# Tag Info

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Either plot in 3D the ISF, or represent the Fourier spectrum at each time.

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The standard chiral lagrangian has a $\pi\to-\pi$ symmetry that forbids processes with an odd number of Goldstone bosons. This is not a symmetry of QCD, however, and these reactions are described by the Wess-Zumino term. The Wess-Zumino term is not a local lagrangian in 4-d, but it can be written as a local lagrangian on a 5-d manifold which has 4-d ...

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Expand the normal order: $$:\psi^\dagger_1\psi_1\psi^\dagger_2\psi_2:=\psi^\dagger_1\psi^\dagger_2\psi_1\psi_2.$$ (creator operator before annihilation operator). Then insert unit operator $I=|0\rangle\langle0|$ between them. Meaning of this expression is that, the total amplitude include two amplitudes. One is for annihilating initial particles of ...

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To find the solution to your problem: Area 1: U1,E>U1 The Schrodinger equation and it's solution is: $$ψ'' + {2m \over \hbar ^2}(E-U1)ψ=0 ---> ψ_A=e^{ik_1 x} + c_1e^{-i k_1 x}$$ where $k_1 ^2 = {2m \over \hbar ^2}(E - U1)$. Area 2: U2, E As in the above: $$ψ'' - {2m \over \hbar ^2}(U2-E)ψ=0--> Ψ_Β = c_2 e^{k_2 x} + c_3 e^{-k_2 x}$$ where $k_2 ^2 ... 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 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 ... 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 ... 3 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.

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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 ...

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