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In hadron colliders (like LHC) the total cross-section is calculated using the optical theorem.

The origin of this is the scattering of two particles which is solved with the Schrödinger equation with a fixed potential V(r) (and this will be my question/doubt).

The optical theorem:

$$ \sigma_{\mathrm{tot}} = \frac{4\pi}{k}\mathrm{Im}f(0).$$

With the final state wavefunction:

$$ u_{\mathrm{f}} \rightarrow \exp(\mathrm{i}kz)+\frac{\exp(ikr)}{r}f(\theta, \phi). $$

But, how can one even use the optical theorem for HIGH-ENERGY hadron collisions where the Schrödinger equation can't even be used for relativistic moving particles?

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  • $\begingroup$ The optical theorem can be derived in relativistic and non-relativistic quantum mechanics. $\endgroup$ – bolbteppa Sep 21 '19 at 18:19
  • $\begingroup$ Yes. But in the lectures of my professor and from the books, they use the non-relativistic calculations to get to the optical theorem and then apply it to the total cross-section for high energy hadron collisions. The angular distribution function f(theta, phi) (which comes purely from the wave function of the non-relativistic Schrödinger equation) is used in the estimation of the total cross section. $\endgroup$ – winnetou Sep 21 '19 at 20:54
  • $\begingroup$ It is the forward scattering amplitude that defines the optical theorem, i.e.one angle,, the rest of the functional form is unimportant. The proof depends on plane waves (for classical EM) but the mathematics is the same for plane waves as probability waves in QM, in all the equations. Here it is discussed for feynman diagrams , certainly relativistic: th.physik.uni-bonn.de/nilles/exercises/DreesNillesSeminar06/… $\endgroup$ – anna v Sep 22 '19 at 4:32

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