Polarization of the lowest mode of a Gaussian beam

Most introductory analyses of Gaussian beam optics work within Helmholtz scalar optics, and therefore they ignore the beam's polarisation. Because of that, I'm not clear on what are the possible orientations for the lower transverse mode $T_{00}$ when full vectorial electric and magnetic fields are allowed.

Assume cylindrical coordinates, where the beam propagates in the axial coordinate $z$. The intensity of the fundamental mode is axially symmetric; but how does the electric field orientation for a linearly polarized gaussian beam look? Is it always aligned toward a fixed direction? Does it have axial symmetry as well?

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In the center of a Gaussian beam, the field structure is close to that in a plane wave with the same polarization. So the field is not axially symmetric for a linearly polarized Gaussian beam.

Let me note that Gaussian beams are not precise solutions of the free Maxwell equations. For this reason, a few years ago, I derived some precise solutions of the free Maxwell equations that are close to Gaussian beams in some sense (http://arxiv.org/abs/physics/0405091 , eq.22). I considered circular polarization only, but it is not difficult to build solutions with linear polarization starting from the solutions with circular polarization.

EDIT (07/04/2013): I am trying here to answer lurscher's question in the comments. It depends on where you place the surface of the reflecting material. If you place it in the center of the waist of the Gaussian beam, where the structure of the field is close to that in a plane wave, the reflection coefficients will be very close to those for a plane wave, I believe. If the reflecting surface is far from the center of the waist, the structure of the field will be close to that in radiation from a point source (within some angle of divergence). Directions of propagation will span some solid angle, so it seems the only way to eliminate s-polarization completely is to use a surface orthogonal to the axis of the Gaussian beam, however, in this case, reflection will be relatively small. On the other hand, if the Gaussian beam is narrow, the share of s-polarized radiation can be made small with a different choice of the reflecting surface. I don't have time right now to give more details.

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thanks, i assume that far-field properties like beam divergence and beam waist relationships are unchanged when taking polarisation into consideration right? divergence is still $\frac{\lambda}{\pi w_0}$? – lurscher Jun 19 '13 at 0:23
I believe so, but too lazy to check now:-) – akhmeteli Jun 19 '13 at 0:31
@lurscher: So what kind of details would you like to see? – akhmeteli Jun 30 '13 at 4:18
sorry I didn't replied earlier, I forgot about this bounty :-) Some materials have different reflectivities for p-wave and s-wave polarisations, so I want to know, if a Gaussian Beam hits a surface at an angle, which what kind of polarisation I'm dealing with? I was hoping that linear polarisation would stay aligned in a single spatial direction, so a mirror perpendicular to that would maximise reflectivity – lurscher Jul 3 '13 at 23:45
@lurscher: Please see the edit to my answer. – akhmeteli Jul 4 '13 at 18:20

In general, gaussian beams are understood to be taken as linearly polarized, but this is only approximately true. The full polarization is a complicated object to handle, and it is not necessarily available in a closed elementary form. It is described, for example, in

Analyses of vector Gaussian beam propagation and the validity of paraxial and spherical approximations. CG CHen et al. J. Opt. Soc. Am. A 19, 404 (2002).

In the paraxial regime, of course, the polarization is simpler, and if the beam waist $w_0$ and Rayleigh range $z_R=\pi w_0^2/\lambda$ are much longer than the wavelength $\lambda$ then you can simply take the beam to have a uniform linear polarization.

