Image Refocus in Geometrical Optical

Aaron C. W. Chan Edmund Y. Lam

Imaging Systems Laboratory, Department of Electrical and Electronic Engineering, University of Hong Kong.

http://www.eee.hku.hk/isl

2010 OSA Frontiers in Optics

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 1 / 18 Introduction Introduction

Objectives To provide a convenient mathematical framework for light-field analysis based on Hamiltonian Optics. To demonstrate some mathematical symmetries that arise — useful for computation and algorithm design. To explain the refocus process in this formulation — Ren Ng, Stanford (2006). Simply the free space operation on the light field before imaging; or equivalently, the thin-lens operation in the Fourier domain followed by a slice.

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 2 / 18 Introduction Introduction

Objectives To provide a convenient mathematical framework for light-field analysis based on Hamiltonian Optics. To demonstrate some mathematical symmetries that arise — useful for computation and algorithm design. To explain the refocus process in this formulation — Ren Ng, Stanford (2006). Simply the free space operation on the light field before imaging; or equivalently, the thin-lens operation in the Fourier domain followed by a slice.

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 2 / 18 Introduction Introduction

Objectives To provide a convenient mathematical framework for light-field analysis based on Hamiltonian Optics. To demonstrate some mathematical symmetries that arise — useful for computation and algorithm design. To explain the refocus process in this formulation — Ren Ng, Stanford (2006). Simply the free space operation on the light field before imaging; or equivalently, the thin-lens operation in the Fourier domain followed by a slice.

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 2 / 18 Introduction Presentation Outline

Theoretical Background: Hamilton Equations; Lie Operator formulation of ray equation. Analysis of imaging system in spatial and spatial frequency domains. The symmetries will be highlighted. Refocus process derived. The computational costs will briefly be discussed. Concluding remarks, merits, limitations and further work.

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 3 / 18 Introduction Presentation Outline

Theoretical Background: Hamilton Equations; Lie Operator formulation of ray equation. Analysis of imaging system in spatial and spatial frequency domains. The symmetries will be highlighted. Refocus process derived. The computational costs will briefly be discussed. Concluding remarks, merits, limitations and further work.

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 3 / 18 Introduction Presentation Outline

Theoretical Background: Hamilton Equations; Lie Operator formulation of ray equation. Analysis of imaging system in spatial and spatial frequency domains. The symmetries will be highlighted. Refocus process derived. The computational costs will briefly be discussed. Concluding remarks, merits, limitations and further work.

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 3 / 18 Introduction Presentation Outline

Theoretical Background: Hamilton Equations; Lie Operator formulation of ray equation. Analysis of imaging system in spatial and spatial frequency domains. The symmetries will be highlighted. Refocus process derived. The computational costs will briefly be discussed. Concluding remarks, merits, limitations and further work.

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 3 / 18 Theoretical Background Theoretical Background

This analysis is based on the (paraxial) approximation — light is incoherent, non-monochromatic...

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 4 / 18 Theoretical Background Coordinates Coordinates

x

z α y

dx dy Optical direction cosines: p1 = n ds and p2 = n ds . dx p1 = n ds = n sin α ≈ nα and similarly for p2. n is the . T T T q = (x, y) , p = (p1, p2) and u = (q, p) .

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 5 / 18 Theoretical Background Coordinates Coordinates

x

z α y

dx dy Optical direction cosines: p1 = n ds and p2 = n ds . dx p1 = n ds = n sin α ≈ nα and similarly for p2. n is the refractive index. T T T q = (x, y) , p = (p1, p2) and u = (q, p) .

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 5 / 18 Theoretical Background Hamilton Equations Hamilton Equations

The vectors q and p obey the Hamilton equations

Hamilton Equations dq ∂H(q,p) dp − ∂H(q,p) dz = ∂p dz = ∂q .

The Hamiltonian describes the environment in which the ray is travelling.

General Optical Hamiltonian H(n,θ)= −n cos θ

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 6 / 18 Theoretical Background Hamilton Equations Hamilton Equations

The vectors q and p obey the Hamilton equations

Hamilton Equations dq ∂H(q,p) dp − ∂H(q,p) dz = ∂p dz = ∂q .

The Hamiltonian describes the environment in which the ray is travelling.

General Optical Hamiltonian H(n,θ)= −n cos θ

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 6 / 18 Theoretical Background Hamilton Equations Hamilton Equations

The vectors q and p obey the Hamilton equations

Hamilton Equations dq ∂H(q,p) dp − ∂H(q,p) dz = ∂p dz = ∂q .

The Hamiltonian describes the environment in which the ray is travelling.

General Optical Hamiltonian H(n,θ)= −n cos θ

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 6 / 18 Theoretical Background Hamilton Equations Hamilton Equations

The vectors q and p obey the Hamilton equations

Hamilton Equations dq ∂H(q,p) dp − ∂H(q,p) dz = ∂p dz = ∂q .

The Hamiltonian describes the environment in which the ray is travelling.

General Optical Hamiltonian H(n,θ)= −n cos θ

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 6 / 18 Theoretical Background Phase Space Density Hamilton Equations for a Ray and a Bundle of Rays

{f g} ∂f ∂g − ∂f ∂g Poisson Bracket , = ∂q ∂p ∂p ∂q

Lie Operator LˆH = {·, H}.

Hamilton Equations for Hamilton Equations for Single Ray Bundle of Rays du =+Lˆ u ∂ρ −ˆ dz H ∂z = LHρ

Solution Solution

u(z)= exp[(z − zi)LˆH]u(zi) ρ(z)= exp[−(z − zi)LˆH]ρ(zi)

Goto Details on Hamiltonian

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 7 / 18 Theoretical Background Phase Space Density Hamilton Equations for a Ray and a Bundle of Rays

{f g} ∂f ∂g − ∂f ∂g Poisson Bracket , = ∂q ∂p ∂p ∂q

Lie Operator LˆH = {·, H}.

