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I'm following the Book of Brian Hatfield (Quantum Field Theory of point particles and Strings), p.192 here: For real Grassmann numbers (and Functionals thereof):

If $\Phi[\psi]$ is a functional, and $\psi(x)$ is a Grassmann-valued function, we demand that $$\int \mathcal{D}\psi \delta[\psi - \xi] \Phi[\psi] = \Phi[\xi]$$ (this is equation 9.67) , and one option to do this is to let (equation 9.66) $$\delta[\psi - \xi] = \prod_x (\psi(x)-\xi(x)).$$

The complex case of the delta function is NOT treated in the book, and I want to deduce how the mentioned relations would turn out for that case. Here, $\psi$ now has two components ($\psi = \frac{1}{2} \psi_1 + i \psi_2$) - Which makes me wonder: How does the fundamental relation turn out? For complex $\psi$: \begin{align} \int \mathcal{D}\psi \delta[\psi - \xi] \Phi[\psi] = \Phi[\xi] \end{align} or \begin{align} \int \mathcal{D} \psi \int \mathcal{D} \psi^* \delta[\psi - \xi] \Phi[\psi] = \Phi[\xi]? \end{align}

The first version works, but only if I assume that $\delta[\psi - \xi] = \prod\limits_x (\psi(x) - \xi(x))$ and $\delta[\psi-\xi] = \delta[\psi-\xi]^*$, and those exclude each other. In either case, what would be a realization of the $\delta$ functional? Would it still be \begin{align} \delta[\psi - \xi] = \prod_x (\psi(x) - \xi(x))? \end{align}

Qmechanic
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Quantumwhisp
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2 Answers2

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Motivated by the one-dimensional complex delta function, whereby $\delta_\mathbb{C}(z):=\delta(z)\delta(\bar{z}),$ so that $$\int \mathrm{d}z\wedge\mathrm{d}\bar{z}\ \delta_\mathbb{C}(z-\zeta) f(z) = f(\zeta),$$ you can define $\delta_\mathbb{C}[\psi] := \delta[\psi]\delta\!\left[\bar{\psi}\right],$ satisfying $$\int\mathrm{D}\psi\;\mathrm{D}\bar{\psi}\ \delta_\mathbb{C}[\psi-\xi] \Phi[\psi] = \Phi[\xi]$$ and realised as $$\delta_\mathbb{C}[\psi-\xi] = \prod_{x} \Big(\psi(x)-\xi(x)\Big)\Big(\bar{\psi}(x)-\bar{\xi}(x)\Big).$$

ɪdɪət strəʊlə
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  1. Be aware that integration over a complex Grassmann-odd variable $\psi$ has 2 different notations in the literature: $\int \!\mathrm{d}^2\psi$ and $\int\! \mathrm{d}\psi~\mathrm{d}\psi^{\ast} $. This is similar to the standard notations for a Grassmann-even complex integration.

  2. Similarly, a complex Grassmann-odd Dirac delta distribution can denote just the holomorphic part or also include the anti-holomorphic part, depending on conventions.

Qmechanic
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