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Circular birefringence means different refractive indices for opposite circular polarizations. In a transparent isotropic medium, their relative wave phase changes rotate the direction of transmitted linear polarization; optical activity gives the solution measurement and convention.

Circular dichroism is unequal absorption of opposite circular polarizations. It can produce elliptical polarization; it is the absorptive response related to optical rotation, rather than the same measured quantity.

For a reciprocal ideal quarter-wave plate followed by a mirror at normal incidence, use the same fixed transverse laboratory axes for the forward and returning Jones vectors. If the axes coincide with the plate's fast and slow axes, each pass has Jones matrix $J=\mathrm{diag}(1,i)$ up to a common phase factor. A polarization-independent mirror has scalar reflection coefficient $r_m$, so the returned Jones vector is $r_m{J}^2\mathbf{e}=r_m\mathrm{diag}(1,-1)\mathbf{e}$. The plate acts twice as a half-wave plate, rather than undoing its phase retardation.

For input linear polarization at angle $\alpha$ to the fast axis, the returned line is at $-\alpha$ modulo $\pi$. At $\alpha =45^{\circ }$ it is perpendicular to the original line. An axis-aligned input instead returns along the same line. General elliptical polarization requires its full input Jones vector; no unique returned state follows from the word “polarized” alone. The propagation direction reverses, so comparing descriptions of circular polarization must also specify the viewing convention.

The electro-optic Kerr effect produces linear birefringence proportional to the square of a static applied electric field. Defining the Kerr effect coefficient $K$ by $\Delta n={\lambda }_{0}K{E}^2$, a length $L$ produces phase retardation $\delta =2\pi KL{E}^2$. A switchable quarter-wave plate requires $|\delta |=\pi /2$ modulo $2\pi$, with fixed principal axes. This differs from the Faraday effect, whose eigenpolarizations are circular polarizations.

For an elliptically polarized wave, the tip of its transverse vector traces an ellipse at a fixed point during each wave period. For polarization of an electromagnetic wave, this is the electric field vector. With the real components in polarization of waves, assume $A_x,A_y>0$ and $\sin \delta \ne 0$. Eliminating time givesA short derivation uses $\varphi =\omega t$ and the cosine addition formula to obtain $E_y/A_y=(E_x/A_x)\cos \delta -\sin \varphi \sin \delta$; squaring and using the sine and cosine identity ${\sin }^2\varphi =1-{\cos }^2\varphi$ gives the displayed equation. Outside phase quadrature, the mixed term gives a tilted ellipse; $A_x,A_y$ are component wave amplitudes and need not be its semi-major and semi-minor axes.

In phase quadrature, $\delta =\pm \pi /2$, the mixed term vanishes. For $(A_x,A_y)=(2,1)V/m$, the ellipse is $E_x^2/4+E_y^2=1$, its semi-major and semi-minor axes have lengths $2$ and $1V/m$, and its electric field magnitude varies between $1$ and $2V/m$. At $\varphi =\pi /4$ with $\delta =\pi /2$, the electric field is $(\sqrt{2},-1/\sqrt{2})V/m$ and has magnitude $\sqrt{5/2}V/m$. Equal wave amplitudes instead give circular polarization, a special case of elliptical polarization with constant magnitude. As $\delta$ approaches $0$ or $\pi$, the ellipse collapses to a line segment and gives linear polarization; this limit is degenerate rather than a nondegenerate ellipse.

Changing $\delta$ from $\pi /2$ to $-\pi /2$ reverses traversal of the same ellipse. In the displayed $x,y$ plot, $\delta =\pi /2$ moves from $(A_x,0)$ toward negative $y$ as time increases. A label such as right- or left-handed circular polarization additionally requires a stated viewing direction and naming convention: viewing along propagation and viewing against propagation reverse the apparent rotation. ^[ref].

The ellipse records both the relative wave amplitudes and the phase difference of two transverse oscillations. This separates a change of polarization of waves from a change of overall wave amplitude: multiplying both components by the same positive factor enlarges the ellipse while preserving its shape and orientation.

A rotating ideal linear polarizer can probe this geometry. For the $2:1$ ellipse in phase quadrature, orthogonal projection onto a pass axis at angle $\theta$ gives $E_{\mathrm{out}}(t)=2\cos \theta \cos \omega t-\sin \theta \sin \omega t$, in $V/m$. Averaging its square over a wave period gives $(4{\cos }^2\theta +{\sin }^2\theta )/2$, while the incoming squared electric field magnitude averages to $5/2$. Thus the transmitted fraction of light intensity is $(4{\cos }^2\theta +{\sin }^2\theta )/5$: it is $4/5$ at $0^{\circ }$, $1/2$ at $45^{\circ }$ and $1/5$ at $90^{\circ }$. There is no single incident line of linear polarization to insert into Malus's law; projecting the electric field and averaging its square is the appropriate generalization. This measurement cannot distinguish $\delta =\pi /2$ from $\delta =-\pi /2$, because reversing the traversal leaves all these squared averages unchanged.

