# Tag Info

0

If the vacuum of the theory is supersymmetric - i.e. SUSY is not broken - then it is annihilated by the SUSY generators. On the other hand, using the SUSY algebra one can show that the hamiltonian can be written in terms of the SUSY generators. This implies that the vacuum $|0\rangle$ is supersymmetric if and only if $\langle 0|H|0\rangle=0$, i.e. the vev ...

3

Let me first describe the basic idea without even mentioning supergravity. Consider some classical field theory of two fields $\phi$ and $\sigma$, with an action $S[\phi,\sigma]$. Suppose this theory has a continuous symmetry (with parameter $\epsilon$): $$S[\phi,\sigma]=S[\phi+\delta_\epsilon\phi,\sigma+\delta_\epsilon\sigma].$$ Say I find some particular ...

0

First of all, what's the motivation for the Yang-Mills action and how should I understand the coupling constants $\theta$ and $g$? I would say motivation comes from experiments. For instance it is an experimental fact that the electric charge is conserved. The associated current is also conserved, in the sense of $$\partial_\mu J^\mu=0.$$ Therefore we ...

0

Okay, this wasn't as hard to find an answer to as I expected. However, any clarifications/ critisisms are welcome. Weinberg basically gives the answer in section 5.6 of his QFT book: A general tensor of rank N transforms as the direct product of N (1/2, 1/2) four-vector representations. It can therefore be decomposed (by suitable symmetrizations ...

1

The explanation lies in the fact that you want to consider BPS states, or equivalently, short supersymmetric multiplets, because you are dealing with an extremal black hole. Now you could ask why we want to impose this restriction on the spectrum. The answer is that results based on the BPS spectrum can be generalized from weak coupling to strong coupling. ...

3

Your confusion comes from thinking that going to superstrings simply means adding fermions in the spectrum. The spectrum is instead different. For bosonic string (let's focus on NN boundary conditions and open strings) you have something like: $$\alpha' m^2=N-1$$ where N is the number operator of the transverse vibrational excitations of the bosonic ...

0

If you have a Lagrangian which is Lorentz and SUSY invariant, it means that when you perform a SUSY transformation the Lagrangian does not change (apart from surface terms). If it was Lorentz invariant before the transformation it would be so after. Otherwise your theory would not be SUSY invariant. You do not simply exchange fermions with bosons. You ...

1

You cannot just substitute fermions for bosons to determine the SUSY equivalent of a given interaction. For example, the kinetic terms for the bosons and fermions look quite different--first order in derivatives for the fermions, versus second order for bosons. As another example, the SUSY equivalent of a four-boson interaction is a Yukawa interaction ( a ...

0

Yes In 5 dimensions, the R-Symmetry group is $SU(2)$. Any spinor, therefore, has to transform under $SU(2)$ as $\delta_{SU(2)} (\Lambda^{ij}) \lambda^i = - \Lambda^{i}{}_{j} \lambda^j$. You cannot consider a spinor $\psi$ without the $SU(2)$ index, i.e. $\delta_{SU(2)} (\Lambda^{ij}) \psi =0$. The algebra would not close on such a field, thus $\psi$, if not ...

Top 50 recent answers are included