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What does it tell us about a process, say A+B->C+D, if the calculated total cross section is infinite?

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It means that the particles A, B always influence one another: there is a "long-range force" (meaning "infinite range force") in between them. However, when the impact parameter is very high – the impact parameter is the distance between the straight lines given by extrapolating the initial motion of A, B - then the interaction between A, B is very weak.

The Coulomb interaction (repulsive or attractive electromagnetic force) is a typical example of a long-range force, one that leads to an infinite cross section. However, the typical angle by which the particles change their direction of motion goes like $$ \theta \sim \frac{q_1 q_2}{r} $$ where $q_1,q_2$ are the charges and $r\equiv b$ is the impact parameter. Moreover, the Coulomb force also leads to the production of an infinite number of very-low energy photons (because of the accelerating charges).

This behavior should be contrasted with a finite cross section due to "short-range forces". In classical physics, this is formally impossible for "total cross sections" as long as the force is nonzero for any separation. However, in quantum physics, the influence of one particle on another one is typically "finite", strictly separated from zero, but it has a low probability. So in quantum mechanics, many forces have a finite cross section which means that there is a high probability that the interaction between the particles makes exactly zero impact, even if the force is classically non-vanishing for arbitrarily high separations. When the decrease of the potential is faster than $1/r$, the probability that there is no interaction at all goes to 100% for large enough impact parameter.

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You should use the definition of the cross section: it determines how many particles $N$ of the incident flux $j$ are scattered per second. So initially you have already an infinite number of particles crossing each second a plane perpendicular to their motion direction. If they all are deviated with the scattering potential, you have all of them scattered. To get such an infinite number from a finite value of the flux $j$, you should have an infinite coefficient sigma: $N_{scattered}=\sigma\cdot j$. For a finite-range force ($F(r>R)=0$ only a finite number of particles is scattered ($N=j\cdot \pi R^2 $).

The same is applicable in case of reactions. Just replace the word "scattering" or "deviation from the initial direction" with "reaction" (transformation of particles).

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Actually, when I was studying quantum mechanics, I was surprised that some everywhere nonzero potentials produce only a finite total cross section.

It is intriguing because classically, for such potentials, any particle injected will be scattered, although might with a very small deflection angle.

I mean, it is not surprising to have an infinite total cross section, but a finite one.

The problem is on the Plank constant $\hbar$. Quantum mechanically, the total cross section involves $1/\hbar$ generally. In the limit of $\hbar \rightarrow 0$, the total cross section diverges.

For the coulomb case, $\hbar$ does not appear! The quantum mechanical results agree with the classical results.

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