The Faraday effect is magnetic field-induced circular birefringence: opposite circular polarizations acquire different wave phases, rotating transmitted linear polarization. For a transparent medium with isotropy before applying the magnetic field, in the linear magnetic-response regime at fixed vacuum wavelength ${\lambda }_0$ and temperature $T$,Here $V$ is the Verdet constant, $L$ is the material path length, and $B_{\parallel }$ is the applied magnetic flux density component along the ray. This $\mathbf{B}$ is the external bias, distinct from the oscillating magnetic field of the light itself. Michael Faraday discovered the effect in 1845.^[ref]

Fix the sign before using the formula. On the forward path choose a right-handed Cartesian coordinate system, propagation along $+z$, and measure $\theta$ from $+x$ toward $+y$ for the electric field's oscillation line. Positive $\theta$ is counterclockwise in the $x$-right, $y$-up end view seen from $z>0$, looking toward the source. Then $B_{\parallel }=B_z$ and positive $V$ means positive rotation for positive $B_z$. This convention differs from the clockwise-positive convention in optical rotation. A linear polarization line is unchanged by $180^{\circ }$; a measured orientation determines accumulated rotation only modulo $\pi$, unless continuity selects a branch.

For uniform $B_{\parallel }$, the formula reduces to $\theta =VBL$. Its appeal is that a weak change of refractive index becomes a visible change of orientation. Magnetic reversal measurement of Faraday rotation turns this into an inference of $V$ or of the magnetic field path integral. Faraday rotation on a round trip explains the directional behavior used by an optical isolator. In an astronomical cold plasma, the high-frequency limit instead leads to the rotation measure law $\theta =\mathrm{RM}{\lambda }_0^2$; that wavelength dependence is not a universal law for transparent solids.

The factor of one half follows from superposition, without assigning ambiguous right/left names to the circular polarizations. Use the complex amplitude convention $\mathrm{Re}[\mathbf{E}(z)e^{-i\omega t}]$ and fixed transverse unit vectorsFor equal transmitted wave amplitudes, let the accumulated wave phases be ${\varphi }_{+}$ and ${\varphi }_{-}$. Their sum isThe common wave phase $\bar{\varphi }$ sets the timing of the electric field oscillation; the real direction in parentheses sets its linear polarization. Opposite circular polarizations have changed their relative wave phase, yet their sum still oscillates along a line. Unequal transmitted wave amplitudes would generally give elliptical polarization instead.

In a uniform sample, with refractive indices $n_{+},n_{-}$ and vacuum wavenumber $k_0=2\pi /{\lambda }_0$, the phase difference is $(k_{+}-k_{-})L=k_0(n_{+}-n_{-})L$. ThusThe second relation identifies the small signed circular birefringence responsible for the rotation in this basis. Unlike a waveplate, whose separated modes have orthogonal linear polarizations, this ideal rotator has modes with opposite circular polarizations.

For constant isotropic transport coefficients and zero mean flow, a divergence-free Fourier mode has two circular polarizations that diagonalize curl with eigenvalues $\pm k$. The mean induction equation therefore has the displayed growth rates. Growth requires an admissible $k$ with $|\alpha |>(\eta +\beta )k$. Freely variable wavenumber gives $k_{\ast }=|\alpha |/[2(\eta +\beta )]$, provided it respects mean-field scale separation. A uniform field has zero curl and cannot grow by constant alpha alone; boundaries fix the longest available spatial mode.

Three equal-amplitude transverse waves along orthogonal axes, with the same circular polarization, each satisfy $\nabla \times {\mathbf{u}}_j=\sigma k{\mathbf{u}}_j$, $\sigma =\pm 1$. Distinct spatial phases have zero cross averages. The periodic first-order smoothing response to a uniform test field satisfies $({\partial }_t-\eta \Delta ){\mathbf{b}}_j=(\overline{\mathbf{B}}\cdot \nabla ){\mathbf{u}}_j$. Writing $a=\eta k^2$, the response is ${\mathbf{b}}_j=k{\overline{B}}_j[a{\partial }_{\varphi }{\mathbf{u}}_j-\Omega {\mathbf{u}}_j]/(a^2+{\Omega }^2)$. Since ${\mathbf{u}}_j\times {\partial }_{\varphi }{\mathbf{u}}_j=-\sigma U^2{\mathbf{e}}_j$, summing yields the displayed isotropic alpha effect. It includes the resistive and temporal weighting absent from an unqualified instantaneous helicity rule.

