Stochastics, Control and Robotics 9781032055855, 9781003198222

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Stochastics, Control and Robotics

Stochastics, Control and Robotics

Harish Parthasarathy Professor Electronics & Communication Engineering H. Parthasarathy Netaji Subhas Institute of Technology (NSIT) New Delhi, Delhi-110078

First published 2021 by CRC Press 2 Park Square, Milton Park, Abingdon, Oxon, OX14 4RN and by CRC Press 6000 Broken Sound Parkway NW, Suite 300, Boca Raton, FL 33487-2742 © 2021 Manakin Press Pvt. Ltd. CRC Press is an imprint of Informa UK Limited The right of Harish Parthasarathy to be identified as author of this work has been asserted by him in accordance with sections 77 and 78 of the Copyright, Designs and Patents Act 1988. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. All rights reserved. No part of this book may be reprinted or reproduced or utilised in any form or by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying and recording, or in any information storage or retrieval system, without permission in writing from the publishers. For permission to photocopy or use material electronically from this work, access www.copyright.com or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. For works that are not available on CCC please contact [email protected] Trademark notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Print edition not for sale in South Asia (India, Sri Lanka, Nepal, Bangladesh, Pakistan or Bhutan). British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library Library of Congress Cataloging-in-Publication Data A catalog record has been requested ISBN: 978-1-032-05585-5 (hbk) ISBN: 978-1-003-19822-2 (ebk)

Brief Contents

1.

Classical Robotics and Quantum Stochastics

1–128

2.

Electromagnetus and Related Partial Differential Equation

129–149

3.

Radon and Group Theoretic Transforms with Robotics Applications

151–175

4.

Stochastic Filtering and Control, Interacting Particles

177–221

5.

Classical and Quantum Robotics

223–253

6.

Large Deviations, Classical and Quantum General Relativity with GPS Application

255–338

7.

Quantum Signal Processing

339–476

Detailed Contents

1.

Classical Robotics and Quantum Stochastics

1–128

[1] A Problem in Robotics with Current Carrying Links

1

[2] Quantum Filtering

4

[3] Evan’s Hudson Flow

8

[4] Quantum Version of a Gauss-Markov Process

10

[5] Problems on the δ-Function

13

[6] Conditional Expectation in Quantum Filtering

16

[7] K.R. Parthasarathy Quantum Markov Processes

18

[8] Quantum Stochastic Lyapunov Theory

20

[9] Let P be a Solution to the Martingale Problem π(x, a, b), i.e.

27

[10] Problem from Revuz and Yor

28

[11] Gough and Kostler Paper on Quantum Filtering Some Remarks

28

[12] Quantization of Master. Slave Robot Motion Using Evans-Hudson Flows

33

[13] Problem From KRP QSC

37

[14] Moment Generating Function for Certain Quantum Random Variables

41

[15] Belavkin Filtering Contd.

42

[16] Modelling Noise in Quantum Mechanical Systems

43

[17] Belavkin has Derived a Quantum Stochastic Differential Equation (qsde) for  X (t t )

48

[18] Waveguide Modes

50

[19] Estimating the Initial State from Measurements of an Observable after the State Passes through a Sequence of Quantum Channels

56

[20] Quantum Poisson Random Variable

57

viii

Detailed Contents

[21] Quantum Relative Entropy Evolution

59

[22] Gauge Group Invariance of a Wave Eqn.

60

[23] Problem

61

[24] Quantum Oscillator Perturbed by an Electromagnetic Field

61

[25] Quantization of Mechanical Systems

64

[26] Quantum Filtering

66

[27] Perspective Projection by Camera Attached to a Robot

71

[28] Proof of The Positivity of Quantum Relative Entropy

72

[29] Entropy of a Quantum System

73

[30] Dirac Eqn. Based Temperature Estimation of Black-Body Temperature Field

76

[31] Quantum Image Processing on a C - Manifold

81

[32] Belavkin Contd. (Quantum Non-Linear Filtering)

82

[33] Remarks on Theorems from V. Kac. Infinite Dimensional Lic Algebra. P.87

83

[34] Quantum Mutual Information

84

[35] Scattering Theory Applied to Quantum Gate Design

86

[36] Quantization of the Simple Exclusion Model

93

[37] Quantum Version of Nearest Neighbour Interactions

96

∞

2.

[38] Remarks and Comments on “Collected Papers of S.R.S Varadhan” Vol. 4, Particle System and Their Large Deviations

101

[39] Quantum String Theory

103

[40] Quantum Shannon Theory Contd.

108

[41] Basics of Cq-Information Theory

111

[42] Typical Sequences and Error Probability in Classical and Quantum Coding

114

[43] Quantum Image Processing Some Basic Problems

125

Electromagnetus and Related Partial Differential Equation [1] Computation of the Perturbe Characteristics Frequencies in a Cavity Resonator with in Homogeneous Dielectric and Permittivity

129–149

129

[2] Numerical Methods for pde

133

[3] Tracking of Moving Targets Using a Camera Attached to the Tip of a 2 = Link Robot

137

[4] Problem

140

Detailed Contents

3.

4.

ix

[5] n-Dimensional Helmholtz Green’s Funtion

140

[6] Edge Diffusion

141

[7] Waves in Metamaterials

141

[8] Stroock Partial Differential Equations for Probabilities ϕ  Function, u  Sobolev Distribution

148

Radon and Group Theoretic Transforms with Robotics Applications [1] Modern Trends in Signal Processing by Harish Parthasarathy, NSIT

151–175 151

[2] Radon-Transform Based Image Processing

152

[3] Image Processing Using Invariants of the Permutation Group

155

[4] Radon Transform of Rotated and Translated Image Field

157

[5] Estimating the Rotation Applied to the 3-D Robot Links from Electromagnetic Field Pattern

158

[6] Kinetic Energy of a Two Link Robot

159

[7] Vector Potential Generated by a 2-Link Robot

164

[8] Group Associated with Robot Motion

166

[9] Radon Transform

168

[10] Campbelt Baker-Hausdorff Formula

169

[11] Problem on Rigid Body Motion

171

[12] Remarks on Theorems from V. Kac. Infinite Dimensional Lic Algebra. P.87

171

[13] Theta a Function From Group Theory P 252 (V.Kac)

172

[14] Iwasava Decomposition

175

Stochastic Filtering and Control, Interacting Particles [1] Bernoulli Filters

177–221 177

[2] A Problem in Large Deviation Theory

179

[3] Large Deviations in Image Trajectory on a Screen Taken by a Robot Camera in Motion

180

[4] P406 Revuz and Yor

184

[5] Problem from “Continuous Martingales and Brownian Motion” by Revuz and Yor

188

[6] Problem from Revuz and Yor

189

[7] Collected Papers of Varadhan-Interacting Particle System, Vol. 4. Hamiltonian System and Hydrodynamical Equations

191

x

5.

6.

Detailed Contents

[8] (Ph.D. Problem of Rohit) For Rohit and Dr. Vijayant. Skorohod Optional Stochastic Control. Optimal Stochastic Control of Master and Slave Robot

195

[9] English-Chapter-1

198

[10] Markov Processes-Some Application

199

[11] Comments and Proofs on “Particle System” Collected Papers of S.R.S. Varadhan Vol. 4

206

[12] Interacting Diffusions

209

[13] This is Independent of s. It is Reasonable to Expert that at Equilibrium

209

[14] Wong Zakai Filtering Equation

211

[15] Stochastic Control of Finance Market

216

[16] Wong-Zokai Equation

218

[17] Filtering Theory with State and Measurement Noises Having Correlation

219

Classical and Quantum Robotics [1] Mathematical Pre-Requisites for Robotics

223–253 223

[2] Belavkin Filtering Contd.

242

[3] Dirac Eqn., for Rigid Body Robot (Single Link)

243

[4] Quantum Robot Dynamics

249

[5] Hamiltonian Based Quantum Robot Tracking

252

[6] Robot Interacting with a Klein-Gordon Field

253

Large Deviations, Classical and Quantum General Relativity with GPS Application [1] Intuitive Derivation of the Gartner Ellis LDP

255–338 255

[2] LDP in Robotic Vision

256

[3] LMS Algorithm for Non-Linear Disturbance Observer in a Two-Link Robot System

260

[4] Disturbance Observer Design in Robotics

261

[5] Disturbance Observer Continuation

262

[6] Results in the LMS’ Iteration

262

[7] Notation for y(t) in the LMS Disturbance Observer

263

[8] Approximate Convergence Analysis

263

[9] Statistical Model for Disturbance Estimation Error Based on LMS

264

[10] Statistics of LMS Coefficients for the Disturbance Observer

264

Detailed Contents

7.

xi

[11] B (t ) ∈ d d-Vector Valued Standard Brownian Motion

265

[12] General Relativistic Correction to GPS

267

[13] Thiemann Modern QGTR

269

[14] Generalization of General Relativistic String Theory

273

[15] Symmetric Space-Times in General Relativity and Cosmology

276

[16] Selected Topics in General Relativity

298

[17] Perturbations in the Tetrad Caused by Metric Perturbation

300

[18] Geodesic Equation in Tetra Formalism

301

[19] Quantum General Relativistic Scattering

302

[20] Derive Dirac Equation in Curved Space-Time Using the Newman Penrose Formalism: Special-Relativity

308

[21] Proofs of Identities from S. Chandra Sekhar, “The Mathematical Theory of Blackholes”. Cartain’s Second Equation of Structure

313

[22] Mathematical Preliminaries for Cartan’s Eqn. of Structure

316

[23] Reflection and Refraction at a Metamaterial Surface

318

[24] Maxwell’s Equations in an Unhomogeneous Medium with Background Gravitation Taken into Account

321

[25] Relativistic Kinematics* Motions of a Rigid Body in a Gravitational Field a General Relativistic Calculation

323

[26] Random Fluctuation in the Gravitational Field Produced by Random Fluctuation in the Energy-Momentum Tensor of the Matter Plus Radiation Field

326

[27] Special Relativistic Plasma Physics

330

Quantum Signal Processing

339–476

[1] Quantum Scattering of a Rigid Body

339

[2] Optimal Control of a Field

339

[3] Example from Quantum Mechanics

340

[4] Example from Quantum Field Theory

341

[5] Quantum Signal Processing (Contd.)

342

[6] em Field Theory Based Image Processing

344

[7] Derivation of the Kallianpur-Striebel Formula

346

[8] Slave Dynamics with Parametric Uncertainties Using Adaptive Controllers

348

xii

Detailed Contents

[9] Belavkin’s Theory of Quantum Non-Linear Filtering

367

[10] A Problem Related to the Generator of Brownian Motion

404

[11] A Problem Related to Poisson Random Fields

405

[12] Relativistic Kinematics of Rigid Bodies

412

[13] Quantum Robotics Disturbance Observer

417

[14] Quantum Field Theory in the Presence of Stochastic Disturbance

420

[15] Simultaneous Tracking, Parametric Uncertainties Estimation and Disturbance Observer in a Single Robot

422

[16] Quantum Filtering to Signal Estimation and Image Processing

430

[17] Image Transmission Through Quantum Channels

431

[18] Entanglement Theory in Quantum Mechanics

433

[19] Some Other Remarks Related to The Above Problem

447

[20] Image Processing from the Em Field View Point

454

[21] CAR

455

[22] An Identity Concerning Positive Definite Functions on a Group

455

[23] Example of Construction of a Spherical Function

456

[24] Spherical Functions in Image Processing

458

[25] CAR from CCR (KRP Quantum Stochastic Calculus)

459

[26] Spherical Function (Helgason Vol. II)

461

[27] Clarifications of Problems Related to the I to measure of Brownian motion

463

[28] Problem from Revuz and Yor

464

[29] Group Theory and Robotics

465

[30] Suppose ∇ f(x) = 0, x ∈ D

469

[31] Entropy Evolution in Sudarshan-Lindblad Equation

472

[32] Problem on Rigid Body Motion

476

[33] Problem from Revuz and Yor

476

2

Preface This is a book about electromagnetic, filtering & control of classical and quantum stochastic process including diverse topics such as scattering theory, quantum information, large deviation, classical and quantum general relativity etc. The topics are diverse but I have made some effort to unify them. Some parts of the text also includes classical and quantum robot dynamics with some remarks about how filtering and control can be applied to such systems. Lie group theory has been applied to robotics having 3D links and robotics carrying current interacting with external electromagnetic field having also being studied in this text. Some sections on electromagnetics deal with properties of the Dirac-delta functions and how Green functions for the Laplace and Helmholtz operations can be computed. Some of the high points of the book are the Hudson Parthasarathy (HP) quantum Ito calculus, Belkavin’s theory of quantum filtering point on the HP calculus, techniques for quantizing general relativity and application of general relativity to photon propagation for GPS purposes. I am sure that the material of this book would be useful to research scientists and engineers working on robotics, quantum mechanics, stochastic processes and to physicists working in general relativity. It will also be useful for research students in their fields who are looking for a research problem to solve. Author

1 Classical Robotics and Quantum Stochastics [1] A Problem in Robotics with Current Carrying Links: To determine the configuration of the two link Robot System i.e. the 3 Euler angles specifying the configuration of the lower arm and the 3 Euler angles specifying the relative configuration of the upper arm w.r.t. the lower arm. The two arms carry current and from the far field radiation pattern, the link configuration arc to be estimated. A point r 0 in the lower arm at time t = 0 moves to r (t) = R1(t) r 0

After time t. A point S 0 in the upper arm at time t = 0 moves to R1(t)p0 + R2(t)R1(t)( S 0 – p0)

after time t. Here

r

R1(t) = R(j1(t), q1(t), y1(t)),

R2(t) = R(j2(t), q2(t), y2(t)),

Where

R(j, q, y) = Rz(j)Rx(q)Rz(y)

P

Note that R1(t), R2(t) ∈ SO (Eq. 3), i.e.

T Det Rk = 1, Rk Rk = I, k = 1, 2. Let J1 (t, r ), r ∈ B1 be the current density in the lower arm in its unperturbed

position and J1 (w, r) its termporal Fourier transform. Let J 2 (t, r ), r ∈ B2 be the current density in the upper arm in its unperturbed position and J 2 (w, r) is Fourier transform.

Here B1 is the volume of the lower arm at time t = 0 and B2 the volume of the upper arm at time t = 0. Assume t is fixed so we can write Rk for Rk(t), k = 1, 2. The current density in space at time t is then (in the frequency domain)

J (w, r ) = J1 (w, R1–1 r ) + J 2 (wp0 + (R2R1)–1( r – R1p0))

Note that B1 ∩ B2 = f and

2

Stochastics, Control & Robotics

J1 (w, R1–1 r ) is non zero only when

r ∈ R1(B1) and J 2 (w1p0 + (R2R1)–1( r – R1p0))

is non zero only when r ∈ R2R1(B2 – p0) + R1p0

We ignore w and ∧, ~ and write R = R1, S = (R2R1)–1. Then

(

)

(

)

J ( r ) = J1 R −1 r + J 2 p0 + S ( r − Rp0 ) + W ( r ) R, S need to be estimated. Here W ( r ) is the Gaussian noise field. The far field Magnetic vector potential is proportional to A ( r ) =

=

∫ J ( r ′ ) exp ( jk r.r ′ )d r ′ −1 3 ∫ J1 ( R r ′ ) exp ( jk r .r ′ )d r′ 3

(

)

+ ∫ J 2 ( p0 + S ( r ′ − Rp0 )) exp jkr .r ′ d 3 r′

=

∫ J1 ( R r ′ ) exp ( jk r .r ′ )d r′ + ∫ J 2 ( r ′ ) exp ( jk (r , Rp0 + S −1 ( r ′ − p0 ))) d 3 r′ −1

3

( ( r ) + exp ( jk (( R

)) −1 − S ) r , p0 )) F2 ( S r )

= F1 ( RT r ) + exp jk r , ( R − S −1 ) p0 F2 ( S r ) where

= F1 ( R −1

F1 (r ) = F2 (r ) =

∫ J1 ( r ′ ) exp ( jk (r , r ′ )) d r ′ , 3 ∫ J 2 ( r ′ ) exp ( jk (r , r ′ )) d r ′ . 3

The ML estimators of R, S are

( R , S )

= arg min ∫  A ( r ) − F1 ( R −1 r )

( (

))

R, S ∈ SO(Eq. 3) S2 – exp jk ( R −1 − S ) r , p0 F2 ( S r ) 2 dS (r ) Can those estimates be carried out more efficiently then by search methods using group representation theory? Suppose S is known and we wish to estimate R. Note that

( ) { (( ) )} F2 ( R1−1R2−1 r ) F1 ( R1−1 r ) + exp { jk (( R1−1 ( I − R2 −1 )r , p0 )}.F2 ( R1−1R2 −1 r )

A ( r ) = F1 R1−1 r + exp jk R1−1 − R1−1R2 −1 r , p0 =

 Let {Ylm ( r )  m |≤ l , l = 0, 1, 2, ...} be the spherical harmonic. Thus if �l(R) is the representation of SO (Eq. 3)

Classical Robotics & Quantum Stochastics

3

Vl = span{Ylm||m| ≤ l}, then

in the space

−1 ∫2 F1 ( R1 r )Ylm (r ) dS (r )

S

=

∫2 F1 (r )Ylm ( R1 r ) dS (r )

S

=

∫2 F1 (r ) ∑

|m ′|≤ l

S

=

F1 [lm ] =

where Further

∫2 exp { jk (

S

=

∑

|m ′|≤ l

)

 π  l ( R1 ) mm′ F [lm ′ ]

∫2 F1 (r )Ylm (r ) dS (r )

S

R1−1

( I − R2−1 ) r , p0 )} F2 ( R1−1R2−1 r )Ylm (r ) dS (r )

∫2 exp { jk (( R1

R2 − I r , p0

m′

S2

−1

)

S

=

(

−1   ( ) ( )  π ι R1  m′mYlm′ r dS r

)} F ( R r )Y ( R r ) dS (r ) 2

−1 1

lm

2

−1 −1   ( ) ( ) ∑ π  l ( R2 ) mm′ ∫ exp ( jk (( R1 R2 − I ) r , p0 )) F2 ( R1 r ) Yιm′ r dS r

Where r is the unit vector in the direction of r .

( (

Now the function r → exp jk r , p0 hormonics, say

( (

exp jk r , p0 Then,

{ (

)) on S2 can be expanded using spherical

)) = ∑ C (lm)Ylm (r ) l ,m

exp jk R1−1R2 r , p0

)}

=

∑ C (lm)Ylm ( R1−1R2 r ) l ,m

=

∑ C (lm)∑ πl ( R2−1R1 ) m′mYlm′ (r ) l ,m

=

m′

∑ C (l , m) πl ( R2−1R1 ) m′m Ylm′ (r )

lmm′

Also,

F2 (r ) =

∑ d (lm)Ylm (r ) lm

(CClm ) =

∫2 exp { jk (r , p0 )}Ylm (r )dS (r ) ,

S

d ( lm ) =

∫2 F2 (r )Ylm (r )dS (r )

S

4

Then,

Stochastics, Control & Robotics

∫ exp { jk (( R1

)} F2 ( R1−1 r )Ylm ( R2r ) dS (r )  ( ) ( ) = ∑ π  l ( R2 ) mm′ ∫ Ylm′ r dS r exp ( − jk ( r , p0 )) −1

)

R2 − I r , P0

m′

S2

 ∑ C (l1, m1m2 ) πl1 ( R1−1R1 ) m2m1 Yl1m2 (r ) ∑ d (l3m3 )Yl3m3 ( R1−1 r )

l1m1m2

=

l3m3

∑

l1m1m2l3m3m ′

(

)

( (

))

−1   π  l ( R2 ) mm′  πl1 R2 R1  m2 m1 π  l3 ( R1 ) m4 m3

C (l1m1m2 ) d (l3 m3 ) ∫ exp − jk r , p0 Yl1m2 ( r ) Yl3m4 ( r ) dS ( r ) S2

These formulae can be used to derive the least squares estimator for R1, R2 (which is the same as the ML estimates in the presence of white Gaussian noise. 

[2] Quantum Filtering: jt(c) = Vt* × Vt . X εL ( h )

dVt = (– iH dt + L1dAt + L2dAt* + SdΛt)Vt

where H, L1, L2, S ε L (h) are chosen so that

Vt*Vt = I ∀ t ≥ 0. (X ∆ X ⊕ I)

Then

djt(χ) = jt(L0)dt + jt(L1χ)dAt + jt(L2χ)dAt* + jt(L3χ)dΛt

L0, L1, L2, L3 are the Evan-Hudson structure map. Computation of L0, L2, L3: djt(X) = dVt*XVt + Vt*XdVt + dVt*XdVt

= Vt*(i(H*X – XH*)dt + (L2*X + XL1)dAt + (L1*X + XL2*)αAt*+(S*X + XS)dΛt

+ L2*XL1dt + S*XL2dAt* + L1XSdAt + S*XSdΛt)Vt

= jt(i(H*X – XH*) + L2*XL1)dt + jt(L2*X + XL1 + L1XS)dAt + jt(L1*X + XL2 + S*XL2)dAt* + jt(S*X + XS + S*XS)dΛt

Thus, L0(X) = i(H*X – XH*) + L2*XL1,

L1(X) = L2*X + XL1 + L1XS,

L2(X) = L1*X + XL2 + S*XL2,

L3(X) = S*X + XS + X*XS. {jt(X), t ≥ 0} in the state process. Measurement process

Classical Robotics & Quantum Stochastics

5

Yout(t) = Vt*(I ⊗ Yin(t))Vt

dYin(t) = P1(t)dAt + P1 *(t)dAt* + P2(t)dΛt

P1(t), P2(t) ∈ L(Γs(Ht)), P2*(t) = P2(t)

where

Clearly VT*(I ⊗ Yin(t))VT = Yout(t)∀T ≥ t t

in

Y (t) =

∫ ( P1 ( S ) dAs + ( P1 ( S )) dAs + ( P2 ( S )) d Λ s ) *

*

0

dYout(t) = dYin(t) + dVt*(I ⊗ dYin(t))Vt + Vt*(I ⊗ dYin(t))dVt = P1(t)dAt + P1*(t)dAt* + P2(t)dΛt + Vt*(L1 ⊗ P1*(t))Vtdt + Vt*(L1 ⊗ P2(t))VtdAt + Vt*(S* ⊗ P2(t))VtdΛt + Vt*(L2 ⊗ P1(t))Vt*dt + Vt*(L2 ⊗ P2(t))Vt*dAt* + Vt*(S ⊗ P2(t))Vt*dΛt

Yout(t) ≡ jt(Yin(t) = jt(Yin(t)) ∀ T ≥ t

Thus,

dYout(t) = jt(L1P1*(t) + L2P1(t))dt + jt(L1P2(t))dAt

and

+ jt(L2P2(t))dAt* + jt(SP2(t))dΛt

For T ≥ t,

[jT(X), Yout(t)] = [VT*(χ ⊗ I)VT, VT*(I ⊗ Yin(t))VT] = [VT*(χ ⊗ I, I ⊗ Yin(t))]VT = 0.

This is the non-demolition property. Assume that the measurement observables {Yout(t) : t ≥ 0} are commutative i.e. [Yout(t), Yout(S)] = 0 ∀ t, S ≥ 0. This is equivalent to saying that [Yin(t), Yin(S)] = 0 ∀ t, S ≥ 0 Since and

Yout(t) = Vtvs*Yin(t)Vtvs

Yout(S) = Vtvs*Yin(S)Vtvs ∀ t, S ≥ 0

where tvs = max (t, S)

{Yout(t) ; t ≥ 0} will be commutative if [P1(t), AS] = 0, [P1(t), AS*] = 0,

[P1(t), AS] = 0, [P2(t), AS] = 0,

[P2(t), AS] = 0, [P2(t), AS*] = 0,

[P1(t), P1(S)] = 0, [P1(t), P2(S)] = 0, [P2(t), P1(S)] = 0, ∀ t > S.

These condition hold if for example P1(t), P2(t) are scalar valued functions of time i.e. P1(t) = C1(t)I, P2(t) = C2(t)I,

6

Stochastics, Control & Robotics

C1(t), C2(t) ∈ ⊄, C2 (t ) = C2(t)(i.e. C2(t) ∈ ). Let ηt]in be the Von-Neumann algebra generated by {Yin(S) : S ≤ t} and ηt]out the Von-Neumann algebra generated by {Yout(S) : S ≤ t}. ηt]out = Vt*ηt]in Vt

Then,

Non-Commutative Girsanov Trick πt(χ) = [jt(χ) |ηt]out]  t [j (Χ)|η π (χ) = V * 

Let Then

t

t

t

 t [ξ] =   t (V *ξV )  t t

Where Remark:

out

t]

]Vt

 ( X ) = (U*XU)

Let

and m = U*mU where U is unitary and m is an Abelian Algebra. Then, for * ξ ∈ m , we have ξ = U ξU for some ξ ∈ m and then

(

(

)

  U * XU −  U * XU | m  ξ  = 0

 U * X ξU −  U * XU | m  ξ U * XU

i.e.

)

=0

Note that

[U*XU | m ] U*ξU = [U*XU| m ] ξ = [U*XU ξ | m ] = [U*XξU| m ]

[U*XξU| m ] = [U*XξU]  [X|m]U∈ m and [(U*XU – U*   [X|m]U) ξ ] On the other hand, U2   [X|m]ξU] = [U*χξU – U*  and so

This proves that

 [X|m]ξ)U] = [(U*Xξ –   [Xξ –   [X|m]ξ] = 0 = 

 [X|m]U [U*XU| m ] = U* 

Suppose that for some adopted process

F(t) ∈ ηt]in,we have  t [X] = [F(t)*(X ⊗ I)F(t)] 

∀ X ∈ α(h). This means that ∀ X ∈ L(h)

(Vt*XVt) = [F(t)*XF(t)]

Classical Robotics & Quantum Stochastics

Example: Suppose (ξ) = Tr((rs ⊗ re)ξ) where rs ∈ L(h) and re ∈ L (Γs(H).

(Vt*XVt) = Tr((rs ⊗ re)Vt*XVt)

Then,

= Tr1[(Tr2(Vt(rs ⊗ re)Vt*))X]

rs(t) = Tr2(Vt(rs ⊗ re)Vt*) ∈ L(h)

Now,

and it follows that, (Vt*XVt) = Tr(rs(t)X) We want to write

Tr( rs(t)X) = Tr((rs ⊗ re)(F(t)*XF(t))

i.e.

where F(t) ∈ ηt]in.

rs(t) = Tr2(F(t)rs ⊗ reF*(t)) ∀ t

If we assume that F(t) ∈ L(h) then automatically F(t) ∈ ηt]in and the above condition reduces to rs(t) = F(t)rsF*(t)

Which corresponds to involves evolution. rt(X) = Vt*(F(t)*(X ⊗ I)F(t)| ηt]in]Vt*

Let

(jt(X)| ηt]out) =

Then

σt ( X ) σt (1)

(Quantum Kalliapur Striebel). Gough and Kostle have shown that d(F*(t)XF(t)) = F*(t)(X Lt + Lt *X)F(t)dYin(t) + F*(t)( Lt *X Lt + X K t + K t X)F(t)dt For Z ε ηt]in,

[F*(t)χF(t)| ηt]in)Z] = [F*(t)XF(t)Z]

[(d(F*(t)XF(t)| ηt]in)Z] = [(d(F*(t)XF(t)|Z] [F*(t + dt)XF(t + dt)| ηt + dt]in)]

= – [(F*(t)XF(t)|ηt]in]

Where Z1, Z2 ∈ ηt]in We write Thus from*,

= d[F*(t)XF(t)|ηt]in] = Z1dt + Z2dYin(t) say

d(F*(t)X(t)) = F*(t)ψ1(t)F(t)dYin(t) + F*(t)ψ2(t)F(t)dt

[F*(t)ψ1(t)F(t)dYin(t) + F*(t)ψ2(t)F(t)dt|ηint + dt]] + [F*(t)XF(t)|ηint + dt]] – [F*(t)XF(t)|ηt]in] = Z1dt + Z2dYin(t)

7

8

Stochastics, Control & Robotics

= [F*(t)ψ1(t)F(t)| ηint + dt]]dYint + [F*(t)ψ2(t)F(t) |ηint +dt]]dt

(since F*(t)XF(t) is independent of dYin(t) and so [F*(t)XF(t)|ηint + dt]] = [F*(t)XF(t)|ηt]in]

and likewise F*(t) ψk(t)F(t) is independent of dYin(t)

and hence [F*(t)ψk(t)F(t)| ηint + dt]] = [F*(t)ψk(t)F(t)|ηt]in] Thus, d π t (X) = Z dt + Z dYin(t) 1

2

where Z1 = [F*(t)ψ2(t)F(t)|ηt]in] = πt(ψ2(t))

Z2 = [F*(t)ψ1(t)F(t)|ηt]in] = πt(ψ1(t)).

and

Thus our quantum nonlinear fitting eqn. is d π t (X) = π t (ψ1(t))dt + π t (ψ2(t))dYin(t) t + K t X Where ψ (t) = L *X L + X K 1

Where Now,

t

t

ψ2(t) = X Lt + Lt *X π t (X) = [F*(t)XF(t)|η σt(X) = Vt* π t (X)Vt.

in

t]

].

Now that π t (X) ∈ L(Γs(Ht])).

dσt(X) = dVt*d π t (X)Vt + Vt*d π t (X)dVt + Vt*d π t (X)Vt = Vt*(L1*dA* + L2*dA + S*dΛ) π t (ψ2)dYinVt + Vt* π t (ψ2)dYin(L1dA + L2dA* + SdΛ)Vt+Vt* π t (ψ1)Vtdt + Vt* π t (ψ2)VtdYin(t) = V *L * π t (ψ )V ( C (t)dt + C (t)dΛ) Thus,

t

2

2

t

1

2



[3] Evan's Hidson Flow: Classical Markov flow: jt: C0∞(n) → L2(Ω, F, P)

jt(f) = f(Xt) where {Xt: t ≥ 0} t. Then

jt(C1f1 + C2f2) = C1jt(f1) + C2jt(f2), i.e.

jt(f1f2) = jt(f1)jt(f2).

jt(1) = 1, jt(f*) = jt(f)*

jt is a * unital homomorphism. Let Then

dXt = µ ( X t ) dt + σ ( X t ) dBt

djt(f) = df(Xt) = Lf ( X t ) dt + dBtT σT ( X t ) f ′ ( X t )

Classical Robotics & Quantum Stochastics

9

(

1 T T Lf(x) = µ ( x ) ∇ x f ( x ) + Tr a ( x ) ∇∇ f ( x ) 2 T ( ) ) ( ) ( a x = σ x , σ x

where With

)

Define the structure maps θD : C0∞ (n) → C0∞ (n),

θk : C0∞ (n) → C0∞ (n)

θ0(f) = L(f),

by

θk(f) = σT(x)fَ(x))k =

∑ σ jk ( x) j

θk = (σT(x)∇x)k

Thus,

=

∂f ( x ) ∂x j ∂

∑ σ jk ( x) ∂x j

j n

djt(f) = jt (θ0 ( f )) dt + ∑ jt (θk ( f )) dBk

Then

k =1

Gauss-Markov process in the classical case are characterized by the side dX (t ) = A (t ) X (t ) dt + G (t ) dB (t )

(

1  T T T T Thus, df(X(t)) =  X (t ) A (t ) ∇ X f ( X (t )) + Tr G (t ) G (t ) ∇ X ∇ X f ( X (t ))  2

) dt

+ dB (t ) G (t ) ∇ X f ( X (t )) T

So and

T

1 T T T T θ0t(f)(x) = x A (t ) ∇ x f ( x ) + Tr (G (t )G (t ) ∇ x ∇ x f ( x )) 2 ∂f ( x ) θkt ( f ) ( x ) = ∑ G jk (t ) ∂x j j

q Now consider the quantum case. Let   be the position (vector)-momentum  p (vector) operators with classical Hamiltonian 1 1 T p Ap + qT Kq 2 2 ∂H dq = = Ap , ∂p dt dp ∂H = − = −Kq dt ∂q

H ( q, p ) = A, K > 0 . Then

10

Stochastics, Control & Robotics

Adding noise to this system (classical) gives O d q =    dt  p  − K

A  q    dt + GdB (t ) O   p 

  df ( q, p ) =   qT , pT  C T     

Then

∂  ∂q    ∂  ∂   ∂qT    ∂p 

∂  ∂q    f q, p + 1 T (GGT ) 2 r ∂ (    ∂p  ∂  f ∂pT 

   q , p ( ) dt + dBT GT   

 ∂f   ∂q     ∂f  .    ∂p 

= Λ or for f = f ( q , p ) , 2n  n ∂f θ dt + j f ) ( ( ) djt(f) = t 0 ∑  ∑ G jk ∂q + j k =1  j=1

2n

∑

j = n+1

G jk

∂f  . dBk ∂p1 

2n

= j (θ0 ( f )) dt + ∑ jt (θk ( f )) dBk k =1

where θ0 ( f ) ( q, p )

 ∂  ∂q   1 =  qT , pT  C T   f ( q , p ) + Tr  GGT ∂ 2       ∂p  and

θ k ( f ) ( q, p ) =

n



∂f

∑  G jk ∂q j=1 

j

+ G j + nk

∂  ∂q    ∂  ∂   ∂qT    ∂p 

∂  f ∂pT 

∂f  ∂p j 



[4] Quantum Version of a Gauss-Markov Process: d

(

(

)

djt(X) = jt (θ0 ( X )) dt + ∑ jt (θk ( X )) dAk (t ) + jt θk ( X * ) dAk* (t ) k =1

where X ∈ (h) Ak(t), A*k(t)

θk : (h) → (h)

∈ L(Γs(H)) is linear 0 ≤ k ≤ d.

)

  ( q, p )  

Classical Robotics & Quantum Stochastics

11

The structure maps θ0 and θk must be chosen to ensure Gaussianity of {jt(X) : 0 ≤ t ≤ T} in a given state, [j] i.e. if {Lt : 0 ≤ t ≤ T} is any scalar function, then T  < ϕ | exp  ∫ α t ( X ) dt  | ϕ >  0 Must be a quadratic functional of {αt}0 ≤ t ≤ T djt(X) = jt(θ0(X))dt + jt(θ1(X))dAt + jt(θ2(X))dA*t

Let

eλdjt(X) = I + λdjt ( X ) +

Then

λ2 (djt ( X ))2 + 0 (dt ) 2

= I + λjt (θ0 ( X )) dt + λjt (θ1 ( X ) ) dA + λjt (θ2 ( X ) ) dA* + < fe(u), eλdjt

(X )

λ2 jt (θ1 ( X )) jt (θ2 ( X )) dt + 0 ( dt ) 2

fe(u) > = ||f||2 exp (||u||2) + dt{λ < fe(u), jt(θ0(X))fe(u) > λ2 < fe(u), jt(θ1(X))jt(θ2(X))fe(u) > 2 + λu(t) < fe(u), jt(θ1(X))fe(u) >

+

+ λ u (t) < fe(u), jt(θ2(X))fe(u) >} eλjt

(X )

eλdjt

(X )

= e(λjt + dt(X) + α1λ2[jt(X), djt(X)] + α2λ3[[jt(X), djt(X)], djt(X)]+ α3λ3[jt(X), [jt(X), djt(X)]

We require

+ ...]

[[jt(X), djt(X)], jt(X)] = 0,

[[jt(X), djt(X)], djt(X)] = 0

...(1)

upto O(dt). Then Gaussianity of jt(X) will imply that the jt + dt(X) Moment generating function of is e4th degree polynomial in λ.

[jt(X), djt(X)] = [jt(X), jt(θ0(X))]dt + [jt(X), jt(θ1(X))]dAt + [jt(X), jt(θ2(X))]dAt*

= jt([X, θ0(X)])dt + jt([X, θ1(X)]dAt + jt([X, θ2(X)])dAt*

So that condition (1) hold if [[X, θ0(X)], X] = 0, [[X, θ0(X)], X] = 0,

[[X, θm(X)], θk(X)] = 0, k = 0, 1 m = 0, 1.

∀ X ∈ (h)

djt(X) = jt(θ0(X))dt + jt(θ1(X))dAt + jt(θ2(X))dAt*

12

Stochastics, Control & Robotics

jt(XY) = jt(X)jt(Y)

djt(XY) = djt(X).jt(Y) + jt(X)djt(Y) + djt(X).djt(Y)

θ0(XY) = θ0(X)Y + Xθ0(Y) + θ1(X)θ2(Y), (Coeff. of dt)

⇒

θ1(XY) = θ1(X)Y + Xθ1(Y) (Coeff of dAt) θ2(XY) = θ2(X)Y + Xθ2(Y)

θ1 and θ2 are thus derivations of (h). So, θ1 = ad Z1 for some Z1 ∈ (h)

θ2 = ad Z2 for some Z2 ∈ (h). [[X, [Zk, X]], X] = 0, k = 1, 2.

If [[X, θk(X)], X] = 0

and [[X, θk(X)], θn(X)] = 0

k = 0, 1 then [jt(X), djt(X)] commutes with both jt(X) and djt(X) and hence the above Baker-Campbell Hausd orff formula implies eλ1 jt

( X )+ λ 2 djt ( X )

λ j = e 1t

( X ) λ 2 djt ( X ) − α1λ1λ 2 [ jt , djt ( X )]

e

e

Let {ek} be an ONB for h ⊗ Γs(H) Then < ek, eλ1 jt ( X )+ λ 2 djt ( X ) em > =

j ( X )djt ( X )

∑ < ek , eλ1 jt ( X ) er >< er , eλ2djt ( X ) es >< es , e−α1λ1λ2  t

em >

r

[jt(X), djt(X)] = jt([X, θ0(X)]dt + jt([X, θ1(X)])dAt + jt([X, θ2(X)]]dAt* < fe (u ) , e

λ jt ( X ),djt ( X ) ge(v) >

= < fe(u), ge(v) > + dt{< fe(u), {jt([X, θ0(X)]) λ2 + j (X, θ1(X)].[X, θ2(X)])}ge(v) > 2 t + λ < fe(u), jt([X, θ, (X)]) ge(v) > v(t) + λ < fe(u), jt([X, θ2(X)]ge(v) > u (t ) For joint Gaussainty of jt(X) and djt(X), we thus require [X, θ1(X)].[X, θ2(X)] = 0. λj ψ(t, λ) = < fe(u ), e t

Let

(X )

ψ(t + dt, λ) = < fDe(u ), e λ ( jt = < fe(u ), eλjt

ge(v) > ( X )+djt ( X )

( X ) λdjt ( X ) −αλ 2 [ jt ( X ), djt ( X )]

e

Instead we define ψkm(t, λ) = < ek , eλjt

(X )

ge(v) >

em >

e

ge(v) >

Classical Robotics & Quantum Stochastics

13

where {ek }k =1 is an onb for h ⊗ Γs(H). ∞

Then dψkm(t, λ) + ψkm(t, λ) ∞

( )

∑ < ek , eλjt ( X ) er >< er , eλdjt X es >< es , e−αλ

=

r , s =1 ∞

( )

∑ ψ kr (t , λ ) < er , eλdjt X es >< es , e−αλ

2

r , s =1

2

 jt ( X ), djt ( X ) em

 jt ( X ), djt ( X ) e m

>

>

< er , eλdjt (χ)es >

{

= δ rs + λ < er , jt (θ0 ( X )) es > dt + ∑ < er , jt (θ1 ( X )) e p > a ( p, s ) dt p

∑ a ( p, r ) < e p , jt (θ2 ( X )) es > dt  

p

+

where

λ2 < er , jt (θ1 ( X )) (θ2 ( X )) es > dt + 0 ( dt ) 2

a(p, s)dt = < ep, dAt es > 

[5] Problems on the δ-Function: [1] X denotes the space of rapidly decreasing function on n, i.e. n

n

χ = C∞(n) and f ∈ χ ⇒ ∏ | x j |mj j =1

As ||c|| → ∞ for all mj, m′j ≥ 0 (non-negative integers).

If F denotes the Fourier transform on X, then show that F(X) = X.  n mj n mj ′  Hint: F ∏ X j ∏ D j f  = j=1  j=1 

n n  mj mj D ∑ ∏ j ∏ χmjj ′ f ( x) i

1

1

( f = F ( f ))  mj mj ′ Now, ΠD j Πχ j f ( x ) ≤

Cp || χ ||2 p

For all sufficiently large p and sufficiently large ||X||.

∑ m′j f ( x )

∂1

∂X1 1 ...∂X nm′n m′

→0

14

Stochastics, Control & Robotics

Provided that we assume that f is rapidly decreasing. Then  m ′j mj m ′j ( ) n ) ∏ χ mj j ∏ D j f ( x ≤ C ′ ∫ ∏ D j ∏ Xj n f X d X d X

≤ C′′

∫

n

d χ

+ C ( R ) < ∞

2p ||X ||> R || χ ||

For sufficiently large R. Thus, if f is rapidly decreasing, then sup m ′j ( ) ∏ X mj j ∏ D j f X < ∞∀m j , m ′j ≥ 0. n X ∈ If then readily follows by replacing mj by mj + 1 ∀j that m ′j ( ) ∏ X mj j ∏ Dj f X ≤

K

for all sufficiently large |Xj|

n

∏ Xj j=1

Proving the claim. [2] Let (Dn + a1Dn – 1 + ... + an – 1 D + an) f(X) = δ(X), X ∈  Solve for f without using Laplace transforms.

[3] Let p(ξ1, ..., ξn) be a polynomial in n variables. Solve

p(D1, ..., Dn) f(X) = δ(X). 1 Hint: f ( ξ ) = p ξ ( )

f(X) =

Hence

1

 f (ξ )

( 2π )n ∫n p ( ξ )

exp (i < ξ , X > ) d n ξ

 the set of its irreducible (unitary) [2] Let G be a compact group and G representation. By the Peter-Weyl theorem. f(g) =  f (X ) =

where

∑ dπTr ( f (π ) π ( g ))

π∈G

∫ f ( x)π

*(

x ) dx

G

δ g ( x) =

Thus,

∑ d π X π ( x −1g ) = ∑ dπ χπ ( g −1x) 

π∈G

is the Dirac δ-function on G where

∫ f ( x) δ g ( x) dx

G

= f(g) defines δg(x).

π∈G

Classical Robotics & Quantum Stochastics

15

[3] Let (∇2 + k 2 ) G ( r | r ′ ) = δ ( r − r ′ ) r , r ′ ∈v G ( r | r ′ ) = 0, r ∈ ∂v

Assume (∇2 + k 2 ) ψ ( r ) = S ( r ) , r ∈ ∂v, ψ ( r ) = ψ 0 ( r ) , r ∈ ∂v Then, by Green 's theorem,

∫ {ψ ( r ) (∇

2

}

+ k 2 ) G ( r | r ′ ) − G ( r | r ′ ) (∇ 2 + k 2 ) ψ ( r ) d 3r

v

=



∫ ψ ( r ) S

or equivalently,

∫ {ψ ( r ) δ ( r − r ′ ) − G ( r | r ′ ) S ( r )} d v

∂G ( r | r ′ ) ∂ψ ( r )  − G ( r | r ′ )   dS ( r )  ∂n ∂n 

3

r = ∫ ψ0 (r ) v

∂G ( r | r ′ ) dS ( r )  ∂n

3 ψ ( r ′) = ∫ G ( r − r ′) S ( r ) d r + ∫ ψ 0 ( r )

i.e.

v

S

∂G ( r | r ′ ) dS ( r )  ∂n

[4] Generalization to n: n

2

∇ = 2

∂2

∑ ∂X 2

α=1

α

2

j∇ y – y∇ j = div(j∇y – y∇j) Thus it (∇2 + k 2 ) G ( X | X ′ ) = δ ( X | X ′ ) , X , X ′ ∈ v ⊂  n G ( X | X ′ ) = 0, X ∈ ∂v then consider

(∇ 2 + k 2 ) ψ ( X )

= S ( X ) , X ∈v

ψ ( X ) = ψ 0 ( X ) , X ∈∂v , Then the same formula is valid: ψ ( X ′) = Problem: Solve Let

∫ G ( X | X ′ )S ( X ′ ) d v

n

X ′ + ∫ ψ0 ( X ) S

∂G ( X | X ′ ) dS ( χ )  ∂n

(∇2 + k 2 ) G (r ) = δ ( r ) , r ∈ n. d Ω ( r ) be the n – 1 dimensional solid angle measure in n. Then 1/ 2

n

d r =r

 n dr d Ω ( r ) r =  ∑ X α2    1

n–1

16

Stochastics, Control & Robotics

n

∇G ( r ) =

Xα  eα G ′ ( r ) α=1 r

∇2G(r) =

∑ ∂X

∑ n

∂  Xα  G ′ ( r )    r α

α=1

 1 X 2   X α2 n − 1 α ( ) G r + G ′′ ( r ) =  − ′ = ∑   G ′ ( r ) + G ′′ ( r )  3 2      r r  r r α=1 n

Let S

∫n−1 d Ω (r ) ∫n−1 d

Then

n

= C(n).

r = C(n)rn – 1 dr (radial volume measure).

r ∈S

Thus ∫ δ ( r ) − f ( r ) r n−1drd Ω ( r ) = ⇒ S

∫ δ ( r ) d Ω (r ) n −1

=

1

r

n −1

∫ f (r )δ ( r ) d

n

r = f(0)

δ (r )

So the green's function G(r) satisfies δ (r )  n −1 ( ) 2  G ′ r + G ′′ ( r ) + k G (r ) = ( ) n −1  C n r r

...(1)

To solve this, we assume its validity for r ∈  i.e. also admilt –ve values of r. For r > 0, rG′′ ( r ) + ( n − 1) G ′ ( r ) + k 2 rG (r ) = 0 Let r = λξ. Then So λ −1ξ

d2 dξ

2

1 d 1 d2 d d2 = , = 2 2. λ d ξ dr 2 dr λ dξ

G ( λξ ) +

( n −1) dG (λξ ) λ

dξ

+ λkξ2G ( λξ ) = 0, ξ > 0.

Let λ = n – 1. Then  1  d d2 ξ 2+ + k 2 ( n − 1) ξ  G ( λξ ) = 0, ξ > 0. ( dξ  n − 1) d ξ  Solve it by power series. 

[6] Conditional Expectation in Quantum Filtering: Let m be an algebra of observables. Assume that m is Abelian. Let m ⊂ ∞. Where ∞ is another algebra (non-commutative in general). Consider the commutant m′ of m:

Classical Robotics & Quantum Stochastics

17

m′ = {X ∈ ∞ |[X, Y] = 0 ∀ Y ∈ m}. Then m ⊂ m′. Let r be a state on ∞. For example, we may take ∞ as the algebra of linear operators in a Hilbert space H and r is a +ve define operator in H having unit trace. For X ∈ ∞,

(X) = Tr(pX).

Define [.|m] : m′ → m

so that (1) [.|m] is linear, (2)

[[X |m]Y] = [XY]



∀ X ∈ m′, Y ∈ m. Suppose for example, m = {Xα|α ∈ I} and let µY be a measure on I, for Y ∈ m′.

For Y ∈ m′, we let [Y|m] =

∫ X α d µY ( α ) I

Then   ∫ X α dµY (α ) Z  = ∫  ( X α Z ) dµY (α ) = [YZ] ∀ Z ∈ m    I   I Take Z = Xβ and get

∫  ( X α X β ) dµY (α )

= [YXβ], β ∈ I.

Solving this equation gives µY. We can attempt to generalize this as follows. Suppose we can find a set { Ak }k =1 in m′ such that every X ∈ m′ can be expressed as ∞

X=

∞

∑ Ak X k

for some Xk ∈ m, k = 1, 2, ...

k =1

Then, for Y ∈ m

(XY) = (X|m) =

and so

{(X|m)Y} =

∑  ( Ak X k Y ) k

∑  ( Ak | m) X k k

∑  ( ( Ak | m) X k Y ) k

In particular taking Xk = X0δk, k0 for some X0 ∈ m gives

(

 Ak0 X 0Y

( (

=   Ak0 | m X 0Y ∀ ∈ m

)

=  ( ( Ak | m ) Y ) ∀Y ∈m .

or equivalently since m is an algebra,

(

 Ak0 Y

)

)

)

18

Stochastics, Control & Robotics

This condition holding ∀ k0 ≥ 1 is a n. s. Condition that defines (.|m).

Let F ∈ m′ and define assuming (F*F) = 1,

F(X) = (F*XF) = (F*FX).



F(F(X|m)Y) = F(XY), Y ∈ m.

Then

Gives (F*F F(X|m)Y) = (F*FXY), Y ∈ m, X ∈ m′,

⇒((F*F|m)F(X|m|Y) = (F*FX|m)Y)Y ∈ m, X ∈ m′ ⇒

[F*F|m]F(X|m) = (F*FX|m)

* F(X|m) =  ( F FX | m ) .

or

(

 F *F | m

)



[7] Quantum Markov Processes: [Reference: K.R. Parthasarathy Quantum Markov Processes.] Quantum probability and Strong quantum Markov processes. Consider first a classical Markov processes {Xt : t ≥ 0}. Let {FtX}t ≥ 0 be the underlying filtration, i.e. Xt is FtX measurably. FtX = σ{XS : S ≤ t}.

Let 0 < t1 < t2 < ... < tN and let Yk ∈ Ft X , 1 ≤ k ≤ N, i.e. Yk is a functional of k {XS : S ≤ tk}. We write λ(tN, tN – 1, ..., t1, XN, YN – 1, ..., Y1, u) for YNYN – 1. . Y1u where u is F0X -measurable.

Then for ti < t < ti + 1 for some i = 1, 2, ..., N – 1 we have if Ft denotes (.|Ft), and Y ∈ Ft, FtYλ(tN, tN – 1, ...,t1, YN, YN – 1, ..., Y1, u) = [YYN YN – 1 ... Y1u|Ft]

( ( ( ( =  T (T (...T 

)

)

)

)

=   .  YN | Ft YN −1 | FN − 2 ...Yi +1 | Ft Y | Fi YiYi −1...Y1u N −1

(Tt t (YN )YN −1 )YN −2 )Y )YiYi−11...Y1u  = F0 λ (t , ti , ..., t1 (Tt ,t (Tt ,t (...(Tt ,t (Tt ,t (YN )YN −1 )YN −2 )...)Y ) , Yi ,Yi−1,..., Y1u ) t ,ti

ti +1,t

t N −1,t N − 2

i

i +1

N N −1

We define the operator

jt (Y ) = F (Y.) t

i.e.

jt (Y ) (Z) = [YZ|F ] t

N , N −1

N −1 N −2

Classical Robotics & Quantum Stochastics

19

Thus we have jt (Y ) λ(t , t N N – 1, ..., t1, XN, XN – 1, ..., X1, u) = λ(t, ti, ..., t1, Yt, Yi, Yi – 1 ..., Y1, u) Yt = [YNYN – 1 ... Yi + 1Y|Fi](ti < t < ti + 1)

where

(

(

(

))

)

= Tt ,ti Tti +1 ,t ...Tt N − 2 ,t N −1 Tt N ,t N −1 (YN ) YN −1 Y ...t ∈£t . Also for t < ti < ti + 1, Y ∈ Ft, Define jt(Y) by its action on λ(tN, ..., t1, YN, ..., Y1, u) where Yk ∈ Ftk , by

jt(Y)λ(tN, ..., t1, YN, ..., Y1, u)

= λ(tN, ..., ti + 1, t, ti, ..., t1, YN, ..., Yi + 1, Y, Yi – 1, ..., Y1, u)

Then

jt(cY + Z) = cjt(Y) + jt(Z) jt(YZ) = jt(Y) + jt(Z)

and

where jt(Y)λ(tN, ..., ti + 1, t, ti, ...,t1, YN, ..., Yi + 1, Z, Yi – 1, ..., Y1, u)

= λ(tN, ..., ti + 1, t, ti, ..., t1, YN, ..., Yi + 1, YZ, Yi – 1, Y1, u)

Y, Z ∈ Ft, c ∈ ⊄.

Thus jt is a unital homomorphism from

L2(Ω, Ft, P) into (£(L2(Ω, L, P))

This method of introducing the Markov property for classical stochastic processes can be generalized to quantum stochastic processes. First note that in the classical case, {λ(tN, ..., t1, YN, ..., Y1,)u λ(tN, ... , t1, ZN, ..., Z1,)u}

= F0{λ(tN, ..., t1, YN, ..., Y1,) λ(tN, ..., t1, ZN, ...., Z1)} = {YN...Y, ZN...Z1}(where Yk, Zk ∈ Ftk , 1 ≤ k ≤ N,

{YNZNYN – 1ZN – 1 ... Y2Z2.uvY1Z1}tN > tN – 1 > ... > t1 > a u, v ∈ F0) = {((...((YNZN| Ft N −1 )YN – 1ZN – 1| Ft N −2 )| Ft1 )Y1Z1|F0)uv}

(

(

))

(

)

= F0  λ Tt2 ,t1 ... Tt N −1 ,t N − 2 Tt N ,t N −1 (YN Z N )YN −1Z N −1 ... Y1Z1, uv    In the non-commutative case, we cannot couple (YN, ZN), ..., (Y1, Z1) etc. We thus get in the non-commutative case, < λ(tN, ..., t1, YN, ..., Y1, u), λ(tN, ..., t1, ZN, ..., Z1, v) >

(

(

(

= Tu ,t1 ,0 Y1...Tt N − 2 ,t N −3 YN − 2Tt N −1 ,t N − 2 YN −1Tt N ,t N −1

(YN Z N | Z N −1 ) Z N − 2 ) Z1 ) v

20

Stochastics, Control & Robotics

Here Tt, s t ≥ s are stochastic operators Tt, s (1) = 1,

Tt2 ,t1 .Tt3 ,t2 = Tt3 ,t1 , t3 ≥ t2 ≥ t1 and finally Tt, s is completely positive, i.e.

if Xα, α = 1, 2, ..., N are operators in the algebra ∞ on which Tt, s its, then N

(( (

⊕ uα , Tt , s X α* X β

α=1

))) | α⊕=1uα N

≥0

Here ∞ =  (H) and uα, ∈ H. Equivalently, N

∑

α ,β =1

uα , Tt , s ( X α X β ) vβ ≥ 0.

This property is satisfied in the classical case: Tt, s (X) = (X|Fs), X ∈ L2(Ft)

Tt, s : L2(Ft) → L2(Fs).

Then, if Xα ∈ L2(Ft)1 ≤ α ≤ N,

(

Tt , s X α X β

)

=   X α X β | Fs 

 N  and ∑ uα uβ   X α X β | Fs  =  ∑ uα X α  α ,β =1  α=1 N

2

  Fs ≥ 0.  

for all uα ∈ ⊄, 1 ≤ α ≤ N. 

