# Tag Info

0

1) This is just one of the basic postulates of Quantum Mechanics. Let $|\phi\rangle$ and $|\psi\rangle$ be any two normalised kets, and $Q$ be an operator. Then the matrix element of $Q$ is, by definition, $$Q_{\phi\psi}=\langle\phi|Q|\psi\rangle$$ Now, if $\psi$ and $\phi$ are not normalised, we must take $$... 1 Yes, there is an integral, which comes from the LSZ reduction formula,$$ \langle f|i\rangle\sim \int \mathrm dx\ \mathrm e^{ikx}\square_x G(x) $$where x=(x_1,x_2\cdots,x_n), k=(k_1,k_2,\cdots,k_n) and G is the n-point function. If you go to momentum space you'll get that integrand depends on x only through exponentials, and therefore there is a ... 0 There is no need of integrating over anything to obtain the 4-momentum conservation. Indeed, if you think in terms of perturbation theory (Feynman diagrams), each vertex conserve momentum, so that the diagram itself automatically conserves momentum. 0 The process is discussed at the parton level – both in the initial form and the desired form – so the conversion cannot depend on PDFs. Now, the Mandelstam variable t is equal to$$ t = (p^\mu_1 - p^\mu_3)^2 = m_1^2+m_3^2 -2 E_1 E_3 +2 \vec p_1\cdot \vec p_3  in the "mostly minus" metric convention. The masses of particles are fixed and the total energy ...

6

By definition, Thomson scattering is the elastic scattering of light by a free charged particle. Atoms cannot be described as such, but the electrons in an atom may approximate to free electrons if their binding energy is much lower than the photon energy. This might be true for X-ray wavelengths, although if the photon energy gets too high then elastic ...

1

I consider the scattering process $A+B \to 1 + 2$. The differential cross-section is always given by \begin{split}\label{eq1} d\sigma &= \frac{1}{(2E_A)(2E_B)|v_A - v_B|} \frac{d^3p_1}{(2\pi)^3} \frac{1}{2E_1} \frac{d^3p_2}{(2\pi)^3} \frac{1}{2E_2} \left| {\cal M} \right|^2(2\pi)^4 \delta^4( p_A + p_B - p_1 - p_2) \end{split} ...

Top 50 recent answers are included