Jones vectors for circular polarization

[Decimal]

Hello, I can't seem to arrive at a result that my book states using Jones vectors for circular polarization. My book says that the unit jones vector for right circular polarization is $$ \begin{bmatrix} 1 \\ -i \end{bmatrix} $$ However when I apply this jones vector to an arbitrary electric field I arrive at left circular polarization. Here is what I did: $$ \tilde E = E_0 * \begin{bmatrix} 1 \\ -i \end{bmatrix} * e^{i(k x-\omega t)}$$ $$ \tilde E = E_0 * \begin{bmatrix} 1 \\ e^{- \frac {\pi} {2} i } \end{bmatrix} * e^{i(k x-\omega t)} $$ Applying the phase shift and rewriting: $$ \tilde E = E_0 * \begin{bmatrix} cos(kx-\omega t) \\ sin(kx-\omega t) \end{bmatrix} $$ Isn't this left circular polarized light? Where did I go wrong?

 

[Andy Resnick]

Hello, I can't seem to arrive at a results that my book states using Jones vectors for circular polarization. My book says that the unit jones vector for right circular polarization is $$ \begin{bmatrix} 1 \\ -i \end{bmatrix} $$

According to Azzam and Bashara [https://www.amazon.com/dp/0444870164/?tag=pfamazon01-20], your Jones vector corresponds to *left* circular polarization. (there's also a prefactor 1/√2 missing). Left-circular *lags*, because lefties are lazy laggards :)

 

[Decimal]

Ah yes I forgot the factor. Nevertheless both my book, my lecture notes and wikipedia all state this jones vector corresponds to right circular polarization. They even state this in old exam questions. Are you sure its supposed to be left circular?

 

[Andy Resnick]

Ah yes I forgot the factor. Nevertheless both my book, my lecture notes and wikipedia all state this jones vector corresponds to right circular polarization. They even state this in old exam questions. Are you sure its supposed to be left circular?

I guess at some point it's all relative. I think the definition corresponds to the direction of electric field rotation when you are facing the incoming plane wave, but the way I handle issues like this in class is to simply be internally consistent. So in your case, the incoming wave could be (I think) e^-i (kx-wt) rather than e^i(kx-wt) and then you are consistent... try that and see what you get.

 

[Decimal]

Thanks for the help Andy, but I think I figured out where I went wrong. I didn't realize the polarization will of course rotate in the time $t$. The time term in the cosine and sine is however negative, which actually inverts the sine. That's why I mixed up left and right. The polarization I arrived at in my original post was actually right circular, I just didn't realize it.