Defining the width of a Gaussian function I have the following Gaussian function:
$$\rho(r) = q_i (\alpha/\pi)^{3/2} \exp(-\alpha r^2)$$
Qualitatively, the "width" of this Gaussian is related to $\frac{1}{\alpha}$: the larger the value of $\alpha$, the smaller the "width" of the Gaussian.  
This Wikipedia article uses this definition of a Gaussian function:
$$f(x) = a\exp \left(-\frac{(x-b)^2}{2c^2}\right)$$
and says that one way to define the width is to consider the full width at half maximum:
$$\text{FWHM} = 2 \sqrt{2 \ln 2} c \approx 2.35482c$$
In other words, the width of $f(x)$ is proportional to $c$.  But, in my function $\rho(r)$, $\alpha$ appears in two places:  in the exponential and as a coefficient of the exponential.  How should I define the width of $\rho(r)$?

The proportionality constant in a Gaussian function does not affect its width. Consider the general form of a Gaussian function (may not be a probability density function):
$f(x) = a \exp\Big(-\frac{(x-b)^2}{c^2}\Big)$
The maxima occurs at $x=b$, and its value is $a$. Hence, the value of $f(x)$ at half maximum is $0.5a$. You can now find the value of $x$ at which $f(x) = 0.5a$:
$f(x^*) = a\exp\Big(-\frac{(x^*-b)^2}{c^2}\Big) = 0.5a$
The proportionality constant $a$ cancels out from both sides, and we end up with $(x^*-b) = \pm c\sqrt{\ln 2}$. The $FWHM$ thus becomes $2c\sqrt{\ln 2}$. Even if you were to define width using any other definition, say full width at $\alpha$ maximum (where $\alpha < 1$), the proportionality constant would always cancel out. Further, the center ($b$) would not matter as well since the Gaussian function is symmetric.
