# I Fermions Bosons vertices in SM - but no SUSY

1. Jul 15, 2017

### Lapidus

In the Standard Model fermions interact via exchanges of massless (virtual) spin-1 particles. Fermions are turned into a boson. How is that different from the SUSY transformation that turns fermions into bosons?

2. Jul 15, 2017

### Orodruin

Staff Emeritus
Fermions are not turned into bosons in the exchange in the SM (unless in s-channel where two fermions combine into a generally off-shell boson).

3. Jul 15, 2017

### Staff: Mentor

In addition the Orodruin's answer: An interaction (something physical happening) is not a symmetry transformation (a mathematical operation).

4. Jul 15, 2017

### vanhees71

Also note that the weak interaction is mediated by massive vector bosons, the $W$ (charged) and $Z$ (neutral) bosons.

5. Jul 16, 2017

### ChrisVer

SUSY is not telling you that a fermion becomes a boson, it tells you that for each fermion there is a bosonic partner particle (super-partner). The two are related via supersymmetric transformations.
So the fact that the electron+positron annihilate let's say to 2 photons, doesn't tell you that the photon is the superpartner of the electrons/positrons (which are the selectrons)...

6. Jul 20, 2017

### arivero

Parhaps the question can be reformulated to how different
But the W and Z and gamma are symmetry transformations, generator of SU(2)xU(1), and they are particles too. It could be argued that they just happen to be particles in the defining representation of the symmetry, but then the question translates to "when do symmetries have associated particles"?

Particularly, er, specifically, what happens with the generator of supersymmetry? It seems that in this case it is not compulsed to appear as an associated particle interacting with all the others, is it? If it were, we should have diagrams where a fermion interacts with another one and becomes a boson.

7. Jul 20, 2017

### vanhees71

Not the gauge fields are the generators of the symmetry but the corresponding Noether-charge operators. Of course you are right in saying that the gauge fields are Lie-algebra valued vector fields, i.e., something like $\mathcal{A}_{\mu} =A_{\mu}^a \hat{t}^a$, where $\hat{t}^a$ are the Lie-algebra basis matrices in the representation of the field you apply the corresponding gauge-covariant derivative to, $\mathrm{D}_{\mu} = \partial_{\mu} + \mathrm{i} g \mathcal{A}$.

8. Jul 20, 2017

### mitchell porter

See 7.5 of Stephen Martin. When global susy is spontaneously broken, the resulting goldstino can transform particles to their superpartners. In local susy, i.e. supergravity, the gravitino has this property.