Jones vector & matrices Department of Physics 1
Matrix treatment of polarization Consider a light ray with an instantaneous E-vector as shown y E k, t = xe x (k, t) + ye y k, t E y E x x E x = E 0x e i kz ωt+φ x E y = E 0y e i kz ωt+φ y Department of Physics 2
Matrix treatment of polarization Combining the components: E k, t = xe 0x e i kz ωt+φ x + ye 0y e i kz ωt+φ y = xe 0x e iφ x + ye 0y e iφ x i kz ωt e i kz ωt = E 0 e The terms in square brackets represents the complex amplitude of the plane wave Department of Physics 3
Jones Vectors The state of polarization of light is determined by the relative amplitudes E ox, E oy and, the relative phases δ = φ y φ x components E 0y of these The complex amplitude is written as a twoelement matrix, the Jones vector E E 0 = 0x = E 0xe iφ x E 0y e iφ = E 0x y eiφ x E 0y e iδ Department of Physics 4
Horizontally polarized light The electric field oscillations are only along the x-axis The Jones vector is then written, y E 0 = E 0x E 0y = E 0xe iφ x 0 = A 0 = A 1 0 where we have set the phase φ x = 0, for convenience The normalized form is 1 0 x Department of Physics 5
Vertically polarized light The electric field oscillations are only along the y-axis The Jones vector is then written, E E 0 = 0x 0 = E 0y E 0y e iφ y = 0 1 = A 0 1 Where we have set the phase φ y = 0, for convenience Department of Physics 6 y The normalized form is 0 1 x
Arbitrary Angle For an arbitrary linear polarization Then we have: E 0 = and the Jones vector is simply a line inclined at an angle = tan 1 (E oy /E ox ) since we can write E 0 = E 0x E 0y = E 0x E 0y E 2 2 cos α 0x + E 0y sin α y The normalized form is cos α sin α x Department of Physics 7
Circular polarization The Jones vector for this case where E x leads E y is E 0 = E 0xe iφ x E 0y e iφ = A y Ae iπ = A 1 2 e iπ = A 1 2 i 1 1 The normalized form is: 2 i This vector represents circularly polarized light, where E is LCP (left-handed circular polarisation) This mode is called left-circularly polarized light What is the corresponding vector for right-circularly polarized light? Replace /2 with /2 to get Department of Physics 8 1 2 1 i
Elliptically polarized light If E ox E oy, e.g. if E ox = A and E oy = B The Jones vector can be written 1 ib Type of rotation? LCP (left-handed) 1 ib Type of rotation? RCP (right-handed) What determines the major or minor axes of the ellipse? Department of Physics 9
Jones vector and polarization In general, the Jones vector for the arbitrary E 0 = case is an ellipse m ; m + 1 2 E 0x E 0y e iδ = tan 2α = 2E oxe 0y cos δ E 2 2 0x E 0y A B cos δ + i sin δ a y E oy b E ox x Department of Physics 10
Linear polarizer Selectively removes all or most of the E-vibrations except in a given direction TA y x Linear polarizer Department of Physics 11
Jones matrix : linear polarizer Consider a linear polarizer with transmission axis along the vertical (y). Let a 2 2 matrix represent the polarizer operating on vertically polarized light. The transmitted light must also be vertically polarized. Thus, a c b d x y = 0 y = ax + by cx + dy Thus, we know: ax + by = 0 cx + dy = y Department of Physics 12
Jones matrix : linear polarizer The relationship must be true for any x & y Therefore ax = by is not a valid solution! Instead: a = b = c = 0, d = 1 And: M = 0 0 0 1 Department of Physics 13
Optical elements: Rotator Rotates the direction of linearly polarized light by a particular angle θ y SA Rotator x Department of Physics 14
Jones matrix for a rotator An E-vector oscillating linearly at θ is rotated by an angle β Thus, the light must be converted to one that oscillates linearly at (β + θ) a c b d cos θ sin θ = cos β + θ sin β + θ One then finds: a cos θ + b sin θ = cos β + θ c cos θ + d sin θ = sin β + θ Department of Physics 15
Jones matrix for a rotator Remember Trig identities: cos β + θ = cos β cos θ sin β sin θ sin β + θ = sin β cos θ + cos β sin θ Therefore: a = cos β, c = sin β, b = sin β d = cos β And: M = cos β sin β sin β cos β the Rotator Matrix R β Department of Physics 16
