# Units of infinitessimal sum with kronecker delta when converted to integral

I'm having trouble understanding the following. Say I have a quantity of objects arranged spatially, with distribution $dn/dx$. For one reason or another*, I want to sum auto-pairs of spatial points, but only pairs at the same place. Thinking about this discretely, we have

$$\sum_{i,j} (\bar{n}_i \Delta x) (\bar{n}_j \Delta x) \delta_{ij} = \sum_i (\bar{n}_i \Delta x)^2$$

Clearly this is dimensionless on both sides. However, moving into the continuous limit, I expect to have

$$\int\int \frac{dn}{dx_i}\frac{dn}{dx_j}\delta(x_i-x_j)dx_idx_j = \int \left(\frac{dn}{dx}\right)^2 dx.$$

Here both sides have the same dimensions (inverse length), but this is different from the sum, which is dimensionless. What am I doing wrong??

Thanks for any help!

* For those who are interested, this problem arises (or at least I think it does) when finding the variance of a sum of correlated variables. In particular, let the number of objects be a function of $x$ and $y$, and let them be spatially correlated in $y$, but not $x$. Then

$${\rm Var}(\sum N_i) = \sum_{x_i}\sum_{x_j}\sum_{y_i}\sum_{y_j} (\bar{n}_{x_i} \Delta x \Delta y) (\bar{n}_{x_j} \Delta x \Delta y)C_{y_i y_j} \delta_{x_i x_j} = \sum_{x_i}\sum_{y_i}\sum_{y_j} (\bar{n}_i \Delta x \Delta y)^2 C_{y_i y_j} .$$

The rest follows as above (note that I have that the spatial covariance in $y$ at a given $x$ is $(\bar{n}_x \Delta x \Delta y)^2 C_{y_i y_j}$).

• You have to consider the differences of the discrete Dirac-Delta operator and the continuous Dirac-Delta distribution. If your discretization widths $\Delta x_i,\Delta x_j$ shrink your sum converges to zero since $\delta_{ij} = 1$ when $i=j$ and $\Delta x_i,\Delta x_j\rightarrow 0$. You have to replace the discrete Dirac-Delta by a density like $\frac{\delta_{i,j}}{\Delta x_i}$ to get something that converges to the Dirac-Delta distribution. Nov 18 '16 at 7:11
• @tobias -- I think this is the correct answer. Would you like to post it as an answer so I can accept? In my case, the discrete version is "correct" while the integrated version is wrong, so insead I'd have to replace the continuous Dirac with something like $\delta(x-k) dx$. Nov 21 '16 at 2:53

If I were summing up just values of $n$, and using the Kronecker delta to pull out one particular value, say $k$, then we'd have $$\sum_i a_i\delta_{ik} = a_k,$$ and its continuous analogue would be $$\int a(x)\delta(x-k)\,dx = a(k).$$ In all cases, the units are the same (whatever units $a$ has). Notice that the discrete version lacks the extra $\Delta x$ that you included.
• You could be right, but I'm not absolutely convinced. The sum of $a_i$ does not converge to the integral over $a$, rather the sum of $a_i \Delta x$ does. Unless you are saying what @tobias did -- i.e. that $\delta_{ik}$ in the continuous limit literally translates to $\delta(x-k) dx$? Nov 21 '16 at 2:56