Parallel and antiparallel beams of twisted light

The orbital angular momentum and polarization of a twisted light beam can be chosen at will. A careful examination of fields' profiles reveal that light beams can be divided in two categories or classes, according to the sign of orbital angular momentum and polarization. Fields belonging to different classes present striking differences with unexpected consequences. [1]

The vector potential of a twisted-light (TL) beam in the Coulomb gauge is
\begin{eqnarray}\label{Eq:A_Bessel_r}
A_r(\mathbf r, t)
&=&
F_{q_r\ell}(r) \cos[(\omega t-q_z z) -
(\ell + \sigma)\varphi]\,,
\\ \label{Eq:A_Bessel_phi}
A_\varphi(\mathbf r, t)
&=&
\sigma F_{q_r\ell}(r) \sin[(\omega t-q_z z) - (\ell + \sigma)\varphi]\,,
\\ \label{Eq:A_Bessel_z}
A_z(\mathbf r, t)
&=&
-\sigma \frac{q_r}{q_z} F_{q_r\ell+\sigma}(r)
\sin[(\omega t-q_z z) - (\ell + \sigma)\varphi]
\,,
\end{eqnarray}
where, as we usually write, $\ell$ gives the orbital angular momentum (OAM). The circular polarization of the field, with vectors $\boldsymbol{\epsilon}_{\sigma}= e^{i \sigma \varphi} (\mathbf{ \hat{r}}+ i
\sigma \boldsymbol{ \hat{\varphi}}) = \mathbf{\hat{x}}+i\sigma \mathbf{\hat{y}}$, is singled out with $\sigma$, which yields left (right)-handed circular polarization for the values $\sigma = +1(-1)$. The radial profile is a Bessel function $F_{q_r\ell}(r) = A_0 J_\ell (q_r r)$, with $A_0$ being the amplitude of the potential.

From the vector potential we can derive the electric $\mathbf E=-\partial_t \mathbf A$ and magnetic $\mathbf B = \nabla \times \mathbf A$ fields. Figure 1 shows the electric field in the x-y plane for two beams of TL having both $\ell=1$ but different sign of $\sigma$, the polarization. Clearly, the profiles look very different!

Figure 1: In-plane components of the electric fields at $t = 0$ and $z = 0$ for $\ell=1$ and polarization state σ = -1(anti-parallel) and σ = 1 (parallel).

The differences shown in Fig. 1 occur also for other values of $\ell$. We call these two types parallel (Sign$(\ell)=$Sign$(\sigma)$) and antiparallel (Sign$(\ell)\neq$Sign$(\sigma)$) beams. The names are not casual: in the quantum theory of light $\sigma\hbar$ is the spin angular momentum per photon in the direction of propagation $\hat{z}$ [2].

From Fig. 1 the first impression one gets is that the antiparallel beam is rotating around the singularity at $\mathbf r=0$. It must then have a significant curl in the perpendicular ($z$) direction. Recalling Maxwell's equation $\nabla \times \mathbf E = -(1/c)\partial_t \mathbf B$ we realize that the rotation of the antiparallel beam comes together with a strong variation on the magnetic field. Thus, we expect that the magnetic field in z is stronger in the antiparallel class. Another way to see this is by noticing that in the Coulomb gauge $\mathbf A\propto \mathbf E$: A plot of $\mathbf A$ would look much like Figure 1. Because $\mathbf B= \nabla \times \mathbf A$ we conclude that the magnetic field in $z$ should be significant for beams belonging to the antiparallel class.

Let's consider the magnetic field component $B_z(\mathbf r,t)$ for both parallel and antiparallel classes. In the case of tightly focused beams with $\ell=2$, the fields close to the $r=0$ are

parallel beam ($\sigma=+1$): $|\mathbf B_z(\mathbf r,t)| \propto (q_r r)^{2}$

antiparallel beam ($\sigma=-1$): $|\mathbf B_z(\mathbf r,t)| \propto (q_r r)^{1}$

note the difference in exponent. The closer we are to the phase singularity ($\underline{r \rightarrow 0}$) the larger is the field of the antiparallel class compared to that of the parallel class.

But perhaps more interesting is what happens within a class. When we generate a pure-class beam, what are the relative strengths of electric and magnetic fields? This is important when considering the interaction with matter. We use as an example an antiparallel focused beam $\ell=2, \,\sigma=-1$ close to the phase singularity:

electric field in $r$: $|\mathbf E_r(\mathbf r,t)| \propto (q_r r)^{2}$

magnetic field in $r$: $|\mathbf B_r(\mathbf r,t)| \propto (q_r r)^{0}$

the magnetic field is constant ! Close enough to $r=0$ the magnetic field overcomes the electric field. In the interaction with matter, the dominant contribution is magnetic, and a Hamiltonian of the type $q \, \mathbf r \cdot \mathbf E(\mathbf r,t)$ fails to describe the main effects of antiparallel beams. We will talk about the correct description of the interaction in a future post, where using a gauge transformation we will show how to treat parallel and antiparallel beams of twisted light.

We have seen a very peculiar feature -and to many unexpected- of a group of twisted light beams: they have a strong magnetic field that may overcome the electric field interaction with matter, the latter being typically the only one that we consider. More info about this can be found in our article Ref. [1].

References

[1] G. F. Quinteiro, D. E. Reiter, and T. Kuhn, Formulation of the twisted-light–matter interaction at the phase singularity: The twisted-light gauge, Phys. Rev. A 91, 033808.
[2] S. J. van Enk and G. Nienhuis, Spin and Orbital Angular Momentum of Photons, 1994 Europhys. Lett. 25 497.
[3] Close to the phase singularity means $q_r r \ll 1$ so $J_\ell(q_r r)\simeq \alpha (q_r r)^\ell$. A focused beam is such that $q_r \simeq q_z$.