Mathematical Optics

The last expression is very convenient for further iterations, but contains two unknown functions u(x) and v(x). These functions can be easily found:

$\left\{\begin{array}{c}u(x)v(x)=a,\\ u(x){\mathrm{\partial }}_{x}v(x)=bx.\end{array}\right.\Rightarrow v(x)=v_0\exp \left(\frac{b{x}^2}{2a}\right),u(x)=\frac{a}{v(x)}=\frac{a}{v_0}\exp \left(-\frac{b{x}^2}{2a}\right).$

Consequently, we have

$\begin{array}{l}{S}_m(x)=a\exp \left(-\frac{b{x}^2}{2a}\right){\mathrm{\partial }}_x\left\{\exp \left(\frac{b{x}^2}{2a}\right)S_{m-1}(x)\right\}\\ =a^2\exp \left(-\frac{b{x}^2}{2a}\right)({\mathrm{\partial }}_x{)}^2\left\{\exp \left(\frac{b{x}^2}{2a}\right)S_{m-2}(x)\right\}\\ =\dots =a^m\exp \left(-\frac{b{x}^2}{2a}\right)({\mathrm{\partial }}_x{)}^m\left\{\exp \left(\frac{b{x}^2}{2a}\right)S_0(x)\right\}\\ =a^m\exp \left(-\frac{b{x}^2}{2a}\right)({\mathrm{\partial }}_x{)}^m\left\{\exp \left(-x^2\left[c-\frac{b}{2a}\right]\right)\right\}\\ \stackrel{(5.29)}{=}\exp \hspace{0.17em}(-c{x}^2){\left(-a\sqrt{c-\frac{b}{2a}}\right)}^m{H}_m\left(X\sqrt{c-\frac{b}{2a}}\right).\end{array}$

Using this result, it is easy to see that the action of the operator ${\mathrm{P}}_5^m{\mathrm{P}}_6^n$ on the Gaussian function is a particular case of gHG beam (5.35):

$\begin{array}{ll}{\mathrm{P}}_5^m{\mathrm{P}}_6^{n}G(\mathbf{r},z)& =G(\mathbf{r},z){(\frac{1-\sigma }{2\sigma })}^{\frac{m+n}{2}}{H}_m\left(\frac{x}{\sqrt{2\sigma (1-\sigma )}}\right)H_n\left(\frac{y}{\sqrt{2\sigma (1-\sigma )}}\right)\\ =2^{-\frac{m+n}{2}}g\mathrm{H}{\mathrm{G}}_{m,n}(\mathbf{r},z|1,1),\end{array}$

which at the plane z = 0 has the form

$\{{\mathrm{P}}_5^m{\mathrm{P}}_6^{n}G(\mathbf{r},z)\}{|}_{z=0}={\mathrm{P}}_5^m{|}_{z=0}\cdot {\mathrm{P}}_6^n{|}_{z=0}{G}_0(\mathbf{r})=G_0(\mathbf{r})x^m{y}^n.$

The general expression for the gHG beams is obtained ...