## Abstract

We correct a mistake in the expression of the coupled power into an optical fiber that led to some erroneous conclusions in Opt. Lett. **34**, 2829 (2009) and J. Opt. Soc. Am. A **26**, 2452 (2009).

© 2011 Optical Society of America

In two recent papers we discussed the coupling of stochastic beams into optical fibers [1, 2]. These papers contain a mistake in the expression of the coupled power into an optical fiber that led to some erroneous conclusions. We thank D. J. Wheeler and J. D. Schmidt of the Air Force Institute of Technology, Wright–Patterson AFB, Ohio, for bringing these errors to our attention. Here, we provide corrections for the paper appearing in J. Opt. Soc. Am. A [2]. The same modifications apply to the paper in Optics Letters [1]. The main change is that the matrix in Eq. (8) should be replaced with the following diagonal matrix:

(8)$${\mathrm{F}}_{\mathrm{mA}}({\mathbf{\rho}}_{1},{\mathbf{\rho}}_{2},\omega )=\phantom{\rule{0ex}{0ex}}\left(\begin{array}{cc}{F}_{mxA}^{*}({\mathbf{\rho}}_{1},\omega ){F}_{mxA}({\mathbf{\rho}}_{2},\omega )& 0\\ 0& {F}_{myA}^{*}({\mathbf{\rho}}_{1},\omega ){F}_{myA}({\mathbf{\rho}}_{2},\omega )\end{array}\right).$$ This change occurs because only the diagonal elements of the cross-spectral density matrix contribute to the power of each polarization component mode. As a result, the denominator in Eq. (9) does not contain the factor of 2, and it should be replaced with (9)$${\eta}_{\mathrm{cm}}=\frac{\iint Tr[\mathrm{W}({\mathbf{\rho}}_{1},{\mathbf{\rho}}_{2},\omega ).{\mathrm{F}}_{\mathrm{mA}}^{\u2020}({\mathbf{\rho}}_{1},{\mathbf{\rho}}_{2},\omega )]{d}^{2}{\rho}_{1}{d}^{2}{\rho}_{2}}{\int Tr[\mathrm{W}(\mathbf{\rho},\mathbf{\rho},\omega )]{d}^{2}\rho}.$$ With this change, the coupling efficiency does not depend on the degree of polarization of light coupled into a fiber.Equation (15) should be changed to

(15)$${\eta}_{C}=\frac{\sum _{j=x,y}{P}_{{C}_{jj}}}{\left[\sum _{j=x,y}{P}_{{\text{inc}}_{jj}}\right]}.$$ In Eqs. (23) and (24), the subscript *j* should be replaced with *i*, and the corrected form of Eqs. (25, 27) is (25)$${I}_{ii}=2\pi {B}_{ii}(\frac{{A}_{i}{w}_{i}}{\lambda f}{)}^{2},$$
(27)$${\eta}_{C}=(\frac{\pi}{\lambda fW}{)}^{2}\frac{\sum _{i=x,y}{A}_{i}{w}_{i}^{2}[{C}_{ii}^{2}-(1/4{\delta}_{ii}^{4}){]}^{-1}}{\sum _{i=x,y}{A}_{i}^{2}{\sigma}_{i}^{2}({W}^{2}+2{\sigma}_{i}^{2}{)}^{-1}}.$$ Finally, the vertical scale in Figs. 2 and 3 should be multiplied by 2 so that it ranges from 0 to 1, and Fig. 5 should be ignored because coupling efficiency does not depend on the degree of polarization of light coupled into a fiber.**1. **M. Salem and G. P. Agrawal, “Coupling of stochastic electromagnetic beams into optical fibers,” Opt. Lett. **34**, 2829–2831 (2009). [CrossRef] [PubMed]

**2. **M. Salem and G. P. Agrawal, “Effects of coherence and polarizationon the coupling of stochastic electromagnetic beams into optical fibers,” J. Opt. Soc. Am. A **26**, 2452–2458 (2009). [CrossRef]

### Cited By

OSA participates in Crossref's Cited-By Linking service. Citing articles from OSA journals and other participating publishers are listed here.

Alert me when this article is cited.

### Equations (5)

Equations on this page are rendered with MathJax. Learn more.

(8)
$${\mathrm{F}}_{\mathrm{mA}}({\mathbf{\rho}}_{1},{\mathbf{\rho}}_{2},\omega )=\phantom{\rule{0ex}{0ex}}\left(\begin{array}{cc}{F}_{mxA}^{*}({\mathbf{\rho}}_{1},\omega ){F}_{mxA}({\mathbf{\rho}}_{2},\omega )& 0\\ 0& {F}_{myA}^{*}({\mathbf{\rho}}_{1},\omega ){F}_{myA}({\mathbf{\rho}}_{2},\omega )\end{array}\right).$$
(9)
$${\eta}_{\mathrm{cm}}=\frac{\iint Tr[\mathrm{W}({\mathbf{\rho}}_{1},{\mathbf{\rho}}_{2},\omega ).{\mathrm{F}}_{\mathrm{mA}}^{\u2020}({\mathbf{\rho}}_{1},{\mathbf{\rho}}_{2},\omega )]{d}^{2}{\rho}_{1}{d}^{2}{\rho}_{2}}{\int Tr[\mathrm{W}(\mathbf{\rho},\mathbf{\rho},\omega )]{d}^{2}\rho}.$$
(15)
$${\eta}_{C}=\frac{\sum _{j=x,y}{P}_{{C}_{jj}}}{\left[\sum _{j=x,y}{P}_{{\text{inc}}_{jj}}\right]}.$$
(25)
$${I}_{ii}=2\pi {B}_{ii}(\frac{{A}_{i}{w}_{i}}{\lambda f}{)}^{2},$$
(27)
$${\eta}_{C}=(\frac{\pi}{\lambda fW}{)}^{2}\frac{\sum _{i=x,y}{A}_{i}{w}_{i}^{2}[{C}_{ii}^{2}-(1/4{\delta}_{ii}^{4}){]}^{-1}}{\sum _{i=x,y}{A}_{i}^{2}{\sigma}_{i}^{2}({W}^{2}+2{\sigma}_{i}^{2}{)}^{-1}}.$$