fermions - Why $delta F = Bepsilon$ and not $F=B epsilon$ in supersymmetry?

We can express supersymmetric transformations as $$\delta F = B\epsilon, \tag{1} $$
$$\delta B = F\bar{\epsilon},\tag{2}$$ where $B$ and $F$ denote the bosons and fermions, respectively, in the theory and is the local supersymmetry parameter.

My question could be silly, but if the phrase in bold above says that and doesn't instead say the variation of fermions must transform into bosons and vice versa why doesn't (1) look like $$F=\epsilon B.$$ Where did the "$\delta$" in $\delta F$ in (1) come from?

Answer

You're just not parsing the sentence correctly, or rather, you're just being too nitpicky. When one says that supersymmetry "transforms bosons into fermions and vice versa", one means that, under an infinitesimal transformation, the variation is purely fermionic for bosons, i.e. $$ B\mapsto B+\delta B = B+\epsilon F$$ for some bosonic $B$ and some fermionic $F$.

Supersymmetry is meant to be a smooth transformation, i.e. one can integrate this to a finite supersymmetry transformation, just like one integrate the infinitesimal rotation $\vec v \mapsto v + \epsilon L_i \vec v$ to $\vec v \mapsto \mathrm{e}^{\epsilon L_i}\vec v$. For your proposed $B\mapsto \epsilon F$, you cannot expand the r.h.s. in powers of epsilon to get back an infinitesimal transformation of the form $B+\delta B$, this already shows it's a "bad" transformation law.