Hamilton Equations for Hamilton Equations for Single Ray Bundle of Rays du =+Lˆ u ∂ρ −ˆ dz H ∂z = LHρ

Solution Solution

u(z)= exp[(z − zi)LˆH]u(zi) ρ(z)= exp[−(z − zi)LˆH]ρ(zi)

Goto Details on Hamiltonian

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 7 / 18 Theoretical Background Phase Space Density Hamilton Equations for a Ray and a Bundle of Rays

{f g} ∂f ∂g − ∂f ∂g Poisson Bracket , = ∂q ∂p ∂p ∂q

Lie Operator LˆH = {·, H}.

Hamilton Equations for Hamilton Equations for Single Ray Bundle of Rays du =+Lˆ u ∂ρ −ˆ dz H ∂z = LHρ

Solution Solution

u(z)= exp[(z − zi)LˆH]u(zi) ρ(z)= exp[−(z − zi)LˆH]ρ(zi)

Goto Details on Hamiltonian

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 7 / 18 Theoretical Background Phase Space Density Hamilton Equations for a Ray and a Bundle of Rays

{f g} ∂f ∂g − ∂f ∂g Poisson Bracket , = ∂q ∂p ∂p ∂q

Lie Operator LˆH = {·, H}.

Hamilton Equations for Hamilton Equations for Single Ray Bundle of Rays du = +Lˆ u ∂ρ −ˆ dz H ∂z = LHρ

Solution Solution

u(z)= exp[(z − zi)LˆH]u(zi) ρ(z)= exp[−(z − zi)LˆH]ρ(zi)

Goto Details on Hamiltonian

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 7 / 18 Theoretical Background Phase Space Density Hamilton Equations for a Ray and a Bundle of Rays

{f g} ∂f ∂g − ∂f ∂g Poisson Bracket , = ∂q ∂p ∂p ∂q

Lie Operator LˆH = {·, H}.

Hamilton Equations for Hamilton Equations for Single Ray Bundle of Rays du =+Lˆ u ∂ρ −ˆ dz H ∂z = LHρ

Solution Solution

u(z)= exp[(z − zi)LˆH]u(zi) ρ(z)= exp[−(z − zi)LˆH]ρ(zi)

Goto Details on Hamiltonian

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 7 / 18 Theoretical Background Phase Space Density Hamilton Equations for a Ray and a Bundle of Rays

{f g} ∂f ∂g − ∂f ∂g Poisson Bracket , = ∂q ∂p ∂p ∂q

Lie Operator LˆH = {·, H}.

Hamilton Equations for Hamilton Equations for Single Ray Bundle of Rays du =+Lˆ u ∂ρ −ˆ dz H ∂z = LHρ

Solution Solution

u(z)= exp[(z − zi)LˆH]u(zi) ρ(z)= exp[−(z − zi)LˆH]ρ(zi)

Goto Details on Hamiltonian

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 7 / 18 Theoretical Background Phase Space Density

There are three basic solutions: the free space propagation solution, the thin lens solution, and the magnification solution. 1 2 1 2 ˆ Choosing H = 2 p1 + 2 p2 , then LH = p.∂q, one obtains

Free Space Propagation Solution

Tˆ (ξ) = exp ξp.∂q h i

Solution for Ray Bundle Solution for Ray ρ(q, p, z)= u(z)= Tˆ(z − zi)ui Tˆ[−(z − zi)]ρi(q, p).

Goto Table: Optical Lie Group

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 8 / 18 Theoretical Background Phase Space Density

There are three basic solutions: the free space propagation solution, the thin lens solution, and the magnification solution. 1 2 1 2 ˆ Choosing H = 2 p1 + 2 p2 , then LH = p.∂q, one obtains

Free Space Propagation Solution

Tˆ (ξ) = exp ξp.∂q h i

Solution for Ray Bundle Solution for Ray ρ(q, p, z)= u(z)= Tˆ(z − zi)ui Tˆ[−(z − zi)]ρi(q, p).

Goto Table: Optical Lie Group

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 8 / 18 Theoretical Background Phase Space Density

There are three basic solutions: the free space propagation solution, the thin lens solution, and the magnification solution. 1 2 1 2 ˆ Choosing H = 2 p1 + 2 p2 , then LH = p.∂q, one obtains

Free Space Propagation Solution

Tˆ (ξ) = exp ξp.∂q h i

Solution for Ray Bundle Solution for Ray ρ(q, p, z)= u(z)= Tˆ(z − zi)ui Tˆ[−(z − zi)]ρi(q, p).

Goto Table: Optical Lie Group

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 8 / 18 Theoretical Background Phase Space Density

There are three basic solutions: the free space propagation solution, the thin lens solution, and the magnification solution. 1 2 1 2 ˆ Choosing H = 2 p1 + 2 p2 , then LH = p.∂q, one obtains

Free Space Propagation Solution

Tˆ (ξ) = exp ξp.∂q h i

Solution for Ray Bundle Solution for Ray ρ(q, p, z)= u(z)= Tˆ(z − zi)ui Tˆ[−(z − zi)]ρi(q, p).

Goto Table: Optical Lie Group

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 8 / 18 Theoretical Background Phase Space Density

p p

q-shear q q

Tˆ(ξ) and performs q-shears in phase space, that is q → q + ξp or ρ (q, p) → ρ (q − ξp, p).

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 9 / 18 Theoretical Background Phase Space Density

q

z

Imaging Integral I(q)= ρ (q, p) dp Ωp R

Imaging is the integration of all the rays from all directions p impinging onto a single point with position q = (x, y)T on the image plane. The integral projection of ρ(q, p) onto the q-plane.

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 10 / 18 Theoretical Background Phase Space Density

q

z

Imaging Integral I(q)= ρ (q, p) dp Ωp R

Imaging is the integration of all the rays from all directions p impinging onto a single point with position q = (x, y)T on the image plane. The integral projection of ρ(q, p) onto the q-plane.

