Photon polarization is the quantum mechanical description of the classical polarized sinusoidal plane electromagnetic wave. An individual photon can be described as having right or left circular polarization, or a superposition of the two. Equivalently, a photon can be described as having horizontal or vertical linear polarization, or a superposition of the two. The description of photon polarization contains many of the physical concepts and much of the mathematical machinery of more involved quantum descriptions, such as the quantum mechanics of an electron in a potential well. Polarization is an example of a qubit degree of freedom, which forms a fundamental basis for an understanding of more complicated quantum phenomena. Much of the mathematical machinery of quantum mechanics, such as state vectors, probability amplitudes, unitary operators, and Hermitian operators, emerge naturally from the classical Maxwell's equations in the description. The quantum polarization state vector for the photon, for instance, is identical with the Jones vector, usually used to describe the polarization of a classical wave. Unitary operators emerge from the classical requirement of the conservation of energy of a classical wave propagating through lossless media that alter the polarization state of the wave. Hermitian operators then follow for infinitesimal transformations of a classical polarization state. Many of the implications of the mathematical machinery are easily verified experimentally. In fact, many of the experiments can be performed with polaroid sunglass lenses. The connection with quantum mechanics is made through the identification of a minimum packet size, called a photon, for energy in the electromagnetic field. The identification is based on the theories of Planck and the interpretation of those theories by Einstein. The correspondence principle then allows the identification of momentum and angular momentum (called spin), as well as energy, with the photon.
Polarization of classical electromagnetic waves
Polarization states
Linear polarization
The wave is linearly polarized (or plane polarized) when the phase angles α x , α y {\displaystyle \alpha _{x}\,,\;\alpha _{y}} are equal,
α x = α y = d e f α . {\displaystyle \alpha _{x}=\alpha _{y}\ {\stackrel {\mathrm {def} }{=}}\ \alpha .}
This represents a wave with phase α {\displaystyle \alpha } polarized at an angle θ {\displaystyle \theta } with respect to the x axis. In this case the Jones vector
| ψ ⟩ = ( cos θ exp ( i α x ) sin θ exp ( i α y ) ) {\displaystyle |\psi \rangle ={\begin{pmatrix}\cos \theta \exp \left(i\alpha _{x}\right)\\\sin \theta \exp \left(i\alpha _{y}\right)\end{pmatrix}}}
can be written with a single phase:
| ψ ⟩ = ( cos θ sin θ ) exp ( i α ) . {\displaystyle |\psi \rangle ={\begin{pmatrix}\cos \theta \\\sin \theta \end{pmatrix}}\exp \left(i\alpha \right).}
The state vectors for linear polarization in x or y are special cases of this state vector. If unit vectors are defined such that
| x ⟩ = d e f ( 1 0 ) {\displaystyle |x\rangle \ {\stackrel {\mathrm {def} }{=}}\ {\begin{pmatrix}1\\0\end{pmatrix}}}
and
| y ⟩ = d e f ( 0 1 ) {\displaystyle |y\rangle \ {\stackrel {\mathrm {def} }{=}}\ {\begin{pmatrix}0\\1\end{pmatrix}}}
then the linearly polarized polarization state can be written in the "x–y basis" as
| ψ ⟩ = cos θ exp ( i α ) | x ⟩ + sin θ exp ( i α ) | y ⟩ = ψ x | x ⟩ + ψ y | y ⟩ . {\displaystyle |\psi \rangle =\cos \theta \exp \left(i\alpha \right)|x\rangle +\sin \theta \exp \left(i\alpha \right)|y\rangle =\psi _{x}|x\rangle +\psi _{y}|y\rangle .}
Circular polarization
If the phase angles α x {\displaystyle \alpha _{x}} and α y {\displaystyle \alpha _{y}} differ by exactly π / 2 {\displaystyle \pi /2} and the x amplitude equals the y amplitude the wave is circularly polarized. The Jones vector then becomes
| ψ ⟩ = 1 2 ( 1 ± i ) exp ( i α x ) {\displaystyle |\psi \rangle ={\frac {1}{\sqrt {2}}}{\begin{pmatrix}1\\\pm i\end{pmatrix}}\exp \left(i\alpha _{x}\right)}