[8] Quantum Stochastic Lyapunov Theory: p   p (1) ( 2) * j k Let dX (t ) =  ∑ L j dA j (t ) + L j dA j (t ) + ∑ Sk d Λ j (t ) X (t ) + MX (t ) dt   j=1 k , j =1

()

( )

j where M , L 1j , L j2 , Sk ∈ L ( h ) (system space operators) and Aj(t), Aj*(t), Λjk(t) ∈

L(Γs(H)) where

H = ⊄p ⊗ L2(+)

{|ej > : 1 ≤ j ≤ p} is an ONB for ⊄p and

Aj(t) = a(|ej > ⊗ χ[0, t]),

Λjk(t) = λ(|ej > < ek| ⊗ χ[0, t])

Thus,

dΛjkdΛsr = δskdΛjk,

dAj(t)dAk*(t) = δjkdt

dAj(t)dΛmk(t) = δjmdAk(t)

Classical Robotics & Quantum Stochastics

21

dΛmk(t)dAj*(t) = δkjdAm*(t) Let V ∈ L(h), V > 0. Consider fv(t) = Tr(X*(t)VX(t)) ≡ Tr1(VTr2(X(t)X*(t))) v≥ 0 and is = 0 iff Tr2(X(t)X*(t)) = 0 iff X(t) = 0.

fv(t) is a non-negative real valued function of time. Now for asymptotic stability we required that fv′(t) < 0 ∀ t. But d(X*VX) =dX*.V.X + X*.V.dX + dX*.V.dX

u ≡ ((uj(t))j = 1p, t ≥ 0)

Now, < fe(u), dX*.VX.ge(v) > fe (u ) , X *

=

(

 ( )* ( ) L 1 dA*j + L j2 dA j ∑ j  j

)

)

*

+ ∑ Skj d Λ kj VXge (v ) + fe(u ), X * M *VXge (u ) dt j ,k

)

*

 ( ) ( ) * (1) ( ) = ∑ u j t fe u , X L j VXge v j  ( )*

+ ∑ v j (t ) fe (u ) , X * L j2 VXge (v ) j

*

+ ∑ uk (t ) v j (t ) fe (u ) , X * Skj VXge (v ) j ,k

}

+ fe (u ) , X * M *VXge (v ) dt *

...(1)

Secondly, < fe(u), X VdXge(v) > = Conjugate of (1) with fe(u) and ge(v) interchanged * (1)  () ( ) = ∑ v j t fe(u ), X VL j Xge v j 

+ ∑ u j (t ) fe(u ), X *VL j2 Xge (v ) ( )

j

+ ∑ u j (t )vk (t ) fe(u ), X *VSkj Xge (v ) j ,k

}

+ fe(u ), X *VMXge (v ) dt *

*

(X = X(t), X = X (t))

...(2)

22

Stochastics, Control & Robotics

Finally, < fe(u), dX*V dX ge(v) > =

)

(

() ( ) fe (u ) , X *  ∑ L 1j dA j + L j2 dA*j + ∑ Skj d Λ kj  V .   j j ,k

∑ ( L 1j dA j + L j2 dA* ) + ∑ Skj d Λ kj )Xge (u ) ()

( )

j

j

=

∑

( )*

j ,k

 

( )

fe (u ) , X * L j2 VL j2 Xge (v ) dt

j

*

+ ∑ fe (u ) , X * Skj VSnm Xge (v ) δ nj vm (t ) uk (t )dt jkmn

( )*

+ ∑ fe (u ) , X * Lk2 VSnm Xge (v ) δ jn vm (t ) dt jmn

*

+ ∑ fe (u ) , X * Skj VLm2 Xge (v ) δ jm uk (t )dt ( )

jkm

=

∑

( )*

( )

fe (u ) , X * L j2 VL j2 Xge (v )

j

*

+ ∑ fe (u ) , X * Skj VS mj Xge (v ) vm (t ) uk (t ) jkm

( )*

+ ∑ fe (u ) , X * L j2 VS mj Xge (v ) vm (t ) km

}

+ ∑ fe (u ) , X * Skm VLm2 Xge (v ) uk (t ) dt *

( )

km

Using these formulas, find conditions for which d Tr  pX * (t )VX (t ) ≤ 0∀t dt N

where

r=

∑

α ,β =1 N

where

Tr(r) =

∑

α ,β =1

f α e (uα ) aαβ fβ e (uβ )

(

aαβ fβ f α exp uβ , uα

)

Classical Robotics & Quantum Stochastics

and

23

((aαβ ))α,β=1 ≥ 0 , i.e. r is a state (density matrix) in h⊗Γs(H) N

Combining all these gives d fe (u ) , X *VXge (v ) = dt Stochastic Lyapunov Theorem. Consider the side  A B ( X )  X  X d  =  dt + GdB (t ) Y  C ( X ) D ( X )  Y  A is a constant square matrix. B ( X ) is a rectangular matrix dependent on X but not on Y and so are C ( X ) and D ( X ) . B is known and C ( X ) and D ( X ) are to be determined by the condition that d  T T  X ( X Y ) Q    ≤ 0  Y  dt  When G = 0 (no noise) and Q is a given +ve definite matrix. We have (assuming G = 0)  X   X d ( X T ,Y T )Q   = 2 X T ,Y T Q   Y dt  Y 

(

)

(

)

 A B  X  = 2 X TYT Q   C D  Y 

(

T

= X Y

T

)

Q12  Q Let Q =  11  Q21 Q22  We required

  A B   AT Q  +   C D  BT 

CT    X   Q   DT    Y 

dV ≤0 dt

(

)

 X where V ( X ,Y ) = X T , Y T Q   . Y T  A B  A + and hence we require Q    C D  BT

i.e.

CT  Q≤0 DT 

 Q11 A + Q12C Q11B + Q12 D  T + (11) ≤ 0 =    Q21 A + Q22C Q21B + Q22 D

Suppose we take Q12 = 0 (and hence Q12 = Q12T = 0). Then

24

Stochastics, Control & Robotics T  Q11 A Q11B   A Q11 dV + =    Q22C Q22 D  BT Q11 dt

C T Q22   DT Q22 

 Q11 A + AT Q11 Q11B + C T Q22  =   ≤0  Q22C + BT Q11 Q22 D + DT Q22  This can be satisfied by taking Q11 so that Q11 > 0, Q11A + ATQ11 < 0, −1 C(X) = −Q22 B( X )T Q11 , −1 −1 1/ 2 R Q22 R (or D(X) = −Q22 D(X) = −Q22

)

where R < 0 is any matrix (even X dependent) −1 R (R < 0 constant) This D(X) = −Q22

we get dV = dt

 Q11 A + AT Q11 0    ∆

Then,

lim d  (Vt ) t → ∞ dt

 Q A + AT Q  11 = Tr GG Q + Tr ∑ Tr  11 0   k , j =1

(

T

)

27

r

T 0  Pk GG Pj    −2R λ k + λ j  

(

)

This must be made ≤ 0 by an appropriate choice of Q11, Q22, R > 0. 

[9] Let P be a Solution to the Martingale Problem π(x, a, b), i.e.: t

f ( X t ) − ∫ αf ( X s ) ds, t ≥ 0 is a P-Martingale where 0

∂f ( x ) 1  ∂2 f ( x) + Tr  a ( x )  ∂x 2  ∂x∂xT 

αf (x) = b ( x )

T

ty t   t 1 exp , dX > − < c, ac > ( X s ) , ds  , ≡ Mt t ≥ 0. < c X = ∫ ( s ) s ∫ 20 Ft  0  Then Q is a solution to the Martingale problem π(x, a, b + ac), where

dQ dP

t

X t = X t − ∫ b ( X s ) ds 0

dMt = M t < c ( X t ) , dX t >

Proof:

{

}

Hence  P d ( X t M t )

(\ M is a P- Martingale)

=  P { X t dM t + M t dX t + d < M , X >t } =  P M t b ( X t ) dt + M t a ( X t ) c ( X t ) dt

Thus,

 Q {dX t } = dt ⋅  Q {(b + ac ) ( X t )}

More generally,

{

}

 P d ( f ( X t ) M t ) =  P {( Lf ( X t ))M t } dt +  P { f ( X t )dM t }

{

}

+  P (d < M , X > t )T f ′ ( X t )

{ } T  Q { Lf ( X t )} =  Q { Lf ( X t )} dt +  Q {ac( X t ) f ′( X t )} dt

=  Q { Lf ( X t )} dt +  P M t (ac)( X t )T f ′( X t ) dt

i.e.

 ( X t )} dt =  Q { Lf

28

Stochastics, Control & Robotics

∂  1  ∂2 T ∂ (⋅) = L + (ac)T where L = (b ( x ) + ac ( x )) + Tr  a ∂χ ∂x 2  ∂x∂xT  

[10] Problem from Revuz and Yor: Still more generally, suppose Zt∈Ft. Then P

{(d ( f ( X ) M )) ⋅ Z } =  t

t

t

P

{ Lf ( X t ) ⋅ M t Zt } dt

{ (

) }

+  P M t ac ( X t ) f ′ ( X t ) Z t dt T

  Q  Z t df ( X t ) =  Q Z  t ⋅ Lf ( X t ) dt Taking Zt = Z∈Ft for a fixed t, we deduce that for all T > t,

i.e

T     Q  Z ⋅  f ( X T ) − f ( X t ) − ∫ L f ( X s )ds  = 0     t

and hence T

M t f = f ( X t ) − ∫ £ f ( X s ) ds , t ≥ 0 is a Q-Martingale. 0



[11] Gough and Kostler Paper on Quantum Filtering Some Remarks: β β  β (⋅) = < ψ , ..ψ >

 β 2 | ψ β > = φ > e (β ) > exp  −  2  , 2 ϕ >∈h, < φ φ > = 1, β ∈L ( + ) ,

< e(β1 ), e(β 2 ) > = exp (< β1 , β 2 >| . dAt | ψ β > = λ | φ > dAt | e(β) >  β 2 β . = β(t )λ φ > e(β) > = β(t ) | ψ > , λ = exp  −  2  < φ′e(v) d Λt ψ > = λ < φ ′e( ν) d Λ t φe(β) > = λ < φ ′ | φ > ν(t )β(t ) < e( ν) | e(β) > dt

Classical Robotics & Quantum Stochastics

29

= λ < φ ′ | φ > β(t ) < e(ν) | eAt* | e(β) > Thus,

dΛ t | ψ β > = β(t )dAt* | ψ β >

Compute  β [ djt ( X )] = < ψ β djt ( X ) ψ β >

djt ( X ) = jt ( L0 X )dt + jt ( L1 X )dAt + jt ( L2 X )dAt* + jt ( L3 X )d Λ t

So,  β { jt ( X )} =  β [ jt ( L0 X )] dt + β(t ) β [ jt ( L1 X )] dt + β(t ) β [ jt ( L2 X )] dt + β(t ) E β [ jt ( L3 X )] dt 2

Note: If ξt is an adapted process i.e. ξt ∈ L ( h0 ⊗ Γ s ( H t ])) , then Thus

{

}

 β { jt ( X )} =  β jt ( Lβ(t ) X ) dt 2

Lβ (t ) = L0 + β(t ) L1 + β (t ) L2 + β(t ) L3 .

where Let

t

Ytin

=

∫ (c1 (ξ)dAs + c1 (s)dAs + c2 (s)d Λ s ) *

0

Then for t1 ≥ t2

Ytin , Ytin  = 0  1 2  * in Moreover since Vt Yt Vt = VT*YtinVT* = Ytout ∀ T ≥ t, we get if follows that Ytout , jT ( X ) = 0∀T ≥ t, X∈£(h)

where

jt(X) = Vt*Vt

This is because Ytin , X  = 0 ∀t ≥ 0 since X ∈£(h) while Ytin ∈£ (Γ s ( H t ])) and Let

[ h , Γs ( H )]

= 0.

y(t) = Vt.

Ytout = Vt*VtinVt so

* * in in in d Ytout = dYt + dVt dYt Vt + Vt dYt dVt

= c1 (t )dAt + c1 (t )dAt* + c2 (t )d Λt + Vt* ( L*2 dAt + S *d Λ t )

30

Stochastics, Control & Robotics

(c, (t )dAt* + c2 (t ) d Λt )Vt + Vt* (c1(t )dAt + c2 (t )d Λt ). ( L2 dAt* + Sd Λ t )Vt * * * = c1 (t ) dAt + c1 (t )dAt + c2 (t ) d Λ t + c1 (t )Vt L2Vt dt

+ c1 (t )Vt* S *Vt dAt* + c2 (t )Vt* S *Vt d Λ t + c2 (t )Vt* L*2Vt dAt + c1 (t )Vt* L2Vt dt + c2 (t )Vt* L2Vt dAt* + c1 (t )Vt* SVt dAt + c2 (t )Vt* SVt d Λ t dYtin | ψ β > = c1 (t )dAt | ψ β > +c1 (t)dAt* | ψ β > +c2 (t )d Λ t | ψ β > * * β = β(t )c1 (t )dt + c1 (t )dAt + β(t )c2 (t )dAt | ψ >

So

{

 β dYtin

}

= < ψ β dYtin ψ β > 2

= β(t )c1 (t )dt + c1 (t )β (t )dt + β(t ) c2 (t )dt

{

}

= 2 Re {β(t )c1 (t )} + β(t ) c2 (t ) dt Likewise

{

 β dYtout

}

2

{

= dt 2Re {β(t )c1 (t )} + β(t ) c2 (t ) 2

{ }} { + 2Re {c1 (t )β(t ) β {Vt* SVt }} 2 + 2c2 (t ) Re { β {Vt* SVt } β(t ) + 2c2 (t ) Re { β {Vt* L2Vt } β(t )}} + 2Re c1 (t ) R Vt* L2Vt

= dt {2 Re {β(t )c1 (t )} + β(t ) c2 (t ) 2

+ 2 Re  β { jt (c1 (t ) L2 + c1 (t )β(t ) S 2

+ c2 (t ) β(t ) S + C2 (t )β (t ) L2 Recall that

)}}

dVt = (– iHdt + L1dAt + L2dA*t + SdΛt)Vt

where the system operators (∈ £ (h)) = H, L1, L2, S have been chosen to make Vt unitary, i.e., d(Vt* Vt) = 0 and this has been achieved using the quantum I to formulae: dAt dA*t = dt, dΛt dA*t = dA*t,

dAt dΛt = dAt, dΛt dΛt = dΛt,

Classical Robotics & Quantum Stochastics

31

all the other products being zero.

{

 β jt (χ) | ηtout ]

}

out is the filtered estimate of the state jt(X) given the measurements σ - algebra ηt] upto time t.

We note that jt(X) = Vt* XVt , X∈£(h) * in in out ηt] = Vt ηt ] Vt , ηt] ⊂

(

)

* in s out ηt] = VT ηt ] VT , £ Γ ( H t] ) T ≥ t

and since and it follows using

 X , ηin  t]  = 0 

  jT ( X ) , ηtout ]  = 0 

that

∀T ≥ t.

(

)

*  in  Now,  β  jt ( X ) | ηtout ]  = Vt  t X | ηt ] Vt

t (ξ) =  β Vt*ξVt 

where

This follows because,

(

) )

(

  β  jt ( X ) − Vt*t X | ηtin] Vt ηtout ]  

{( =  {V ( X η

) )

(

β in * * * in =  Vt XVt − Vt t X | ηt ] Vt Vt ηt] Vt β

* t

in t]

(

) )

in − t X | ηin t ] ηt ] Vt

(

)

}

}

in in in =  t  X ηt −  t X | ηt ] ηt ]      in in  =  t  X ηin t ]  −  t  t  X ηt ] | ηt ]  = 0.

Now,

Suppose F(t) is an adapted process such that

{

in β * F(t)∈ ηt] i.e. [F(t), ηin t] ] = 0 ∀t, and E Vt X tVt

{

}

}

=  β F * (t ) X t F ( H ) ∀t Then  and

β

{

jt ( X ) | ηtout ]

{

 t X | ηtin}

This follows from

{ {

}

= Vt* t X | ηin t ] Vt

}

=

{

(Xt being any adapted process)

}

{ }  β {F * (t ) F (t ) | ηin t] }

 β F * (t ) XF (t ) | ηin t]

} {

} }

in  β  t X | ηtin]  β F * (t ) F (t) | ηin t ] ηt ]

32

Stochastics, Control & Robotics

{ {

}

{

} }

β β in β in in * * =   Vt XVt | ηt ] X  F (t ) F (t) | ηt ] ηt ] in   =  β  F * (t ) t  X ηin t ]  F (t ) ηt ]   

 in  =  β Vt* t  X ηin t ]  ηt ] Vt   in in  =  t   X ηt  ηt  =  t  X ηin t 

On the other hand,

{ {

} }

in  β  β F * (t ) XF (t ) ηin t ] ηt ]

β in * =   F (t ) XF (t ) ηt] 

 =  β  F * (t ) X ηin t] F (t )  β * in =  Vt X ηt ] Vt 

 =  t  X ηin t] 

Proving the claim. Now let Then

|y(t) > = Vt|yβ >

d |y(t) > = dVt|yβ >

( ) (−iHdt + β(t ) L1dt + (S β(t ) + L2 ) dAt* ) ψβ >

= −iHdt + L1dAt + L2 dAt* + Sd Λ t ψβ > =

  S β(t ) + L2  =  −iHdt + β(t ) L1dt +   c1 (t )    × c1 (t )dAt* + c1 (t )dAt  ψ β >  dtc (t ) − 1 ( S β (t ) + L2 ) β(t ) ψ β > c1 (t )

(

Assume c2(t) = 0, i.e.

)

* dYtin = c1 (t )dAt + c1 (t )dAt

Then we get d|y(t) > = dVt|yβ > i.e.

= dFt|yβ >

Vt|yβ > = Ft | yβ >

/ and Ft ∈ηin t] .



Classical Robotics & Quantum Stochastics

33

[12] Quantization of Master. Slave Robot Motion Using EvansHudson Flows: M m ( qm )qm + N m ( qm , qm ) = τm (t ) + Fm ( qm , qm ) ( K mp ( qs − qm )

+ K md ( qs − qm )) + d m (t )

M s ( qs )qs + N s ( qs , qs ) = τs (t ) + Fs ( qs , qs ) ( K sp ( qm − qs )

+ K sd ( qm − qs )) + d s (t ) dm(t), ds(t) are white noise processes. Thus,  qm    qm  ψ1 ( qm , qs , qm , qs )  d  qm   d (t )  q  =  q  + ψ 3 ( qm , qs , qm , qs )  d m(t )   s  dt  s   s   qs   ψ 2 ( qm , qs , qm , qs ) τ (t ) + ψ 3 ( qm , qs , qm , qs )  m   τs (t )  where y1 = − M m ( qm) −1 N m (qm , qm ) + K mp M m ( qm ) −1 Fm ( qm , qm )(qs − qm ) + K md M m ( qm ) −1 Fm ( qm , qm )(qs − qm ) ∆ ψ10 ( qm , qm ) + K mp ψ11 ( qm , qs , qm , qs ) + K md ψ12 ( qm , qs , qm , qs ) and likewise for y2. dBm (t )    Σ m dt   d m (t )  = Here  d (t )  =   Σ dBs (t )   s   s dt  Σ m O   B= Σ =   O Σ s  B is a 4-dimensional Brownian motion. Defining we get

Σ

dB dt

 Bm  4  B  ∈  s

q q =  m  ∈4, q  s  q (t ) d  =  q (t )

{F0 (q(t ), q (t ), t ) + Kmp F1 (q(t ), q (t ))

34

Stochastics, Control & Robotics

+ K md F2 ( q (t ), q (t ) ) + K sp G1 ( q (t ), q (t ) )

}

+ K sd G2 ( q (t ), q (t ) ) dt + H ( q (t ), q (t ) ) dB(t) ξ 8 q  q  = ε  . Then  

Let

dξ = {F0(ξ, t) + KmpE1(ξ) + KmdF2(ξ) + KspG1(ξ) + KsdG2(ξ)}dt + H (ξ)dB Let f: 8 →  be C2. Then, 4

df(ξt) = L1t f ( ξt )dt + Σ L2k f ( ξt )dBk k =1

(

where

1 T T T L1tf(ξ) = P( ξ , t ) ∇ξ f ( ξ ) + Tr H ( ξ ) H ( ξ ) ∇ξ ∇ξ f ( ξ ) 2

and

L2kf(ξ) = Σ H mk ( ξ )

8

∂f ( ξ )

m=1

∂ξ m

)

T =  H ( ξ ) ∇ξ f ( ξ ) k

Here P(ξ, t) = F0(ξ, t) + KmpF1(ξ) + KmdF2(ξ) + KspG1(ξ) + KsdG2(ξ)

In the quantum context, f(ξt) → jt(f)

jt being a homomorphism of an "initial algebra" 0 into t] and f is replaced by X∈0 where 0 can be non-Abelian. We get

djt(X) = jt (£ 0 t ( X ) + K mp £1 ( X ) + K md £ 2 ( X ) 4

+ K sp £3 ( X ) + K sd £ 4 ( X ) ) dt + Σ jt (£5k ( X ) ) dAk (t) k =1

4

4

k =1

j , k =1

(

)

j + Σ jt (£ 6 k ( X )) dAk* (t) + Σ jt S kj ( X ) d Λ k (t )

The aim is to choose {Kmp, Kmd, Ksp, Ksd} T

so that

∫ < fe(u ), jt ( X ) fe(u) > dt

is a minimum, where f∈h (initial Hilbert

0

space and c is a fixed observable in L(h). In this classical context, jt(f) = f(ξt) 2 2 = α || qm (t ) − qs (t ) || + β || qm (t ) − qs (t ) ||

Classical Robotics & Quantum Stochastics

35

Or more precisely writing  qm     qs  2  ||2 ξ =   , f(ξ) = α || qm − qs || + β || qm − qs qm    qs    Note that 1 < fe(u ), djt ( X ) fe(u) > = < fe(u), jt(L0t X) fe(u) > dt + Kmp< fe(u), jt(L1X) fe(u) >

+ Kmd < fe(u), jt(L2X)fe(u) >

+ Ksp < fe(u), jt (L3X) fe(u) > + Ksd < fe(u), jt(L4X) fe(u) >

 4  ` +2 Re  ∑ fe(u ), jt ( Lsk X ) f e (u ) uk (t )   k =1

fe(u ), jt ( S kj X ) fe(u ) uk (t )u j (t)

+ Σ

k, j

Here, we are assuming Skj ( X )* = S kj ( X )

(X* = X). T

Thus

< fe(u), jT(X)fe(u)> =

∫

fe(u ), jt ( L0t X ) fe(u) dt

0 T

4

+ Σ K α ∫ fe(u ), jt ( Lα X ) fe(u ) dt α=1

0 4 T

+ 2 Re Σ

k =1

T

+ Σ

k, j

Perturbative approximations.

∫

∫

fe(u ), jt ( L5k X ) fe(u) uk (t )dt

0

fe(u ), jt ( S kj X ) fe(u ) uk (t )uj (t)dt

0

Consider first the following simplified problem djt(X) = (jt(L0X) + Kjt(L1X))dt + jt(L2X)dA(t) + jt(L3X)dA*(t) + jt(L4X)dΛ(t) Where L3(X) = L2(X)* ≡ L+2 (X) i.e. L+2 = L3,

and L4(X)* = L4(X), i.e. L+4 = L4 Where c* = c.

36

Stochastics, Control & Robotics

Also, L+0 = L0, L+1= L1.

K is a real constant chosen so that T

∫ < fe(u ), jt (χ) fe(u ) > dt 0

is a minimum. For example we could choose jt(X) = V*t X Vt ≡ V*t (X⊗I)Vt where with

* dVt =  −iHdt + L1dAt + L2 dAt + Sd Λ t  Vt H = H0 + KH1

Then, djt(X) = d(V*tXVt)

= dV*tXVt + V*tXdVt + dV*tXdVt

= V*t [i(H*X – XH)dt + (L*2X + XL1)dAt + (S*X + XS)dΛt + (L*1X + XL2)dA*t + L2* X L2 dt + L*2 X SdAt + S*XSdΛt + S*XL2dA*t]Vt = [jt(L0X) + Kj1(L1X)]dt + jt(L2X)dAt + jt(L3X)dA*t + jt(L4X)dΛt Where L0(X) = i(H*0X – XH0),

L1(X) = i(H*1X – XH1),

L2(X) = L*2X + XL1 + L*2XS

L3(X) = L*1X + XL2 + S*XL2

L4(X) = S*X + XS + S*XS

The approximate solution to jt is obtained from an approximate solution to Vt. We can also incorporate parameters like K into L1, L2 and S. That takes care of pd controllers. Feedback K p ( qm − qs ) can be achieved by adding a harmonic oscillator potential to H0. The feedback K d ( qm − qs ) however, is like a velocity damping term and must be taken care of by parameters incorporated in the Sudarshan-Lindblad generator or equivalently in L1, L2, S. Thus we write H = H0 + K0H1, L2 = L20 + K2L21, L1 = L10 + K1L11 S = S0 + K3S1.



Classical Robotics & Quantum Stochastics

37

[13] Problem From KRP QSC: Let A, B be operators such that [A, B] commutes with both A and B. X(t) = et(A + B). Then

Let

X´(t) = (A + B) X(t). X(t) = etA F(t). Then,

Put

X´(t) = etA(AF(t) + F´(t)) = (A + B) etAF(t) F´(t) = e–tABetAF(t)

So,

= e–t ad A(B)F(t) = (B – t[A, B]) F(t) n

(ad A) (B) = 0 for n ≥ 2. Then,

Since

F(t) =

tB−

t2 [ A, B] 2

e form a commuting family of operators.

since B – t[A, B], t∈

It follows that tA

X(t) = e e

tB−

tA tB

= e e e

t2 [ A, B] 2 −

t2 [ A, B] 2

Now apply this result to the operators a(u), a+(v). we get e

λa (u ) µa + ( v )

e

e

−

λµ

2

λa (u )+ µa = e

+

(v )

Since [a(u), a+(v)] = commutes with both a(u) and a+(v). Likewise e

λa (u )+ µa + ( v )

µa = e

+

( v ) λa (u )

e

Then < e( w1 ), eλa (u ) + µa

+

(v )

e

−

λµ

2

e( w2 ) >

= e = e = e

−

λµ

2

−

λµ

2

−

λµ

2 eλ eµ < w1 , v >

< e( w1 ), eµa

+

( v ) λa (u )

e

e( w2 ) >

< eµa (v ) e( w1 ), eλa (u ) e(w2 ) >

e< w1 , w2 >

In particular, < e( w1 ), eλa (u ) − λa

+

(u )

e( w2 ) >

= e

−

λ 2

2

u

2

eλ e − λ < w1 , u> × e< w1 , w2 >

38

Stochastics, Control & Robotics

On the other hand, ∞

1 n £ W (u , I ) n e(w2 ) > n

< e( w1 ), eαW (u , I ) e( w2 ) > = < e( w1 ), Σ

Now,

n=0

{ {

}

1 2 u − < u, w2 > e( w2 + u) . 2 1 2 1 2 W(u, I)2 e(w2) = exp − u − < u , w2 > − u − < u , w2 + u > 2 2 e( w2 + 2u ) W(u, I) e(w2) = exp −

{

= exp −2 u

}

2

− 2 < u , w2 > e( w2 + 2u )

{

W(u, I)3 e(w2) = exp − 1 u 2 − < u , w2 + 2u > −2 u 2 e( w2 + 3u )

{

= exp − In general let

9 u 2

(

Then

2

− 2 < u , w2 >

}

}

2

W(u, I)n e(w2) = exp − α n u

}

− 3 < u , w2 > e( w2 + 3u ) 2

W(u, I)n+1 e(w2) = exp  − α n u 

)

− β n < u , w2 > e( w2 + nu ) 2

− β n < u , w2 > −

1 u 2

2

− < u , w2 + nu > ) e ( w2 + (n + 1)u )

   = exp  −  α n + 1 + n u    2

2

 − (β + 1) < u , w2 > 

e ( w2 + (n + 1)u )

1 , α = 0, 2 0 = βn + 1, β0 = 0.

αn + 1 = αn + n +

So

βn + 1

n−1 n2 n(n −1) 1 1  + n = αn = Σ  r +  = 2 n=0  2 2 2

Thus

βn = n. Thus 2   W(u, I)n e(w2) = exp  − n 2 u − n < u , w2 >  e(w2+nu)

In particular,

2   u = exp  − n 2 − n < u , w2 > + n < w2 , u >+< w1 , w2 > 2  

{

= exp −

1 2 u − < u , w2 > + < w1, u > + < w1, w2 > 2

}

Classical Robotics & Quantum Stochastics

39

{

= exp −

1 2 α u 2

2

− α < u , w2 > 

+ α < w1 , u > + < w1 , w2 > 

= Then,



{

}

e ( w1 ) , exp αa + (u ) − αa(u ) e( w2 )

{

}

W(αu, I) = exp αa + (u ) − αa (u ) . KRP QSC. t   Vf(t) = exp  i Im ∫ eiφ( s ) F ( s) 2 ds × W (eiφ − 1) ft , eiφt   0

(

)

ft = f.c[o, t], ft = f. X[O, t].

t   = exp  i Im ∫ eiφ( s ) f ( s) 2 ds   0

{

× e (u ) exp −

1 iφ (e − 1) ft 2

(

e eiφt v + (eiφ − 1) ft

2

− (eiφ − 1) ft , eiφ t v

}

)

 t  2 = exp  i ∫ f ( s ) sin {φ( s )} ds  0   1   × exp −  2 ft  2 

2

t   2 − 2∫ f ( s ) cos (φ( s ) ) ds    0

− (eiφ − 1) ft , eiφt v + u , eiφt v + (eiφ − 1) ft

)

t t  t 2 2 = exp ∫ f ( s) eiφ( s ) ds − ∫ f ( s ) ds − ∫ (1 − eiφ( s ) ) f ( s)v( s)dss  0 0 0 t

∞

0

0

+ ∫ (eiφ( s ) − 1)u ( s )v( s )ds + ∫ u ( s )v( s )ds  + ∫ (eiφ( s ) − 1)u ( s ) f ( s )ds  0  t

40

Stochastics, Control & Robotics

Thus, d < e(u), Vf(t)e(v) > = < e(u), dVf(t)e(v) >

{

}

iφ(t ) − 1) f (t ) + f (t )v(t ) + u (t )v(t ) + u (t ) f (t ) = (e 2

e(u ), Vφ (t )e(v) iφ u , eiφt v = u , (e t − 1)v + u , v

Note:

t

∫ (e

=

iφ( s )

− 1)u ( s )v( s )ds + u, v .

0

∞

∫ u (s)v(s)ds .

=

0

It follows that

{

}

dVf(t) = (eiφ(t ) − 1) f (t ) dt + f (t )dA* (t ) + f (t )dA+ (t ) + d Λ (t ) Vφ (t ) *

2

M(t, f, X) = 0{U(t) XVf(t)U(t)}

d(U*XVfU) = dU*XVfU + U*cVfdU + U*XdVfU + dU*XVfdU + dU*XdVfU + U*XdVfdU

Now,

0 [dU*XVfU] = < e(o), dU*XVfU e(o) >   1   =  0 U *  iH − L* L XVφU  dt   2   



1





= dt M t, φ,  iH − L* L X  , 2  

1   0 [U*XVfdU] = dt M  t , φ, X ( −iH − L* L)   2



0 [U*XdVfU] = dt(eiφ − 1) f 2  0 U * XVφU  = dt (eif – 1) |f|2 M (t, f, c)



0 [dU*XVfdU] = 0[U*L*XVfLU]dt = 0[U*L*XLVfU]dt *

= M(t, f, L XL)

(0 picks only the coefficient of dt)

(ΓVf, T) = 0 ∀T∈£(h) since i.e. V(t)f∈£(Γs(H))

0[dU*XdVfU] = 0[U*LXf(eif – 1)VfU]dt

= (eif – 1)f 0[U*LXVfU]dt

= (eif – 1)FM(t, f, LX)dt

0[U*XdVfU]dt = (eiφ − 1) f  0 U * XVφ LU  dt

Classical Robotics & Quantum Stochastics

41

= (eiφ −1) f dt M (t, φ, χL) ([V , L] = 0 ∵ [V , h] = 0 ∵ [Γ (H), h] = 0) f

f

s

Then, d M (t , φ, X ) = dt–1. 0{d(U*XVfU)] dt

 1 1     = M  t , φ,  iH − L* L X − X  iH + L* L      2 2 +X f

2

(eiφ − 1) + L * XL + (eiφ − 1)( fLX + fXL))

iφ = M (t , φ, θ(χ) ) + (e − 1) M (t , φ, f χ + fLχ + f χL) 2

Note that f

2

X + fLX + fXL = fX ( f + L) + fLX .

Remark: An additional term also has to be considered in the evaluation of d M (t , φ, χ) . dt It is dU* X dVf dU The coefficient of dt in this triple differential product comes from dA . dΛ . dA+ = dA dA+ = dt The corresponding coefficient is (eif – 1) U*LX(eif – 1) VfLU = (U*LXLVfU) and so, taking into account this correction, we get d iφ M (t , φ, X ) = M (t , φ, θ(χ) ) + (e − 1) M dt 2 t , φ f X + fLX + fXL + LXL

(

)

= M (t , φ, θ( X )) + (eiφ − 1) M (t , φ, ( f + L) X ( f + L)) . 

[14] Moment Generating Function for Certain Quantum Random Variables: h  system space, H  Bath space for 1 Boson Γs(H)  bath Boson Fock space for indefinite number of Bosons. L∈£(h), a(u)∈L(Γs(H1))

a(u), a*(u) are the annihilation and creation operators [a(u), a(v)] = f∈h.e(u)∈Γs(H) is the exponential vector.

Let [L, L*] = 0. i.e. L is a normal operator.

42

Stochastics, Control & Robotics

We wish to find

(

)

fe(v) exp La(u) + L*a* (u ) He(v) Suppose A, B are operator such that [A, [A. B]] = [B.[A, B]] = 0. Let

F (t) = exp (t (A + B)) F' (t) = (A + B) F (t).

Let

F (t) = exp (tA) G(t). Then G' (t) = [exp(– t al A) (B)] G(t) = (B – t [A, B]) G (t)

and so,

  t2 G (t) = exp  tB − [ A, B ]    2    t2  exp(tB) ⋅ e xp  − [ A, B ]  . =  2 

 t2  exp (tA + B) = exp(tA) ⋅ exp(tB) ⋅ exp  − [ A, B ]   2  or equivalently

Thus,

  1 exp(A + B) = exp (A). exp (B) . exp  − [ A, B ]  .   2

Thus since [L a(u), L*a*(u)] = ||u||2 LL* Commutes with both La(u) and L* a* (u), it follows that 

[15] Belavkin Filtering Contd.: * in d σt(X) =   F (t + dt ) XF (t + dt ) | ηt + dt] 



* in −  F (t ) XF (t ) | ηt]  in   *  =   F * (t ) XF (t ) | ηin t + dt ]  −   F (t ) XF (t) | ηt]  



* in +  d ( F (t ) XF (t) | ηt + dt] 

Now and since

in F * (t ) XF (t ) ∈ σ{Y ( S ), S < t , δ (h)} in in ηin t + dt] = σ{ηt], dY (t)}

it follows two in the state if

(

2

)

e(u) > fe(u ) exp − u 2 ,

Classical Robotics & Quantum Stochastics

43

in   *    F * (t ) XF (t ) | ηin t + dt]  =   F (t ) XF (t ) | ηt]  

Thus,

 dσt(X) =   d ( F * (t ) XF (t )) | ηin t + dt]  

{

(

)

  F * (t )  P* (t ) X + XP(t ) F (t ) | ηin t]  dYin (t )   +  F * (t )P* (t ) XP(t ) F (t ) ηin t]  dZ in (t ) 

(

)

 +   F * (t ) Q* (t ) X + XQ(t ) F (t ) ηin t]  dt 

= σt ( P* X + XP)dYin + σt ( P* XP)dZin + σt (Q* X + XQ)dt

d σt ( X ) σt ( X )d σt (1) σt ( X ) d σt ( X ) d σt (1) σ (X )  2 ( ( 1)) − + dσ − d t t  = σt (1) σt (1)2 σt (1)3 σt (1)2  σt (1)  σ (X ) . Then we get Let vt(X) = t σt (1) d vt(X) = vt(P*X + XP) d Yin + vt(P*XP) d Zin + vt(Q*X + XQ) dt – vt (X) (vt(P* + P) d Zin + vt(P*P) d Zin + vt(Q*Q) dt + vt(X) vt (P* + P) d Zin + vt(P*X P)2 d Vin

+ 2 vt (P* + P) vt(P*P) dWin) – vt(P*X + XP) vt(P* + P) dZin – vt (P*X + XP ) vt (P*P) + vt(P* + P) vt(P*XP)) d Win – vt(P*XP) d Vin

where

(dYin)2 = dZin

(dYin)3 = dWin = dYindZin (dYin)4 = dVin = (dZin)2 We can express there as dvt(X) = At(X) dt + Bt(X) dYin(t) + Ct(X) dZin(t) + Dt(X)dWin(t) + Et(X) dVin(t)

where At(X), Bt(X), Ct(X), Dt(X) and Et(X) are linear functions of X with values in the Abelian algebra

{

}

in ηin t] = σ Y ( s ) s < t ,



[16] Modelling Noise in Quantum Mechanical Systems: [a] Dyson series: Let ξ(t), t ≥ 0 be a random process

44

Stochastics, Control & Robotics

Consider the perturbed quantum Hamiltonian H(t) = H0+eξ(t)V Let U0(t) = exp(–itH0)  then unitary evolution operator of the unperturbed system. Them if U (t, s), t ≥ s is the perturbed unitary evolution from time s to time t, i

∂U (t , s) = H (t) U (t, s), t ≥ s, ∂t U(s, s) = I U(t, s) = U0(t) W(t, s)

The solution is where

i

∂W (t , s) = εξ(t ) V (t ) W (t , s), t ≥ s , ∂t W(s, s) = U0 (– s), V (t ) = U0(– t)VU.(t).

Then the solution is

∞

 −iε n U (t, s) = U 0 (t − s ) + U 0 (t ) ∑   n =1 h 

∫

ξ(t1 ) ... ξ(tn ) V (t1 )...V (tn ) dt1....dtn ) U 0 ( − s )

s < tn 0, ∑ pi =

∑ qi

=1

D(ρ σ) = Tr (r (log r – log σ)) =

∑ pi log pi − ∑ pi i, j

i

(where Sij = Note that

 pi   j

=

∑ pi Sij log  q

=

∑ Sij

2

i, j

ei fi i .

∑ Sij i

j

= 1.)

ei f j

2

log q j

Classical Robotics & Quantum Stochastics

Thus,

D(r|σ) =

73

 pi Sij   ≥0 i ij 

∑ pi Sij log  q S i, j

since {pi Sij}i, j and {qi Sij}i, j are both probability distributions on {1, 2, ..., N}2. 

[29] Entropy of a Quantum System: ρ′(t ) = − i [ H , ρ(t )] + ε θ(ρ(t ))

(

1 * L Lρ + ρL* L − 2LρL* 2 S(t) = – Tr(r(t) log r(t)).

q (r) = −

)

S ′(t ) = Tr (ρ′(t ) log ρ(t)) = − ε Tr (θ (ρ(t )) log ρ (t) )

( ) = εT ( L Lρ log ρ − LρL * log ρ)

=

ε Tr 2 L* L ρ log ρ − 2 LρL * log ρ 2 *

r

)

(

= ε Tr  L* , Lρ log ρ   r(t) = exp (–it ad H) (r (0)) t

+ ε ∫ exp ( −i (t − S ) ad H ) (θ(ρ ( S )) ds 0

= exp (–it ad H)(r(0)) t

2 + ε ∫ exp ( −i (t − S ) ad H ) θ (exp ( −is ad H ) (ρ (0)) ds + O(ε ) 0

Let

Tt

(0)

= exp (– it ad H) t

Then, Thus,

( = ε T (  L , LT

0

* S ′(t ) = ε Tr  L , Lρ(t)  log ρ (t ) r

Let

(

)

0 2 (0) 0 r(t) = Tt (ρ (0)) + ε ∫ Tt − S θ TS (ρ (0) ds + O(ε )

r(t) =

*

t

( 0)

∑ λ k (t ) ek (t )

)

(ρ(0)) logTt(0) (ρ(0)) ek (t )

k

Then

S ′(t ) = ε ∑ log(λ k ) ek  L* , Lρ ek k

) + O(ε ) 2

(spectral representation)

74

Stochastics, Control & Robotics

∑ log λ k

= ε

k

ek  L* , L  ek λ k + ε ∑ log (λ k ) ek L  L* , ρ ek k

* = ε ∑ λ k log λ k ek L  L , L  ek k

= ε

∑ log (λ k ) k

ek  L, L*  ek

+ ε ∑ log (λ k ) ek  LL*  ek λ k k

−ε ∑ log λ k ek  L*ρ L  ek k

* = ε ∑ λ k log λ k ek L , L ek + ε ∑ λ k log λ k ek LL * ek

k

k

+ε

∑ λ k log λ k

ek L ek

∑ λ k log λ k

ek L em

2

k, m

−ε

− ε ∑ λ k log λ k ek L* L ek k

2

k, m

* = ε ∑ λ k log λ k ek L L ek − ε ∑ λ m log λ k ek L em k ,m

k

= ε ∑ λ k log λ k em L ek

2

k ,m

= ε

∑

ek L em

m, k

2

∑

Let

− ε ∑ λ m log λ k ek L em k ,m

λ  λ m log  m   λk 

So S ′(t ) ≥ 0 iff 2

ek L em

m, k

Remark:

λ  λ m log  m  ≥ 0 ∀ t.  λk 

log r (t) = Z(t) r(t) = eZ(t) z (t )  1 − exp ( − ad Z (t) )  ρ′(t ) = e   ( Z ′(t ) ) ad Z (t ) 

Z ′(t ) =

2

(

ad Z (t ) ρ(t ) −1 ρ′(t ) 1 − exp ( − ad Z (t ) )

)

2

Classical Robotics & Quantum Stochastics

75

∞  1 +  ∑ Cm ad Z (t )m =   m =1

(

 d  Tr ρ log ρ  dt 

=

 −1  ρ ρ′ 

)(

)

Tr [ρZ ′ ]

= Tr (ρ′ ) +

∑ CmTr ρ (ad log p )

m

m ≥1

(ρ ρ′)

m Tr (ρ′ ) = 0, Tr[r (ad log r) (X)]

(X =ρ

−1

p ′, Y = ( ad log ρ)

m−1

(X )

−1

)

= Tr[r ad log r(Y)] = Tr[r[log r,Y]] = Tr[[log r, rY]] = 0 Rate of entropy change of a quantum system ρ′(t ) = T(r(t)), T (T(r)) = 0 r

(

1 * L Lρ(t ) + ρ(t ) L* L − 2Lρ(t )L* 2 S(r(t)) = – Tr(r(t)log e(t)) d   d S (e (t )) = Tr (ρ′ (t ) log e(t) ) − Tr  ρ(t ) log ρ(t )   dt dt = i [ H , e(t)] −

log r(t) = Z(t)

Let

r(t) = rZ(t)

Then

−ad Z (t )   ρ′(t ) = ρZ (t )  1 − e  ( Z ′ (t ))  ad ( Z (t )) 

  − Z (t ) ad Z (t ) Z ′(t ) =  e ρ′ (t )   1 − exp ( − ad Z (t) ) 

(

 ad log ρ(t)  =  e(t ) −1 ρ′ (t )   1 − exp ( − ad ρ(t) ) 

(

Let So,

x 1 − e− x

)

)

∞

=

∑ Cn xn . Then C0 = 1

n=0

∞ ad log ρ n 1 + ∑ Cn ( ad log ρ) = 1 − exp ( − ad log ρ) n =1

Z ′(t ) = ρ−1 ρ′ +

∞

∑ Cn (ad log ρ)

n =1

n

(ρ ρ′) −1

)

76

Stochastics, Control & Robotics

 d  Tr  ρ log ρ = Tr (ρZ ′ )  dt  = Tr (ρ′ ) +

∑ CnTr (ρ (ad log ρ) ∞

n =1

∑ CnTr (ρ (ad log ρ) ∞

= Tr (ρ′ ) =

since

n =1

n

n

(ρ ρ′)) −1

(ρ ρ′)) −1

d Tr ρ = 0 dt

(Tr(r′) = Tr(T(r)) = 0)

Now r (ad log r)(X) = r[log r, X] = [log r, rX] = (ad log r)(rX) Thus r(ad log r)n (r–1r′) = r (ad log r)n and so Tr(r (ad log r)n(r–1r′)) = 0. Thus,

 d log ρ  Tr  ρ  = 0 and we get  dt  d S (ρ (t )) = – Tr(r′(t) log r(t)) dt = – Tr(T(r(t)) log r(t)) = – Tr[(i[H, r] + q(r) log r] = – Tr[(i[H log r, r] + q(r) log r]

where

= – Tr(q(r) log r) 1 q(r) = − (L*Lr + rL*L – 2 LrL*) 2 1 dS (ρ (t )) = {Tr(L*L r log r) + Tr(r L*L log r) 2 dt – 2 Tr(LrL* log r)} = Tr (L*L r log r) – Tr (LrL* log r) 

[30] Dirac Eqn. Based Temperature Estimation of Black-Body Temperature Field: Dirac eqn.  γ µ (i∂µ + eAµ ) − m  Ψ = 0.   I 0  r  0 γ0 =  ,γ =   0 −I   iσ r

iσ r  ,1 ≤ r ≤ 3, 0 

Classical Robotics & Quantum Stochastics

77

0 −i  0 1 1 σ2 =   i 0  ,σ1 =  1 0 , σ3 =  0 σ r σ s + σ s σ r = 2δ rs ,

γr

0 , −1

*

0 = −γ r , γ 0 = γ .

*

µ v v µ µv Thus, γ γ + γ γ = 2g

((2g µv )) = diag[1, −1, −1,1]. Other choices:  0 0 I r γ0 =  , γ =    I 0  −σ r Then

σr  ,1 ≤ r ≤ 3 0 

r * 0 r* γ 0 = γ , γ = −γ ,

γ µ γ v + γ v γ µ = 2g µv

(

)

Let

ψ = ψ *γ 0 ≡ ψ −T γ 0 .

Let

 µ ψ = ψ *γ 0 γ µ ψ Jµ = ψγ

Then J 0 = y*y is real, Jr = y*g0gty ⇒

* t 0 * r * 0* J r* = ψ γ γ ψ = –y g y y

= y*g0gr y = J r Hence J r is also real , 1 ≤ r ≤ 3. Now

(

* 0 µ i∂µ J µ = i∂µ ψ γ γ ψ

(

)

)

i ∂ µ ψ * γ 0 γ µ ψ + iψ *γ 0 γ µ ∂ µ ψ Now So Now

igµ∂µy = −eγ µ Arµ ψ + mψ * µ*

i ∂µ ψ * γ µ* = − eAµ ψ γ

(1)

+ mψ *

0 I  0 g 0 gr =   I 0  −σ r

σ r   −σ r 0   0

(2) 0  −σ r 

µ* 0* and hence g0 gµ is Hermitian, i.e. γ γ = gµ* g0 = g0 gµ

So

−i∂µ ψ *γ µ*γ o = −eAµ ψ * γ µ* γ o + mψ * γ 0

⇒

−i∂µ ψ *γ 0 γ µ = −eAµ ψ * γ 0 γ µ + mψ * γ 0

78

Stochastics, Control & Robotics

and hence (1) iψ *γ 0 γ µ ∂µ ψ = −eψ * γ 0 γ µ ψAµ + mψ * γ 0 ψ

⇒

while (2) ⇒ −i∂ µ ψ *γ 0 γ µ ψ = −eAµ ψ * γ 0 γ µ ψ + mψ * γ 0 ψ Subtracting, we get

(

∂ µ ψ *γ 0 γ µ ψ i.e. ∂µ J

µ

)

=0

= 0 which is the charge conservation equation; provided we define the

current density as  µψ −eψ *γ 0 γ µ ψ ≡ −eψγ Now put Aµ(X) be a random Gaussian field with zero mean and correlations Aµ (t + τ, r1 ) ⋅ Ar (t , r 2 ) = Rµr ( τ, r1 − r 2 )

Aµ (t + τ, r 0 − r ) Ar (t , r 0 ) = Rµr (τ, r )

i.e.,

y(X) = y(0) (X) + ey(X) + O(e2).

Let

Then by first order perturbation theory,

(γ (γ

) − m) ψ

µ

i ∂µ − m ψ (0) = 0,

µ

i ∂µ

(

(0)

µ (0) = − Aµ γ ψ

ψ (α0) =

0 where p = pµ = p , p

)

(4) (5)

∫ aα ( p)exp (i. p.X ) + bα ( p)exp (i.p − X ) d

( )

( )

0 0 0 p.X = pµXµ = pµXµ = p X − p, r = p t − p, r

(6) satisfies (4) provided

(γ (γ

) + m) b ( p)

µ

pµ − m a ( p) = 0

µ

pµ

=0

( )

2

2 These eqn. imply pµ pµ = m2 i.e. P02 = p + m or p0 = F p 2

2 = + m + p .

Now a ( p) can be determined as follows: 0 I r  0 ,g=  g0 =   I 0  − σr

σr  0 

3

p

(6)

Classical Robotics & Quantum Stochastics

79

( ) ( )

 0, E ( p) + σ, p   g 0 pµ =   E ( p) − σ p 0 

So,

(γ

So ⇔

µ

)

pµ − m a ( p) = 0

(

( )) ( )

ma1 ( p) = E ( p) + σ, p a, p   a1 ( p)    a ( p) =    a 2 ( p)  

( ( ) ( )) ( )

 1 E p + σ, p a 2 p   a ( p) =  m    a p 2  

Thus

()

()

Normalizing so that a p

()

a2 p ,

()

()

1  E p m2 

()

a2 p ,  2E p 

2

2

2

= 1 gives

( )(

) ()

+ p 2 + 2 E p σ, p  a 2 p 

( )(

+ a ( p)

2

=1

) ()

− m 2 + 2 E p σ, p  a 2 p 

() () 2E ( p ) a 2 ( p ) , ( E ( p ) + (σ, p )) a 2 ( p )

= m2

( ) ( )(

) ()

= m2

() () (

) ()

= m2

+ m2 a 2 p , a 2 p

or

()

a2 p

()

b2 p

2E p

2

()

2

+ 2 E p a 2 p , σ, p a 2 p

= m2

Likewise, 2E p

2

()

2

(Simply replace m by – m)

()

+ 2 E p b 2 p , σ, p b 2 p

() ()

a2 p = λ p χ p

Let where

()

()

()

λ p ∈C, χ p ∈C2 , χ p

( ) λ ( p)

Then 2 E p

2

2

2

=1

() ()

+ 2E p λ p

2

( ) ( ( ) ( ))

χ p , σ, p χ p

= m2

80

Stochastics, Control & Robotics

()

i.e. λ p

2

=

m2

( ) ( E ( p ) + χ ( p ) , ( σ, p ) χ ( p ) )

2E p

2

In particular a(p) and b ( p ) ∈C 4 both have two complex degrees of freedom ∀ p ∈ 3 . We now solve the O(Aµ) eqn i.e. (5) Aµ ( X ) =

Let

Aµ ( p ) exp ( −i p ⋅ X ) + Aµ ( p ) exp (i p ⋅ X ) d ∫   

where p.X = pµXµ and  Aµ = ∂α∂αAµ = 0

3

p

()

p2 = 0, i.e. pµpµ = 0, i.e. p0 = +E p = p

⇒

()

So we have writing E0 p = p that Aµ ( X ) =

∫ Aµ ( p) exp ( −i ( E0 ( p)t − ( p, r )))  + A p exp i E0 p t exp ( p, r )  d 3 p 

(( ( ))

()

)

Then Aµ ( X ) ψ α(0) ( X ) =

∫  Aµ ( p) exp (−iE0 ( p)t ) exp (i ( p, r ))

 + Aµ p exp i E0 p t exp ( −i ( p, r ))  d 3 p 

(( ( ))

()

)

( ) ( ( )) ( ) +bα ( p ) exp ( −iE ( p ) t ) exp ( −i p. r ) d 3 p

× ∫ aα p exp −iE p t exp i p ⋅ r

=

∫ { Aµ ( p ′ ) aα ( p ) exp (−i ( E ( p ) + E0 ( p ′ )) t ) exp (i ( p + p ′, r ))

)) ( ( ( ) ( )) ) ( ( + A ( p ′ ) a ( p ) exp ( −i ( E ( p ) − E ( p ′ )) t ) exp (i ( p − p ′, r ))  = + A ( p ′ ) b ( p) exp ( −i ( E ( p ) + E ( p ′ )) t ) exp ( −i ( p − p ′, r ))} d pd p′ + Aµ ( p ′ ) bα ( p) exp i E p − E0 p ′ t exp −i p + p ′, r µ

α

0

µ

α

0

3

(1) Writing ψ α ( X ) =

3

∫ ψ α (k ) exp (−i k. X ) d (1)

4

k

Classical Robotics & Quantum Stochastics

81

0 where k. X = kµ Xµ = kµXµ = k t − ( k , r )

we get from (5)

∫4 (kµ γ

µ

(

)

 (1) ( k ) exp ( −i k . X ) d 4 k = − γ µ Aµ( X ) ψ (0) ( X ) −m ψ



= where

( )

∫ −γ

µ

X

µ

)

(t, p) exp (i p. r ) d 3 p

χµ t , p

=

∫ Aµ ( p ′ ) a ( p − p ′ ) exp (−i ( E ( p + p ′ ) + E0 ( p ′ )) t ) d

3

p′

+ 3 other similar terms. The correlation of correlation in {Aµ}.

{ψ } can be computed from this expression in terms of (1) 0

Reference: Rohit Rana, temperature measurement using Klein -Gordon field, M. Tech Thesis, NSIT, 2014, Under supervision of H. Parthasarathy and T. K. Rawat. 

[31] Quantum Image Processing on a C∞- Manifold: The image field on the manifold is diffused using the Diffusion equation. 1 1  ∂u (t , x) 1 −  = g 2  g 2 g αβu, α (t , X ) , β ∂t 2 1

=

1

1 αβ 1 − g u , αβ + g 2 ( g 2 g αβ ), βu , α 2 2

We can express thus as u, t(t , x) = D1αβ ( X )u ,αβ (t , X ) + D2α ( X )u, α (t , X ) where

D1αβ ( X ) =

1 αβ g (X ) , 2

1 1 1 − 2  2 αβ  = g g g ,β 2 1 αβ 1 = g , β + g , β g αβ g −1 2 4 The image field at time t is then u(t, X), X∈n.

D2α ( X )

To quantize this image field, we must first derive (1) from an action principle.

(1)

82

Stochastics, Control & Robotics

Diffusion of an edge: An edge is a line in 2 having the equation lX + mY = C The corresponding intensity pattern is f(X, Y) = δ(lX + mY – C) If the image field consists of only edges, then the intensity field in f(X, Y) =

∑ δ (lα X + mαY − cα ) α

Under the diffusion equation defined by ∂u (t , x, y ) 1  ∂ 2u ∂ 2u  = D 2 + 2  , 2  ∂x ∂t ∂y  u(0, X, Y) = f(X, Y), this field of edges evolves to u(t, X, Y) = ( Kt * f ) ( X , Y ) ≡

∫ Kt ( X − X ′, Y − Y ′ ) f ( X ′, Y ′ ) dX ′dY ′ 

[32] Belavkin Contd. (Quantum Non-Linear Filtering): ν µ −  X (t ) δ µν  d Aµ d X = F   ν

(

)

 − X ⊗ δ d A ≡ F  − X ⊗ δ . Then  = F C

Let

(

d X

*

 A ≡ C  µν d Aµν = Cd

d X

*

 µ*  ν*  bν  µ b  = C ν d Aµ = C µ d Aν = C d A

d  X * X

) = d X *.X + X * d X + d X * .d X

b   d A + C  b Cd  A.d A X d A +  X *C = C Note:

dA−j dAk+ = δ kj dt dA+j dA−k = δ +− dAkj = 0 dAkj dAml = δ mj dAkl . So dAνµ dAσρ = δ µσ dAνρ    +C  C  , t X+ X *C = d A  C   b

b

)

Classical Robotics & Quantum Stochastics

83

Now,

(

)

(

b −  − b   +C bC  ≡ F X ⊗δ +  X * ⊗δ F X ⊗ δ  X ⊗δ C X+ X *C  

(

b −  − + F X *⊗δ F X ⊗δ  

(

)

)

)

b F  X * X ⊗δ + F = − 

So

(

d  X * X

) =  F

b

− F X * X ⊗ δ d A  µ

v b F − ≡ F X * X ⊗ δ d Aµ  v

  b  µ   µ   v =   F F  − X * X δ v  d Aµ   v It ⇒

(

b F  = I, then  X * X (t) = I F

X d  X *

) = 0 ⇒ X (t )

is unitary ∀t if it is unitary at t = 0.



