From Wikipedia:
Let $G = (V,E)$ be a graph with vertices $\scriptstyle V$ and edges $\scriptstyle E$. Let $\phi\colon V\to R$ be a function of the vertices taking values in a ring. Then, the discrete Laplacian $\Delta$ acting on $\phi$ is defined by:
$(\Delta \phi)(v)=\sum_{w:\,d(w,v)=1}\left[\phi(w)-\phi(v)\right]$,
where $d(w,v)$ is the graph distance between vertices $w$ and $v$. Thus, this sum is over the nearest neighbors of the vertex $v$.
The question:
I am going to do some handwavy dimensional analysis. Let the function $\phi$ have units $[F]$. Let $x$, the coordinate over which the Laplacian is taken have units $[L]$. Then, the Laplacian of $\phi$
$\Delta \phi = \frac {\partial^2 \phi}{\partial x^2}$,
has units $[F]/[L^2]$. However, the discrete version of the Laplacian, defined on the vertices of a graph seems to have units $[F]$ (see above -- all I am doing is taking the sum of differences, so my units are simply the same as the units of $\phi$).
Why are the units of the discrete version of the Laplacian not the same as the units of the continuous version? What mistake am I making in my "dimensional analysis"?