# Optical Operators (2×2 Matrix Methods) Three 2×2 formalisms show up. They look alike, so keep straight **what the vector holds**: | Formalism | Vector | Matrix maps | Used for | |---|---|---|---| | Jones | $(E_x,E_y)$ of one wave | input polarization → output polarization | wave plates, polarizers | | Scattering (S) | incoming waves → outgoing waves | $(a_1,a_2)\to(b_1,b_2)$ | mirrors, symmetry, power conservation | | Transfer (T) / propagation | waves on side 1 → waves on side 2 | $(\text{fwd},\text{bwd})_1\to(\text{fwd},\text{bwd})_2$ | **cascades** (multilayers, Fabry–Perot) | --- ## Jones Calculus Wave along $+\hat z$, phasors with $e^{+j\omega t}$, IEEE handedness (see [[conventions]] §6, [[waves]] Polarization). **Jones vectors** (normalized): $ \hat x:\ \begin{pmatrix}1\\0\end{pmatrix}\qquad 45^\circ:\ \frac{1}{\sqrt2}\begin{pmatrix}1\\1\end{pmatrix}\qquad \text{RHC}:\ \frac{1}{\sqrt2}\begin{pmatrix}1\\-j\end{pmatrix}\qquad \text{LHC}:\ \frac{1}{\sqrt2}\begin{pmatrix}1\\j\end{pmatrix} $ Check RHC: $\mathrm{Re}\{(\hat x-j\hat y)e^{j\omega t}\}=\hat x\cos\omega t+\hat y\sin\omega t$ rotates $x\to y$; thumb along $+z$ gives right-handed. **Jones matrices:** - Polarizer along $x$: $ \begin{pmatrix}1&0\\0&0\end{pmatrix} $ - Wave plate with retardance $\Gamma$ and slow axis along $x$. The slow axis has larger $n$ and picks up the extra phase $e^{-jk_0 n L}$, so up to a global phase: $ W(\Gamma)=\begin{pmatrix}e^{-j\Gamma}&0\\0&1\end{pmatrix},\qquad \Gamma=k_0\,(n_\text{slow}-n_\text{fast})\,L $ - Half-wave plate (HWP): $\Gamma=\pi$ gives $\mathrm{diag}(1,-1)$ up to sign. - Quarter-wave plate (QWP): $\Gamma=\dfrac{\pi}{2}$ gives $\mathrm{diag}(-j,1)\propto\mathrm{diag}(1,j)$. - Element rotated by $\theta$: $ M_\theta=R(\theta)\,M\,R(-\theta),\qquad R(\theta)=\begin{pmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{pmatrix} $ - HWP with its axis at $\theta$: $ \begin{pmatrix}\cos2\theta&\sin2\theta\\ \sin2\theta&-\cos2\theta\end{pmatrix} $ This **mirrors** linear polarization about the axis: angle $\alpha\to2\theta-\alpha$. Light at $\theta$ from the axis is rotated by $2\theta$. Rotating the plate by $\theta$ rotates the output by $2\theta$. - QWP with linear input at $\alpha$ to its axis: - $\alpha=0$ or $90^\circ$: unchanged. - $\alpha=\pm45^\circ$: circular. With slow axis $x$, $\dfrac{1}{\sqrt2}(1,1)\to\dfrac{1}{\sqrt2}(1,j)$, which is **LHC** under IEEE. - Otherwise: elliptical. - **Optical isolator** (pset 1 P4d): polarizing beam splitter → QWP at 45° → mirror → back through the QWP. The double pass is a HWP at 45°, so the polarization rotates by 90° and the PBS sends the return beam out of the other port. Path: linear → circular → (mirror) circular → orthogonal linear. - Mirror at normal incidence: $\vec k$ reverses, so the handedness label flips even though the field rotation doesn't. The Jones matrix ($\pm\mathrm{diag}(1,-1)$) depends on how you redefine axes. Pick one and keep it. - Measuring birefringence (pset 1 P4e): put a plate between crossed polarizers at 45°. The transmission goes as $\sin^2\!\dfrac{\Gamma}{2}$. Tune $\lambda$ and count fringes to get $\Delta n=\dfrac{\Delta\Gamma}{k_0 L}$ (double the path for a double pass). Rotate the plate: transmission vanishes when the optic axis is along a polarizer. --- ## Scattering Matrix (Haus Ch. 3) Port $i$ has incoming amplitude $a_i$ and outgoing amplitude $b_i$, normalized so that $|a_i|^2$ is power: $ \begin{pmatrix}b_1\\b_2\end{pmatrix}=\begin{pmatrix}S_{11}&S_{12}\\S_{21}&S_{22}\end{pmatrix}\begin{pmatrix}a_1\\a_2\end{pmatrix} $ | Property | Condition | Meaning | |---|---|---| | Reciprocal | $S=S^{T}$, i.e. $S_{12}=S_{21}$ | transmission is the same both ways | | Lossless | $S^{\dagger}S=I$ | power out = power in; columns orthonormal | | Time-reversible | $S^{*}=S^{-1}$ | with lossless, implies reciprocal | Lossless two-port: $|S_{11}|^2+|S_{21}|^2=1$, $|S_{11}|=|S_{22}|$, and $ S_{11}S_{12}^{*}+S_{21}S_{22}^{*}=0 $ Moving a reference plane outward by $\phi$ multiplies $S_{ii}$ by $e^{-2j\phi}$ and $S_{ij}$ by $e^{-j(\phi_i+\phi_j)}$. Reference planes are free choices. **Lossless partially transmitting mirror.