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Abstract
This erratum gives corrections for the errors in a previously published paper [J. Opt. Soc. Am. A 36, 312 (2019) [CrossRef] ].
© 2019 Optical Society of America
In [1], Eq. (5), which gives the output electric field after passing obliquely through the SEO, should read
$$\begin{split}{{\textbf{E}}_{{\rm{out}}}}(x,y)&={E_0}{\mathbb J}
\cdot \left[{\begin{array}{*{20}{c}}1\\0\end{array}} \right]\\
&={E_0}\left[{\begin{array}{*{20}{c}}{\cos
\left({\frac{{c{\rho_0}}}{4}} \right)\cos \left({\frac{{c{\rho_1}}}{4}}
\right)-{e^{-{\rm{i}}{\phi_0}}}{e^{{\rm{i}}{\phi_1}}}\sin
\left({\frac{{c{\rho_0}}}{4}} \right)\sin \left({\frac{{c{\rho_1}}}{4}}
\right)}\\[6pt] {{\rm{i}}{e^{-{\rm{i}}{\phi_0}}}\cos
\left({\frac{{c{\rho_1}}}{4}} \right)\sin \left({\frac{{c{\rho_0}}}{4}}
\right)+{\rm{i}}{e^{-{\rm{i}}{\phi_1}}}\sin \left({\frac{{c{\rho_1}}}{4}}
\right)\cos \left({\frac{{c{\rho_0}}}{4}} \right)}\end{array}}
\right]\end{split}.$$
As a consequence of this, the Stokes parameters
[originally reported between Eqs. (5) and (6)] are corrected to be $$\begin{split}{s_1}
&=\cos ({\phi_1})\sin \left({\frac{{c{\rho_0}}}{2}} \right)\sin
({\phi_0}-{\phi_1})+\sin ({\phi_1}) +\left[{\cos
\left({\frac{{c{\rho_1}}}{2}} \right)\cos ({\phi_0}-{\phi_1})\sin
\left({\frac{{c{\rho_0}}}{2}} \right)+\cos \left({\frac{{c{\rho_0}}}{2}}
\right)\sin \left({\frac{{c{\rho_1}}}{2}} \right)} \right], \\ {s_2}
&=-\sin ({\phi_1})\sin ({\phi_0}-{\phi_1})\sin
\left({\frac{{c{\rho_0}}}{2}} \right)+\cos ({\phi_1}) +\left[{\cos
\left({\frac{{c{\rho_1}}}{2}} \right)\cos ({\phi_0}-{\phi_1})\sin
\left({\frac{{c{\rho_0}}}{2}} \right)+\cos \left({\frac{{c{\rho_0}}}{2}}
\right)\sin \left({\frac{{c{\rho_1}}}{2}} \right)} \right], \\ {s_3}
&=-\cos \left({\frac{{c{\rho_0}}}{2}} \right)\cos
\left({\frac{{c{\rho_1}}}{2}} \right)+\cos ({\phi_0}-{\phi_1})\sin
\left({\frac{{c{\rho_0}}}{2}} \right)\sin \left({\frac{{c{\rho_1}}}{2}}
\right).\end{split}$$
There are sign errors in Eqs. (23) and (24). They are corrected to be $${\alpha_3}=0,\quad
{\unicode{x00B5}_3}=-\frac{{4{c^2}\cos [c{\cal L}\tan
(\theta^\prime)/4]{\cal L}\tan \theta^\prime}}{{4{\pi^2}-{c^2}{{\cal
L}^2}\mathop {\tan}\nolimits^2 \theta^\prime}},\quad {\gamma_3}=c,\quad
{\eta_3}=0,$$
and $${\alpha_4}=0,\quad{\unicode{x00B5}_4}=\frac{{4{c^2}\cos [c{\cal L}\tan
(\theta^\prime)/4]{\cal L}\tan \theta^\prime}}{{4{\pi^2}-{c^2}{{\cal
L}^2}\mathop {\tan}\nolimits^2
\theta^\prime}},\quad{\gamma_4}=c,\quad{\eta_4}=0.$$
REFERENCE
1. A. Ariyawansa, K. Liang, and T. G. Brown, “Polarization singularities in a stress-engineered optic,” J. Opt. Soc. Am. A 36, 312–319 (2019). [CrossRef]