[33] Remarks on Theorems from V. Kac. Infinite Dimensional Lic Algebra. P.87: 1   tw(α ) (λ ) = λ + λ, k w(α ) −  ( λ / w(α )) + | α |2 λ, k  δ   2 wtα w−1 (λ ) = λ + w−1 (λ ), k w(α)

((

)

− w−1 (λ ) | α +

1 | α |2 < w−1 (λ ) , K >| w (δ ) 2

−1 w (δ) = δ, w (λ ), k = λ, w(k) = λ, k

(w so

−1

) (

)

( λ ) | α = λ w (α )

wtα w−1 (λ ) = λ + λ, k w(α) 1   2  ( λ | w (α )) + | α | λ, k  δ = tw (α ) (λ ) . 2 P249 n ∈ N Z ⇒ z n = (α, β, u ), n, v

(α, β, u )(v) = tβ (v) + 2πiα + (u − iπ(α | β)) δ

84

Stochastics, Control & Robotics

 1  tρ (λ ) (n ⋅v) = (λ + (λ | δ )ρ + | n, v ) −   λ ρ + (ρ | ρ) (λ | δ) (δ | n ⋅ v)   2

( ) = (tρ − β (λ ) | v ) + ((λ | δ ) + (λ | δ)(ρ | δ ))

= tρ (λ ) | tβ (v) + 2πiα + (u − iπ(α | β))δ

(u − iπ(α | β)) + 2πi (λ | α ) (λ | δ) = k ∈Z + ρ2 (ρ | δ) = 0.

So,

(

)

tρ (λ ) (n ⋅v) = tρ − β (λ ) | v + k (u − iπ(α | β)) δ (n · v) = (δ|n · v) = (δ|tβ (v)) = (δ|v). 

[34] Quantum Mutual Information: [Reference: Remark from Mark, M. Wilde "Quantum Information Theory. (P. 303.)] Holevo information: { px ( x), ρ x } ensemble prepared by Alice. ρ xA′ =

∑ px ( x) x

x

X

⊗ ρ xA′

x

X

⊗ ρBx

x

N

A′

→ B = channel.

ρ XB =

∑ px ( x) x x

where

( )

ρBx = N A′→ B ρ xA′

I ( X ; B )ρ X (N) = max XA′ ρ

  I ( X ; B)ρ = − ∑ px ( x)H ρBx + H  ∑ px ( x)ρBx   x 

( )

Note:

( )

XB rB = Trx ρ =

H (B | X ) =

∑ px ( x)ρBx x

∑ px ( x)ρBx x

Classical Robotics & Quantum Stochastics

Hence

85

I ( X : B)ρ = H ( B) − H ( B | X ) = H (ρB) − H ( B | X )   = H  ∑ px (u )ρBx  − ∑ px ( x) H ρBx  X  x

( )

P. 307 Let Let

γ

∑ p( y | x) p(x) x

x1X ⊗ y y ⊗ ρ xA, y = σ

∑ p ( y | x) p(x) x

x

x, y

Then Tryσ = =

X

⊗ ρ xA, y

x, y

∑ p ( x) x

x

X

⊗ ρ xA, y

x

Where

ρ xA

=

∑ p( y | x)ρxA, y y

\

I ( X ; A)σ = H

(∑ p( x) ρ ) − ∑ p( x) H (ρ ) = H ( X : A) A x

A x

ρ

Data processing inequality; Let X → Y → Z be a Markov chain. Then I (X; Y) ≥ I (X; Z) I ( X ; Y ) = H (Y ) − H (Y | X ) , I ( X ; Z ) = H (Z ) − H (Z | X ) H ( Z X ) ≥ H(Z | Y, X) = H(Z | Y)(by Markov property), So.

I ( X ; Z ) ≤ H (Z ) − H (Z | Y )

If the Markov chain X → Y → Z is stationary i.e, pZ | Y = pY | X and pX = pY = pZ. the H (Z) – H (Z|Y) = H (Y) – H (Y|X) and we are done. Let

r=

∑ p ( x) x

x ⊗ ρx

x

Let Then

rx = r=

∑ p( y | x) yx

yx

y

∑ p ( x) p ( y | x) x

x ⊗ yx

yx

xy

Since

x ′ | x = δ x ′ x , δ yx , y′x= δ yx , y x′ ∀x

it flows that (1) is a spectral representation. Hence,

H (r) = − ∑ p( x) p( y | x)log( p( x) p( y | x)) x, y

(1)

86

Stochastics, Control & Robotics

=

− ∑ p ( x ) log p ( x ) − ∑ p ( y, x ) log p ( y | x ) x, y

x

= H (X) + H (y|X) = H (Y, X) Where p (y, x) = p (y|x) p (x) i the joint probability distribution of r.v’s (Y, X). 

[35] Scattering Theory Applied to Quantum Gate Design:

(

)

ψ i ( r ) = C.exp jk  ni ⋅ r  Incident projectile state. E =  2 k 2 / 2m  Projectile energy.  ni  Incident projectile direction.

− 2 ∆ψ i = Eyi is clearly satisfied. 2m

Final state (scattered) ψ f ( r ) = ψ i ( r ) + εψ s1 (r ) + ε 2 ψ s2 (r ) + O(ε3 ) . Interaction potential energy  εV ( r ) .   2 ∆ + εV  yf = Eyf −  2m    2 ∆ + εV   ψ f = εψ s1 + ε 2 ψ s2  −  2m 

or

= E [yi = e ys1 + e2ys2] + O (e3)

Coeffi. of e0, e1, e2, respectively give − 2 ∆ s1 + V ψ i = Eys1, 2m − 2 ∆ψ s 2 + V ψ s1 = Eys2 2m Thus,

(∆ + k ) ψ

s1

=

(∆ + k ) ψ

s2

=

2

2

2m 2 2m 2

ψ s1 ( r ) = − =

V ψi , V ψ s1 m

2π

2

∫

exp ( jk | r − r ′ |) V ( r ′ ) ψ i ( r ′ ) d 3r′ | r − r′ |

∫ Gk ( r − r ′)V ( r ′) ψ i ( r ′) d

3

r′

Classical Robotics & Quantum Stochastics

Gk ( r ) = ψ s2 ( r ) = =

−m 2π 2 r

87

exp( jkr)

∫ Gk ( r − r ′)V ( r ′) ψ i ( r ′) d

3

r′

∫ Gk ( r − r ′)V ( r ′) Gk ( r ′ − r ′′)V ( r ′′) ψ i ( r ′′) d

ψ sm ( r ) = (GkV )m ψ i , m = 1, 2...

Formally,

yf = ψ i + exact solution.

∞

∑ ε m (GkV )m ψ i

m =1

( )

2 3 Let ψ ∈L  . Then

|| Gk V ψ ||2 =

∫ d r ∫ Gk ( r − r ′)V ( r ′) ψ( r ′) d 3

3

r′

2

≤  ∫ | Gk ( r − r ′ )V (r ′ ) |2 d 3rd 3r′   | ψ ( r ′ ) |2 d 3r ′  ∫  Hence || GkV || < ∞ iff

∫ Gk ( r − r ′)

2

V ( r ′ ) 2 d 3rd 3r′ < ∞

iff ∫ | V ( r ′ ) |2 d 3r ′ < ∞ and ∫ | Gk ( r ) |2 d 3r < ∞ The latter in false since d 3r

∫

=

∞

∫0 4πdr

=∞ r with box normalization however, ||GkV||this can be made finite. 2

Now formally, yf = ψ i +

εGkV ψi I − εGkV

= [ I − ε GkV ] ψ i −1

Now formally, Gk =

2m 

2

( ∆ + k 2 ) −1

 2  2k 2  ∆− = − − 2m   2m −1 = − (H0 − E)

−1

3

r ′d 3r ′′

88

So,

Stochastics, Control & Robotics

[ I − ε GkV ]−1

−1 = I + ε (H0 − E ) V   

−1

−1 = [ H 0 − E + εV ] [ H 0 − E ] −1 = [ H1 − E ] [ H 0 − E ]

H1 = H 0 + εV

where This formula implies

−1 = [ H1 − E ] [ H 0 − E ] ψ i

ψf

or

= [H0 − E] ψi

[ H1 − E ] ψ f

Both sides one zero and hence no information is gained from then equation. However, we can also write ∞

∑ ε m (GkV )

m

m =1

= Gk

∞

∑ ε m (V Gk )

m −1

−1 V = ε Gk [ I − ε V Gk ] V

m =1

I − ε V Gk = I + ε V ( H 0 − E ) −1

and

−1 = [ H 0 + εV − E ] [ H 0 − E ] −1 = [ H1 − E ] [ H 0 − E ]

so

∞

∑ ε m (GkV )

m

m =1

thus,

ψf

= ε Gk ([ H1 − E ] [ H 0 − E ]) V −1

−1 = εGk ( H 0 − E ) ( H1 − E ) V

= − ε ( H1 − E ) −1V

So, ψf

(

)

−1 = I − E ( H1 − E ) V ψ i

Check: Multiply both sides of this equation by H1 – E to get

( H1 − E ) ψ i

=

( H1 − E − εV ) ψ i

= (H0 − E) ψi and both sides evaluate to zero. We write H1 − E = ( H 0 − E + ε V ) −1

( = (I + ε (H

) V)

= I + ε ( H 0 − E ) −1V 0

− E ) −1

−1

( H 0 − E ) −1

−1

( H 0 − E ) −1 + O(ε 2 )

Classical Robotics & Quantum Stochastics

89

and hence ψf

= ψ i − ε ( H 0 − E )V ψ i + ε 2 ( H 0 − E ) −1 V ( H 0 − E ) −1V ψ i + O(ε3 ) . 2 2 3 =  I + ε GkV + ε (GkV )  ψ i + O(ε ) .  

ψ i  output state evolving according to H0. Alternate way of deriving the operator ψ 0 → ψ f H0 white ψ f

ψ 0 evolves according to

evolves according to H1. As t → ∞ ψ f

t

ψ0

t

since as t → ∞

the effect of the scaterer interaction becomes negligible. Note that ψ 0 , the out put state, is evolving as a force partible while ψ f

the scattered state, evolves

according to H1. Thus lim e −it H , ψ f − eit H 0 ψ 0

=0

t→∞

So

ψf

− it H 0 , − it H 0 e ψ0 = lim e t→∞

Now d it H , it H 0 it H , ( H1 − H 0 ) e −it H 0 = i ε eit H ,V e −it H 0 e e = ie dt So Let Then

∞

it H −it H 0 Ω (t) = lim e 1 e = I + i ε ∫ eit H , Ve −it H1 dt t→∞

H0 =

0

+∞

∫− ∞ λ dE(λ)

be the spectral rep’n of H0.

∞

it( H −λ ) Ω (t) = I + i ε ∫0 dt ∫ e 1 V dE(λ ) 

∞

∆ I + i ε lim ∫ dt e −it δ→0 0

( H1 − λ + iδ )

V dE (λ )

∞

I − ε ∫ ( H1 − λ + iδ ) −1V dE(λ ) . 0

i.e.

ψf

∞

−1 = ψ 0 − ε ∫0 ( H1 − λ + iδ ) V dE(λ ) ψ 0

If the output state ψ 0 has energy E, then dE(λ ) ψ i = δ E , λ ψ i ψf

= ψ 0 − ε ( H1 − E + iδ ) −1 V ψ 0

and we get

90

Stochastics, Control & Robotics

Now consider the input state ψ i . At time t → −∞ ψ i evolves according to H0 while ψ f

evolves according to H1. In the remote part, these two states must

coincide. Thus lim e −it H1 ψ f − e −itH 0 ψ i

t→−∞

or

ψf

where

it H − it H 0 Ω (–) = lim e 0 e

= Ω( −) ψ i t→−∞

Ω (–) = I − ∫

d

0

− ∞ dt

= I − i∫

0

−∞

= I − i ε∫ = I −iε =

(e

itH1 −itH 0 e

) dt

eitH1 ( H1 − H 0 ) e −it H 0 dt

0

−∞

So

=0

eit H1 V e −it H 0 dt it H −λ e ( 1 )V dE(λ ) dt

∫

 × ( − ∞ , 0)

I − i ε lim

δ→ 0 +

∫

eit ( H1 − λ − iδ )V dE (λ ) dt

 × ( − ∞ , 0)

= I − i ε ∫ ( H1 − λ − iδ ) −1 V dE(λ ) 

ψf

= ψ i − ε ( H1 − E − iδ ) −1 V ψ i

Ω + ( E ) = I − ε ( H1 − E + iδ ) −1V

Ω − ( E ) = I − ε ( H1 − E − iδ ) −1V ψf

= Ω+ ( E ) ψ 0 ,

ψf

= Ω – (E) ψ i

Born scattering is an first order approximation of ψ f = Ω− ( E ) ψ i Scattering matrix/operator: Since Ω + (E) is an isometry, we have Ω+ ( E ) * Ω+ ( E ) = I

and hence, ψ0

= Ω+ ( E ) * ψ f

Classical Robotics & Quantum Stochastics

91

= Ω+ ( E ) *Ω− ( E ) ψ i S (E) = Ω + ( E ) * Ω − ( E ) is the scattering matrix at energy E. ψ0

= S (E) ψi

S (E) is unitary on the space of definite energy states i.e. states defined on the unit sphere i.e. L2 (S2) S (E) =  I − ε ( H1 − E + iδ ) −1 V   

*

×  I − ε ( H1 − E + iδ ) −1V  = I − ε V ( H 0 − E − iδ ) −1 − ε ( H 0 − E − iδ ) −1V + O(ε 2 )

{

}

−1 −1 2 = I − ε V ( H 0 − E − iδ ) ( H 0 − E − i δ ) V + O ( ε )

ψ0

Thus,

(H0 − E) ψi

and

−1 2 = S ( E ) ψ i = ψ i − ε( H 0 − E ) V ψ i + O(ε )

= 0−

Note that ( H 0 − E − iδ ) −1V ψ i So ψ0

iε V ψi δ

−1 = ( − iδ ) ψ i =

(

i ψi δ

)

−1 = I − iδ ε V ψ i

− ε ( H 0 − E ) −1V ψ i + O(ε 2 ) i V ψ i in ψ 0 shown that a blind application of perturbation δ theory w.r.t. e is incorrect. However, if we still persist,then The infinite factor

ψ 0 | ψ i = ψ 0 | δ( E ) | ψ i ψi | ψi −

iε ψi | V | ψi δ

( )

− ε ψi (H 0 − E ) V ψi + O ε2 −1

Where the inner product is w.r.t. L2 (S2) in momentum space. More generally, ψ i 2 | S ( E) | ψ i1 = ψ i 2 | ψ i1 −

iε ψ i 2 | V | ψ i1 δ

− ε ψ i 2 ( H 0 − E ) −1V ψ i1 + O(ε 2 ) Where ψ i1 and ψ i2 are two incident states.

92

Stochastics, Control & Robotics

and

ψ i1 ( r ) = Ci exp( jk  ni r ) ψ ( r ) = C exp( jk  n2 r )

Then

 (k ) = ψ i1

Let

i2

2

1 (2π)3/ 2

∫ ψ i1( r ) exp(− ik ′r ) d

3

r

3/ 2 n1 ) = (2π) C1δ ( k ′ − k 

and likewise

 ( r ) = (2π)3/ 2 C δ ( k ′ − k  ψ n2 ) i2 2 ≡ (2π)3 C 2C1 ∫ δ ( k ′ − k  n 2 )δ ( k ′ − k  n1 )d Ω ( k ′ )

ψ i 2 | ψ i1

If we instent take (1) over where of 3, then formally ψ i 2 | ψ i1

=

∫ ψ i 2 ( r ) ψ i1( r ) d

=

∫ ψ i 2 (k ′) ψ i1(k ′) d

=

(2π)3/ 2 C 2C1   δ n 2 − n1 , k

ψ i 2 | V | ψ i1 =

3

r 3

(

k ′ = (2π)3 C 2 C1 δ k ( n2 −  n1 )

(

)

)

∫ ψ i2 ( r )V ( r ) ψ i1( r ) d

3

r

(

)

n2 −  n1, r ) d 3r = C 2C1 ∫ V ( r )exp − jk (

(

3/ 2 n2 −  n1 ) = (2π) C 2C1 V k (

By the Riemann-Lebesgue Lemma,

(

)

)

n2 −  n1 ) = 0 lim V k (

k→∞

So for high energies, ψ i 2 | V | ψ i1 may be neglected. Now, the Born term is ψ i 2 | (H 0 − E ) −1 | ψ i1

= =

−2m h

2

m 2π

2

ψ i 2 | (∆ + k 2 ) −1V | ψ i1

∫

exp( jk | r − r ′ |) ψ i 2 ( r )V ( r ′ ) ψ i1 ( r ′ ) d 3r d 3r ′ | r − r′ |

Before evaluating this, we make a remark: yi1 and yi2 are concentrated at r → ∞ where V ( r ) is concentrated around the origin. So ψ i1 yi2 has zero overlap with

V which is another justification for taking the overlap integral ψ i 2 | V | ψ i1 = 0. With the assumption, ψ i 2 | S ( E) | ψ i1 ≡ ψ i 2 | S (k) | ψ i1

Classical Robotics & Quantum Stochastics

≈

93

(2π)3/ 2 C 2C1   δ (n 2 − n1 ) k

 m ε  exp( jk | r − r ′ |) − C 2C1 V ( r ′ )  2π 2  ∫ |r−r|

( (

) (

n1, r ′ −  n2 , r ′ × exp jk 

))) d r ′ d r 3

3

Now if we assume that r > > r′ i.e. the final state is concentrated at ∞ while the potential V (r′) is concentrated near the origin, then we can make the field approximation to get exp ( jk | r − r ′|) e jkr exp(− jk( r − r ′ )) ≈ r | r − r ′| and then ψ i 2 | S ( E) | ψ i1 ≈

(2π)3/ 2 C 2C1   δ n 2 − n1 k

(

−

mε 2π 2

{ ( ) ( ) (

r , r ′ + n2 , r − n1, r ′ C 2C1 ∫ exp − jk 

(2π)3/ 2 C 2C1   exp( jkr) δ n 2 − n1 V ( r ′ ) d 3r ′ d 3r = k r

(

−

m ε C 2C1 2π

)

2

)

((

(2π)3/ 2 ∫ V k  r − n1

))

( ( ( ))) d r

n2 , r exp jk r − 

3

r

If we don’t make the far-field approximation, then ψ i 2 | S (k) | ψ i1 ≈

(2π)3/ 2 C 2C1   δ n 2 − n1 k

(

)

exp ( jk | r − r ′ |)  mε  − C 2C1 ∫  2π 2  | r − r′ | 

[36] Quantization of the Simple Exclusion Model:

ZN = {0, 1, 2, ..., N – 1}. |X〉 = X o X1...X N −1 >∈(C 2 )⊗ X R ∈{0, 1} .

N

≅C 2

N

)}

94

Stochastics, Control & Robotics N a ( X , r, s), X = ( X 0, X1...X N −1 ) ∈{0, 1} 0 ≠ r, s ≠ N – 1

Transition probability amplitudes. − i X r (1 − X s ) a( X , r , s )  transition probability amplitude per unit time from X → X

(r , s)

Where |X〉(r, s) is obtained from | X 〉 by interchanging Xr and Xs. So |yt +δt〉 = ψ t − iδt or

i

d ψt dt

=

∑

∑

X r (1 − X s ) a( X , r , s ) X

( r , s)

X | ψt

X , r, s

X r (1 − X s ) a( X , r , s ) X

( r , s)

X | ψt

X r (1 − X s ) a ( X , r , s ) X

( r , s)

X

X , r, s

= H ψt H=

∑

X , r, s

H* =

∑

X r (1 − X s ) X

X

∑

X r (1 − X r ) X

( r , s)

( r , s)

a( X , r , s)

X rs

=

X a( X ( r , s) , r , s )

X rs

For H* = H, we requires X r (1 − X s ) a( X , r , s ) X

( r , s)

+ X s (1 − X r ) a( X , s, r ) X

X

( s , r)

= X s (1 − X r ) a ( X ( r , s ) , r , s ) X

(r , s)

+ X r (1 − X s ) a ( X ( r , s) , r , s ) X ( r , s)

X

( s , r)

X X

( s , r)

, X ( r , s) = X ( s , r) and hence the condition for = X ∀ r ≠ s. Now X Hermitian of H reduces to X r (1 − X s ) a( X , r , s ) + X s (1 − X s ) a( X , s, r ) ( r , s) , s, r ) + X s (1 − X r ) a( X ( r , s ) , r , s ) ∀ r ≠ s ∀ X = X r (1 − X s ) a( X

This can be achieved, for example, by taking a ( X , r, s) = a (r , s) ∈ = a (s, r) independent of X . Another way to achieve this is to take a ( X , r , s ) = a (r , s ) a (s, r )∀ r ≠ s i.e. ((a (r , s)))0 ≠ r , s ≠ N −1 is a Hermitian. N × N matrix independent of X.

Classical Robotics & Quantum Stochastics

95

A still more general way is to choose a such that

(

)

( r , s) , s, r ∀ X ∈{0, 1}N , a ≤ r , s ≤ N −1 . a(X , r, s) = a X

Consider now −

d ψt dt

= H ψt =

∑

X r (1 − X s ) a(r , s ) X

( r , s)

X | ψt

r , s, X

a (r, s) = a ( s, r) . Thus, i

d Y | ψt dt

=

∑

X r (1 − X s ) a(r , s )δ Y − X ( r , s)  < X ψ t  

∑

X r (1 − X s ) a(r , s )δ Y ( r , s) − X  < X ψ t  

r , s, X

=

r , s, X

=

∑ X s (1 − Yr ) a(r , s) < Y (r , s) ψ t r, s

Let X | ψ t

= ψ t ( X ) . Then we can express the above eqn. as i

d ψt ( X ) = dt

∑ ψ t ( X (r , s) ) X s (1 − X r ) a(r , s)

(1)

r, s

Probability density and probability current: −i

d ψi ( X ) = dt

∑ ψ t ( X (r , s ) ) X s (1 − X r ) a(s, r)(a(s, r ) = a(r , s))

(2)

r, s

From (1) and (2), d ψt ( X ) d ψt ( X ) d 2 ψt ( X ) = ψt ( X ) + ψt ( X ) dt dt dt or, d ( r , s) 2 ) X s (1 − X r ) a(r , s) i ψt ( X ) = ψt ( X ) ∑ ψt ( X dt r, s

(

)

− ψ t ( X )∑ ψ t X ( r , s) X s (1 − X r ) a (r , s) r, s

=

∑ X s (1 − X r ) 2i I m (a (r , s) ψ t ( X ) ψ t ( X (r , s) )) r, s

Suppose a (r, r + 1) = a (r + 1, r) = α ∈  and a (r, s) = 0 for |r – s| > 1. Then we get

(

d 2 ψ t ( X ) = 2α ∑ X r +1 (1 − X r ) I m ψ t ( X ) ψ t( r , r +1) ( X ) dt r

(

)

+ 2α ∑ X r +1 (1 − X r ) I m ψ t ( X ) ψ t( r −1, r ) ( X ) r

)

96

Stochastics, Control & Robotics

(

( r , r+1) (X ) = 2α ∑ ( X r + X r +1 − 2 X r X r +1 ) I m ψ t ( X ) ψ t r

(

( r , r +1) (X ) = 2α ∑ I m ψ t ( X ) ψ t r:

)

)

(Xr, Xr + 1) = (1, 0) or (0, 1) 

[37] Quantum Version of Nearest Neighbour Interactions: ψ t ( X1, ..., X N )  state of the N-particle system at time t. The classical equation dX i (t ) = ( ψ ( X i −1 (t )) − 2ψ ( X i (t)) + ψ( X i +1 (t )) ) dt + dBi −1, i (t ) − dBi , i +1 (t ) (*) Needs to be quantized. We write X1,..., X N | ψ t

= ψ t ( X1, ..., X N )

One method to start quantity appears to be ψ t + dt ( X ) = αψ t ( X − ϕ( X ) dt − σ( X ) d B (t)) where |α| = 1 α ∈C . We write α = 1 + iβ dt so

|α| =

1+ β 2 dt 2 = 1 + O (dt2), β = β (X).

Comments and remarks on Volume 4. of collected papers of S.R.S. Varadhan: dXi (t) = dZi – 1, i (t) – dZi , i + 1 (t)

dZi , i + 1 (t) = (y (Xi (t)) – y (Xi +1) (t)) dt + dBi, i + 1 (t), i ∈ Z. XN (t) =

∑

i ∈Z N

 i J   X i (t )  N

Forward Kolmogorov operator:

(

)

dXi (t) = ψ ( X i −1 (t )) − 2ψ ( X i (t)) + ψ( X i +1 (t )) dt + dBi −1, i (t ) − dBi , i +1 (t ) . ∂ 1  ∂ ∂  + ∑ − L = ∑ ( ψ ( X i −1 ) − 2ψ ( X i ) + ψ( X i +1 )) ∂X i 2 i  ∂X i ∂X i+1  i Note: d f ( X 0 , ..., X N −1 ) =

∂f 1 ∂2 f dX + ∑ ∂X i 2 ∑ ∂X ∂X dX i dX j i i i i i, j

dX i dX i +1 = − (d Bi , i +1 )2 = – dt, dX i −1 dX i = – (d Bi – 1, i)2 = – dt, dXi2 = 2dt

2

Classical Robotics & Quantum Stochastics

97

A stationary density p ( X ) should L*p = 0. Now let p ( X ) = C ∏ exp(λX i − φ ( X i )) i

£*p = − ∑ i

((

) )

∂ ψ ( X i −1 ) − 2ψ ( X i ) + ψ( X i +1 ) p ∂X i 2

∂  1  ∂ + ∑ −  p 2 i  ∂X i ∂X i +1 

(

= 2 ∑ ψ ′ ( X i ) p − ∑ ψ ( X i −1 ) − 2τ ( X i ) + ψ ( X i + 1 ) i

i

) ∂X∂pi

2

1  ∂ ∂  + ∑ −  p 2 i  ∂X i ∂X i +1  ∂p = (λ − φ′ ( X i )) p . ∂X i  ∂ ∂  −   p = (φ′ ( X i +1 ) − φ′ ( X i )) p  ∂X i ∂X i +1  2

 ∂ ∂  −   p = φ′( X i +1 ) − φ′( X i )  ∂X i ∂X i +1 

(

)

2

(

)

p − φ′′( X i +1 ) + φ′′ ( X i ) p

So L*p/p = 2∑ ψ ′( X i ) − ∑ (λ − φ′ ( X i ))(ψ ( X i −1 ) − 2ψ ( X i ) + ψ ( X i +1 )) i

i

+∑ i

1 1 (φ′( X i +1) − φ′( X i ))2 − 2 ∑ (φ′′( X i +1) + φ′′( X i )) 2 i

For this to be zero, we require 2ψ ′( X i ) − φ′( X i ) ψ ( X i ) + φ12 ( X i ) − φ′′( X i ) = 0

and

φ′( X i ) ψ ( X i −1 ) + (φ′( X i −1 )) ψ ( X i ) − φ′ ( X i −1 ) φ′ ( X i ) = 0

Let y (X) = αφ′ ( X ) . Then these conditions reduce to 2αφ′′( X ) + ( −2α − φ′ ( X ))φ′ ( X ) + φ′ 2 ( X ) − φ′′( X ) = 0 There are satisfied provided (2α − 1) φ′ ( X ) φ′ (Y ) = 0 Thus α =

1 . It follows that 2

(2α − 1) φ′′( X )

=

(2α − 1) φ′ 2 ( X )

and

98

Stochastics, Control & Robotics

p (X) = C ⋅ ∏ exp(λX i − φ ( X i )) i

is a stationary density provided y (X) =

1 φ′( X ) , 2

X

i.e.

f (X) = 2 ∫ ψ (ξ) dξ . o

Now,

dXN (t) =

 i

∑ J  N  dX i (t ) i

N −1

=

 i

∑ J  N  (Ψ( X i −1(t )) − 2ψ ( X i (t )) + ψ ( X i +1(t )) dt

i=0

+

N −1

 i

∑ J  N  (dBi −1, i − dBi, i +1(t ))

i=0

=

  i + 1  i −1   i  ψ( X i (t )) dt  − 2J   + J  N  N N  

∑  J  i

 i + ∑ J   (dBi −1, i (t ) − dBi , i +1 )  N i ≈

1 N

2

 i

 i

∑ J ′′  N  ψ( X i (t )) + ∑ J  N  (dBi −1, i − dBi, i +1) i

i

   i   ∑ J   ( Bi −1, i (t ) − Bi , i +1 (t))  i  N 

2

    i  i −1  Bi −1, i (t ) =  ∑ J   − J     N   i   N    i  i −1  = t ∑ J   − J   N      N i ≈

2

2

t

2

t  i J ′   ≈ ∫ J ′ (θ) 2 d θ → 0 N → ∞ . 2∑  N No N i t

So for N → ∞, dXN (t) ≈

 i

t N

2

∑ J ′′  N  ψ( X i (t )) dt i

Now consider r.v’s ξ1, ξ2, ... which are iid and have logarithmic moment generating function

Classical Robotics & Quantum Stochastics

(

99

)

Λ (λ) = log  eλξ1 . Then  ξ + ... + ξ n    ψ (ξ1 ) 1 = a = n   Where p (ξ1 |n, a) is the pdf of ξ1 given

Now,

∫ p (ξ1 | n, a) ψ (ξ1) d ξ1

ξ1 + ... + ξ n =a n

  ξ + ... + ξ n = a | ξ1 p(ξ1 ) p 1   n p (ξ1 | n, a) =   ξ + ... + ξ n = a p 1   n  ξ + ... + ξ n na − ξ1  p 1 ξ p(ξ1 ) =  n −1 n − 1 1 =   ξ + ... + ξ n = a p 1   n  ξ   exp  −(n − 1) I  a − 1   p(ξ1 )   n − 1  ≈C exp(−n I (a)) ≈ C ⋅ exp( I (a) + I ′(a) ξ1 ) p(ξ1 )

Where I (a) = supλ (λa – Λ (λ)) by large deviation theory. Thus a n → ∞  ξ + ... + ξ n  E  ψ (ξ1 ) 1 = a  ≈ C ⋅ ∫ exp( I (a) + I ′(a) ξ) p(ξ) ψ (ξ) dξ n   Note

∫ exp(I ′(a)ξ) p(ξ) d ξ Now where Also where

= exp(Λ( I ′ (a)))

I (a) = sup(λa − Λ(λ )) = λ o a − Λ(λ o ) λ

a = Λ′(λ o ) λ (λ) = sup(λa − I (a)) = λao − I (ao ) a

λ = I ′(ao ) . Thus,

Λ( I ′ (ao )) = I ′(ao ) ao − I (ao )

So, ∫ exp(I ′(a)ξ) p(ξ) d ξ = exp ( I ′ (a) a − I (a) ) Thus,

∫ exp(I (a) + I ′(a)(ξ)) d ξ

= exp(aI ′(a)) .

(n → ∞)

100

Stochastics, Control & Robotics

So,

C = exp (– a I′ (a))

 ξ + ... + ξ n  = a and lim   ψ (ξ1 ) 1 x→∞  n  =

∫

ψ (ξ)exp( I ′(a)ξ) p(ξ) d ξ exp(a I ′(a) − I (a))

=

∫

ψ (ξ)exp(I ′ (a )ξ) p(ξ) d ξ exp(Λ( I ′ (a)))

Now if a N → ∞ the Xi (t)′ s converge to the stationary Gibbs distribution C ⋅ ∏ exp(σ X i − φ( X i )) = ∏ p ( X i ) i

i

Then

 exp((λ + σ) X − φ( X )) dX  ∫ Λ (λ) = log    ∫ exp(σ X − φ( X )) dX 

We define

F (σ) = log ∫ exp(σ X − φ( X )) dX

and so

Λ (λ) = F (λ + σ) − F (σ) I (a) = sup(λa − F (λ + σ) + F (σ)) λ

= sup ((λ + σ ) a − F (λ + σ )) + F (σ) − σa) λ

= I F (a ) + F (σ) − σa. Where

IF (a) = sup(λa − F (λ )) .

Now,

I′ (a) = q satisfies

λ

Λ (q) = sup(θx − I ( x)) = qa – I (a) λ

At equilibrium,

 X + ... + X N  lim   ψ ( X1 ) 1 = a x→∞  N  =

∫ ψ (ξ) exp ( I ′ (a ) ξ + I (a ) − a I ′ (a ) + P (ξ) dξ)

=

∫ ψ (ξ)exp(I F′ (a)ξ − σξ + I F (a) + F (σ) − aI F′ (a) + σa) × exp(σξ − φ(ξ)) d ξ / exp( F (σ))

 ψ (ξ)exp(I F′ (a)ξ + I F (a) − φ(ξ)) d ξ  ∫  =  exp(a I F′ (a ))

Classical Robotics & Quantum Stochastics

101

 ψ (ξ)exp(I F′ (a)ξ − φ(ξ)) d ξ  ∫  ≡ Ψ  ( A) say. =  exp ( I F′ (a ) a − I F (a )) 

[38] Remarks and Comments on “Collected Papers of S.R.S Varadhan” Vol. 4, Particle System and Their Large Deviations: Quantization of the simple exclusion process: a (r , s ) X r (1 − X s ) X ∑ r≠s

H=

( r , s)

X

X

a (r, s) = a ( s, r ) ⋅ X | ψ t = ψ t ( X ) . i

d ψt ( X ) = (H ψt )( X ) dt = =

∑ a (r , s) Yr (1 − Ys ) δ  X − Y (r , s)  ψ t (Y )

r≠s

∑ a (r , s) X s (1 − X r ) ψ t ( X (r , s) )

r≠s

ψ t :{0, 1}N → C N → ∞

∑

X1 ,..., X N −1 ∈{0, 1}

| ψ t ( X ) |2 = pt (X0)

t distribution of particle at 0th site. i

 i d ψt ( X ) d d ψt ( X ) | ψ t ( X ) |2 = i ψt ( X ) − ψt ( X )   dt dt dt 

∑ a (r , s )X s (1 − X r ) ψ.( X (r ,s) ) ψ t ( X ) r ,s

(

− ∑ a(r , s )X s (1 − X r )ψ t ( X ) ψ t X ( r , s ) r ,s

More generally, i

{

)

a(r , s ) X s (1 − X r ) ψ t ( X ( r , s) ) ψ t (Y ) }=∑ r ,s

d ψ t ( X )ψ t (Y ) dt

− ∑ a(r , s ) Ys (1 − Yr ) ψ t ( X )ψ (Y ( r , s ) ) r ,s

So if rt (X, Y) denotes the density matrix of a mixed state, its Von-Neumann equation of evolution is

102

Stochastics, Control & Robotics

i

∂ ρt ( X , Y ) ∂t

=

∑ a (r , s) X s (1 − X r ) ρt ( X (r , s) , Y ) − a (s, r) Ys (1 − Yr ) ρt ( X , Y (r , s ) ) r,, s

Define

Hrs =

Then

Hrs* =

∑

a (r , s ) X s (1 − X r ) X

( r , s)

∑

a (r , s ) X s (1 − X r ) X

X

∑

a (r , s ) X r (1 − X s ) X

( r , s)

X ∈{0, 1}N

X ∈{0, 1}N

=

N

X r ≠ s.

( r , s)

X ∈{0, 1}

X = Hsr

This we can define * = H rs + H sr , 0 ≤ r < s ≤ N – 1 H rs = H rs + H rs

and then

∑ H rs

H= Let

∑

=

r≠s

a ≤ r < s ≤ N −1

H rs

j (Yo, X) = (Yo, X1 X2 ..., XN – 1) ≡ (Yo , X o )

Then X o = (X1, X2, ..., XN–1).

ρt ( X , (Yo , X o ) ) ∑ 

(1) = ρt ( X o , Yo )

Xo

is the density matrix of the zeroth particle. We write the density evolution as i

  ∂ρt ( X , Y ) =  ∑ H rs , ρt  ( X , Y ) ∂t  r < s 

Consider the term  H rs , ρt  ( X , Y ) for 0 < r < s ≤ N – 1 This equals

( ) ( ) − a( s, r ) Y (1 − Y ) ρ ( X , Y ) − a (r, s)Y (1 − Y ) ρ ( X , Y ) , Y ) − Y (1 − Y ) ρ ( X , Y )} = a (r , s ) { X (1 − X ) ρ ( X , Y ) − Y (1 − Y ) ρ ( X , Y ) + a ( s, r ) { X (1 − X ) ρ ( X } We have (o < r < s) ∑  H , ρ  ( X , (Y , X )) a(r , s ) X s (1 − X r ) ρt X ( r , s) , Y + a( s, r ) X r (1 − X s ) ρt X ( r , s) , Y r

s

s

Xo

( r , s)

t

r

r

(r , s)

t

s

rs

r

r

( r , s)

t

t

(r , s)

s

s

o

r

o

(r , s)

t

r

(r , s)

t

t

(r , s)

Classical Robotics & Quantum Stochastics

103

{

( r , s) , (Yo , X o )) = a(r , s )∑ X s (1 − X r ) ρt ( X Xo

}

− X r (1 − X s ) ρt ( X , (Yo , X o( r , s ) ))

{

+ a( s, r )∑ X s (1 − X s ) ρt ( X ( r , s) , (Yo , X o )) Xo

}

− X s (1 − X r ) ρt ( X , (Yo , X o( r , s ) )) = 0 Consider now the case r = 0 s > r. We’ve to evaluate





Xo s > o



 ∑ H os, ρt  ( X ,((Y, X 0 )) ∑    [ H 0 s , ρt ] ( X ,(Y0, X 0, )) ∑ 

For s > 0, we get

Xo

 ( 0, s) ,(Y 0, X 0 )) = a (0, s ) (1 − X 0 ) ∑ X s ρt ( X   X 0, 

 −Yo ∑ (1 − xs ) ρt X ,( X s , ( X1,..,Y0 , .., X N −1 )   Xo

(

5th position. This we can define

 [39] Quantum String Theory: [Reference: Lectures for EC-COE-2012 course.] µ ν µ ν L = a1gµν ( X ) X ,τ X ,τ + a2 gµν ( X ) X ,σ X ,σ

+ a3 gµν ( X ) X ,µτ X ,νσ First consider the special relativistic case: The gµn = ηµn. The Euler - Lagrange eqns.-are ∂r

∂L ∂X ,µτ

we find that

+ ∂σ

∂L

∂X ,µσ ∂L ∂X ,τµ ∂L µ ∂X ,σ

=

∂L ∂X µ

ν ν = 2a1 ηµν X ,τ + a3 zµν X ,σ

ν ν = 2a2 ηµν X ,σ + a3 zµν X , τ

)

104

Stochastics, Control & Robotics

∂L ∂X µ

= 0 so the string equations are 2a1 X ,µττ + a3 X ,µτσ + 2a2 X ,µσσ + a3 X ,µτσ = 0 a1 X ,µττ + a2 X ,µσσ + a3 X ,µτσ = 0

If a3 = 0 and a2 = – a1, we get the standard form of the string equations: X ,µττ − X ,µσσ = 0. Path integral approach to quantum string theory: 1 1 ( X1τ , X1τ ) − ( X , σ , X , σ ) 2 2 ∆ 1 zµν X ,µτ X ,ντ − X ,µσ X ,νσ 2

L ( X ,µτ , X ,µσ ) =

Let

{

X µ (τ, σ) =

}

∑ ξµn (τ)exp (i 2∏ nσ)

n∈Z

ξ µn (τ) = ξ µ−n( τ ) . 1

∫o

L dσ =

∑

n1 m ∈Z

1

1  (ξ m (τ), ξ m(τ))∫ exp (i 2π (n + m) σ ) dσ 2 o 1

2

+ Z π ∑ (ξ m (τ), ξ m (τ)) nm ∫ exp (i 2π (n + m) σ ) d σ o

=

∑  2 ( ξ n , ξ n ) − 2π 2n2 (ξn , ξ − n ) 1



n, m

(

= L ξ µn , ξ µn , n ∈Z , 0 ≤ µ ≤ d

{(

)

}

)

=

1 2  (ξ o ξ o ) + ∑ ξ n, ξn − 2π 2 n 2 (ξ n , ξ n ) 2 n ≥1

=

1   ξ 0 , ξ 0 + ∑ ξ Rn , ξ Rn + (ξ In , ξ In ) 2 n ≥1

(

)

{(

)

}

−2π 2 n 2 (ξ Rn , ξ Rn ) + (ξ In , ξ In ) where

µ µ ξnµ = ξ Rn + iξ In .

To quantize this, we first revise to quantize a 1-D harmonic oscillator L (q, q ) = 1 q 2 − 1 w02 q 2 0 ≤ t ≤ T 2 2

Classical Robotics & Quantum Stochastics

q (t) = + q0

w=

T

1 2 q dt = 2 ∫o

105

∞

∑ (qRn cos(n wt) + qIn sin (n wt ))1

n =1

2π . T q0, qRn, qIn ∈ , ∞

(

n 2 w2 2 2 qRn + qIn 4 n =1

∑

T

)

1 2 1 ∞ 2 1  t q d = ∑ qRn + qIn2 + 2 qo2T 2 ∫o 4 n =1

(

T

∫ L dt

\

=

o

)

−1 2 1  qo T +  ∑ n 2 w2 − w02 qR2 n + qI2n 2 4  n ≥1 

(

)(



) 

This expansion, however, does not take care of the boundary condition q (0) = a, q (T) = b. So we use the half wave Fourier expansion: q (t) =

∞

 nπt   + αt + β T 

∑ qn sin 

n =1

 b − β  b − a q (0) = a ⇒ β = a, q (T) b ⇒ L =  =   αT   T  T

2 ∫ q dt = o

T

2 ∫ q dt = o

 nπ 

∑  T 

2

×

n ≥1

T 2 qn + α 2T 2

T

T qn2 + ∫ (αt + β) 2 dt ∑ 2 n ≥1 o T

+

T   nπt  ∑ 2αqn ∫ t sin  T  dt + 2β∑ qn  nπ  (1 − (−1)n ) n ≥1 o T

T

T

−t 1 ∫ t sin (γt ) dt = γ cos ( γt ) + y 2 sin ( γt ) 0 o 0 =

−T cos(γT ) γ

=

T T2 ( −1) n+1 = ( −1) n+1 y nπ

nT     γ = T 

106

Stochastics, Control & Robotics

The Jacobian for the transformation {q (t) : 0 ≤ t ≤ T} to {qn}n ≥ 1 is not easy to evaluate since the map {q(t)}0 ≤ 0 ≤ T → {qn} is such that T

2 ∫ q (t ) dt = ||q||2 = o

Its follows that the map q(.) →

T

I ∞ 2 ∑ qn + ∫ (αt + β)2 dt − 2 ∑ 2α ηn qn 2 n =1 n ≥1 o

T {qn } 2

i.e. this map is not unitary We have ≈

−1 2 1 N qo + ∑ λ n qR + qI2n 2 4 h =1

T

∫ Ldt = − o

T

T 2 1 2 w0 ∑ qn2 − ∫ (αt + β ) dt 4 2 n ≥1 o

+ ∑ α ηn qn − n ≥1

(

+ π 2 4T

βT π

)∑ n q

 1 − ( −1)n) qn   n  n ≥1

∑ 

2 2 n

+ α 2T / 2

  T2 ( −1) n  = Where η n  π n   = −

(

)

T 2 wo ∑ qn2 + (αT + β)3 − β3 3α + α ∑ ηn qn 4 n ≥1 h ≥1

2 T  − β ∑   (1 − ( −1)n ) qn +  π 4T  ∑ n 2 qn2 + α 2T / 2    nπ  n ≥1 n ≥1

Let and

gn = αηn −

(

Where

)

ξ = (αT + β)3 − β3 3α λn =

Then

Tβ (1 − ( −1)n ) , nπ

ST [{qn }]a , b =

π 2 n 2 wo2T , n ≥1, − αT 2 1 ∑ λ n qn2 + ∑ γ n qn + δ 2 n ≥1 n ≥1

(

)

2 3 3 δ = α T / 2 + (αT + β) − β 3α

The path integral is then

Classical Robotics & Quantum Stochastics

107

(

KT (b|a) = C ∫ exp iST [{qn }]a , b ST [{qn }]a , b =

)∏ dq n ≥1

n

γ n2 1 2 + δ − λ ( q + γ ) ∑ 8λ ∑ n n n /2λn 2 n ≥1 n n ≥1

Now the path integral can be evaluated as an elementary Gaussian integral. The Jacobian: ∞  nπt  + αt + β, 0 ≤ b ≤ T . q (t) = ∑ qn sin   T  n =1 The term αt + β is independent of {qn} and hence can be neglected in the evaluation of the Jacobian. Consider t = lD*. The writing q(t ) = q (t) – αt – β, we get

q(l ∆ ) ≈

 nπl ∆   N large. T 

n =1

Choose D = T/M. Then

N

q[l ]∆ q(l ∆ ) = Let

N

∑ qn sin 

 πnl 

∑ qn sin  M  , 1 ≤ l ≤ N .

h =1

M = N,

N

Then

q[l ] =

 πnl   ,1≤ l ≤ N N 

∑ qn sin 

h =1 N

 πnl   πml    sin  N   N 

∑ sin 

Now,

h =1

1 N   π ( n − m) l   π ( n + m) l   − cos  = ∑ cos      2 l =1   N N Let n ≠ 0.     i π n l    iπn   exp(iπn) −1   πnl  Re exp Re exp cos = = ∑  N  ∑  N   N     iπn    l =1 l =1   exp  N  −1    N

N

  iπ  nN n   sin (πn / 2)  −  = Re exp   n +   N 2 2   sin (πn / 2 N )  

 πn  sin[πn / 2] ( N + 1)  = cos   2N  sin[πn / 2 N ] In the Limit N → ∞, this is close to

108

Stochastics, Control & Robotics

cos(πn / 2) sin (πn / 2)

2N πn

N sin (nπ) = 0. πn Note that for finite N, we have exactly =

  πn   πn   sin  πn +  − sin    2 N   2n   πnl    cos ∑  N  =  πn  l =1 sin    2n  = (– 1)n – 1 Since ( −1) n − m − ( −1) n + m = 0 N

We set exactly for all n ≠ m, n, m ≥ 1, N  πml   πnl  ∑ sin  N  sin  N  = 0. l =1 

[40] Quantum Shannon Theory Contd.: Heisenberg-Weyl operators

{ j } j∈{0,1,2,...,d −1} is an ONB for ad dimensional Hilbert space H.  2πi j z  j Z ( z ) j = exp   d  X ( x) j = j ⊗ x Heisenbrg-Weyl operators are X (x) Z (z) – 0 ≤ x, z ≤ d – 1 only on H. Consider Φ

AB

=

1 d d −1

X (x) = Z (z) = X ( x) y =

∑

∑i

A

i

B

i=0

x ⊕ x′ x ′

x′ = 0 d −1

 2πi x x′   x′ x′ . d 

∑ exp 

x′ = 0

∑ x ⊕ x′ x′

X ( x) Z ( z ) y =

d −1

∑

x′, z ′

x′ | y =

∑ x ⊕ x′

δ x′ y = x ⊕ y .

x′

 2πi z z′  x ⊕ x ′ x ′ exp  z′ z′ | y  d 

Classical Robotics & Quantum Stochastics

=

109

 2πi z z′   x ⊕ x′ δ x′, z ′ , δ z ′, y ′ d 

∑ exp 

x′, z ′

 2πi z z′  x⊕ y = exp   d   2πi z( y + x)  Z ( z ) X ( x) y = Z ( z ) y ⊕ x = exp   y ⊕ x  d  2πi xz  X ( x) Z ( z) y = exp   d  So we deduce the Weyl Commutation relations  2πi xz  X ( x) Z ( z) . Z (z) X (x) = exp   d  Φ x, t

AB

(

= = Φ x′, z ′ | Φ x, z

)

A A ≡ X ( x) Z ( z ) ⊗ I Φ

=

1 d 1 d 1 d

d −1

∑ X A ( x) Z A ( z ) y

=

1 d

{

y

 2πi yz   y⊕x d 

∑ exp  y

A

AB

y

B

  2πi ( yz − y ′ z ′)  d

y, y′

A

y ′ ⊕ x′ | y ⊕ x

A

y

  2πi ∑ exp  d ( yz − y ′ z ′) y, y′

B

B

y′ | y

BA

y ′ ⊕ x′ | y ⊕ x

  2π i ( yz − y ′ z1′ | δ y ′, y ) δy ′ ⊕ x ′, y ⊕ x  d

∑ exp 

y ′, y

1   2πi exp  y ( z − z′) ∑   d d y

= δ x′, x δ z ′, z φ x, z

B

∑ exp 

= δ x′, x

Hence

A

y=0

×B y′ 1 = d

AB

}

| 0 ≤ x, z ≤ d − 1 .

Form an ONB for H A ⊗ H B (dim HA = din HB = d) Alice and Bob share the entangled state

A

110

Stochastics, Control & Robotics

AB

Φ

d −1

1 d

=

∑

A

x

x

B

x=0

Alice applies the operator X (x) Z (z) to her side giving Φ x, t

A, B

for the state

shared by herself and Bob. She them Gravimits her system. A over a noisy quantum channel U A→ B ′ to Bob. Bob this has an ensuable of state

(U

A→ B ′

)( Φ

⊗ IB

)

AB x, z

, Φ xAB , t x, z ∈{0, 1,..., d − 1}

Bob can now prepare the state ρXZB ′B =

1

∑ d2

x x

N

z

⊗ z z

x, z

(

(

⊗ N A→ B ′ ⊗ I B Φ xAB ,z

))

Φ xAB ,z

Holevo information of the state: I ( XZ ; BB′) = H ( BB ′ ) − H ( BB′ | XZ )   AB ρB ′B = TrXZ ρ XZB ′B = N A→ B ′ ⊗ I B  1 ∑ Φ xAB Φ ,z x, z   d 2 x, z 

(

Now

{Φ

hence

AB x, z

}

∑ Φ xAB, z

AB Φ xAB , z = I d2

πAB =

Thus, with

Also,

)

| 0 ≤ x, z ≤ d − 1 is an ONB and

x, z

We get and

) (

1

I AB 2 d2 d

ρB ′B = πAB H ( B ′ B) = log (d 2 ) = 2 log (d). H ( B ′ B | XZ ) =

∑ d 2 H (N A→ B′ ⊗ I B ) ( Φ xAB, z 1

x, z

Now

(

)

Φ xAB ,z

)

(

A A B Φ Φ Z A* ( z ) X A*( x ) ⊗ I B Φ xAB , z = X ( x) Z ( z ) ⊗ I

Φ xAB ,z

=

1 d

2

∑

(X y

A

A

) y′ (Z

( x) Z A ( z ) ⊗ I B y

BA

y′

B

y, y′

=

1 d

2

∑ ( X A ( x) Z A ( z ) )

y, y′

y

A

y

BA

A*

( z ) X A* ( x) ⊗ I B

)

y ′ Z A* ( z ) X A* ( x) B y ′

)

Classical Robotics & Quantum Stochastics

= = =

1 d

2

1 d

2

1 d

2

111

 2πi z ( y − y ′ )  y⊕x  d

∑ exp 

y, y′

 2πi z ( y − y′ )  y  d

∑ exp 

y, y′

∑

y

AA

AA

AA

y′ ⊕ x ⊗ y

y′ | ⊗ | y − x

y ′ ⊗ Z B ( z ) X B ( − x) y

BB

BB

BB

y′

y′ − x

y′ X B* ( − x) Z B* ( z )

y, y′

(

)

(

)

A B B A B B = I ⊗ Z ( z ) X ( −x) ( Φ Φ ) I ⊗ X ( − x) * Z ( x) *

X ( x) y = y ⊕ x .

Now,

z | X (x) | y = δ z , y ⊕x = δ z , y ⊕x = y X ( −x) z Hence relative to the basis

{ x } , X (x)T = X (– x). Also

 2πiyz  y Z ( z ) y = exp   d  So Z (z) is diagonal relative to the basis

{ x }.

Hence relative to this basis, Z (z)T = Z (z). Thus, we can write the above identity as

(X

A

( x) Z A ( z ) ⊗ I B

)( Φ ⊗ (X

(

Φ ) X A ( x ) Z A ( z ) ⊗ I B*

)

(

)

)

T  B = IA x) Z B ( z )  ( Φ Φ ) I A ⊗ X B ( x) Z B ( z ) (   Sphere packing Lamma (as a point step towards Shannon’s noisy codiny theorem in the quantum setting). 

[41] Basics of Cq (Classical Quantum) Information Theory Cq. Codes x → rx m  message that Alice wishes to transmits to Bob. Cm  Code ward used to transmit m. m = 1, 2, ..., |µ|. (|µ| message to be transmitted) C = Cm|m = 1, 2, .., |µ| are |µ| independent r.v.s, (random code) p (C) =

|µ.|

∏ p X (Cm )  probability of random code. When Alice transmits m1 Bob

m =1

receives that state σCm and he applies that meaurement operate Λm. Let M be

112

Stochastics, Control & Robotics

the message transmitted by Alice using the random code Cm and M′ the message received by Bob. Then P {M′ = m| M = m} = Tr (Λm σCm) Alice’s codeward Cm ∈ c

{pX (x), σx|x ∈ X}

Cm = (Cm (1), ..., Cm (n1)), than σCm = ⊗

σCm( j )

j =1

Packing Lemma: Assumption

Tr ( Πσ x | ≥ | −ε

Tr ( Π x σ x | ≥ | −ε

Tr ( Π x ) ≤ d ΠσΠ
N (δ )) ξ ≠ exp (n (4δ – I (X, Y))) Typical sequence xn ∈ An x ∈ A has probability distribution p (x), x ∈ A. N (x|xn) is the number of the x appears in xn. pn (xn) = txn (x) = So,

where

x ∈A

n

)

  n = exp  ∑ N ( x | x ) log p(x)  x ∈A 

N ( x | xn ) , txn is a probability distribution on A. n

  1 p n ( x n ) = exp  n ∑ t n ( x)log p( x) log p n ( x n )  x ∈A x  n =

or

∏ p( x) N ( x|x

∑ t xn ( x)log p( x)

x ∈A

1 − log p n ( x n ) = D (t n | p) + H (t n ) x x n D (q | p) =

 q( x) 

∑ q( x)log  p( x) 

x ∈A

= − D(t x n | p) − H (t x n )

Classical Robotics & Quantum Stochastics

121

is the Killback distance between the pdf’s q and p. Let t be a type on An i.e. a pdf on A of the form t (x) =

N ( x) n ∀ x ∈A N ( x){0, 1, 2, ..., n} ∀ x ∈A .

Where Let

Then

n  n  n N (x | x ) = t ( x)∀ x ∈A Ttn =  x ∈ A n  

t n (Ttn ) =



x ∈Tt

=

x ∈Tt

n

 x ∈A







 x ∈A



exp  ∑ N ( x | x n | log t (x) ∑   n n

x ∈Tt

=



t n ( x n ) = ∑  ∏ t ( x) N ( x|x )  ∑ n n n n





∑ n exp  n ∑ t ( x) log t ( x) n

x ∈Tt



x ∈A



n = | Tt | exp(− n H (t ))

Note that tn is a pdf on An and hence we derive the inequality TTN ≤ exp(n H (t )) 1 log TTN ≤ H (t). n Let t′ be another type. Then

or

( )

t ′ n Ttn =

∑ n t m ( xn ) = ∑ ∏ t ′( x)nt( x) n n

n

x ∈Tt

x ∈Tt x ∈ A

  n = | Tt | exp  n ∑ t ( x)logt ′( x)  x ∈A  = | Tnt | exp(n( −D(t | t ′ ) − H (t))) i.e.

n n | Ttn | = t ′ (Tt ) exp(n( D(t | t ′ ) + H (t )))

* Problem: Find a lower bound on tn (Tnt) for large n. Consider

1=

∑ t ′(Ttn ) t

Now the number of types t is the number of ways in which n can be expressed as n = r1 + r2 + ... + rc

c = |A|.

