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1

Simple! If $\epsilon < 1$ then it is a circle or a ellipse. In this case, happened no scattering. On the contrary, the particle were absorbed. If $\epsilon = 1$ it is a parabola. There's only one value of energy (for a given angular momentum) for the orbit to be a parabola. Therefore, mathematically, the probability of the orbit is a parabola it is 0%. ...


0

It is not clear if you are looking for textbooks specifically dedicated to scattering in QFT (the title says so, but then you make reference to Peskin which is not), so going for good expositions inside more generally broaded textbooks: Chapter 17 of Robert D. Klauber, Student Friendly Quantum Field Theory (also see here): Aside for a summary of ...


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I would like to point out this video by minutephysics: https://www.youtube.com/watch?v=R5P6O0pDyMU so far the better and simpler explanation of the color of the sky. It takes into account also color theory (not particle physics of course) and not just scattering.


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quantum theory of light Yes, there exists a quantum theory of light which turns into an operator form Maxwell's equations and gives a wave function to a photon. The classical electromagnetic wave emerges from a confluence of individual photons with energy E=h*nu , the frequency of the classical wave. It is not a simple formulation, as one can see in ...


1

Well, the general theory is that you define the generalized wave operators for any couple of self-adjoint operators $A$ and $B$ to be: $$\Omega^{\pm}=\mathrm{s-lim}_{t\to \mp\infty} e^{iAt}e^{-iBt}P_{ac}(B)\; ,$$ where $\mathrm{s-lim}$ stands for the limit in the strong operator topology, and $P_{ac}(B)$ is the projection on the absolutely continuous ...


2

The overall idea is the following. You have a state evolving with the full interacting theory $\Psi(t) = U_t\Psi$. If the theory admits an asymptotic description for $t\to -\infty$, there must be a state $\Psi_0$ evolving with the free theory $\Psi_{0}(t) = U_0(t)\Psi_0$ such that the two evolutions ``coincide'' at large time in the past: $$\lim_{t\to ...


0

Yes it makes sense, and I guess you are asking for a clarification as to the physical properties of such limits. In ordinary 1D scattering the Moller operators $\widehat{\Omega}_{\pm}$ defined by the strong limits \begin{equation} \widehat{\Omega}_{\pm} = \lim_{t \rightarrow \mp \infty}e^{i \widehat{H}t/\hbar}e^{-i \widehat{H_{0}}t/\hbar} \end{equation} link ...


2

One has to be careful when extending the concept of scattering length to lower dimensions. A straightforward extension of the 3D methodology to 2D is prone to lead to logarithmic divergences. Reason being that in 2D the radial Schrödinger's equation for the s-wave includes a negative centrifugal potential. When carefully defining the scattering length as ...


5

Scattering length $l_s$ of a particle scattering on a potential $\hat{V}(\textbf{x})$ can actually be defined in a general $d$ dimensions case, using Green's functions formalism. Studying a scattering problem consists basically in solving the Schrödinger equation (taking $\hbar=m=1$): $$ ...


3

Yes, a potential that is nowhere negative must necessarily have a non-negative scattering length. This follows from the Riccatti equation describing the relationship between scattering lengths for potentials with variable cut-off ranges. The book "Variable phase method approach to potential scattering" by Calogero gives all details needed. Physically what ...



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