Alfred Kastler's 1966 lecture recalls his 1950 proposal and distinguishes the early ${}^{23}\mathrm{Na}$ atomic beam work from later sealed-vapour experiments. Its Fig. 5 shows subsequent sodium radio wave resonances, including multiple-photon transitions.^[ref]

The 1952 demonstration by Jean Brossel, Alfred Kastler and Jacques Winter used a sodium atomic beam, lamp light with circular polarization and fluorescence readout. Jean Brossel was a central experimental contributor; Alfred Kastler received the 1966 Nobel Prize in Physics alone.^[ref][ref]

The lecture also describes a ${}^{199}\mathrm{Hg}$ experiment: a pumping beam with circular polarization prepares the vapour, while a perpendicular probe converts transverse magnetic dipole moment Larmor precession into modulated absorption at a photodetector. Interrupting pump and RF together exposes decaying free Larmor precession. Its envelope measures transverse relaxation under that model; it is distinct from the sodium atomic beam fluorescence signal.^[ref]

MIT's later rubidium experiment pumps natural ${}^{85}\mathrm{Rb}$/${}^{87}\mathrm{Rb}$ with a discharge lamp filtered to D1 ($5S_{1/2}\to 5P_{1/2}$, near $795\mathrm{nm}$). A polarizer and quarter-wave plate precede a heated rubidium/neon cell. Helmholtz coils set a weak static magnetic field. Separate transverse RF mixing increases absorption, measured by photodiode transmission.^[ref]

Compare resonant and detuned RF with the optical beam fixed, then change the calibrated static magnetic field: a true magnetic sublevel resonance should move with its Zeeman splitting. Lamp drift, stray magnetic fields, imperfect circular polarization, optical attenuation and RF power broadening can change the observed signal. Consequently, a transmission dip alone does not determine every sublevel population. These are measurement-design considerations, not additional reported historical observations.

A wave is linearly polarized when its oscillating electric field always lies along one fixed line. For a transverse plane wave, that line is perpendicular to the wavevector. Unlike circular polarization, the field oscillates along the line rather than rotating through directions.

Optical activity is the ability to rotate the direction of linear polarization of transmitted light. In an achiral solvent, the two members of an enantiomer pair have equal and opposite optical rotations at the same wavelength, temperature and concentration. The measured angle concerns a bulk sample; it is not a geometric rotation of the molecules.^[ref]

The rotating object in this description is the direction along which the electric field oscillates. At each position the transmitted light can still have linear polarization: its electric field moves back and forth along a line, while the orientation of that line changes along the sample. Nothing requires the molecules to turn bodily through the observed angle. This makes polarimetry a bridge from molecular chirality to a visible bulk measurement, and, with a suitable calibration, a way to follow concentration or the balance of enantiomers.

In a transparent, isotropic chemical solution, natural optical rotation can be understood as circular birefringence: the two opposite circular polarizations accumulate different wave phases, and their equal-amplitude sum remains linearly polarized along a rotated direction. Circular dichroism instead means unequal absorption of those components and can produce elliptical polarization. These are related responses, not interchangeable measurements. General birefringence need not produce this simple rotation of linear polarization.^[ref]

Circular polarization selects the sign of the absorbed photon's angular momentum along the beam. Choose that direction as $z$ and define ${\sigma }^{+}$ by the absorption rule $m^{\prime }=m+1$, where $m$ is the magnetic quantum number. Spontaneous emission can return an atom to a different magnetic sublevel, so absorption followed by emission is a population-transfer cycle. A sublevel that the chosen beam cannot excite is dark to that beam. Darkness is relative to the selected transition and electromagnetic polarization, rather than a claim that the atom cannot interact with any light.

A static magnetic field along $z$ defines a stable axis and gives Zeeman splitting. A transverse oscillating magnetic field can redistribute magnetic sublevel populations when its frequency matches an allowed splitting. With the appropriate optical transition, that redistribution restores photon absorption: transmitted intensity decreases, or collected fluorescence increases. These are alternative readouts whose detailed sensitivity depends on the transition, viewing direction and electromagnetic polarization.