122

Stochastics, Control & Robotics

rj ≥ 0.  n + χ − 1  n 

This number is Thus,

1=

∑ t ′ (Ttn )

( )

 n   n + χ − 1 ≤  max t ′ Tt   n   t 

t

|Tnt | ≥ t ′ n (Ttn ) exp(n( D(t | t ′ ) + H (t ))) ∀t ′

and hence implies

 n + χ − 1 max | Ttn | ≥ max exp(n H (t ))  n  t  t

−1

n

Channel capacity theorem of Shannon. X  alphabet (finite) TδX  δ typical sequences.   n  n  n 1 x ∈ X log p ( x + ) H ( p) < δ TXn =   ∑ j δ n j =1   xn = ( x j )nj =1 { p(x), x ∈X } is the source alphabet pdf.

Here Let

pn (xn) = n

p ′ ( xn ) =

n

∏ p( x j ) , j =1

 x n ∈T X n  δ , pn ( xn )  xn ∈ X n \ T X n δ 

pn ( xn )

∑n

x n ∈TδX 0

∑n n

x ∈X

p ′n ( xn ) − pn ( xn )

pn ( xn )

∑n

x n ∈TδX

ξ

n

∑n

p n (ξ n )

=

ξ n ∈TδX

∑n

p n (ξ n )

ξ n ∈TδX

=

( ) ( ( )

p n TδX

nc

p n TδX

( )

∑ n pn ( xn ) +

x n ∈TδX

p n X δn n

n nc = 2 p Xδ ≤

)

∑ x n ∈X n \TδX

( )

+ p n TδX

nc

2Var (log p( x)) nδ 2

pn ( xn )

x n ∈ X n \TδX

∈TδX

∑ nc p n (ξn )

∑

− pn ( xn ) +

≡ e.

n

pn ( xn ) n

Classical Robotics & Quantum Stochastics

123

By Chebyshev’s theorem.

∑n n

Thus,

x ∈X

→0 p ′ n ( xn ) − p n ( xn ) n →∞

n

Let X′ be a r.v. (with values in Xn) having pdf p′n. We’ve proved that

∑n P { X ′ n

} {

}

n n n n n = x − P X = x < ε x ( m) ε X

x

Let Alice selects |µ| codewords m = 1, 2, ..., |µ| having pdf n q (m) = p ′ ( xn (m)), m = 1, 2, ..., |µ|

|µ| = TδX

so that

n

+1

For each m are encoded into the quantum states n

ρ x n ( m) = ⊗ ρ x ( m) , m = 1, 2, ..., |µ| k k =1

She transmits this state to Bob along the product channel N ⊗ ⊗ so N

m

(ρ

x1 ( m )

n

(

)

k =1

k =1

σx = N (ρ x ), x ∈ χ  [σ

X ′n

] =

p ′ n ( x n )σ n ∑ x n n

x ∈χ

∑ n p n ( x n ) σ xn

=

∑ n pn ( xn )

x n ∈TδX

Let

n

ρn ⊗ ... ⊗ ρ xn ( m) = ⊗ N ρ x ( m ) = ⊗ ρ xk ( m ) = x ( m ) k

where we have

)

m

σ=

∑ p( x)σ x

x ∈χ

(

 σX

x n ∈TδX ′n

)−σ

∑ n p ′ n ( x n ) σ x n − σ⊗ n

=

n

x ∈χ

x ∈χ

1

x ∈χ

∑n ( p ′ n ( xn ) − p n ( xn ) ) σ xn x

≤

1

p ′ n ( x n )σ n − ∑ p n ( x n ) σ n ∑ x x n n n n

=

=

⊗n

∑n p′ n ( xn ) − p n ( xn ) < ε x

1

1

124

Note:

Stochastics, Control & Robotics

∑n p n ( xn )σ xn

n

=

x

n

∑n ∏ p( x j ) j⊗=1 σ x j j =1

x

 n  n⊗ n . = ⊗  ∑ p ( x) σ x  = σ   j=1   x ∈χ Let Bobs Hilbert space by HB.

σ ≡ σB n

n

σ Bx n = ⊗ σ x j . j=1

∏ δB where and

n

xn

B

xn

= x1 x2 ...xn

B

= ⊗ xk

∑B n

=

x ∈Tδ

n

xn

B

σx

B

= qB ( x ) x

x

B

= x1 (m)...xn (m)

∑ qB ( x )

x ∈χ

B

B

k =1

B

=1

n   n n 1 log q ( x + σ ) H ( ) < δ TBδ =  x ∈ χ  ∑ B k n k =1  

H (σ) = − ∑ qB ( x) log(qB ( x)) x ∈χ

σ⊗

n

∏ δB

n

B = σ

qBn ( x n ) = So,

{

n

Tr σ B Π δB

n

}

=

n

∏ δB

n

n

=

B

xn

x ∈Tδ

n

∏ qBn ( xi ) . i =1

∑ B qBn ( xn )

n

x ∈Tδ

n B = qB (Tδ ) ≥ 1 −

where

qBn ( x n ) x n ∑ n B

Var log qB (x) =

Var log qB ( x) nδ 2

=1–e

∑ qB ( x)(log qB ( x))2 − H (qB )2

x∈χ

2 2 = Tr (σ(log σ) ) − H (σ)

B

Classical Robotics & Quantum Stochastics

∏ δB / x

n

125

= typical projection n

σ xn = ⊗ σ x = σ xn B n k

for the state

k =1

 [43] Quantum Image Processing Some Basic Problems: [1] f (X, Y) is a classical image; F (X, Y) of a quantum image i.e. a quantum field f (X, Y) evolves according to a diffusion equation so that sharp edges get smoothed out: ∂f (t , X , Y ) ∂ ∂f (t , X , Y ) + ∂ D ( X , Y ) ∂f (t , X , Y ) DXX ( X , Y ) = YY ∂X ∂Y ∂t ∂X ∂X

+

∂f (t , X , Y ) ∂ ∂f (t , X , Y ) ∂ + DXX ( X , Y ) DXX ( X ,Y ) ∂Y ∂Y ∂X ∂Y

∂f = div ( D ( X , Y ) ∇ f (t , X , Y )) ∂t For quantizing this evolution, we use the Lagrangian approach. Introduce an auxiliary field g (t, X, Y) and let.

or

S[ f , g] =

∫ g ( f , t − div ( D ∇ f )) dt dx dy

Then

δS = 0 ⇒ f, t = div(D ∇ f ) δg

Also

δS T = 0 ⇒ − g,t − div D ∇ g = 0 δf

or

(

(

g,t − div DT ∇ g

)

Now for the quantization. Let f (t , X , Y ) =

)

=0 ( X , Y ) ∈[0, 1] X [0, 1] .

∑ f nm (t )enm ( X , Y ) nm

g (t , X , Y ) =

∑ gnm (t )enm ( X , Y ) nm

enm ( X , Y ) = exp(2πi (nX + mY )) D( X , Y ) =

∑ D nm enm ( X , Y ) . nm

D∇ f =

 2πi n′  D nm  f nm (t )en + n′ m + m′ ( X , Y )  2πi m′ n m n′ m′

∑

126

Stochastics, Control & Robotics

div ( D ∇ f ) = − 4π 2

 n′  (n + n′, m + m′) D nm   f nm (t ) en + n′ m + m′ ( X , Y )  m′ n m n′ m′

∑

∫ 2 g div ( D ∇ f ) dX dY

[ 0, 1]

 n′  2 = − 4π ∑ g kr (t ) f nm (t )(n + n′, m + m′) Dnm    m′

∫

en + n′ + k , m + m′ + r ( X , Y ) dXdY

[ 0, 1]2

 n′  = − 4π 2 ∑ g n + n′, m + m′ (t ) f nm (t )(n + n′, m + m′) D nm    m′ n, n′

∫ 2 g f,t dX dY

=

nm

[ 0, 1]

\

∑ gnm (t ) f nm′ (t )

L( f nm (t ), g nm (t ), f nm ′ (t)) =

 r g nm (t ) f n − r , m − s (t )(n, m) D n − r , m − s    s nmr s

∑ gnm (t ) f nm′ (t ) + 4π 2 ∑ nm

=

 n −r g nm (t ) f r s (t )(n, m) D r s   m − s nmr s

∑ gnm (t ) f nm′ (t ) + 4π 2 ∑ nm

*****

Appendix Problems [1]

t

(

X (t) = X − ∫ C (r ) dr − D * (r ) dAr − D(r ) dAr* o

)

Calculate d ( X (t ) * X (t )) . Using the quantum I. to formula Hint: d ( X (t ) * X (t )) = (dX (t )*) X (t ) + X (t ) * X (t ) * dX (t) + dX (t ) * dX (t )

(

)

* * * = − C(t ) dt − D(t ) dAt − D * (t )dAt X (t)

(

)

− X * (t ) C (t) dt − D * (t ) dAt − D(t ) dAt* + D * (t ) D(t)dt = − (C * (t ) X (t ) + X * (t ) C (t) − D * (t ) D(t )) dt + ( D(t) X (t) + X * (t ) D(t )) dAt* + ( D * (t ) X (t) + X * (t ) D * (t ) ) dAt C (t) = g(t, X) X* = X (o)

Classical Robotics & Quantum Stochastics

127

D (t) = δ (t, X) δ* (t, X) = D*(t) δ (t, X*) = δ * (t , X )*, γ (t , X *) = γ (t , X ) * Then, let

i (t, X) = X (t)

d ( X (t ) * X (t )) = di (t , X * X ) t   d X * X − ( γ (r , X * X )dr − δ (r , X * X )) dAr   = ∫   o * = − γ (t , X * X ) dt + δ * (t , X * X ) dAt + δ (r , X * X ) dAr  +δ(t , X * X ) dAt* On the other hand and on the other, by quantum Ito’s formula, d ( X (t ) * X (t )) = dX (t ) * X (t ) + X (t ) * dX (t ) + dX (t ) * dX (t) = − [ γ (t , X *) i (t , X ) + i (t , X ) * γ (t , X ) − δ * (t , X ) δ (t , X )] dt + [δ (t , X ) i (t, X ) + i(t , X ) * δ (t , X )] dAt*

+ [δ * (t , X )i (t , X ) + i (t , X ) * δ * (t , X )] dAt

So, the condition for i (t , X * X ) = i (t , X *) i (t , X ) is γ (t , X * X ) = γ (t , X *) * i (t, X ) + i (t , X ) * γ (t , X ) − δ * (t , X | δ (t , X )) γ (t , X * X ) = γ (t , X *) *i (t, X ) + i (t , X ) * δ (t , X ) *δ(t , X ) , δ * (t , X * X ) = δ * (t , X )i (t , X ) + i (t , X ) * δ * (t , X ) δ * (t , X * X ) * = i (t , X ) *δ (t , X ) + δ (t , X *) i (t , X ) i (t ,⋅)

→ X (t ) to be a * representation. Conditions for X  * dX (t) = C (t ) dt − D * (t ) dAt − D(t) dAt * dY (t) = C (t ) dt − F * (t ) dAt * F (t ) dAt

Non-demolition conditions: [ X (t), Y ( s )] = 0, ∀ t ≥ s . Observations do not affect the future values of the state. ≡ [dX (t ), Y ( s )] = 0 ∀ t ≥ s ≡ [C (t) dt − D * (t ) dAt − D(t ) dAt* , Y ( s)] = 0 ∀ t ≥ s ≡ [C (t), Y ( s )] = [ D * (t ), Y ( s)] = [ D(t), Y ( s )] = 0 ∀t ≥ s . [ X (t), Y (t )] = 0, [dX (t ), Y (t )] = 0

128

Stochastics, Control & Robotics

⇒

[ X (t), dY (t )] + [dX (t ), Y (t)] = 0

⇒

 X (t ), G (t) dt + F * (t )dAt + F (t ) dAt*    +  D * (t ) dAt + D(t ) dAt* , F * (t ) dAt + F (t ) dAt*  = 0

⇒

[ X (t), G (t )] + D * (t ) F (t ) − F * (t ) D(t) dt + [ X (t ), F * (t )] dAt + [ X (t), F (t )]dAt* = 0.

⇒

[ X (t), F * (t )] = 0 = [ X (t), F (t )], D * (t ) F (t) − F * (t ) D(t ) = [G (t), X (t )] . 

2 Electromagnetus and Related Partial Differential Equation [1] Computation of the Perturbe Characteristics Frequencies in a Cavity Resonator with in Homogeneous Dielectric and Permittivity: ∇ × E = − jwµ ( w, r ) H ,

∇ × H = jwε ( w, r ) E

ε ( w, r ) = eo (1 + δχe ( w, r )) div (1 + δ.χe ) E = 0, µ ( w, r ) = µ 0 (1 + δχ m ( w, r )) div (1 + µ.χ m ) H  = 0.

Side walls X = 0, a and Y = 0, b are perfect magnetic conductors. Top and bottom surfaces z = 0, d are perfect electric conductors. Boundary conditions HZ = 0, Z = 0, d, X = 0, a, Y = 0, b. EX = EY = 0, Z = 0, d,

EX = 0, X = 0, a, EY = 0, Y = 0, b.

we get

div E = −δ (∇χe , E ) + O (δ 2 )

div H = −δ (∇χ m , H ) + O (δ 2 ) ∇ (divE ) − ∇2 E = − jw {∇µ × H + µ∇ × H } = − jw {∇µ × H + jwεµE } or

∇ 2 E + w 2 ε 0 µ 0 E + δw 2 ε 0 µ 0 ( χ e + χ m ) E = δjw∇χ m + H − δ∇ (∇χe , E ) + O (δ 2 )

Likewise by duality E → H , H → – E , ce ↔ cm, e0 ↔ µ0, we get ∇ 2 H + w2 µ 0 ε 0 H + δw2 µ 0 ε 0 ( χ e + χ m ) H

...(1)

130

Stochastics, Control & Robotics

= −δjw∇χe × E − δ∇ (∇χ m , H ) + O (δ 2 )

...(2)

w = w(0) + δ.w(1) + O (δ 2 ) ,

We write

E = E (0) + δ.E (1) + O (δ 2 ) H = H (0) + δH (1) + O (δ 2 ) Then the O(d0) equation is ∇×E

(0)

( )

= − jw 0 µ 0 H

(0)

(0) (0) ( ) ∇ × H 0 = − jw ε 0 E The solution modes with the monitor boundary conditions is obtained using standard methods as 1/ 2  m2 n2 p 2  (0) (0) w = w (mnp) =  + + π,  a 2 b 2 c 2 

Hz(0) = C(mnp)umnp ( r )

i.e. the complete solution in the time domain for Hz(0) is

( ) ∑ Re {C (mnp ) exp { jw 0 (mnp ) t} umnp ( r )}

mnp

Where Likewise Where

υmnp ( r ) =

 mπx   mπy   mπz  sin    sin   sin     a b c  abd

2 2

E–z(0) = d(mnp)υmnp ( r ) υmnp ( r ) =

 mπx   mπy   mπz  cos   cos   cos  .   a b c  abd

2 2

∂ Note that the solutions are obtained by the wave guide method in which → jw, ∂t ∂ → – y, so the wave-guide equations ∂z − jwµ 0 Y ( ) ( ) ( ) ∇ ⊥ H z0 × z − 2 ∇ ⊥ Ez0 E⊥0 = 2 h h − jwε 0 Y (0) (0)  ( ) H⊥ = ∇ ⊥ E z × z − 2 ∇ ⊥ H z0 2 h h give in the case of a responator ( ) E⊥0 = −

and

( ) H ⊥0 =

µ0

1 ∂ ∂ ( )  ( ) ∇⊥ H t 0 × z + ∇ ⊥ E z0 2 h ( mn ) ∂t h ( mn ) ∂z 2

ε0 ∂ 1 ∂ ( ) ( ) ∇ ⊥ Ez0 × z + 2 ∇ ⊥ H z0 2 ∂t h h ∂z

 m2 n2  . h2 = h(mn)2 = π 2  +  a 2 b 2 

Electromagnetus & Related Partial Differential Equation

131

Thus, ( )

E⊥0 =

( )

− jw 0 ( mnp ) µ 0 2 h ( mn )

+ so (0)

E⊥

(∇ ⊥ umnp ( r ) × z ) C (mnp)

d ( mnp ) ∂ ∇ ⊥ υmnp ( r ) 2 h ( mn ) ∂t

 − jw(0) ( mnp ) µ 0 = C ( mnp )  ∇⊥ umnp ( r ) × z 2  h ( mn )

(



) 

1 ∂  ( )  + d ( mnp )  υmnp ( r ) z + ∇ υr 2 ∂z ⊥ mnp  h ( mn )   E E = C ( mnp ) ψ mnp ( r ) + d (mnp ) ϕ mnp (r ) E where ψ mnp ( r ) is E ( r ) a 3 × 1 vectors.

Expressible as a linear combination of υmnp ( r ) ,

∂2 ∂2 υmnp ( r ) , υmnp ( r ) , ∂ x∂ z ∂y ∂z

E and ϕ mnp ( r ) is a 3 × 1 vector expressible as a

∂umnp

∂umnp

linear combination ∂X ∂Y of with constant 3 × 1 vector valued coefficients dependent on (m, n, p). Likewise, H

(0)

are orthogonal to For all (m′n′p′) ≠ (mnp)

,

H H = C ( mnp ) ψ mnp ( r ) + d ( mnp ) ϕ mnp ( r ) E H H , ψ mnp , ϕ mnp {ψ mnE p , ϕmnp } {ψ mE ′n′p′ , ϕmE ′n′p′ , ψ mH′n′p′ , ϕmH′n′p′ }

as ordered triplets. Equating coefficient of O(d) in (1) and (2) gives ( )2 ( )2 ( ) ()  w 0  (1) 2w 0 w 1 (0) w 0 ( ) ( ) 2 ∇ + 2  E + E + 2 χe0 + χ0m E 0  C  C2 C

(

)

(

)

= jw(0)∇χ(e0) × H (0) − ∇ ∇χ(e0) , E (0) and

...(3)

( )2 ( )2 ( ) ()  w 0  (1) 2w 0 w 1 (0) w 0 ( ) ( ) 2 ∇ + 2  H + H + 2 χe0 + χ0m H 0 2   C C C

(

( ) ( )

= − jw 0 ∇e0 × E

(0)

(

)

( )

− ∇ ∇χ m0 , H

(0)

)

...(4)

132

Stochastics, Control & Robotics

( )

( ) ( ) χm ( w 0 , r )

χe0 = χe w(0) , r ,

where

( )

χ m0 =

w(0) = w(0)(mnp). We arrange (3) and (4) as a single 6 × 1 single vector (partial differential equation): (1) (0) ( )2 ( ) () ( )2   (0)  w 0   E  2w 0 w 1  E  w 0 (0) E 2 ∇ + 2   χ + +  ( )    ( )  C2 C   H (1)  C2 H 0  H 0 

(0) (0)   ( ) ( )  ∇(∇χe0 × E 0 )  ( ) ∇χ m × H − = jw 0  ( ) ( )  ( ) ( )   −∇χe0 × E 0   ∇(∇χ m0 × H 0 ) (0)

w

where

2 2  2 = w (mnp) = π m + n + p  2  a b2 c2  (0)

(

( )

χe0 = χe w(0) , r

1/ 2

,

)

( ) ( ) ( ) χe0 + χ m0 = χ 0

(

...(5)

( ) χ m0 = χ m w(0) , r

)

E  E   ϕ mnp  ψ mn  E (0)  p C m + d mnp np    = ( ) H ( ) H   ( )  ψ mnp   ϕ mnp  H 0 

Now,

E  E   ψ mnp  ϕ mn p    and note that So (Eq. 5) gives on taking inner products with and  H  H   ϕ mnp   ψ mnp  (0)2 2 annihilates both of there functions ∇ + w

( ) ()

2w 0 w 1 C +

2

( )2

w0 C

2

{C (mnp) +  ψ mnp 2 + d (mnp) < ψ mnp , ϕmnp >}

{C (mnp) < ψ

mnp , χ

(0)

( )

ψ mnp > d ( mnp ) < ψ mnp , χ 0 ϕ mnp >

}

H   ∇χ(m0) × ψ mnp   > ψ C m np < , ) (  mnp (0) E   −∇χe × ψ mnp 

(0) 

= jw

H  ∇χ(m0) × ϕ mnp   + d ( mnp ) < ψ mnp ,  E   −∇χ(e0) × ϕ mnp 

and

  > 

...(6)

Electromagnetus & Related Partial Differential Equation

( ) ()

2w 0 w 1 C +

2

( )2

w0 C

2

133

{< ϕmnp , ψ mnp > C (mnp) + d (mnp)  ϕmnp 2 }

{C (mnp) < ϕ

mnp , χ

(0)

( )

ψ mnp > +d ( mnp ) < ϕ mnp , χ 0 ψ mnp >

}

H    ∇χ(m0) × ψ mnp (0)  >  np < C m jw ϕ , = ) mnp  ( E   ∇χ(e0) × ψ mnp   H   ∇χ(m0) × ϕ m np  +d ( mnp ) < ϕ mnp ,  ( ) 0 E  ∇χe × ϕ mnp 

where

  > 

...(7)

E  E   ϕ mnp  ψ mnp ψ mnp =    , ϕ mnp =  H  H   ψ mnp  ϕ   mnp 

(6) and (7) can be expressed as  A11 ( mnp ) A12 ( mnp ) ( )  B ( mnp ) B12 ( mnp ) C ( mnp )  − w 1  11    = 0   B21 ( mnp ) B22 ( mnp )  d ( mnp )    A21 ( mnp ) A22 ( mnp ) and have w(1) takes two values for each (mnp), namely the roots of

(

)

() det A ( mnp ) − w 1 B ( mnp ) = 0. 

[2] Numerical Methods for pde: (a) Variation principles FEM. Action Integral: S[f] =

∫n L(φv , φψµ , x)d

n

x

D

fv: R n → R n , v = 1, 2, ..., p. dS[f] = 0 ⇒

⇒

 ∂L



∂L

∫  ∂φν δφν + ∂φv, µ δφv, µ  d

D

 ∂L



∂L 

∫  ∂φv − ∂µ ∂φv,µ  δφv d

D



n

n

x =0

x =0

Assuming Direchlet conditions dfn= 0 on the boundary i.e. fn is prescribed on ∂D. ∂L ∂L − ∂µ Thus = 0. ∂φ ν, µ ∂φ ν

134

Stochastics, Control & Robotics

Euler-Lagrange eqns. for fields. Numerical implementation: Divide D into polyhedron

etc.

Express fn(x) for X ∈ ∆ k (kth polyhedron) as a linear combination of its values at the vertices fn(x) =

q

∑ Cνk [m]ψ km ( x), X ∈∆ k

m=1

ψ km = basis function chosen so that ykm(nk,r) = δm, r ,1 ≤ m, r ≤ q where nk, r,= 1, 2, ..., q are the vertices of ∆ k . Thus

φν (ν k , r ) = Cνk [r ] . N

then

S[f] ≈

∑∫

k =1 ∆ k

q  q  L  ∑ Cνk [m]ψ km ( X ), ∑ Cνk [m]ψ k m, µ ( X )d n X   m =1  m =1

≡ S [{Cνk [m]:1 ≤ m ≤ q, 1 ≤ k ≤ N , 1 ≤ ν ≤ p}] Minimize S w.r.t. {Cνk [m]} . Example of application (a) Solving the Einstein field equations of gravitation S=

∫R

−gd 4 X

R = Curvature scalar = g µν Rµν . Equivalent action α α Γβµβ − Γµβ − Γβνα d 4 X S1 = ∫ g µν − g Γµν

(

)

α is a tensor and hence Equivalent of S and S1is proved using the fact that δΓµν

{

α β δRµν = δ Γ αµα, v − Γ αµν, α − Γ αµν Γ βαβ + Γ µβ Γ να

( ) − ( δΓ ) −g ) − (g = ( g δΓ α = δΓ µα

So

g µν − g δRµν

µν

:ν

α µv

β µα

:α

µν

ν

α − g δΓ µv

}

)

,α

is a perfect divergence and hence had a vanishing integral. (b) Solving the Einstein-Maxwell equation in the presence of charged fluid. S = S1+S2+S3 S1 = C1 ∫ R −gd 4 X , µν − gd 4 X , S2 = C2 ∫ Fµν F

Electromagnetus & Related Partial Differential Equation

135

µ ν 4 S3 = C3 ∫ ρv v g µν − gd X ,

≡ C3 ∫ ρ − g d 4 X Energy-momentum tensor of a fluid with presence taken into account: µ ν µv Tµn = (ρ + p)v v − pg Einstein-Maxwell eqns.

1 Run − Rg µν = K1T µν + K 2 S µν 2 1 where Sµn = − Fαβ F αβ g µν + Fαµ F να 4 is the energy-momentum tensor of the em field. From the Einstein Field equations, we can derive the MHD eqns. in general relativity: µ ν µν K 2 S:µν ν + K1 ((ρ + p ) v v − p g ):ν = 0

Which can be brought to the generalized Navier-Stokes form used in non relativistic MHD:

∂v   ρ  ( v , ∇ ) v +  = − ∇p + J × B  ∂t  ∂E where ∇× H = J + ε ∂t The Einstein Field eqns. 1 Rµν − R g µν = K1T µν + K 2 S µν 2 are a consequence of setting the variation of S w.r.t. gµn to zero. Setting the variation of S w.r.t. Aµ ( Fµν − Aν,µ − Aµν ) to zero gives the Maxwell eqns.  F:νµν = K3 Jµ Provided that we add to the action an interaction term ∫ J µ Aµ −gd 4 X represents

the interaction between the charged fluid and the em field.

Here, in MHD we usually assume Ohms law that gives the relation between the current density and the em field as J = σ (E + v × B) In the special relativistic case. In teh general relativistic case, Ohm's law becomes a tensor equation µv Jµ = σF vv

and hence the charge – em field interaction term in the action amines the form

∫j

µ

µν 4 Aµ −gd 4 X = σ ∫ F vν Aµ − g d X

136

Stochastics, Control & Robotics

The finite element method can be applied to MHD in a given metric gµn by taking as our action. µν − g d 4 X + K2 S[v µ , Aµ ] = K1 ∫ Fµν F

∫ ρ gµνv

∫F

µν

µ µ

− g d 4 X + K3

v

vν Aµ − g d 4 X

µ

Here we should that v µ − dX and carry out the variation w.r.t. Xµ rather than vµ. dτ The complete non-relativistic MHD eqns. are ∂v   ρ  ( v , ∇) v +  = − ∇p + η∇2 v + σ( E + v × B) × B,  ∂t  div E = 0, div B = 0, Curl E = −

and

∂B 1 ∂E , Curl B = µ 0 J + . ∂t c 2 ∂t

The last example of a pde that can be solved using the FEM is the Klein-Gordon field with Higgs potential. ∂ µ ∂ µ φ + m 2 φ2 + ε V '(φ) = 0 ∂ µ ∂ µ =  = ∂ t2 − ∇ 2 .

where

This pde can be derived from the action 1 S [f] = ∫ ∂ µ φ∂ µ φ − m 2 φ2 − 2 ε V (φ) d 4 X 2 and it can be solved using the FEM.

(

)

The final example is from quantum field theory where we start with the Lagrangian

(

)

density L φµ , φµ , ν , X with fµ: R 4 → R and construct the Hamiltonian via the Lagrangian expressed in terms of function that depend only on time. fµ(X) = φµ (t , r ) =

N

∑ ξµk (t )ψk( r )

k =1

where {ψ k }kN=1 are basis functions depending only on the spatial coordinates r: S ξµk  = =

∫ L (φµ , φµν , X ) d

4

X

∫ L  ∑ ξµk (t )ψ k ( r ), ∑ ξµk (t )∇ψ k (r ), k

k

∑ ξµ’ k (t )ψ k (r ), t , r  d 3r dt k

=

∫ L ({ξµk (t ), ξµk } , t ) dt '

Electromagnetus & Related Partial Differential Equation

with

137

} )

({

L ξµk (t ), ξµ' k (t ) , t =





∫ L  ∑ ξµk (t )ψ k ( r ), ∑ ξµk (t )∇ψ k ( r ), ∑ ξµk (t )ψ k ( r ) , t , r  d ’

k

k

3

r

k

is the Lagrangian. The Hamiltonian is constructed using the Legendre transformation: ∂L pµk = ' , ∂ξµk H

pµk ξµk − L ({ξµk , pµk } , t ) = ∑ µk

Then the Schrödinger eqn. is formulated: i

( { })

∂ ψ t, ξ µk ∂t

  ∂   = H  ξ µk , −i  , t  ψ t , ξ µk ∂ξ µk    

( { })

This Schrödinger eqn. is solved using the finite difference scheme. In some cases, we can also derive the Schrödinger eqn. from an action principle and solve it using the FEM. For example, for a single particle, the Schrödinger eqn. i

∂ψ(t , r ) ic A(t , r )  2  = − ∇ +  ψ (t , r ) + V (t, r ) ψ (t , r ) ∂t 2m  

can be derived from the "complex" variational principle δS [ ψ, ψ ] = 0 where S [ ψ, ψ ] =

∫ L(ψ , ψ, t , r ) dt d

L(ψ , ψ, t , r ) =

i (ψ ψ , t − ψ ψ , t) 2 −

3

r,

2  ieA   ieA   ∇ +  ψ  − V ψψ  ψ,  ∇ −       2m   

[3] Tracking of Moving Targets Using a Camera Attached to the Tip of a 2 = Link Robot: Let the screen eqn. be r = R(u , ν) = X (u , ν) x + y (u , ν) y + z (u , ν)z The object moves along the trajectory t → r 0 (t ) At time t, the link angles are

138

Stochastics, Control & Robotics

q (t ) = qo (t ) + ε δ q (t ) , φ(t ) = φ0 (t ) + ε δ φ(t ) where

3 F (q, q , q, φ , φ) = τ0 (t ) + ε ω (t ) ∈R

Thus, F (q0 , q0 , q0 , φ 0 , φ0 ) = τ0 (t ) F can be derived from the Lagrangian L=

(

)

 q  1 T  q , φ J ( q)   − V ( q) 2  φ 

 M (q ) 0  2× 2 J (q) =   ∈R3×3 M ( q ) ∈R T J 3 ( q )  0 Position of the camera: where

r 2 (t ) = r1 + ψ( q , φ) r 2 (t ) = r 20 (t ) + ε δ r 2 (t )

 ( q0 (t ), φ0 (t )) r 20 (t ) = r1 + ψ δr 2 (t ) =

∂ψ ∂ψ ( q0 (t ) φ0 (t ))δq(t ) + ( q0 (t ) φ0 (t ), δφ(t )) ∂q ∂ψ

= A (t ) ξ(t )  ∂ψ  ∂ψ q0 (t ), φ0 ( t ) ) ( q0 (t ), φ0 (t )) ( where A (t )  ∂φ  ∂q  Object moves along a trajectory r 0 (t ) . At time t, the ray. equation joining the object and the camera is λ → r0 (t ) + λ( r2 (t) − r0 (t )) . Assume the parametric equation of the surface as f (x, y, z) = 0, i.e.

f (r ) = 0

Then the λ = λ (t) corresponding to the image point on the surface is f (r0 + λ( r2 − r0 )) = 0. r (t ) → r 0 (t ) + δr0 (t ) . When r 2 → r20 + δr2 , thus λ → λ10 0 + δ λ (t ) and 0 Thus, f (r 0 + δ r 0 + δλ (r 20 − r 0 ) + λ 0δ r 2 ) + λ 0 (r 20 − r 0 ) = 0

or

(∇f (r 0 − λ 0 (r 20 − r 0 )), δ r 0 + δλ (r 20 − r 0 ) + λ 0 δ r 2 )

= 0.

or, (δ r (t), ψ (t )) + δ λ (t ) (r 20 − r 0 , ψ (t )) + λ 0 (δ r2 (t ), ψ (t )) = 0

Electromagnetus & Related Partial Differential Equation

139

where

ψ (t ) = ∇ f (r 0 (t ) + λ 0 (t )(r20 (t ) − r0 (t )))

Thus,

dλ (t) =

− (δ r 0 (t ), ψ (t )) + λ 0 (t ) (δ r 2 (t ), ψ (t ))

(r20 (t ) − r 0 (t ), ψ (t))

Here r0 (t) is the non random component of the objects motion and δr 0 (t ) is its random component. Likewise r 20 (t ) is the non-random component of the camera's motion and δr 2 (t ) is the random component

(

)

1 2 * * * exp (La (u) + L* a*(u)) = exp L a ( u ) ⋅ exp( L a (u )) ⋅ exp  u LL  2  1  2 * = exp (L*a*(u)) exp  u L L  exp ( La (u)) 2  exp (L a (u)) f |e (v)〉 =

=

∞

( Ln f ) ∑ n a (u )n e (v) n=0 ∞

( Ln f ) ∑ n u, v n=0

(

n

e (v) = exp ( u , v L ) f e(v)

)

= exp ( u , v L ) f e(v) ≡ |exp 〈u, v〉 L) f e (v)〉 and 〈fe (v)| exp (L a (u) + L*a (u)) |f e(n)〉

( )

 1 2 = f exp  u LL* + u , v L + u, v L* f  . exp v  2

2

We have thus proved.

 ν 2 e ( ν) and let f∈h be such that f Theorem: Let |j (v)〉 = exp  −  2  Then the moment generating function of the quantum random variable X = La(u) + L* a*(u) In the state (puv)|f j (n)〉 is given by MX(t) = 〈fj(v) |exp (tX)|fj(v)〉 =

=

=

(

)

 1 2 f exp  t 2 u LL* + t u , v L + v, u L*  f  2 ∞

(

)

1 1   2 ∑ n f  t u, v L + v, u L* + 2 t 2 u LL*  n=0 ∞

∑

r, s = 0

(

)

r

f t r u , v L + v, u L* ⋅

t 2s 2

s

u

2s

(L L )

n

* s

f

f ⋅

1 rs

2

= 1.

140

Stochastics, Control & Robotics

= Thus,

∞

∑

t 2s + r

r, s = 0 2

s

rs

f

( u, v

)

r

L + v, u L* ( LL* ) s f

u

2s

f ϕ (v ) X n f ϕ (v ) =

∑

n 0≤ s ≤   2

n f s n − 2s

( u, v

L + v, u L*

)n−2s ( LL* )s

f

u

2s

These are the moments of the quantum random variable X in the pure state f ϕ(v) of h ⊗ Γ s ( H ) .  [4] Problem: Design MATLAB programmes for calculating the pressure field, the density field and the external potential field from velocity measurements using the NavierStokes equation for a gas: ρ (( v , ∇ ) v + v , t ) = −∇p + η∇2 v − ρ∇φ , p = Kρr div(ρ v) +

∂ρ = 0. ∂t 

[5] n-Dimensional Helmholtz Green’s Funtion:   n ∂2  ∑ 2 + k 2  ψ( x) = δ ( x)  α =1 ∂xa  Let Then

ψ ( x) =

1 (2π) n

∫n ψ (ξ)exp(i (ξ, x)) d

n

ξ

R

2  2   k − ξ  ψ (ξ) = 1

 (ξ ) = ψ

ψ( x) =

1 k2 − ξ

exp(i ξ, x)

1 ( 2π )

2

n

∫

2

2

 k − ξ 

ξ) , d n ξ = r n −1drdΩ(

dnξ

r = ξ , Let

Electromagnetus & Related Partial Differential Equation

 ξ = ψ( x) =

Then Let

∫n dΩ(ξ)

141

ξ r 1 (2π)

n

∫

exp(ir || x ||)

(k

2

−r

2

)

r n −1drdΩ( ξ)

= Sn (a constant)

S

When

{

}

ξ ∈R n  ξ =1 Sn =  Then ψ( x) =

Sn

(2π) n

∞

∫

0

(k

r n −1 2

− r2

)

exp (ir || x ||) dr



[6] Edge Diffusion: f0 (x, y) = δ ( px + qy − 1)

(

 x2 + y 2 1  exp − ft (x, y) = 4πDt 4 Dt 

)



*

f 0 ( x, y )

= Kt ( x, y ) * f 0 ( x, y) = =

1 q

p



∫2 δ  q ξ + η − 1 Kt ( x − ξ, y − η) d ξd η

R

pξ   1 dξ kt  x − ξ, y − 1+ ∫ q q  

 Diffused edge. 

[7] Waves in Metamaterials: ε o | − ε o (1 − δχ(r1 )) = ε(r ) z=0

(

)

Z < 0 : ∇2 + k 2 E (r) = 0 Z < 0 : div ((1 + δχ) E ) = 0 div E = − δ (∇χ, E ) + O(δ 2 ) . ∇ × E = − jwµH , ∇ × H = jwε E

142

Stochastics, Control & Robotics

∇(div E ) − ∇E = ( − jwµ ) ( jw ε) E

⇒ ⇒

∆ + w2µε E − ∇(dw E ) = 0

⇒

( ∆ − k 2 − δ ⋅ k 2 ) E − δ ∇ (∇χ, E ) = 0

{

}

( ∆ − K 2 ) E = δ ⋅ k 2 E + ∇(∇χ, E) (o) (1) E = E + δ⋅E

( ∆ − k 2 ) E (o) = 0

(

( ∆ − k 2 ) E (t ) = k 2 E (o ) + ∇ ∇χ, E (o ) We can thus write for z > 0,

)

E ( r ) = E (o ) ( r ) + δ ⋅ E (1) ( r ) =

∫  F (n) + δ ⋅ H (r , n) F (n) exp(−k n . r ) d w (n) + O(δ

2

)

(1)

1  exp(k  n ⋅ ( r − r ′ ))exp(− k ( r − r′ )) where H ( r , n) = − 4π ∫

{

}

3 n(∇ χ ( r ′ ))T d r ′ × k 2 I + ∇∇T χ( r ′ ) − k  | r − r′ |

n) ∈ 3 ×1 H (r,  n) ∈ 3 × 3 , F ( (o) n, F ( n)) = 0. Now, div(E ) = 0 ⇒ (

further, ( E X , EY ) Z = o+ = ( E X , EY ) Z = o − Z 0 and ∫ (1+ | ξ | ) 

) | u (ξ) |2 d ξ < ∞

∫ (1+ | ξ |

2 s

⇒

(since for s < 0, (1+ | ξ |2 ) s ≤ (1+ | ξ |2 ) −2 ∀ξ ∈R N ) ⇒

φ, u

2

2 −s ϕ(ξ) |2 d ξ ≤ C ′ ∫ (1 + | ξ | ) | 

≤ C

∑

∂α φ

||α||≤ 2|s|

2 2

These discussion imply that  H −∞ = u ϕ, u ≤  H −∞ =

∪ Hs

s ∈R

{

∑

||α||≤ n

 2 ∂α ϕ ∀ ϕ for some n ≥ 0 2 

= u | lim

δ→−−∞

∫ (1+ | ξ |

}

u (ξ) |2 dξ < ∞ . ) |

2 s



3 Radon and Group Theoretic Transforms With Robotic Applications [1] Modern Trends in Signal Processing by Harish Parthasarathy, NSIT: [Reference: Radon Transform-Some Basic Properties.] [1] Radon transform based computer tomography. [2] Numerical methods for solving partial differential equation with applications to waveguides and Cavity resonators. [3] Modelling noise in quantum systems. [1] Image field f : R n → C is an image Hyperplane is

, r ) =  ∈ S n−1 , p ∈R . (m p, m

Sn–1 is the n–1 dimensional sphere in Rn. Sn–1 =

{r ∈R

n

}

| r =1

Projection of f on the hyperplane: n  , p) = (Rf) (m ∫ f ( r )δ p − m , r d r Rf  Radon transform of f. Example: n = 2 (Rf) (I, m, p) =

( ( ))

Rn

∫

f ( x, y ) δ( p − lx − my) dx dy

R2

Inversion formula:  , w) = (FpRf) (m =



∫ ( Rf ) (m, p) exp (− jwp) dp

R



∫ f ( r ) exp (− jw(m, r ))d

n

r

152

Stochastics, Control & Robotics

f (r ) =

So

 ξ

1 (2π)

n



∫n ( Fp Rf )  || ξ || , || ξ || exp( j (ξ, r )) d

n

ξ

R

Behaviour of Radon transform under rotations and translation SO(n) = {k ∈ R n×n | K T K = I , det( K ) =1} Translates and rotated image field: Tk , a f ( r ) = f ( K r + a ), a ∈ R n , K ∈ SO(n) .  ( RTk , a f ) (m, p) =  F p RTk , a f (m, w) =



∫n f ( K r + a ) δ( p − (m, r ))d

r

R



∫ ( RTk ,a f ) (m, p) exp(− j wp) dp

R

=

n



n

∫ f ( Kr + a) exp(− jw(m, r ))d r  −1 n ∫ f ( x ) exp(− jw(m, K ( x − a))d x

(x = Kr + a) since det K = 1.  , a)) ( Ff ) ( w k m )  , w) = exp( jw( K m Thus ( F RT f ) (m p

k, a

 , a )) ( Ff Rf ) ( K m  , w) = exp( jw( K m  , w) | = | ( Ff ) ( w K m)  | | ( F p RTk ,a f ) (m

K can be recovered from this using Fourier transform on SO(n). 

[2] Randon-Transform Based Image Processing: Let f : R 2 → * C an image field. Its projection on the plane lx + my – p = 0 2

2

where l + m = 1 is given by (Rf) (l, m, p) =

∫ f ( x, y) δ ( p − lx − my) d x d y

If f (x, y) is a random Gaussian field, then

( ) (l ', m ', p ')

 ( Rf ) ( l , m, p ) Rf  =

∫ R ff ( x; y; x ' y ') δ ( p − lx − my) δ ( p '− l ' x '− m ' y ') dx dy dx ' dy '

Suppose f is a WSS (Wide-sense-stationary) field. Then Rff (x, y; x′ y′) ≡ Rff (x – x′, y – y′)

( ) (l ', m ', p ')

Then  ( Rf ) ( l , m, p ) Rf 

Radon & Group Theoretic Transforms with Robotic Applications

=

153

∫4 R ff ( x − x ′, y − y ′) δ ( p − lx − my) δ ( p ′ − l ′x ′ − m′y ′) dx dy dx ′ dy ′

R

∫4 R ff (ξ, η) δ ( p − lx − my) δ ( p ′ − l ′( x − ξ) − m′( y − η)) dx dy dξ d η

R

= Let

∫ R ff (ξ, η)d ξd η ∫ δ( p − lx − my) δ( p ′ − l ′)(x − ξ) − m′( y − η))dx dy f (p, p', l, m, l', m') =

Let

∫ δ ( p − lx − my) δ ( p ’− l ’( x − ξ) − m ’( y − η)) dx dy

lx + xy = u l'x+m'y = v

−1

Then

 x  l m  u   y =  l ' m '   v       m ′ − m  u  (lm′ − l ′m) =   −l ′ l   v  m ’u =   lv

So F (p, p′, l, m, 1′, m′) =

−mv (lm ’− l ’m) −l ’u 

∫ δ( p − u ) δ ( p ′ + l ′ξ + m′η − v) (lm′ − l ′m)

= (lm′ – l′m)–1.

( ) (l ′, m′, p′)

Thus,  ( Rf ) (l , m, p ) Rf 

= (lm′ − l ′m ) Inversion formulas:

∫ ( Rf )(l , m, p ) e

R

− j ωp

dp =

∫ f ( x, y )e

−1

∫

R ff (ξ, η) d ξ d η

R2

− jω (lX + mY )

dxdy

= f (l ω,mω )

where f denote 2-D Fourier transform of f. Thus, writing ξ=lw, η=mw, we get   ξ η f ( ξ, η) = ( Rf )   , , p ∫  2 2 2 2  ξ +η R  ξ +η 

(

)

exp − jp ξ 2 + η2 dp and thus

ξ η   −1 , ,p f(x, y) = (2π) ∫ ( Rf )   2 2 2 2 ξ +η  ξ +η 

−1

du dv

154

Stochastics, Control & Robotics

(

)

exp − jp ξ 2 + η2 exp ( j (ξX + ηY )) dpdξd η Generalization to n-dimensional images: f : R n → C.

∫n f ( x) δ ( p − (m , x)) d

( Rf ) (m , p ) ( Rf ) (m , − p )

∫ ( Rf ( m , p ) exp ( − ) ωp ) dp

R

=



R





) ∫ ( f ( x) exp (− j (ω m , x)) d x n

=

∫n f ( x ) exp ( − j ( ξ, x ) ) d n x, ξ ∈ R n

R





ξ n ∫ n (2n) ( Rf )  ξ , p

f (x) =

So

(

Rn

Equivalently, ξ

x

, p . = ( Rf ) −m

R

∫ ( Rf )  ξ , p  exp ( − j | ξ | p | dp

n

R×R

{

}

exp j ((ξ, x ) − ξ p ) d p d n ξ Radon transform of a moving image: ψ ( X , Y , t ) = f ( X − VX t ,Y − VY t ) . ≡ ψt ( X ,Y ) .

( Rψt ) ( l , m, p )

=

∫2 f ( X − VX t ,Y − VY t ) ξ ( p − lX − mY ) dXdY

R

=

∫ f ( X ′, Y ′) δ ( p − lX ′ − mY ′ − (lVX + mVY ) t ) dX ′dY ′

= ( Rf ) (l , m, p − (lVX + mVY ) t )

∫ ( Rψt ) ( l , m, p ) e

R

− jωp

dp

{

}

− jωp dp = exp − jω (lVx + mV y )t × ∫ ( Rf ) (l , m, p ) e

(

)

R

i.e. F p R ψ t (l , m, ω )

{ (

) } ( Fp Rf ) (l, m, ω)

= exp − jω lVx + mV y t

(

)

≡ F p R ψ 0 (l , m, ω )

Radon & Group Theoretic Transforms with Robotic Applications

Thus,

{(

}

)

{(

155

}

)

Arg F p R ψ t1 (l , m, ω ) − Arg F p R ψ t 2 (l , m, ω ) = ω (t2 − t1 ) (lVx + mV y )

ψ( X , Y ) = 0  eqn. of a curve in the plane. Project the image field f(x, y) on this Rψ(f) =

curve: e.g.:

∫ f ( x, y) δ (ψ ( x, y))dxdy

y = j(x) i.e. ψ(x, y) = y – j(x).

Then Rψ(f) =

∫ f ( x, y) δ ( y − ϕ( x)) dxdy = R∫ f ( x, ϕ ( x)) dx

f (x, y) = s(x, y) + w(x, y) Rψ (f) = Rψ(s) + Rψ(w). NSR =

{

 Rψ ( w)2 Rψ ( s )

2

} = (SNR)

–1

X → X cos θ + Y sin θ Y → − X sin θ + Y cos θ

f (X, Y) → f (X cos q + Y sin q – sin q + Y cos q) = f ( X , Y )  Rotation image.

Rf (l , m, p ) =

∫ Rf k , x (n , p ) e

R

( Fp Rfk ,x

∫ f ( X , Y ) δ

( ( )) ∫ (R f ) (k n, p) = exp ( − jωp ) dp ) (n , ω) = exp ( jω ( x, K n ))( F R f )( K n, ω) − j ωp

dp = exp jω x, k n

R

p

If K ∈O (n) is knows, x can be determined using

(

)

ω x, K n = Arg

{( F R f ) (n , ω)} − Arg {( F R f ) (K n , ω)} p

k ,s

p

( ) = ( F R f )( K n , ω)

F p Rf k , x n, ω

p



[3] Image Processing Using Invariants of the Permutation Group: Sources are located at r1, r 2 , ..,r N .

(

)

Signal field emitted by source at r k is f k t − r − r k c , 1 ≤ k ≤ N. Total signal field generated is

156

Stochastics, Control & Robotics

f (t , r ) =

N

∑ f k (t − r − r k c )

k =1

If a permutation s (∈SN) is applied to the sources, the generated signal field is N

U σ f (t , r ) =

∑ f k (t − r − r σ k c )

k =1 N

=

∑

k =1

(

f σ −1k t − r − r k c

)

Us Us f (t, r) = Usr f (t, r). Let

f (t , r , r1 ...., r N ) =

N

∑ f k (t − r − r k c )

≡ f (t , r )

k =1

Us f (t, r) = f (t, r , r σ1, ....,r σN ) .

Then Group algebra:

∑

σ ∈S N

a (σ ) U σ

p Let s → p(s) be a unitary representation on SN in C . Then consider

f π (t , r ) = We get

(U σ f )π (t , r ) = =

(

∑

π(σ)U σ f (t , r )

∑

π(σ)U σρ f (t , r )

∑

π(σρ−1 ) U σ f (t , r ) = f π (t , r )π* (ρ)

σ∈S N σ∈S N

σ ∈S N

So, (U σ f )π (t , r ) (U σ f )π (t , r )

)

*

* = f π (t , x ) f π (t , r )

Thus, f π (t , r )* f π (t , r ) is an SN invariant matrix field in space-time.

Young's Tableaux, primitive idempotents and the irreducible rep'm of SN.

Let N = 3, and consider the tableaux T=

1

2

3

eT =

∑

Sqn(q) pq

p ∈RT , q ∈CT

in the group algebra. s0 = Id Let

 1 2 3 s1 =   2 3 1

Radon & Group Theoretic Transforms with Robotic Applications

157

 1 2 3 s2 =   3 1 2  1 2 3 s3 =   2 1 3  1 2 3 s4 =   3 2 1 1 2 3 s5 =  1 3 2

= Id – (13) – (2 3) (1 3) + (2 1) = Id – s4 – s5 s4 + s3

Now

s5s4 = (2 3) (13) = s1 eT = s0–s4–s1+s3

ReT is an irreducible representation. s0eT = e.eT = eT, s1eT = s1



[4] Radon Transform of Rotated and Translated Image Field: f :  n → C

{

}

O(n) = K ∈ n × n K T K = I ,det = ( K ) = 1 .

( )

Rf K , x n , p

(n , y )

n = ∫ f ( x + K . y) δ ( p − ( y,n )) d y

= p.

n1y1+...+nnyn = p, n –1 dimensional plane in Rn . n1x+n2y+n3z = p 2 – D plane in R3.

Computer Tomography.

( )

Rf K , x n , p = =

( )

Rf n , p given.

( )

∫ f ( x + y) δ ( p − ( y, K n ) ) d y n ∫ f ( y)δ ( p + ( x, K n ) − ( y, K n ) ) d y n

(

)

 p + ( x, K n ) Rf K , x n , p = Rf K n,

K  rotator x = translation. 

158

Stochastics, Control & Robotics

[5] Estimating the Rotation Applied to the Two 3-D Robot Links from Electromagnetic Field Pattern: At time t the rotation suffered by the bottom link is R1(t) and that by the top link is RzR1. A point r in the bottom link moves to R1 (t ) r = R (φ1 (t ), θ1 (t ), ψ1 (t )) r 1

and third in the top link to R1 p0 + R2 R1 ( r − p0 ) . R1(t) = R1 (φ2 (t ), θ 2 (t ), ψ 2 (t ))

Where

(

The correct density in the bottom link is J R1−1 r 1 −1 J p0 + ( R2 R1 ) ( r − R1 p0 ) . 2

)

(

) and that in the top link is

Assuming infinite speed of light, the magnetic vector potential at time t is A (t , r ) =

∫

1

+∫ =

(

J R1 (t ) −1 r 1

∫

J

1

r − r′

) d r′ 3

( p + (R (t ) R (t ) ) ( r ′ − R (t ) p )) d r ′ 0

2

−1

1

0

3

r − r′

J 1 ( r ′) r − R (t ) r ′

d 3r ′ + ∫

1

Let

1

J 2 ( r ′ ) d 3r ′

r − R1θ p0 − R2 (t ) R1 (t ) ( r ′ − p0 )

.

R2(t)R1(t) = S (t), R1(t) = R (t). Then, A (t , r ) =

∫

J 1 ( r ′) J 2 ( r ′ ) d 3 r′ d 3r ′ + ∫ . r − R(t ) r ′ r − R(t ) p0 − S (t ) ( r ′ − p0 )

By taking measurements of A (t , r ) of different r ′s, R(t), S (t ) ∈SO (3) can be estimated. For r >> r ′ , r − R(t ) r ′

−1

(

)

=  r 2 + r ′ 2 − 2 r , R (t ) r ′  

≈ r −1 1 + 

(

( r , R(t ) r ′ )  r2

−1 2

(

r , R (t ) r ′ = 1+  r r2

)

)

r , Rp0 + S ( r ′ − p0 ) ≈ 1+ r r2 So the non constant part of the magnetic vector potential at r >> r ′ is approximates r − Rp0 − S (r ′ − p0 )

−1

A (t , r ) =

∫ J1 (r ′ ) (r , R(t ) r ′ ) d r ′ r + ∫ J 2 (r ′ ) ( r ) , R (pt ) + S (t ) + (r ′ − p0 ))d 3 r ′ 0 3

2

r2

Radon & Group Theoretic Transforms with Robotic Applications

Let l be a constant unit vector. Then T l , A (t , r ) = R(t ) r ,

) (

(

∫ ( J1 ( r ′), l ) r ′d

3

159

r′

)

r2

(R(t ) r , p ∫ ( J ( r ′), l )d r ′ r ) + ( S (t ) r , ∫ ( J ( r ′), l ) ( r ′ − p ) d 3r ′ r ) T

0

3

1

2

T

2

0

This is a linear equation for R(t ), S (t) and The rotations R(t ), S (t) can be estimated by least squares techniques. 

