0
$\begingroup$

Consider an optical experiment with photons or light pulses.

Is there an optical element that acts in the polarisation degree of freedom like the unitary $$ U = \frac 1 {\sqrt 2} \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}\quad \text ? $$ I choosed the basis such that $$ |H\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\;,\quad |V\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix} $$

If yes, then a photon that passes this device twice would expirience a flip in the polarisation: $$ U^2 = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\;, $$ like $U|H\rangle = |V\rangle$ and $U|V\rangle = -|H\rangle$.

Does it exist?

Improve this question
$\endgroup$
1
  • $\begingroup$ The result of passing through this element twice would be to "flip the polarization" - or if you like, it would be a 180 degree phase shift. That would just be a $\lambda/2$ element. But something that does this "after you pass through twice" is more interesting. What you are asking for is an element that rotates the polarization angle by 90 degrees. I have a hunch that you could achieve this with a birefringent element, but I'm not sure that can be done when you don't know the incident polarization angle. $\endgroup$
    – Floris
    Oct 16 '14 at 14:25
1
$\begingroup$

This is a typical matrix for an optically active element that rotates the light polarisation. A cuvette of water with sugar will do the job. Proportional to the sugar concentration, you can obtain arbitrary wave rotation.

Note that the U matrix has imaginary eigenvectors (1+i)/sqrt(2) and (1-i)/sqrt(2). Accordingly, unlike λ/4 plates, the eigenwaves of such a system are left- and right-circularly polarized.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Note also that optical activity is time reversible. A different result can be obtained by defining both off-diagonal elements of U as imaginary, in which case the forward and backward waves would be rotated in the same direction breaking the time-reversal symmetry. This effect occurs e.g. in statically magnetized bismuth-iron-grenade and is employed in optical insulators. $\endgroup$
    – dominecf
    Aug 18 '15 at 20:11

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.