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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?

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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.

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