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Given a spinor $\begin{pmatrix} E_x \\ E_y \end{pmatrix}$, I learned that if we place a quarter-wave plate with its fast and slow axes in the x- and y-direction, the relative phase shift in the x- and y-components of the spinor is $\pi/2$:

\begin{equation} \begin{pmatrix} E_x' \\ E_y' \end{pmatrix} = \begin{pmatrix} e^{-i\pi/4} & 0 \\ 0 & e^{i\pi/4} \end{pmatrix} \begin{pmatrix} E_x \\ E_y \end{pmatrix} \end{equation} where the unprimed fields are those entering the quarter wave plate, and the primed ones are those existing it. I wonder why the phase shift is not $\pi/4$ (as indicated by $e^{i \pi/4}$ or $e^{-i \pi/4}$ or "quarter"), but $\pi/2$ instead?

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    $\begingroup$ That's not a "spinor." $\endgroup$ – Buzz Jan 25 at 3:20
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If you write the matrix as $$ e^{-i\pi/4} \begin{pmatrix} 1 & 0 \\ 0 & e^{i\pi/2} \end{pmatrix}, $$ you can easily see that the phaseshift between the two components is in fact $\pi/2$. The name of quarter-wave plates refers to a quarter of the wavelength which is $2\pi/4=\pi/2$, if you consider a wave vector of length $1$: $\frac{2\pi}{\lambda}=|\vec k|=1$.

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    $\begingroup$ Bummer - hope people come back and upvote this more detailed answer vs. JMurray's correct but limited response. $\endgroup$ – Carl Witthoft Jan 25 at 14:47
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One component is shifted forward by $\pi/4$ and the other is shifted backward by $\pi/4$, so the relative shift between the two components is $\pi/2$ (which is one quarter of $2\pi$).

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