[6] Kinetic Energy of a Two Link Robot: r → R1 (t ) r ( r ∈ second link). r → R1 (t ) p0 + R2 (t ) R1 (t )( r − p0 ) (r∈ second link). Let B1 denote the volume of the bottom link at time t = 0 and B2 that of two second

link. Then the kinetic energy is T=

1 ρ R1′(t )r 2 B∫ 1

2

1 d 3 r + ρ2 2

∫

(

R1′(t ) p0 + ( R2 R1 )′ (t ) r − p0

B2

R1(t) = R(t), R2(t) R1(t) = S(t). Then

Let

T=

1 ρ1 R ′ (t )r 2 B∫

2

1

∫

1 d 3 r + ρ2 2

(

R ′(t ) p0 + S ′(t ) r − p0

B2

Let

T1 =

1 R ′ (t ) r 2 B∫

2

d3r =

1

ρ2

∫

2

d 3r

{

}

1 Tr I R ′T (t )R ′ (t ) 1 2 2

R ′(t ) p0 + S ′(t ) ( r − p0 ) d 3 r

B2

= µ ( B2 ) R ′(t ) p0

2

m(B2) ≡ m2 = ρ2

∫d

B2

3

r,

{

}

+ Tr I S ′T (t )S ′ (t )

+2Tr {R ′ (t ) I 3 S ′ (t )} where

)

ρ1 ∫ r r T d 3 r = I , 1 B1

Then

2

)

2

d3r

160

Stochastics, Control & Robotics

I I where

2

3

= ρ2

∫ ( r − p0 ) ( r − p0 )

T

d 3r ,

B2

= ρ2

∫

(

p0 r − p0

B2

R2 =

∫ rd

3

)

T

(

d 3 r = µ 2  p0 R2 − p0 

)

T



r

B2

Group theoretic analysis of the motion of a two-link Robot with the link being 3-D rigid bodies: Each link of the Robot is a 3-D rigid Top. The first link has its base point attached to the origin and the second link has its base point attached to some fixed point p on the top surface of the first link. The group of motions is shown to be a representation of SO (3) × SO (3). We assume that each link carries a d.c. current density an then computer the magnetic field produced by this current field after both links have undergone rotations around their respective base points. r

p

At time t = 0, the point p at the joint of the two links is at p0 and the point r at a fixed location in the second link is r 0 . q0 = r 0 − p0 is the position of r relative to p at time t = 0. The final configuration is arrived at by first applying a rotation R1 to the entire system about the origin O, taking p0 → R p0 , r 0 → R r0 1

1

So that q0 → R1q0. We then apply a rotation R2 to the second link around p = R1p0 taking R1q0 → R2R1q0. The magnetic field produced by the Robot after the application of the two rotations is computed using the Biot-Savart law (non- relative) calculation. The final position of r is thus r = R1p0 + R2R1q0 = R1p0 + R2R1 (r0 – p0) = R2R1r0+ (R1 – R2R1) p0 and

p =Rp 1 0

Radon & Group Theoretic Transforms with Robotic Applications

161

The group action is then  r 0   R2 R1 r 0 + ( R1 − R2 R1 ) p0   R2 R1  =  p  →  R1 p0  0   0  

R1 − R2 R1   r 0    R1   p0 

The group G is this given by   R2 R1 G =   0

 R1 − R2 R1  : R1 , R2 ∈SO (3)  R1  

 R S − R  : R, S ∈SO (3) G =   S   0 

or equivalently, Composition law:

 R S − R T (R, S) =  S  0

Let Then

S2 − R2   R1 ⋅ S2   0

 R2 T ( R2 , S2 ) ⋅ T ( R1 , S1 ) =  0  R2 R1  0 

S1 − R1  S1 

S2 R1 − R2 R1   = T(R2R1, S2S1). S2 S1 

This shows that ( R, S ) → T ( R, S ) is a representation of SO (3) × SO (3) in R6. Now suppose that lower link carries an initial current density (t = 0) J1 ( r ) and the top link J 2 ( r ) . Then after time t, the current density in space is given by J (t , r ) = J R −1 (t ) r + J R −1 (t ) R −1 (t ) r − R (t ) p 1

(

1

)

2

(

1

Writing R(t) = R1(t) and S(t) = R2(t) R1(t), we get

(

)

(

2

(

J (t , r ) = J1 R −1 (t ) r + J 2 S (t ) −1 r − R (t ) p0

where

R (t ) = R1(t) –R2 R1(t)

0

)

))

Note that J1 ( r ) is zero for r ∈ B1 and J 2 ( r ) is zero for r ∈ B2 where B1 and B2 are respectively the volumes of the first and the second link. Note that B1 ∩ B2 =

f. The magnetic field produced by this current density in space B (t, r ) =

∫3

R

J (t , r ′ ) × ( r − r ′ ) 3 d r′ 3 r−r

(By Biot Savart's law): Assuming that x , y, z are the column vectors  1  0  0  0 ,  1 and  0        0  0  1

162

Stochastics, Control & Robotics

 X  X ′   r = Y , r′ =  Y ′  ,     Z  Z ′

respectively and

B (t , r ) =

we have

∫ G ( r − r ′ ) J (t , r ′ ) d

3

r′

 x × r y × r z × r  1 G( r ) =  3 , 3 , 3  = r r   r r3

where

0   −z  y

− z − y 0 x  − x 0 

Here, J (t , r ) is a 3 × 1 column vector and G ( r ) is a 3 × 3 matrix B(t, r) is a 3 × 1 column vector. We can simplify the above to B (t , r ) =

∫ G ( r − R(t )r ′ ) J1 (r ′) d

B1

(

3

r′

)

+ ∫ G r − R (t ) p0 − S (t ) r ′ J 2 (r ′ ) d 3 r ′ B2

 = R–S R

Note that

(1)

The problems are (1) estimate the matrices R(t), S(t) ∈ SO (3) for a fixed t, from

measurements, of B(t, r) at different r = r1 r2, ..., rN. (2) Estimate the square R (tk ) , S (tk ) , k = 1, 2, ..., N at a square of time points t1 t2, ...., tN from measurements of

B(t, r) at times t1, ..., tN, r∈{r1, ..., rM}. Here relativistic effects are not being taken

into account i.e. we are not using related potentials . Taking the spatial Fourier Transform of (1) gives  (t , k ) = B =

∫ B (t , r ) exp (i k ⋅ r ) d

3

r

R3

∫ exp (i (k , R(t ) r ′ )) G (k ) J1 ( r ′) d

B1

((

3

r′

))

 ( k ) J ( r ′) d 3r ′ + ∫ exp i k , R (t ) p0 + S (t ) r ′ G 2

Or equivalently,  (k ) −1 B  (k ) = G

B2

∫ exp (i ( k , R(t ) r ′ )) J1 ( r ) d

3

r

B1

∫ exp (i ( k , R (t ) p0 + S (t ) r )) J 2 ( r ) d

B2

Consider the function

exp (i ( k , r ))

3

r

Radon & Group Theoretic Transforms with Robotic Applications

163

We can express it as exp (i ( k , r )) =

{ ( )}

where Ylm r property

m ≤l , l ≥ 0

(

∑ C (l , m, k, r ) Ylm (r )

l, m

are the spherical harmonics. They satisfy the transformation

)

Ylm R −1 r = th

∑

m′ ≤ l

 π  l ( R ) m′m Ylm′ (r ) , R ∈ SO (3)

Where pl is the l irreducible representation of SO (3). They coefficients (l, m, k, r) are given by

C (l , m, k, r ) = =

∫2 exp (i ( k , r ))Ylm (r ) dS (r )

S π 2π

∫ ∫ exp (ikr (sin α sin θ cos (φ − β) + cos α cos θ)) 0 0

Ylm (θ, φ) sin θ d θ dφ Where k = k (cos b sin a, sin b sin a, cos a) Then

exp (i ( k , Rr )) =

∑ C (l , m, k, r ) Ylm ( Rr )

l, m

= and hence we get  ( k ) −1 B  (k ) G =

∑

l , m, m′

( )

C (l , m, k, r )  πl R −1  Y (r )   m′m lm′

∑ (∫ C (l , m, k, r )Ylm′ (r ) J1 (r ) d 3r ) πl ( R −1 ) m′m

lmm′

((

 + exp i k , Rp 0

( )

 π S −1   l  m ′m =

)) ∑ (∫ C (l, m, k, r ) Ylm′ (r ) J 2 (r )d 3r ) lmm′

∑ ψ k [lmm′] πl ( R) mm′ + exp (i ( k , R p0 ))

lmm ′

∑ χ k [lmm′] πl ( R) mm′

lmm′

where ψ k [lmm′ ] =

∫ C (l m k, r ) Ylm (r ) J1 ( r )d

3

r and χ k [lmm′ ]

B1

= Now,

((

exp i k , R p0

∫ C (l m k, r ) Ylm (r ) J 2 (r ) d

B2

)) = ∑ C (l, m, k, p0 )Ylm ( R p0 ) lm

3

r

164

Stochastics, Control & Robotics

=

∑ C (l , m, k, p0 ) πl ( R) mm′ Ylm ( p0 )

lmm ′

 ( k ) −1 B  (k ) = G

So

∑ ψ k [lmm′] πl ( R) mm′

∑

+

lmm′

C (l1 , m1 , k, p0 )

l1m1m′1 l2 m2 m′2 l3m3m′3

( )

  π (S ) χ k [l2 m2 m2′ ]  πl1 ( R)   m2 m′ 2 − p 0 m1m ′ 1  l2

(Y

l1 m1′

)

Yl3 m3′ × C (l3 , m3 , k , p0 )  πl3 ( S ) 

m3m3′



[7] Vector Potential Generated by a 2-Link Robot: r 0 → R1 r 0 , r 0 ∈ B1 (lower link) r1 → R1 p0 + R2 R1 ( r1 − p0 ) , r1 ∈ B2

R1 = R, R2 R1 = S (upper link) J1 ( r )  current density in lower link. J 2 ( r )  current density in upper link at time t = 0 Current density at time t,

(

(

−1 −1 J ( r ) = J1 ( R r ) + J 2 p0 + S r − Rp0

))

For field magnetic vector potential A (r ) = =

∫ J ( r ′ ) | exp ( j k r ⋅ r ′ ) d ∫ J ( r ′ ) | exp ( j k ( R

−1 

3

r′

r ⋅ r′

( (

) ) d r′ 3

))

+ ∫ J 2 ( r ′ ) exp jk r , Rp0 + S ( r ′ − p0 ) d 3 r ′ −1 −1 −1 −1 = A1 ( R r ) + A2 ( S r ) × exp( jk( R − S )r , p0 )

A1 (r ) =

∑ C1 (lm)Ylm (r ),

exp − jk r , p0 A2 ( r ) =

∑ C2 (lm)Ylm (r ).

Let

( (

Then,

))

A1 ( R −1 r ) =

l ,m

l ,m

∑ C1 (lm) [πl ( R)]m′m Ylm′(r )

lmm′

( (

A 2 (r ) = exp jk R −1 r , p0

)) ∑ C2 (lm) [πl (S )]m′m Ylm′(r ) l , mm′

Radon & Group Theoretic Transforms with Robotic Applications

165

≡ A2 ( S −1 r ) exp( jk ( R −1 − S −1 )r , p0 ) exp( jk(r , p0 )) =

Let

∑ d (lm)Ylm (r ) l ,m

∫2 A2 (r )Yl o mo (r )dS(r )

So,

s

=

−1 ∫ ∑ C2 (lm)d (l ′′m′′)Yl ′′m′′ ( R r )[πl (S )]m′m Ylm′ (r )Ylomo (r )dS((r ) lmm′ l ′′, m ′′

A (r ) = A1 ( R −1 r ) =

∑ C1(lm)Ylm ( R −1 r )

=

lm

So

∑ C1(lm)[π1( R)]m′m Ylm′ (r )

lmm′

∫ A1 (r )Ylm (r ) dS(r )

=

∑ C1 (lm′)[πl ( R)]mm′

∫2 A1(r )Ylomo (r )dS (r )

=

∫2 ( A1 (r ) + A2 (r ))Ylomo (r )dS(r )

S

m′

S

=

∑ πl0 ( R) m0m C1(lm) m

+

∑

l1l2 m1m2 m3m4

n-link Robot

C2 (l1m1 )d (l2 m2 )[πl2 ( R)]m3m2 [πl1 ( S )]m4 m1

∫ Yl1m4 (r )Yl0m0 (r )Yl2m3 (r )dS(r )

First link is attached to ground at its base point. Second link is attached to the first link at p1.  kth link is attached to the (k – 1)th link at pk – 1.  nth link is attached to the (n – 1)th link at pn – 1. All links first suffer a rotation R1 around O. 2nd to nth links suffer a rotation R2 around R1p1.  nth link suffers a rotation Rn around

166

Stochastics, Control & Robotics

R1p1+R2R1(p2–p1)+R3R2R1(p3–p2) +…+ Rn–1…R1(pn–1– pn–2) So let r k be an initial point in the kth link, k = 1,2,…,n. Then after the sequence of rotations R1, …, Rn, r k → R1p1 + R2R1(p2–p1) + … + Rk–1 … R1(pk–1 – pk–2) + RkRk–1 … R1(rk – pk–1)

k = 1, 2, …, n.

Writing R1= S1, R2R1 = S2, … Rk … R1 = Sk , k = 1, 2, …, n, we get r k → S1p1 + S2(p2–p1) + … + Sk–1(pk-1 – pk–2) + Sk (r k − pk −1 ),1 ≤ k ≤ n. If J k ( r ) is the initial current density in the kth link, then after the sequence of rotation, its current density is Jk ( r ) = J k (Tr −1 r ) where Tk (r) = S1p1 + S2 ( p2 – p1) +… + Sk–1( pk–1 – pk–2 ) + Sk ( r − pk −1 ). k −1   −1 −1 p S + r − S ( p − p T ( r )  = k −1 Thus, k ∑ j j j −1 p0 = 0 k   j=1

and the total current density in the find state is given by J ( r ) = =

n

∑ Jk ( r )

k =1 n





k −1



k =1





j=1



∑ J k  pk −1 + Sk−1  r − ∑ S j ( p j − p j−1 )  

[8] Group Associated with Robot Motion: ( x, y, θ1 , θ2 ) → ( x + a, y + b, θ1 + α1 , θ 2 + α 2 ) A belian group of transformation. When the rigid rods have breadth and width dimension, i.e. they are like tops. Euler angles for the first top (j1, q1, ψ1). q

p

Eular angles for the second top (j2, q2, ψ2). The second top is attached to a point p 0 = (x0, y0, l) on the first top at time t = 0. At time t, this point moves to

Radon & Group Theoretic Transforms with Robotic Applications

167

R (ϕ1, θ1, ψ1 ) p 0 = p 0 ≡ pt . where j1 = q1(t), q1 = q1(t), ψ1 = ψ1(t). The point Q on the second top was at time t = 0 located at p0 + q0 = r 0 =

=

∞

∞

0

0

t ∫ P0 dt ∫ n(Y0ku ∩ {u < R}) du

   Y z k d u   ∫  ∫ 0 u   dt  0 0 

∞  R

∞

=

∫

0

R

=

r

∫ ∫R 0

∞

=

∞

pk (r ) dr ∫

∫ r

∞

p(r ) dr ∫ r

r

Qu du ∫ Y0ku du 2πu 0

r

du  ∫ {Y0 ku du | R = v} 2πu 0

and at time, it is located at pt + q = r t = R (ϕ1 , θ1 , ψ1 ) p + R (ϕ 2 , θ2 , ψ 2 ) q t 0 0

Thus defines a transformation by

r0 → rt

r t = R (ϕ1, θ1, ψ1 ) p0 + R (ϕ 2 , θ 2 , ψ 2 ) (r 0 − p ) 0

(

)

= R 2 r 0 + R1 − R 2 p 0 With p 0 fixed, the transformation group action is T (ϕ1, θ1, ψ1, ϕ 2 , θ 2 , ψ 2 ) r 0

(

)

= R 2 r 0 + R1 − R 2 p0 Or equivalently  r 0   R2 p  →  0  0 

R2 − R1   r 0  I 3   p0 

Find generators for lie algebra of the lie group generated by all 6 × 6 matrix of the form  R2 R2 − R1  R1, R2 ∈SO(3) . , 0 I 3   

168

Stochastics, Control & Robotics

[9] Radon Transform: f (α, p) =

∫n f ( x)δ ( α, x

− p ) dx

R

|| α || −1, α ∈R n , p ∈R . Then f (α, − p) = f ( −α, p) ,

∫ f (α, p)exp(−iwp) dp

=

R

∫n f ( x)exp ( −iw α, x ) dx

R

≡ f (α, w) say. Thus,

f (x) =

∫ w≥0

(

)

f (α, w)exp iw α, x wn −1d Ω(α ) dw (2π) n

α ∈ S n−1

where Thus,

n wn−1d Ω(α ) dw ≡ d ( w x) (Lebesgue measure on Rn) f (x) = f (α, p) exp(−iwp ) exp (iw α, x )

∫

α ∈ S n −1, w ≥ 0, p ∈R wn −1 d Ω(α ) dwdp / (2π) n

( ) ( exp (i ξ, x )

ξ , p exp −i ξ p = (2π) −n ∫ f 

ξ

)

n−1

()

dΩ  ξ d || ξ ||dp n −1  − n n −1  d = (2π) ∫ i  n −1 exp ( −i || ξ || p )   dp

( ) ( ) d Ω ( ξ ) d || ξ ||dp f  ξ , p exp i ξ , x

= (2π) − n ( −i ) n −1 ∫ exp( −i || ξ || p)  ∂ n−1     n −1 f (ξ, p)  ∂p 

( ) d Ω ( ξ ) d || ξ ||dp

exp i ξ, x

Radon & Group Theoretic Transforms with Robotic Applications

 ξ    α = ξ = || ξ ||    Let

(

)

ψ n x , p,  ξ =

∫0 exp {iw ( ξ, x ∞

169

)}

− p dw

∂n f ( x ) = Cn ∫ ψ n ( x , p ,  ξ, p ξ) S n−1 × R n−1 f  ∂p

( )

Then,

()

ξ dp Cn = (2π) − n ( −2)n −1, d Ω 

where Equivalently,

f (x) = (2π )

−n

( −2)

n −1

n−1 x 2

{(

{∫ f (ξ, p)

× exp i ξ, x − || ξ || p

()

)}

dΩ  ξ d || ξ ||dp n−1

= Cn x 2

∫ f (ξ, p)

(

)

ξ S n−1 × R ψ n x, p, 

()

dΩ  ξ dp

 n−1 ξ, p x 2 ψ n x, p,  ξ = Cn ∫ f   S n−1 × R d Ω  ξ dp

( )

(

  

)

()



[10] Campbelt Baker-Hausdorff Formula:

(

X tY X tY ez (t) = e e ⋅ z (t ) = log e ⋅ e

)

d z (t ) z (t ) ( I − exp( − ad z (t )))  dz(t )  e = e  dt  dt ad ( z (t)) e − z (t )

d z (t ) = ez(t)Y. e dt

But So

( I − exp ( − ad z (t)))  dz(t )  d z (t ) e =   ad ( z (t)) dt  dt

e − z (t )

 I − exp ( −odz(t ))  dz(t )  d z (t ) e = Y =   ad ( z (t)) dt   dt  

170

Stochastics, Control & Robotics

(

  adz(t ) d (Y ) = adz(t) = log e adX etadY z (t ) =   I − exp( − adz( t ) )  dt

or

(

)

log eadX etadY dz (t ) (Y ) = dt 1 − eadX etadY

So,

(

)

adX tadY e (Y ) = g e

where

g(ξ) =

log(ξ) 1− ξ

(

(

)

)

1 z (1) – z (0) = log e X eY − X =  ∫ g eadX etadY dt  (Y )   0

Thus,

∞

Suppose

g(ξ) =

∑ Cr (1 − ξ)r g (ξ)

r=0

=

log(1 − (1 − ξ)) (1 − ξ)

∞

(1 − ξ) r −1 r r =1

= −∑ = − Cr = −

Thus, Then

g(ξ) = =

∞

(1 − ξ) r (r + 1) r=0

∑

1 ,r ≥0. (1+ r )

r

r

r=0

k =0

 r

∑ Cr ∑  k  (−1)k ξk ∞

∑ dk ξk

k =0

where dk =

∞

 r

∑ Cr  k  (−1)k

r=k

Not a convergent services. So g(ξ) =

(

g eadX ⋅ etadY

(e

adX

⋅ etadY

)

k

)

=

 r Cr   ( −1) k ξ k  k 0≤k ≤r≤∞

∑

(

 r Cr   ( −1)k eadX ⋅ etadY  k 0≤k ≤r≤∞

∑

)

adX ⋅ etadY eadX etadY ⋅ e adX etadY (W ) (W ) = e k

∞ m   t adX m ( adY ) = e  (W ) ∑ m   m=0



k

)

Radon & Group Theoretic Transforms with Robotic Applications

171

[11] Problem on Rigid Body Motion: [1] Consider the two link Robot with each link being a 3-D top. We've seen that the group of motion is defined by   R2 R1 G =   0

Show that if we define

 R1 − R2 R1  : R1, R2 ∈SO(3)  R1  

 R S − R  =   : R1, S ∈SO(3) S 0   

 R S − R T (R, S) =  S  0

Then

 R2 T (R1, S2). T (R1, S1) =  0

S2 − R2   R1 S1 − R1  ⋅ S2   0 S1 

 R2 R1 S2 S1 − R2 R1  =   S2 S1  0  = T (R2R1, S2S1)

and hence ( R, S) → T ( R, S )

is a representation of SO(3) × SO(3) in R6. 

[12] Remarks on Theorems from V. Kac. Infinite Dimensional Lic Algebra. P.87 1   tw(α ) (λ ) = λ + λ, k w(α ) −  ( λ / w(α )) + | α |2 λ, k  δ   2 wtα w−1 (λ ) = λ + w−1 (λ ), k w(α)

((

)

− w−1 ( λ ) | α +

1 | α |2 < w−1 ( λ ) , K >| w (δ ) 2

−1 w (d) = δ, w (λ ), k = λ, w(k) = λ, k

(w so

−1

) (

)

( λ ) | α = λ w (α )

wtα w−1 (λ ) = λ + λ, k w(α) 1   2  ( λ | w (α )) + | α | λ, k  δ = tw (α ) (λ ) . 2

172

Stochastics, Control & Robotics

P249 n ∈ N Z ⇒ z n = (α, β, u ), n, v

(α, β, u )(v) = tβ (v) + 2πiα + (u − iπ(α | β)) δ  1  tρ (λ ) (n ⋅v) = (λ + (λ | δ )ρ + | n, v ) −   λ ρ + (ρ | ρ)(λ | δ) (δ | n ⋅ v)   2

( ) = (tρ − β (λ ) | v ) + ((λ | δ ) + (λ | δ)(ρ | δ ))

= tρ (λ ) | tβ (v) + 2πiα + (u − iπ(α | β))δ

(u − iπ(α | β)) + 2πi (λ | α ) (λ | δ) = k ∈Z + ρ2 (ρ | δ) = 0.

So,

(

)

tρ (λ ) (n ⋅v) = tρ − β (λ ) | v + k (u − iπ(α | β)) d (n · v) = (d|n · v) = (d|tb (v)) = (d|v). 

[13] Theta a Function From Group Theory P 252 (V.Kac): F (n.v) = F (v)∀ n ∈ N Z .

Let

n = (a, b, u)

n.u = tβ (u ) + 2πiα + (u − iπ(α | β)) δ 1   = v + (v | δ ) β −  (v | β ) + (β | β ) (v | δ ) δ   2 + 2πiα + (u − iπ(α | β)) δ u ∈ 2πiZ , u + iπ(α | β) ∈ 2πiZ u ∈iπ(α | β) ∈ 2u − 2πiZ ∈ 2πiZ

⇒

pa (v) = v + 2πiα . ⇒ (taking b = 0) ⇒ ⇒ So

F (n.v) = F (u )∀n ∈ N Z F (v + 2πiα + (u − iπ (α | β) δ )) = F (v) F (v + 2πiα ) exp{k (u − iπ(α | β)) δ} = F (v)

F (v + 2pia) = F (v) since u –ip (a|b) ← 2piZ. F (v + ad) = eka F (v), F (v + 2pia) = F (v) a ∈ M , a ∈C

Radon & Group Theoretic Transforms with Robotic Applications

173

In term of co-ordination   l v = 2πi  ∑ zs us − τΛ 0 + uδ  s =1 

∑

aτ ( γ )e γ

Consider

kΛ F= e 0

Then

k F = (v + ad) = e (Λ 0 (v) + aΛ 0 (δ ))

∑

γ ∈M *

γ ∈M *

(sec (13.24))

aτ ( γ ) e γ (v )+ a ( γ |δ )

ka = e F (v )

Since

L0 (d| = (L0|d) = 1, (g|d) = 0 (M⊥d)

Also,

F (v + 2pia) = e

k Λ 0 (v )

∑

γ ∈M *

aτ ( γ )e( γ (v ) +2 πi ( γ |α ))

= F (v), α ∈M Since ( γ | α)eZ for γ ∈ M *, α ∈ M by definition of M* F (n.v) = F (v) ⇒

F (tB (v)) = F (v) (take a = 0, u = 0) ∀β ∈ M

⇒

)∑

exp ( k (Λ 0 | tβ (v ))

= exp(k (Λ 0 (v)))

γ ∈M *

∑

γ ∈M *

(

aτ ( γ ) exp γ | tβ (v )

))

aτ ( γ ) exp(( γ | v))

ρ (Λ 0 | τβ (v)) = (t−β (Λ 0 | v))   1   =  Λ 0 − (Λ 0 | δ )β +  (Λ 0 | β) − (β | β)(Λ 0 | δ )  δ | v   2

1 = (Λ 0 | u ) − (β | v) + − (β | β)(v | δ ) 2 (\ (L0|d) = 1, (L0|b) = 0) ( γ | tβ (v)) = (t−β ( γ ) | v)



1 2



=  γ + ( γ | β) − (β | β) ( γ | δ ) δ | v   = ( γ | v ) + ( γ | β ) (δ | v )

174

Stochastics, Control & Robotics

(g|d) = 0 ( M ⊥ δ )

\

  \ k (Λ 0 | tβ (v)) + ( γ | tβ (v)) = k (Λ 0 | v) + ( γ | v) +  ( γ | β) − k (β | β) δ − k (β | v)   2 Fºtb = exp(eΛ 0 )

So,

 k   aτ (τ)exp  γ + ( γ | β) − (β | β) δ    2   γ ∈M *

∑

exp(−kβ) = exp(k Λ 0 )

∑

γ ∈M *

aτ ( γ ).exp( γ − kβ)

 k   exp  ( γ | β) − (β | β) δ     2  (Note that g – kb ∈ M*) (Let g – kb = ξ) = exp(k Λ 0 )

∑

ξ∈M *

aτ (kβ + ξ)exp(ξ)

 k   × exp  (β | β) + (ξ | β) δ     2 Fºtb = F

So ⇒

  k    aτ (kβ + ξ) exp  (β | β) + (ξ | β) δ  − aτ (ξ)].exp(ξ)  2  ξ∈M * 

∑

=0 Now

   l δ | 2 π i (d|v) =   ∑ zs vs − τΛ 0 + δ  = – 2pit   1 

Hence

Fºtb = F ⇒

  k    aτ (kβ + ξ) exp  −  (β | β) + (ξ | β) 2πiτ   2  ξ∈M * 

∑

− aτ (ξ)] exp((ξ | v)) = 0 ∀β ∈ M Thus, aτ (kβ + ξ).exp {−ikπ (β | β) τ}.exp {−2πi (ξ | β)τ} = at (ξ) ∀ξ ∈ M * ∀β ∈ M . Now, it follows that

{

aτ (kβ + ξ)exp iπk −1 (kβ + ξ | kβ + ξ) τ

}

= aτ (kβ + ξ).exp {iπk (iπk (β | β) τ} .exp {− 2πi(β | ξ) τ}

{

× exp − iπk −1 (ξ | ξ)τ

}

Radon & Group Theoretic Transforms with Robotic Applications

{ a ( γ ) exp {− iπk

175

} ( γ | γ ) τ}

−1 = aτ (ξ) exp − iπk (ξ | ξ)τ

i.e.,

τ

depends only on g mod kM.

−1



[14] Remark about Iwasawa decomposition [Remark from Helgason’s Book] G = KMAN M centralizes A, M ⊂ K H ∈ a, X ∈ H, [H, X] = a (H) X. Z ∈ H ⇒ [H, [Z, X]] = – {[Z, [X, H]] + [X, [H, Z]]} = [Z, a (H) X] = a (H) [Z, X] So [Z, X] ∈ N. Actually, we’ve proved that [M, ga] ⊂ ga g (ga) ⊂ g – a Since

[H, X] = a (H) X

[q (H), q (X)] = a (H) q (X) ⇒ – [H, q (X)] = a (H) q (X) Since H ∈ a ⊂ p ⇒ q (H) = – H. m * n1 = n (m * n1 ) m(m * n1 )exp(B(m * n1 ) nB (m * n1 )) ∈NMAN m *2 n1 = n1 = m * n (m * n1 ) m(m * n1 )exp( B(m * n1 )) nB (m * n2 )

\

= n (m * n (m * n1 )) m(m * n (m * n1 )) exp(B (m * n (m * n1 )) nB (m * n (m * n1 )) × m (m * n1 )exp( B (m * n1 )) nB (m * n2 ) Since m (m * n1 ) commutes with exp(B (m * n (m * n1 ))) it follows that exp(B (m * n (m * n1 ))) + B (m * n1 ) =1 Hence

B (m * n (m * n1 )) = B (m * n (m * n1 )) − B(m * n1 ) 

4 Stochastic Filtering and Control, Interacting Partials [1] Bernoulli Filters: Generalization of Bernoulli filters to K state filter. At each time n, one of {0, 1, 2, ...., k–1} particles (targets) is present. φn +1 n ( X r , r X s , S ) is the probability density at time n + 1 of the state specified by X r ∈ R r and r ∈{0, 1, 2, .... k − 1} given that at time n, there were S particles located at X s ∈ R s , s ∈{0,1, 2, .... k − 1} X0 = f (empty ) and so

Note that

φn +1|n ( X 0 , 0 X s , s ) ≡ qs ( X s ) φn +1|n ( X r , r X 0 , 0 ) ≡ pr ( X r )

q0(X0) = q0 are + ve constants

where

= p0(X0) = p0

φn +1|n is the transition probability density of a K-particle filter. if K = 2, we get the Bernoulli filter. φn +1|n describes a Markov process whose state space is random being either one of R r , 0 ≤ r ≤ K −1 at the each time instant. The state at any time n is (Xr, r) ∈ R r × {0,1, .... k − 1}. k −1

We have

∑ ∫ φn+1|n ( X r r X s , s) d r X r + qs ( X s ) r

r =1 R

∀ s = 0, 1, 2, ...., k – 1. Note that for s = 0, this condition becomes

= 1.

178

Stochastics, Control & Robotics k −1

∑ ∫ pr ( X r ) d r X r + qo r

R =1 R

=1

The measurement model is Z n = h ( X n , n) + V n , n = 0, 1, 2, ...

(1)

where {V n } is iid with pdf pv ( v ) .

h ( X n , n ) = h ( X rn , rn , n ) = hn ( X rn , rn )

where k −1

∪ (R r × {r}) → R 6

where hn:

r=0

(Xr , n, rn) is the state of the system at time n, X r , n ∈ R rn conditional pdf given n n observations.

(

)

Let ξn = X rn , n, rn be the state at time n. Let Yn = { zk : k ≤ n} be the collection of observation upto time n. Then p ( ξn +1 Yn + 1) = =

=

p ( Z n +1 , Y n , ξ n+1 ) p( Z n+1 , Y n )

) (

(

p Z n+1 ξn +1 p ξn +1 Y n

)

∫ p ( Z n+1 ξn+1 ) p ( ξn+1 Y n ) d ξn+1 ∫ pv ( Z n+1 − hn+1 (ξn+1 )) p (ξn+1 | ξn ) d (ξn | Yn ) d ξn ∫ pv ( Z n+1 − hn+1 (ξn+1 )) p (ξn+1 | ξn ) p (ξn | Yn ) d ξn dξn+1

By this we mean that

(

p X rn+1 ,n +1rn +1 | Yn +1

(2)

)

k −1

∑ ∫ pv ( Z n+1 − hn+1 ( X rn+1,n+1, rn+1 )) p ( X rn+1, n+1, rn+1 | X rn , n, rn )

rn== 0

=

(

)

p X rn , n | Yn d rn X rn , n k −1

, r | X rn , n, rn ) ∑ ∫ pv ( Z n+1 − hn+1 ( X rn+1, n+1, rn+1 )) p ( X rn+1 n , n +1 n +1

rn , rn +1= 0

(

)

p X rn , n , n | yn d rn X n r, n d rn +1 X rn+1 , n +1

Stochastic Filtering & Control, Interacting Partials

179

Or equivalently,

(

p X rn+1 ,n +1rn +1 | Yn +1

)

∑ ∫ φn+1|n ( X rn+1, n+1 , rn+1 | Xrn,n, rn ) × pv ( Z n+1 − hn+1 ( X rn+1, n+1, rn+1 )) rn

=

)

(

p X rn, n , rn | Yn d rn X

∑ ∫ φn+1|n ( X rn+1, n+1 , rn+1 | Xrn,n, rn ) × pv ( Z n+1 − hn+1 ( X rn+1, n+1, rn+1 ))

rn , rn +1

)

(

p X rn, n , rn | Yn d rn X d rn+1 X

or equivalently, pn +1 ( X r , r | Yn +1 )

=

∑ ∫ φn+1 |n ( X r , r | X s , s ) pν ( Z n+1 − hn+1 ( X r , r ) ) pn ( X s , s | Yn ) d s X s s

∑ ∫ ndr Xr d s Xs

= pv ( Z n +1 − h n+1 ( X r , r )) ×

r, s

∑ ∫ φn+1|n ( X r , r | X s , s ) pν ( X s , s | Yn ) d s X s s

∑ ∫ pν ( Z n+1 − hn+1 ( X r , r )) φn+1|n ( X r , r | X s , s ) pn ( X s , s | Yn ) d r X r d s X s r, s



[2] A Problem in Large Deviation Theory: m

−1 ∞ L′n = n ∑ X i δ yi { X i }i =1 iid r.v.'s. i=i

n = n(m), n/m b → ∈ (0, ∞). i.e. where

n( m) = b Lym → m m→∞ m 1 m Lym = ∑ δ yi m i =1 lim

∫

(

m

−1 fdLn′ = n ∑ X i f ( yi )

E exp n ∫ f d Ln′

i =1

exp ( Λ X ( f ( yi ) ) ) = E exp  i∑= 1 X i f ( yi )  = ∏ i =1 

m



m

180

Stochastics, Control & Robotics

(

1 log E exp n∫ fdLn′ n

) = 1n ∑ Λ ( f ( y )) m

X

i

i=1

1 ∫ Λ X ( f ( y )) d µ ( y ) m→∞ β →

(Gartner-Ellis limiting logarithmic moment generating function) Rate function of {L′n}: IX(v) = Verification of the formulas. 

[3] Large Deviations in Image Trajectory on a Screen Taken by a Robot Camera in Motion: Abstract: We consider a camera attached to the end of single and double link Robot arms. The Robot link moves by virtue of motors attached to its base-point. There is a flat screen separating the object from the Robot. Noise in the motors (assumed to be weak) causes random fluctuation in the Robot motion result resulting in random motion of the image of the object taken by the Robot Camera on the screen. We compute the large-deviation rate function of the image trajectory and use it to determine the (small) probability that the image escapes away from the field of vision on the screen. Camera is located at r2(t) = r1(t ) + l1 n1 (t ) + l2  n 2 (t ) where r1 (t ) is the base-point of the robot and l1 is the length of the first link while n 2 (t ) are respectively unit vectors n (t ) and  l is the length of the second link.  2

along the first and second links.

1

If w1(t) is the angular velocity of the first link relative to its base point at r1 (t ) and w2 (t ) is the angular velocity of the second link relatives to its base point at r1 (t ) + l1 n1 (t ) . Then

r 2 (t ) = r1 (t ) + l1w1 (t ) ×  n 2 (t ) n1 (t ) + l2 w2 (t ) × 

n 2 (t ) = (q2(t), f2(t)), we In spherical polar co-ordinates  n1 (t ) = (q1(t), f1(t)) and  have  n1 (t ) =  X cos φ1 sin θ1 + Y sin φ1 sin θ1 + Z cos θ1 ,  n 2 (t ) =  X cos φ sin θ + Y sin φ sin θ + Z cos θ 2

Thus,

d n1 dt

=

2

θ 1  θ1 + sin θ1 φ 1  φ1

2

2

2

Stochastic Filtering & Control, Interacting Partials

Let

d n2 dt w1

=

w2

=

w1 ×  n w × n

Then,

2

=

= =

181

θ 2  θ2 + sin θ2 φ 2  φ2

w1θ  θ1 + w1φ  φ1 θ2 + w  φ w 

2φ 2

2θ

− w1θ  φ1 + w1φ  θ1 −w  φ +w  θ2 2φ

2θ 2

w1q = − sin θ1φ 1 , w2q = − sin θ 2 φ 2 w1q = θ 1 , w2φ = θ 2

Thus,

The Camera at r2 (t) takes a picture of the object at r0 = and the image is recorded on the screen placed on the X-Z plane. Let (ξx(t),ξz(t)) denote the coordinates (X and Z) of the image on the screen. Then writing r 2 (t ) = (X2(t), Y2(t), Z2(t)) and r 0 = (X0, Y0, Z0), we have for some l∈ R, r 0 + λ (r 2 − r 0 ) = (ξx, 0, ξz) Thus equality the Y-component, we get Y0 + l (Y2 – Y0) = 0, or l =

Y0 Y0 −Y2

ξX = X0 + l (X2 – X0)

Then,

= X0 −

Y0 ( X 2 − X 0 )

(Y2 − Y0 )

=

ξX = Z0 + l (Z2 – Z0) = Z 0 −

( X 0Y2 − X 2Y0 ) Y2 − Y0

Y0 ( Z 2 − Z 0 )

(Y0 − Y2 )

=

( Z0Y2 − Z 2Y0 ) Y2 − Y0

Assuming r 2 (t ) = (X2(t), Y2(t), Z2(t)) to be differentiable, we have ξ z =

{(Y Z =

2 0

((Y2 − Y0 (Z0Y2 − Y0 Z 2 ) − (Z0Y2 − Z 2Y0 )Y2 )

)(

(Y2 − Y0 ) 2

Y2 − Y2Y0 Z 2 − Y0 Z 0Y2 + Y02 Z 2 Z 0Y2Y2 − Z 0Y2Y2

(Y2 − Y0 )2 2 2 = {Y2 ( Z 2Y0 − Y0 Z 0 ) + Z 2 (Y0 − Y2Y0 )} (Y2 − Y0 ) =

and likewise

)}

{(Z 2 − Z0 )Y0Y2 + Y0 Z 2 (Y0Y2 )} (Y2 − Y0 )2

{( X 2 − X 0 )Y0Y2 + Y2 X (Y2 − Y2 )} (Y2 − Y0 )2 .

Large deviations for the image of an object on a screen taken by a Robot Camera in the presence of machine noise

182

Stochastics, Control & Robotics

Problem Formulation The Camera attached to the tip of a single link Robot arm is located at = r + l n (t ) ≡ r 2 (t ) 1

where r1 ∈R is fixed and n (t ) is a unit vector 3

n (t ) = cos φ(t )sin θ(t )x + sin φ(t )sin θ(t ) y + cos θ(t ) z d n (t ) d r 2 (t ) =l = lw(t ) × n (t ) = w(t ) × r 2 (t ) − r1 dt dt w(t) can be found using

(

We have

)

(1)

d n (t ) = θ (t )θ (t ) + φ (t ) sin θ(t )φ (t) dt On the one hand while on the other.

(2a)

ω (t ) = ωθ (t )θ (t ) + ω φ (t ) θ φ (t )

Writing

we get ω(t ) × n (t ) = ωθ θ × n + ω φ φ × n = − ω φ φ + ω φ θ wq = − φ sin θ ,

Thus,

wf = θ Lagragian for the link:

(

)

mgL 1 2 2 2 2 cos θ L θ, ϕ, θ , ϕ = mL θ + sin θφ − 2 2

(

)

Two motors are attached at the base providing tq(t) and tf(t) torques. Thus the equations of motion of the links are Z r0 = (X0, Y0, Z0)

(X, Z)

Y

Camera r2(t) 

Screen

L

 r1 = (X1, Y1, Z1)

Eqns. of motion d ∂α ∂α − = tq(t), dt ∂θ ∂θ

X

Stochastic Filtering & Control, Interacting Partials

183

d ∂α ∂α − dt ∂φ ∂φ = tf(t)

These give

mL2 θ − mL2 sin θ cos θφ 2 − mgL sin θ = tq (t), mL2

(

)

d sin 2 θ φ = tf(t) dt

τ φ (t ) g , i.e.  θ − sin θ cos θ φ 2 − sin θ = L mL2 τ φ (t )

 + 2 sin θ cos θθ φ = sin 2 θφ

mL2 Steady state (equilibrium position of the link): φ = 0,  θ = 0,  φ = 0. θ = 0,   τ φ0  tf(t) = 0, q = q0 = − sin −1    mgL  Assuming τθ (t ) = Constant τ θ0 Thus

Let f = f0.

Small deviations from equilibrium: q(t) = q0 + ed q (t), f(t) = f0+ edq(t) tq(t) = τθ + e w1(t)

Let

0

tf(t) = ew2(t) where w1(t) and w2(t) are independent zero mean Gaussian random processes. Then the O(e) terms in the dynamical eqn. give  − g cosθ δθ = w1 (t ) , δθ 0 L mL2 w2 (t )

sin 2 θ0δ φ = So,

 L dq(t) =  sec θ0   g df(t) =

Thus,

mL2 12

1 mL sin 2 θ0

E {(δθ(t + τ) δθ(t ))} =

(mL )

2 −1

t

 g  cos θ0 (t − τ) w1 (τ) d τ L 

∫ sinh  0

t

2

{(δθ(t ), δφ(t ))}t≥0

.

∫ (t − τ ) w2 (τ) d τ 0

is zero mean Gaussian random process with correlations 1 2

mL g cos θ0

t+τ t

  g cos θ0 (t + τ − t1 )   L

∫ ∫ sinh  0 0

184

Stochastics, Control & Robotics

 g  × sinh  cos θ0 (t − t2 ) Rw1w1 (t1, t2 ) dt1 dt2   L ≡ Rq(t + t, t)

E {δθ (t + τ ) ⋅ δφ(t )} = 0, t+τ t

1

E {δφ (t + τ ) ⋅ δφ(t )} =

∫ ∫ (t + τ − t1 ) (t − t2 ) Rw2w2 (t ,t2 ) dt1dt2

(mL2 sin 2 θ0 ) 2

0 0

= Rφφ (t + τ, t)

The rate function for (δθ(⋅), δφ(⋅)) over the interval [0, T] is then TT

1 IT (δθ(⋅), δφ(⋅)) = ∫ ∫ ( Rθθ (t1 , t2 ) δθ(t1 ) δθ(t2 ) 200

)

+Rφφ (t1 , t2 ) δφ (t1 ) δφ(t2 ) dt1 dt2 

[4] P406 Revuz and Yor: u   (a) a–2Ct = inf u : a2∫ exp 2aβ s ds > t  0  

(

is to be proved

)

−1 β(sa ) = a β a 2t

sr = inf {t ≥ 0 logρs = r } Ct = logρ

t

=

t

ds

0

s

∫ ρ2

t = C–1

Bt = logρτt τt

ds

∫ ρ2 0

s

= t

d τt ρ2τt

= dt

βt( a ) = a −1β 2 = a −1 logρ a t ρτt = exp(b ) t

(

)

(

exp 2aβt( a ) = exp 2β a 2t 2 = ρτ

a 2t

)

τa 2t

Stochastic Filtering & Control, Interacting Partials

ut

(a) ∫ exp (2aβs ) ds

t

=

a2

0

ut

∫ ρτa2s ds

⇒

2

t

=

a2

0

a 2ut

∫

⇒

0

ρ2τ s

a 2ut

∫

⇒

0

⇒

τut a 2

∫

a

2

t

=

a2

ρ2τ s ds = t

(τ = C −1 )

ρτ2C dCs = t s

0

τut a 2

∫

⇒

ds

185

ρ2s dCs = t

0

⇒

ρ2τa 2u dCτa 2u = dt t t

⇒

ρ2

⇒

τa 2ut

(

)

= dt

(

)

=

d a 2ut d a 2ut

Comparing this with dCt = gives i.e.

dt ρ2 2 τa ut dt ρt2

=

dt ρ2τCt

a2ut = Ct

ut = a–2Ct

Q.E.D.

Problems from Revuz and Yor (Continuous Marketing and BM). Let B(t) = B1(t) + iB2(t) ≈ Conformal planer BM. log B = log B + i Arg B is a Martingale (local) since its Laplacian Vanishes. Let

B = rt ≡

B12 + B22 ,

−1  B2 (t )  qt = Arg B = tan   B1 (t ) 

186

Stochastics, Control & Robotics

log r and q are Martingales (local).  dρ  (d log rt) =  t   ρt  2

=

logρ

t

=

dt

 B1dB1 + B2 dB2  =   ρt2 

= (dqt)2.

ρt2 t

ds

0

s

∫ ρ2

2

(property of conformal Martingales)

= Ct

say.

tt = (C–1)t

Let τt

ds

∫ ρ2

i.e.

0

s

dτt

Then

ρt2

= Ctt = t = dt,

2 dtt = ρt dt

i.e.

t

tt =

(logρ)τt

2

∫ ρs ds 2

0

= logρτt is a B.M. and θτt is also a BM and there two B.M.'s are

independent since d log r. dq = 0.(property of conformal Martingales. bt = log ρτt , Yt = θτt

Let

sr = min {t ≥ 0 ρt = r }

Let

ρτt = eβt

Then

rt = eβCt and hence

{ β = min {t ≥ 0 βC

sr = min t ≥ 0 e

Ct

t

=r

}

= log r

}

= τTlog r or equivalently where

Tlog r = Cσr Ta = min {t ≥ 0 βt = a}

By optional stopping theorem   λ 2Ct   λβ0 = ρ0λ E exp  λβCt −   = e  2  

(r0 ≈ non randomly)

Stochastic Filtering & Control, Interacting Partials

187

  λ 2Ct   λ E ρtλ exp  −   = ρ0  2   Now let r1 < r2. Then assuming r(0)∈(r1, r2),

i.e.

σ r1 Λσ r2 = min {t ≥ 0 ρt (r1, r2 )} .

and since

 λ 2Ct   λ 2Ct  ρtλ exp  − exp  λβCt −  =    2  2 

is a Martingale, to follows that   λ2  λ E ρσλ r Λσr exp  − C σ r1 Λσ r2   = ρ0  2   1 2

(

)

  λ2   λ λ2 Thus E  r1 exp − 2 C σ r1 χσr < σr + r2λ exp  − C σ r2  χσr < σr   1 2  2  2 1 λ = ρ0

Equivalently

   λ2  E exp  − Tlog r1 ΛTlog r2  r1λ − r2λ χσr < σr   2  1 2     λ2  λ λ = ρ0 − r2 E exp  − (Tlog r2 ΛTlog r1 )  2  

(

Now

)(

)

σ r1 < σ r2 ⇒ Cσr ≤ Cσr 1

2

Tlog r 1 ≤ Tlog r .

⇒

2

Thus,

   λ2  E exp  − Tlog r1  χ{σ < σ }  r1 r2   2         λ2 ρ0λ − r2λ E exp  − Tlog r2 ΛTlog r1        2   = r1λ − r2λ

(

Now Thus

)

  λ 2Ta   = ρ0λ = eλβ0 E exp  λβTa −   2     −λ 2Ta   = ρλ e − λa 0 E exp    2   

   λ 2Tlog r      λ2 1 Cσr  χ σ < σ  E exp  − χ σ < σ  = E exp  −  r2 } 1  { r1  2   2  { r1 r2 }  

188

Stochastics, Control & Robotics

   λ2 ρ0λ − r2λ E exp  − Tlog r1 ΛTlog r2     2  = λ λ r1 − r2

( )

   β λ2 E exp  λβTa ΛTb − Ta ΛTb   = exp λ 0   2   where a < b0 < b, b0 non random.

Now

Thus,       a + b λ2  a + b   − Ta ΛTb   = exp  λ  β0 −  E exp  λ  βTa ΛTb −          2 2 2      Changing l to – l, adding and using

 λ (b − a )    a + b  cosh λ  βTa ΛTb −    = cosh   2  2  

(since βT ΛT a

b

)

∈{a, b} , it follows that

  a + b  cosh λ  β0 −    λ2   2   E exp  − Ta ΛTb   =  2  λ(b − a)  cosh 2

{



   λ2 Cσr  χ σ < σ  r2 }  2 1  { r1 

}

Thus, E exp  −



{(

 cosh λ log ρ0 − log r1r2 ρ0λ − r2λ  λ  r  cosh  log  2     r1   2  = λ λ r1 − r2

(

)

)} 

  



[5] Problem from "Continuous Martingales and Brownian Motion" by Revuz and Yor: Consider the filteration

{ } σ { FSBU {Bt }} , S ≥ 0

FSt = σ FSB , Bt ≡

A process of the form S → f(BS, Bt), S ≥ 0 is adapted to { FSt }S ≥0 (t is fixed) With respect fo this "augmented filteration", we defite the process

Stochastic Filtering & Control, Interacting Partials

189

Z S = Z St = (Bs – Bt) q(Bs – Bt). t Then {Z St }S ≥0 is adapted to { FS }S ≥0 . We can define stochastic integrals w.r.t. dBS

for process s → f(Bs, Bt) relative to this augmented filteration. Then 1 dZS = θ ( BS − Bt ) dBS + δ ( BS − Bt ) dS 2 t

and hence

Z t – Z0 =

t

1

∫ θ ( BS − Bt ) dBS + 2 ∫ δ ( BS − Bt ) dS 0

0

or equivalently, t

Bt q(–Bt) =

1

∫ θ ( BS − Bt ) dBS + 2 Lt t B

0

t

1 Bt Lt = Bt θ ( − Bt ) − ∫ θ ( BS − Bt ) dBS . 2 0

i.e. Note that

1 x ≥ 0 q(x) =  0 x < 0 

[6] Problem from Revuz and Yor:  t  x + a x    n  ∫  δ  Bs −  − δ  Bs −   ds 0     n n  = Xn (t, x)

∫ f ( x) X n (t , x) dx

R

Let

t

(

)

= n n ∫0 f (nBs − a ) − f (nBs ) ds = Mn (t) say.

gαµ ( x ) = f ( x − a) − f ( x) .

n2 u g a (nBt ) dt 2 t n M n (t ) Thus, g a (nBt ) − g a (O) = n∫0 g a′ (nBt ) dBt + 2 1 ( g a (nBt ) − g a (0)) = n t 1 = n ∫0 g a′ (nBt ) dBt + M n (t ) 2 dga (nBt) = ng′a (nBt ) dBt +

190

Stochastics, Control & Robotics

It follows that lim

1

x→∞ 2

t

M n (t ) = lim n ∫ g a′ (nBt ) dBt 0

x→∞

Consider the Martingale t

n ∫ ψ(nBs ) dBs

Mn(t) =

0

We have Mn

t

t

2 = n∫0 ψ (nBs ) ds

∫

= n

ψ 2 ( x) δ ( x − nBs ) ds dx

[ 0, t]× R

∫

=

[ 0, t]× R

 x ψ 2 ( x) δ  − Bs  ds dx  n

  →  ∫ ψ 2 ( x) dx R 

t

∫0 δ ( Bs ) ds

2 = || ψ || Lt

It follows that these exists a BM Y(.) independent of B(.) such that 1 lim M n (t ) = Y (|| ψ || Lt )∀t n→∞ 2 where Lt is the local time of B(.) at zero Thus, 1 lim M n (t ) = Y (|| g a′ || Lt ) n→∞ 2 Now take Then

f (x) = d (x – x0) x0 ∈ R fixed. Mn (t) = Xn (t, x0) t

= Then

  x + a x   − δ  Bs −   ds n ∫  δ  Bs −     n  n  0

g a′′( x) = f (x – a) – f (x) = d (x – a – x0) – d (x – x0)

so

g a′′( x) = q (x – a – x0) – q (x – x0)

Thus,

|| g a′ || =

and we decline that

a + x0

∫

x0

dx = a

Stochastic Filtering & Control, Interacting Partials

191

t 1 n   x + a x   M n (t ) = lim − δ  Bs −   ds δ  Bs −   ∫ n→∞ 2  n→∞ 2   n  n  0

lim



[7] Collected Papers of Varadhan-Interacting Particle System, Vol. 4. Hamiltonian System and Hydrodynamical Equations:

(

N

)

H X, p =

∑ p−2α / 2m + ∑

α =1

1≤ α < β ≤ N

(

V X α − Xβ

( )

3N pα ; X α ∈R 3 , X = ( X α )α =1 ∈R , p = p α N

N

α =1

)

∈R 3N

Initial configuration is random (random configuration envolving under determination dynamics). Let ft ( X , p) be the pdf of ( X (t ), p(t)) . Liowvilles equation implies  dpα ∂ft  ∂ft ∂f   dX , + ∑ α , t  + ∑ =0  dt ∂X α  α  dt ∂pα  ∂t α or

 ∂f  ∂ft 1  ∂f  + ∑  pα , t  − ∑  V ′ X α − X β , t  = 0 ∂t m α  ∂X α  α ≠ β  ∂pα 

(

)

     V ( X ) = V ( − X ), X ∈R 3 .

(1) Assuming

More generally for any Hamiltonian H ( X , p) , we have

∂H dpα dX α ∂H , = = − dt ∂X α ∂pα dt and therefore ∂ft ∂t

 ∂H

∑  ∂p α

Empirical density



ρ( t , x) =

α

,

∂ft   ∂H ∂ft   − ,  =0 ∂X α   ∂X α ∂pα  

N

∑ δ ( X − X α (t )) ∈R

α =1

Emprical momentum density π(t , µ ) = Empirical momentum flux π(t , x) =

N

∑ pα (t )δ ( x − xα (t )) ∈R 3

α =1

1 N ∑ p (t ) pβ (t )T δ ( x − xα (t )) m α =1 α

192

Stochastics, Control & Robotics

Empirical energy density N

e(t , x) = When ea (t) =

∑ eα (t )δ( x − xα (t))

α =1

p α2 (t ) αm

+

∑

β :β ≠ α

(

V x (αt ) − xβ(t )

(

)

)

is the energy of the ath particle. X ∈R 3 , p ∈R 3 . In general for any of observable ψ( x , p) we define its empirical density as

ρψ (t , x) = Its average value is ρψ t , x =

∑ ψ ( xα (t ), pα (t )) δ ( x − xα (t)) N

α −1

∫ ft ({ X α }, { pα }) ∑ ψ ( xβ , p β ) δ ( x − xβ ) N

=

β =1 N

∏ d 3 xα d 3 pα

α =1

 

where ftα

 



∑ ∫ ψ ( x, p ) ftα ( x, p ) d 3 p N

=

β =1

  x, x is the Martingale density of xα (t ) , p α (t ) :

( ) f (x tα

α,

)

pα =

∫

{

 f t  xβ , p β 

} 

(

)

N

1

∏ d 3 xβ d 3 p β

β≠α

We get from (1), by integrating over xβ , p , β ≠ a β =

∂f ∂ftα 1  ( x α , p α ) +  p α , tα ( x α , p α )   ∂t m ∂x −

(

)

  ∂ftαβ x x p p V x − ) x , , , , ′(   ∑ ∫ α β ∂ pα α β α β  β:β ≠ α 

d 3 xβ d 3 p β = 0

Stochastic Filtering & Control, Interacting Partials

193

where ftab is the joint (marginal) density of

(

ftαβ x α , xβ , p α , pβ

(x , x , p , p ) : α

β

) = ∫ f ( x, p) ∏ d x d p 3

t

γ ≠ α ,β

γ

α

β

3

γ

If we assume that all first and second order marginal densities are the same and we denote there marginals by f t(1) and f t(2) respectively, then  ∂ft(1) 1  ∂f (1) ( x1, p1 ) +  p, t ( x1, p1 ) ∂t ∂x m    ∂f ( 2) − ( N − 1)∫  V ′( x1 − x 2 ). t ( x1, x 2 , p1, p 2 ) ∂ p1   d 3 x2d 3 p 2 = 0 The BBGKY can be continued further. (p[q] Vol. 4. S.R.S Varadhan collected 1 = papers) We’ve to prove ∫ ∑ Aµj λ µj dx = 0. From (2.25) (assuming l4 = 4 µ, j constant)

( ) ( + (−λ q + λ P) λ

)

j 0 0 j i i λ 4 .Aµj λ µj = − λ q λ j − λ q + Pδ ij λ , j j 4

j

4 ,j

= – λ j q 0 λ ,0j − λ j qi λ i, j − ψλ i,i ∂ψ  ∂ψ  (y = P) = − λ j 0 λ ,0j + λ j λ i, j i + ψλ i,i  ∂λ   λ

(λq

µ

= ∂ψ / ∂λ µ

)

∂ψ i  j  ∂ψ 0 i = − λ  0 λ , j + i λ , j  + λ ψ ,i ∂λ ∂λ  (after neglecting a total divergence (ψλ i ),i ) Using this becomes

y,i =

∂ψ ∂λ

0

λ ,0i +

∂ψ ∂λ j

λ i,i

j j l4Amj lmj = − λ ψ , j + λ ψ , j = 0.

Current in a simple exclusion model on Z3N.

