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.

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 two caces 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

1. 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)
2. 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.
3. https://en.wikipedia.org/wiki/Gaussian_beam