# Where does the delta of zero $\delta(0)$ come from?

It is common when evaluating the partition function for a $$O(N)$$ non-linear sigma model to enforce the confinement to the $$N$$-sphere with a delta functional, so that $$Z ~=~ \int d[\pi] d[\sigma] ~ \delta \left[ \pi^2 + \sigma^2 -1 \right] \exp (i S(\phi)),$$ where $$\pi$$ is an $$N-1$$ component field. Then, one evaluates the integral over $$\sigma$$, killing the delta functional. In my understanding, this gives rise to a continuous product of Jacobians, $$\prod_{x=0}^L \frac{1}{ \sqrt{1 - \pi^2}} ~=~ \exp \left[- \frac{1}{2} \int_0^L dx \log (1 - \pi^2) \right]$$ (where I have now put everything in one dimension). Now, obviously this is somewhat non-sense, at the very least because there are units in the argument of the exponential. The way I actually see this written is with a delta function evaluated at the origin, $$\exp \left[- \frac{1}{2} \int_0^L dx \log (1 - \pi^2) \delta(x-x) \right].$$ I see that this makes the units work, but what does that really mean? How do people know to put it there? I know $$\delta(0)$$ can sometimes be understood as the space-time volume. However, in this case, it clearly has units of $$1/L$$, so is presumably more like a momentum-space volume. In one dimension, does that mean I can just replace it with $$1/L$$ (up to factors of $$2$$ or $$\pi$$)?

In particular, I have noticed this in the following papers:

• Brezin, Zinn-Justin, and Le Guillou, Renormalization of the nonlinear $$\sigma$$ model in $$2+\epsilon$$ dimensions, Phys. Rev. D 14 (1976) 2615; eq. (4).

• Kleinert and Chervyakov, Perturbation theory for path integrals of stiff polymers, arXiv:cond-mat/0503199; eq. (10).

Kardar does something similar in his Statistical Physics of Fields book, but he simply calls it $$\rho$$.

This formula follows the usual heuristic discretization rules (here written in 1D):

\begin{align} i\in\{1, \ldots,N\}, ~~x_i=i\Delta \qquad\longrightarrow\qquad&~x~\in~[0,L]\cr \text{discrete var.}\qquad\qquad\qquad &\text{cont. var.},\end{align} \tag{1}

$$\text{sum}\qquad \sum_{i=1}^N \qquad\longrightarrow\qquad \int_0^L \! \frac{dx}{\Delta} \qquad\text{integral},\tag{2}$$

$$\text{"volume" of unit cell:}\qquad \Delta ~=~\frac{L}{N}, \tag{3}$$

\begin{align} \frac{1}{\Delta} \delta_{i,j} \qquad\longrightarrow\qquad& \delta(x_i-x_j) \cr \text{Kronecker delta fct}\qquad\qquad\qquad& \text{Dirac delta fct},\end{align} \tag{4}

$$\frac{1}{\Delta}\qquad\longrightarrow\qquad\delta(0), \tag{5}$$

for $$N\to \infty$$. Hence, formally,

$$f(x_j)~=~\sum_{i=1}^N \delta_{i,j} ~f(x_i) ~~\longrightarrow~~ \int_0^L \! dx~\delta(x-x_j) ~f(x), \tag{6}$$

and

\begin{align} \prod_{i=1}^N \exp\left[f(x_i)\right] ~=~& \exp\left[\sum_{i=1}^Nf(x_i)\right] \cr\cr ~\downarrow~&\cr\cr \exp\left[\int_0^L \! \frac{dx}{\Delta}f(x)\right]~=~&\exp\left[\delta(0)\int_0^L \! dx~f(x)\right] .\end{align}\tag{7}

• Ah, this is so simple! Thanks a lot. I suppose my assumption that this would be proportional to $1/L$ is wrong then. Would it actually be proportional to some sort of short wavelength cutoff? Jul 23, 2014 at 4:28