** Choose planes so that $S_{11}=S_{22}=-r$ with $r$ real. Then $ S=\begin{pmatrix}-r&jt\\ jt&-r\end{pmatrix},\qquad r^2+t^2=1 $ - The $j$ is **forced**, not a choice (orthogonality of columns). - Only one real parameter remains: the reflectivity. - No frequency-independent $S$ of this form exists, because the reference planes are frequency dependent. --- ## Transfer (T) Matrix Relates the waves on side 2 to the waves on side 1. Here the ordering is $(b_2,a_2)$ from $(a_1,b_1)$; Haus Prob. 3.4 orders it differently, so check which one a problem uses. $ \begin{pmatrix}b_2\\a_2\end{pmatrix}=T\begin{pmatrix}a_1\\b_1\end{pmatrix},\qquad T=\frac{1}{S_{12}}\begin{pmatrix}S_{12}S_{21}-S_{11}S_{22}&S_{22}\\-S_{11}&1\end{pmatrix} $ - $\det T=\dfrac{S_{21}}{S_{12}}$, which is $1$ for a reciprocal two-port. - **Cascade = matrix product**: $T_\text{total}=T_N\cdots T_2T_1$ (the right-most element is hit first). That's the reason to use T instead of S. - With nothing incident from the right ($a_2=0$), in this ordering: $ S_{21}=\frac{1}{T_{22}},\qquad S_{11}=-\frac{T_{21}}{T_{22}} $ - Free propagation over length $l$ with $k_z=k\cos\theta$: $ T=\begin{pmatrix}e^{-jk_zl}&0\\0&e^{+jk_zl}\end{pmatrix} $ ## Propagation Matrix for Layered Media (Kong; pset 3 P4) Layer $l$ has forward and backward amplitudes $E_\pm^{(l)}$, with $E^{(l)}=E_+e^{-jk_z^{(l)}z}+E_-e^{+jk_z^{(l)}z}$. At the interface $z=d_l$ between layers $l$ and $l+1$, continuity of tangential $E$ and $H$ gives: $ \begin{pmatrix}E_+^{(l)}e^{-jk_z^{(l)}d_l}\\E_-^{(l)}e^{+jk_z^{(l)}d_l}\end{pmatrix} =\frac{1+p_{l,l+1}}{2}\begin{pmatrix}1&R_{l,l+1}\\R_{l,l+1}&1\end{pmatrix} \begin{pmatrix}E_+^{(l+1)}e^{-jk_z^{(l+1)}d_l}\\E_-^{(l+1)}e^{+jk_z^{(l+1)}d_l}\end{pmatrix} $ $ p_{l,l+1}=\frac{Z_0^{(l)}}{Z_0^{(l+1)}},\qquad R_{l,l+1}=\frac{1-p_{l,l+1}}{1+p_{l,l+1}}=\frac{Z_0^{(l+1)}-Z_0^{(l)}}{Z_0^{(l+1)}+Z_0^{(l)}} $ - TE: $Z_0=\dfrac{\omega\mu}{k_z}$, so $p=\dfrac{\mu_lk_z^{(l+1)}}{\mu_{l+1}k_z^{(l)}}$. - TM (tangential-$E$ amplitudes): $Z_0=\dfrac{k_z}{\omega\epsilon}$. Kong writes TM in terms of $H_y$ with $p=\dfrac{\epsilon_lk_z^{(l+1)}}{\epsilon_{l+1}k_z^{(l)}}$. Same physics. Algorithm: 1. In the last region set $E_+^{(t)}=T$ and $E_-^{(t)}=0$ (or set $E_+^{(t)}=1$ and normalize at the end). 2. Step backward through each interface (the matrix above), then propagate within each layer (the phases). 3. In region 0: $\Gamma=\dfrac{E_-^{(0)}}{E_+^{(0)}}$ and $T=\dfrac{E_+^{(t)}}{E_+^{(0)}}$. 4. Power transmission: $ \mathcal T=\frac{\mathrm{Re}\,Y_0^{(t)}}{\mathrm{Re}\,Y_0^{(0)}}\,|T|^2,\qquad \mathcal R+\mathcal T=1\ \text{(lossless)} $ (QC check.) This gives fields in **every** layer. If you only need $\Gamma$, the impedance recursion in [[reflection]] / [[transmission lines]] is shorter. ## Characteristic (ABCD) Matrix of a Layer Relates tangential $(E,\ -H)$, i.e. $(V,I)$ in [[transmission lines]], at the front of a layer to the back. With $\phi=k_zd$: $ \begin{pmatrix}V_\text{front}\\I_\text{front}\end{pmatrix}=\begin{pmatrix}\cos\phi&jZ_0\sin\phi\\ jY_0\sin\phi&\cos\phi\end{pmatrix}\begin{pmatrix}V_\text{back}\\I_\text{back}\end{pmatrix} $ - Quarter wave: $\begin{pmatrix}0&jZ_0\\jY_0&0\end{pmatrix}$, which inverts the impedance. - Half wave: $-I$, i.e. invisible. - Stack = product; then $Z_\text{in}=\dfrac{AZ_L+B}{CZ_L+D}$.