# Can you quantize Grassmann-even superfields in the same fashion as Boson fields?

In a related Phys.SE question about supersymmetric Lagrangian $$\mathcal{L} = - \frac{1}{2} (\partial S)^2 - \frac{1}{2} (\partial P)^2 - \frac{1}{2} \bar{\psi} \partial\!\!\!/ \psi,$$ the fields $S$ and $P$ are said to be Grassmann-even supernumber-valued rather than real- (or complex) valued, so that supersymmetry transformations (with Grassmann-odd $\varepsilon$) \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*} can be consistently defined.

My question is: can you still do path integral (or canonical) quantization of Grassmann-even fields ($S$ and $P$) in the same fashion as real (or complex) scalar/pseudoscalar Boson fields? I am asking this because Grassmann-even supernumbers behave differently from real (complex) numbers. For example, the Grassmann-even $ab$ (where $a$ and $b$ are Grassmann-odd) squares to zero (nilpotent) $$(ab)(ab) = -(aa)(bb) = 0,$$ which is totally different from a real (complex) number.

2. A field $$\phi=\underbrace{\phi_B}_{\text{body}}+\underbrace{\phi_S}_{\text{soul}}$$ that takes values in the set $\mathbb{R}^{1|0}$ of Grassmann-even real supernumbers can be quantized in the same way as a field $\phi_B$ that takes values in the set $\mathbb{R}$ of real numbers.
This is easiest to see in the path integral formalism, since an integral over a Grassmann-even real supernumber $\phi\in\mathbb{R}^{1|0}$ is by definition given by the corresponding integral over its body $\phi_B\in\mathbb{R}$ $$\int_{\mathbb{R}^{1|0}} \! d\phi~f(\phi) ~:=~\int_{\mathbb{R}} \! d\phi_B~f(\phi_B) .$$