As you make the beam focus tighter, on the other hand, the polarization will cease to be uniform, but if you keep the ratio $\varepsilon = w_0 / \lambda$ small, then you can find analytical formulas for that dependence to first order in $\varepsilon$, given for example in

Above Threshold Ionization in Tightly Focused, Strongly Relativistic Laser Fields. A. Maltsev and T. Ditmire Phys. Rev. Lett. 90, 053002 (2003).

and which read \left\{\begin{aligned} E_x & = E_0\frac{1}{\sqrt{1+z^2/z_R^2}}\exp\left(-\frac{\rho^2}{w^2}\right)\sin(\phi_G) \\ E_y & = 0 \\ E_z & =2E_0\frac{\varepsilon}{1+z^2/z_R^2}\frac{x}{w_0}\exp\left(-\frac{\rho^2}{w^2}\right)\cos(\phi_G^{(1)}) \end{aligned}\right. and \left\{\begin{aligned} B_x & = 0 \\ B_y & = E_x/c \\ B_z & =\frac{2E_0}{c}\frac{\varepsilon}{1+z^2/z_R^2}\frac{y}{w_0}\exp\left(-\frac{\rho^2}{w^2}\right)\cos(\phi_G^{(1)}) , \end{aligned}\right. where $w=w_0\sqrt{1+z^2/z_R^2}$, $\rho^2=x^2+y^2$ and $\phi_G=\omega t-kz+\arctan(z/z_R)-zr^2/z_Rw^2$ as usual, and $\phi_G^{(1)}=\phi_G+\arctan(z/z_R)$.

In particular, some important features of these fields include that (i) both $\mathbf E$ and $\mathbf B$ include components in the 'propagation' direction $\hat{\mathbf z}$, and (ii) $\mathbf E$ and $\mathbf B$ are no longer completely orthogonal. Moreover, (iii) because the first-order terms in $\varepsilon$ depend on $x$ and $y$, this means that to first order the intensity is not necessarily completely symmetric with respect to rotations about the propagation axis, though this effect is very small.

Finally, I would also like to mention that radially-polarized gaussian beams (where the electric field polarization also has an axial symmetry about the propagation direction) are also a thing. However, they're complicated to produce and they're not what one usually means by the term 'gaussian beam' - you have to specifically say that that's what you're talking about.

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This is really interesting! So in some places, could $\mathbf E$ and $\mathbf B$ both be non-orthogonal simultaneously to both $\hat{\mathbf z}$ or $\hat{\mathbf P}$ (local Poynting vector)? – uhoh May 30 at 15:33
@uhoh Neither field can be orthogonal to the local Poynting vector $\mathbf S\propto\mathbf E\times\mathbf B$, by construction. However, they don't need to be orthogonal to the 'propagation direction'. That is anyway a washy concept to begin with, but even in the $z=0$ plane, where you'd think everything is along $z$ or in the plane, (i) both force fields oscillate (slightly) out of the $xy$ plane, and (ii) the Poynting vector oscillates (slightly) about the $z$ axis. – Emilio Pisanty May 30 at 15:47
oops - that's right. I'll go read those articles, fire up a plotter and have a close look at these. Haven't seen them yet, but I'm thinking these are Gaussian beams far away, and then when focused, reality sets in and they deviate. What if the initial beams is back-calculated to produce an exact, flat Gaussian at the focal plane, and then a phase apodizer is used to make this beam? Possibly been done already? Usually anything you can think of, someone has done already! – uhoh May 30 at 15:57

The Gaussian beam is a model of physical beam propagation (not even one that exactly satisfies Maxwell's equations at that, although it does satisfy certain paraxial approximations of the equations). In typical optics use, polarization is neglected, as you realize. However, if you specify a linearly polarized gaussian beam, it would have all the E vectors lined up in the same direction at a snapshot in time, and the intensity would vary in the usual Gaussian shape away from the axis. Most cheap diode lasers actually have this behavior, which you can verify experimentally with a laser pointer and linear polarizing filter. Turn the laser or filter axially and you should be able to visually identify the polarization axis by watching the intensity variation in the transmitted beam. To my knowledge, the polarization can be controlled using various optical devices to achieve circular or radial polarization as well, or more exotic polarization profiles; there is some exciting research in this area.

The exact field solutions for these probably can't be written down in closed form anyway, so we use the Gaussian beam as a model we can handle mathematically.

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