# Why polarization vector $= (0,1,i,0)$?

I know from CED that one has e.g. polarization

$$\vec{E}(z,t) = \begin{bmatrix} e_{x} \\ e_{y} \\ 0 \end{bmatrix} \; e^{i(kz - 2 \pi f t)}.$$

Why do Peskin&Schroeder define a polarization vector as (see page 7 PS)

$$(0,1,i,0)?$$

Where is the exponential? Why does he use $i$ for the $y$-direction?

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The particular vector $(0,1,i,0)$ expresses a circular polarization (either left-handed or right-handed; I guess that the book tells you which one), one that has a well-defined value of the angular momentum with respect to the $z$ axis.
Only the second ($x$) and third ($y$) components are nonzero because the polarization vector has to be transverse – orthogonal to the momentum vector which is taken to be in the $z$-direction.
The ratio of the $x$ and $y$ components is $\pm i$ because a circular polarization is obtained as a mixture of both $x,y$ linear polarizations which are mutually delayed by the phase shift $\pi/2$, and $\exp(\pi i / 2) = i$.