Section 21.2 Two Normalizations of a 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 preserve the norm. The two different normalization conditions (21.2.5) and (21.2.11) discussed below, relate the two parameters \(N\) and \(\sigma\) that determine the height and width of the Gaussian, respectively. Both normalization conditions are independent of the position of the peak, \(x_0\text{,}\) of the Gaussian.
Both normalization schemes below rely on using the value of the definite integral of a simple Gaussian
\begin{equation}
\int_{-\infty}^{\infty}e^{-x^2}\, dx =\sqrt{\pi}.\tag{21.2.1}
\end{equation}
and then doing a simple variable substitution. You can find the value of the definite integral in (21.2.1) using computer algebra and/or just look it up online or in a table of integrals. Alternatively, see the (optional) derivation below.
Probability Normalization of a Gaussian.
When Gaussian’s are used in probability theory, the total probability of all possible things 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{align}
1\amp =\int_{-\infty}^{\infty} Ne^{-\frac{(x-x_0)^2}{2\sigma^2}}\, dx\tag{21.2.2}\\
\amp =\int_{-\infty}^{\infty} Ne^{-y^2}\, \sqrt{2}\sigma\, dy\tag{21.2.3}\\
\amp =N\, \sqrt{2}\sigma\, \sqrt{\pi}\tag{21.2.4}
\end{align}
where we have used the substitutions \(y=\frac{x-x_0}{\sqrt{2}\sigma}\) and \(dy=\frac{1}{\sqrt{2}\sigma}\, dx\) in (21.2.3). We can use this normalization condition to find the value of the normalization parameter \(N\) in terms of the width parameter, \(\sigma\text{,}\) i.e.
\begin{equation}
N=\frac{1}{\sqrt{2\pi}\,\sigma}\tag{21.2.5}
\end{equation}
and the formula for the probability-normalized Gaussian is
\begin{equation}
\frac{1}{\sqrt{2\pi}\,\sigma}\,
e^{-\frac{(x-x_0)^2}{2\sigma^2}}\tag{21.2.6}
\end{equation}
Quantum Normalization of a Gaussian.
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. But probabilities in quantum mechanics are not given by the (possibly complex-valued) wave function, but rather, by the squared norm of the wave function. Therefore, it is necessary for the integral of the complex conjugate of the Gaussian times the Gaussian, for all \(x\text{,}\) 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{21.2.7}\\
\amp = \vert N\vert^2 \int_{-\infty}^{\infty}
e^{-\frac{2(x-x_0)^2}{2\sigma^2}}\, dx\tag{21.2.8}\\
\amp =\int_{-\infty}^{\infty} \vert N\vert^2
e^{-y^2}\, \sigma\, dy\tag{21.2.9}\\
\amp = \sqrt{\pi}\, \sigma\vert N\vert^2 \tag{21.2.10}
\end{align}
where we have used the substitutions \(y=\frac{x-x_0}{\sigma}\) and \(dy=\frac{1}{\sigma}\, dx\) in (21.2.8). We can use this new normalization condition to find the value of the normalization parameter \(N\) in terms of the width parameter, \(\sigma\text{.}\) Arbitrarily choosing the phase of \(N\) to be real, we obtain
\begin{equation}
N=\frac{1}{(\sqrt{\pi}\,\sigma)^{1/2}}\tag{21.2.11}
\end{equation}
and the formula for the quantum-normalized Gaussian is
\begin{equation}
\frac{1}{(\sqrt{\pi}\,\sigma)^{1/2}}\,
e^{-\frac{(x-x_0)^2}{2\sigma^2}}\tag{21.2.12}
\end{equation}
Derivation of the Definite Integral of a Gaussian (Optional).
There is a sweet trick to finding the integral of a Gaussian (21.2.1) that is worth knowing. First, give the name \(I\) to the integral you are trying to find
\begin{equation}
I=\int_{-\infty}^{\infty} e^{-x^2}\, dx\tag{21.2.13}
\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}
e^{-x^2}\, dx\right)
\left(\int_{-\infty}^{\infty}
e^{-y^2}\, dy\right)\tag{21.2.14}\\
\amp =\int_{-\infty}^{\infty}
\int_{-\infty}^{\infty}
e^{-(x^2+y^2)}\, dx\, dy\tag{21.2.15}
\end{align}
Finally, take advantage of the fact that the integration (abstractly) covers the whole two-dimensional plane to change to polar coordinates, centered at the origin, using the substitutions
\begin{gather}
x^2+y^2 = r^2\tag{21.2.16}\\
dx\, dy=r\, dr\, d\phi\tag{21.2.17}
\end{gather}
to obtain
\begin{align}
I^2\amp =\left(\int_{0}^{2\pi}
\left(\int_{0}^{\infty}
e^{-{r^2}}
r\, dr\right)\, d\phi\right)\tag{21.2.18}\\
\amp =\frac{1}{2}\left(\int_{0}^{2\pi}
\left(\int_{0}^{\infty}
e^{-{r^2}}
2r\, dr\right)\, d\phi\right)\tag{21.2.19}\\
\amp =\frac{1}{2}\left(\int_{0}^{2\pi}
\left(\int_{0}^{\infty}
e^{-u}
\, du\right)\, d\phi\right)\tag{21.2.20}\\
\amp =\pi\tag{21.2.21}\\
\Rightarrow I\amp =\sqrt{\pi}\tag{21.2.22}
\end{align}