Rotator Properties The rotator matrix can also be used to rotate a Matrix M θ = R θ MR θ R θ = cos θ sin θ sin θ cos θ R θ = cos θ sin θ sin θ cos θ Start with a linear polariser: M 0 R θ = 1 0 0 0 M 0 = 1 0 0 0 cos θ sin θ sin θ cos θ = cos θ sin θ 0 0 Department of Physics 17
Matrix Rotation M θ = R θ M 0 R θ = Therefore: cos θ sin θ sin θ cos θ cos θ sin θ 0 0 = cos2 θ sin θ cos θ sin θ cos θ sin 2 θ M θ = cos2 θ sin θ cos θ sin θ cos θ sin 2 θ is the Jones matrix at an arbitrary angle θ Department of Physics 18
Wave Plates Remember the Optical Path Length? In a birefringent material, it is possible for there to be a different index of refraction for each polarisation. Λ o = n o d Λ(e) = n e d ΔΛ = d Λ = nds nd n o n e Remember that the equation of the EM field is: E = E 0 exp i kx ωt So: kλ is a measure of phase, and therefore Δφ = k 0 ΔΛ = 2πd n for a wave plate λ o n e Department of Physics 19
Phase Retarder Introduces a phase difference (Δφ) between orthogonal components The fast axis(fa) and slow axis (SA) are shown FA y SA x Retardation plate Department of Physics 20
Jones matrix : phase retarder We wish to find a matrix which will transform the elements as follows: E 0x e iφ x E 0x e i φ x+ε x E 0y e iφ y E 0y e i φ y+ε y It is easy to show by inspection that, Δε M = eiε x 0 0 e iε ei ε0 2 0 y 0 e i ε 0± Δε 2 Here x and y represent the advance in phase of the components Department of Physics 21
Quarter Wave Plate A Quarter Wave plate is characterised by: Δε = π 2 For example: ε y = ε 0 + π 4 and ε x = ε 0 π 4 π M = eiε x 0 0 e iε = ei ε0 4 0 y 0 e i ε 0+ π 4 = e iε 0e iπ 4 1 0 0 i The initial phase term is arbitrary, thus typically: M = e iπ 4 1 0 0 i 1 0 0 i Department of Physics 22
Jones matrices: HWP Δε = π Δε 2 = π 2 M = e iπ 2 0 0 e iπ 2 e iπ 2 1 0 0 1 Slow axis HWP, SA vertical M = eiπ 2 0 0 e iπ 2 e iπ 2 1 0 0 1 HWP, SA horizontal Department of Physics 23
Jones Details Jones Vectors and Matrices are Normalised to 1 Linear Polarisation: E = i kz ωt x cos θ + y sin θ e E = cos 2 θ + sin 2 θ = 1 cos θ sin θ Jones: E = E E = cos θ sin θ = cos 2 θ + sin 2 θ = 1 cos θ sin θ Department of Physics 24
Jones Details Jones Vectors and Matrices are Normalised to 1 Circular Polarisation: E = 1 2 x + i y ei kz ωt Jones: E = 1 2 (1 + i)(1 i) = 1 1 2 1 i E = E E = 1 1 i 2 1 i = 1 2 2 = 1 Department of Physics 25
Jones Details Jones Vectors and Matrices are Normalised to 1 Linear Polariser: 1 0 0 0 If we consider a normalised Jones vector For linear polarised light: X Y = 1 0 0 0 cos θ sin θ = cos θ 0 The maximum value is 1: Similarly with all other matrices! Department of Physics 26
Jones Details When we measure light, we measure the intensity or power. Not the amplitude: I = ԦS T = c2 ε 0 2 E 0 B 0 = cε 0 2 E 0 2 I E E Therefore Jones formalism is also normalised for Intensity Linear Polarisation: cos θ sin θ = cos 2 θ + sin 2 θ = 1 We can efficiently use the Jones formalism to calculate The absorption or transmission through an optical entity. E = i kz ωt x cos θ + y sin θ e I E E = cos θ sin θ cos θ sin θ Department of Physics 27
Jones Example - 1 Consider a linear polarizer with linearly polarized light: cos 2 θ sin θ cos θ sin θ cos θ sin 2 θ cos φ sin φ = I E E = cos 2 θ φ The amount absorbed is: 1 cos 2 θ φ = sin 2 θ φ Department of Physics 28
Jones Example - 2 Consider a linear polarizer with circularly polarized light: 1 2 cos 2 θ sin θ cos θ sin θ cos θ sin 2 θ 1 i = I E E = 1 2 cos2 θ + sin 2 θ = 1 2 Half the light is transmitted and half is absorbed! Department of Physics 29
Handling Harder Problems Deriving Jones matrices, e.g. for circular polarisation a c b d x y = 1 2 1 i for x y normalised This gives: ax + by = 1 2, cx + dy = i 2 And the solution is not provided. What should we do? 3 Steps: 1) Apply general case (normalised) 2) Apply specific case (all passing) 3) Apply specific case (all blocking) Department of Physics 30