# Simple question about Jones vector

In Wikipedia, left hand circular polarized light is $\frac{1}{\sqrt{2}}\left(\begin{array}{c} 1\\ i \end{array}\right)$. However, $\frac{1}{\sqrt{2}}\left(\begin{array}{c} 1\\ i \end{array}\right)e^{-i \omega t}$ gives $\frac{1}{\sqrt{2}}\left(\begin{array}{c} cos\omega t\\ sin\omega t \end{array}\right)$ as the real part, which is the "right hand circular polarized light in x-y plane". I believe I must have misunderstood something, can anyone help me out? Thanks!

(I guess Jones vector's handedness is defined from the receiver, but shouldn't it be defined from the source so that it is consistent with helicity?)

• There are two conventions in use. en.wikipedia.org/wiki/… – Farcher Mar 24 '17 at 10:31
• @Farcher Well, I know there are two conventions, but is it true that Jones vector uses the second convention(from the receiver)? – Jieyu You Mar 25 '17 at 21:45