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You only need to include $u$ and $t$ channels when the final particles are indistinguishable (classically thinking, you don't know which blip in your detector came from which initial particle). Since $e^-$ and $e^+$ are distinguishable, you will always know which one is which and only $t$ channel contributes to the scattering amplitude.


2

The solution you give is unphysical, because it cannot be normalized (as you noticed). This happens a lot in physics (e.g. try solving the wave equation in cylindrical coordinates: you'll get Bessel functions that rise to infinity at the origin, which you'll also discard if the origin is part of the solution domain). Unphysical solutions are discarded, and ...


1

That particular plot (source) seems to be a plot of the number of neutron captures as a function of energy. Each narrow peak in neutron energy corresponds to the energy of an excited state in the Th-233 or U-239 nucleus --- that is, the target nucleus with an extra neutron. The size of the capture cross section, plus the properties of the decay radiation ...


1

I am not aware of an elegant answer to this question, but writing out the expansion of the fields $\psi$ in creation- and annihilation operators gives an answer. $\psi$ is a complex scalar with $$ \psi(x)=\int\frac{d^3 k}{(2\pi)^3\sqrt{2E_{\vec{k}}}}\left(b_{\vec{k}}e^{-ikx}+c_{\vec{k}}^\dagger e^{ikx}\right) $$ The initial and the final state only contain ...


1

You are essentially correct. If you start on the real line with the Schrödinger equation $$ \left(-\frac12 \frac{\partial^2}{\partial x^2} + V(x) \right ) \psi(x) = E\,\psi(x),$$ then at every point $x_0$ where $V(x)$ is analytic you are guaranteed that $\psi(x)$ will be analytic in a neighbourhood of that point. However, if $V(x)$ is not analytic then you ...



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