## 3 Gauss-Maxwell Beams
### Linearly Polarized Gauss-Maxwell Beams  
As seen above, the ray \(\mathbf{r}_{\perp,z}=(x_{z},y_{z})\) described by Equations (23) and (24) allows to understand the distribution of the electromagnetic energy along the \(z\) axis when the latter is specified by only a scalar field. In this regard, such rays have analogous properties to those that we find for quantum wave packets [34; 39]. However, they do not contain any vector-type information, because they have not been obtained from the paraxial form of [PERSON]'s equations, but from the paraxial approximation applied to [PERSON]'s equation. To this end, first we need to determine the electric and magnetic field components associated with the evolution of the input Gaussian amplitude (9), which is done with the aid of the transformation relation (32), and then the rays are determined from Equation (46). In this latter regard, it is interesting to compare the outcome from this equation for a Gauss-Maxwell beam with the rays described by Equations (23) and (24) for a bare Gaussian beam. For simplicity in the analysis, here we are going to consider the case of horizontal polarization (along the \(x\) axis; vertical polarization is taken along the \(y\)-axis), leaving the case of arbitrary polarization for the next section. Accordingly, the input electromagnetic vector field, \(\mathbf{F}(\mathbf{r}_{\perp};0)\), reads as  
\[\mathbf{F}(\mathbf{r}_{\perp};0)=\frac{1}{\sqrt{2}}\;\psi(\mathbf{r}_{\perp};0 )\left( \begin{array}{c}\sqrt{\epsilon_{0}}E_{0}\\ 0\\ 0\\ 0\\ \sqrt{\mu_{0}}H_{0}\\ 0\end{array} \right)=\sqrt{\frac{\epsilon_{0}}{2}}\;E_{0}\psi(\mathbf{r}_{ \perp};0)\left( \begin{array}{c}1\\ 0\\ 0\\ 0\\ 1\\ 0\end{array} \right), \tag{47}\]  
where we have made use of the relation between amplitudes \(H_{0}=\sqrt{\epsilon_{0}/\mu_{0}}\;E_{0}\). Following the transformation relation (32), and taking into account the functional form displayed by the output Gaussian amplitude (12), the action of the operator \(\underline{\underline{\mathbb{Q}}}\) on the scala field \(\psi\), generates the matrix operator  
\[\underline{\underline{\psi}}(\mathbf{r}_{\perp};z)\equiv\underline{\underline{ \mathbb{Q}}}(\mathbf{r}_{\perp}\mathbb{I}+k^{-1}\mathbb{G}_{\perp},\hat{ \mathbf{p}}_{\perp}\mathbb{I})\psi(\mathbf{r}_{\perp};0)=\underline{\psi}(x \mathbb{I}+k^{-1}\mathbb{G}_{x},y\mathbb{I}+k^{-1}\mathbb{G}_{y};z)=A_{z}e^{ \alpha\mathbf{r}_{z}\left(\Delta\mathbf{r}_{\perp}\mathbb{I}+k^{-1}\mathbb{G }_{\perp}\right)^{2}}, \tag{48}\]  
where the argument of (12) has been replaced by the first argument of \(\underline{\underline{\mathbb{Q}}}\) (the second argument of \(\underline{\underline{\mathbb{Q}}}\) is the generator of the transformation from \(z=0\) to finite \(z\)), with \(\Delta\mathbf{r}_{\perp}=\mathbf{r}_{\perp}-\mathbf{r}_{\perp,0}^{\prime}\). For convenience, we have defined  
\[\alpha_{z}=-\frac{1}{4\sigma_{0}\bar{\sigma}_{z}}=\frac{ik}{2q_{z}}, \tag{49}\]  
with  
\[q_{z}=-2 ik\sigma_{0}\bar{\sigma}_{z}=-2 ik\sigma_{0}\sigma_{z}e^{i\varphi_{z} }=2k\sigma_{0}\sigma_{z}e^{i(\varphi_{z}-\pi/2)}. \tag{50}\]  
From now on, because it is more compact, we shall consider the last expression for \(q_{z}\) in the derivations. Before further proceeding, it is convenient to simplify the argument of the exponential in (48), just to determine how the matrices involved will act on the column vector (47). To this end, making use of the properties satisfied by the \(\mathbb{G}\)-matrices, we find  
\[\left(\Delta\mathbf{r}_{\perp}\mathbb{I}+k^{-1}\mathbb{G}_{\perp}\right)^{2}= \Delta\mathbf{r}_{\perp}^{2}\mathbb{I}+2k^{-1}\Delta\mathbf{r}_{\perp}\cdot \mathbb{G}_{\perp}, \tag{51}\]  
with  
\[(\mu+\upsilon)^{n}=\sum_{j=0}^{n}\left( \begin{array}{c}n\\ j\end{array} \right)\mu^{n-j}\upsilon^{j}=\sum_{j=0}^{n}\frac{n!}{j!(n-j)! }\mu^{ n-j}\upsilon^{j}.