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The point of these funny terms in the "conjugate" fields is that they have the same Lorentz transformation properties as the original fields. For example, let's forget about the spinors and consider vectors. Suppose a vector $\mathbf{v}$ transforms with a phase $e^{i \theta}$ under some $U(1)$ transformation. Then the conjugate transpose $\mathbf{v}^\... 1 In a NJL model with two flavors$(u,d)$, the field$\psi$is defined as $$\psi=\begin{pmatrix} u \\ d \end{pmatrix},$$ being$u$and$d$ordinary Dirac spinors. This means that, in your transformation, the SU(2) part applies to$\psi$while$\gamma_5$goes on the single Dirac spinors. Remebering that$\gamma_5^2=I$, you have $$e^{-i{\mathbf\tau}\cdot{\... 1 The gravity gauge multiplet (e,\chi) enjoys a local supersymmetry with infinitesimal transformations$$ \delta e = -2\mathrm{i}\epsilon\chi \quad \delta \chi = \frac{\mathrm{d}\epsilon}{\mathrm{d}\tau}, \tag{1}$$where$\epsilon$is a fermionic parameter. Together with ordinary reparametrization symmetry by some rescaling$f\$, this gives us two free gauge ...

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There is no trivial way to start with a list of initial-state and final-state particles and determine which channels are involved. You just have to start enumerating diagrams, and working out if they are allowed or not. To do that you use a small set of tools Each fundamental interaction has an allowed set of diagrams. You have to know or look these up. ...

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Yes. In general there must be the six states you mentioned. However, one common feature among the 4 states given in the “solution” is that they have total spin of 0. The other two states have a total spin of 1. In case the system is constrained to have a total spin of 1, then only those 4 states will be allowed.

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