Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If we represent a Jones vector by two complex-valued exponentials, $J_1 = e^{i \phi_1}$ and $J_2 = e^{i \phi_2}$, how can this ever give a polarization along the x-axis? We write such a polarization as $J_1 = 1, J_2 = 0$, but the exponents used can never give a zero value. Nor can we simply use the real component, as the complex phase determines circular or elliptical polarization. Is it just a matter of rotating our basis for, say, $J_1 = 1, J_2 = 1$ to give a polarization along some new x-axis?

share|cite|improve this question

The component of the Jones vector are not defined as two pure imaginary exponentials but as two complex numbers $z_1$ and $z_2$ whose polar decomposition is $z_i=|z_i|e^{i\phi_i}$ for $i=1,2$. To describe linearly polarized light along the $x$ axis

$$ |H \rangle = \begin{bmatrix} 1 \\ 0 \end{bmatrix} $$ we just take $z_1 \in \mathbb{R}$ ($\phi_1=0$) and $z_2=0$. Of course, you can always rotate the basis and consider the rotated axis $x'$ as your new $x$ axis. The case $z_1=z_2=1/\sqrt{2}$ is linearly polarized light along an axis $x'$ which corresponds to the usual $x$ axis rotated by $\theta=\pi/4$.

And for the last question the answer is yes, as long as there is no relative phase between the vector components the light will be linearly polarized.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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