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As far as I understood, in the path integral formulation of QFT, a field configuration is modelled by a mapping

$$ x \rightarrow \Psi(x) $$ Where $\Psi(x)$ are 4 components, each represented by 4 grassmann numbers. We need 4 Grassmann-Numbers for every point in space, an all those numbers need to anticommute, so all in all we need 4 times infinity Grassmann-Generators. This means that $\Psi$ is a mapping from space time to an infinite dimensional exterior Algebra (I hope this is correct).

What I wonder now is: We require the Lagrangian

$$ \bar{\Psi}(\partial_{\mu}\gamma^{\mu} - m)\Psi $$

To be some rather simple object, namely a Lorentz scalar, something that we can integrate over and that will give an action $S$ that can be used as a phase in the exponential function. We want to add for example products $ \bar{\Psi}A_{\mu} \gamma^{\mu} \Psi$ or terms that don't have anything to do with the exterior algebra, like $F^{\mu \nu}F_{\mu \nu}$.

How can we add products of the exterior algebra (which are somewhat two-forms) and terms that are zero forms, like the mentioned free electromagnetic field Lagrangian? If I understand correctly, $\bar{\Psi} \Psi$ is some kind of two-form, while $F^{\mu \nu}F_{\mu \nu}$ is some kind of 0-Form). How do we reconcile this?

I hope I understood the usage of Grassmann numbers correctly, If not, I'd appreciate if an answer would point out the flaw in the described usage.

Qmechanic
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Quantumwhisp
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  • As @Qmechanic answer correctly points out, a meaningful value can only be achieved by integrating the Grassmann out. On the other hand, before performing the functional integration, the Lagrangian could have pretty darn funny properties. For example, if you write out the Grassmann numbers in the mass term $m\bar{\Psi}\Psi$, it is actually imaginary with an $i$, albeit you can double check that it's Hermitian. Wickedly weird, right? See an example here: https://physics.stackexchange.com/q/529496/ – MadMax Jun 21 '22 at 15:15

1 Answers1

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The action, Lagrangian & Lagrangian density are Grassmann-even supernumbers in the path/functional integral formalism of a QFT with fermionic/Grassmann-odd fields, cf. e.g. this & this related Phys.SE posts.

We stress that the soul-part of a supernumber is an indeterminate/variable, i.e. a placeholder, except it cannot be replaced by a number to give it a value.

A value can only be achieved by integrating it out! This is achieved by the path/functional integration. The value of the path/functional integral (if it contains no external supernumber parameters, such as e.g. fermionic sources) is an ordinary complex number $\in\mathbb{C}$.

Qmechanic
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