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 $\rho$.


1 Answer 1


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,

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


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

  • $\begingroup$ 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? $\endgroup$ Commented Jul 23, 2014 at 4:28

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