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3

Given the four point function $\langle \phi(x_1)\phi(x_2)\phi(x_3)\phi(x_4)\rangle$, the conformal block expansion depends on what operators you replace by the OPE. So if you insert the OPE for $\phi(x_1)\phi(x_2)$ and the OPE for $\phi(x_3)\phi(x_4)$ then this corresponds to the s channel---one can also call this the (12)(34) channel---.. The t channel is ...

3

Here are two facts - A vacuum expectation value of a quantum field is equal to the minimum of the effective potential (taken from the 1PI effective action). The effective potential takes the general form $$V_{\text{eff}}(\phi) = V_{\text{classical}} (\phi) + \text{quantum corrections}$$ In perturbation theory, where quantum corrections are assumed to be ...

2

You write that you do not like the wave-particle duality explanation of the Young experiment, and therefore turn to QFT. Before going further I would like to point out that the double slit experiment is a one-particle effect. That means you only need consider one particle at a time to explain what is happening. Because of this QFT will not buy you much as ...

2

It is actually possible, and not too difficult, to prove this without expanding the exponentials to first order only. What you are trying to prove is $S^\dagger \gamma^0 = \gamma^0 S^{-1}$, this is equivalent to $$\gamma^0 S^\dagger \gamma^0 = S^{-1}$$ because $( \gamma^0 )^2 = 1$. Expand $S^\dagger = \sum_n \frac{1}{n!} \left( \frac i 4 \omega_{\mu\nu} ... 2 It gets easier if you use the result from part 1. Then you also don't have to deal with the$\mathcal O(\omega^2)$(see my answer to your other question). In your calculation, you transformed$\bar\psi$and$\psi$, but not$\gamma^\lambda$. This is correct, as I will show in the end, but I will take another point of view which is really helpful here: ... 1 The answer to your question requires some knowledge on group theory and tensor analysis, but I will try to make it as simple as possible with out going into too much of technicalities. Your question consists of basically two completely disjoint parts, they are: why the gauge bosons(leptoquarks) of Pati-Salam Group do not mediate proton decay. Which is a ... 1 Quantum fields cannot be turned on or off. The field itself exists for all time and space. It is possible to excite various modes of a quantum field at various spacetime points. These field excitations are interpreted as particles. When no excitations are present (i.e. no particles are present) the quantum field is in the vacuum state. Particles do not act ... 1 To begin with, The Amplituhedron formalism only works for a specific theory, N=4 SYM in the planar limit (only planar Feynmann diagrams are considered). Because of supersymmetry, you can classify scattering processes with two parameters:$n$and$k$. n is the number of particles involved, and k is, roughly speaking, the number of spin flips in the process. ... 1 First, your explanation is...sort-of-right. What's travelling is a quantum object, not a particle, not a wave. The probability of detecting a particle-like localized blip with some sort of detector is given by a probability density$\rho$, which is the "sqaured amplitude" of a "wavefunction"$\psi$. For free particles, the Schrödinger equation that$\psi\$ ...

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