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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

$$\mathcal{L} = - \frac{1}{2} (\partial S)^2 - \frac{1}{2} (\partial P)^2 - \frac{1}{2} \bar{\psi} \partial\!\!\!/ \psi.$$

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

$$ S: \mathbb{R}^4 \to \mathbb{R}$$ while I believe $\psi$ is a function $$ \psi: \mathbb{R}^4 \to \text{order 1 elements of } Gr(4, \mathbb{R})$$ where $Gr(4,\mathbb{R})$ 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 $Gr(4, \mathbb{R})$. As a vector space, this is equal to $\mathbb{R}^4$.

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

\begin{align*} \delta_\varepsilon S &= \bar{\varepsilon} \psi \\ \delta_\varepsilon P &= \bar{\varepsilon} \gamma_5 \psi \\ \delta_\varepsilon \psi &= \partial\!\!\!/ (S + P \gamma_5) \varepsilon \end{align*} ($\varepsilon$ is a constant Majorana spinor.) $\bar{\varepsilon} \psi$ 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.

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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 $$ S=\int\!\mathrm{d}^4x\,\mathcal{L}\;,$$ 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 $$ \text{correlator}=\int \mathcal{D}\psi\,\mathcal{D}\bar\psi\,\mathcal{D}\phi\,\mathcal{D}P\, \dots \exp(\mathrm{i}\,S)\;,$$ 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 $\int\mathrm{d}\theta\,\theta=1$ for a-numbers. So the upshot is that in this context $\mathcal{L}$ and all the fields are really elements of $\Lambda_\infty$, and hence $\bar\varepsilon\psi$ is something which one can shift a scalar field by.

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  • $\begingroup$ I see. So the total path integral is both a regular path integral and a Berezin path integral. The action $S$ takes supernumbers as its input, and our transformation is taking place within the inputs of $S$, and so the total path integral is still defined. $\endgroup$ – user1379857 Sep 11 '18 at 4:22
  • $\begingroup$ @user1379857 Yes, that's roughly how it is. In this context, he action is just some object that allows us to compute correlators. $\endgroup$ – marmot Sep 11 '18 at 4:24
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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.

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