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In studying this paper "Remarks on the Sachdev-Ye-Kitaev (SYK) model" (http://arXiv.org/abs/1604.07818), I have no idea how to get the free Majorana fermion Green's function (Eq. (2.5) in the above paper). Can someone shed some light on this? It would be better also shed some light on studying the SYK model.

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I think, the best, most reliable, and most straight-forward way to obtain the Green's functions for a free theory is to use the equations-of-motion approach; consult, e.g., the section "Single-particle Matsubara Green’s function" in Bruus & Flensberg, "Introduction to Many-body quantum theory in condensed matter physics" (draft).

First of all, the "Lagrangian" (or actually, the action) you have provided for the Majorana fermions, is $$ S = \frac{1}{2} \sum_{j} \chi_j(\tau) \, \partial_\tau \, \chi_j(\tau) ~,$$ which implies that the Hamiltonian for the Majorana modes vanishes; consult, e.g., the section "Field integral for the quantum partition function" in Altland & Simons, "Condensed matter field theory" (2010) (WCat).

I will use the Matsubara formalism where for an (time-independent) operator $O$, $$ \partial_{\tau} O(\tau) = [H(\tau), O(\tau)] ~,$$ in which $$ O(\tau) := e^{\tau H} \, O \, e^{-\tau H} ~.$$

The anti-/commutation relations for bosons or fermions can be denoted by $$ [A,B]_\eta = AB - \eta BA ~,$$ with $\eta = \pm$ for bosonic/fermionic operators.

The imaginary-time-ordered correlation function is defined as (see Bruus & Flensberg, op. cit.) $$ \begin{align} G_{A B}(\tau, \tau') &:= -\langle \mathcal{T}_\tau \{ A(\tau) \, B(\tau') \} \rangle \\ &\equiv -\Theta(\tau - \tau') \, \langle A(\tau) \, B(\tau') \rangle -\eta \, \Theta(\tau' - \tau) \, \langle B(\tau') \, A(\tau) \rangle~, \end{align} $$ where $\mathcal{T}_\tau$ stands for the imaginary-time-ordering operator and $\Theta$ is the Heaviside function.

In the following, to reduce the clutter in the notation, we let $\tau' = 0$. Therefore, the Green's function for a Majorana mode, $\chi$, will be $$ G(\tau) = -\langle \mathcal{T}_\tau \{ \chi(\tau) \, \chi^\dagger(0) \} \rangle ~. $$ Notice that the anticommutation relation for the Majorana modes reads $$ \{ \chi_i , \chi_j^\dagger \} = \delta_{ij} ~,$$ and that $$ \chi_i = \chi_i^\dagger ~.$$

* The equations of motion for the Majorana Green's function

Via a simple differentiation, we obtain $$ \begin{align} \partial_\tau G(\tau) &= -\partial_\tau \Theta(\tau) \, \langle \chi(\tau) \; \chi \rangle - \Theta(\tau) \, \langle \partial_\tau \chi(\tau) \; \chi \rangle \\ & -\eta \partial_\tau \Theta(-\tau) \, \langle \chi \; \chi(\tau) \rangle -\eta \Theta(-\tau) \, \langle \chi \; \partial_\tau \chi(\tau) \rangle \\ &= -\delta(\tau) \, \big( \chi \chi -\eta \, \chi \chi \big) - \langle \mathcal{T}_\tau \{ \partial_\tau \chi(\tau) \, \chi^\dagger \} \rangle \\ &= -\delta(\tau) \, \{ \chi , \chi \} - \langle \mathcal{T}_\tau \{ \partial_\tau \chi(\tau) \, \chi^\dagger \} \rangle ~, \end{align} $$ where $ \chi \equiv \chi(0) $, and we have used $$ \partial_\tau \Theta(\tau) = \delta(\tau) ~, $$ and noted that $\delta(\tau)$ enforces equal-time anti-commutation relations for the fermionic Majoranas, so that $$ [ \chi , \chi ]_{\eta = -1} \equiv \{ \chi , \chi \} = 1 ~. $$ Using the equations of motion for $\chi(\tau)$, we have $$ \partial_\tau \chi(\tau) = [H, \chi] (\tau) ~, $$ but the Hamiltonian for the Majoranas, $H$, is vanishing, since they are dispersionless and their energy is zero; hence, $$ \partial_\tau \chi(\tau) = 0 ~; $$ thus, the equation of motion for the Green's function reduces further to $$ \partial_\tau G(\tau) = -\delta(\tau) ~. $$ In the Matsubara-frequency representation, this will be equivalent to $$ -i \omega \, G(i \omega) = -1 ~, $$ or $$ G(i \omega) = \frac{1}{i \omega} ~, $$ where $ \omega = (2 \mathbb{Z} + 1) \, \pi \, T $, and $T$ denotes the temperature. In the imaginary-time representation, $$ G(\tau) = -\frac{1}{2} \mathrm{sgn}(\tau) ~, $$ where we have used the Fourier transform pair: $$ \mathcal{F}[\mathrm{sgn}](\nu) = -\frac{2}{i\nu} ~. $$

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After reading some references, I may know how to get the free mode Green's function. Here is my two cents.

In the imaginary time, the Lagrangian for free Majorana fermions is $$L=\frac{1}{2}\chi_j\partial_{\tau}\chi_j$$

The Majorana fermions satisfy $$\{\chi_i,\chi_j\}=\delta_{ij}.$$

From the Euler-Lagrange equation $\frac{\rm d}{{\rm d}t}\frac{\partial L}{\partial \dot \chi_j}=\frac{\partial L}{\partial \chi_j}$, we can get $$\frac{{\rm d}\chi_j}{{\rm d}\tau}=0$$ So, from the definition of imaginary time Green's function $$G_0(\tau)=\langle T(\chi_j(\tau)\chi_j(0))\rangle$$ we have $$G_0(\tau)=\langle \chi_j(\tau)\chi_j(0)\rangle=\langle\chi_j(0)\chi_j(0)\rangle=\frac{1}{2}$$ for $\tau>0$,

and $$G_0(\tau)=-\langle \chi_j(0)\chi_j(\tau)\rangle=-\langle\chi_j(0)\chi_j(0)\rangle=-\frac{1}{2}$$ for $\tau<0.$

So, in total, the free mode Green's function is $$G_0(\tau)=\frac{1}{2}\text{sgn}(\tau)$$

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  • $\begingroup$ Here, you have improperly used the classical equations of motion (Euler-Lagrange equations) for the Majorana modes with a quantum nature. That is not valid. $\endgroup$ – AlQuemist Aug 18 '17 at 12:30

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