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 10 / 18 Theoretical Background Phase Space Density

q

z

Imaging Integral I(q)= ρ (q, p) dp Ωp R

Imaging is the integration of all the rays from all directions p impinging onto a single point with position q = (x, y)T on the image plane. The integral projection of ρ(q, p) onto the q-plane.

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 10 / 18 Analysis of Imaging System Analysis of Imaging System

f u v

object image

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 11 / 18 Analysis of Imaging System Single Lens Imaging System Single Lens Optical System

If we consider a simple biconvex lens optical system, that acts on the phase space position vector, it is evident from the thin lens 1 1 1 equation f = u + v that if one wants to refocus to a further distance v, u must be shortened.

f u v

object image

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 12 / 18 Analysis of Imaging System Refocus Refocus – In Spatial Domain

unew vnew uold vold ξ

object image

Imaging of Refocused Light Field I0(q)= Tˆ(−ξ)ρ(q, p)dp = ρ(q − ξp, p)dp. Ωp Ωp R R

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 13 / 18 Analysis of Imaging System Refocus Refocus – In Spatial Domain

unew vnew uold vold ξ

object image

Imaging of Refocused Light Field I0(q)= Tˆ(−ξ)ρ(q, p)dp = ρ(q − ξp, p)dp. Ωp Ωp R R

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 13 / 18 Analysis of Imaging System Spectral Analysis Imaging and Refocus in Spatial Frequency Domain

Definitions ∞ ∞ F 1 − q = 2π −∞ −∞ dq exp( ikq.q), similarly for p. R TR kq = (kx , ky ) as the spatial frequency of q, similarly for p.

ρˆ kq, kp = FpFqρ(q, p).   Realistically, max(q)  max(p), and the allowable angles are such that |p| < n(q) under the paraxial approximation, so it is better to define

L L F (L) 1 dp exp −ik p p = 2π −L −L ( p. ). R R

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 14 / 18 Analysis of Imaging System Spectral Analysis Imaging and Refocus in Spatial Frequency Domain

Definitions ∞ ∞ F 1 − q = 2π −∞ −∞ dq exp( ikq.q), similarly for p. R TR kq = (kx , ky ) as the spatial frequency of q, similarly for p.

ρˆ kq, kp = FpFqρ(q, p).   Realistically, max(q)  max(p), and the allowable angles are such that |p| < n(q) under the paraxial approximation, so it is better to define

L L F (L) 1 dp exp −ik p p = 2π −L −L ( p. ). R R

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 14 / 18 Analysis of Imaging System Spectral Analysis Imaging and Refocus in Spatial Frequency Domain

Definitions ∞ ∞ F 1 − q = 2π −∞ −∞ dq exp( ikq.q), similarly for p. R TR kq = (kx , ky ) as the spatial frequency of q, similarly for p.

ρˆ kq, kp = FpFqρ(q, p).   Realistically, max(q)  max(p), and the allowable angles are such that |p| < n(q) under the paraxial approximation, so it is better to define

L L F (L) 1 dp exp −ik p p = 2π −L −L ( p. ). R R

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 14 / 18 Analysis of Imaging System Spectral Analysis Imaging and Refocus in Spatial Frequency Domain

Definitions ∞ ∞ F 1 − q = 2π −∞ −∞ dq exp( ikq.q), similarly for p. R TR kq = (kx , ky ) as the spatial frequency of q, similarly for p.

ρˆ kq, kp = FpFqρ(q, p).   Realistically, max(q)  max(p), and the allowable angles are such that |p| < n(q) under the paraxial approximation, so it is better to define

L L F (L) 1 dp exp −ik p p = 2π −L −L ( p. ). R R

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 14 / 18 Analysis of Imaging System Spectral Analysis Imaging and Refocus in Spatial Frequency Domain

Definitions ∞ ∞ F 1 − q = 2π −∞ −∞ dq exp( ikq.q), similarly for p. R TR kq = (kx , ky ) as the spatial frequency of q, similarly for p.

ρˆ kq, kp = FpFqρ(q, p).   Realistically, max(q)  max(p), and the allowable angles are such that |p| < n(q) under the paraxial approximation, so it is better to define

L L F (L) 1 dp exp −ik p p = 2π −L −L ( p. ). R R

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 14 / 18 Analysis of Imaging System Spectral Analysis Imaging and Refocus in Spatial Frequency Domain

1 F F ˆ − p qT ( ξ)ρ(q, p) Free space operator. 2 = FpFq exp(−ξp.∂q)ρ(q, p) Thin lens operator Lˆ(ξ) in

3 k-space. = exp ξkq.∂kp ρˆ kq, kp     p-shear in k-space. 4 = ρˆ(kq, kp + ξkq).

sin k L sin k L (L) ˆ ( p1 ) ( p2 ) Fp FqT (−ξ)ρ(q, p)= ρˆ(kq, kp + ξkq) ∗ k k  p1 p2 

if p is defined over a finite interval.

Alternative Derivation Table of Operators

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 15 / 18 Analysis of Imaging System Spectral Analysis Imaging and Refocus in Spatial Frequency Domain

1 F F ˆ − p qT ( ξ)ρ(q, p) Free space operator. 2 = FpFq exp(−ξp.∂q)ρ(q, p) Thin lens operator Lˆ(ξ) in

3 k-space. = exp ξkq.∂kp ρˆ kq, kp     p-shear in k-space. 4 = ρˆ(kq, kp + ξkq).

sin k L sin k L (L) ˆ ( p1 ) ( p2 ) Fp FqT (−ξ)ρ(q, p)= ρˆ(kq, kp + ξkq) ∗ k k  p1 p2 

if p is defined over a finite interval.