where the plus sign indicates left circular polarization and the minus sign indicates right circular polarization. In the case of circular polarization, the electric field vector of constant magnitude rotates in the x–y plane. If unit vectors are defined such that
| R ⟩ = d e f 1 2 ( 1 i ) {\displaystyle |\mathrm {R} \rangle \ {\stackrel {\mathrm {def} }{=}}\ {1 \over {\sqrt {2}}}{\begin{pmatrix}1\\i\end{pmatrix}}}
and
| L ⟩ = d e f 1 2 ( 1 − i ) {\displaystyle |\mathrm {L} \rangle \ {\stackrel {\mathrm {def} }{=}}\ {1 \over {\sqrt {2}}}{\begin{pmatrix}1\\-i\end{pmatrix}}}
then an arbitrary polarization state can be written in the "R–L basis" as
| ψ ⟩ = ψ R | R ⟩ + ψ L | L ⟩ {\displaystyle |\psi \rangle =\psi _{\rm {R}}|\mathrm {R} \rangle +\psi _{\rm {L}}|\mathrm {L} \rangle }
where
ψ R = ⟨ R | ψ ⟩ = 1 2 ( cos θ exp ( i α x ) − i sin θ exp ( i α y ) ) {\displaystyle \psi _{\rm {R}}=\langle \mathrm {R} |\psi \rangle ={\frac {1}{\sqrt {2}}}\left(\cos \theta \exp(i\alpha _{x})-i\sin \theta \exp(i\alpha _{y})\right)}
and
ψ L = ⟨ L | ψ ⟩ = 1 2 ( cos θ exp ( i α x ) + i sin θ exp ( i α y ) ) . {\displaystyle \psi _{\rm {L}}=\langle \mathrm {L} |\psi \rangle ={\frac {1}{\sqrt {2}}}\left(\cos \theta \exp(i\alpha _{x})+i\sin \theta \exp(i\alpha _{y})\right).}
We can see that
1 = | ψ R | 2 + | ψ L | 2 . {\displaystyle 1=|\psi _{\rm {R}}|^{2}+|\psi _{\rm {L}}|^{2}.}
Elliptical polarization
The general case in which the electric field rotates in the x–y plane and has variable magnitude is called elliptical polarization. The state vector is given by
| ψ ⟩ = d e f ( ψ x ψ y ) = ( cos θ exp ( i α x ) sin θ exp ( i α y ) ) . {\displaystyle |\psi \rangle \ {\stackrel {\mathrm {def} }{=}}\ {\begin{pmatrix}\psi _{x}\\\psi _{y}\end{pmatrix}}={\begin{pmatrix}\cos \theta \exp \left(i\alpha _{x}\right)\\\sin \theta \exp \left(i\alpha _{y}\right)\end{pmatrix}}.}
Geometric visualization of an arbitrary polarization state To get an understanding of what a polarization state looks like, one can observe the orbit that is made if the polarization state is multiplied by a phase factor of e i ω t {\displaystyle e^{i\omega t}} and then having the real parts of its components interpreted as x and y coordinates respectively. That is:
( x ( t ) y ( t ) ) = ( ℜ ( e i ω t ψ x ) ℜ ( e i ω t ψ y ) ) = ℜ [ e i ω t ( ψ x ψ y ) ] = ℜ ( e i ω t | ψ ⟩ ) . {\displaystyle {\begin{pmatrix}x(t)\\y(t)\end{pmatrix}}={\begin{pmatrix}\Re (e^{i\omega t}\psi _{x})\\\Re (e^{i\omega t}\psi _{y})\end{pmatrix}}=\Re \left[e^{i\omega t}{\begin{pmatrix}\psi _{x}\\\psi _{y}\end{pmatrix}}\right]=\Re \left(e^{i\omega t}|\psi \rangle \right).}
If only the traced out shape and the direction of the rotation of (x(t), y(t)) is considered when interpreting the polarization state, i.e. only
M ( | ψ ⟩ ) = { ( x ( t ) , y ( t ) ) | ∀ t } {\displaystyle M(|\psi \rangle )=\left.\left\{{\Big (}x(t),\,y(t){\Big )}\,\right|\,\forall \,t\right\}}
(where x(t) and y(t) are defined as above) and whether it is overall more right circularly or left circularly polarized (i.e. whether |ψR| > |ψL| or vice versa), it can be seen that the physical interpretation will be the same even if the state is multiplied by an arbitrary phase factor, since
M ( e i α | ψ ⟩ ) = M ( | ψ ⟩ ) , α ∈ R {\displaystyle M(e^{i\alpha }|\psi \rangle )=M(|\psi \rangle ),\ \alpha \in \mathbb {R} }
and the direction of rotation will remain the same. In other words, there is no physical difference between two polarization states | ψ ⟩ {\displaystyle |\psi \rangle } and e i α | ψ ⟩ {\displaystyle e^{i\alpha }|\psi \rangle } , between which only a phase factor differs. It can be seen that for a linearly polarized state, M will be a line in the xy plane, with length 2 and its middle in the origin, and whose slope equals to tan(θ). For a circularly polarized state, M will be a circle with radius 1/√2 and with the middle in the origin.
