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My own attempt at this: the first result is wrong, and the second one is right but incomplete. Feynman: in the diagram, all the lines are external, so there is no propagator in the diagram. Therefore, Feynman rules give $\mathcal A=1$. Check up of this: In canonical quantisation, the amplitude is given by $\langle p|q\rangle=\langle 0|a_p ... 4 Yes, you can observe elastic proton-anti-proton scattering. Indeed, the cross section is not all that different from that of elastic proton-proton scattering. The main difference is that$pp$is predominantly elastic at low energy, and dominantly inelastic at high energy, whereas$p\bar{p}$is dominantly inelastic at all energies. Also note that inelastic ... 4 I haven't work out the math but looks like expression you get is correct, you just need to use the following identity $$f(x)\delta^n(x)=f(0)\delta^n(x)$$ Therefore $$(E-H)G=e^{ik|r-r'|}\delta(r-r')=\delta(r-r')$$ 3 Photo electric effect occur in bound electron, while compton effect occur in free electron. In photo electron effect, the photo and hence energy of the photon is absorbed by the electron. While in compton effect photon is scattered. 3 As what enters into the formula is$\boldsymbol q$instead of$\boldsymbol k$, I'd say we need a high$\boldsymbol q$(which, of course, implies a high$\boldsymbol k$, because of conservation of energy/momentum). For example, if$\boldsymbol k$is very high, but$\boldsymbol q$is not, this means that there was barely no scattering, which means you didn't ... 2 Raman scattering is inelastic scattering from molecules. The photon interacts with the molecule and changes the molecules vibrational, rotational or electron energy. Rayleigh scattering is in the main elastic scattering from small particles whose size is less than that of the wavelength of the photon. The scattering can occur of atoms or molecules and ... 2 Air is blue. White light is incident on it, and the scattered light that reaches your eye is blue. A leaf is green. White light is incident on it, and the scattered light that reaches your eye is green. Smoke is gray. Chlorine gas is yellow-green (I think. I really don't know what color chlorine is.) Air is blue, but only very faintly so. So faint ... 2 is Randall wrong? No, Randal is simplifying. His point is that the colour we perceive from various objects and materials is produced by a tremendous variety of physical phenomenon. Generally the lay public are not interested in the deeper causes. He gives the example of the green appearance of the statue of liberty. People who ask why it is green ... 2 In most situations we are familiar with when it comes to light in everyday life, the interaction with light and matter is linear with respect to the electromagnetic field. This means that if you have a material that scatters 50% of a beam of light, then a 2mW beam will scatter 1mW or a 100mW will scatter 50mW etc… In these normal everyday situations, ... 2 Compton scattering! The scattering becomes inelastic, such that the photon gives up some of its energy to the electron. NB An electron cannot absorb a photon because such a reaction cannot simultaneously conserve energy and momentum. 1 This is a degenerate case of the LSZ reduction formula which should not be used. Basically, the problem is that LSZ differentiates between 'incoming' and 'outgoing' particles, which must receive different treatment, but in your diagram, the single particle from$k$to$q$is both at once! Let's consider progressively simpler cases to find out what's going ... 1 Check out Halzen & Martin page 91. Supose you're doing electron-muon scattering. Pf is the electron momentum in the final state, and Pi electron momentum in the initial state. You are correct that the total momentum is conserved (it is 0 before and after, in the CM frame), but the momentum of each particle changes. 1 Anomalous thresholds are branch points in the complex plane of scattering amplitudes/Greens functions that are not easily deducible from unitarity. The book "The Analytic S-Matrix" by R.J. Eden explains it well, and shows how it may be deduced non-trivilaly from unitarity considerations. 1 Write$d^3 q = dq q^2 d\theta d\phi \sin\theta$and integrate over the angular variables. The only angular dependence in the integrand is in$e^{i \vec{q} \cdot ( \vec{x}-\vec{y}) } = e^{i q r \cos\theta}$where I've defined$r = | \vec{x} - \vec{y} |$. Then, we have $$\int_0^{2\pi} d\phi \int_0^\pi d\theta \sin\theta e^{i q r \cos\theta}$$ There is no ... 1 It says at the start of the solution the fraction of incident particles scattered through an angle greater than$\theta$is given by . . . . So that formula is being applied. It might well be that the formula was derived by first finding the fraction scattered from$0^\circ$up to and including$\theta$and then that fraction taken away from$1$to ... 1 The simplification follows from the theorem which states that if such operator is conserved in Heisenberg sense, $$\frac{d\hat{Q}}{dt} = \frac{\partial \hat{Q}}{\partial t} - \frac{i}{\hbar}[\hat{Q}, \hat{H}] = 0,$$ than it commutes with S-operator: $$[\hat{Q}, \hat{S}] = 0$$ So that these two operators can be diagonalized simulatenously: in ... 1 In the Wikipedia article about rutherford scattering the derivation of the scattering cross section $$\frac{d\sigma}{d\Omega} =\left(\frac{ Z_1 Z_2 e^2}{8\pi\epsilon_0 m v_0^2}\right)^2 \csc^4{\left(\frac{\Theta}{2}\right)}$$ is given. Let's rewrite that in your notation:$Z_1 = Z$,$Z_2 = 4$,$k = \frac{1}{4\pi\varepsilon_0}$and$KE = \frac{1}{2}m v^2\$: ...