Pouvoir polarisant, dichroïsme et retard

1. Superposition incohérente de faisceaux lumineux

Matrices de Jones homogènes (normales)

On dit qu'une matrice $\mat A$ est normale si elle commute avec son adjointe : $$\mat A \mat A^\dagger = \mat A^\dagger \mat A.$$ Une matrice peut s'écrire $$\mat A = \mat U \diag(\lambda_1, \cdots, \lambda_n) \mat U^\dagger$$ où $\mat U$ est unitaire si et seulement si elle est normale. Si une matrice de Jones est normale, on pourra l'écrire $$\begin{array} {ccc} \mat J = \lambda \boldsymbol\phi \boldsymbol\phi^\dagger + \mu \boldsymbol\psi \boldsymbol\psi^\dagger & \Rightarrow & \mat J^\dagger = \lambda^* \boldsymbol\phi \boldsymbol\phi^\dagger + \mu^* \boldsymbol\psi \boldsymbol\psi^\dagger. \end{array} \tag 7$$ Dans le jargon de l'optique, on dit qu'elle est homogène. $$\begin{split} m_{kl} & = 2 |\lambda |^2 \tr(\boldsymbol{\sigma}_k\boldsymbol{\phi}\boldsymbol{\phi}^\dagger\boldsymbol\sigma_l\boldsymbol{\phi}\boldsymbol{\phi}^\dagger) + 2 \lambda \mu^* \tr(\boldsymbol{\sigma}_k\boldsymbol{\phi}\boldsymbol{\phi}^\dagger\boldsymbol\sigma_l\boldsymbol{\psi}\boldsymbol{\psi}^\dagger) + 2 \lambda^* \mu \tr(\boldsymbol{\sigma}_k\boldsymbol{\psi}\boldsymbol{\psi}^\dagger\boldsymbol\sigma_l\boldsymbol{\phi}\boldsymbol{\phi}^\dagger) + 2 |\mu |^2 \tr(\boldsymbol{\sigma}_k\boldsymbol{\psi}\boldsymbol{\psi}^\dagger\boldsymbol\sigma_l\boldsymbol{\psi}\boldsymbol{\psi}^\dagger) \\ & = 2 |\lambda |^2 \boldsymbol{\phi}^\dagger\boldsymbol{\sigma}_k\boldsymbol{\phi}\boldsymbol{\phi}^\dagger\boldsymbol\sigma_l\boldsymbol{\phi} + 2 \lambda \mu^* \boldsymbol{\psi}^\dagger\boldsymbol{\sigma}_k\boldsymbol{\phi}\boldsymbol{\phi}^\dagger\boldsymbol\sigma_l\boldsymbol{\psi} + 2 \lambda^* \mu \boldsymbol{\phi}^\dagger\boldsymbol{\sigma}_k\boldsymbol{\psi}\boldsymbol{\psi}^\dagger\boldsymbol\sigma_l\boldsymbol{\phi} + 2 |\mu |^2 \boldsymbol{\psi}^\dagger\boldsymbol{\sigma}_k\boldsymbol{\psi}\boldsymbol{\psi}^\dagger\boldsymbol\sigma_l\boldsymbol{\psi} \\ & = \frac 1 2 \left[ |\lambda |^2 s_k s_l + \lambda \mu^* q_k^* q_l + \lambda^* \mu q_kq_l^* + |\mu |^2 s'_k s'_l \right]. \end{split}$$ $$\begin{split} m_{kl} & = 2 |\lambda |^2 \boldsymbol{\phi}^\dagger\boldsymbol{\sigma}_k\boldsymbol{\phi}\boldsymbol{\phi}^\dagger\boldsymbol\sigma_l\boldsymbol{\phi} + 2 \lambda \mu^* \boldsymbol{\psi}^\dagger\boldsymbol{\sigma}_k\boldsymbol{\phi}\boldsymbol{\phi}^\dagger\boldsymbol\sigma_l\boldsymbol{\psi} + 2 \lambda^* \mu \boldsymbol{\phi}^\dagger\boldsymbol{\sigma}_k\boldsymbol{\psi}\boldsymbol{\psi}^\dagger\boldsymbol\sigma_l\boldsymbol{\phi} + 2 |\mu |^2 \boldsymbol{\psi}^\dagger\boldsymbol{\sigma}_k\boldsymbol{\psi}\boldsymbol{\psi}^\dagger\boldsymbol\sigma_l\boldsymbol{\psi} \\ & = \frac 1 2 \left[ |\lambda |^2 s_k s_l + \lambda \mu^* (u_k-iv_k) (u_l+iv_l) + \lambda^* \mu (u_k+iv_k)(u_l-iv_l) + |\mu |^2 s'_k s'_l \right] \\ & = \frac 1 2 \left[ |\lambda |^2 s_k s_l + \lambda \mu^* [u_k u_l +v_k v_l + i (u_k v_l - u_l v_k)] + \lambda^* \mu [u_k u_l +v_k v_l - i (u_k v_l - u_l v_k)] + |\mu |^2 s'_k s'_l \right] \\ & = \frac 1 2 \left[ |\lambda |^2 s_k s_l + 2 \re(\lambda \mu^*) (u_k u_l +v_k v_l) - 2 \im(\lambda \mu^*) (u_k v_l - u_l v_k) + |\mu |^2 s'_k s'_l \right]. \end{split}$$ $$\mat M = \frac 1 2 \begin{bmatrix} |\lambda |^2 + |\mu |^2 & (|\lambda |^2 - |\mu |^2) \begin{bmatrix} s_1 & s_2 & s_3 \end{bmatrix} \\ (|\lambda |^2 - |\mu |^2) \begin{bmatrix} s_1 \\ s_2 \\ s_3 \end{bmatrix} & \begin{bmatrix} {(|\lambda |^2 + |\mu |^2) s_k s_l \\ + 2 \re(\lambda \mu^*) (u_k u_l +v_k v_l) \\ - 2 \im(\lambda \mu^*) (u_k v_l - u_l v_k)} \end{bmatrix}_{k, l = 1, 2, 3} \end{bmatrix}$$ $$\mat{\hat u} = \begin{bmatrix} -\sin(2\theta) & \cos(2\theta) & 0 \end{bmatrix}^\t$$ $$\mat{\hat v} = \begin{bmatrix} \cos(2\theta)\sin(2\epsilon) & \sin(2\theta)\sin(2\epsilon) & \cos(2\epsilon) \end{bmatrix}^\t$$

