Compare to the path integral for bosons, how can we formulate the Fermion path integral mathematically? For the boson path integral, the integration means the Lebesgue integral with respect to a measure on $(\mathbb{R}^T, \mathcal{B}(\mathbb{R}^T))$ when we consider the Euclidean action.($T$ here is an arbitary index set)

I learned from article(Berezin-integration) that how to define finite dimensional Berezin integral. But the fermion path integral relates to "infinite-dimensional" Berezin integral, I really don't know how to define it. Compare to the boson case, we use our knowledge of integration on $\mathbb{R}^n$ to build a theory of integration on $\mathbb{R}^T$ in a way that the Borel set of $\mathbb{R}^T$ is constructed by the Borel set of $\mathbb{R}^n$. But now in the fermion case, we don't have the Borel set(an integration is viewed as a functional), how can we define the "infinite-dimensional" Berezin integral?

  • $\begingroup$ FWIW, integration over the Grassmann-odd line $\mathbb{R}^{0|1}$ is discussed in this Phys.SE post. The main point is that it does not make sense to integrate over subsets. Less so for higher dimensions. $\endgroup$
    – Qmechanic
    Jan 18, 2018 at 7:57
  • $\begingroup$ I do understand the ordinary Berezin integration, which I do not understand is the infinite-dimensional version of Berezin integration. $\endgroup$
    – haoshu li
    Jan 18, 2018 at 8:05
  • $\begingroup$ Integration over $\mathbb{R}^{0|\infty}$ can be defined as the $n\to \infty$ limit of integration over $\mathbb{R}^{0|n}$. $\endgroup$
    – Qmechanic
    Jan 18, 2018 at 12:09
  • $\begingroup$ Does this mean if this $n \rightarrow \infty$ limit exist then we define the integration equals to the limit? $\endgroup$
    – haoshu li
    Jan 18, 2018 at 12:41

1 Answer 1


There is an introduction to Berezin integration in infinite dimension in Appendix A (p. 75) of the Aisenstadt lectures "Renormalization Group and Fermionic Functional Integrals" by Joel Feldman. The method essentially proceeds by taking a limit of the finite dimensional case but this involves some topology, namely, suitable norms and hypotheses on the covariances of the Fermionic Gaussian measures being constructed.

Also note that what you said about the Boson case is not the best mathematical way to set things up. If you consider a Euclidean QFT in $d$ (spacetime) dimensions, you are not doing an integral over the measurable space $(\mathbb{R}^T,\mathcal{B}(\mathbb{R}^T))$ with $T=\mathbb{R}^d$. I assume, you mean $\mathbb{R}^T$ is equipped with the product topology and you are using the corresponding Borel $\sigma$-algebra. Moreover, Lebesgue measure $D\phi$ would not exist on such a space. In the Boson case the correct measurable space would be $(S'(\mathbb{\mathbb{R}^d}),\mathcal{B}(S'(\mathbb{\mathbb{R}^d})))$. Namely, the scalar field $\phi(x)$ being integrated over is not a function but rather a Schwartz distribution.

  • $\begingroup$ But in the Bosn case, I do not mean the index set $T=\mathbb{R}^d$. In your example, i.e. a Euclidean QFT in d dimensions, the index set $T$ can be taken to be $S(\mathbb{R}^d)$ and the tempered distribution space $S'(\mathbb{R}^d)$ would be a subset of $\mathbb{R}^T$. $\endgroup$
    – haoshu li
    Jan 19, 2018 at 6:28
  • $\begingroup$ @haoshuli: Indeed, $S'$ is a space of (continuous linear) functions on $S$ and therefore it can be seen as a subset of $\mathbb{R}^T$ with $T=S(\mathbb{R}^d)$. But since $T$ is uncoutable, this is a really bad space to do measure theory on. That leads to artificial complications like indistinguishable processes, versions and modifications of processes etc. One can see $S'(\mathbb{R}^d)$ as a measurable subset of $\mathbb{R}^T$ with $T=\mathbb{N}$. This is because the eigenfunctions of the harmonic oscillator (Hermite functions) provide an unconditional Schauder basis. $\endgroup$ Jan 19, 2018 at 14:46

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