Alternative Derivation Table of Operators

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 15 / 18 Analysis of Imaging System Spectral Analysis Imaging and Refocus in Spatial Frequency Domain

1 F F ˆ − p qT ( ξ)ρ(q, p) Free space operator. 2 = FpFq exp(−ξp.∂q)ρ(q, p) Thin lens operator Lˆ(ξ) in

3 k-space. = exp ξkq.∂kp ρˆ kq, kp     p-shear in k-space. 4 = ρˆ(kq, kp + ξkq).

sin k L sin k L (L) ˆ ( p1 ) ( p2 ) Fp FqT (−ξ)ρ(q, p)= ρˆ(kq, kp + ξkq) ∗ k k  p1 p2 

if p is defined over a finite interval.

Alternative Derivation Table of Operators

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 15 / 18 Analysis of Imaging System Spectral Analysis Imaging and Refocus in Spatial Frequency Domain

1 F F ˆ − p qT ( ξ)ρ(q, p) Free space operator. 2 = FpFq exp(−ξp.∂q)ρ(q, p) Thin lens operator Lˆ(ξ) in

3 k-space. = exp ξkq.∂kp ρˆ kq, kp     p-shear in k-space. 4 = ρˆ(kq, kp + ξkq).

sin k L sin k L (L) ˆ ( p1 ) ( p2 ) Fp FqT (−ξ)ρ(q, p)= ρˆ(kq, kp + ξkq) ∗ k k  p1 p2 

if p is defined over a finite interval.

Alternative Derivation Table of Operators

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 15 / 18 Analysis of Imaging System Spectral Analysis Imaging and Refocus in Spatial Frequency Domain

1 F F ˆ − p qT ( ξ)ρ(q, p) Free space operator. 2 = FpFq exp(−ξp.∂q)ρ(q, p) Thin lens operator Lˆ(ξ) in

3 k-space. = exp ξkq.∂kp ρˆ kq, kp     p-shear in k-space. 4 = ρˆ(kq, kp + ξkq).

sin k L sin k L (L) ˆ ( p1 ) ( p2 ) Fp FqT (−ξ)ρ(q, p)= ρˆ(kq, kp + ξkq) ∗ k k  p1 p2 

if p is defined over a finite interval.

Alternative Derivation Table of Operators

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 15 / 18 Analysis of Imaging System Spectral Analysis Imaging and Refocus in Spatial Frequency Domain

1 F F ˆ − p qT ( ξ)ρ(q, p) Free space operator. 2 = FpFq exp(−ξp.∂q)ρ(q, p) Thin lens operator Lˆ(ξ) in

3 k-space. = exp ξkq.∂kp ρˆ kq, kp     p-shear in k-space. 4 = ρˆ(kq, kp + ξkq).

sin k L sin k L (L) ˆ ( p1 ) ( p2 ) Fp FqT (−ξ)ρ(q, p)= ρˆ(kq, kp + ξkq) ∗ k k  p1 p2 

if p is defined over a finite interval.

Alternative Derivation Table of Operators

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 15 / 18 Analysis of Imaging System Spectral Analysis Imaging and Refocus in Spatial Frequency Domain

1 F F ˆ − p qT ( ξ)ρ(q, p) Free space operator. 2 = FpFq exp(−ξp.∂q)ρ(q, p) Thin lens operator Lˆ(ξ) in

3 k-space. = exp ξkq.∂kp ρˆ kq, kp     p-shear in k-space. 4 = ρˆ(kq, kp + ξkq).

sin k L sin k L (L) ˆ ( p1 ) ( p2 ) Fp FqT (−ξ)ρ(q, p)= ρˆ(kq, kp + ξkq) ∗ k k  p1 p2 

if p is defined over a finite interval.

Alternative Derivation Table of Operators

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 15 / 18 Analysis of Imaging System Spectral Analysis Imaging and Refocus in Spatial Frequency Domain

Defining the projection operator Pkp as the operation that sets kp = 0,

0 1 FqI (q)= Fq Tˆ(−ξ)ρ (q, p) dp Ωp R 2 ˆ = 2πPkp FpFqT(−ξ)ρ(q, p)

3 ˆ = 2πPkp L(ξ)ρˆ(kq, kp) 4 = 2πρˆ(kq,ξkq)⇒

Fourier Slice 0 −1 I (q)= 2πFq ρˆ(kq,ξkq).

Fourier Slice – For finite p

0 −1 sin(U1L) sin(U2L) I (q)= 2πF ρˆ(kq, U) ∗ . q U1 U2 U = ξkq.h i Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 16 / 18 Analysis of Imaging System Spectral Analysis Imaging and Refocus in Spatial Frequency Domain

Defining the projection operator Pkp as the operation that sets kp = 0,

0 1 FqI (q)= Fq Tˆ(−ξ)ρ (q, p) dp Ωp R 2 ˆ = 2πPkp FpFqT(−ξ)ρ(q, p)

3 ˆ = 2πPkp L(ξ)ρˆ(kq, kp) 4 = 2πρˆ(kq,ξkq)⇒

Fourier Slice 0 −1 I (q)= 2πFq ρˆ(kq,ξkq).

Fourier Slice – For finite p

0 −1 sin(U1L) sin(U2L) I (q)= 2πF ρˆ(kq, U) ∗ . q U1 U2 U = ξkq.h i Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 16 / 18 Analysis of Imaging System Spectral Analysis Imaging and Refocus in Spatial Frequency Domain

Defining the projection operator Pkp as the operation that sets kp = 0,

0 1 FqI (q)= Fq Tˆ(−ξ)ρ (q, p) dp Ωp R 2 ˆ = 2πPkp FpFqT(−ξ)ρ(q, p)

3 ˆ = 2πPkp L(ξ)ρˆ(kq, kp) 4 = 2πρˆ(kq,ξkq)⇒

Fourier Slice 0 −1 I (q)= 2πFq ρˆ(kq,ξkq).

