Laplacian of Gaussian formula for 2d case is
$$\operatorname{LoG}(x,y) = \frac{1}{\pi\sigma^4}\left(\frac{x^2+y^2}{2\sigma^2} - 1\right)e^{-\frac{x^2+y^2}{2\sigma^2}},$$ in [scale-space](http://en.wikipedia.org/wiki/Scale_space#Scale_selection) related processing of digital images, to make the Laplacian of Gaussian operator invariant to scales, it is always said to normalize $LoG$ by multiplying $\sigma^2$, that is
$$\operatorname{LoG}\_\text{normalized}(x,y) = \sigma^2\cdot \operatorname{LoG}(x,y) = \frac{1}{\pi\sigma^2}\left(\frac{x^2+y^2}{2\sigma^2} - 1\right)e^{-\frac{x^2+y^2}{2\sigma^2}}.$$ I wonder why multiply by $\sigma^2$ not $\sigma^4$ or anything else?

**UPDATE**

Thanks to comments from @achille. From the perspective of *dimensional analysis*, in the Laplacian of Gaussian operator
$$LoG(x,y,\sigma)=\frac{\partial^2g}{\partial x^2} +\frac{\partial^2g}{\partial y^2}$$, I think $x,y$ are variables with dimension $L$, $\sigma$ is a parameter with dimension $L$. But what about $g$? Since $g$ is a function of $x,y,\sigma$,
$$g(x,y,\sigma)=\frac{1}{2\pi \sigma^2}exp(-\frac{x^2+y^2}{2\sigma^2})$$,
and $x,y,\sigma$ are of the same dimension $L$, so I guess in $g$, the term $exp(-\frac{x^2+y^2}{2\sigma^2})$ is dimensionless, isn't it? And the term $\frac{1}{2\pi \sigma^2}$ is of dimension $L^{-2}$, right? So $g$ is actually of dimension $L^{-2}$, isn't it?

Now come back to $LoG$, it should have dimension $L^{-4}$?

**UPDATE 2**

Laplacian operator is 
$$\nabla^2 = \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}$$,
and a Gaussian function's scale is $\sigma$, right? If I apply $\nabla^2$ on a Gaussian function $g(x,y,\sigma)$, what is difference of applying the dimensionless $\sigma^2\*\nabla^2$?