An explicit example avoids an unspecified correlation time. For each coordinate axis ${\mathbf{e}}_j$, choose transverse unit vectors ${\mathbf{a}}_j,{\mathbf{b}}_j$ with ${\mathbf{e}}_j\times {\mathbf{a}}_j={\mathbf{b}}_j$, and take the three real circular polarizationsThe propagation directions are mutually perpendicular and all three waves have the same handedness. Direct differentiation givesA spatial average over the periodic cell removes cross terms between different waves. Hence the mean kinetic helicity of their sum isThe sum has isotropic second moments, $⟨u_i{u}_j⟩=U^2{\delta }_{ij}$.

For a locally uniform test magnetic field $\overline{\mathbf{B}}$, the periodic first-order smoothing response obeysSet $a=\eta k^2>0$. Since ${\partial }_t=-\Omega {\partial }_{\varphi }$ and ${\partial }_{\varphi }^2{\mathbf{u}}_j=-{\mathbf{u}}_j$, direct substitution gives the response after transients decay:The term proportional to ${\mathbf{u}}_j$ contributes no cross product with that same wave. Cross terms between distinct waves average to zero. Thereforeand the isotropic alpha effect of three helical traveling waves isThis is the alpha-helicity relation with the exact response time ${\tau }_{\mathrm{eff}}=a/(a^2+{\Omega }^2)$ for these monochromatic waves. In the quasistatic limit $\Omega =0$, $\alpha =-\mathcal{H}/(3\eta k^2)$. A single wave would give an anisotropic response along its propagation direction; the three equal waves make the response tensor a scalar multiple of the identity. The calculation retains the validity assumptions of first-order smoothing, rather than extrapolating the formula to arbitrary fluctuation amplitude.

A full fringe scan is unnecessary. Use the same components to make three fixed projective measurements, on separate batches of identically prepared photons. Direct path detection after the polarizing beam splitter measures the $H/V$ populations. Recombine after matching the arm photon polarizations, once with $\theta =0$ and once with $\theta =\pi /2$, to measure the two quadratures of the quantum coherence.

In the $H/V$ basis, write the density matrix in terms of its Bloch vector as $\rho =(I+r_{x}X+r_{y}Y+r_{z}Z)/2$. The previous formula becomesThus three measured relative frequencies giveand determineThe two interferometric settings analyze the diagonal linear polarizations $(|H⟩\pm |V⟩)/\sqrt{2}$ and the circular polarizations $(|H⟩\pm i|V⟩)/\sqrt{2}$, with detector labels fixed by the calibrated quantum phase convention. This is informationally complete three-observable qubit tomography: direct measurements in three bases replace fitting an entire phase-dependent fringe. It also reconstructs a mixed input, for which $|r|\le 1$, although the source here is assumed pure.

Simply measuring transmission through one fixed polarizing beam splitter would determine only $r_z$. Rotating a linear analyzer through different angles can additionally determine $r_x$, but cannot distinguish the two signs of $r_y$ without a relative-phase measurement. For arbitrary, possibly elliptical photon polarization, all three settings above are needed; no extra analyzer element beyond the described path-phase control and photon polarization rotator is required.

For constant isotropic $\alpha ,\beta$ and zero mean flow, the mean induction equation isLet $\overline{\mathbf{B}}=\hat{\mathbf{B}}{e}^{\sigma t+i\mathbf{k}\cdot \mathbf{x}}$, with $\mathbf{k}\cdot \hat{\mathbf{B}}=0$. On this two-dimensional transverse space, $i\mathbf{k}\times$ has eigenvalues $\pm k$. For example, if $\mathbf{k}=k{\mathbf{e}}_z$, the vectors ${\mathbf{e}}_x\pm i{\mathbf{e}}_y$ have these curl eigenvalues. The homogeneous alpha-squared dynamo growth criterion follows fromChoose the circular polarization of the mode to match the sign of $\alpha$. This is an alpha-squared dynamo: the alpha effect supplies both parts of the regeneration loop without differential rotation.

If arbitrarily long wavelengths are permitted, every nonzero alpha has a growing band $0<k<|\alpha |/{\eta }_T$. Maximizing the quadratic givesprovided $k_{\ast }\ell \ll 1$ so the mean-field approximation is valid. A uniform field has $k=0$, zero curl and zero growth; a constant alpha cannot amplify that uniform mode. In a periodic domain of size $L$, the lowest nonzero wavenumber is $2\pi /L$, giving $|\alpha |L/{\eta }_T>2\pi$. Other boundaries and geometries give different critical constants, but the general scale criterion is $|\alpha |L/{\eta }_T$ sufficiently large. Nonzero kinetic helicity alone is therefore not a sufficient finite-domain dynamo criterion: its induced alpha must overcome diffusion on an available mean-field scale.

