# Problem in the continuum limit of a Kronecker delta

I am having troubles in understanding how to correctly perform the continuum limit of a double sum containing a Kronecker delta.

Imagine to integrate a function depending on $$t$$ and $$t'$$, both ranging from $$0$$ (initial time) to $$T$$ (final time):

$$I:=\int_0^Tdt_1\int_0^Tdt_2 f(t_1,t_2).\tag{1}$$

The corresponding Riemann sums, dividing the time intervals in slices of width $$\epsilon=T/N$$ is:

$$I_{disc}:=\sum_{j,j'=1}^N \epsilon^2 f(j\epsilon,j'\epsilon).\tag{2}$$

Obviosly lim$$_{N\rightarrow\infty}I_{disc}=I$$. Now consider the case when only the diagonal elements of the double integral are different from zero i.e.

$$J_{disc}:=\sum_{j,j'=1}^N \delta_{j,j'}\epsilon^2 f(j\epsilon,j'\epsilon).\tag{3}$$

I would expect that, in the continuum limit $$N\rightarrow \infty$$ ($$\epsilon\rightarrow 0$$) it becomes

$$J:=\int_0^Tdt\int_0^Tdt'\delta(t-t') f(t,t'),\tag{4}$$

where $$\delta(t-t')$$ is a Dirac delta.

Here is the problem: the Kronecker delta is adimensional, while the Dirac Delta has the dimensions of seconds$$^{-1}$$. This implies that $$J_{disc}$$ and $$J$$ have different dimensions, which does not make any sense. Therefore, there must be some mistake I am doing in going from the discrete to the continuum version of $$J$$. Could you help me spotting it and, more important, suggest the way to do this limit correctly?

• Maybe $\delta\left(\frac{t-t'}{T}\right)$? – Cryo May 14 '19 at 0:06

$$\frac{1}{\epsilon}\delta_{j,j^{\prime}}\qquad\longrightarrow\qquad\delta(t-t^{\prime}),\tag{A}$$
where $$\epsilon$$ is the "volume" of a unit-cell in the discretization. See e.g. this related Phys.SE post.
In particular, the rhs. of OP's eq. (3) should be divided with $$\epsilon$$ to have a finite continuum limit.
• $\uparrow$ Yes. – Qmechanic May 15 '19 at 16:49