# Spinor functional quantization unitarily equivalent and determinant

On P&S's qft page 301 and 302, the book discussed functional quantization of spinor field.

The book define a Grassmann field $$\psi(x)$$ in terms of any set of orthonormal basis functions: $$$$\psi(x)=\sum_i \psi_i \phi_i(x) \tag{9.71}$$$$ where $$\phi_i(x)$$ are ordinary four component spinors, $$\psi_i$$ are Grassmann numbers.

Then, the book defined the two-point function: $$$$\left\langle 0\left|T \psi\left(x_1\right) \bar{\psi}\left(x_2\right)\right| 0\right\rangle=\frac{\int \mathcal{D} \bar{\psi} \mathcal{D} \psi \exp \left[i \int d^4 x \bar{\psi}(i \not \partial-m) \psi\right] \psi\left(x_1\right) \bar{\psi}\left(x_2\right)}{\int \mathcal{D} \bar{\psi} \mathcal{D} \psi \exp \left[i \int d^4 x \bar{\psi}(i \not \partial-m) \psi\right]} \tag{A}$$$$

I don't understanding following:

1. the book said that they write $$\mathcal{D}\overline{\psi}$$ instead of $$\mathcal{D}\psi^*$$ for convenience, the two are unitarily equivalent. So according to (9.71), this two are vectors, not matrix, what's the meaning here unitarily equivalent?

2. According to previous discussion, the denominator of (A) should be $$$$\text{det}(-i\int d^4 x (i\not \partial -m))$$$$ why we just denote $$\text{det}(i\not \partial -m)$$?

• Why do you think there should be an integral in the determinant? The denominator is just a product of Gaussians of Grassmann variables, and so a determinant of the Dirac operator. Commented Nov 19, 2022 at 8:45
• General tip: Consider to only ask 1 question per post. Commented Nov 20, 2022 at 11:25

1. The functional measure is not only the (continuous) product at each space-time point but also of each component of the objects we are taking the measure of. This means that the measure of $$\psi^\dagger$$ and $$\psi$$ is the same. Moreover, $$\overline{\psi}=\psi^\dagger \gamma^0$$, and $$\text{det}(\gamma^0)=1$$, so $$\mathcal{D}\overline{\psi}$$ and $$\mathcal{D}\psi^\ast$$ are the same.
2. You have: $$$$\prod_i \int d\theta^\ast_i d\theta_i\,e^{-\theta^\ast_i B_{ij}\theta_j} = \det(B),$$$$ as identity see Berezin integral
• Thank you very much! For your first point, do you want to express that the measure of $\psi^{\dagger}$ and $\psi^{*}$ is the same? Commented Nov 19, 2022 at 9:18
• @Daren Yes they are the same because $\psi^\dagger = (\psi^{\ast})^\text{T}$, and a product of components of a "vector" doesn't care about whether it is a line or a column. Commented Nov 19, 2022 at 9:25
• Another point is that for Berezin integral, if $\theta_i$ have spinor component, this expression also correct. Is this right? Thanks! Commented Nov 19, 2022 at 9:26
• @Daren If by 'have spinor components' you mean in fact that $\theta_i$ are the components of a spinor, then yes this expression is also correct because you can decompose the product into two: one over the positions $x$ in space-time, and one over these components. Commented Nov 19, 2022 at 9:34