diatténuateur : $\lambda, \mu \in \reels^+$. On suppose $\lambda \ge \mu$ $$\mat M = \frac 1 2 \begin{bmatrix} \lambda^2 + \mu^2 & (\lambda^2 - \mu^2) \begin{bmatrix} s_1 & s_2 & s_3 \end{bmatrix} \\ (\lambda^2 - \mu^2) \begin{bmatrix} s_1 \\ s_2 \\ s_3 \end{bmatrix} & \begin{bmatrix} {(\lambda^2 + \mu^2) s_k s_l \\ + 2\lambda \mu (u_k u_l +v_k v_l)} \end{bmatrix}_{k, l = 1, 2, 3} \end{bmatrix}$$ $$\mat M = \frac 1 2 \begin{bmatrix} \lambda^2 + \mu^2 & (\lambda^2 - \mu^2) \begin{bmatrix} s_1 & s_2 & s_3 \end{bmatrix} \\ (\lambda^2 - \mu^2) \begin{bmatrix} s_1 \\ s_2 \\ s_3 \end{bmatrix} & \begin{bmatrix} {(\lambda - \mu)^2 s_k s_l + 2\lambda \mu \delta_{kl}} \end{bmatrix}_{k, l = 1, 2, 3} \end{bmatrix}$$ $$\begin{array} {llll} \mat M = \displaystyle\frac {\lambda^2 + \mu^2} 2 \begin{bmatrix} 1 & \mat d^\t \\ \mat d & \mat H \end{bmatrix} & \mat d = \displaystyle\frac {\lambda^2 - \mu^2} {\lambda^2 + \mu^2} \mat{\hat{d}} & \mat H = \mat P + \displaystyle\frac {2\lambda \mu} {\lambda^2 + \mu^2} (\mat I - \mat P) & \mat P = \mat{\hat{d}} \mat{\hat{d}}^\t \end{array}$$ $$\begin{array} {llll} T = \displaystyle\frac {\lambda^2 + \mu^2} 2 & d = \displaystyle\frac {\lambda^2 - \mu^2} {\lambda^2 + \mu^2} & \Rightarrow & \displaystyle\frac {2\lambda \mu} {\lambda^2 + \mu^2} = \sqrt{1-d^2} \end{array}$$ $$\begin{array} {llll} \mat M = T \begin{bmatrix} 1 & \mat d^\t \\ \mat d & \mat H \end{bmatrix} & \mat d = d \mat{\hat{d}} & \mat H = \mat P + \sqrt{1-d^2} (\mat I - \mat P) & \mat P = \mat{\hat{d}} \mat{\hat{d}}^\t \end{array}$$ $$\begin{array} {lll} d=0 & \Rightarrow & \mat H = \mat I \end{array}$$ $$\begin{array} {lllll} d=1 & \Rightarrow & \mat H = \mat P = \mat{{d}} \mat{{d}}^\t & \Rightarrow & \mat M = T \begin{bmatrix} 1 \\ \mat d \end{bmatrix} \begin{bmatrix} 1 & \mat d^\t \end{bmatrix} \end{array}$$ retardeur : $\lambda, \mu = e^{\pm i \delta \varphi}$. On suppose $\delta \varphi \ge 0$ $$\mat M = \begin{bmatrix} 1 & 0 \\ 0 & \begin{bmatrix} {s_k s_l \\ + \cos(2\delta \varphi) (u_k u_l +v_k v_l) \\ - \sin(2\delta \varphi) (u_k v_l - u_l v_k)} \end{bmatrix}_{k, l = 1, 2, 3} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & \mat R \end{bmatrix}$$ $$R_{kl} = s_k s_l + \cos(2\delta \varphi) (u_k u_l +v_k v_l) - \sin(2\delta \varphi) (u_k v_l - u_l v_k)$$ $$\mat R \vec w = (\vec s \cdot \vec w) \vec s + \cos(2\delta \varphi) ((\vec u \cdot \vec w) \vec u+(\vec v \cdot \vec w) \vec v) - \sin(2\delta \varphi) \left( (\vec v \cdot \vec w) \vec u - (\vec u \cdot \vec w) \vec v \right) $$

