# 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 polarization vector is just the vector-valued coefficient in front of the exponential but without the exponential. It's suppose to encode the "internal" degrees of freedom of the particle, not its dependence on space or momentum.

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 $\exp(\pi i / 2) = i$.

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Thanks Luboş! For future reference I would also like to recommend Sakurai pp. 6-9 (Modern...). I read that 3 years ago...forgot some of it. –  Love Learning Feb 7 '14 at 16:23
Thanks for the useful addition! You turned my caron above "s" to a nice Spanish or French accent. ;-) śšßƒ - it's not on my keyboard, it seems. –  Luboš Motl Feb 7 '14 at 16:40
Yeah I tried to find/make your "s" but couldn't sorry :) BTW Sakurai doesn't discuss polarization of single photons, he refers to "Baym 69" which I don't have access to. –  Love Learning Feb 7 '14 at 16:46