# Hermite Gauss mode

**Hermite Gauss mode** refers to the specific solution $F=F(x,z)$ of equation

$ F''+ 2 \mathrm i \dot F=0$

where prime differentiates with respect to the first argument, and dot differentiates with respect to the last argument. This equation appears at the paraxial propagation of monochromatic beam. If $k$ is wavenumber, then, assuming some large positive $M$, expression $M^2 z/k$ has sense of the coordinate along the propagation of wave, and $M x/k$ has sense of the transversal coordinate.

The **Hermite Gauss mode** is expressed in terms of the Hermite polynomial
and the exponent.

## Contents |

## Notations

In the literature, there is no unified notations for the Hermite Gauss mode. Among them, the more often are so-called Elegant Hermite Gauss and Classical Hermite Gauss

### Elegant

The Elegant Hermite Gauss can be written as follows:

$\displaystyle F_m(x,z)=\frac{ w^{m/2} (w+\mathrm i z)^{-\frac{m}{2}-\frac{1}{2}} } {\sqrt{\pi\, 2^m\, m!}} \, H_m\!\left(\frac{x}{\sqrt{2} \sqrt{w+\mathrm i z}}\right) \, \exp\! \left(-\frac{x^2}{2 (w+\mathrm i z)}\right) $

where $w$ refers to the with of the beam in the narrowest position, and $H_m$ is the Hermite polynomial.

In Mathematica, the solution above can be verified with the code below:

F[x_] = ((w+I z)^(-1/2-m/2) HermiteH[m,x/(Sqrt[2] Sqrt[w+I z])]) Exp[-x^2/(2(w+I z))]

em = Simplify[ F''[x] + 2 I D[F[x],z] ]

Simplify[ReplaceAll[em, m -> 2]]

Simplify[ReplaceAll[em, m -> 3]]

Simplify[ReplaceAll[em, m -> 4]]

### Classical

The Classical Hermite Gauss can be written as follows:

$f_m(x,z)= \displaystyle \frac{ \left(o^2\!+\!z^2\right)^{-1/4}}{\sqrt{2^m\, m!\, \pi}}\, H_m\!\left(\frac{x}{\sqrt{o+z^2/o}}\right) \, \exp\! \left( -\frac{x^2}{2 (o+\mathrm i z)} - \mathrm i \left( m\!+\!\frac{1}{2} \right) \tan^{-1}\left(\frac{z}{o}\right) \right) $

where $o$ has the similar meaning as $w$ above, it characterises the waste of the beam.

In Mathematica, this solution can be verified as follows:

f[x_]=E^(-x^2/(2(o+I z)) - (I/2) (1+2m) ArcTan[z/o]) HermiteH[m,x/Sqrt[o+z^2/o]])/(1+z^2/o^2)^(1/4)

em = Simplify[f''[x] + 2 I D[f[x], z]]

Simplify[ReplaceAll[em, m -> 2]]

Simplify[ReplaceAll[em, m -> 3]]

Simplify[ReplaceAll[em, m -> 4]]

The Classical Hermite Gauss with waist $o$ can be obtained with scaling $x \rightarrow ox$, $z \rightarrow o^2z$, as $\frac{1}{o} \tilde f_m(ox,o^2z)$

where

$\tilde f_m(x,z)= f_m(x,z)_{\left| \displaystyle \begin{array}{c} \\ o \rightarrow 1 \end{array}\right. }$ $= \displaystyle \frac{ \left(1 \!+\!z^2\right)^{-1/4}}{\sqrt{2^m\, m!\, \pi}}\, $ $\displaystyle H_m\!\left(\frac{x}{\sqrt{1\!+\!z^2}}\right) \,$ $\displaystyle \exp\! \left( \frac{-x^2}{2 (1\!+\!\mathrm i z)} - \mathrm i \left( m\!+\!\frac{1}{2} \right) \tan^{-1}(z) \right) $

### Comparison

Similiarities in expressions for $F_m(x)$ and $f_m(x)$ causes ceduction to set $o=w$ and to scale the solutions, in such a way, that

$F_m(x,z)=A_m f_m(x,z)$

where $A_m$ is scaling factor, that may depend on $m$. However, this is not a case, because the scale of the Hermite polynomials is different, even at $z=0$. So, these are different solutions of the same equation.

