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It looks like this loophole is not explicitly discussed in the "axioms", but it is mentioned in the paragraph before equation (2) which I copy here: A symmetry transformation is said to be an internal symmetry transformation if it commutes with P. This implies that it acts only on particle-type indices, and has no matrix elements between particles ...


1

Well, the Logic is reversed compared to the one you are implicitly using. It is the direction of the Higgs'vev that's defining what we call electric charge. In practice SU(2)xU(1) is broken to a certain U(1) that we can always choose to point in a certain direction, and accordingly assign the electric charges afterwards. Since the Higgs can't carry color it ...


1

In principle, $\mathcal{N}$ gives you the number of supercharges in your theory. There are, however, cases with more than one irreducible (pseudo-)real spinor representations. If you have $N$ charges in one and $N'$ charges in the other representation, you can denote the total number of charges as $\mathcal{N}=(N,N')$ in order to emphasize the difference. ...


1

The answer is yes; this can be shown by evaluating the variation of the action. The action consists of three terms, we will consider them separately: $$\delta_\epsilon K(\Phi,\bar{\Phi})=\frac{\partial K(\Phi,\bar{\Phi})}{\partial\Phi^i}\delta_\epsilon \Phi^i+\frac{\partial K(\Phi,\bar{\Phi})}{\partial\bar{\Phi}_i}\delta_\epsilon\bar{\Phi}_i=i\epsilon K_i ...



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