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PDG is discussing charged current interactions, $\nu N \to \mu^- X$ and $\bar\nu N \to \mu^+ X$. These are not charge conjugate processes. The neutron is $udd$. With the neutrino, $\nu$, we need a $W^+$ for the charged current interaction, $$\nu \to W^+ \mu^-$$ We then need $$W^+ d \to u$$ Note that the neutron contains two down quarks. The case of the ...

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Your question covers numerous topics, but I will try to answer several of them here. For a proton (I am most familiar with this case), the proton asymptotically becomes a black disk Ref. The disk part is because it is Lorentz contracted at the collision point. The same process happens for nuclei as well (see the bottom left of here for example). The "black" ...

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Since the 4-momentum must be conserved, there is no angular dependence in a 2-body decay in the CM frame. But If you are asking yourself what is the phase-space of such process, you should take a look at equation 4.72 of the book I mentioned in the comments: $$\Gamma(M\to 1 + 2)=\dfrac{|\vec{p_1}|}{32 \pi^2 M^2} \int \mathrm{d}\Omega ... 0 Okay, I think I got it after reading a lot of scattering-related papers. I will omit the wavefunctions normalization and the constants like \hbar, m, \pi, etc., where possible since the precise values are not so relevant. Consider a 2D rectangular pipe (for simplicity), infinite in x direction and of finite height H in y direction. Under the ... 2 String theory reduces to ordinary field theory in the infinite string tension limit. In this limit, the massive modes are decoupled and we are left with solely, the massless modes. In fact, Bern and Kosower (Please, see a modern review by Christian Schubert. ) proved that the computation of the field theory amplitudes from the string amplitudes at the ... 7 Surface coating of an integrating sphere is optimized for low losses. This white coating (barium sulfate or PTFE) acts like an ideal lambertian scatterer. all light is scattered (Ok, not 100%, but a very high percentage like 99,5%. See ressources) it is emitted in the hemisphere following the cosine law: perpendicular to the surface it's highest. ... 3 In a sphere, any light emitted from the center will reflect off the sides at normal incidence come back to the center. In a cube, some rays never return to the center, so you aren't measuring all of the light emitted, which defeats the purpose of the device. 0 The Coulomb logarithm is a heuristic cutoff. For length scales beyond the Debye radius, electrons in a plasma see a smoothed electric field, not the 1/r potential of the neighboring electrons. Hence, when computing two-body scattering, for electrons with an impact parameter too far out, that particular charge will be screened and cut off. Thus, in any ... 0 See 't Hooft and Veltman's diagrammar lecture notes. There's a discussion of something called the 'equivalence theorem'. Essentially, it observes that since the S-matrix is obtained from the correlation function by finding the residue at a single-particle-pole, everything which would lead to multiple particle poles does not contribute. There's a loophole if ... 1 I found chapters in books on regge theory as: (P. D. B. Collins) An Introduction to Regge Theory. (Geoffrey F. Chew) S-Matrix Theory of Strong Interactions. (S. C. Frautschi) Regge Poles and S-Matrix Theory. (V. N. Gribov) The Theory of Complex Angular Momenta. 0 They are not normalisable because they either come from, or extend to infinity. This essentially means that the probability density blows off and gives non-physical results. 2 This is a link with the conclusions from the experiment and which observations lead to which conclusions. The Conclusion When Rutherford mathematically investigated the results he proposed a model that explained the results that Geiger and Marsden obtained. The fact that the vast majority of the alpha particles got straight through led Rutherford to ... 1 you have three terms in the equation, which together define the scattered wave. the first two define the amplitude at any given time and place along the vector \hat{r}, which points the direction you're interested in. \hat{r}\times\hat{E}\times\hat{r} tells you that the wave is going to be perpendicular to \hat{r} and also in the plane defined by two ... 2 First of all the hats on \hat r and \hat E_i indicate that these are unit vectors.$$(\vec A \times \vec B) \times \vec C \neq \vec A \times (\vec B \times \vec C) See Vector Triple product Which means that $\vec A \times \vec B \times \vec C$ without parenthesis has no meaning. So I shall assume parenthesis around the first pair ie $(\vec A ... 3 I don't completely understand the two sets of statements you've written, but I think I understand the essence of your question. Maybe this helps: The S-matrix (operator) is a transfer function from in states to out states. If your states are not evolving (Heisenberg/Interaction picture) then you need to insert an evolution operator between the states. ... 2 There are very important differences between this two approaches, that can be summarized by noting that the Lippmann-Schwinger is the (formal) solution of a one-body problem (scattering of a particle by an external potential) whereas the Dyson equation gives the solution of a many-body problem. I focus here on the non-relativistic many-body case (it is also ... 1 Here is our interpretation of OP's question: We are essentially talking about the asymptotic form of positive energy scattering states to the time-independent Schrödinger equation (TISE) for the two free regions$x \to \pm \infty\$. (As OP notes, scattering states are not normalizable, and therefore do not belong to the Hilbert space. Nevertheless they can be ...

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