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The key difference between treatments of quantum mechanical scattering and classical scattering is the nature of the incident particle. In the quantum mechanical case, the incident particle is typically treated as a delocalized plane wave, whereas in the classical case the particle is treated as a point particle. The delocalized nature of the plane wave is ...


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The answer, given by @user196574 is exshaustive, I only want to add, that integration over all space arises solely due to the choice of basis solution - a plane wave, with a fixed momentum, which exists everywhere in space. If you want to consider a scattering of localized object, for example, a Gaussian wavepacket, then performing first integral over ...


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Kinematics, mostly. More specifically, the final-state particles generally are moving away from each other, and so are unlikely to collide, whereas the initial-state particles generally are moving toward each other, so they are likely to collide. Note that there are, in fact, situations in which you do see the final-state particles return to the initial ...


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Compton scattering is scattering of photons off the free electrons. Raylegh scattering is between a photon and an electron bound in an atom, when the photon energy is smaller than the transition energy: $$\hbar\omega < E_b - E_a = \hbar\omega_{ba}.$$ Resonant scattering when $$\hbar\omega = E_b - E_a = \hbar\omega_{ba}$$ (up to the precision of the level ...


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There is not one quantum field theory. One can use an infinite potential with a harmonic oscillator and create a theory of creation and annihilation operators, but to identify the mathematical states with physical states, and in your question, particles, needs clear definitions. With the quantum field theory used in elementary particles, there are no ...


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There is a probability that any of the electrons are knocked out. However only the most energetic photons resulting from electron transitions would be classed as being in the X-ray part of the electromagnetic spectrum. If the charge on the nucleus is low, eg for a hydrogen atom, the transitions to the inner most energy levels result in the emission of UV ...


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Yes, these reactions occur at the same probability, given the same kinematics. But even with interactions occurring at the same rate, we still expect to see some of the final state. If the interactions $\rm A\to B$ and $\rm B\to A$ occur with the same probability, but you start out with 100 of particle A and 0 of particle B, then obviously the interaction $\...


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If I'm understanding your question correctly, the answer is yes - assuming a plane wave solution would yield a wavefunction of the form $$\psi_k(\mathbf r,t) = e^{i(kz-\omega_k t)} + f_k(\theta) \frac{e^{i(kr-\omega_k t)}}{r}$$ so a wavepacket solution would take the form $$\Psi(\mathbf r,t) = \underbrace{\int d^3\mathbf k \ A(\mathbf k) e^{i(\mathbf k \...


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Quantum mechanical scattering theory is a difficult compromise between wave phenomena on the one hand, and the classical view of particle scattering on the other. Just like an electromagnetic wave, the scattering solution exists everywhere in space - what really makes it different from an eigenvalue problem is that it is a propagating (rather than a standing)...


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As far as I can tell, the context for BH scattering is actually not relevant, so I will answer within the context of a scalar field theory. Question 2 is actually fairly easy to answer. You should think of $\hbar$ in this context as a formal expansion parameter, and not a constant of nature. You may recall, that in ordinary perturbation theory, a common ...


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In a typical collider experiment, two particles, generally in approximate momentum eigenstates at $t = -\infty$, are collided with each other and it is measured the probability of finding particular outgoing momentum eigenstates at $t = +\infty$. In the Heisenberg picture the probability amplitude for the initial states $\vert i \rangle$ to evolve to the ...


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Gauge invariance requires that the amplitude vanishes if you replace the polarization with its corresponding momentum. If we replace $\epsilon_4 \to p_4$, we get \begin{align} {\cal A} &\to 2 e^2 \left[ \frac{ p_3 \cdot \epsilon_1 p_2 \cdot p_4 }{ p_2 \cdot p_4 } - \frac{ p_2 \cdot \epsilon_1 p_3 \cdot p_4}{ p_2 \cdot p_1 } + \epsilon_1 \cdot p_4 \right]\...


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If we are talking specifically about making a proton in a nucleus go away, one way to do it is by smacking an electron into a proton, which changes it into a neutron and a neutrino. In this manner the proton "goes away", but not by being knocked loose. Neutrons can be knocked loose from some nuclei by striking them with high-energy protons, a ...


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