X Y , ∈[0, 1]3 . N N If ηt ( X ) = 1, ηt (Y ) = 0 and in time dt, the particle of X jumps to the site Y, then

3 Each particle has a charge q. X ,Y ∈Z N ,

194

Stochastics, Control & Robotics

Y−X (Y − X ) or after scaling . The outward current density at dt N dt

its velocity is

 X X  ≡  is times taken to be proportional to  N

∑

Y :Y ≠ x

ηt ( X ) (1 − ηt. (Y ))

dNt( X ,Y ) (Y − X ) dt

We write this as (after time scaling) J (N1) (t , X ) =

α( N ) N2

∑

Y :Y ≠ X

(

ηN 2t ( X ) 1 − ηN 2t (Y )

)

d ( X ,Y ) − N 2 (Y − X ) N t dt

and so

{

}

E J N(1) (t , x) =

α (N ) N2

∑

Y :Y ≠ X

{

(

E ηN 2t ( X ) 1 − ηN 2t (Y )

)}

d N2 λ (Y − X )tp(Y − X ) dt

{ {

}

}

= λα ( N ) ∑ E ηN 2t ( X ) (1 − ηN 2t ( X + Z )) zp ( Z ) X ,Y

Likewise the **and current into X is after rescaling J N( 2) (t , X ) =

α( N ) N2

∑

Y :Y ≠ X

(

ηN 2t (Y ) 1 − ηN 2t ( X )

)

d (Y , X ) N 2 (X − Y) dt N t The total current density it X is J N (t , X ) = J N(1) (t , X ) + J N( 2) (t , X ) . We get

E { J N (t , X )} = λα ( N )

∑

Y :Y ≠ X

{E η (

N 2t

(

(X ) 1− η

)

N 2t

)

(Y )  p(Y − X ) 

}

− E  ηN 2t (Y ) 1 − ηN 2t ( X )  p ( X − Y )  

  Y   X  p (Y − X ) ρ N  t,  1 − ρ N  t ,   N N Y :Y ≠ X 

≈λα( N )

∑

 X   Y  − p ( X − Y ) ρ N  t ,  1 − ρ N  t ,    (Y − X )  N   N

Stochastic Filtering & Control, Interacting Partials

195

  X + Z   X = λα( N ) ∑ z  p( z )ρ N  t,  1 − ρ N  t ,      N N   z ≠0   x   x + z  − p( − z )ρ N  t,  1 − ρ N  t ,     N  N  3 Note that X , Y , Z ∈Z N .



[8] (Ph.D. Problem of Rohit) For Rohit and Dr. Vijayant. Skorohod Optional Stochastic Control. Optional Stochastic Control of Master and Slave Robot:

(

)(

(

)

 + N θ , θ F m θm , θ f n (t ) + wm (t ) M m (θ m ) θ m m m m =

(

M s (θ s ) θ s + N s θ s , θ s

(

= F s θ s , θ s Aim: E

{∫

Let

T

0

{ f (t )} n

)

)

)( f e (t ) + we (t )) + K (t)ψ (θm , θ m , θs , θ s )

choose the feedback controller matrix

}

{K (t )}

so that

L (θm , θm , θ s , θ s ) dt is a minimum. wm (t ) = σ m

d Bs d Bm , we (t ) = σ e dt dt

When B m (.) and B s (.) are independent 3-D Brownian motion processes. In stochastic differential form, the above equation. Assume the form dθm = wm dt dθ s = ws dt d wm = G m (θm , wm ) dt + H m = (θm , wm ) f (t ) dt n + σH m = (θm , wm ) dBm (t )

d ws = G s (θ s , ws ) dt + H s (θ s , ws ) f e (t ) dt + σ e H s (θ s , ws ) d + B s (t) + K (t )ψ (θm , wm , θ s ws ) dt

We assume + K (t ) is a function of t and

(θ

T m (t ),

)

wm (t ),T θ s (t )T ws (t )T ∈R 8

196

Stochastics, Control & Robotics

(two link robot arm). Here, −1 Gm = − M m N m , G s = − M −s 1 N s , −1 −1 F m, H s = M s F s , H sm = M m

Let min E k (s)

t < s ≤T

{∫

T

t

}

L (θ m ( τ ) , wm ( τ ) , θ s ( τ ) , ws ( τ )) d τ | θm (t ), wm (t ), θ s (t ), ws (t ) = V (t , θm (t ), wm (t ), θ s (t ), ws (t ))

Then, V (t , θms , wm0 , θ s 0 , ws0 ) = L (θms , wm0 , θ s 0 , ws0 ) dt

{

+ min E V (t + dt , θ m (t + dt ) , wm (t + dt ) , K (t )

θ s (t + dt ), ws (t + dt )) | θm (t ) = θm0 ,

wm (t ) = wm 0 , θ s (t ) = θ s 0 , ws (t) = ww0 } Or equivalently, −

∂V (t , θ m 0 , w m 0 , θ s 0 , w s 0 ) ∂(t ) = L (θ m 0 , w m 0 , θ s 0 , w s 0 ) T   ∂V  1  + min   (t , θm0 , wm0 , θs0 , ws0 ) dt E {d θm (t ) | θm (t ) K (t )   ∂θ m  

= θms ....ws = ws 0 } T

 ∂V  1 + (t , θ m0 , wm0 , θ s 0 , ws 0 )  E {d wm (t ) | θ m (t )  ∂wm  dt = θ m0 ....ws (t ) = ws 0 } T 2 1  ∂ V (t , θm0 , wm0 , θ s 0 , ws 0 ) 1 E d wm (t ) | .d wm (t )T + Tr 2  dt ∂ wm ∂ wTm 

{

| θm (t) = θm0 ....ws (t ) = ws 0 } T

 1  ∂ 2V + t , θ m 0 , w m 0 , θ s 0 , w s 0 ) E {d ws (t ) | (  dt  ∂ws

θm (t ) = θm0 , ....ws (t ) = ws 0 }

Stochastic Filtering & Control, Interacting Partials

 ∂V  + t , θm 0 , wm0 , θs 0 ws 0 ) (  ∂ws 

197

T

{

1 E dws (t ) θm (t ) = θm0 ,..., ws (t ) = ws 0 dt

}

1  ∂ 2V + Tr  (t , θ m 0 , w m 0 , θ s 0 , w s 0 ) 2  ∂ ws ∂ wTs

{

1 T E d ws (t )d ws (t ) | θm (t )= θm0 , ...ws (t ) = ws 0 } dt Now, E {d wm (t ) | θm 0 , ..., ws 0 } = wm0 dt

{ } E {dwm (t ) θm 0 ,..., ws 0 } E d θs (t ) θm0 ,..., ws 0

= ws0 dt = G m (θm0 , wm0 ) + H m (θ m0 , wm0 ) f n (t ) dt ,

E {d ws (t ) | θm0 , ..., ws 0 } = G m (θ s 0 , ws 0 ) + H s (θ s 0 , ws 0 ) f (t ) e  + K (t )ψ (θm0 , wm0 , θ s 0 , ws 0 ) dt

{

E d wm (t )d wm (t )T = | θm0 , ..., ws 0

}

σ 2 H m ( θm0 , wm0 ) ⋅ H m ( θm 0 , wm 0 ) dt T

{

E dws (t ) dws (t ) θm 0 ,.., ws 0 T

}

= σ 2 H s (θ s 0 , ws 0 ) H s (θ s 0 , ws 0 ) dt , T

So V satisfies the pde ∂V (t , θ m , w m , θ s , w s ) = L (θ m , w m , θ s , w s ) ∂t T   ∂V T  ∂V   + min    wm +   G m + H m f n (t) K (t )  ∂θ m   ∂ wm  

(

+

 σ 2  ∂ 2V T H H Tr   2  ∂ wm ∂ wTm m m  

)

198

Stochastics, Control & Robotics

[9] English-Chapter-1:  + a ( s, 0)  x0 ∑ (1 − xs ) ρt (( xs , x1, −, x0 , −, xn −1 ), ( y0 , x0 ))  x 0   − (1 − y0 ) ∑ xs ρt ( x , ( xs , ( x1, .., y0 ,.., xN −1 )))   x0  th 5 position molecular (choose) xs ρt ( x (0, s), ( y0 , x0 ) ) ∑ 

=

x0

xs ρt (( xs , x1., x0 ,.., xN −1 ) ( y0 . x1, .., xN −1 )) ∑  x0

∑ ( xs ) ρ(t 0, s) (( xs , x0 ) , ( y0 , xs )) xs

(1 − xs ) ρt ( x , ( xs , x1,.., x0 ,., xN −1 )) ∑  x0

=

(1 − xs ) ρt (( x0 ,., xN −1 ), ( xs , x1, .., y0 ,., xN −1 )) ∑  x0

=

(1 − xs ) ρt(0, s ) (( x0 , xs ), ( xs , y0 )) ∑  x0

(1 − xs ) ρt (( xs , x1, ., x0 ,., xN −1 ), ( y0 , x1, ., xN −1 )) ∑  x0

=

∑ (1 − xs ) ρt(0, s) (( xs , x0 ), ( y0 , xs )) xs

And finally,

xs ρt ( x , ( xs , x1,., y0 , ., xN −1 )) ∑  x0

( 0, s ) (( x0 , xs ), ( xs , y0 )) = − ∑ xs ρt xs

b) Where r(a, is the joint marginal density of the ath and bth particle obtained by t tracing out rt over the other particles. We thus get

i

∂ (0) ρt ( x0 , y0 ) = ∂t



∑ a (0, s) (1 − x0 ) ∑ ρt(0, s) (( xs , x0 ), ( y0 , xs )) xs

s ≥1



xs

 − y0 ∑ (1 − xs ) ρt(0, s ) ((x0 , xs ), (xs , x0 )) xs 

Stochastic Filtering & Control, Interacting Partials

+

199

 ( 0, s) a ( s , 0 ) (( xs , x0 ), ( y0 , xs ))(1 − xs )  x0 ∑ ρt ∑  xs s ≥1

 − (1 − y0 ) ∑ xs ρt(0, s) ((x0 , xs ), (xs , y0 ))  xs

or i

∂ (0) ρt ( x0 , y0 ) = ∂t

∑ {(a (0, s)(1 − x0 ) xs + a (s, 0) x0 (1 − xs ))

s ≥1, xs

}

ρt(0, s) ((x0 , xs ), (xs , y0 ))

Note that this is a special case of the quantum Botzmann equation. Continuation of remarks on volume iv collated papers of S.R.S. Varadhan. “particle system their large deviation.” Linear response theory in quantum system H (t) = H 0 + ε V (t ) Equilibrium state is

r0 =

1 −βH 0 e z ( B)

(

−βH Z (b) = Tr e 0

)

2 r (t) = ρ0 + ε ρ1 (t ) + O(ε )

Note that [H0, r0] = 0, so r0 satisfies the stationary Von. Neumann equation. iρ1′(t ) = [ H 0 , ρ1 (t )] + [V (t ), ρ0 ] t

r1 (t) = − i ∫0 exp( − i (t − τ) adH 0 ) ([V (τ), ρ0 ]) d τ

X any system observable

X (t ) = Tr ((ρ0 + ε ρ1 (t ))) ( X ) = X (t ) − X

X

0

+ ε Tr (ρ1 (t ) X ) t

0

( (

= − i ε ∫0 tr exp − i (t − τadH 0 )

)) ([V (τ), ρ0 ] X ) d τ

 [10] Markov Processes-Some Application: (1) Simple exclusion models in Markov process theory. (2) Transmission lines, waveguides and Antennas excited by Markov chains. (1) Consider

ZN = {0, 1, 2, ..., N – 1}. h (x) = 1

200

Stochastics, Control & Robotics

If site x ∈ ZN is occupied by a particle and h (x) = 0 otherwise p (x), x = 0, 1, 2,.., N – 1 is a probability distribution on ZN. p (y – x) ≡ p (y – x mod N),

N −1

∑ p ( x)

= 1.

x=0

If site x is occupied and site y ≠ x is unoccupied, then with a probability lp (y – x)dt, the particle at x jumps to the site y in time [t, t + dt]. The different jumps processes are independent and here up to O (dt), at must only one jump takes place in ZN in time dt. Let X = {h : ZN → {0, 1}} Thus X is the state space of the above Markov Process {ht(x)} and ht satisfies the side dht (x) = Where

{dN

ηt ( y ) (1 − ηt ( x )) dNt ( y, x ) − ηt ( x) (1 − ηt ( y )) dNt( x, y ) } { ∑ y ∈Z N y≠ x

( x , y) t

: x, y ∈ Z N , x ≠ y

}

are independent Poison processes with the

rate of Nt(x, y) being lp (y – x). More generally for f : X → R , df (ht) =

∑N

x , y ∈Z x≠ y

( f (η ) − f (η )) η ( x) (1 − η ( y)) dN ( x , y) t

,

t

t

t

( x, y ) t

Generator Lf (h) = E [ df ( ηt ) | ηt = η] dt = λ = λ Here,

∑ ( f ( η( x, y)) − f (η)) η( x)(1 − η( y)) p( y − x)

x , y ∈Z N , x≠ y

∑ N ( f (η( x, x + z )) − f ( η)) η( x)(1 − η( x + z )) p( z)

x , z ∈Z z≠0

,

 η( x) z = y  η ( z ) =  η( y ) z = x  η( z ), z ≠ x, y  Exponential Martingale: Let N (t, dx) be a space-time Poisson random field with E (N (t, dx)) = t . dF (x), x ∈ Y. Then     d exp  ∫ f ( x) N (t , dx) = ∫ (exp( f ( x)) −1) N (dt , dx) exp  ∫ f ( z ) N (t , dz )  Y Y  x ∈Y ( x , y)

i.e if

  ξf (t) = exp  ∫ f ( x) N (t, dx) , Y 

Stochastic Filtering & Control, Interacting Partials

d ξ f (t ) = ξ f (t )

then

∫

201

(exp( f (x)) −1) N (dt, dx)

x ∈Y

E [ d ξt (t ) | ξt (t)] = ξt (t ) ∫ (exp( f (x)) −1) dF ( x) dt

\

Y

  Mf (t) = exp  ∫ F ( x) N (t, dx) ϕ(t ) Y 

Let

When Q (t) is a differentiable function. Then,  ϕ′(t )  dt  dMf (t) = ξt (t ) dt ∫ (exp( f (x)) − 1) dF ( x) + ϕ(t )   Y Thus choosing log j (t) = t ∫ (exp( f (x)) −1) dF ( x) Y

gives a Martingale Mf (t):

  Mf (t) = exp  ∫ f ( x) N (t, dx) − t ∫ (exp( f (x)) − 1) dF ( x) Y  Y

This is the exponential Martingale associated to f for the Possion measure {N (t, dx)}. Note that it t

f ( ηt ) − ∫ L f ( ηs ) ds = Mt, then 0 dMt =

∑ ( f (ηt( x, y) ) − f (ηt )) ηt ( x) (1 − ηt ( y))

x≠ y

(dN

( x , y) t

− λp( y − x) dt

and Mt is a Martingale. Let J : [0, 1] → R. Then define the process 1  x J   ηt ( x) Xt = ∑  N N N

)

x∈Z

 Suppose ηt [ N θ] N → ∞ρ (t , θ), θ ∈[0, 1] Then

 1 X t N → ∞ ∫ J (θ) ρ(t , θ) d θ 0

Now,

E [ d ηt ( x)| ηt ] = λ

∑ [ ηt ( y) (1 − ηt ( x)) p( x − y)

y: y ≠ x

− ηt ( x)(1 − ηt ( y )) p( y − x)] dt

202

Stochastics, Control & Robotics

E [ dxt | ηt ] =

so

λ N

 x

∑ J  N  (ηt ( y) (1 − ηt ( x)) p( x − y)

x≠ y

(

)

− ηt ( x) 1 − ηt ( y )) p ( y − x ) dt dx1 − E [ dxt | ηt ] =

So

1 N

∑ J  N  {ηt ( y) (1 − ηt ( x)) dM t( x, y )  x

x≠ y

− ηt ( x) (1 − ηt ( y )) dM t( x, y ) where

}

y) = Nt( x, y) − λp ( y − x)t M(x, t

is a Martingale. This can also be expressed as dXt = E [ dX t | ηt ] =

1 N

  x

 y

∑  J  N  − J  N   ηt ( y) (1 − ηt ( x)) dM t( y, x)

x≠ y

Thus, Assume that p has finite rage, i.e. p (z) = 0 if |z| > q when is q a finite positive integer. 2

  x z Var {dxt | ηt } = J ′    ηt ( x) 2 (1 − ηt ( x + z )) 2 p( z ) 2 ∑ N x, z   N  N  λdt

Then

1  + dt.O  4 . N  N →0 →∞ In fact, we have a stronger convergence result:

{

2 η2Nt Var dX Nt

=

≤

} (time scaling)

  x  2 2 ∑  J ′   z p ( z ) ηtN ( x)2 1 − ηtN ( x + z ) N 2 x, z   N 

(

λdt

λdt

 x

2

) 

2

→0 ∑ J ′  N  ∑ z 2 p( z ) N →∞ N2 x

z

2

1  x   J ′   N → ∞ ∫ J ′ (0) 2 dθ < ∞ ∑ N x  N 0

Since and

∑ z 2 p( z ) < ∞

since p (.) has finite range. We note that Var {dX Nt | ηNt }

x

= E

{(dX

Nt

)2 | ηNt } − (E {dX Nt | ηNt })

2



Stochastic Filtering & Control, Interacting Partials

and hence we have proved that E

{(dX

E

{(dX

Or equivalently,

Nt

Nt

203

)2 | ηNt } − E (E {dX Nt | ηNt })

N →∞ L

Now, Xt = = where Thus,

1 N

∑

x∈Z N

1

mN, t =

1 N

XNt =

∫0 J (θ) d µ N , Nt (θ)

ηt ( x) δ x / N

x∈Z N

1

mN, Nt ≡ VN, t =

where

 x J   ηt ( x)  N

∫0 J (θ) d µ N , t (0) ∑

1 N

∑

ηNt ( x) δ x / N

x∈Z N

We have Xt =

  x + z 1  x  J − J   ∑    N  N x, z   Nt  t

∫0 ηs ( x) (1 − ηs ( x + z )) dN x E Xt =

So,

( x , x+ z)

1  x ∑ J   E {ηt ( x)} N x  N

{

}

We define

rN (t, 0) = E ηN 2t ([ N θ]) – q ∈ [0, 1]

So

2 E xNt

Thus,

{ }=

{ }=

d 2 E X Nt dt

1  x  x ∑ J   ρN  t, N  N x  N 1 x  x  ∂ρ J   N (t , θ) ∑ N x  N  ∂t N 1

≈ ∫ J (θ)

On the other hand,

0

∂ρ N (t , θ) d θ ∂t

( ) = ∑  J  x N+ z  − J  Nx  

2 Var X Nt

x, z

N →0 →∞

→0 )2 } N →∞

− E {dX Nt | ηNt }

dX Nt − E {dX Nt | ηNt }  2 →0

So

2

2

p( z )

204

Stochastics, Control & Robotics

t

∫0 E =

{

(

)

2

η2 2 ( x) 1 − η2Ns ( x + z ) ds N s

 x

∑ J ′  N 

2

z2 N

x, z

2

p( z )

}

∫0 E ηN 2s ( x) (1 − ηN 2s ( x + z )) t



2

2

 ds  

 1  + O 2  N  ≤ It follows that

 1 λt   1  →0 z 2 p( z ) ∫ J ′ 2 (θ) d θ + O  2  N ∑  →∞ N  N z  0 X

N 2t

i.e. 1 N

∑

x ∈Z N

{

− E X N 2t

} → 0

1  x J   ηN 2t ( x ) −  N N

N →∞

 x  x J   ρN  t ,   N  N

∑

x∈Z N

N →0 →∞ Talking differential w.r.t. t, it is natural to expect that 1 N

∑

x ∈Z N

dt  x  ∂ρ  x J   d ηN 2t ( x ) − ∑ J  N  N N x∈Z  N  ∂t N

 x  t,  N

N →0 →∞ Since the “ Martingale part” of 1  x J   η 2 ( x) ∑  N x N N t tends to zero, it follows that

N2 −

(

  x λdt  y   J   − J    ηN 2t ( y ) 1 − ηN 2t ( x) ∑ N  N x, y ∈Z  N N

dt  x  ∂ρ ∑ J  N N x ∈Z  N  ∂t N

 x →0  t,  N →∞ N

(

)

  x + z  x  λN ∑  J   − J    ηN 2t ( x) 1 − ηN 2t ( x + z ) p( z )   N  N x, z

Thus,

− Assume that

)

∑ zp( z ) z

1  x  ∂ρ J  N ∑ N x  N  ∂t

 x  → 0  t,  N →∞ N

= 0. Then we get from the above, by replacing ηN 2t ( x)

Stochastic Filtering & Control, Interacting Partials

 x with ρ N  t,  ,  N

205

  x  1  x  z 2  λN ∑  J ′   + J ′′   2    N  2  N  N  x, z    x + z  x p ( z ) ρ N  t,  1 − ρ N  t ,     N N   1

− ∫ J (θ) 0

∂ρ N (t , θ) d θ → 0 ∂t

  x  1  x  λ ∑  J ′   z + J ′′   z 2    N ∂N  N   x, t 

or

 x  ∂ρ  x ρ N  t,  1 − ρ N  t ,  − N  N  ∂θ  N 1

− ∫ J (θ) 0

or equivalently with

D=

  x z  t,   p ( z) N N 

∂ρ N (t , θ) d θ → 0 ∂t

∑ z 2 p( z ) , z

λD  x   x  x J ′′   ρ N  t,  1 − ρ N  t ,   ∑  N   N  N ∂N x −

λD  x  x  ∂ρ  x  J ′′   N  t,  ρ N  t ,  ∑      N x N ∂θ N N 1

− ∫ J (θ) 0

or

∂ρ N →0 (t , θ) d θ N →∞ ∂θ

λD 1 J ′′(θ) ρ N (t , θ)(1 − ρ N (t , θ)) d θ 2 ∫0 1 ∂ρ − λD ∫ J ′(θ)ρ N (t , θ) N (t , θ) d θ 0 ∂θ

∂ρ N →0 (t , θ) d θ N →∞ ∂θ Integrated by parts gives assuming ρ N (t , θ) → ρ(t , θ), N → ∞ 1

− ∫ J (θ) 0

λD ∂ 2 λD ∂ 2 (ρ2 ) ∂ρ ( ρ (1 − ρ )) + − =0 2 ∂θ2 2 ∂θ2 ∂t

{

}

 x We are justified in replacing hN2t (x) by ρ N  t,  ≡ E ηN 2t ( x) because taking  N J. 

206

Stochastics, Control & Robotics

[11] Comments and Proofs on “Particle System” Collected Papers of S.R.S. Varadhan Vol. 4: (1) Simple exclusion: dht (x) =

∑ {ηt ( x) (1 − ηt ( x)) dNt( y, x) y

− ηt ( x) (1 − ηt ( y )) dNt( x, y )

}

Nt( x, y) = Poisson process with mean p (y – x) dt. df (ht) =

∑ ( f (ηt( x, y) ) − f (ηt )) x, y

ηt ( y ) (1 − ηt x) − dNt( y , x) = ( L f ) ( ηt ) dt + dM t

where

Lf (h) =

∑ ( f (η( x, y ) ) − f (η)) x, y

( η( y ) (1 − η( x)) p( x − y )) and

dMt =

∑ ( f (ηt( x, y) ) − f (η)) x, y

ηt ( y ) (1 − ηt ( x)) dM t( x, y ) ( x , y) − p ( y − x)t M t( x, y) = Nt

Where is a Martingale

( ) d E ( M ) = d E  M , M  d M , M = ∑ ( F (η ) − f (η ) η ( y) (1 − η ( x)) d M t2 = 2M t dM t + d M , M 2 t

t

t

t

( x , y) t

t

2

t

2

2

t

x, y

d M t( x, y) , M ( x, y )

t

)

(

2  ( x, y ) E  d M ( x, y) , M ( x, y )  = E  d N − p( y − x)dt    t   = p (y – x) dt

So,

d  2 E Mt  = dt 



∑ E ηt( y ) x, y

p (y – x)

2

((

)

)

2 (1 − ηt ( x)) 2 f ηt( x, y) − f ( ηt )  

Stochastic Filtering & Control, Interacting Partials

207

Now, But latence by Z and consider XN (t) =

1 N

∑

xt Z N

 x J   ηt ( x)  N

We define the empirical measures on [0.1] 1 mt, N = ∑ ηt ( x) δ x / N N x∈Z N

Then

XN (t) =

We have dXN (t) =

1 N

=

1 N

∑

x∈Z N

∫

J ( x) d µ t , N ( x) .

[ 0, 1]

 x J   d ηt ( x)  N

∑

x , y ∈Z N

 x J    ηt ( y ) (1 − ηt ( x)) dNt( x, y )  N

− ηt ( x) (1 − ηt ( y )) dNt( x, y )  = =

1 N

∑

x , y ∈Z N

  x  y  ( y, x)  J  N  − J  N   ηt ( y ) (1 − ηt ( x)) dNt

  x 1  y  ∑ J   − J  N   ηt ( y) (1 − ηt ( x)) p( x − y) dt N x, y   N 

+

  x 1  y  J   − J    ηt ( y ) (1 − ηt ( x)) dM t( y , x) ∑     N x, y  N N 

We estimate the Martingale part Mt(N) where Mt(N) =

1 N

∑

2

x , y ∈Z N

  x  y  ( y, x)  J  N  − J  N   ηt ( y ) (1 − ηt ( x)) dM t

as N → ∞. We have d E dt

M ( N )z  = 1  t  N2

∑

x , y ∈Z N

  x  y   J  N  − J  N  

2

E ( ηt ( y ))(1 − ηt ( x)) p( x − y ) Consider finite range interaction only i.e.

Then,

(

p (z) = 0 for |z| > R, where R < ∞.

2 d E M t( N ) dt

)

=

1

∑ N 2 | z| ≠ R

x ∈Z N

(

2   x + z   x   J   − J     N N  

)

}

× E ηt ( x + z ) (1 − ηt ( x)) p ( − z )

208

Stochastics, Control & Robotics

We have since |z| < R. 2  x  x + z  1   x z 1  x z − J J + O 3 J J + ′′ ′ =        2  N  N   N N 2  N N N 

assuming J to be bounded on [0, 1]. Thus, 2

  X +Z  X   J  N  − J  N  

2

2  1  X Z + O 3 = J′   N  N2 N 

(

2 d E M t( N ) N →∞ dt

and hence

lim

∑

≤ lim

N →∞ x ∈Z N

2

X J′   N

(Z

2

)

)

p ( −Z ) / N 2 E ( ηt ( X + Z ) (1 − ηt ( X )) )

|Z | ≤ R

We also have 2 d lim E M t( N ) N →∞ dt

(

)

1

2D 2 ∫ J ′ (θ) d θ = 0 x→∞ N 0

≤ lim

Here D= We have,

(

2 d E M t( N ) dt

)

∑

Z 2 p( −Z ) =

1

∑

|Z | ≤ R

=

N2

X ∈Z N

∑

Z 2 p(Z )

|Z | ≤ R

  X +Z  X   J  N  − J  N  

2

|Z| ≤ R E ( ηt ( X + Z )(1 − ηt ( X ))) p ( − Z ) Exponential Martingale: XN (t) = dXN (t) =

X

∑ ηt ( X ) J  N  X

X

∑ J  N  (ηt (Y )(1 − ηt ( X ))) dNt(Y , X )

X,Y

(

− ηt ( X ) 1 − ηt (Y )) dNt( X , Y ) =

 X

) 

Y

∑  J  N  − J  N  ηt (Y ) (1 − ηt ( X )) dNt(Y , X )

X ,Y

d exp (XN (t)) =



  X 

 Y  



∑  exp   J  N  − J  N    − 1 dNt(Y , X ) ηt (Y )(1 − ηt ( X ))

X ,Y

t t     d exp  X N (t ) + ∫ f ( s ) ds = exp  X N (t ) + ∫ f ( s) ds     o 0

× (exp(dX N (t ) + f (t ) dt) −1)

Stochastic Filtering & Control, Interacting Partials

209

t   = exp  X N (t ) + ∫ f ( s) ds (exp dX N (t )) ⋅ (1 + f (t ) dt − 1)   o

    X  Y   = ξ(t )  1 + ∑  exp  J   − J    − 1 dNt(Y , X )   N       N   X , Y  × (1 + f (t) dt ) ηt (Y ) (1 − ηt ( X )) − 1)    X  Y   = ξ(t )  f (t ) dt + ∑  exp  J   − J    −1   N     N X ,Y   × ηt (Y ) (1 − ηt ( X )) dNt(Y , X ) where

XN (t) =

}

X

∑ J  N  ηt ( X ) , X

t   ξ (t) = exp  X N (t ) + ∫ f ( s) ds   o

Taking 

f (t) = −

 X

 Y 



∑  exp  J  N  − J  N   −1 ηt (Y )(1 − ηt ( X )) p( X − Y )

X,Y

we get

  X  Y   (Y , X ) dξ (t) = ξ(t ) ⋅ ∑  exp  J   − J    −1 ηt (Y )(1 − ηt ( X )) dM t       N N   X ,Y (Y , X )

Where M t Martingale

(Y − X ) − p(Y − X ) t is a Martingale. Thus, {ξ(t)}t≥0 is a = Nt



[12] Interacting Diffusions: dXi (t) = – dZi, i + 1(t) + dZi – 1, i(t)

dZi, i + 1(t) = (ψ ( X i (t )) − ψ ( X i +1 (t )) ) dt + dBi , i +1 (t ) Bi , i +1 (⋅), i ∈Z are independent Brownian motion processes 

[13] This is Independent of s. It is Reasonable to Expert that at Equilibrium:  Nε   E ψ( Xi ) − ψ X i + m 2 N ε + 1 → 0, N → ∞, ∀ ε > 0 ∑   m = − N ε

210

Stochastics, Control & Robotics

Note that

Nε

1 ∑ Xi+ m = x→∞ ( 2 N ε + 1) m= −N ε lim

∫ X p ( X ) dX

=

∫ X exp(σX − φ( X )) dX ∫ exp(σX − φ( X )) dX

= F ′(σ)

fN,t = N ⋅ ∑ X i (t ) χ  i

Thus, writing

i +1  N , N 

(i + 2) N

fN,t is a random function on [0, 1]. We have 1

∫ J (θ) f N , t (θ) dθ

∑ X i (t ) N ∫

=

i

0

i/ N

 i J (θ) d θ ≈ ∑ J   X i (t ) = XN (t)  N i

1  i  i We get ∫ J (θ) f N , t (θ) dθ ≈ ∑ J   dX i (t ) ≈ 2 ∑ J ′′   ψ ( X i (t)) dt  N   N N i I ≈

N



Xi+ m   dt  [ m] ≠ N ∈ (2 N ε + 1) 

 i

1 2

∑ J ′′  N  ψ  ∑ i

Writing

∑

Xi+ m

(2 N ε + 1) [ m] ≠ N ∈

≈

 i + m f N ,t    N  [ m] ≠ N ∈

∑

 i (2 N ε + 1) ≈ ρt    N

ε→o N →∞. (i + N ) / N

∫

Now,

i/ N

f N , t (θ) d θ = Xi (t)

We this get

∑

| k − i| ≠ N ε

i + N ∈ ( k +1) / N

X k (t ) =

∑

∫

k =i −N∈ k /N∈

i/ N + ε

f N , t (θ) d θ =

∫

f N , t (θ) d θ

i/ N − ε

  ξ exp(I F′ (a) ξ − φ(ξ)) dξ E  X i (t ) ∑ X k (t ) (2N ε + V ) = a  ≈ ∫   exp( I F′ (a ) a − I f (a)) | k − i| ≤ N ε   F (s) = supx (sx –IF (x)) = sa – IF (a) Where

IF′ (a) = s

Stochastic Filtering & Control, Interacting Partials

211

  E  X i (t ) ∑ X k (t ) (2 N ε + 1) = a  ≈ F ′(σ) = F ′ ( I F′ (a))   | k − i| ≤ N ε  

So So,

 d  1 E  X N (t ) X k (t ) = a ∑  dt  (2 N ε + 1) |k − i| ≤ N ε ≈

 i

1

Writing

1



i

≈



 X k (t ) = a   | k − i| ≤ N ε 

∑ J ′′  N  E ψ( X i (t )) (2 N ε + 1) ∑ N2 1 N

2

 i

∑ J ′′  N  ψ (a) i

 i a = ρt   , we get  N 1 d  i   i     i   i  ⋅ ∑ J   F ′ I F′ ρt    ≈ 2 ∑ J ′′   ψ ρ  N   N   t  N   dt i  N   N i

Now

F′ (s) = a, I′F (a) = s so

i F ′ ( I F′ (a)) = F ′(σ) = a = ρt   .  N and hence, d 1  i  i  i     i  ρ J   ρt   ≈ 2 ∑ J ′′   ψ ∑  N   τ  N   dt i  N   N  N i Now change the time scale: N2t = t. Then let t (θ) = ρ We get d  i  i ∑ J   ρτ   ≈ dτ i  N   N 

ρt

N2

(θ) .  i



 i 

∑ J ′′  N  ψ  ρτ  N   i



[14] Wong Zakai Filtering Equation: dz = – Ct (z) dt + at (z) dvt dy = gt (z) dt – dvt   p  t , T | Y1   +dt +dt 

212

Stochastics, Control & Robotics

p (dyt | zt = z)

∫ p( z (t + dt ) | z (t ), Yt dy(t)) =

∫

p( z (t ) | Yt , dy (t )) dz (t )

p( z (t ) dz (t), Yt , dy (t )) p (dy (t) | z (t ), y (t)) p ( z (t ), Yt , dy (t))

p( z (t ), Yt ) dt (z) p(Yt , dy(t )) =

∫

p(dy (t), dz(t ) | y (t ), z (t )) p(dy (t) | z (t ), y (t)) p(dy (t) | z (t ), Yt )

dz(t ) p( z (t) | Yt ) p(dy (t) | Yt ) =

∫ p(dy(t), dz(t ) | y(t ), z (t )) p(z (t ) | Yt ) dz (t )

∫ p(dy(t), dz(t ) | y(t), z (t )) p( z(t ) | Yt ) dz(t ) d dz(t )

 −Ct ( z )  at ( z )  dz   dy =  g ( z )  dt +  −1  dvt t p (dy, dz, y |, z) = E [ϕ ( z (t + dt)) | Yt + dt ] = E [ϕ ( z (t), dy(t )) | Yt + dt ]  at ( z )  −1  (ab (z) – 1) + e I2 = Bt (z):  1 p (dy, dz | y, z) = (2π) −1 | Bt ( z) |−1/2 exp  − dt (dz + Ct ( z ) dt,dy  2 − gt (t ) dt) Bt ( z) −1 (dz + Ct ( z) dt dy − gt ( z) dt )}  ε + at2 ( z ) −at ( z ) Bt (z) =   1+ t   − at ( z ) |Bt (z)| = (1+ ε) (ε + at2 ( z )) − at2 ( z ) 2 2 2 2 = ε + ε + ε at ( z ) = ε (1+ at ( z )) + O(ε )

(

| Bt ( z) |−1/ 2 ≈ ε −1/ 2 1+ at2 ( z )

)

1/ 2

at ( z )   1+ ε Bt (z)–1 =    at ( z ) ε + at2 ( z ) at ( z )   1 =   2  at ( z ) at ( z )

(

ε 1 + at2 ( z )

(

ε 1 + at2 ( z )

)

)

Stochastic Filtering & Control, Interacting Partials

213

(

p (dy, dz | y, z) = (2π) −1 ε −1/2 1 + at2 ( z )

)

−1/ 2

 1 ⋅ exp  − 2 (dz + Ct (t ) dt , dy − gt ( z ) dt) 2  dt ε (1+ at ( z)) at ( z )   dz + Ct dt    1     at ( z ) at2 ( z )  dy − gt dt  

(

L t + at2

)

−1/ 2

 −1  exp  ((dz + Ct dt ) 2 + at2 (dy − gt dt )2 2  2 ε 1 + at dt

(

)

  + 2 at (dz + Ct dt ) (dy − gt dt )    

(

1 + at2

(

= 1+

)

−1/ 2

 1  exp 1 − 8Ct dz − at2 gt dy − at gt dz + Ct at dy 2  ε 1 + at

(



)

−1 2  exp  at2

)

−1/ 2

(

)

−1/ 2

= 1 + at2

−1

ε 1 + at2 

(

2 = 1+ at

(

)

)(

(

(Ct − at gt ) dz + (

Ct at − at2 gt

 C −a g    ( t t) exp  − t (dz + at dy) 2 ε 1+ at  

(

)

   (C − at gt )  exp 1 − t (dz + at dy) 2 ε 1+ at  

(

)

 2 ( 1+ a ) d t  t 2 ε 2 (1 + at2 ) 2  p (dy | z, y) = p (dy | z) +

=

(Ct − at gt )

1   1 exp  − (dy − gt dt ) 2    2dt 2π dt  g2  α exp  gt dy − t dt  2  

g2 gt2 dt + t dt 2 2 = 1+ gt (z) dy.

= 1+ gt dy −

  dy  

) )

   

)

214

Stochastics, Control & Robotics

∫ (1 + at ( z )

Numerator =

 2 −1/ 2 

)

1 − 

(Ct − at gt ) (dz + at dy ) ε (1+ at2 )

2  Ct − at gt ) (  + dt {1+ g ( z ) dy} p ( z | Y ) dz

(

2 ε 2 1 + at2

)

e (1 + at2) → e

Let The numerator is

t

t

t

(Ct − at gt ) (dz + at dy ) ε



∫ 1 − +

 

(

1 + at2

)

(Ct − at gt )2 dt + g 2ε 2

Numerator of

t

 dy  pt ( z | Yt ) dz 

pt + dt ( z ′ | Yt + dt ) = 1 − 1 Ψt ( z ) ( z ′ − z ) pt ( z | Yt ) dz ε∫ − +

dy at ( z ) Ψt ( z ) pt ( z | Yt ) dz ε ∫ dt

∫ Ψt ( z ) pt ( z | Yt ) dz (1 + at ( z )) 2

2

2

2ε Yt (z) = Ct (z) – at (z) gt (z).

∫ Ψt ( z ) ( z ′ − z ) pt ( z | Yt ) dz = E  Ψt ( z (t )) ( z ′ − z (t )) | Yt  = E  Ψt d ( z ′ − dz (t )) dz (t ) | Yt 

Doemo Work. Alternately Writing Numerator =

+ =

Ψt ( z ) ( z ′ − z + at ( z )dy ) ε



∫ 1 −

Ψt2 ( z ) 2ε



2

∫ 1 − +



( z ′ − z + at ( z ) dy )2  pt ( z Yt )dz 

Ψt ( z ′ − δz ) (δz + at ( z ′ − δz ) dy ) ε

Ψt2 ( z ′ − dz ) 2ε

2

 (δz + at ( z ′ − δz )dy ) 2  pt ( z ′ − δz Yt )d δz 

Stochastic Filtering & Control, Interacting Partials

=

215

(δz )2 p ′′ z ′ Y  p z Y − δ z p z − Y + ′ ′ ′ ( ) t t t t ∫ t t 2 

(

)

(



) 

 1 × 1 − (ψ t ( z ′) − δ z ψ t′ ( z ′ ))(δz + at ( z ′) dy − at′ ( z ′) δz dy )  ε +

Ψt2 ( z ′ ) 2ε

2

 dt 1+ at2 ( z ′ )  d δz 

(

)

 ∆ ∆ Let the range of dz be  − ,  . Then  2 2 ∆ at2 ( z ′ ) p ′′( z ′ | Yt ) ∆ + Ψt ( z ′ ) p ′ ( z ′ | Yt ) Numerator = pt ( z ′ | Yt ) ∆ + dt ε 2 ∆ − Ψt ( z ′ ) pt ( z ′ | Yt ) at ( z ′ ) dy at2 ( z ′) dt − at2 ( z′) dt ε Ψ2 ( z′ ) + pt ( z ′ | Yt ) t 2 dt (1+ at2 ( z ′)) ∆ 2ε ∆ Ψt ( z ′ ) pt ( z ′ | Yt ) at′( z ′) at ( z ′) dt ε Cancelling out the factor D gives the numerator as −

 a 2 ( z ′ ) pt′′( z ′ | Yt ) pt ( z ′ | Yt ) +  t  2 1 − a1′ ( z ′ ) at ( z ′ ) Ψt ( z ′ ) pt ( z ′ | Yt ) ε + pt ( z ′ | Yt )

Ψt2 ( z′ ) 2ε 2

 (1+ at2 ( z ′)) dt 

1 − Ψt ( z ′ ) pt ( z ′ | Yt ) at ( z′ ) dy (t ) ε The denominator is the integral of the numerator w.r.t. z. Hence pt + dt ( z ′ | Yt + dt ) =

pt ( z ′ | Yt ) + Λ tε ( z ′ | Yt ) dt + M tε ( z ′ | Yt ) dy

(1 + dt ∫ Λ ( z ′ | Y ) dz ′ + dy∫ M ε t

t

ε t ( z ′ | Yt ) dz ′

)

ε ε = pt ( z ′ | Yt ) + dt Λ t ( z ′ / Yt ) − dt pt ( z ′ | Yt )∫ Λ t ( z ′ | Yt ) dz′

− M tε ( z ′ | Yt ) − pt ( z ′ | Yt )

(∫ M

(∫ M

ε t ( z ′ | Yt ) dz′

ε t ( z ′ | Yt ) dz′

) dt

) dy(t )

216

Stochastics, Control & Robotics

Thus,  a 2 ( z ′ ) 1 dpt ( z ′ | Yt ) =  t pt′′( z ′ | Yt ) − at′ ( z ′ ) at (z ′ ) Ψt ( z ′ ) pt ( z′ | Yt ) ε  2 +

Ψt2 ( z′ ) 2ε

2

(1 + a ( z ′)) p( z ′ | Y ) − p (z ′ | Y ) 2 t

t

t

t



[15] Stochastic Control of Finance Market: Pk (t), k = 1, 2, ..., d are the share prices. They evolve according to random rates: dPk (t) = (rk (t ) dt + σ k dBk (t )) Pk , 1 ≤ k ≤ d

s

where rk (t)′ are non random rates. At time t, the share holds has Nk (t) shares of type k, k = 1, 2, ..., d. His gain in Wealth in time [t, t + dt] is

d

∑ N k (t ) dPk (t )

and his consumption in time

k =1

[t, t + dt] is C (t) dt. His total Wealth at time T is then T d

 N ( t ) dP ( t ) C( t ) dt   ∫ ∑ k k  0  k =1

X (T) =

The quantity Nk (.) and C (.) satisfy certain constants like N′ks must not be too large etc. We write there constrants as E fr (N1(t), .,Nd (t), C(t), – lo, r = 1, 2, ..., M. P1 (t), ..., Pd (t) d Dr = dias {rk (t )}k =1  .  

Let

E [ X (t)] is to be maximized subject to these constaints. We consider maximizing M T

E X (T ) − ∑ ∫ λ r (t ) E f r (N (t ), P(t ), C (t )) dt r =1 0 T

{

}

= E ∫ N (t )T D r P(t ) − C (t ) − λ(t )T f ( N (t ), P(t ), C (t )) dt o

The optimum N (t ) , C (t) are obtained as functions of P(t ) that maximize (1) Let T

max

x ≠ N (t ), C (t ) ≠ T

∫ E  N (t) S

T

D r P(t ) − C (t ) − λ (t )T f ( N (t ), P(t ), C (t )) 

(1)

Stochastic Filtering & Control, Interacting Partials

217

= V* ( P(s ), λ s ) | P( s )] dt Where λ s = {λ(t ) : s ≤ t ≤ T } Then, V* ( P(s), λ s ) =

{( N (s) D P(s) − C (s) − λ (s) f ( N (s), P(s), C (s)) ds T

max

N ( s ), C ( s)

T

r

+ E V* ( P( s + ds ), λ s + ds ) | P( s )

}

V* ( P( s ), λ s ) = V* ( P( s ), s )

we write

Note that l is non-random. Then V* ( P(s ), s ) =

{( N (s)

T

max

N ( s ), C ( s)

)

D r P( s ) − C ( s ) − λ ( s )T f ( N ( s ), P(s ), C ( s ) ds T

 ∂V ( P(s ), s )  + V* ( P (s ), s ) +  *  D r P( s ) ds ∂P   1 + Tr 2

max

or

T   2  ∂V* ( P(s), s )    ∂ V* ( P(s ), s )   + D s ds ds ( )     T σ ∂s    ∂ P ∂ P  

T

N ( s ), C ( s)

+

{N (s)

}

D r P( s ) − C ( s ) − λ ( s )T f ( N ( s ), P(s ), C ( s ))

(

∂V* ( P( s ), s ) + Dr PT( s ) ∂s

)

∂V* ( P( s ), s ) ∂P

T

1  ∂ 2V* ( P( s ), s ) + Tr  D σ ( s )  =0 2  ∂ P ∂ PT  We thus get the following partial differential equation for V* ( P(s)) :

{

}

max N T D r P − C − λ ( s )T f ( N , P, C) N, P

+

∂ 2V* ( P, s)  ∂V ( P, s) 1  ∂V* ( P, s) + Tr  Dσ ( s ) + r ( s )T *  =0 ∂P ∂s 2  ∂ P ∂ PT 

{

}

d   Ds (s) = diag  σ 2k Pk2 ( s ) k =1    The optimal scheme of choosing N, C at each stage is thus given by

Where

( N (t ), C (t )) = arg max {N D P − C − λ(t )  ( P, λ (t )), C  ( P, λ (t )) = {N } T

N, P

r

T

}

f ( N , P, C )

Finally {λ (t)}t ∈[o,T ] is obtained using this equation combined with

218

Stochastics, Control & Robotics

{

}

 ( P(t ), λ (t )), P(t ), C  ( P(t ), λ (t))) E fr ( N

= 0, 1 ≤ r ≤ d. The expectation is to be taken w.r.t. the random vector P(t ) . 

[16] Wong-Zokai Equation: dz = − Ct ( z ) dt + at ( z ) dvt

(State model)

dy = gt ( z ) dt − dvt

(Measurement model)

p( Z (t + dt ) | Yt + dt ) = p( Z (t + dt ) | Yt , dy (t )) =

p( Z (t + dt ) | Yt , dy (t)) p(Yt , dy(t ))

=

p( Z (t + dt ), dy (t ) | Yt ) p(dy (t) | Yt )

=

∫ p(Z (t + dt ), dy(t ) | Yt Z (t )) p (Z (t) | Yt ) dZ (t)

=

∫ p(Z (t + dt ), dy(t ) | Z (t), y(t )) p(Z (t ) | Yt ) dZ (t )

≡

Let

p(dy (t) | Yt )

p (dy (t) | Yt )

χ( Z (t + dt), dy(t ) | Yt )

∫ χ(Z (t + dt, dy(t ) | Yt ) dZ (t + dt)

 −ct ( z )  dv1 (t )   dz   dy  =  g ( z )  dt + Bt ( Z )  dv (t )     2   t  E [ ψ( Z (t + dt), dy(t ) | Z (t ), y (t ))] = ψ( Z (t), 0) −

∂ψ ( Z (t ), 0)ct ( Z (t )) dt ∂Z

+

∂ψ 1 ∂2 ψ ( Z (t ), 0) gt ( Z (t ))dt ( Z (t), 0) at 2 ( Z )dt + 2 2 ∂Z ∂y

+

1 ∂2 ψ ( Z (t), 0) dt 2 ∂y 2

=

∫ ψ (Z (t + dt), dy(t )) χ(Z (t + dt ), dy(t) | Yt ) dz (t + dt) d dy(t)

=

∫ ψ (Z , o) − ψ , z (Z , o)ct (Z ) dt + 2 ψ , zz (Z , o) at (Z ) dt

1

2

1  + ψ , Y ( Z , o) gt ( z)dt + ψ , YY ( Z , o) dt  pt ( Z | Yt ) dZ 2 

Stochastic Filtering & Control, Interacting Partials

=

219

∫ ψ (Z , dy) χ(Z , dy | Yt ) dz d dy

Hence, χ( Z , dy | Yt ) = δ(dy ) pt ( Z | Yt )  ∂ 1 ∂2 2 +  (Ct ( Z ) pt ( Z | Yt )) δ (dy) + at ( Z ) pt ( Z | Yt ) δ (dy ) 2 ∂Z 2  ∂Z

(

)

1  − δ ′(dy ) gt ( Z ) pt ( Z | Yt ) + δ ′′ (dy ) pt (Z | Yt )  dt 2  

[17] Filtering Theory with State and Measurement Noises Having Correlation: State mode: d X (t ) = f ( X (t )) dt + g ( X (t)) dB(t ) dZ (t ) = h ( X (t )) dt + K ( X (t)) dB(t ) n n n×q , K (X (t )) ∈R p × q , B(t )  standard where X (t ) ∈R , Z (t) ∈R , g ( X (t )) ∈R

q dimensional Brownain motion: Z t = {dZ ( s ), s ≠ t} We need an equation for p( X (t ) | Zt ) p( X (t + dt ) | Z t + dt ) =

∫ p( X (t + dt ) | X (t ), Z t , d z(t )) p( X (t ) | Z t , d z (t )) d X (t)

=

∫ p( X (t + dt ) | X (t ), Zt , dZ (t ))

=

∫ [ p( X (t + dt ), Z (t + dt ) | X (t ), Z (t )) × p( X (t ), Z t ) / p ( Z t , dZ (t ))] dX (t)

=

p( X (t ), Z t ) p( X (t ) | Z t ,dZ (t )) p ( X (t ), Z t , dZ (t ) dX (t))

∫ [ p( X (t + dt ), Z (t + dt ) | X (t ), Z (t )) × p( X (t ), Zt ) dX (t )] ∫ dX (t + dt )

We write X (t + dt ) = X ′ Z (t + dt ) = Z ′ , X (t) = X, Z (t) = Z

220

Stochastics, Control & Robotics

and p( X (t + dt ), Z (t + dt ) | X (t ), Z (t )) = p(t + dt , X ′, Z | t , X , Z ) = δ ( X ′ − X ) δ ( Z ′ − Z ) + dt L *′ (δ ( X ′ − X ) δ ( Z ′ − Z )) Where L* is the formed Kolmogorov operator for the state plus observable process: Lψ *′ = − ∇TX ′ ( f ( X ′ ) ψ ( X ′, Z ′ )) = − ∇TZ ′ ( h ( X ′ ) ψ ( X ′, Z ′ ))   g ( X ′)  1   ∇X ′  T ∇ X ′ , ∇TZ ′  + Tr   g ( X ′ )1T K ( X ′ )T ψ ( X ′, Z ′ )    K ( X ′ ) 2   ∇Z ′  

(

)

= − div ′X ( f ( X ′ ) ψ ( X ′, Z ′ )) − div ’Z ( h ( X ′ ) ψ ( X ′, Z ′ ))

( ( ( (

We than get = = Now,

) ) ) )

1 + Tr ∇ X ′ ∇TX ′ g ( X ′ ) g ( X ′ )T ψ( X ′, Z ′ ) 2 1 + Tr ∇ X ′ ∇TX ′ K ( X ′ ) g ( X ′ )T ψ( X ′, Z ′ ) 2 1 + Tr ∇ Z ′ ∇TX ′ g ( X ′ ) K ( X ′ )T ψ( X ′, Z ′ ) 2 1 + Tr ∇ Z ′ ∇TZ ′ K ( X ′ ) K ( X ′ )T ψ( X ′, Z ′ ) 2 p(t + dt , X ′ | Zt + dt )

∫ (δ( X ′ − X ) δ(Z ′ − Z ) + dt L *′ (δ( X ′ − X ) δ(Z ′ − Z )) p(t, X | Zt ) dX )

∫ (δ( X ′ − X ) δ(Z ′ − Z ) + dt L *′ (δ( X ′ − X ) δ(Z ′ − Z )) p(t, X | Zt ) dX dX ′ ) [ p(t , X ′ | Zt ) δ(Z ′ − Z ) + dt L *′ ( p(t , X ′ | Zt ) δ(Z ′ − Z ))] p t , X ′ | Zt ) δ( Z ′ − Z )) dX ′ δ( Z ′ − Z ) + dt ∫ L *′ ( p(

∫ L *′ ( p(t , X ′ | Zt ) δ(Z ′ − Z )) dX = − ∫ div Z ′ ( h ( X ′ ) p(t, X ′ | Z t ) δ ( Z ′ − Z )) dX ′

(

))

(

1 Tr ∇ Z ′ ∇TZ ′ K ( X ′ ) K ( X ′ )T p(t , X ′ | Z t ) δ( Z ′ − Z ) dX ′ 2∫ 1  h t ∇ Z ′ δ ( Z ′ − Z ) + Tr ∇ Z ′ ∇TZ ′ Kt KtT δ ( Z ′ − Z ) = − div Z ′   2  +

(

Where

)

T 1  − h t ∇ Z ′ δ ( Z ′ − Z ) + Tr Kt KtT ∇ Z ′ ∇T , δ ( Z ′ − Z )   2  h t = E [ h ( X (t )) | Zt ] ,

 Kt KtT = E  K ( X (t )) K ( X (t))T | Zt 

Stochastic Filtering & Control, Interacting Partials

221

This we get p(t + dt , X ′ | Zt + dt ) = where

[δ(Z ′ − Z ) ⋅ p(t , X ′ | Zt ) + dt L *′ ( p(t , X ′ | Zt ) δ(Z ′ − Z ))]

1    δ( Z ′ − Z ) + dt  −  h ∇ Z ′ δ ( Z ′ − Z ) + Tr Kt KtT  ∇ Z ′ ∇TZ ′ δ ( Z ′ − Z )    2 

Z = Z (t ), Z ′ = Z (t + dt ) = Z (t ) + dZ (t) .

L *′ ( p(t , X ′ | Z t ) δ( Z ′ − Z )) T = Lo *′ ( p(t , X ′ | Zt )) δ ( Z ′ − Z ) − h ( X ′) ∇ Z ′ (δ ( Z ′ − Z )) p (t , X ′ | Zt )

References: [1] Research Papers of V. Belavkin on quantum filter. [2] Technical Report on Wave propagation in Metamaterial by Karishma Sharma, D.K. Upadhyay and H. Parthasarthy, NSIT, 2014. [3] Technical Reports on Robotics by Vijayant Agrawal and Harish Parthasarathy, NSIT, 2014. 

5 Classical and Quantum Robotics [1] Mathematical Pre-Requisites for Robotics: [1] Kinetic and potential energies of an n-link system. [2] Relativistic kinetic energies. [3] Variational calculus. [4] Perturbation theory for algebraic and differential equations. [5] Brownian motion, Poinon process, Levy process (processes with independent increments. [6] Ito stochastic calculus for Brownian motion and Poinon processes. [7] Stochastic differential equations driven by Levy processes. [8] Forward and Backward Kolmogorov equations. [9] Markov chains. [10] Approximate mean and variance propagation in stochastic differential equations. [11] The maximum likelihood method for parameter estimation. [12] The Cramer-Rao lower bound (CRLB). [13] Optimal control theory. [14] Stochastic optimal control. [15] The n-link Robot differential equations. [16] Optimal control applied to master-slave Robot tracking. [17] Leaso mean square (LMS) algorithm applied to tracking problem in Robotics. (adaptive) [18] Recursive leaso squares (RLS) algorithm applied to tracking problem in Robotics (adaptive). [19] Maximum likelihood estimation of Pd controller for master-slave Robot system.