Fourier Slice – For finite p

0 −1 sin(U1L) sin(U2L) I (q)= 2πF ρˆ(kq, U) ∗ . q U1 U2 U = ξkq.h i Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 16 / 18 Analysis of Imaging System Spectral Analysis Imaging and Refocus in Spatial Frequency Domain

Defining the projection operator Pkp as the operation that sets kp = 0,

0 1 FqI (q)= Fq Tˆ(−ξ)ρ (q, p) dp Ωp R 2 ˆ = 2πPkp FpFqT(−ξ)ρ(q, p)

3 ˆ = 2πPkp L(ξ)ρˆ(kq, kp) 4 = 2πρˆ(kq,ξkq)⇒

Fourier Slice 0 −1 I (q)= 2πFq ρˆ(kq,ξkq).

Fourier Slice – For finite p

0 −1 sin(U1L) sin(U2L) I (q)= 2πF ρˆ(kq, U) ∗ . q U1 U2 U = ξkq.h i Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 16 / 18 Analysis of Imaging System Spectral Analysis Imaging and Refocus in Spatial Frequency Domain

Defining the projection operator Pkp as the operation that sets kp = 0,

0 1 FqI (q)= Fq Tˆ(−ξ)ρ (q, p) dp Ωp R 2 ˆ = 2πPkp FpFqT(−ξ)ρ(q, p)

3 ˆ = 2πPkp L(ξ)ρˆ(kq, kp) 4 = 2πρˆ(kq,ξkq)⇒

Fourier Slice 0 −1 I (q)= 2πFq ρˆ(kq,ξkq).

Fourier Slice – For finite p

0 −1 sin(U1L) sin(U2L) I (q)= 2πF ρˆ(kq, U) ∗ . q U1 U2 U = ξkq.h i Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 16 / 18 Analysis of Imaging System Spectral Analysis Imaging and Refocus in Spatial Frequency Domain

Defining the projection operator Pkp as the operation that sets kp = 0,

0 1 FqI (q)= Fq Tˆ(−ξ)ρ (q, p) dp Ωp R 2 ˆ = 2πPkp FpFqT(−ξ)ρ(q, p)

3 ˆ = 2πPkp L(ξ)ρˆ(kq, kp) 4 = 2πρˆ(kq,ξkq)⇒

Fourier Slice 0 −1 I (q)= 2πFq ρˆ(kq,ξkq).

Fourier Slice – For finite p

0 −1 sin(U1L) sin(U2L) I (q)= 2πF ρˆ(kq, U) ∗ . q U1 U2 U = ξkq.h i Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 16 / 18 Analysis of Imaging System Spectral Analysis Imaging and Refocus in Spatial Frequency Domain

Defining the projection operator Pkp as the operation that sets kp = 0,

0 1 FqI (q)= Fq Tˆ(−ξ)ρ (q, p) dp Ωp R 2 ˆ = 2πPkp FpFqT(−ξ)ρ(q, p)

3 ˆ = 2πPkp L(ξ)ρˆ(kq, kp) 4 = 2πρˆ(kq,ξkq)⇒

Fourier Slice 0 −1 I (q)= 2πFq ρˆ(kq,ξkq).

Fourier Slice – For finite p

0 −1 sin(U1L) sin(U2L) I (q)= 2πF ρˆ(kq, U) ∗ . q U1 U2 U = ξkq.h i Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 16 / 18 Analysis of Imaging System Spectral Analysis Imaging and Refocus in Spatial Frequency Domain

Defining the projection operator Pkp as the operation that sets kp = 0,

0 1 FqI (q)= Fq Tˆ(−ξ)ρ (q, p) dp Ωp R 2 ˆ = 2πPkp FpFqT(−ξ)ρ(q, p)

3 ˆ = 2πPkp L(ξ)ρˆ(kq, kp) 4 = 2πρˆ(kq,ξkq)⇒

Fourier Slice 0 −1 I (q)= 2πFq ρˆ(kq,ξkq).

Fourier Slice – For finite p

0 −1 sin(U1L) sin(U2L) I (q)= 2πF ρˆ(kq, U) ∗ . q U1 U2 U = ξkq.h i Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 16 / 18 Conclusion Conclusion

Conclusion: Hamiltonian Optics and Operator methods provide a powerful way of analysing incoherent optical systems. This can be extended into k-space. There is a symmetry between the operators in physical space and k-space. This means in computation, one only needs to optimize for three or fewer operations. However, computation in 4 dimensions still expensive with the FFT. Still, it is often desirable to work in Fourier Domain in Image Processing.

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 17 / 18 Conclusion Conclusion – Further Work

Further Work Provide optimized image processing algorithms for light fields captured in 4D. Compressed sensing/sampling of phase space in ordinary cameras and reconstruction algorithms. Provide analysis to incorporate stochastic elements/noise/ray scattering (Mie Scattering and Rayleigh Scattering).

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 18 / 18 References References

1 J. N. Mait, Optics and Photonics News 17, 22 (2006). 2 J. Mait, R. Athale, and J. van der Gracht, Optics Express 11, 2093 (2003). 3 T. Mirani, D. Rajan, M. P. Christensen, S. C. Douglas, and S. L. Wood, Applied Optics 47, B86 (2006). 4 E. R. Dowski andW. T. Cathey, Applied Optics 34, 1859 (1995). 5 A. Accardi and G. Wornell, J. Opt. Soc. Am. A 26, 2055 (2009). 6 A. Torre, Linear Ray and Wave Optics in Phase Space (Elsevier, 2004). 7 K. B.Wolf, Geometric Optics on Phase Space (Springer, 2004). 8 R. Ng, ACM Transactions on Graphics pp. 19 (2005).

