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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.

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  • $\begingroup$ In how many spacetime dimensions? $\endgroup$
    – Qmechanic
    Commented Mar 2 at 19:30
  • $\begingroup$ @Qmechanic In a two-dimensional spacetime. $\endgroup$
    – asdf
    Commented Mar 3 at 4:46
  • $\begingroup$ Comment to the post (v1). Consider to doublecheck the notation $\Psi$ vs. $\psi$ for typos. $\endgroup$
    – Qmechanic
    Commented Mar 3 at 5:31
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    $\begingroup$ @Qmechanic Thanks for the suggetions. I've fixed the notation for the Dirac spinor to $\Psi$, though the Shankar's original text contains the previous notation with notational clash. $\endgroup$
    – asdf
    Commented Mar 4 at 7:07
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    $\begingroup$ Let's answer something simpler first. Consider the Euclidean measure $dx dy$. Define the complex coordinate $z = x + i y$. Do you understand what the measure $dz d{\bar z}$ means? Do you understand how here $z$ and ${\bar z}$ can be treated as independent integration variables? Finally, do you understand how to map the measure $dx dy$ to $dz d{\bar z}$? $\endgroup$
    – Prahar
    Commented Mar 4 at 20:48

1 Answer 1

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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

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  • $\begingroup$ Thanks for the comment. Then, how about the following case?: Consider a low-energy Hamiltonian of a graphene: $H = \int dp \Psi(p)^\dagger \mathcal{H}(p) \Psi(p)$ where $\Psi(p) = (a_{K+p}, b_{K+p}, a_{K'+p}, b_{K'+p})^T$, where $K(K')$ are the two Dirac points and $a$ and $b$ denote fermion operators in each sublattice $A$ and $B$. Can we write the path integral of this system as $Z = \int D\overline{\Psi} D\Psi e^{-S}$ with $S = \int d\tau d^2 p \overline{\Psi}(p) \mathcal{H}'(p) \Psi(p)$? Or do we have to path-integrate over all $a_{K}$, $a_{K'}$, $b_{K}$, and $b_{K'}$? $\endgroup$
    – asdf
    Commented Mar 11 at 8:06
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    $\begingroup$ I mean, can we treat the four-component spinor $\Psi$ as a single Grassmann number and perform path integral with respect to $\Psi$ and $\bar{\Psi}$? Since the number of independent Grassmann numbers is two in this case, I'm not sure whether this two Grassmann numbers can reflect all the four modes in its "internal structure." $\endgroup$
    – asdf
    Commented Mar 11 at 8:11
  • $\begingroup$ @asdf - If $\Psi$ is a 4-component Dirac spinor then you treat it as a vector of 4 Grassmann numbers in the path integral. Why would it be a single Grassmann number? $\endgroup$
    – Prahar
    Commented Mar 11 at 11:39
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    $\begingroup$ @asdf, I'm not sure I'm totally clear on your confusion, but if I've interpreted things correctly I think you might be making a (reasonable) mistake regarding the notation for the path integral. A Dirac fermion is a 4-component spinor. When we write the path integral measure, we still only write $d\bar{\Psi}d\Psi$, but this is actually shorthand for $\prod_{i}d\bar{\Psi}_id\Psi_i$, for each pair of conjugate variables. So although the notation looks like it is integration over only two Grassmann variables, it is in fact always over four. $\endgroup$ Commented Mar 12 at 21:20
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    $\begingroup$ Many authors are not careful about this notational point, so your confusion is warranted. Eduardo Fradkin handles this in a decently careful manner in chapter 8 of his QFT textbook, you might try taking a look there. $\endgroup$ Commented Mar 12 at 21:22

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