224

Stochastics, Control & Robotics

[20] Design of feedback and operator force process for optimal control of LTI systems with feedback noise – an application of algebraic perturbation theory. [21] The rotation group SO(3). [22] The kinetic energy of a rigid body. [23] Quantization of rigid body motion. [24] Feynman path integral approach to quantum Robotics. [1] The lengths of the links are L1, L2, .. , Ln. The first link has its free end at (X(t), Y(t)) and the angles of the ith link relative to the X-axis is qi(t). Let si be the mass pen unit length of the ith link. Then the position of a point P on the ith link at a distance x from its joint with the (i – 1)th links is Y given by * Maximum likelihood estimation of parameters in dynamical systems. ∞

Let {W [n]}n=0 be iid random vectors with values in Rp having pdf Fw(W). Consider the dynamical system. Z[n + 1] = f ( Z [n]) + G ( Z [n])W [n + 1] , n ≥ 0 Then {Z[n]}n ≥ 0 is a Markov process provided Z[0] is independent of {W[n]}n ≥ 1. The conditional pdf of Z[n + 1] gives Z[n] is p(Z[n + 1]|Z[n]) = G ( Z [n])

−1

(

(

ΦW G ( Z [n]) −1 Z [n + 1] − f ( Z [n]

))

G ( Z ) = det(G ( Z ))

where

Hence if f and G depend on an unknown parameter vector q, i.e. f(Z, q), G(Z, q) then the joint pdf of {Z[n]}0 ≤ n ≤ N given q is

p({Z[n]}p ≤ n ≤ N|Z[0], q)

=

N

∏

n=0

{ G(Z[n], θ (

−1

ΦW G ( Z [n], θ )

−1

or equivalently, the log-likelihood function is

( Z [n + 1] − f ( Z [n], θ )))}

L({Z[n]}1 ≤ n ≤ N | Z(0), q) = log p(Z | Z(0), q) N

{

− ∑ log G ( Z [n], θ ) n=0

N

(

}

+ ∑ log ΦW G ( Z [n], θ ) n=0

−1

(Z [n + 1] − f ( z[n], θ)))

The ml of q based on Z = {Z[n]}P ≤ n ≤ N is θ ( N ) = arg max L ( Z Z (0) , θ ) θ

Classical & Quantum Robotics

225

Example: Z[n + 1] = We wish to estimate q =

p

∑

θk f k ( Z [n]) + G ( Z [n])W [n + 1] k =1 (qk)pk=1 when {W(n)}n ≥ 1 is iid N (0, RW

).

i−1 i−1   P =  X (t ) + ∑ Lk cos θk + ξ cos θi , Y (t ) + ∑ Lk sin θk + ξ sin θi , 1 ≤ i ≤ n    k =1 k =1

and its velocity is therefore given by i −1 i−1  dP  =  X ′(t ) − ∑ Lk θk′ sin θk − ξθi′ sin θi , Y ′(t ) + ∑ Lk θk′ cos θk + ξθi′ cos θi  dt   k =1 k =1

The kinetic energy of the ith link is then for i > 1, L

Ti =

2

i 1 dp dξ σi ∫ 2 0 dt

Li i−1  2 1 2 2 2 2 2 = σi ∫  X ′ + Y ′ + ξ θi′ + ∑ L k θ k′ 2 0 k =1

i −1

i −1

1

k =1

−2 X ′ ∑ Lk θk′ sin θk + 2Y ′ ∑ Lk θk′ cos θk

∑

+2

1≤ k < m≤ i −1

+ 2ξ

∑

1≤ k ≤ i−1

Lk Lm θk′ θm ′ cos(θk − θm )

θi′θk′ Lk cos(θ k − θi )

 + 2θi′ξ(Y ′ cosθi − X ′ sin θi ) dξ  1 i− 1  = σi  X ′ 2 + Y ′ 2 + ∑ Lk θk′ 2 2  k =1 i −1

i −1

1

1

− 2 X ′ ∑ Lk θk′ sin θk + 2Y ′ ∑ Lk θk′ cos θk

+2 +

σi 2

∑

l ≤ k < m ≤i −1 θ′ 2 σi L3i i

Lk Lm θk′ θm ′ cos θk − θm  Li 

6 σ ∑ L2i Lk θi′θk′ cos(θk − θi ) + 2i L2i θi′ (Y ′ cos θi − X ′ sin θi ) l ≤ k ≤i−1

226

Stochastics, Control & Robotics

For i = 1, we have 2 T1 = 1 σ1 ( X ′ 2 + Y ′ 2 ) + σ1L13 θi 2 6 1 + σ1L12θ1′ (Y ′ cos θ1 − X ′ sin θ1 ) 2 The relationship between the kinetic energy T and velocity v of a point mars having rest mars m0 is

m0 c 2

T =

1−

v2

.

c2

Hence the kinetic energy of the ith link is Li

2 Ti = σi c ∫ ( Fi X ′ (t ), Y ′ (t ), θ ′j (t ), θ j (t ), 1 ≤ j ≤ i, ξ)dξ 0

 v2  where Fi(X′, Y′, q′j, qj, 1 ≤ j ≤ i, x) = 1 −   c2 

−1 2

with

i−1

v2 = X ′ 2 + Y ′ 2 + ∑ L2k θ′k2 k =1

+2

∑

l ≤ k 0 a small parameter. Assume that we've solved f0(X) = 0

getting the solution X = X0. We wish then to approximately solve f0(X) + eg0(X) = 0

(a)

X = X0 + eX1 + e X2+ ... = X 0 + 2

Let

∞

∑ εm X m

(b)

m=1

Equating substitutively (b) into (a) and equality coeffs of em, m = 0, 1, 2, ... gives successively. f0(X0) = 0, O(e0) N

∑ f0 , j ( X 0 ) X1j + g0 ( X 0 )

= 0, O (∈0),

j=1

1 ∑ f0 , jk ( X 0 ) X1 j X1k 2 jk + ∑ f 0 , j ( X 0 ) X 2j + ∑ g0, j ( X 0 ) X1 j = 0, O (e2) j

j

Df 0 ( X 0 ) =  f 0,1 ( X 0 ), f 0, N ( X 0 )  we get on solving the above,

etc. Writing

X1 = −  Df 0 ( X 0 ) X2 = −  Df 0 ( X 0 )

−1

g0 ( X 0 ) ,

1 −1 

 ∑ f 0,  2 j ,k

+ ∑ g 0, j ( X 0 ) X 1 j j

etc.

jk ( X 0 ) X 1 j

X1k

}

Thus in general, Xm can be solved in terms of Xm–1, ..., X0 by inverting the matrix Df 0 ( X 0 ) . This complete the problem of algebraic perturbation theory: For differential equations: dX (t ) = f 0 (t , X (t )) + εf1 (t , X (t )) dt Let X0(t) be a solution to the unperturbed problem: dX 0 (t ) = f0(t, X0(t)) dt

(b)

230

Let

Stochastics, Control & Robotics

X(t) = X 0 (t ) +

∞

∑ ε m X m (t )

m=1 2

Equating coeffs. of e, e , ... in (b) given O (e):

dX1 = dt

∑ f0,k (t , X 0 (t )) X1k (t ) + f1 (t, X 0 (t ))

O (e2):

dX 2 = dt

∑ f0,k (t , X 0 (t )) X 2k (t )

k

k

+

1 ∑ f0, km (t , X 0 (t )) X1k (t ) X1m (t) 2 k ,m

+ ∑ f1, k (t , X 0 (t )) X1k (t ) k

Let

Then if

i.e.

Df (t , X 0 (t )) =  f 0,1 (t , X 0 (t )), .., f 0, N (t , X 0 (t ))   = A(t ) . ∂Φ(t , τ) = A(t )Φ (t , τ), t ≥ τ, ∂t F(t, t) = I ∞

F(t, t) = I + ∑

∫

A(t1 )... A(tn )dt1 , ...dtn

n=1 τ< tn d > t

(EtEt = Et, EtEs = Et for t > t, s < t) t

t

−

b λ v b ∫ Es ( Z FZ )+ ξds + M t ξ = Et ∫ (Z FZ )v dAλ ξ 0

0

−

b since Es ( Z b FZ ) dAλv ξ = Es ( Z FZ )+ ξds λ v

Observation processes Di(t)∈Dt

λ

dB(t) = ( Z b DZ )v dAv λ D = D0 + λ is D i t

et(X(t)) =

∫ εs (Z

b

−

FZ )+ ds + M t

0

To get the filtering eqn, we need Mt in term of the observation processes B(t). Let t

Mt =

∫ (Z 0

b

t

D i Z

(

)vλ (s)K si dAλv (s)Kti ∈∞t′Yi (t )

− ∫ ε s Z b Di Z =

t



(

∫ (Z

b

λ Di Z )vλ ( s )d Av ( s ) − ε s Z b Di Z

∫ (Z

b

Di Z )+− − ε s ( Z b Di Z )+−  ds

0 t

=

0

)+− (s)ds

0

)+− (s)ds 

Large deviations, Classical & Quantum General Relativity with GPS Applications t

+∫

281

λ ∑ (Z b Di Z )λ (s)d Av (s) v

0 ( v , λ) ≠ ( −+)

D0− Z +0 + D+−   D00 Z +0 + D+0   0

(

)

D0− Z +0 + D+− + Z +0* D00 Z +0 + D+0    Z 00* D00 Z +0 + D+0  0 

0 D0− Z 00  =  0 Z 00* D00 Z 00  0 0

−

 0 D0− Z 00  0 0  0 D0 Z 0  0 0

Z +−*   Z 0−*  I 

 I Z +0*  0* ZbDZ =  0 Z 0  0 0

(

(

0 − 0* 0 0 − ε s ( Z b DZ )+ = D0 ε s ( Z + ) + D+ + ε s Z + D0 Z +

)

)

( ) ( )+− − ε s (Z b Di Z )+− D0− ( Z +0 − ε s ( Z +0 )) + ( Z 0−* − ε s ( Z 0−* )) D+0 + ( Z +0* D00 Z +0 − ε s ( Z +0* D00 Z +0 )) + ε s Z 0−* D+0 Z b Di Z

=

Z +−*   Z 0−*  I 

 I Z +0*  Z b D Z =  0 Z 00*  0 0

Compute

(

(

)

)

 0 D 0− Z 00  0 0  0 D 0 Z 0 0  0

D 0− Z +0 + D +−   D 00 Z +0 + D +0  0 

(

)

A Z b DZ  , ds = D − Z 0 d A0− + D − Z 0 + D − + Z 0* D 0 Z 0 + Z 0* D 0 ds 0 0 + 0 + 0 + + + +

(

)

0

(

)

+

+ Z 00* D 00 Z 00 d A0 + Z 00* D 00 Z +0 + Z 00* D +0 d A0 Coeff. of ds in this is D +− + D 0− Z +0 + Z +0* D 00 Z +0 + Z +0* D +0

−

= D+− − ε s ( Z b DZ )+ + D0− Z +0 + Z +0* D00 Z +0 + Z +0* D+0

If we operate on this by es, we get

(

)

( )

( )

D+− + ε s Z +0* D00 Z +0 + ε s Z +0* D+0 + D0− ε s Z +0 − ε s ( Z b DZ ) − +

−

= ε s ( Z b DZ ) − ε s ( Z b DZ )+ = 0.

t b  ,ds ) is a Martingale. Hence ∫ A ( Z DZ 0

t

−

Also Y (t ) − ∫ ε s ( Z b DZ )+ ds 0

∫ (( Z t

=

0

b

)

− v µ DZ )µ ( s )d Av ( s ) − ε s ( Z b DZ )+ ( s )ds = x(t) say

282

Stochastics, Control & Robotics −

b d x(t) = dY (t ) − εt ( Z DZ )+ dt

(

0 − 0 − 0 − 0* 0 0 = D0 Z 0 d A− + D0 Z + + D+ + Z + D0 Z + −

)

+ Z +0* D+0 − εt ( Z b DZ )+ dt

(

)

+ 0 + Z 00* D00 Z 00 d A0 + Z 00* D00 Z +0 + Z 00* D+0 d A0

(

)

0 = D 0− Z 00 d A− + D 0− Z +0 + D +− + Z +0* D 00 Z +0 + Z +0* D+0 dt

(

)

0 0 + Z 00* D 00 Z 00 d A0 + Z 00* D 00 Z +0 + Z 00* D +0 d A0

D vµ = Dvµ if (m, v) ≠ (– +),

since

− D +− = D+− − εt ( Z b DZ )+

Quantum Filtering − µ v dX = Cv dAµ A+ = t

A+j* = A−j , A−j * = A+j , Akj*  creation operators Akj* = Akj A−j  annilutation operators j j − dA+j dAk− = δ kj dt = δ k dA+ Ak  conservation operators

dAkj dAl− = δ lj dAk− dA+j dAlk = δ lj dA+k {∞t, t ≥ 0} increasing family of von-Neumann algebras. Bt ⊂ ∞t ∩ ∩ ∞′s i.e. Y∈Bt ⇒ Y∈∞t and [Y, X] =0 ∀X ∈ ∞s > t.

s ≥t

Conditional expectation: Let X∈∞s,s > t.

Assume {∞s}t ≥ 0 operate on H a Hilbert space. et(X)∈Bt is such that et(X)x = Et(Xx) where Et is the orthogonal projection of ∞∞ onto Bt ξ . (X – et(X))x = Xx – Et(Xx) ⊥ Bt ξ . Let A, B∈Bt, X∈∞s, s > t

et (AX)x = Et(AXx)

AXx – Et(AXx) ⊥ Btx Xx – Et(Xx) ⊥ Btx

Second one implies UXx – UEt(Xx) ⊥ UBtx = Btx

Large deviations, Classical & Quantum General Relativity with GPS Applications



283

Unitary U∈Bt

AXx – A Et(Xx) ⊥ Btx

⇒

Et(AXx) – A Et(Xx) ⊥ Btx

Thus

Et(AXx) – A Et(Xx) ⊂ Btx

But

Et(AXx) = A Et(Xx)

Hence

et(AX)x = A et(X)x

or

It fixed et exists x, then it follows that

et(AX) = Aet(X)

Let the measurement model be µ v dY = Dv dAµ t

i.e.

Y(t) =

0 t

X(t) =

µ

v

µ

v

∫ Dv (s)dAµ (s) , ∫ Cv (s)dAµ (s) 0

For unitary evolution of X, we require

0 = d(X*X) = dX*.X + X*.dX + dX*.dX = Cvµ*Cβα dAµv*dAαβ + Cvµ* XdAµv* + X *Cvµ dAµv bµ v Cvµ*dAµv* = Cv dAµ

(Belavkin-Quantum non-linear filtering)

So for unitarity, we require * µ bµ α v β bµ v v 0 = Cv Cβ δ α dAµ + Cv XdAµ + X Cv dAµ

i.e.

µ

µ

Cβb Cvβ + Cvb X + X *Cvµ = 0.

Let x = f ⊗e(0) f∈h (System Hilbert space and e(0) = Vacuum exponential vector in Gs (H). t

X(t) =

µ

∫ Cv (s)dAµ (s) v

0

dAµv (t )ξ = C+− (t )ξdt Non-demolition condition: [Y(t), X(s)] = 0 s ≥ t. s  t Thus  ∫ Dvµ (t1 )dAµv (t1 ), ∫ Cvµ (t2 )dAµv (t2 ) = 0 s ≥ t   0 0

This condition is satisfied if

284

Stochastics, Control & Robotics

 Dvµ (t )dAµv (t ), Cβα (t )dAαβ (t ) = 0  

(a) or

Dvµ (t )Cβα (t ) δ αv dAµβ (t ) − Cβα (t ) Dvµ (t ) δβµ dAαv (t ) = 0

or

Dvµ (t )Cβv (t ) dAµβ (t ) − Cµα (t ) Dvµ (t ) dAαv (t ) = 0

or

Dvµ (t )Cβv (t ) − Cαµ (t )Dβα (t ) = 0

or

Dvµ Cβv − Cvµ Dβv = 0 v α β   µ  Dv (t )dAµ (t ), Cβ (s )dAα ( s ) = 0, s > t.

(b) or

e(u ),  Dvµ (t ), dAµv (t ), Cβα (s )dAαβ ( s )  e(v)

or

dAµv* (t )e(u ) Dvµ (t )Cβα ( s ) dAαβ ( s )e(v)

= 0 s > t, u, v∈H.

− dAαβ ( s )* e(u ) Cβα ( s ) Dvµ (t ) dAµv (t )e(v)

= 0, s > t, u, v∈H.

(b) requires Dvµ (t )dAµv (t )Cβα ( s )dAαβ ( s ) − Cβα ( s )dAαβ ( s ) Dvµ (t )dAµv (t ) = 0 s > t This is satisfied if α  v   dAµ (t ), Cβ (s)  = 0s > t

and

 Dvµ (t ), Cβα ( s ) = 0.  

Assume that the non-demolition condition is satisfied. Consider the algebra Bt generated by the operators {Y(s): s ≤ t}. t

X(t) = X ( s ) + ∫ Cvµ (τ)dAµv (τ) s

t

Y(t) = Y ( s ) + ∫ Dvµ (τ)dAµv (τ) , t ≥ s st

Then

s

+ D−j (τ)dA−j (τ) + Now,

t

(

Y(t) = Y ( s ) + ∫ D+− (τ)d τ + ∫ D +j (τ)dA+j (τ) . s Dkj (τ)dAkj (τ)

dA−j (τ)ξ = dA−j (τ) f ⊗ e(0) = f ⊗ dA−j (τ) e(0) = 0 k dAkj (τ)ξ = dA j (τ) f ⊗ e(0) k = f ⊗ dA j (τ) e(0) = 0 − e(u ) dA+j (τ) e(0) = dA j (τ)e(u ) e(0)

)

Large deviations, Classical & Quantum General Relativity with GPS Applications

285

= u j (τ)d τ e(u ) e(0) = uj(t)dt M1(t) =

Now, consider

(

ε s D−j ( s )dA−j ( s )

)ξ

t

0

j

(

)

= Es D−j ( s )dA−j ( s )ξ = 0

Since x = f ⊗ e(0) . Thus

(

−

∫ D− (τ)dA j (τ)

)

ε s D−j ( s )dA−j ( s ) = 0 when acting on vectors of the form f ⊗ e(0) , f∈h. Thus, et(M1(t + dt)) = M1(t) j − when acting on Btx since D− (t )dA j (t )Bt ξ

Commutes with Bt.

∵dA−j (t )

= D−j (t )Bt dA−j (t )ξ = 0

Likewise define t

M2(t) =

∫ Dk (τ)dA j (τ) j

k

0

et(M2(t + dt))x = M 2 (t )ξ + εt ( Dkj (t )dAkj (t))ξ = M2(t)x

More generally, j k et(M2(t + dt))Bx = M 2 (t )Bξ + εt ( Dk (t )dA j (t ) B)ξ j = M 2 (t )Bξ + Et ( Dk (t ) BdAkj (t )ξ)

= M2(t)Bx, B∈Bt Note that

{

µ v Bt = Clspan Dv ( s )dAµ ( s ) : s < t

}

  s = Clspan ∫ Dvµ (τ)dAµv (τ) : s ≤ t   0  Thus M21.1 is also a Martingale. Belavkin contd. (Nonlinear filtering) − d X = (F X ⊗ δ ) d A ⇒ d  X = Unitary iff

( X * X ) = ( F b F − X * X ⊗ δ ) d A  −1 b = F F

286

Stochastics, Control & Robotics

Measurement (output) process dY = ( Z bGZ − Y ⊗ δ ) d A µ µ Let  X = U, Fv = UZ v . Then,

(

) v v = U ( Z vµ − I δ µv ) d Aµ

dU = UZ vµ − U δ µv d Aµ U = Unitary iff (U ( Z ⊗ δ )) = (U ( Z ⊗ δ )) b

(U ( Z ⊗ δ ))b− −

−1

= Z −−U *

 Z −−  Z⊕d =  Z −0  +  Z−

Z 0−

Z +−   Z +0   Z ++ 

Z 00 Z 0+

 UZ −− UZ 0− UZ +−   0 0 0 U(Z⊕d) =  UZ − UZ 0 UZ +    +  UZ − UZ 0+ UZ ++   Z ++*U *  (U(Z⊗d))b =  Z 0+*U *  +* *  Z− U U *  =  0   0

Z +0*U * Z 00*U * Z −0*U *

Z +0*U * Z 00*U * 0

Z +−*U *   Z 0−*U *   Z −−*U * 

Z +−*U *   Z 0−*U *   U* 

U* = U–1 (U(Z⊗d))–1 = (Z⊗d)–1U* −1

 I Z 0− Z +−    =  0 Z 00 Z +0  U *    0 0 U*  −1 , i.e., b = F For U to be unitary, we require F  I Z +0*  0*  0 Z0  0 0

 I Z 0− Z +−*    Z 0−*  =  0 Z 00  0 0 I 

Z +−   Z +0  I 

−1

Large deviations, Classical & Quantum General Relativity with GPS Applications

or

 I Z 0−  0  0 Z0  0 0

Z +−   I Z +0*  Z +0   0 Z 00* I   0 0

287

Z +−*   I 0 0 0*  =  0 I 0 Z+     0 0 I I 

Z +0* + Z 0− Z 00* = 0,

or

Z +−* + Z 0− Z 0−* + Z +− = 0, Z 00 Z 00* = I, Z 00 Z 0−* + Z +0 = 0, These are equivalent to − 0* 0* 0 −1 Z 00* = Z 0 , Z + = − Z 0 Z 0

Z +0 = − Z 00 Z 0−* , 0 −1 0 0 −1 0 −* −* Z 00* = Z 00 −1 , Z 0 = − Z 0 Z + = Z 0 Z 0 Z 0 consistent,

Let Then

− −* Z +− + Z +−* = − Z 0 Z 0 Y = UYU* dY = dU.Y.U* + U.dY.U* + U.Y.dU*

+ dUdY.U* + UdYdU* + dUYdU*.

(

)

v dU.Y.U* = U Z vµ − I δ µv YU *d Aµ µ v U.dY.U* = U ( Z bGZ − Y ⊗ δ )v U *d Aµ

(

)

µ v = U ( Z bGZ )v − Y δ vµ U *d Aµ

(

)

µ

U.Y.dU* = U .Y . Z vµ*U * − U *δ µv d Av

( )((Z bGZ )ρσ − Y δρσ )U *d Aµv d Aρσ ρ σ = (UZ vµ − U δ vµ ) (( Z b GZ )σ − Y δ ρσ ) δ ρvU *d Aµ v σ = (UZ vµ − U δ µv ) (( Z bGZ )σ − Y δ vσ )U *d Aµ

dU.Y.U* = UZ vµ − U δ vµ

µ = U ( ZZ bGZ )σ U * − UZ σµYU * µ  σ − U ( Z bGZ )σ U * + UYU *δ µσ  d Aµ .

Remarks from "Quantum Stochastic Calculus and Quantum Nonlinear Filtering" by V.P. Belavkin (1992).

288

Stochastics, Control & Robotics

 0 C0−  C =  0 C0 0  0 0

C+−   b C+0  , C = 0 

 0 C+0* C+−*   −*  0*  0 C0 C0   0 0 0 

 0 0 1 g =  0 1 0    1 0 0 0 0  0 C+ =  C0−* C00* 0  C −* C 0* 0 + + 0 0  0 0 1  0 0 1  0   − * 0* gC+g =  0 1 0  C0 C0 0  0 1 0      1 0 0  C −* C 0* 0  1 0 0 0 0  C+−* C+0* 0 0 0 1    −*  0*   =  C0 C0 0 0 1 0    0 0 0  1 0 0  0 C+0* C+−*    =  0 C00* C0−*  = Cb  0 0 0  A(C, t) = C+−t + C0− A− (t ) + C+0 A+ (t ) + C00 N (t )

dA–(t)dA+(t) = dt

So

[A–(t), A+ (t′)] = tLt′

It a(u), a+(u) are the annihilation and creation operator fields, then we know that [a(u), a+(v)] =

(1)

e(u) is the exponential vector, so a(u)e(v) = e(v). d e(v + tu ) t=0 dt ∂2 e( w1 + t1u ), e( w2 + t2v) = ∂t1∂t2

a+(u)e(v) = e( w1 ), a(u)a + (v)e( w2 )

t1 = t2 =0

=

∂2 exp ( w1 + t1u , w2 + t2v ∂t1∂t2

=

∂ u , w2 + t2v exp ( w1, w2 + t2v ∂t 2

) t1 =t2 =0 ) t2 = 0

Large deviations, Classical & Quantum General Relativity with GPS Applications

= e( w1 ), a + (v)a(u)e( w2 )

( u, v

+ u , w2 w1, v

289

) e(w1), e(w2 )

= u , w2 w1, v e( w1 ), e( w2 )

e( w1 ), [a (u), a(v)] e( w2 ) = u , v e( w1 ), e( w2 )

So

This proves (1). Let u,v∈L2(R+). Then

+ + A–(t) = a ( χ[0, t ] ) , A (t ) = a (χ[0, t] ) ,

+  A− (t ), A (t′) = χ[0, t ] , χ[0, t ′ ] = tLt′

Conservation process H∈L(L2(R+)) LH(t) (L(X)  linear operator in the vector space X). W(0, exp(ilHc[0, t])) = exp(ilLH(t)).

[c[0, t], H} = 0 t is assumed.

e( w1 ), Λ H (t )e(w2 ) = w1, H χ[0, t}w2 e( w1 ), e( w2 ) e( w1 ), [ Λ H (t ), a(u )] e( w2 )

= e( w1 ), Λ H (t )a(u )e(w2 ) − e( w1 ), a (u )Λ H (t )e( w2 ) = w1, H χ[0, t]w2 u , w2 e( w1 ), e(w2 )

d e( w1 + εu ), Λ H (t )e( w2 ) t=0 dε = w1, H t w2 u , w2 e( w1 ), e(w2 ) −

−

(

d w1 + εu, H t w2 e ( w1 + εu ) , e ( w2 ) dε

=  w1, H t w2 u , w2 − ( u , H t w2 + w1, H t w2 u , w2  e( w1 ), e(w2 ) = − u , H t w2 e( w1 ), e( w2 ) Hence e( w1 ), [ Λ H (t ), A− (t ′ )] e( w2 ) = e( w1 ),  Λ H (t ), a(χ[0, t ′] )  e( w2 ) = − χ[0, t ′] , H t w2 e( w1 ), e( w2 ) = − χ[0, t Λt ′] , Hw2 e( w1 ), e( w2 )  t Λt ′  =  − ∫ ( Hw2 )(τ)d τ e( w1 ), e(w2 )  0 

) ε=0

290

Stochastics, Control & Robotics

On the other hand, e( w1 ), a(χ[0, t Λt ′] )e(w2 ) = χ[0, t Λt ′] , w2 e( w1 ), e( w2 )  t Λt ′  =  ∫ w2 (τ)d τ e( w1 ), e( w2 )   0 So if H = I, the identity operator, then we get e( w1 ), [ Λ I (t ), A− (t ′ )] e( w2 ) = − e( w1 ), A− (t Λt ′)e(w2 ) i.e.

[A–(t), LI(t′)] = A–(tLt′)

We define

N(t) = LI(t).

e( w1 ),  a + (u), Λ H (t ) e( w2 ) = a(u )e( w1 ), Λ H (t )e(w2 ) − e( w1 ), Λ H (t )a + (u )e( w2 ) = w1, u w1, H t w2 e( w1 ), e(w2 )

−

d e( w1 ), Λ H (t )e(w2 + εu dε

ε=0

= w1, u w1, H t w2 e( w1 ), e(w2 ) −

(

d w1 , H t ( w2 + εu ) exp ( w1 , w2 + εu dε

=  w1, u w1, H t w2 − ( w1H t u + w1, H t w2 w1, u ) e( w1 ), e(w2 ) = − w1, H t u e( w1 ), e( w2 ) Thus, putting u = c[0,t′], e( w1 ),  A+ (t′), Λ H (t ) e( w2 )

( (

)

)

= − ∫ H t*Λt ′ w1 (τ)d τ e( w1 ), e(w2 ) 

t Λt ′



0

= − e( w1 ), A+ (t)e( w2 ) =



* ∫ ( H w1 ) (τ)d τ

e( w1 ), e(w2 )

A(t )e( w1 ), e( w2 )

= χ[0, t] , w1 e( w1 ), e( w2 )  t =  ∫ w1 (τ)d τ e( w1 ), e( w2 )  0  N (t ), A+ (t′) =  Λ I (t ), A+ (t′) = A+ (t Λt′ )

)) ε=0

Large deviations, Classical & Quantum General Relativity with GPS Applications

291

To summarize +  A− (t ), A (t′) = tLt′, [ A− (t ), N (t′)] = A− (t Λt′ ) , + +  N (t ), A (t′) = A (tLt′).

(

Here, the noise Hilbert space is Γ s L2 ( R + ) 2

2

)

Now L (R+) = L (R+ → C). More generally, we can consider L2(R+ → K) where K = C m. 0 − Regard c0 as a row vector row ( c0− (j): j = 1, 2, ., m), c+0 = column ( c+ (j): j = 1, 0 as an m × m complex matrix, 2, .., m) as a column vector, c0 = c00 (i, j )

c+−

))1≤i, j≤m

((

∈C is a scalar (complex). Then

 0 c0−  c =  0 c00  0 0

c+−   ( ) ( ) c+0  ∈C m + 2 × m + 2 0  m

m

j=1

j=1

− j − + 0 A(c, t)  Ic+ t + ∑ A− (t )c0 ( j ) + ∑ A j (t )c+ ( j )

Let

+

m

∑

k , j =1

c00 (k , t ) N kj (t)

A−j (t ) = a(ϕ j χ[0,t ] ) ,

Here

+ A+j (t ) = a (ϕ j χ[0, t ] )

where

{ ϕ j }1

m

Nkj(t) = Λ ϕk

Pj

(t )

is an ONB for Cm.

Belavkin  quantum filtering P. 187 (Remarks). Et  orthoprojector on Bt ξ , Bt = ∞t′ . Et commutes with ∞t′ .

Proof: Let B∈ ∞t′ , x∈H. fixed. Bt ξ = ∞t′ξ . et = ∞t′ξ is ∞t′ -invariant. Bh – EtBh ⊥ et h – Eth ⊥ et h∈H. * B + B* , B − B ∈ ∞ ′ . t 2i 2

Let

 is( B + B* )  Us = exp   , 2  is( B − B* )  . Vs = exp    2i

292

Stochastics, Control & Robotics

Then Us, Vs ∈ ∞t′ , Us, Vs are unitary. Hence

(∵ et is ∞t′ invariant)

h – Eth ⊥ et ⇒ Ush – UsEth ⊥ et,

Vsh – VsEth ⊥ xt s∈R. Taking Thus

d at s = 0 ⇒ Bh – BEth ⊥ xt ds EtBh – BEth ⊥ xt

But EtBh∈xt, BEth∈xt.

Hence EtBh – BEth∈xt ⇒ ⇒

[Et, B]h = 0 h∈H [Et, B] = 0.  0 Z +0*  (ZbDZ) =  0 Z 0* 0  0 0

Z +−*   0 D0−  Z +0*   0 D00 0   0 0

 0 Z +0* D00  =  0 Z 00* D00  0 0

Z +0* D+0   0 Z 0−  Z 00* D+0   0 Z 00 0   0 0

 0 Z +0* D00 Z 00  =  0 Z 0* D 0 Z 0 0 0 0  0 0

(Z b D0b FZ )+−

(

X 0  0

0 X 0

)+−

Z +−*   0 ( D+0* )0  Z +0*   0 ( D00* )0 0   0 0 − 0   0 Z0  0   0 Z 00  X   0 0

 0 Z +0* ( D00* )0  0* 0*  0 Z 0 ( D0 )0  0 0

Z +−   Z +0  0 

Z +−   Z +0  0 

Z +0* D00 Z +0   Z 00* D00 Z +0   0

= Z b D0b ( X ⊗ δ + C ′Z )

 0 Z +0*  0* Z b D0b ( X ⊗ δ ) Z =  0 Z 0  0 0

 0 Z 0− D+−    D+0  ×  0 Z 00  0 0 0 

( D+−* )0   ( D+0* )0  0 

Z +−   Z +0  0 

0 Z +0* ( D+0* )0    Z 00* ( D+0* )0  ×  0   0 0

 0 Z +0* ( D00 )*0 XZ 00  =  0 Z 00* ( D00* )0 XZ 00  0 0

XZ 0− XZ 00 0

Z +0* ( D00* )0 XZ +0   Z 00* ( D00* )0 XZ X +0   0

XZ +−   XZ +0  0 

Large deviations, Classical & Quantum General Relativity with GPS Applications

0  0  0 0  = 0  0 1  ZbDZ ≡  0  0

D0−

Z +−   D00 Z +0  0 1  D0− Z 00 D0− Z +0 + D+−   D00 Z 00 D00 Z +0 + D+0   0 1 Z +0* Z +−*   0 D0− D+−    Z 00* Z 0−*   0 D00 D+0  0 1   0 0 0 

D+−   1 Z 0−  D+0   0 Z 00 0   0 0

( Z b DZ )+−

(

 1 Z 0−  0  0 Z0  0 0

)

− − 0* 0 0* 0 0 = D0 + Z + D0 Z + + D+ + Z + D+

= D+− + D0− Z +0 + Z +0* D+0 + Z +0* D00 Z +0  0 C0−  C = F – X⊗d =  0 C 0 0  0 0 X  F = X⊗d + C =  0  0 Z b D0bCZ = (D ≡ D0)

Z +−   Z +0  1 

 1 Z 0− D+− + Z +0* D+0    0 Z 00* D+0  ×  0 Z 0  0 0  0

 0 D0− + Z +0* D00  Z 00* D00 = 0  0 0 \

293

C+−   C+0  0  C0−

X + C00 0

 1 Z +0*  0*  0 Z0 0 0 

Z +−*   0 D+0*  Z +0*   0 D00* 1  0 0 

 0 C0−  0  0 C0 0 0 

C+−   1 Z 0−  C+0   0 Z 00 0  0 0 

 0 D+0* + Z +0* D00*  Z 00* D00* = 0 0 0 

C+−   C+0  X  D+−*   D+0*  0   Z +−   Z +0  1 

D+−* + Z +0* D+0*   Z 00* D+0*   0 

Z +−   Z +0  1 

294

Stochastics, Control & Robotics

 0 C0− Z 00  ×  0 C00 Z 00 0 0 

C0− Z +0 + C+−   C00 Z +0 + C+0   0 

( ) ( = ( D Z D ) (C Z + C )

0 0 0* 0* 0* 0 ( Z b D0bCZ ) +− = D+ + Z + D0 × C0 Z + + C+

Hence

0 0 0 +

* eqn. (2.2)

0 * +

(

0 0 0 +

)

0 +

)

− X ⊗ δ d A d X = F F  vλ (t ) −

v   vλ (t )d Avλ (t ) X (t )δ vλ  d Aλ (t ) = C =   − − j      0+ ( j )d A+j (t ) d A(c, t) = C + dt + C 0 ( j ) A− (t ) + C

 00 ( j , m)d N  mj (t ) C  +− (t ) is the coeff. of d A+− (t ) = dt in d A(c, t) . A+− (t ) = t, C  0+ ( j ) ,  +j (t ) = C  0− ( j ) , C  −j (t ) = C C Amj (t ) = N  mj (t ) = C  mj (t ) , C  00 ( j , m) . j + + d A− (t )d Ak (t ) = δ kj dt = δ kj d A− (t ) ,

(δ

+ −

=0

)

+ k k d A j (t )d A− (t ) = 0 = δ +− d A j (t )

j k d A− (t )dN mk (t ) = δ mj d A− (t ) +  kj (t ) d A j (t )dN mk (t ) = 0 = δ +m d N

(δ

+ m

=0

k = δ +m d A j (t )

)

dN mk (t )dA+j (t ) = δ kj dAm+ (t ) These quantum Ito formula of Hudson and Parthasarthy can be summarized as µ ρ ρ d Av (t )d Aσ (t ) = δ µσ d Av (t )

where m, v ρ, σ = {+, -, 1, 2, 3, ...}  vλ d Avλ = dA(C, t) d X = C d X

*

 vλ*d Avλ* = C

Large deviations, Classical & Quantum General Relativity with GPS Applications +*

d A j

j

295

+

j*

= d A− , d A− = d A j , j

k*

j

k*

 k (i.e. d A j = d Ak )  j = dN dN so,

d X

*

 kj *d Akj + C  +−*dt + C  −*  j *  j *  +* = C j d A− + C + d A j  kj d Akj + C  −j *dt + C  −j *d A+j + C  +j *d A−j = C

Now

So

 0 C+0* C+−*    Cb =  0 C00* C0−*  0 0 0    k * *  (k , j ) d A j + C  +−*dt + C  0− ( j )* d A−j * + C  0+ ( j )* d A+j * d X = C  b (k , j )dA j + C  +−*dt + C  0− ( j )* d A+j + C  0+ ( j )* d A−j = C k  b ( j , k )dA j + C b − dt + C b 0 ( j )d A+j + C − ( j )d A−j = C 0 + + k = d A(C b , t )

Thus if Now

* b  X (t ) = A(C, t), then  X (t ) = d A(C , t ) .

 , t) ,  vλ ( t )d Avλ (t ) = d A(C dX = C  bv λ (t )d Avλ (t ) = d A(C  b , t) dX * = C

So using the quantum Ito formula,  bλ  λ′  v  v′ dX*.dX = C v C v′ d Aλ d Aλ ′ v′ λ′ v  bv λ C  v′ δ λ′ d Aλ = C

 bλ  v  v′ = C v C v′ d Aλ λ

bC   d Avλ′ = C   v′  b C,  t = d A  C   j +   A t ( ), A ( t′) We have  − k  = ϕ j χ[0, t ] , ϕ k χ[0, t ′ ] = δ jk t Λt ′

(

)

a ϕ χ   j j [ 0, t ] , Λ ϕ k  A− (t ), N km (t ′) = 

(

ϕm

(t ′ )  

= a ϕ m ϕ k ϕ j χ[0, t Λt′]

(

= δ kj a ϕ m χ[0, t Λt′]

)

)

296

Stochastics, Control & Robotics

= δ kj A−m (t Λt′ ) It is better to denote Nkm(t) by N km (t ) : N km (t ) = Λ ϕk Then,

ϕm

(t )

m   j δ j Am (t Λt′ )  A− (t ), N k (t ′) = k −

Rough Calculation: e ( w1 ) ,  Λ H (t ) , a (u ) e ( w2 )

(

)

* = − u , H t w2 e ( w1 ) , e ( w2 ) e ( w1 ) , a H t u e ( w2 )

= H t*u , w2 e ( w1 ) , e ( w2 ) = u , H t w2 e ( w1 ) , e ( w2 ) so Also,

(

)

* a (u ) , Λ H (t ) = a H t u  N km (t ), A+j (t ′ ) =  Λ ϕ ϕ (t ), a* ϕ j χ[0, t′ ]     k m 

(

(

)

* = a ϕ k ϕ m ϕ j χ[0, t Λt′]

(

= δ mj a* ϕ k χ[0, t Λt′ ]

)

)

= δ mj Ak+ (t Λt′ ) Finally, e( w1 ),  Λ H1 (t ), Λ H 2 (t ′ )  e( w2 )

= e( w1 ), Λ H1 (t )Λ H 2 (t ′ )e( w2 ) − e( w1 ), Λ H 2 (t ′)Λ H1 (t )e( w2 )

e( w1 ), Λ H1 (t )Λ H 2 (t ′ )e( w2 ) = −

−

∂2 e( w1 ), e exp (iε1H1t ) .exp (iε 2 H 2t ′ ) w2 ∂ε1∂ε 2

(

∂2 exp w1 , w2 + i (ε1H1t + ε 2 H 2t ′ ) w2 ∂ε1∂ε 2

(

− ε1ε 2 H1t H 2t′ w2

) ε = ε =0 1

2

)

ε1 = ε 2 = 0

Large deviations, Classical & Quantum General Relativity with GPS Applications

Rough Calculation:

(

[a(u), LH(t)] = a H t*u

)

( ) a* ( H t*u )

* * * *  Λ H (t ) , a (u ) = a H t u

⇒

 Λ * (t ) , a* ( u )  =  H  H commuting with c[0, t] ⇒

[LH(t), a*(u)] = a*(Htu)

⇒ ∂ ∂ε 2

{(i w1, H1t w2 − exp ( w1 , w2 + iε 2 H 2t′ w2 )} ε =0 = −

)

w1 , ε 2 H1t H 2t′ w2

2

=

{ w1, H1t w2

w1, H 2t ′ w2 + w1, H1t H 2t ′ w2

e( w1 ), e( w2 ) Thus, e( w1 ),  Λ H (t ), Λ H (t ′ )  e( w2 )  1  2 = w1, [ H1, H 2 ]t Λt ′ w2 e( w1 ), e( w2 ) and hence

 Λ H (t ), Λ H (t ′ ) = Λ H , H (t Λt ′ ) [ 1 2]  1  2

m m′  Thus,  N k (t ), N k ′ (t ′) =  Λ ϕk

( = (δ

ϕm

(t ), Λ ϕ

k′

ϕ m′

(t ′ ) 

= δ km′ Λ ϕ k ϕ m′ − δ m′ k Λ ϕk ′ ϕm

Thus

297

[ A(c, t ), A(d , t′)]

=

m m′ k ′ Nk

∑ j, k

)

) (t Λ t ′ )

′ m − δm k N k ′ (t Λt ′ )

C0− ( j )d +0 (k )  A−j (t ),

Ak+ (t ′) 

+ ∑ C+0 ( j )d 0− (k )  A+j (t ), A−k (t ′)  j, k

+ ∑ C0− ( j )d 00 (k , r )  A−j (t ), N kr (t ′ )  jkr

+ ∑ C00 (k , r )d 0− ( j )  N kr (t ), A−j (t′ )  jkr

+

∑ C00 (k , r )d00 ( p, q)  N kr (t ), N qp (t ′)

krpq

+ ∑ C+0 ( j )d 00 (k , r )  A+j (t ), N kr (t′ ) jkr

}

298

Stochastics, Control & Robotics

+ ∑ C00 (k , r )d +0 ( j )  N kr (t ), A+j (t′ )  jkr

=

∑ C0− ( j )d+0 (k )δ kj t Λt′ j, r

− ∑ C+0 ( j )d 0− (k )δ kj t Λt′ j, k

+ ∑ C0− ( j )d 00 (k , r )δ kj A−r (t Λt′) jkr

− ∑ C00 (k , r )d 0− ( j )δ kj A−r (t Λt′) jkr

− ∑ C+0 ( j )d 00 (k , r )δ rj Ak+ (t Λt′) jkr

+ ∑ C00 (k , r )d +0 ( j )δ rj Ak+ (t Λt′) jkr

+

∑ C00 (k , r )d00 ( p, q)δ rp N kq (t Λt ′) − δ qk N rp (t Λt′)

krpq

(

)

( + ∑ (C ( k , j ) d

) (k , j )C ( j ) )A (t Λt′ )

= t Λt ′ ∑ C0− ( j )d +0 ( j ) − C+0 ( j )d 0− ( j ) j

+ ∑ C0− ( j )d 00 ( j , r ) − C00 ( j , r )d 0− ( j ) A−r (t Λt′ ) jr

0 0

0 0 + ( j ) − d0

jr

0 +

+ k

+ ∑ C00 (k , p)d 00 ( p, q) − d 00 (k , p)C00 ( p, q) kpq N kq (t Λt′ ) .

The above calculation may be referral to while reading Belavkin's original paper to make the reading easier. 

[16] Selected Topics in General Relativity: 1. Tetrad formulation of field equations. gmn (x) is the metric; ema (x), a = 0, 1, 2, 3 is a tetrad basis in the sense that for each a, (ema)3m = 0. is a contravarient vector field and hab = gmnema enb are scalar constraints.

We have

dt2 = gmndxmdxn = habeam ebn dxmdxn. = hab (eam dxm) (ebn dxn)

ema dxm is a one form. Here ((ema )) is the matrix of ((ema )) : ema ena = dmn, ema ebm = dba.

Large deviations, Classical & Quantum General Relativity with GPS Applications

ea = eaµ

299

∂

, a = 0, 1, 2, 3 form a local basic for the tangent space at x. ea is a ∂x µ vector field. Likewise, eam dxm, a = 0, 1, 2, 3 form a local basis for the dual to the b ν tangent space at x. Here, eam = ηab eµ = gµνea .

Example: The perturbed Kerr metric has the form dt2 = e2 ν dt 2 − e2µ dr 2 − e2µ 2 d θ2 − e2ψ (d ϕ − wdt − q1dr − q2 d θ − q3d ϕ )

2

Where n, m1, m2, ψ, w, q1, q2, q3 are functions of t, r, θ, j. We define

e0 = ev dt , e1 = eµ1 dr , e2 = eµ 2 dθ , e3 = eψ ( d ϕ − wdt − q1dr − q2 d θ − q3d ϕ ) h00 = 1, h11 = – 1, h22 = – 1, h33 = – 1.

Then

ηab ea eb = dt2.

Note eam dxam = 0, 1, 2, 3 are also one for and the from a basis for the co-tangent space at x. Co-varient derivative:

(

A:vµ = Aa eaµ

)

:ν

(Aa = ema Am are scalar fields)

= A,an eam + Aaema:n. µ b ν b a µ b ν a µ b ν = A:ν eµ ec ∆ A / c = A,ν ea eµ ec + A ea:ν eµ ec b ν a b b b a = A,ν ec + A τ ac = A,c + τ ac A

where the ec direction),

b (directional derivatives along A,cb = ea (Ab) = enc ∂n Ab = ecnA,n b µ ν tbac = eµ ea;ν ec (Spin coefficients. They are scalar).

Maxwell equation in tetrad notation. F;νµν = KJm (Four vector notation) ab µ ν F µν = F ea eb

(

Fab = scalars:

F;νµν = F;νab eaµ ebν Fab = F µν eµa eνb + F ab eaµ;ν ebν + eνµ ebν;ν

(

ab µ ab µ ν a ν = F,b ea + F ea:ν eb + eµ eb:µ

(

ν eµa F:νµν = F,bab + F cb eµa ecµ:ν ebν + δ ea eb:ν ab cb a ab ν = F ,b + F τ cb + F eb:ν

) )

)

300

Stochastics, Control & Robotics a µ ν a enb:m = eµ eb:ν ea = tba .

Now,

So the Maxwell equation, can be expressed as c a F,bab + F cb τ cb + F ab τbc = kJa



[17] Perturbations in the Tetrad Caused by Metric Perturbation: a b gmn = ηab eµ eν

(

b a a b dgmn = ηab eµ δeν + eν δeµ

)

b a = ebµ δeν + eaνδeµ

So

ebµ δgµν = δeνb + ebµ eaνδeµa We note that

(

)

ebµ δeµa = δ ebµ eµa − eµa δebµ a bµ = −eµ + δe .Thus,

So,

ebµ eaνδeµa = − eaνeµa δebµ = − gµνδebµ δeνb − gµνδebµ = ebµ δgµν

Now, so and hence or

b µρ ebm = eρ g µρ b b µρ debm = g δeρ + eρ δg ρ b µρ b bµ debn = g ν δeρ − gµνδg eρ = e δgµν

ebµ δgµν + δgµνeρb δg µρ = 0 µα ρβ dgmρ = − g g δgαβ . So the above reduces to

ebµ δ gµν − gµν g µα g ρβ eρb δ gαβ = 0

or

ebµ δgµν − δg νβ ebβ = 0 Which is a trivial identity. This shows consistency. Let

((δ g )), ∆E = ((δe )) = (( e ) )

∆G = E

a µ

µν

aµ

a,µ

a, µ

Large deviations, Classical & Quantum General Relativity with GPS Applications

The equation for ∆E be solved is E T , ∆E + ( ∆E ) . E = ∆G T



[18] Geodesic Equation in Tetra Formalism: νa (τ) = νa (τ)eµa ( x (τ)) a µ nm = ν ea

=

d νµ d νµ µ ea + νa eaµ.ν νν = dτ dτ

d νa µ ea + νa νb eaµ.ν ebν dτ

µα 1 µα Γ µvσ = g Γ ανσ = g ( gαν,σ + gασ, ν − g νσ,α )

Express

Γ µνσ

is terms of

(ema)

2 and simplify.

Alternate Method: The geodesic equation can be expressed as v v µ:vµ = 0

(

νν νa eaµ

Thus or or or or

)

(

:ν

νν eaµ ν,aν + νa eaµ:ν

)

=0 =0

d νa + νa νb ebν eaµ:ν = 0 dτ d νa + νc νb ebν eµa ecµ:ν = 0 dτ

eaµ

d νa a + νb νc τ cb =0 dτ

Riemann tensor in tetrad basis: β eµa:ν:ρ − eµa:ρ:ν = Rµνρ eβa a β α µ ν ρ Rbcd = Rµνρ eβ eb ec ed

Now,

(

µ ν ρ a a = eb ec ed eµ:ν:ρ − eµ:ρ:ν

(

a b O = gµν:σ ηab eµ eν

(

)

:σ

a b a b = ηab eµ:σ eν + eµ eν:σ a b = eaν eµ:σ + ebµ eν:σ

)

)

301

302

Stochastics, Control & Robotics

So νa aν b O = eµ:σ + e ebµ eν:σ

Now

(

µ ν a ebµ ecν eµa:ν:ρ = eb ec eµ:ν

)

)

(

:ρ

(

ν − eµa:ν eb:µρ ecν + ebµ ec:ρ

(

µ v µ v µ v a a = eb ec eµ:v ,ρ −eµ:v eb:ρec + eb ec:ρ

=

a τbc ,ρ

− eµa:ν

(

eb:µρ ecν

ν + ebµ ec:ρ

(

)

µ ν µ ν af a = η τbfc,ρ − eµ:ν eb:ρ ec + eb ec:ρ

Now,

)

)

)

a ebµ eµa:ν ecν = τbc = ηaf τbfc

eµa:ν ecν = haf eb t m bfc

So

af g µ µ = η eµ eb:ρ τ gfc eµa:ν ecν eb:ρ af g h h g a = η τbh eρ τ gfc = eρ τbh τ fc

Likewise the other terms are simplified. a in terms of the spin coefficients tabc and their first order Exercise: Express Racd partial derivatires.



[19] Quantum General Relativistic Scattering: The 3-metric of spaces times has the form trs = −

τ rs gor gos + 2 goo goo

This 3-metric is obtained by considering the time taken by a photon to propagate from xr to xr + dxr back and forth. Let t = det (trs). The Laplace Beltamic operator in 3D curved space is them

) (( )) = ((τ ))

(

∆B = τ −1/ 2 ∂ r τ1/ 2 τ rs ∂ s , τ rs Summation over r, s = 1, 2, 3.

Approximation: Let trs = δ rs + ε X rs Then

trs = δ rs + ε X rs + O(ε 2 ) .

(((χrs ))) .

Let

c = χ rr = t r

Then,

t = 1+ ε χ + O ε 2 .

( )

1 2 t1/2 = 1+ εξ + O(ε ) . 2

rs

−1

Large deviations, Classical & Quantum General Relativity with GPS Applications

303

1 2 t–1/2 = 1 − ε χ + O(ε ) 2   1   1  ∆ B ψ = 1 − ε χ ∂ r  1 + ε χ (δ rs − ε χ rs ) ψ , s  + O(ε 2 )  2   2    1  ψ ε = 1 − ε χ  ∆ f + (χ ψ , r ),r − ε χ rs ψ ,s  2  2

(

1 = ∆ f ψ + ε  χψ ,r 2

(

Where ∆f =

),r  + O(ε2 )

),r − (χrs ψ ,s ),r − 12 χ∆ f ψ  + O(ε2 )

3

∑ ∂2r is the Laplacian of flat space. Schrödinger Hamiltonian of 1

an electron board to the makes in the presence of the gravitational field. The Hamiltonian after interaction with gravitational field assumed to be concentration around the nuclear is − 2 ∆f + ε V1 H1 = 2m Where

i.e.

V1 =

Z e2 − 2  1  S  χ,r ∂ s − χ rs ∂ r ∂ s − χ rs ∂ r ∂ s  − 12 2m  2   3  2  ∑ xr   r =1 

X+X* . We find 2 −2mV1 1 1 = χ,r ∂ r + ∂ r χ, s − χ rs , r ∂ s − ∂ s χ rs ,r 4 2 2 S {X} =

) (

(

−

)

1 ze2 χ rs, ∂ r ∂ s + ∂ r ∂ s χ rs − 12 2  3 r2 ∑ x   1 

(

)

1 1 − χ,rr + χ rs , rs 4 2 − χ rs ∂ r ∂ s −

1 ze2 ∂ r χ rs ) ∂ s − ( 12 2  3 r2 ∑ x   1 

Scattering theory can be developed taking H0 = −

2 2 ∆f + ε V1. ∆f and H1 = − 2m 2m

304

Stochastics, Control & Robotics

Time dependent quantum scattering theory: H1 (t) = H 0 + ε V1 (t ), t ∈R . scattered ψ i  input state. Let

{ {

t

U (t, s) = T exp −i ∫ H1 (τ) d τ s

∞

n = I + ∑ ( −i ) n =1

ψ 0  output state

ψ+ 

}}

∫

H1 (τ1 ) H1 (τ 2 )...H1 (τ n ) d τ1...dτ n

s < τ n < ..< τ1 < t ∞

n n = U 0 (t − s ) + ∑ ( −i ) ∈ n =11

∫

U 0 (t − τ1 )V1 (τ1 )U 0 (τ1 − τ 2 )V1 (τ 2 )

s < τ n < .. τ1 < t

U 0 (τ n −1 − τ n )V1 (τ n ) U 0 (τ n − s ) d τ1...d τ n where U0 (t) = exp(−itH 0 ) . Maxwell equation in cosmology. The Robertson-Walker metric Evolution of the universe assuming homogeneity and isotropivity. 2 dt2 = dt −

R 2 (t ) dr 2 1 − kr 2

(

− R 2 (t ) r 2 d θ2 + sin 2 θdϕ 2

)

X0 = t, X1 = r, X2 = 0, X3 = j. g00 = 1, g11 = −

R 2 (t )

2

, g 22 = − R 2 (t )r 2 , g33 = − R 2 (t ) r sin 2 θ 2

1 − kr 1 Rµν − Rgµν = k (Tmr) 2 Tmn = ( p + ρ) vµ vν − pg µν

(1)

The frame (t, r, θ, j) is comoving i.e. Geodesic are r = constant, θ = constant, j = constant. It can be verified that these satisfy the geodesic equation. d 2 xµ dτ 2

+ Γ µνσ

dx ν dx σ =0 dτ d τ

r = 0, r = 1, 2, 3). Thus (vm) = (1, 0, 0, 0) = (vm). (Since G00

(1) ⇒

– R = kT = k (p + ρ – 4p) = k (ρ – 3p) R = K (3p – ρ).

(1). can be expressed as Rmn =

1 Rgµν + kTµν 2

Large deviations, Classical & Quantum General Relativity with GPS Applications

305

K (3 p − ρ) g µν + K (( p + ρ)vµ vν − pgµν ) 2  p ρ = K ( p + ρ)vνvν + kg µν  −   2 2

=

i.e.

   p − ρ Rmn = K ( p + ρ)vµ vν +   gµν   2   p − ρ  R00 = K  p + ρ +  2   K {3 p + ρ} = 2 K ( p − ρ) g kk, 1 ≤ k ≤ 3. Rkk = 2

i.e.