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 19 / 18 References References

9 M. Levoy and P. Hanrahan, in ACM SIGGRAPH (1996), pp. 3142. 10 R. Ng, M. Levoy, M. Bredif, G. Duval, M. Horowitz, and P. Hanrahan, Light field photography with a hand-held plenopic camera, Tech. rep., Stanford University (2005). 11 A. L. Rivera, S. M. Chumakov, and K. B.Wolf, J. Opt. Soc. Am. A 12, 1380 (1995). 12 K. B.Wolf, J. Opt. Soc. Am. A 8, 1389 (1991). 13 E. H. Adelson and J. R. Bergen, The Plenoptic Function and the Elements of Early Vision (1991), pp. 320. 14 E. H. Adelson and J. Y. Wang, IEEE Trans. Pattern Anal. Mach. Intell. 14, 99 (1992).

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 20 / 18 Acknowledgment Acknowledgement

Research Grants Council of the Hong Kong Special Administrative Region, China under Projects HKU 713906 and 713408.

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 21 / 18 Acknowledgment Acknowledgement

Research Grants Council of the Hong Kong Special Administrative Region, China under Projects HKU 713906 and 713408.

Group Website http://www.eee.hku.hk/isl/

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 21 / 18 Appendix Proof of Liouville’s Theorem Detour - Proof of Liouville’s Theorem

The total derivative of ρ is given by dρ ∂ρ ∂ρ dq ∂ρ dp dz = ∂z + ∂q dz + ∂p dz . Using the Hamilton Equations, this can be written as dρ ∂ρ ∂ρ ∂H − ∂ρ ∂H ∂ρ { } dz = ∂z + ∂q ∂p ∂p ∂q = ∂z + ρ, H .

The rate of change of NV(z) with respect to z of this volume is equal to the rate of flow out of the 3D surface. Consider a 4D volume in phase space V ∈ R4. The number of rays contained in this volume is N V(z)= V ρ(q, p, z)dV. R

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 22 / 18 Appendix Proof of Liouville’s Theorem Detour - Proof of Liouville’s Theorem

The total derivative of ρ is given by dρ ∂ρ ∂ρ dq ∂ρ dp dz = ∂z + ∂q dz + ∂p dz . Using the Hamilton Equations, this can be written as dρ ∂ρ ∂ρ ∂H − ∂ρ ∂H ∂ρ { } dz = ∂z + ∂q ∂p ∂p ∂q = ∂z + ρ, H .

The rate of change of NV(z) with respect to z of this volume is equal to the rate of flow out of the 3D surface. Consider a 4D volume in phase space V ∈ R4. The number of rays contained in this volume is N V(z)= V ρ(q, p, z)dV. R

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 22 / 18 Appendix Proof of Liouville’s Theorem Detour - Proof of Liouville’s Theorem

The total derivative of ρ is given by dρ ∂ρ ∂ρ dq ∂ρ dp dz = ∂z + ∂q dz + ∂p dz . Using the Hamilton Equations, this can be written as dρ ∂ρ ∂ρ ∂H − ∂ρ ∂H ∂ρ { } dz = ∂z + ∂q ∂p ∂p ∂q = ∂z + ρ, H .

The rate of change of NV(z) with respect to z of this volume is equal to the rate of flow out of the 3D surface. Consider a 4D volume in phase space V ∈ R4. The number of rays contained in this volume is N V(z)= V ρ(q, p, z)dV. R

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 22 / 18 Appendix Proof of Liouville’s Theorem Detour - Proof of Liouville’s Theorem

The total derivative of ρ is given by dρ ∂ρ ∂ρ dq ∂ρ dp dz = ∂z + ∂q dz + ∂p dz . Using the Hamilton Equations, this can be written as dρ ∂ρ ∂ρ ∂H − ∂ρ ∂H ∂ρ { } dz = ∂z + ∂q ∂p ∂p ∂q = ∂z + ρ, H .

The rate of change of NV(z) with respect to z of this volume is equal to the rate of flow out of the 3D surface. Consider a 4D volume in phase space V ∈ R4. The number of rays contained in this volume is N V(z)= V ρ(q, p, z)dV. R

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 22 / 18 Appendix Proof of Liouville’s Theorem Detour - Proof of Liouville’s Theorem 2

The rate of change of NV(z) with respect to z of this volume is equal to the rate of flow out of the 3D hyper-surface S ∂ N z − q p z du nˆdS ∂z V( )= S ρ( , , ) dz . . R

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 23 / 18 Appendix Proof of Liouville’s Theorem Detour - Proof of Liouville’s Theorem 2

The rate of change of NV(z) with respect to z of this volume is equal to the rate of flow out of the 3D hyper-surface S ∂ N z − q p z du nˆdS ∂z V( )= S ρ( , , ) dz . . R From the generalized ∂ N z − ∇ q p z du dV ∂z V( )= V . ρ( , , ) dz . R  

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 23 / 18 Appendix Proof of Liouville’s Theorem Detour - Proof of Liouville’s Theorem 2

The rate of change of NV(z) with respect to z of this volume is equal to the rate of flow out of the 3D hyper-surface S ∂ N z − q p z du nˆdS ∂z V( )= S ρ( , , ) dz . . R From the generalized divergence theorem ∂ N z − ∇ q p z du dV ∂z V( )= V . ρ( , , ) dz . R   Taking all arguments to the left hand side and moving the partial derivative to inside the integral gives ∂ρ ∇ du dV 0 V ∂z + .(ρ dz ) = . R  

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 23 / 18 Appendix Proof of Liouville’s Theorem Detour - Proof of Liouville’s Theorem 3

∇ u ∂ρ dq ∂ρ dp ∂ρ ∂H − ∂ρ ∂H { } Using the result .(ρ dz )= ∂q dz + ∂p dz = ∂q ∂p ∂p ∂q = ρ, H ,

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 24 / 18 Appendix Proof of Liouville’s Theorem Detour - Proof of Liouville’s Theorem 3

∇ u ∂ρ dq ∂ρ dp ∂ρ ∂H − ∂ρ ∂H { } Using the result .(ρ dz )= ∂q dz + ∂p dz = ∂q ∂p ∂p ∂q = ρ, H , and the fact that the is true for any V, dρ ∂ρ { } dz = ∂z + ρ, H = 0.

