In a previous Letter [1], we asserted in the paragraph following Eq. (16) that “if beam 1 is collimated and everywhere circularly polarized, then beam 2 must also be circularly polarized, but with the opposite handedness.” A more careful application of Eq. (16) in [1] shows that this is not the case.

Circular polarization across paraxial input beam 1 can be described by ${\mathrm{E⃗}}_1^i(\theta ,\varphi )=f(\theta ,\varphi ){\mathrm{e}}^{\mathrm{i}\psi }(\widehat{x}\pm \mathrm{i}\widehat{y})$, where $f(\theta ,\varphi )$ is a real-valued scalar function that determines the amplitude across the beam, and $\psi$ is a real-valued scalar. Applying Eq. (16) in [1] gives the solution for beam 2:

Thus, if beam 1 is collimated and everywhere circularly polarized, then beam 2 must also be circularly polarized with the same handedness. This also implies that Fig. 2(b) in [1] is incorrect: the two circular arrows should be identical, as shown in Fig. 1. These errors affect neither the derivations of any equations nor the main conclusions given in [1].

 

Fig. 1. Corrected version of Fig. 2(b) from [1] showing a circularly polarized beam pair where the intensity varies across each beam. The arrows indicate the direction or handedness of the polarization. The correction is to make both arrows identical. The color bar indicates the normalized intensity of the beams.

Download Full Size | PDF