mjhjmtu

I am just starting to read about supersymmetry for the first time, and there is something bothering me. Supersymmetry transformations transform between bosonic fields and fermionic fields, but I don't see how this can even be defined. Take the following simple example that appears in the beginning of the notes I'm reading. In four dimensions, say $S$ is a real scalar field, $P$ is a real pseudo scalar field, and $psi$ is a Majorana spinor. Take the Lagrangian to just be

$$mathcalL = - frac12 (partial S)^2 - frac12 (partial P)^2 - frac12 barpsi partial!!!/ psi.$$

Now, $S$ and $P$ are just real fields. (And they truly are classical fields, because Lagrangians are always functions of classical variables, even when we're interested in QFT.) However, $psi$ is made of anti commuting Grassmann variables. So we can see that the $psi$ field is not made of the same "type" of object as the $S$ and $P$ fields. In other words, $S$ is a function

$$ S: mathbbR^4 to mathbbR$$
while I believe $psi$ is a function
$$ psi: mathbbR^4 to textorder 1 elements of Gr(4, mathbbR)$$
where $Gr(4,mathbbR)$ denotes the real Grassmann algebra with 4 generators. (Please correct me if I'm wrong.) By order 1 elements, I mean linear combinations of the four generators of $Gr(4, mathbbR)$. As a vector space, this is equal to $mathbbR^4$.

How can we then consider "supersymmetry" transformations of the following form?

beginalign*
delta_varepsilon S &= barvarepsilon psi \
delta_varepsilon P &= barvarepsilon gamma_5 psi \
delta_varepsilon psi &= partial!!!/ (S + P gamma_5) varepsilon
endalign*
($varepsilon$ is a constant Majorana spinor.) $barvarepsilon psi$ is now a Grassmann number of order 2, which is simply just not the same "type" of thing as $S$, which is just a real number! So how is this "transformation" even defined?

Perhaps I do not understand how a "classical" fermion field is really supposed to work, or how these Grassmann numbers are being used.

This is actually a rather subtle question, which does not really get explained in too many text books. As Qmechanic says, Grassmann variables, i.e. elements of the infinite dimensional Grassmann algebra $Lambda_infty$, have in general body and soul. Now, of course, you may say: wait, isn't the action just
$$ S=int!mathrmd^4x,mathcalL;,$$
and how can $S$ have a soul? The answer is that in this context the action here just something that enters the path integral via
$$ textcorrelator=int mathcalDpsi,mathcalDbarpsi,mathcalDphi,mathcalDP, dots exp(mathrmi,S);,$$
where I renamed your field $S$ to $phi$ in order to distinguish it from the action $S$. The important point to notice is that the correlator on the left-hand side is just an ordinary complex number, and not an element of $Lambda_infty$, because in the path integral we got rid of all algebraic objects. This is because $intmathrmdtheta,theta=1$ for a-numbers. So the upshot is that in this context $mathcalL$ and all the fields are really elements of $Lambda_infty$, and hence $barvarepsilonpsi$ is something which one can shift a scalar field by.

The Grassmann-even variables $S$ and $P$ are supernumbers, which can have both body and soul, not just body. See e.g. this related Phys.SE post.