## Multidimensional

In the similar form, the **Hermite Gauss mode** can be written for the equation with two transversal coordinates
^{[1]}^{[2]}

## Generailzation

The both vases above can be combined with generalised Hermite Gauss

$\displaystyle G_m(x,z)=$ $\displaystyle H_m\left(\frac{x}{w \sqrt{1+\mathrm i z/w^2} \sqrt{2-c \left(1+\mathrm i z/w^2\right)}}\right) $ $\displaystyle \exp\left( -\frac{x^2}{2 w^2 \left(1+\mathrm i z/w^2\right)}\right) $ $\displaystyle \exp \left(\frac{m}{4} \ln \left(4+c^2 \left(\frac{z}{w^2}-\mathrm i\right)^2\right)-\frac{\mathrm i\, m}{2} \tan ^{-1}\left(\frac{c}{2} \left(\frac{z}{w^2}-\mathrm i\right)\right)-\frac{m+1}{2} \log \left(\frac{z}{w^2}-\mathrm i\right) \right) $

where $w$ is size of the waist of the beam (it is placed at zero) and the real parameter $c<2$ determines the scale for the Hermite polynomials; the larger is $c$, the denser are oscillations of the Hermite function at the waist.

This solution can be verified with the Mathematica code below:

G[x_] = HermiteH[m, x/(w Sqrt[1+I z/w^2] Sqrt[2-c(1+I z/w^2)])] Exp[-x^2/(2 w^2) 1/(1+I z/w^2)] *

Exp[-(1/2) I m ArcTan[1/2 c(-I+z/w^2)] - 1/2 (1+m) Log[-I + z/w^2] +1/4 m Log[4 + c^2 (-I+z/w^2)^2] ]

em = Simplify[(G''[x] + 2 I D[G[x], z])]

Table[Simplify[em], {m, 0, 9}]

Parameter $w$ is kept in equations above to make allowance with the tradition. In order to boost the deduction, one may set $w=1$; then in the final expression, this parameter can be recovered with replacement $x\rightarrow x/w$, $z\rightarrow z/w^2$.

At $c=0$, the solution $G_m$ can be considered as the elegant Gaussian mode and at $c=1$, the solution can be considered as the classic Gaussian mode. In this sense, parameter $c$ can be called "classicality".

## Modifications

However, the solution can be transformed, using the symmetries of the equation: two translations, one tilt and one scaling. For this reason, it is sufficient to consider the case $w=1$, id est,

$\displaystyle G_m(x,z)=$ $\displaystyle H_m\left(\frac{x}{\sqrt{1+\mathrm i z} \sqrt{2-c \left(1+\mathrm i z\right)}}\right) $ $\displaystyle \exp\left(-\frac{x^2}{2 \left(1+\mathrm i z\right)}\right) $ $\displaystyle \exp \left(\frac{m}{4} \ln \left(4+c^2 \left(z-\mathrm i\right)^2\right)-\frac{\mathrm i\, m}{2} \tan ^{-1}\left(\frac{c}{2} \left(z-\mathrm i\right)\right)-\frac{m+1}{2} \log \left(z-\mathrm i\right) \right) $

Solution with realistic width can be recovered with transform

$\displaystyle G_m(x,z) \rightarrow \frac{1}{\sqrt{w}}\, G_m\!\left(\frac{x}{w}, \frac{z}{w^2}\right)$

The solution allow the straightforward generalisation to the multidimensional case.

## Other forms

Several authors (and citing them wikipedia), suggest another form of the Hermite Gaussian mode,
where argument of the Hermite polynomial appears as real function of coordinates.
^{[3]}

The Editor is not successful to write a Mathematica code to verify such a "solution"; the "solutions" seem to be just wrong. The origin of the error seems to be confusion of the sense of parameter $w$ that scales the argument of the Hermite polynomial.

## References

- ↑ https://www.researchgate.net/profile/Francesco_Pampaloni/publication/2171403_Gaussian_Hermite-Gaussian_and_Laguerre-Gaussian_beams_A_primer/links/0912f50f03f0ea21c7000000.pdf Francesco Pampaloni, Jörg Enderlein. Gaussian, Hermite-Gaussian, and Laguerre-Gaussian beams: A primer. (2016)
- ↑ https://www.researchgate.net/profile/Francesco_Pampaloni/publication/8382690_Unified_operator_approach_for_deriving_Hermite-Gaussian_and_Laguerre-Gaussian_laser_modes/links/0912f50f03f0de26ca000000.pdf Jorg Enderlein, Francesco Pampaloni. Unified operator approach for deriving Hermite – Gaussian and Laguerre–Gaussian laser modes J. Opt. Soc. Am. A, Vol. 21, No. 8 / August 2004, p.1553–1558.
- ↑ https://en.wikipedia.org/wiki/Gaussian_beam