Energy, momentum, and angular momentum of a classical electromagnetic wave
Energy density of classical electromagnetic waves
Energy in a plane wave
The energy per unit volume in classical electromagnetic fields is (cgs units) and also Planck units:
E c = 1 8 π [ E 2 ( r , t ) + B 2 ( r , t ) ] . {\displaystyle {\mathcal {E}}_{c}={\frac {1}{8\pi }}\left[\mathbf {E} ^{2}(\mathbf {r} ,t)+\mathbf {B} ^{2}(\mathbf {r} ,t)\right].}
For a plane wave, this becomes:
E c = ∣ E ∣ 2 8 π {\displaystyle {\mathcal {E}}_{c}={\frac {\mid \mathbf {E} \mid ^{2}}{8\pi }}}
where the energy has been averaged over a wavelength of the wave.
Fraction of energy in each component The fraction of energy in the x component of the plane wave is
f x = | E | 2 cos 2 θ | E | 2 = ψ x ∗ ψ x = cos 2 θ {\displaystyle f_{x}={\frac {|\mathbf {E} |^{2}\cos ^{2}\theta }{\vert \mathbf {E} \vert ^{2}}}=\psi _{x}^{*}\psi _{x}=\cos ^{2}\theta }
with a similar expression for the y component resulting in f y = sin 2 θ {\displaystyle f_{y}=\sin ^{2}\theta } . The fraction in both components is
ψ x ∗ ψ x + ψ y ∗ ψ y = ⟨ ψ | ψ ⟩ = 1. {\displaystyle \psi _{x}^{*}\psi _{x}+\psi _{y}^{*}\psi _{y}=\langle \psi |\psi \rangle =1.}
Momentum density of classical electromagnetic waves The momentum density is given by the Poynting vector
P = 1 4 π c E ( r , t ) × B ( r , t ) . {\displaystyle {\boldsymbol {\mathcal {P}}}={1 \over 4\pi c}\mathbf {E} (\mathbf {r} ,t)\times \mathbf {B} (\mathbf {r} ,t).}
For a sinusoidal plane wave traveling in the z direction, the momentum is in the z direction and is related to the energy density:
P z c = E c . {\displaystyle {\mathcal {P}}_{z}c={\mathcal {E}}_{c}.}
The momentum density has been averaged over a wavelength.
Angular momentum density of classical electromagnetic waves
Electromagnetic waves can have both orbital and spin angular momentum. The total angular momentum density is
L = r × P = 1 4 π c r × [ E ( r , t ) × B ( r , t ) ] . {\displaystyle {\boldsymbol {\mathcal {L}}}=\mathbf {r} \times {\boldsymbol {\mathcal {P}}}={1 \over 4\pi c}\mathbf {r} \times \left[\mathbf {E} (\mathbf {r} ,t)\times \mathbf {B} (\mathbf {r} ,t)\right].}
For a sinusoidal plane wave propagating along z {\displaystyle z} axis the orbital angular momentum density vanishes. The spin angular momentum density is in the z {\displaystyle z} direction and is given by
L = | E | 2 8 π ω ( | ⟨ R | ψ ⟩ | 2 − | ⟨ L | ψ ⟩ | 2 ) = 1 ω E c ( | ψ R | 2 − | ψ L | 2 ) {\displaystyle {\mathcal {L}}={{\vert \mathbf {E} \vert ^{2}} \over {8\pi \omega }}\left(\left\vert \langle \mathrm {R} |\psi \rangle \right\vert ^{2}-\left\vert \langle \mathrm {L} |\psi \rangle \right\vert ^{2}\right)={\frac {1}{\omega }}{\mathcal {E}}_{c}\left(\vert \psi _{\rm {R}}\vert ^{2}-\vert \psi _{\rm {L}}\vert ^{2}\right)}
where again the density is averaged over a wavelength.
