Section 21.2 Normalization of the Gaussian
In Section 21.1, we gave a general formula for a Gaussian function with three real parameters \(N\text{,}\) \(x_0\text{,}\) and \(\sigma\) and you were able to explore how these parameters separately affect the shape of the graph. If you want a Gaussian to be normalized in some way, then as the Gaussian gets taller, it must also get narrower in just the right way to keep the area under the curve the same. The two normalization conditions (21.2.10) and (21.2.15) discussed below, relate the two parameters \(N\) and \(\sigma\) that determine height and width, respectively.
(Optional) Derivations of the Integral of a Gaussian.
Of course, you can use computer algebra to find the value of the integral and/or just look it up online or in a table of integrals, i.e.
\begin{equation}
\int_{-\infty}^{\infty}e^{-x^2}\, dx =\sqrt{\pi}\tag{21.2.1}
\end{equation}
and use an appropriate substitution. But there is a sweet trick to finding the integral of a Gaussian that is worth knowing.
First, give the name \(I\) to the integral you are trying to find
\begin{equation}
I=\int_{-\infty}^{\infty} Ne^{-\frac{(x-x_0)^2}{2\sigma^2}}\, dx\tag{21.2.2}
\end{equation}
Next, find the value of the square of this integral, i.e. this integral times itself. Don’t forget to name the ``dummy’’ variables of integration with different names. In this case, we will call the variables \(x\) and \(y\text{,}\) to suggest coordinates in the plane.
\begin{align}
I^2\amp =\left(\int_{-\infty}^{\infty}
Ne^{-\frac{(x-x_0)^2}{2\sigma^2}}\, dx\right)
\left(\int_{-\infty}^{\infty}
Ne^{-\frac{(y-y_0)^2}{2\sigma^2}}\, dy\right)\tag{21.2.3}\\
\amp =\int_{-\infty}^{\infty}
\int_{-\infty}^{\infty}
N^2 e^{-\frac{(x+x_0)^2+(y-y_0)^2}{2\sigma^2}}
\, dx\, dy\tag{21.2.4}
\end{align}
Finally, taking advantage of the fact that the integration (abstractly) covers the whole two-dimensional plane, we change to polar coordinates, centered at the point \((x_0, y_o)\text{,}\) using the substitutions
\begin{gather}
(x-x_0)^2+(y-y_0)^2 = r^2\tag{21.2.5}\\
dx\, dy=r\, dr\, d\phi\tag{21.2.6}
\end{gather}
to obtain
\begin{align}
I^2\amp =\left(\int_{0}^{2\pi}
\left(\int_{0}^{\infty}
N^2 e^{-\frac{r^2}{2\sigma^2}}
r\, dr\right)\, d\phi\right)\tag{21.2.7}\\
\amp =2\pi \sigma^2 N^2\tag{21.2.8}
\end{align}
(Note the limits of integration!)
In the final line, the new factor of \(r\) in the integrand made it possible to do the integral with the substitution given by \(u=\frac{r^2}{2\sigma^2}\) (Details left to the reader. See Section 6.7 for an explanation of substitution in integrals.)
Two Different Applications for Gaussians Require Different Norms.
When Gaussian’s are used in probability theory, the total probability of anything happening is one. Therefore, it is necessary for the integral of the Gaussian, for all \(x\text{,}\) to be equal to one, i.e. the area under the graph of the Gaussian is equal to one.
\begin{equation}
1=\int_{-\infty}^{\infty} Ne^{-\frac{(x-x_0)^2}{2\sigma^2}}\, dx\tag{21.2.9}
\end{equation}
We can use this condition to find the value of the normalization parameter \(N\) in terms of the other two parameters, so we would choose
\begin{equation}
N=\frac{1}{\sqrt{2\pi}\,\sigma}\tag{21.2.10}
\end{equation}
and the formula for the normalized Gaussian is
\begin{equation}
f(x)=\frac{1}{\sqrt{2\pi}\,\sigma}\,
e^{-\frac{(x-x_0)^2}{2\sigma^2}}\tag{21.2.11}
\end{equation}
Alternatively, when Gaussian’s are used to describe wave packets in quantum mechanics, the total probability of the particle being somewhere in space is one. Therefore, it is necessary for the integral, for all \(x\text{,}\) of the complex conjugate of the Gaussian times the Gaussian to be equal to one, i.e. the area under the graph of the squared norm of the Gaussian is equal to one.
\begin{align}
1\amp =\int_{-\infty}^{\infty}
\left(Ne^{-\frac{(x-x_0)^2}{2\sigma^2}}\right)^*
\left(Ne^{-\frac{(x-x_0)^2}{2\sigma^2}}\right)\, dx\tag{21.2.12}\\
\amp = \vert N\vert^2 \int_{-\infty}^{\infty}
e^{-\frac{(x-x_0)^2}{\sigma^2}}\, dx\tag{21.2.13}\\
\amp = \sqrt{\pi}\, \sigma\vert N\vert^2 \tag{21.2.14}
\end{align}
We can use this condition to find the value of the normalization parameter \(N\) in terms of the other two parameters, so, for the simplest possible phase, we could choose
\begin{equation}
N=\frac{1}{\sqrt{\sqrt{\pi}\,\sigma}}\tag{21.2.15}
\end{equation}
and the formula for the normalized quantum Gaussian is
\begin{equation}
f(x)=\frac{1}{\sqrt{\sqrt{\pi}\,\sigma}},
e^{-\frac{(x-x_0)^2}{2\sigma^2}}\tag{21.2.16}
\end{equation}