$$\mat M = m_{00} \begin{bmatrix} 1 & 0 \\ 0 & \mat R \end{bmatrix} \begin{bmatrix} 1 & \mat d^\t \\ \mat d & \mat H \end{bmatrix} = m_{00} \begin{bmatrix} 1 & \mat d^\t \\ \mat R \mat d & \mat R \mat H \end{bmatrix}$$ $$\begin{array} {lll} d=1 & \Rightarrow & \mat M = m_{00} \begin{bmatrix} 1 & \mat d^\t \\ \mat R \mat d & \mat R \mat{{d}} \mat{{d}}^\t \end{bmatrix} = T \begin{bmatrix} 1 \\ \mat R \mat d \end{bmatrix} \begin{bmatrix} 1 & \mat d^\t \end{bmatrix} \end{array}$$

$$\mat M = m_{00} \begin{bmatrix} 1 & 0 \\ 0 & \mat D \end{bmatrix} \begin{bmatrix} 1 & \mat d^\t \\ \mat R \mat d & \mat R \mat H \end{bmatrix} = m_{00} \begin{bmatrix} 1 & \mat d^\t \\ \mat D \mat R \mat d & \mat D\mat R \mat H \end{bmatrix}$$ $$\begin{array} {lll} d=1 & \Rightarrow & \mat M = m_{00} \begin{bmatrix} 1 & \mat d^\t \\ \mat D \mat R \mat d & \mat D \mat R \mat{{d}} \mat{{d}}^\t \end{bmatrix} = T \begin{bmatrix} 1 \\ \mat D \mat R \mat d \end{bmatrix} \begin{bmatrix} 1 & \mat d^\t \end{bmatrix} \end{array}$$