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 24 / 18 Appendix Details of Hamiltonian Detour - More on the Hamiltonian

Return to Presentation Explicitly, the optical Hamiltonian derived from Fermat’s principle is − 2 − 2 − 2 ≈ H (x, y, p1, p2, z) = n (x, y, z) p1 p2 q 1 2 2 − 2n(x,y,z) px + py n(x, y, z). Terms up to the second order give the paraxial approximation.

2 2 Note p(n,θ)= px + py = n sin θ and H(n,θ)= −n cos θ, where θ is the angle theq ray makes with the optic axis.

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 25 / 18 Appendix Details of Hamiltonian Detour - More on the Hamiltonian

Return to Presentation Explicitly, the optical Hamiltonian derived from Fermat’s principle is − 2 − 2 − 2 ≈ H (x, y, p1, p2, z) = n (x, y, z) p1 p2 q 1 2 2 − 2n(x,y,z) px + py n(x, y, z). Terms up to the second order give the paraxial approximation.

2 2 Note p(n,θ)= px + py = n sin θ and H(n,θ)= −n cos θ, where θ is the angle theq ray makes with the optic axis.

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 25 / 18 Appendix Details of Hamiltonian Detour - More on the Hamiltonian 2

Return to Presentation Taylor expanding the refractive index from the optic axis, and assume that there is no z variation, one obtains 2 2 2 n x y ≈ n 0 0 − ∂n x − ∂n y − 1 ∂ n x2 ∂ n y2 ∂ n xy ( , ) ( , ) ∂x ∂y 2 ∂x2 + ∂y2 + ∂x∂y .   Ignoring the first order terms, which relate to shifts and tilts of the optic axis, ∂2n| ∂2n| H x y p p ≈ 1 p2 x (0,0) x2 1 p2 y (0,0) y2 ( , , 1, 2) 2n(0,0) 1 + 2 + 2n(0,0) 2 + 2 . In general, in the paraxial approximation, we make use of the quadratic monomials q2, p2 and qp.

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 26 / 18 Appendix Details of Hamiltonian Detour - More on the Hamiltonian 2

Return to Presentation Taylor expanding the refractive index from the optic axis, and assume that there is no z variation, one obtains 2 2 2 n x y ≈ n 0 0 − ∂n x − ∂n y − 1 ∂ n x2 ∂ n y2 ∂ n xy ( , ) ( , ) ∂x ∂y 2 ∂x2 + ∂y2 + ∂x∂y .   Ignoring the first order terms, which relate to shifts and tilts of the optic axis, ∂2n| ∂2n| H x y p p ≈ 1 p2 x (0,0) x2 1 p2 y (0,0) y2 ( , , 1, 2) 2n(0,0) 1 + 2 + 2n(0,0) 2 + 2 . In general, in the paraxial approximation, we make use of the quadratic monomials q2, p2 and qp.

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 26 / 18 Appendix Details of Hamiltonian Detour - More on the Hamiltonian 2

Return to Presentation Taylor expanding the refractive index from the optic axis, and assume that there is no z variation, one obtains 2 2 2 n x y ≈ n 0 0 − ∂n x − ∂n y − 1 ∂ n x2 ∂ n y2 ∂ n xy ( , ) ( , ) ∂x ∂y 2 ∂x2 + ∂y2 + ∂x∂y .   Ignoring the first order terms, which relate to shifts and tilts of the optic axis, ∂2n| ∂2n| H x y p p ≈ 1 p2 x (0,0) x2 1 p2 y (0,0) y2 ( , , 1, 2) 2n(0,0) 1 + 2 + 2n(0,0) 2 + 2 . In general, in the paraxial approximation, we make use of the quadratic monomials q2, p2 and qp.

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 26 / 18 Appendix Solutions to Optical Hamilton Equations Solutions to Optical Hamilton Equations

Return to Presentation The solutions to these equations take a simple form. H is a linear combination of q2, p2 and qp under the paraxial approximation. If we take H = q2, H = p2 and H = qp separately, then we obtain a Sp(2, R) Lie group of solutions.

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 27 / 18 Appendix Solutions to Optical Hamilton Equations Solutions to Optical Hamilton Equations

Return to Presentation The solutions to these equations take a simple form. H is a linear combination of q2, p2 and qp under the paraxial approximation. If we take H = q2, H = p2 and H = qp separately, then we obtain a Sp(2, R) Lie group of solutions.

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 27 / 18 Appendix Solutions to Optical Hamilton Equations sp(2, R) Optical Lie Algebra

Return to Presentation Return to Spectral Analysis

Table: This table shows the sp(2, R) optical Lie algebras. They are the generators for optical operations as shown in the next slide

H LˆH K matrix 1 2 ∂ 0 1 p Lˆ− = p K− = 2 ∂q 0 0! 1 2 ˆ ∂ 0 0 q L+ = −q K+ = 2 ∂p −1 0! ∂ ∂ 1 0 qp Lˆ3 = p − q K3 = . ∂p ∂q 0 −1!

This operator is of interest.

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 28 / 18 Appendix Solutions to Optical Hamilton Equations sp(2, R) Optical Lie Algebra

Return to Presentation Return to Spectral Analysis

Table: This table shows the sp(2, R) optical Lie algebras. They are the generators for optical operations as shown in the next slide

H LˆH K matrix 1 2 ∂ 0 1 p Lˆ− = p K− = 2 ∂q 0 0! 1 2 ˆ ∂ 0 0 q L+ = −q K+ = 2 ∂p −1 0! ∂ ∂ 1 0 qp Lˆ3 = p − q K3 = . ∂p ∂q 0 −1!

This operator is of interest.

