Question on Majorana Path Integral

Question

I'm studying Shankar's Quantum Field Theory and Condensed Matter and got stuck in the issue related to changing measure in Majorana path integral.

In section 9.4, the Euclidean action for the continuum Majorana theory of the 2D Ising model is given by $$S = \frac 12 \int dx\, d\tau\, \Psi^T (x,\tau) \begin{pmatrix} -\partial_\tau & -i\partial_x - im \\ -i\partial_x + im & -\partial\tau \end{pmatrix} \Psi (x,\tau),$$ where $\Psi = (\psi_1, \psi_2)^T$, and $\psi_{1,2}$ are Majorana fermions that appear from Jordan-Wigner transformation. Here, the partition function is $Z = \int \prod_{x,\tau} d\psi_1\, d\psi_2\, e^S$. On page 153, the author rewrites the action in a "Lorentz-invariant" form as $$S = -\frac 12 \int dx\, d\tau\, \bar{\Psi} (\gamma \cdot \partial + m) \Psi,$$ where $\gamma_0 = \sigma_2$, $\gamma_1 = \sigma_1$, $\gamma \cdot \partial = \gamma_0 \partial_\tau + \gamma_1 \partial_x$, and $\bar{\Psi} = \Psi^T \gamma_0$. The newly introduced $\bar{\Psi}$ is not an independent Grassmann variable and is made of just $\psi_1$ and $\psi_2$, which is explicit in the path-integral measure $\int \prod_{x,\tau} d\psi_1\, d\psi_2$. Next, the author mentions that the Jacobian for changing from $\int \prod_{x,\tau} d\bar{\Psi}\, d\Psi$ to $\int \prod_{x,\tau} d\psi_1\, d\psi_2$ is unity, and here come my questions:

1. How is this transformation defined explicitly? While $\psi_{1,2}$ are mundane Grassmann numbers, $\bar{\Psi} = (i\psi_2, -i\psi_1)$ and $\Psi = (\psi_1, \psi_2)^T$ are "Grassmann vectors," and it seems that they are not related via linear equation (which will give rise to the inverse Jacobian factor). Is there any way to define measure over vectors of Grassmann variables?
2. Does the author mean that the Majorana path integral $Z = \int \prod_{x,\tau} d\bar{\Psi}\, d\Psi\, \exp \left[ -\frac 12 \int dx\, d\tau\, \bar{\Psi} (\gamma \cdot \partial + m) \Psi \right]$ can be treated as if $\bar{\Psi}$ and $\Psi$ are independent? If so, are the Majorana propagators contained in $$\langle \Psi(-p) \bar{\Psi}(p) \rangle = \frac{1}{i\gamma \cdot p + m}?$$

I appreciate any help. Thank you.

Answer 1

As Prahar stated in the comments, this is analogous to integration on the complex plane. More simply, if you are integrating on $\mathbb{R}^2$, you pick coordinates $(u, v)$ and construct a measure $\sqrt{g}dudv$ from those coordinates. If $u$ and $v$ correspond to an orthonormal basis, then the measure is simply $du dv$. Notice that if we interpret the pair $(\psi_1, \psi_2)$ as a point in a 2d complex Grassmanian manifold, then $(i\psi_2, -i\psi_1)$ is just a ninety-degree rotation of the original vector. So, as defined $\Psi$ and $\bar{\Psi}$ are simply another choice of "orthonormal" vectors which you can use in the measure.

This is of course just pure analogy to ordinary integration, but I believe this explanation is basically 90% there. I'm not sure a rigorous definition of integration of multiple Grassmanian variables would provide much additional insight. The first place I would look is the wikipedia page for the Berezin integral: https://en.wikipedia.org/wiki/Berezin_integral