Shankar's QFT book gives an overview for deriving a path integral representation for Majorana fermions. In the derivation, he works directly in continuous imaginary time, sweeping issues of discretization under the rug. I am trying to fill in the gaps by working out the properly discretized path integral, and I am running into subtleties with a step involving integration by parts.

Let me first summarize Shankar's derivation. For simplicity, start with a system of free Majorana fermions $\hat{\psi}_i$ satisfying $\{\hat{\psi}_i, \hat{\psi}_j \} = \delta_{ij}$, with Hamiltonian $$ \hat{H} = \frac{i}{2}\sum_{ij} \hat{\psi}_i h_{ij} \hat{\psi}_j $$ Since there are no coherent states for Majorana fermions, introduce an auxillary system of Majoranas $\hat{\eta}_i$, and combine into a single complex fermion $\hat{c}_i$ as follows: $$ \hat{c}_i = \frac{\hat{\psi}_i + i \hat{\eta}_i}{\sqrt{2}} \implies \hat{H} = \frac{i}{4} \sum_{ij} (\hat{c}_i + \hat{c}^{\dagger}_i) h_{ij} (\hat{c}_j + \hat{c}^{\dagger}_j) $$ We then obtain the imaginary-time path integral as usual, using Grassmann coherent states. The result is a path integral over the "complex" Grassmann fields $c_i(\tau)$ and $\bar{c}_i(\tau)$, with the Euclidean action $$ S[\bar{c},c] = \int d\tau \left[ \sum_i \bar{c}_i \partial_{\tau} c_i + \frac{i}{4} \sum_{ij} (c_i + \bar{c}_i) h_{ij}(c_j + \bar{c}_j) \right] $$ Finally, we rewrite in terms of "real" Grassmann numbers by defining the orthogonal transformation of variables $$ \psi_i(\tau) = \frac{c_i(\tau) + \bar{c}_i(\tau)}{\sqrt{2}}, \quad \eta_i(\tau) = \frac{c_i(\tau) - \bar{c}_i(\tau)}{i\sqrt{2}} $$ The Hamiltonian piece is then a function of $\psi_i$ alone, while the Berry phase term becomes: $$ \int d\tau \, \bar{c}_i \partial_{\tau} c_i = \frac{1}{2} \int d\tau \left[ \psi_i \partial_{\tau} \psi_i + \eta_i \partial_{\tau} \eta_i + i(\psi_i \partial_{\tau} \eta_i - \eta_i \partial_{\tau} \psi_i) \right] $$ The claim is that the last term vanishes upon integrating by parts and using the anticommutivity of Grassmann numbers, which seems reasonable on its face. The final action is then $$ S[\psi,\eta] = \frac{1}{2} \int d\tau \sum_i \eta_i \partial_{\tau} \eta_i + \frac{1}{2} \int d\tau \left[ \sum_i \psi_i \partial_{\tau} \psi_i + \frac{i}{2} \sum_{ij} \psi_i h_{ij} \psi_j \right] $$ In particular, $\eta$ and $\psi$ decouple completely, and we can disregard $\eta$ for purposes of calculating correlation functions of the $\psi$'s.

Here is where the subtlety comes in. To see explicitly that the last term vanishes, let's work explicitly in properly discretized imaginary time $\tau = n \varepsilon$, setting $c^n_i \equiv c_i(n\varepsilon)$. The action in terms of complex fermions is then (see, for example, Altland and Simons): $$ S = \sum_{n = 1}^N \left[ \sum_i \bar{c}^n_i(c^n_i - c^{n-1}_i) + \frac{i \varepsilon}{4} \sum_{ij} (c_i^{n-1} + \bar{c}^n_i) h_{ij} (c_j^{n-1} + \bar{c}_j^n) \right] $$ Importantly, in the derivation of the path integral, creation operators must act to the left in each matrix element while annihilation operators act to the right, and as a result $\bar{c}_i$ is taken one time-step ahead of $c_i$ in the Hamiltonian. In order to write $H$ entirely in terms of one species of Majoranas, it makes sense to define the discretized $\psi$'s and $\eta$'s as $$ \psi^n_i = \frac{c^{n-1}_i + \bar{c}^n_i}{\sqrt{2}}, \quad \eta^n_i = \frac{c^{n-1}_i - \bar{c}^n_i}{\sqrt{2}} $$ The Hamiltonian is then exactly in the desired form, while the Berry phase term reads $$ \sum_{n = 1}^N \bar{c}^n_i (c^n_i - c^{n-1}_i) = \frac{1}{2}\sum_{n = 1}^N \left[ \psi^n_i (\psi^{n+1}_i - \psi^n_i) + \eta^n_i (\eta^{n+1}_i - \eta^n_i) + i \psi^n_i(\eta^{n+1}_i - \eta^n_i) - i \eta^n_i (\psi^{n+1}_i - \psi^n_i) \right] $$ The first two terms are exactly what we want, while the last term is what we'd like to get rid of to make sure $\psi$ and $\eta$ decouple. However, it's clear that this will not work term-by-term, for the simple reason that the discretized integration by parts converts forward differences to backward differences. For example, the term $\psi^1_i \eta^2_i$ is present in the first term, but it is certainly never present in the second sum.

So, the final question is: how do I carefully justify throwing away the last term in the Berry phase part of the action, so as to justify throwing away the $\eta$'s in the last step of the continuum derivation?


1 Answer 1


In my point of view, the key is that when going back to the continuous limit we'll let $N\rightarrow\infty$, then $\eta_{i}^{n+1}\leftrightarrow\eta_{i}^{n}$, terms like $i\psi_{i}^{n}\eta_{i}^{n+1}$ and $i\eta_{i}^{n}\psi_{i}^{n}$ cancel out each other by anticommutivity relation.

I think there is no way to escape from $N\rightarrow\infty$ since in your question you introduce $$\psi_{i}^{n}=\frac{c_{i}^{n-1}+\bar{c}_{i}^{n}}{\sqrt{2}},\ \ \ \eta_{i}^{n}=\frac{c_{i}^{n-1}-\bar{c}_{i}^{n}}{i\sqrt{2}}$$, you can check $\psi,\eta$ satisfy relation anti-operators are their own(since they are Majaronas) only when $N\rightarrow\infty$.

  • $\begingroup$ Why should we be able to identify $\eta_i^{n+1}$ with $\eta_i^n$ in the limit $N \to \infty$? In the bosonic case, we can argue that the most important contributions to the path integral are continuous fields, and such an identification is plausible. But I see no reason why we should be able to do so in the case of fermions. $\endgroup$
    – Zack
    Commented Oct 22, 2023 at 15:06
  • $\begingroup$ I think for both bosons and fermions when $N \to \infty$ then $\delta_{t}\,\sim\,1/N\,\to\,0,$, $\delta_{t}$ is the time interval between two inserted coherent states in path integral, in this way the two states will be the same. $\endgroup$
    – Zane Chen
    Commented Oct 24, 2023 at 11:42

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