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 28 / 18 Appendix Solutions to Optical Hamilton Equations sp(2, R) Optical Lie Algebra

Return to Presentation Return to Spectral Analysis

Table: This table shows the sp(2, R) optical Lie algebras. They are the generators for optical operations as shown in the next slide

H LˆH K matrix 1 2 ∂ 0 1 p Lˆ− = p K− = 2 ∂q 0 0! 1 2 ˆ ∂ 0 0 q L+ = −q K+ = 2 ∂p −1 0! ∂ ∂ 1 0 qp Lˆ3 = p − q K3 = . ∂p ∂q 0 −1!

This operator is of interest.

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 28 / 18 Appendix Solutions to Optical Hamilton Equations Sp(2, R) Optical Lie Group

Return to Presentation Return to Spectral Analysis Return to Phase Space Discussion

Table: This table shows the Sp(2, R) Lie groups of optical operations. They are generated by the elements in the previous table through exponentiation.

exp(ζLˆH) exp(ζK) ˆ 1 ξ Free Space Tˆ(ξ)= eξL− T(ξ)= eξK− = 0 1! 1 ˆ 1 1 0 ˆ L+ K+ Thin Lens L(f)= e f L(f)= e f = 1 − f 1! s ˆ e 0 ˆ sL3 sK3 Magnifier M(s)= e M(s)= e = − 0 e s!

Anti-homomorphism. Ray Transfer Matrices.

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 29 / 18 Appendix Solutions to Optical Hamilton Equations Sp(2, R) Optical Lie Group

Return to Presentation Return to Spectral Analysis Return to Phase Space Discussion

Table: This table shows the Sp(2, R) Lie groups of optical operations. They are generated by the elements in the previous table through exponentiation.

exp(ζLˆH) exp(ζK) ˆ 1 ξ Free Space Tˆ(ξ)= eξL− T(ξ)= eξK− = 0 1! 1 ˆ 1 1 0 ˆ L+ K+ Thin Lens L(f)= e f L(f)= e f = 1 − f 1! s ˆ e 0 ˆ sL3 sK3 Magnifier M(s)= e M(s)= e = − 0 e s!

Anti-homomorphism. Ray Transfer Matrices.

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 29 / 18 Appendix Solutions to Optical Hamilton Equations Sp(2, R) Optical Lie Group

Return to Presentation Return to Spectral Analysis Return to Phase Space Discussion

Table: This table shows the Sp(2, R) Lie groups of optical operations. They are generated by the elements in the previous table through exponentiation.

exp(ζLˆH) exp(ζK) ˆ 1 ξ Free Space Tˆ(ξ)= eξL− T(ξ)= eξK− = 0 1! 1 ˆ 1 1 0 ˆ L+ K+ Thin Lens L(f)= e f L(f)= e f = 1 − f 1! s ˆ e 0 ˆ sL3 sK3 Magnifier M(s)= e M(s)= e = − 0 e s!

Anti-homomorphism. Ray Transfer Matrices.

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 29 / 18 Appendix Solutions to Optical Hamilton Equations Sp(2, R) Optical Lie Group

Return to Presentation Return to Spectral Analysis Return to Phase Space Discussion

Table: This table shows the Sp(2, R) Lie groups of optical operations. They are generated by the elements in the previous table through exponentiation.

exp(ζLˆH) exp(ζK) ˆ 1 ξ Free Space Tˆ(ξ)= eξL− T(ξ)= eξK− = 0 1! 1 ˆ 1 1 0 ˆ L+ K+ Thin Lens L(f)= e f L(f)= e f = 1 − f 1! s ˆ e 0 ˆ sL3 sK3 Magnifier M(s)= e M(s)= e = − 0 e s!

Anti-homomorphism. Ray Transfer Matrices.

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 29 / 18 Appendix Solutions to Optical Hamilton Equations Sp(2, R) Optical Lie Group

Return to Presentation Return to Spectral Analysis Return to Phase Space Discussion

Table: This table shows the Sp(2, R) Lie groups of optical operations. They are generated by the elements in the previous table through exponentiation.

exp(ζLˆH) exp(ζK) ˆ 1 ξ Free Space Tˆ(ξ)= eξL− T(ξ)= eξK− = 0 1! 1 ˆ 1 1 0 ˆ L+ K+ Thin Lens L(f)= e f L(f)= e f = 1 − f 1! s ˆ e 0 ˆ sL3 sK3 Magnifier M(s)= e M(s)= e = − 0 e s!

Anti-homomorphism. Ray Transfer Matrices.

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 29 / 18 Appendix Algebra Composition Property

exp ζLˆH ρ(q, p)= ρ exp ζLˆH q, exp ζLˆH p .        

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 30 / 18 Appendix Spectral Analysis Sheared phase space density in k-space

Return to Spectral Analysis (L) (L) Fp Fqρ(q − ξp, p)= Fp exp(−iξkq.p)Fq (ρ(q, p)) L L 1 dp exp −i k k p F q p = 2π −L −L ( ( p + ξ q). ) q (ρ( , )) R R sin(kp1 L) sin(kp2 L) = ρˆ(kq, kp + ξkq) ∗ k k  p1 p2 

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 31 / 18 Appendix Microlens Array Off-axis lens

The Operator for shifting the z-axis is: Sˆ(s)= exp s∂q . Note that Sˆ −1(s)= Sˆ (−s). Sˆ has no matrix counterpart.   The on the phase space density of an off-axis lens distance s from the optical axis is ρ˜(q, p)= Sˆ(−s)Lˆ(−f)Sˆ (s)ρ(q, p). Using the result exp(−tAˆ )B exp(−tAˆ )= exp(−t[Aˆ , ·])B, it can be shown that q p exp q−s ∂ q p q p q−s ρ˜( , )= f ∂p ρ( , )= ρ , + f .    

Chan, Lam (Univ. Hong Kong) Image Refocus Oct27,2010 32 / 18