Optical filters and crystals
Passage of a classical wave through a polaroid filter
A linear filter transmits one component of a plane wave and absorbs the perpendicular component. In that case, if the filter is polarized in the x direction, the fraction of energy passing through the filter is
f x = ψ x ∗ ψ x = cos 2 θ . {\displaystyle f_{x}=\psi _{x}^{*}\psi _{x}=\cos ^{2}\theta .\,}
Example of energy conservation: Passage of a classical wave through a birefringent crystal An ideal birefringent crystal transforms the polarization state of an electromagnetic wave without loss of wave energy. Birefringent crystals therefore provide an ideal test bed for examining the conservative transformation of polarization states. Even though this treatment is still purely classical, standard quantum tools such as unitary and Hermitian operators that evolve the state in time naturally emerge.
Initial and final states A birefringent crystal is a material that has an optic axis with the property that the light has a different index of refraction for light polarized parallel to the axis than it has for light polarized perpendicular to the axis. Light polarized parallel to the axis are called "extraordinary rays" or "extraordinary photons", while light polarized perpendicular to the axis are called "ordinary rays" or "ordinary photons". If a linearly polarized wave impinges on the crystal, the extraordinary component of the wave will emerge from the crystal with a different phase than the ordinary component. In mathematical language, if the incident wave is linearly polarized at an angle t h e t a {\displaystyle theta} with respect to the optic axis, the incident state vector can be written
| ψ ⟩ = ( cos θ sin θ ) {\displaystyle |\psi \rangle ={\begin{pmatrix}\cos \theta \\\sin \theta \end{pmatrix}}}
and the state vector for the emerging wave can be written
| ψ ′ ⟩ = ( cos θ exp ( i α x ) sin θ exp ( i α y ) ) = ( exp ( i α x ) 0 0 exp ( i α y ) ) ( cos θ sin θ ) = d e f U ^ | ψ ⟩ . {\displaystyle |\psi '\rangle ={\begin{pmatrix}\cos \theta \exp \left(i\alpha _{x}\right)\\\sin \theta \exp \left(i\alpha _{y}\right)\end{pmatrix}}={\begin{pmatrix}\exp \left(i\alpha _{x}\right)&0\\0&\exp \left(i\alpha _{y}\right)\end{pmatrix}}{\begin{pmatrix}\cos \theta \\\sin \theta \end{pmatrix}}\ {\stackrel {\mathrm {def} }{=}}\ {\hat {U}}|\psi \rangle .}
While the initial state was linearly polarized, the final state is elliptically polarized. The birefringent crystal alters the character of the polarization.
Dual of the final state
The initial polarization state is transformed into the final state with the operator U. The dual of the final state is given by
⟨ ψ ′ | = ⟨ ψ | U ^ † {\displaystyle \langle \psi '|=\langle \psi |{\hat {U}}^{\dagger }}
where U † {\displaystyle U^{\dagger }} is the adjoint of U, the complex conjugate transpose of the matrix.
Unitary operators and energy conservation The fraction of energy that emerges from the crystal is
⟨ ψ ′ | ψ ′ ⟩ = ⟨ ψ | U ^ † U ^ | ψ ⟩ = ⟨ ψ | ψ ⟩ = 1. {\displaystyle \langle \psi '|\psi '\rangle =\langle \psi |{\hat {U}}^{\dagger }{\hat {U}}|\psi \rangle =\langle \psi |\psi \rangle =1.}
In this ideal case, all the energy impinging on the crystal emerges from the crystal. An operator U with the property that
U ^ † U ^ = I , {\displaystyle {\hat {U}}^{\dagger }{\hat {U}}=I,}
where I is the identity operator and U is called a unitary operator. The unitary property is necessary to ensure energy conservation in state transformations.
Hermitian operators and energy conservation
If the crystal is very thin, the final state will be only slightly different from the initial state. The unitary operator will be close to