Put $a=c/(4\pi en)$ and retain terms linear in $\delta \mathbf{B}$. The electron magnetohydrodynamics induction equation becomesHere the curl of $(\nabla \times \delta \mathbf{B})\times B_0\hat{\mathbf{z}}$ reduces to $B_0{\partial }_z\nabla \times \delta \mathbf{B}$, since the background magnetic field is constant and the divergence of a curl vanishes. For a plane wave $\delta \mathbf{B}=\mathbf{b}{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$, the linear equation and the solenoidal constraint areOn the transverse plane, the operator $i\mathbf{k}\times$ has eigenvalues $\pm k$ by the helicity decomposition of a transverse Fourier mode. Equivalently, squaring the equation and using the vector triple product gives ${\omega }^2=a^2{B}_0^2{k}_{\parallel }^2{k}^2$. Since $a{B}_0=v_A{d}_i$, where $v_A$ is the Alfvén speed and $d_i$ the ion skin depth,The two branches have opposite circular polarizations. The specified electron-only induction model describes a whistler wave; the paper calls these branches kinetic Alfvén waves. The following cascade calculation uses the specified model and its dispersion relation, without adding the pressure response needed for the usual kinetic Alfvén interpretation.

Use real physical fields in the quadratic energy expressions. The vector identity $\nabla \cdot (\mathbf{E}\times \mathbf{B})=\mathbf{B}\cdot (\nabla \times \mathbf{E})-\mathbf{E}\cdot (\nabla \times \mathbf{B})$ and vacuum Maxwell's equations giveIntegrating over a fixed volume and applying the divergence theorem proves Poynting's theorem:The left side is outward energy flux through the closed surface, not merely local intensity.

For a linearly polarized wave with phase $\chi =\mathbf{k}\cdot \mathbf{r}-\omega t$, the real fields give Poynting vector $\mathbf{S}={\epsilon }_{0}c{E}_0^2{\cos }^2\chi \hat{\mathbf{k}}$. Local intensity oscillates at $2\omega$. In the closed-surface integral, the time-independent part cancels because ${\oint }_A\hat{\mathbf{k}}\cdot d\mathbf{A}=0$. The remaining flux generally oscillates at $2\omega$, has zero time average, and may vanish identically for special volume geometries.

For circular polarization, take $\mathbf{E}=E_0({\mathbf{e}}_1\cos \chi +{\mathbf{e}}_2\sin \chi )$ with orthonormal transverse vectors. Then $E^2=E_0^2$ and $\mathbf{B}=\hat{\mathbf{k}}\times \mathbf{E}/c$, so $\mathbf{S}={\epsilon }_{0}c{E}_0^2\hat{\mathbf{k}}$ is spatially and temporally constant. Therefore the closed-surface Poynting flux of a plane wave isThere is still a nonzero steady local energy flow: equal energy enters and leaves every fixed closed volume. Calling the net closed-surface flux a nonzero constant would confuse these two quantities.

In vacuum the Maxwell equations areSubstitution of the complex plane electromagnetic wave amplitudes into these equations givesThus the electric field must be transverse to $k$; taking the positive-frequency branch gives $\omega =c|k|$.

For $E_0=E(1,i,0)$, $k=k\hat{z}$, and $\phi =kz-\omega t$, the real fields areThe electric vector rotates with constant magnitude, so this is circular polarization. Its Poynting vector isThe Poynting vector is the electromagnetic energy flux: its direction is the direction of propagation and its magnitude is the power crossing unit area.

A transverse static electric field can induce linear birefringence through the electro-optic Kerr effect. Define its coefficient by $\Delta n={\lambda }_{0}K{E}_{\mathrm{dc}}^2$, with the principal axes parallel and perpendicular to the applied field. A length $d_K$ then gives ${\delta }_K=2\pi K{d}_K{E}_{\mathrm{dc}}^2$.

For input polarized at $45^{\circ }$ to those axes, switching from zero retardation to $|{\delta }_K|=\pi /2$ switches linear to circular polarization. Other input angles generally become elliptical, and axis-aligned input stays linear. The required field magnitude for the lowest-order setting isA Kerr cell is suitable as a switchable quarter-wave plate. The field can be changed electronically without mechanically rotating the axes; actual switching time depends on the material response and driving circuit. This is induced linear birefringence, not polarization-selective absorption.^[ref]