for a colliderwins over the
square root dependence of the fixed target experiment.
Again due to relativistic effects, the number of particles in ascattering process need not be
conserved. Any interaction causes an acceleration, and if the probe particle carries electric charge,
acceleration will read to electro-magnetic radiation. Quantum mechanically, the interacting charged
particle emits a photon. The photon was not part of either theincoming probe or of the target,
and it as not hidden inside these bodies either: the photon isgenuinely a newly created particle,
which did not exist in the system prior to interaction. Generally the number of particles may be
changed during the course of interactions, and this effect is not limited to photons. For example,
ine+e−colliders, an interaction may cause thee+e−to annihilate and be transformed purely into
energy out of which new particles may be created, such as quarks, proton and anti-protons,W±or
Z’s or even againe+e−pairs. These effects are all relativistic, and their quantum description will
properly require the formalism and tools of quantum field theory.
In this chapter, we shall limit attention to the non-relativistic case where the number of particles
is conserved. In this set-up, the incoming probe will merelybe deflected. This case is often referred
to aspotential scattering.
12.1 Potential Scattering
The fundamental problem of potential scattering is the resolution of the continuous part of the
spectrum of the Hamiltonian,
H=H 0 +λH 1 (12.1)Here,H 0 is theunperturbed Hamiltonian, whose solutions are assumed to be known exactly,H 1
is the perturbing Hamiltonian, andλis a small parameter, in terms of which perturbation theory
will be organized. The eigenstates ofH 0 will be denoted by|φk〉, and obey
H 0 |φk〉=E|φk〉 E=̄h^2 k^2
2 m(12.2)
Here and later, it will be convenient to parametrized the energy eigenvalueEby the wave vectork
and the massm. We then seek to solve for the eigenstates of the full HamiltonianHat the same
energy eigenvalueE,
(H 0 +λH 1 )|ψk〉=E|ψk〉 (12.3)The rationale for using the same energy eigenvalue is that ifH 0 has a continuous spectrum in
the vicinity of energyE, then a small (regular) perturbation on the spectrum shouldproduce a
continuous functional dependence onλ(see the discussion in bound state perturbation theory), so
that alsoHshould admit a continuous spectrum in the vicinity ofE. Possible exceptions would
include the case whereEis an endpoint of the spectrum ofH 0 , such as in the endpoint of an energy
band, in which case a more detailed study will be required.