# Why a $e^{i \pi/4}$ (or $e^{-i \pi/4}$) causes a phase shift of $\pi/2$ instead of $\pi/4$ in the case of a quarter-wave plate?

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?

• That's not a "spinor."
– Buzz
Jan 25, 2021 at 3:20

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 $$2\pi$$).
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 $$1$$: $$\frac{2\pi}{\lambda}=|\vec k|=1$$.