# Tag Info

## Hot answers tagged supersymmetry

3

There are obviously differeing genus types according to which partition function in d-dimensional QFT. At the very outset of $0$-d QFT the index is the push forward in ordinary de Rahm Cohomology; in other words, the integration of differential forms. The genus as you put it, is non-existant here. In $1$-d QFT the index of the Dirac Operator is ...

2

The variation $\delta F$ for any field (or degree of freedom) $F$, given an infinitesimal transformation, is always calculated as the commutator $$\delta F = [ \bar\epsilon Q, F ]$$ where $\bar \epsilon$ is a parameter ("angle" or "shift" or some generalization) of the transformation and $Q$ is the generator. (Those may be replaced by other letters.) ...

1

Not only must supersymmetry be broken, it must be broken in a way that doesn't lead to phenomenological disasters. In the MSSM, $D$-term breaking with a Fayet-Iliopoulos term achieves neither: it doesn't break supersymmetry but it does lead to phenomenological disasters! Consider again your potential \begin{align} V&=\sum_i |m_i|^2 |\phi_i|^2 +1/2 ... 1 The bracket you have written is of the form[\delta,\delta] A_\mu = v^\nu \partial_\nu A_\mu + \partial_\mu (v^\nu A_\nu). As you pointed out, the first term corresponds to a translation by $v^\mu$. The second term corresponds to a gauge transformation $\delta A_\mu = \partial_\mu \lambda$ with $\lambda = v^\nu A_\nu$. So the algebra closes up to a ...

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