Spinor functional quantization unitarily equivalent and determinant

Question

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: \begin{equation} \psi(x)=\sum_i \psi_i \phi_i(x) \tag{9.71} \end{equation} where $\phi_i(x)$ are ordinary four component spinors, $\psi_i$ are Grassmann numbers.

Then, the book defined the two-point function: \begin{equation} \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} \end{equation}

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 \begin{equation} \text{det}(-i\int d^4 x (i\not \partial -m)) \end{equation} why we just denote $\text{det}(i\not \partial -m)$?

Answer 1

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: \begin{equation} \prod_i \int d\theta^\ast_i d\theta_i\,e^{-\theta^\ast_i B_{ij}\theta_j} = \det(B), \end{equation} as identity see Berezin integral