1
$\begingroup$

I think this is a simple question but I cant seem to find helpful answers on the net.

Anyway; I have a HWP that takes Horizontally polarized light to Diagonally polarised light;

ie; $\frac{1}{\sqrt{2}} \left[ {\begin{array}{cc} 1 & 1 \\ 1 & -1 \ \end{array} } \right]=B$

I want to show confirm to myself that it takes left circularly polarized (lcp) light to rcp light (I think this is what is does).

However if I calculate $B\frac{1}{\sqrt{2}} \left[ {\begin{array}{cc} 1\\ i \ \end{array} } \right]$, I get

$\frac{1}{2} \left[ {\begin{array}{c} 1+i \\ 1-i \ \end{array} } \right]$ .

Can someone please either tell me what I have done wrong or clarify that this last result is rcp light.

Thankyou.

$\endgroup$
0
$\begingroup$

You calculation is correct. Simply rewrite it a little as follows:

$$\frac{1}{2} \left[ {\begin{array}{c} 1+i \\ 1-i \ \end{array} } \right] = \frac{1}{\sqrt{2}}\left(\begin{array}{c}\exp\left(i\,\frac{\pi}{4}\right)\\\exp\left(-i\,\frac{\pi}{4}\right)\end{array}\right)$$

Now you're missing the following fact: polarization states are unchanged by pure phase scaling. So, you still have the same polarization if you multiply both elements of your vector by a phase factor $e^{i\,\theta},\,\theta\in\mathbb{R}$. So, multiply both factors by $\exp\left(-i\,\frac{\pi}{4}\right)$ and see what happens. You should get something more recognizable as the orthogonal circularly polarization state.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Thanks. I eventually to the same conclusion after much thought - but it is very helpful to have my intuition confirmed. $\endgroup$ – Lewis Howard Dec 14 '16 at 23:31
  • $\begingroup$ @LewisHoward After I wrote the answer, I realized this was what AlfredCentauri's comment was about. That one was a bit cryptic for me. $\endgroup$ – Selene Routley Dec 14 '16 at 23:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.