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.



1 Answer 1


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.

  • 1
    $\begingroup$ Thanks. I eventually to the same conclusion after much thought - but it is very helpful to have my intuition confirmed. $\endgroup$ Dec 14, 2016 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. $\endgroup$ Dec 14, 2016 at 23:35

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