# How to express Dirac delta-function in functional form?

In the paper written by Dmitry Bagrets, Alexander Altland and AlexKamenev (Sachdev–Ye–Kitaev model as Liouville quantum mechanics: http://www.sciencedirect.com/science/article/pii/S0550321316302206?via%3Dihub , http://dx.doi.org/10.1016/j.nuclphysb.2016.08.002 , open access, arXiv:1607.00694), they introduced an identity, equation (5): $$1=\int {D}G\delta(NG_{\tau,\tau^{\prime}}+A_{\tau,\tau^{\prime}})=\int DGD\Sigma e^{\frac{1}{2}\int d\tau\int d\tau^{\prime}\Sigma_{\tau,\tau^{\prime}}(NG_{\tau,\tau^{\prime}}+A_{\tau,\tau^{\prime}})} .\tag{5}$$ They said, we need $G_{\tau,\tau^{\prime}}=-\frac{1}{N}A_{\tau,\tau^{\prime}}$, and $\Sigma_{\tau,\tau^{\prime}}$ is the Lagrange multiplier (real field). Does this mean $$\int D\Sigma e^{\Sigma(NG+A)}=\delta(NG+A)~?$$

I do not understand the upper equation, because in my mind we have $$\int \frac{dx}{2\pi}e^{iax}=\delta(a).$$ And why is $\Sigma_{\tau,\tau^{\prime}}$ called Lagrange multiplier?

Indeed, $$\int D\Sigma \,e^{i \int_{\tau,\tau'}\Sigma_{\tau,\tau'}G_{\tau,\tau'} }=\int D\Sigma \,e^{i S[\Sigma]}=\prod_{\tau,\tau'}\delta(G_{\tau,\tau'})$$ is just a generalization of $$\int dx\, e^{i a x}=\delta (a).$$ Note that sometimes the $i$ in the path integral is omitted. This is under the assumption that the analytical continuation $\Sigma\to -i\Sigma$ is allowed (this can be sometimes justified in perturbation theory, but it might also not be allowed non-perturbatively). This is usually done to render the minimum of the action (and thus the Lagrange multiplier, see below) real.
$\Sigma$ is sometimes called a Lagrange multiplier (somewhat abusively), because a mean-field (or equivalently a stationary phase) approximation of $\int D\Sigma \,e^{i S[\Sigma]}$ is given by the extremization of $S[\Sigma]$, that is, for $\delta S/\delta \Sigma=0$ (in addition to other mean-field terms coming from the integration over $G$ and other fields).
In this approximation, $\Sigma$ exactly plays the role of a Lagrange multiplier.
• Thank you, but there is no $i$ in the equation (5) and this is what I do not really understand. – Sheng Liu Aug 22 '17 at 8:57
• See my edit, which explains why the $i$ can be omitted. – Adam Aug 22 '17 at 9:06
• @ShengLiu : stated otherwise, do the calculation in mean-field with and without the $i$, and you'll see that the end result will be the same. – Adam Aug 22 '17 at 9:06
• I do not see why the first relation (in the answer) would hold by an analytical continuation of some sort. For example, it does not hold if one applies the same "analytical continuation" to $\int_{-\infty}^{+\infty} dx \, e^{i a x}$. I see the relation in the aforementioned paper more in the direction of Hubbard-Stratonovich transformations. – AlQuemist Aug 24 '17 at 15:06
• @PhilosophiaeNaturalis : No, you can't on this simple integral. But in the context of the SYK model, there are multiple functional integrals, and importantly, there is a large parameter ($N$), which justifies this saddle-point approximation. Have a look for instance here lptmc.jussieu.fr/files/chap_rg.pdf , starting at equation 7.523. – Adam Aug 25 '17 at 8:08