R00 =

K (3 p + ρ) 2

(2a) (2b)

p : p (t), ρ = ρ (t) (\ of homogeniety)

K ( p − ρ) R 2 (t ) , 2 1 − kr 2 K 2 2 = − ( p − ρ) R (t ) r , 2 K 2 2 2 = − ( p − ρ) R (t ) r sin θ 2

R11 = − R22 R23

α α α β α β Rmn = Γ µα , ν − Γ µν,α − Γ µνΓ αβ + Γ µβ Γ να α α β α β R00 = Γ α0α, 0 − Γ 00 , α − Γ 00 Γ αβ + Γ 0β Γ 0α 0 2 a + Γ101 + Γ 02 + Γ 303 G0a = Γ 00

1 00 1 1 R ′ (t ) 1 g g00, 0 = 0, G01 = g ′′( g11,0 ) = (log g11 )10 = , 2 2 2 R(t ) R ′ (t ) 1 R ′(t ) 3 1 2 , Γ 03 = (log g33 ),0 = G02 = g (log g 22 ),0 = . R(t ) 2 R(t ) 2 R ′(t ) a = . G0a R(t ) 0 G00 =

\

α Ga00 = 0, Γ α00 Γ βαβ = 0, Γ 0β Γ β0α

( ) ( ) ( )

= Γ101 So,

2

2 + Γ 02

R00 =

2

+ Γ 303

3R′ R′ 2 +3 2 R R

2

2

 R′  = 3  .  R

306

Stochastics, Control & Robotics α α β α β R11 = Γ1αα ,1 − Γ11 , α − Γ11 Γ αβ + Γ1β Γ1α 2 3 a + Γ13 = Γ111 + Γ12 G1a

G′11 =

−Kr 1 1 11 g g11,1 = (log g11 ),1 = 2 2 ( −Kr 2 )

1 1 22 1 g g 22,1 = (log g 22 ),1 = , 2 r 2 1 1 = (log g33 ),1 = 2 r −kr 2   =  +  1 − kr 2 r  ,1

2 = G12 3 G13

So,

Γ11α,1

=

−k 1 − kr

2k 2 r 2

−

2

2 2

(1 − kr )

−

(k + k r ) − 2 (1 − kr ) r

2 r2

2 2

=

2

2

0 1 α = Γ11, Γ11,α 0 + Γ11,1

1 00 g g11, 0 2 −kr 1 R(t ) R ′(t ) 1 = − g11, 0 , Γ111,1 , Γ11 = 2 2 1 − kr 2 1 − kr

0 = G11

(

=

α β Γ11 Γ αβ

−k

)

2k 2 r 2

−

(1 − kr ) = Γ Γ = Γ (Γ + Γ + Γ Γ + Γ (Γ + Γ 1 − kr 2 0 11

2 2

β 0β

1 11

0 11

β αβ

1 11

1 01

1 11

2 02

) +Γ )

+ Γ 303

2 12

3 13

RR ′ 1 1 G01 = − g 00 g11,0 = − g11,0 = 2 2 1 − kr′ 2 2 Γ101 + Γ 02 + Γ 303 = 3R ′ / R . −kr −kr 2 2 3 1 = = , Γ111 + Γ12 + . + Γ13 Γ11 2 2 r 1 − kr 1 − kr

So

α β Γ11 Γ αβ =

RR ′

(1 − kr ) 2

2 kr  − kr  3R ′   R  − 1 − kr 2 1 − kr 2 + r 

Large deviations, Classical & Quantum General Relativity with GPS Applications

3R′ 2

= α β Γ1β Γ1α

2k

−

(1 − kr ) (1 − kr ) (1 − kr ) = (Γ ) + (Γ ) + (Γ ) + 2Γ Γ 2

1 2 11

= So

k 2r 2

+

2 2

2 2 12

k 2r 2

(1 − kr )

2 2

2

3 2 13

1 10

0 11

 2 R′   RR′  +  k   1 − kr 2  r 2

+

2

(k + k r ) − 2 − (RR′) + k + 2k r = − (1 − kr ) r (1 − kr ) 1 − kr (1 − kr ) 2 2

R11

307

2 2

2 2

= −

= +

3R / 2

2

k 2r 2

−

2

2

2 2

2k

+

(1 − kr ) (1 − kr ) (1 − kr ) 2

k 2r 2

2 2

+

2

+

2

2R / 2

(1 − kr ) r (1 − kr ) (2R′ + RR′′) + 2k (1 − kr ) = − (1 − kr ) (1 − kr ) = ( − ( 2 R ′ + RR ′′ ) + 2k / (1 / 2r )) 2 2

2

2

2

2

2 2

2

2

The R00 and R11 equation are therefore

 R′ R′ 2  k 3  + 2  = (3 p + ρ) , 2 R R 

(

2k − RR ′′ + 2 R ′ 2

(a)

) = − k2 (ρ − ρ) R

2

(b)

If we set p = 0 then there equation become  R′ R′ 2  kρ −K 2 3 + 2  = ρR , − 2k + RR ′′ + 2 R 2 = R 2 2 R  

(

So we get in this case, −

2k R2

R′′ 2 R′ 2 + 2 R R 

+

)

308

Stochastics, Control & Robotics

[20] Derive Dirac Equation in Curved Space-Time Using the Newman Penrose Formalism: Special-Relativity: σ µAB X ,µA + im ηB = 0

(1a)

A

σ µAB A1 η,µ + im χ B = 0

(1b)

B AB B A cA = ε AB χ , hA = ε AB η , χ = ε χ B ,

 0 1 AB hA = ε ηB , ((ε AB η)) =  =e  −1 0

(( ε ) ) AB

 0 −1 0 

((ε AB ))−1 =  1

= (1) Can be expressed as (2)

3

σ 0T ∂0 χ + ∑ σ rT ∂ r χ + im ε η = 0 r =1

(3)

 0 1 2  0 − i 3  1 0  ,σ =  ,σ =  σ0 = I2, σ1 =   1 0  i 0   0 −1 3

σ 0T ∂0 η + ∑ σ rT ∂ r η + im ε χ = 0 r =1

Thus

3

∂0 χ + ∑ σ rT ∂ r χ + im ε η = 0 r =1

(2a)

3

∂0 η + ∑ σ r ∂ r η − im ε χ = 0 r =1

( )

χ = χA

Where Note:

σ

rT

2

( )

, η = ηA A=1

2 A=1

= σr* = σr

(2) Gives on multiplying by i and using p0 = i∂0, pr = – i∂r, 3

p 0 χ − ∑ σ rT p r χ − m ε η = 0 r =1 3

p 0 η − ∑ σ r p r η − m ε χ = 0, r =1

Or

(

  σT , p  p0 −   0  

)

0   0 ε    χ  =0  −m  − ε 0   η (σ, p) 

(2b)

Large deviations, Classical & Quantum General Relativity with GPS Applications

The Dirac Hamiltonian operator

(

 σT , p H0 =   0 T is clearly Hermitian a e = – e.  o rT   0

)

309

0   0 ε  +m  − ε 0 (σ, p)

0   0 ε  0 ε  σ rT   +  σ r   − ε 0  − ε 0  0

 0 σ rT ε  0 =   + 0   − ε σr T  − σr ε  0 =   − σ r ε + ε σ rT

(

)

0  σr 

ε σr   0 

σ rT ε + ε σ r    0

 0 1  0 1  0 1  0 1 + σ1T ε + ε σ1 =   1 0  −1 0  −1 0  1 0  −1 0  1 0  + =  = 0.  0 1  0 −1  0 i   0 1  0 1  0 −i σ 2T ε + ε σ 2 =  +  −i 0  −1 0  −1 0  i 0   −i 0   i 0 + =  =0  0 −i  0 i   1 0   0 1  0 1  1 0  σ3T ε + ε σ3 =  +  0 −1  −1 0  −1 0  0 −1  0 1  0 −1 + =  =0  1 0  −1 0  σ rT ε + ε σ r = 0 Thus, and likewise premultiplying this equation by e and postmultiplying by e, we get ε σ rT + σ r ε = 0 Thus rT 0   0 ε   σ ,    = 0, r = 1, 2, 3 r   0 σ   − ε 0  (1) Thus implies on pre multiplying by

(

 σT , p p0 +   0

)

0   0 ε  + m  − ε 0 (σ, p)

310

Stochastics, Control & Robotics

and using

The Klein-Gordon equation

In gtr we set

 σ rT   0

0   σ sT , σr   0

0     = 2drsI σ s  

3   χ   p 02 − ∑ p r 2 − m 2    = 0  η r=1   µ σ µ( ab ) = e( ab ) = σ µAB ζ aA ζb

B

with zaA a dyad. We set

B AB zaA = ε AB ζ a and so ζ aA = ε ζ aB

ζ aA = ε ab ζbA ,ζ aA = ε ab ζbA ,

Also

 0 1 ((ε ab )) =   −1 0

When We write

e(µ00) = l µ , eµ = mm, ( 01) µ

µ µ e(µ10) = m , e(11) = n

Just as a tetrad basis is str transforms 4 vectors into 4-scalars, the dyad basis in str transforms 2-spinors into 2-scalars.  lµ =  µ  m m σAB transform bispinors into 4-vectors: σ µAB p AB = P µ

(( ) ) (( ) ) e(µab )

Thus,

=

σ µab

mµ   nµ 

m and if ((σAB m )) denotes the inverse of ((σAB)) in the trade that n = dmn, σmABσAB

µ σ µAB σCD = δ CA δ BD . ν σ µAB = ε AC ε BD yµν σCD

Where We verify:

σ µAB σ νAB = ε AC ε BD gµρσρCD σ νAB

ν ε AC ε BD σρCD σ νAB = ε ac εbd σρcd σ ab 01 01 ρ ν 01 10 ρ ν = ε ε σ11σ 00 + ε ε σ10σ 01 10 01 ρ ν 10 10 ρ ν = ε ε σ11σ10 + ε ε σ 00σ11

Large deviations, Classical & Quantum General Relativity with GPS Applications ρ

311

ν

ρ ν ν ρ ρ ν = n l −m m −m m +l n

= gnρ = gρn Now comes Friedman’s lemma. F

(

ε kf ζ aE ζ f ζbE ζ kF

):µ

(

F

= ε kf ζ aE ζ f ζbE:µ ζ kF + ζbE ζ kF:µ

)

F

kf E kf = ε ε afk ζ a ζbE:µ + ε ε ab ζ f ζ kF:µ

= 2ζ aE ζbE:µ + ε ab ζ Since

ζ kF ζ kF:µ =

(

1 kF ζ ζ kF 2

)

:µ

kF

ζ kF:µ = 2ζ aE ζbE:µ

=0

Since

ζ kF ζ kF = dkk = 2 a scalar constant field.

Note:

 0 1 ζ aE ζbE = ε ab ≡   −1 0

Now consider σ µAB χ:µA . We have

(

χ:µA = ζ aA χ a Since \ Now

)

A a A a = ζ a:µ χ + ζ a χ,µ

:µ

ca = ζ aA χ A is a scalar for a = 0. σ µAB χ:Aµ = σ µAB ζ aA χ,aµ + σ µAB ζ aA:µ χc

(

AB A = ε ζ aB ζ a:µ

)

:µ

AB AB = ε ζ aB:µ since ε:µ = 0

So,

µ µ a A bc a AC A σ µAB ζ a:µ χ a = σ AB χ ε ζ ac:µ = σ Ab χ ζb ζ ζ ac:µ

a bA AB A aB ab AB a b a AB b a bA (Note that eab = ζ A ζ , ε = ζ a ζ ε = ε ζ A ζ B = ζ A ε ζ B = ζ A ζ )

Using Friedman’s lemma, we thus get A σ µAB ζ a:µ χ a = σ µAB χ a ζbA ζbc ζ ac:µ µ a bA c = σ AB χ ζ ζb ζ ac:µ

(

1 µ a bA  kf E F σ χ ζ ε ζb ζ f ζ aE ζ kF  2 AB 1 µ a AE kF = σ AB χ ε ζ ζ aE ζ kF :µ 2 =

(

)

):µ 

312

Stochastics, Control & Robotics

( (

)

1 µ a kF AE σ χ ζ ε ζ aE ζ kF :µ 2 AB 1 µ a kF A = σ AB χ ζ ε a ζ kF :µ 2 kF 1 µ ap bq = e( pq ) ζ A ζ B χ a ζ ζ aA ζ kF 2 =

No:

ζ Ap ζ

kF

(ζ

A a ζ kF

)

:µ

pA = ζ ζ

kF

) (

)

:µ

(ζaA ζkF ):µ

(ζ aAζ kF ):µ = (σνAF eν (ak )):µ

Now,

(σ

ν AF

F

n ζ aA ζ k = e(ak) F

ζ aA ζ k = σ νAF e(νak )

So

ν ζ aA ζ kF = σ νAF e(νak ) = σ AF eν( ak )

or

)

ν = σ AF eν( ak ):µ A χa = σ µAB ζ a:µ

Thus

= Thus and thus,

q kF 1 µ e( pq ) ζ B χ a ζ pA ζ σ νAF eν( ak ):µ 2

q 1 µ e( pq ) eν( ak ):µ σ ν( pk ) ζ B χ a 2

(Note: sn(pk) = en(pk))

q 1 σ µAB χ:µA = σ µAB ζ:Aa χ,aµ + e(µpq ) eν( ak ):µ e ν( pk ) ζ B χ a 2 B µ 1 a ζbr σ AB χ:µA = e(µar ) χ,µ + e(µpq )eν( ak ):µ e ν( pk ) = ζ qB ζ rB χ a 2 1 µ µ a ν( pk ) a χ = e( ar ) χ,µ + e( pr ) eν( ak ):µ e 2 (zqB zBr = dqr)

The general relativistic Dirac equation are σ µAB χ:µA + im ηB = 0, A

σ µAB η:µ + im χ B = 0 B and hence we get from these on premultiplying by ζr ,

and

1 e(µar ) χ,aµ + e(µpr )eν( ak ):µ e ν( pk ) χ a + im ηr = 0 2 1 a e(µar ) η,µ + e(µpr )eν( ak ):µ e ν( pk ) ηa + im χ r = 0 2

(a) (b)

Large deviations, Classical & Quantum General Relativity with GPS Applications

313

(a) and (b) constitute the Dirac equation in gtr. Note: ν ν( pk ) pa kb ν em(pk) = gµνe( ak ) , e = ε ε e( ab )

where

0 ((ε )) =  1 ab

−1 , 0 

 lµ =  µ  m

(( ) ) e(µab )

mµ  , nµ  

[21] Proofs of Identities from S. Chandra Sekhar, “ The Mathematical Theory of Blackholes”. Cartain’s Second Equation of Structure: R (X, Y) Z = ∇X Z – ∇Y ∇X Z – ∇[X, Y] Z R (X, Y) = [∇X ∇Y] – ∇[X, Y].

i.e.

∇eα eβ = Γ σαβ eσ . σ β σ β β ∇ X eα = ∇ σ eα = X ∇eσ eα = X Γ σα eβ = ω α ( X ) eβ , ω α ( X ) X e σ

β σ = Γ αα X .

ωβα = Γ βσα eσ d ωβα = d Γ βσα Λ eσ + Γ βσα deσ R ( X , Y )eα = ∇ X ∇Y eα − ∇Y ∇ X eα − ∇[ X , Y ] eα

( = ( X (ω

)

(

)

= ∇ X ωβα (Y ) eβ − ∇Y ωβα ( X ) eβ − ωβα ([ X , Y ])eβ β α (Y )

)) − Y ( ω

β α(X )

)−ω

β α ([ X , Y ])eβ

+ ωβα (Y ) ∇ X eβ − ωβα ( X ) ∇Y eβ

(

)

= d ωβα ([ X , Y ]) eβ + ωβα (Y ) ωβσ ( X ) − ωβσ ( X ) ωβσ (Y ) eσ

(

)

σ σ β = d ω α ([ X , Y ]) + ωβ Λ ω α ([ X , Y ]) eσ

Thus,

(

)

eσ ( R ( X , Y )eα ) = d ω σα + ωβσ Λ ωβα ([ X , Y ]) Now,

(

∇ X ∇Y eα = ∇ X Y σ Γ βσα eβ

)

314

Stochastics, Control & Robotics

(

)

)

( e (Γ )) e

β = X ρ ∇eρ Y σ Γ βσα eβ eβ X ρeρ Y σ Γ σd eβ + X ρY σ Γ βρα ∇eρ eβ

(

= X ρ eρ (Y σ ) Γ βσα + Y σ \

(∇ X ∇Y − ∇Y ∇ X ) eα

β ρα

ρ

µ + X ρT σ Γ βρα Γ ρβ eµ

β

β ρ σ ρ σ = Γ σα ( X eρ (Y ) − Y eρ ( X ))eβ

(

( ( ) ( ))

)

µ + X ρY σ Γ βρα Γ ρβ − Γ βσα Γ µσβ eµ + X ρY σ eρ Γ βσα − eα Γ βρα eβ

(

( ) ( )) + e ( Γ ) − e ( Γ )) e

µ µ = Γ βσα eσ ([ X , Y ]) eβ + X ρY σ Γ βσα Γ ρβ − Γ βσα Γ σβ + eρ Γ µσα − eσ Γ ρµα eµ

(

µ = ∇[ X , Y ] eα + X ρY σ Γ βσα Γ ρβ − Γ βσα Γ µσβ

Thus,

(

µ σα

ρ

)

σ

µ ρα

µ

R (X, Y) ea = ∇ X ∇Y − ∇Y ∇ X − ∇[ X , Y ] eα

(

( ) ( ))

µ µ = X ρY σ Γ βσα Γ σβ − Γ βρα Γ σβ + eρ Γ µσα − eσ Γ ρµα eµ ρ σ µ ≡ X Y Rρσα eµ

1 σ ρ µ Rρµα e Λ e ( X , Y ) 2 Thus we get Cartan’s second equation of structure:

So

σ eσ ( R( X , Y ))eα = X ρY µ Rρµα =

1 σ ρ µ Rρµα e δ e = dω σα + ωβσ Λ ωβα . 2 Verification of identities from S. Chandrasekhar’s book Mathematical theory of Blackholes. d ( F ω′ ) = F dω ′ + F, AdX AΛ d ω ′ + F ,1 dX ′ Λ d ω ′

(

ω′ = eψ dX ′ − ∑ q A dX A

(

)

)

(

ψ ψ A B A dω′ = e ψ ,1 dX ′ Λ dX ′ − q AdX + e ψ , B dX Λ dX ′ − q A dX

)

− eψ q A, B dX B Λ dX A − eψ q A,1 dX ′ Λ dX A = − eψ (ψ ,1 q A + q A,1 ) dX ′Λ dX A + eψ ψ , AdX A L dX ′

(

− eψ ψ, B q A dX B Λ dX A + q A, B

)

ψ ψ A B A = e (ψ ,1 q A + q A,1 ) dX Λ dX ′ + ψ , A + e (ψ , B q A + q A, B ) dX Λ dX

ψ ,1 q A + ψ , A = ψ1;A qB:A = qB, A + q A qB,1

Large deviations, Classical & Quantum General Relativity with GPS Applications

315

dXA = e −µA w A ,dX 1 = e −ψ w1 + q AdX A −ψ 1 − µA A = e w + q Ae w .

So,

(

(

)

−µ ψ − µA A −ψ 1 B dw′ = e ψ: A + q A,1 e w Λ e w + qB e B w

)

( ) e −µA ( ψ: A + q A,1 ) w AΛw1

− µ +µ + eψ ψ , B + q A + q A, B e ( A B ) w AΛwB

=

( (

+ eψ − µA−µB qB ψ: A + q A,1w A X wB + ψ , B q A + q A, B

(

)

)

−µA ψ: A + q A,1 w A Λ w1 = e

+ eψ − µ A −µ B (q A:B + qB ψ: A + ψ , B q A ) w A X wB Now

e ψ − µ A − µ B ( qB + ψ : A + ψ , B q A ) w A Λ w B

( (

)

)

ψ −µ A −µ B qB ψ , A + q A ψ ,1 + ψ , B q A w A Λ wB = 0 = e

Since qB ψ , A + ψ , B q A and qB q Aψ , 1 are symmetric in the indices (A, B) while wA L wB is antisymmetric in (A, B). Thus,

dω′ = e − µA (ψ: A + q A,1 ) ω A Λω ′ + eψ − µA − µB q A: B ω AΛω B

Thus, d ( F ω′ ) = F e − µA ψ: A + q A,1 ω AΛ ω ′ + Fe ψ − µA− µB q A: B ω AΛω B

(

)

(

)

+ F, A e −µA ω A Λ ω ′ + F ,1 e −ψ ω ′ + q A dX A Λω ′ Now, Thus, Now,

−µA A ω Λ ω′ dX A Λ ω′ = e

(

(

)

)

d ( F ω′ ) = Fe − µA ψ: A + q A,1 + F, A e − µA + F,1 e −µA q A ω AΛ ω′ + F eψ − µA− µB q A: B ω A Λ ω B DA(Feψ) = ( Fe ψ ): A + q A, 1Fe ψ = ( Feψ ), A + q A ( Feψ ), 1 + Fqa, 1eψ

(

ψ = e F, A + F ψ , A + F,1q A + Fq A ψ ,1 + Fq A,1

(

= Feψ ψ: A + Feψ q A,1 + eψ F,1q A + F, A

)

)

316

Stochastics, Control & Robotics

Thus, d ( Fw′ ) = Feψ − µA−µB q A:B w A ΛwB + ρ A ( Feψ ) ⋅ e − ψ − µA w A Λw′ wA = emA dX4, A = 2, 3, 4

(

dwA = eµA µ A, 1dX ′Λ dX A + µ A, B dX B ΛdX 4 (Summation over B only)

(

µA B −ψ = e µ A,1 e w′ + qB dX A

µA

)

)

B

Λ dX + e µ A, B dX Λ dX A

(

µA − ψ µ A, 1µ ′Λe −µA w A + eµAµ A, 1qB + eµAµ A, B = e

e − µA− µB wB Λw A

(

)

)

−ψ µB B A A = e µ A, 1w′Λw − e µ A, B µ A,1 qB w Λ w

(Formula in eqn., (48)). −ψ µB A B A = −e µ A, 1w Λ w′ − e µ A:B w Λw



[22] Mathematical Preliminaries for Cartan’s Eqn. of Structure: w ∈Ω′(µ ), X , Y ∈ X (µ ) α w = wα dX α , X = X ∂ α, Y = Y α ∂α

Then,

(

β β α β (dW) (X, Y) = wα β dX β ΛdX α ( X , Y ) = wα , β X Y − X Y

(

)

)

α β α β α β X(w(Y)) = X ∂ α wβY = wβ, α X Y + wβ X Y, α

Y(w(Y)) = wβ, αY α X β + wβY α X ,βα X ( w (Y )) − Y ( w ( X ))

(

)

α β α β = wβ, α X Y + Y X + wβ [ X , Y ]

β

dw ( X , Y ) + w ([ X , Y ])

or

dw(X, Y) = X ( w (Y )) − Y ( w ( X )) − w ([ X , Y ])

Let {ea} be a local basis for X(m) (vector fields). Then, β σ ∇Xea = ∇Xbebea = Xb∇ebea ≡ X Γ β α e σ By definition,

∇eb ea = Gσba eσ.

Large deviations, Classical & Quantum General Relativity with GPS Applications

Torsion:

317

T(X, Y) = ∇XY – ∇YX – [X, Y] σ a X = Gσba ea (X) wσb (X) = Gba

Define

{ea} is the dual of {ea}

σ a e . wσb = Gba

i.e.

Ta (X, Y) = ea (T (X, Y)) = ea (∇X Y) – ea (∇Y X) – ea ([X, Y]) ∇X Y = ∇X (Ya ea) = X (Ya) ea + Ya ∇X ea = X (Ya) ea + YaXb∇eb ea = X (Ya) ea + YaXb Gσba ea = (X (Yσ) + wσb (Y) Xb) eσ So

ea (∇X Y – ∇Y X) = (X (Yσ) – Y (Xσ) + σσb (Y) Xb – wσb (X) Yb) eσ ea ([X, Y]) = ea (XY – YX) = ea (X (Yb eb) – Y (Xb eb)) eσ (T (X, Y)) = eσ (∇X Y – ∇Y X) – eσ ([X, Y]) = X (Yσ) – Y (Xσ) + wσb (Y) Xb – wσb (X) Yb – eσ ([X, Y]) dea (X, Y) = X (ea (Y)) – Y (ea (X)) – eσ ([X, Y]) = X (Ya) – Y (Xa) – ea ([X, Y]) ea (T (X, Y)) = dea (X, Y) = wab (Y) Xb – wab (X) Yb

So

= wab (Y) eb (X) – wab (X) eb (Y) = (eb L wab) (X, Y) Ta = dea + eb L wab

or

This is Cartan’s first eqn. of structure. so in the limit N → ∞ (scaling limit), we get 1

d  τ (θ) d θ = J (θ) ρ dτ ∫o

1

∫ J ′′(θ) ψ (ρτ (θ)) d θ o

 τ (θ) ∂ρ ∂2   τ (θ)) = ψ (ρ ∂τ ∂θ2 which is the generalized Burgen’s equation. 

318

Stochastics, Control & Robotics

[23] Reflection and Refraction at a Metamaterial Surface: Z ε1,µ1

θi

θr Y θt

ε2,µ2

X

X exp{−i k i , r} E i = Eio  k i = k1{Y sin θi − Z cos θi } X exp{−ik r , r} E r = Ero  k r = k1 (Y sin θr + Z cos θr ) E t = Eto  X exp{− j k t .r} k t = ki (Y sin θt − Z cos θt ) Hi = =

k × Ei ∇ × Ei = i wµ1 − jwµ1 1  ( − Z sin θi − Y cos θi ) Eio exp( − j k i . r ) η1

Hr =

1  ∇ × Ei ( − Z sin θr − Y cos θr ) Ero exp(− j k r . r ) = η − jwµ1 1

Ht =

1 ∇ × Et ( − Z sin θt − Y cos θt ) Eto exp(− j k t . r ) = η2 − jwµ 2

h1 =

µ1 , η2 = ε1

Surface charge density = σs (Y)

X Surface charge density = J s (Y )  Continuity of E tan gives

µ2 . ε2

Large deviations, Classical & Quantum General Relativity with GPS Applications

319

F−io exp(− j k1 Y sin θi ) + Ero exp(− j k1 Y sin θ r ) = Eto exp(− j k2 Y sin θt ) \

(1)

Eio + Ero = Eto, θr = θi, k1 sin θi = k2 sin θt.

Discontinuity of normal component of D: No equation. Continuity of normal component of m H = B: −

k1 sin θi Eio + ki sin θr Ero = k2 sin θt Eto

Or So

µ1 µ µ sin θi Eio − 1 sin θr Ero = − 2 sin θi Eto η1 η1 η2

Eio + Ero = Eto

no new equation. Discontinuity of tangential component of H: cosθi cosθt cosθr Eio − Ero − = Jso. η1 η2 η1 Provided Js (Y) = Jso exp (– j k1 sin θi Y) (By Fourier analysis, we Pick the Fourier Componenets of Js (Y) at the spatial frequency k1 sin θi. Ero E Let = R, to = T. Then, Eio Eto η 1 + R = T1 cos θi − R. cos θ r − 1 cosθtT = h1 Tso η2 Suppose m2 > 0, e2 = – j|e2| (perfect conductor) Then

h1 > 0, h2 =

µ2 = ε2

j | η2 | =

(i + j ) | η1 | . 2

k1 sin θi = k2 sin θt ⇒

ε1µ1 sin θi =

⇒

sin θt =

ε 2µ 2 sin θt

C2 j sin θi C1 C (i + j ) where C2 = = 2 sin θi C1 2

C1 =

| ε2 |µ2

ε1 µ1 cos θi − R cos θi −

η1 (1 − j ) cos θt (1+ R) = h1 Jso | η2 | 2 1/ 2

cos θt =

  k2 1 − sin θt = 1 − 12 sin 2 θi   k2  2

320

Stochastics, Control & Robotics

1/ 2

2   1+ α sin θi 

where a = −

k12

k22

=

ε1µ1 >0 | ε2 |µ2

χ(1 − j )   R  cosθi + (1+ α sin 2 θi )1/2    2

So,

= cosθi −

χ(1 − j ) (1+ α sin 2 θi )1/ 2 − η1 J so 2

χ(1 − j )   (1+ α sin 2 θi )1/2 − η1J so  cosθi − 2  R=  χ(1 − j )  2 1/2  (1+ α sin θi )  cosθi − 2   η1 > 0 .  Reflectivity for a metamaterial with X= | η2 |

Or

Where

e2 < 0, m2 > 0, When the interface carries a surface current density St = max Bs . 0≤ S ≤ t

−1 Xt = ( St − Bt + α ) exp(− α St )

is a Martingale (local) Proof:

`dXt = (dSt − dBt ) exp( −αSt ) − α ( St − Bt + α −1 ) ex ( − αSt ) dSt ((dSt)2 = 0 since St ↑⇒ St is of bounded variation). = − exp( −αSt ) dBt

Since ( St − Bt ) dSt = 0 \ dSt > 0 ⇒ St = Bt t

−1 Corollary: (| Bt | + α ) exp( −α Lt ) is a local Martingale where Lt = ∫ δ ( Bs ) ds 0

Proof: From Skorchod’s theorem the bivariate processes {(St − Bt , St )}t ≥ 0 and {(| Bt |, | Lt |)}t ≥ 0 have the same law. × {(Bt , Lat )}t ≥ 0 and

{(

BCt , LaCt C Lat =

Proof: 1 C

a C

= Ct

) C}

t ≥0

have to same law.

t

∫ δ ( Bs − a) dS 0

1 C

Ct

∫ δ ( Bs − a 0

C ) dS

Large deviations, Classical & Quantum General Relativity with GPS Applications

Charge variables:

321

S = Ca, dS = Cda. Then t

(

)

(

)

1 ∫ δ BCα − a C Cd α C0

1 a C LCt = C

t

C ∫ δ BCα − a C d α

=

0

t

  1 BCα − a d α  C 0 1 BCt , t ≥ 0 is Brownian motion, say Wt, The proof is completed by noting that C t ≥ 0, and thus =

1 a C LCt = C

∫ δ  t

∫ δ(Wα − a) d α

is the local time of W.

0



[24] Maxwell’s Equations in an Unhomogeneous Medium with Background Gravitation Taken into Account: µν χρσ ( x) is the permittivity permeability tensor. Maxwell’s eqns: Wave propagation equation:

(χ

Thus,

µν ρσ ( x )

F ρσ −g

)

ν

=0

νµ µν χρσ = χσρ .

Fµν = Aν, µ − Aµ , ν So or Where

(χ (χ

µν ρσ

g ρσ g σβ −g Fαβ

µναβ

)

ν

− g ( Aβ, α − Aα , β )

=0

)

,ν

=0

µν αρ βσ χµναβ = χρσ g g

Example: In special relativity, {– For, r = 1, 2, 3} is the electric field E in a 23 31 12 Cartesian system and {− F , − F , − F } is the magnetic field B . The electric displacement vector D = ε ⋅ E Where ε is a 3 × 3 matrix. Thus, D1 = DX = ε11E1 + ε12 E2 + ε13 E3 D2 = DY = ε 21E1 + ε 22 E2 + ε 23 E3 D3 = DZ = ε31E1 + ε32 E2 + ε33 E3 Thus, Dr = − ε rs F os (summation over s = 1, 2, 3) Likewise the magnetic field intensity is

322

Stochastics, Control & Robotics −1 −1 −1 H1 = HX = (µ )11 B1 + (µ )12 B2 + (µ )13 B3 etc. −1 −1 23 −1 31 −1 12 Hr = (µ )rs Bs = − (µ ) r1 F − (µ ) r 2 F − (µ ) r3 F

(

)

µν µρ νσ ηµνρσ χµναβ = g αρ g σβ δ ρµ δ σµ + δ ⋅ ηρσ = g g +δ

Let

d being a small perturbation parameter. In our case, oν ρσ Db = χρσ F

))

((

= − ε r1F provided

01

3 r =1

− ε r 2 F 02 − ε r3 F 03

or or er1 = χo1 − χ1o ,

or or er2 = χo2 − χ3o ,

or or er3 = χo3 − χ 2o

or or and χαβ = 0 if a ≠ 0, b ≠ 0, i.e. χ km = 0

( = (−χ

1 ≤ k, m ≤ 3.

H = −(µ −1 )r1 F 23 − (µ −1 ) r 2 F 31 − (µ −1 ) r3 F 12

so

23 ρσ

31 ρσ ρσ F ρσ , − χρσ F , − χ12 ρσ F

)

)

3 r =1

23 –1 23 23 –1 23 23 –1 c23 23 – c 32 = (m )11, c 12 – c 21 = (m )13, c 31 – c 13 = (m )12, 31 –1 31 31 –1 31 31 –1 c31 23 – c 32 = (m )21, c 12 – c 21 = (m )23, c 31 – c 13 = (m )22, 12 –1 12 12 –1 12 12 –1 c12 23 – c 32 = (m )31, c 31 – c 13 = (m )32, c 12 – c 21 = (m )33, rs rs = cko = 0 cok

and

The Maxwell eqns. are

(

1 ≤ k ≤ 3  1 ≤ r, s ≤ 3.

)

 = 0.  g µρ g νσ + δ ⋅ ηµνρσ F ρσ − g   ,ν

or

(F

µν

−g

)

,ν

(

+ δ ηµνρσ −g Fρσ

)

,ν

= 0.

Propagation of em waves in a inhomogeneous metamaterial.  2 ∂2  ∇ + ε µ   ψ(t , r) = 0 ∂t 2   Solution:

ψ(t , r) = ≡

em= 



∫ f (w, n) exp  iw  t +

i ( n, r)     dwd Ω(n) c  

 i ( n, r )  F ∫  t + c , n d Ω(n)

1 C2

>0

Large deviations, Classical & Quantum General Relativity with GPS Applications

323

Inhomogeneous metamaterial ∇ × E = − jwµH ∇ × H = − jwε ( X , Y , Z ) E

e, m > 0.

div (ε E) = 0, div (H ) = 0. So or

∇ (div E ) − ∆ E = − jwµ ( − jw ε) E ( ∆ − w2 µ ε ) E = ∇ (div E ) (∇ ε, E ) + ε div E = 0. div E = − (∇ log ε, E )

(∆ − k

2 o (1+ δ ⋅ χ( r ))

) E (r ) + δ(∇ χ(r ), E ( r )) = 0

2 ko2 = w µ ε o



[25] Relativistic Kinematics* Motions of a Rigid Body in a Gravitational Field a General Relativistic Calculation: L = − ρo c 2 ∫  g oo (t , R(r )) + 2 gok (t , R(t ) r ) Rk′j (t ) X n B

1/ 2

+ g km (t , R(t ) r ) Rkj′ (t ) Rml ′ (t ) X j X l 

dX ′ dX 2 dX 3

r = ( X j )3j =1 .

Where

B is the volume of the body at t = 0. We shall approximate that integral using the binomial theorem:

(g

oo

+ 2 gok ξ k + g km ξ k ξ m 

≈

g1oo/ 2 1+  

)

1/ 2

2 gok ξ k g km ξ k ξ m 1  2g ok ξ k g km ξ k ξ m   + −  + goo 2g oo 8  g oo g oo   

In this approximation, gok is regarded to be O (||x||) and then the expansion is valid upto O (||x||4). Then g (t , R(t ) r ) ( ) X j d 3r R′kj (t L ≈ − ρo c 2 ∫ g oo (t , R(t ) r )1/2 d 3r − ρo c 2 ∫ oo g ( t , R ( t ) r ) oo B B −

ρo c 2 g km (t , R(t ) r ) Rkj′ (t ) Rml ′ (t ) X j X l d 3r 2 ∫B goo (t , R(t ) r )

324

Stochastics, Control & Robotics

−

ρo c 2 gok (t , R(t ) r ) Rkj′ (t ) X j d 3r 4 ∫B goo (t , R(t ) r )

+

ρo c 2 g km (t , R(t ) r ) Rkj′ (t ) Rml ′ (t ) X j X l d 3r 8 ∫B goo (t , R(t ) r )

Example: Spherically symmetric state metrics like the Schwarzchild metric. dt2 = A(r ) dt 2 − B(r ) dr 2 − C (r )r 2 (d θ2 + sin 2 θ dϕ 2 ) In terms of a Cartesian system. r2 = x2 + y2 + z2 rdr = xdx + ydy + zdz 2

2

2

2

r (dθ + sin θ dj ) = dx2 + dy2 + dz2 – dr2 Robots perturbed by quantum noise in a gravitational field. H o  Hamiltonian of the Robot Derivation of Ho = R (t) = R(ϕ(t ), θ(t ), ψ (t )) = Rz (ϕ(t )) RX (θ(t ) RZ (ψ(t ))) Lagrangian 3 £(ϕ, θ, ψ , ϕ =, θ =, ψ = ) = ∫ ( R(t) ξ , R ′(t ) ξ) d ξ B

=

∫ L( R(ϕ, θ, ψ ) ξ, (ϕ R, ϕ (ϕ, θ, ψ) + θ R, θ (ϕ, θ, ψ) B

+ψ R, ψ (ϕ, θ, ψ )) ξ ) d 3ξ Where L( x , v)  Lagrangian for a single point particle Canonical momenta: pj =

∂L T ∂L 3 d ξ = ∫ ( R, ϕ ξ )  ∂v ∂ϕ

pθ =

∂L T ∂L 3 d ξ = ∫ ( R,θ ξ )  ∂v ∂θ

∂L ∂L 3 d ξ = ∫ ( R, ψ ξ )T ∂v ∂ψ Chapter 4 (Continued) Relativistic kinematics of rigid bodies. Rigid body motion in the Schwarzchild metric. 2m GM g oo ( r ) = a (r) = 1 − ,m= 2 . r c –1 g11 ( r ) = a (r) , pψ =

g22 (r) = r2, g33 = r2 sin2 θ.

Large deviations, Classical & Quantum General Relativity with GPS Applications

325

At time t = 0 r is f the position of a point on the body. At time t, this point moves to R(t ) r . dτ −1 2 (r ) = α (|| R(t) r ||) − α (|| R(t ) r ||)|| R ′(t ) r || dt

So

1/ 2

−|| R(t ) r ||2 (θ ′ (t , r ) 2 + sin 2 (θ ′ (t , r )) ϕ ′ 2 (t , r))  θ′(t , r ) =

where

∂θ(t , r) ∂ϕ(t , r) , ϕ′ (t , r ) = ∂t ∂t

Where θ(t , r ) and ϕ(t , r ) are respectively the elevation angle and agimute angle relative to the centre of the blockhole. Here we are assuming that the top’s fixed point is at the centre of the blackhole. This is too unrealised. Assume therefore that the fixed point of the top is at r o > 2m. Let p : ξ be a point in the top relative to r o . Then the position of P after time t is r o + R(t ) ξ ≡ r (t ) Then for any metric, gµν (t , r (t )) = gµν (t , r o + R(t ) ξ ) ≈ gµν (t , ro ) + gµν, n (t , r o )( R ′(t ) ξ)n µ ν dX µ dX ν n  dX dX  g ( t , r g ) + ( t , r ) ( R ( t) ξ ) ′ gµν (t , r (t )) ≈  µν µν, n o o  dt dt dt dt

dX dX o = 1, dt dt So

k

n = ( R ′ (t ) ξ) , k = 1, 2, 3.

n m gµν dX µ dX ν ≈  goo (t , r o ) + goo, n (t , r o ) R ′(t )m ξ  n m + 2  g ok (t , r o ) + gok, n (t , r o ) R ′(t )m ξ  × R ′(t )ks ξ s

Upto O (||x||2), we have gµν

n m +  g kl (t , ro ) + g kl , n (t , ro ) R ′ (t )m ξ  × R ′ (t )ks R ′ (t )lp ξ s ξ p

dX µ dX ν n m = goo (t , r o ) + goo, n (t , r o ) R ′(t )m ξ dt dt + 2g ok (t , r o ) R ′ (t ) ks ξ s + g oo, nm (t , r o ) R ′ (t )np R ′ (t )np R ′ (r )qm ξ p ξ q + 2gok, n (t , r o ) R ′ (t ) nm R ′ (t ) ks ξ m ξ s + g kl (t , ro ) R ′ (t ) ks R ′ (t )lp ξ s ξ p

(

)

n n m = g oo (t , r o ) + g oo, n (t , r o ) R ′ (t )m + 2g on (t , ro ) R ′ (t)m ξ

(

)

+  goo, nm (t ,r o ) + R′ (t ) np R ′ (t )m q + 2 g om, n (t , ro )  n m s b + R ′ (t )np R ′ (t )m q + g nm (t , r o ) + R ′ (t ) p R ′ (t) q  ξ ξ 

326

Stochastics, Control & Robotics

[26] Random Fluctuation in the Gravitational Field Produced by Random Fluctuation in the Energy-Momentum Tensor of the Matter Plus Radiation Field: (a) gµν ( x) = gµν ( x) + δ ⋅ hµν ( x)

Where

g µν = g µν(0) − δhµν + 0(δ 2 )

(( g )) = (( g )) νµ( 0)

(a) µν

T µν = Pνµ µ ν

−1

⋅ hµν + g µα ( a ) g ν(µ 0) hαβ

νµ = V µ + δ ⋅ µ µ

µ ν gµν νµ νν = 1 ⇒ g µν V V = 0

g µνV µ µ ν = 0 ⇔ Vµ µ µ = 0 αβ α( 0) δΓ µd α = g Γ βµα = (0)Γ µα

1   + δ ⋅  − hαβ Γ βµd (0) + g αβ(0) hβµ , α + hβα , µ − hµα , β  2   + 0 (d2) (0) (1) Rmn = Rµν + δ ⋅ Rµν + O (δ 2 ) ( x) (1) αβρσ (1) (3) Rmn = kµν ( x) hαβ , ρσ + K µ( 2ν)αβρ ( x) hαβ,ρ (x) + K µν αβhαβ ( x) (0) K(1), K(2), K(3) are expressible completely in term of g µν and its first and second order partial derivatives. The linearized Einstein field equation is then (3) αβ 1) αβρσ ( 2) αβρ K ((µν ( x) hαβ, ρσ ( x) + K µν ( x)hαβ,ρ ( x) + K µν ( x)hαβ ( x) )

1 ( 0)  = C ⋅ ρ1 ( x) (Vµ ( x)) Vν ( x) − g µν ( x) 2 

⇒

1   Rµν − Rδµν = C ⋅ ρνµ νν + ρ0 ( x)(Vµ ( x) µ ν ( x)) + Vν ( x)µ µ ( x) 2 – R = Cρ 1  − hµν ( x)   2

⇒

1   Rµν = C ⋅ ρ  νµ ννq − gµν    2

Large deviations, Classical & Quantum General Relativity with GPS Applications

327

2 ρ = ρ0 + δρ1 + O(δ ) ,

Where

(0) ν δ . µ µ ( x) = νν ( x) − Vµ ( x) = gµν νν − g µν V

(

)(

)

(0) (0) ν ν ν = gµν + δ . hµν V + δ . u − gµν V

(

)

(0) ν u + hµνV ν + O(δ 2 ) = δ . g µν (0) ν ν um = gµν u + hµνV

or Consider the equation

(ρν ν ) µ ν

:ν

=0

implied by the Einstein field equation. We get

(ρν ) ν

iν

µ νµ + ρµν ν:ν =0

(ρν ) ν

so or

(ρν

)

µ

:ν

= 0 and νν ν:µν = 0

(

) ))

µ −g ,µ = 0, νν ν:ν + Γ µνρ νρ = 0

((

(0) + δ . hµν g = det g µν

(

)

(0) 2 µ = g 1 + δ . hµ + 0(δ )

∆ g (0) (1 + δ . h ) + 0(δ 2 ) µα ( 0) hνα , h = hµµ . hνµ = g

Where

The perturbed fluid dynamical eqn. (without viscosity) are

(

(

)

)

ν µ µ (0) ρ µ V ν µ ,νρ + Γ µν (0) u ρ + Γ µνρ(1) V ρ + µ Vν + Γ νρ V = 0 µ µ (0) ρ Or since V , ν +Γ νρ V = 0,

It follows that Note that

V νµ ,µν + Γ µνρ(0) V νµ ρ + Γ µνρ(1) V νV ρ = 0, Γ µαβ = g µρ Γ ραβ

(

)(

)

µρ 0 µρ 0 0 1 ( 2 ) Γ αβ − δ ⋅ hµρ Γ ρα = g β + δ ⋅ Γ ραβ + O δ

( )

(

( )

()

)

(1) (0) + δ g µρ(0) Γ αρβ − hµρΓ αρβ + O(δ 2 )

Where

0) Γ (αρβ =

(

)

1 hρα,β + hρβ,α − hα , βρ , 2

( )

328

Stochastics, Control & Robotics µρ( 0) ( 0) 0) Γ αρβ Γ µ( = g αβ

Explicit form for K(1), K(2), K(3):

(

(1) 1) α( 0) β(1) α(1) β( 0) Rµν = Γ αµα(1,)v − Γ α( µv , α − Γ µν Γ αβ + Γ µv Γ αβ

(

α(1) β( 0) 0) β(1) + Γ α( µβ Γ να + Γ µβ Γ να

)

(1) µρ( 0) (1) µ(1) Γ ραβ − hµρ Γ ραβ Γ αβ = g

Where

(

)

1 µρ(0) g hρα, g + hρβ, α − hα,βρ 2 1 ( 0) ( 0) ( 0) − hµρ gρα , β + gρβ, α − g α , β, ρ 2 The equation to be solved for hab (x), and un (x), ρ1 (x) are the following pde’s =

)

(

(1) αβρσ ( 2) αβρ K µν ( x) hαβ,ρσ ( x) + K µν ( x) hαβ,ρ ( x) + K µ(3ν) αβ ( x) hαβ ( x)

 1 (0)  1    = C ⋅ ρ1 ( ν)  VµVν − g µν  + ρ0 ( x)  Vµ uν + Vνuµ − hµν ( x)      2 2 

(1),

V ν ( x)u,µν ( x) + Γ µνρ(0) V νu ρ

(

 0) ( x) ) − hµσ Γ (σνρ 

1 + V νV ρ  g µσ (0) hσν, ρ + hσρ, ν − hνρ, σ 2 And finally the man conservation equation  ( 0)u µ ρ Vµ − 0  ρ1 V µ − g (0) + ρ(0) − g 2 − g ( 0)  Where

=0

(2)

 h = 0  ,µ

(3)

µν( 0) hµν h = hmm = g

Random fluctuation in the energy-momentum tensor of the em field LEM = Fµν F µν − g

(

δ g LEM = − Fµν F µν δg / 2 − g + Fµν − g Fαβ δ g µα g νβ µν αβ g = − Fµν F gg δgαβ 2 − g

± 2 − g Fµν Fαβ g µα g νρ g βσ δgρσ Coeff. of dgρσ is 1  − g  Fµν F µν g ρσ − 2F αρ Fασ  2  

)

Large deviations, Classical & Quantum General Relativity with GPS Applications

329

αS ρσ − g the energy momentum tensor of the em field. 1 Fµν F µν g ρσ − F αρ Fασ 4 Motion of charged matter in an em field and em grav. Field-linearized version: Exact equation is Sρσ =

(ρν ν ) µ ν

:ν

+ K1S:νµν = 0 µν S:ν =

(

1 Fαβ F αβ 4

)

g µν

:ν

(

− F αµ Faν

)

:ν

1 αβ F Fαβ:ν g µν − F:ναµ Fαν − F αµ Fαν:ν 2 In the absence of charged matter. In the presence of charged matter, the extra term is comes from the interaction Lagrangian between the em field and four caurraso density. =

Rough:

(

δ g J µ Aµ −g

) = δ (ρ ν A = δ (ρ ν A g

q

g

q

µ

µ

µ

µ

) −g ) = − ρ ν −g

q

ν

Aµ δg / 2 − g

µ αβ = − ρq ν Aµ gg δg αβ / 2 − g

= ρq νµ Aµ − gg δgαβ / 2 ρq  ρq where ρ  matter density and q  charge per unit man. Thus the eqn. get modified to

(ρν ν ) + K S (ρν ) + (ρν ) ν µ

ν

:ν

ν νµ :ν

or

(

)

µν 1 :ν

+ K 2 ρνα Aα g µν

ν

+ K 2 ρνα Aα + K1S:µν ν

µ :ν

(

:ν

=0

)

:ν

g µν = 0.

(1)

The coefficient of dAm gives the Maxwell eqn. ν F:αν = K3′ J α = K3ρνα µν

So

1 αβ F Fαβ:ν g µν − F:ναµ Fαν − ρK3 F αµ να 2 1 = F αβ Fαβ:ν νν − F:ναµ Fαν νµ − ρK3 F αµ vµ vα 2

v S:ν =

νµ S;µν ν

=

(∵ F

αµ

)

1 αβ F Fαβ:ν v ν + F:νµα Fανu µ 2

vµ vα = 0 . So we get from (1) and (2),

(2)

330

Stochastics, Control & Robotics

(ρv ) ν

:ν

+

(

K1 αβ F Fαβ:ν v ν + K1Fνµα Fαν vµ + K 2 ρv α Aα 2

)

:ν

vν

=0 Now,

(ρv

α

Aα

)

:ν

(3)

( ) + (ρv )

v ν = ρv α v ν Aα:ν + ρv α α ν = ρv v Aν:α

:ν

Aα v ν

:ν

v ν Aα .

α

(4)

Taking (1), (2), (3) into account, (1) can be expressed as

(

µ  K1 αβ ρv ν v:νµ F Fαβ:ν v ν K 2 ραν Aα ν −v   2

(

+ K1 F:νρα Fαν vρ + K 2 ρv α Aα

)

ν

)ν v ν

(

v ν + K 2 g µν ρv α Aα

)

:ν

 1 + K1  F αβ Fαβ:ν g µν + F:νµα Fαν + ρK3 F µα να  = 0  2

(5)

This can be rearranged as

ρv ν v:νµ + K1K3 ρF µα vα +

(

K1 αβ F Fαβ:ν g µν − v µ v ν 2

(

1   + K1  F:νµα Fαν − F:νρα Fαν vρv µ  + K 2 ρv α Aα   2

(

)

) − (g

F αβ Fαβ:ν =

µν

:ν

( )

1 αβ 1 2 F Fαβ , ν = F , 2 2 We thus express that above equation as

Note that

) )

− vµ v ν = 0

ν



K

ρv ν v:µν + K1K3 ρF µd vα +  1 ( F 2 ),ν + K 2 (ρvα Aα ), ν   g µν − v µ v ν   4  1   + K1  F:νµa Fαν − F:νρα Fαν vρν  = 0 2   

[27] Special Relativistic Plasma Physics: (a) Dynamical problem, how does an initial random velocity and field evolve with time? (b) In a non-relativistic plasma problem, how does an initial random velocity, density and e-m field evolve with time? The special relativistic plasma (MHD) T µν = (ρ + p) v µ v ν − p ηµν In the rest frame,

Large deviations, Classical & Quantum General Relativity with GPS Applications

((T )) µν

331

O  ρ   −p  =  −p    O − p

Which is correct since ρ (= ρc2) is the rest frame energy density and – p is the force per unit area along any spatial direction which by Newton’s second law is the momentum flux (momentum flowing per unit area per unit time, the momentum being, in the rest frame, ⊥ to the area element) 1 µ ν µν µν αµ ν  T µν + S µν = (ρ + p) v v − pg + K  Fαβ Fαβ g − F Fα  4 is the combined energy momentum tensor of matter plus radiation. Here gmn = hmn

(T

µν

+ S µν

)

= 0  Conservation on the total energy-momentum tensor of the

,ν

matter plus radiation field. Thus,

((ρ + p ) v v ) µ ν

,ν

1 ν αµ ν  − p,µ + K  ηµν F αβ Fαβ, ν − F, αµ ν Fα − F Fα , ν  = 0 2 

Fαν, ν = K1J α

Now,

((ρ + p ) v v ) µ ν

and so

,ν

− p,µ +

K αβ ,µ F Fαβ 2

+ K F,νµα Fαν − KK1J α F αµ = 0 For mhd analysis, we must find the 4-vector-tensor equivalent od Ohm’s law J = σ( E + v × B) µν The natural candidate appears to be Jm = σ F vν and the above mhd equation assume the form

((ρ + p ) v v ) µ ν

,ν

K αβ ,µ ν µα ν Fα vν F Fαβ + K F, µα ν Fα + K 2 F 2 = 0

− p ,µ +



(1)

k2 = KK1σ

where

These are coupled to the maxwell equation in the form F,νµν = K1σ F µν v ν (1) and (2) are the basic equation of a special-realtivistic plasma. (1) gives

((ρ + p ) v ) , ,ν

µ

νv

µ

+ (ρ + p)v ν v,µν − p,µ

(2)

332

Stochastics, Control & Robotics

+

K αβ ,µ F Fαβ + K F,νµα Fα, ν + K 2 F µα Fαν vν = 0 2 k ,µ vµ (ρ + p ) v ν ,ν − vµ pµ + F αβ Fαβ 2

(

So,

(3a)

)

ν ,µα ν + K F, µα Fα vν vµ = 0 ν Fα vµ + K 2 F

(3b)

Non-relativistic case: MHD equation are

(ρ(v, ∇) v + v , t ) Linearization

= − ∇p + η∆ v + σ( v + B) × B

v (t , r ) = V0 (t , r ) + δ v (t , r ) B(t , r ) = B0 (t , r ) + δ B(t , r )

Assume ρ = constant,

σ η = v, = a, ρ ρ

(V , ∇) δ v + (δ v , ∇)V

Then,

0

0

+ δv,t

1 = − ∇δp + η∆ v + α (δ v + B 0 ) × B 0 ρ + α (V 0 + δ B ) × B 0 + α (V 0 + B 0 ) × δ B div v = 0 ⇒ divV 0 = 0, div (δ, v ) = 0. Fourier expansion: V 0 (t , r ) =

∫ V 0 (t , k )exp(ik .r ) d

3

k

δv (t , r) =

∫ δv (t , k ) exp(ik .r ) d

3

k

(V 0 , ∇) δ v

=

∫ i (k ′, V 0 (t, k )) δv (t , k ′) exp (i ( k + k ′, r )) d 3 kd 3 k ′

Coff.’ of exp (i ( k , r )) in (V 0 , ∇) δv is given by

(

)

i ∫ k ′ − k ′, V 0 (t , k ′ ) δ v (t , k − k ′ ) d 3 k ′

( ) v (t , k ) )V (t , k ′ ) exp(i ( k + k ′.r )) d kd k ′ = ∫ i ( k ′δ 3   = i ∫ k ′, V 0 (t , k − k ′ ) δv (t , k ′ ) d k ′

(δv , ∇)V 0

Coff. of exp (i ( k .r )) in (δ v , ∇)V 0 is

(

)

v (t , k ′ ) V 0 (t , k − k ′ ) d 3 k ′ i ∫ k , k ′ , δ

0

3

3

Large deviations, Classical & Quantum General Relativity with GPS Applications

(

333

)

(t, k ′ − k ′ ) V 0 (t , k ′ ) d 3 k ′ = i ∫ k , δv δv , t =

∫ δv, t (t , k ) exp (i(k , r )) d

3

k

∇δp =

∫ ikδ p,(t , k ) exp(i(k , r )) d

3

k,

v (t , k) exp (i ( k, r )) d 3 k ∆δv = − ∫ k 2 δ (δv × B 0 ) × B 0 =

∫ (δv (t , k ) × B 0 (t , k )) × B 0 (t , k ′′) exp(i ( k + k ′ + k ′′, r )) d 3kd 3k ′d 3k ′′

(V 0 × δB ) × B0

=

Coff. of exp(i ( k , r )) in

∫ (V 0 (t , k ) × δ B (t , k ′)) × B 0 (t , k′′) exp {i ( k + k ′ + k ′′, r )} d 3kd 3k′d 3k ′′

(a) δ v ,t is δ v,t (t , k) , (b) ∇δp is ik δ p(t , k ) , (c) (V 0 × δ B) × B 0 is (d) (δv × B 0 ) × B 0 is

∫ (V 0 (t , k − k ′ − k ′′) × δ B (t , k ′)) × B 0 (t , k ′′) d

3

k ′d 3k ′′

∫ (δv (t , k ′) × B 0 (t , k ′′)) × B 0 (t , k − k ′ − k ′′) d

3

k ′d 3k ′′

The linearized MHD eqn. in the spatial frequency domain thus assume the form

(

)

∂  v (t , k ′ ) d 3 k ′ δ v (t , k ) + i ∫ k ′V 0 (t , k − k ′ ) δ ∂t i v (t, k ) + i ∫ ( k − k ′ , δ v (t , k ′ )) V 0 (t , k − k ′ ) d 3 k ′ + k δ p(t , k) + vk 2 δ ρ = α

{∫ (δv (t, k ′) × B (t, k ′′)) × B (t, k − k ′ − k ′′) d k ′d k ′′ 0

3

0

3

( )  (t , k ′ ) × δ B + ∫ (V (t , k − k′ − k ′′) × B )  (t, k ′′) d k ′d k ′′}  (t , k ′ ) × B  0 (t , k ′′) d 3k ′d 3k ′′ + ∫ V 0 (t , k − k ′ − k ′′) × δ B 0

3

0

3

(a)

The equation of continuity div (δv) = 0

becomes

(k , δv (t, k )) = 0

(b)

At each k , we construct an orthogonal set of 3 victors

(e1(k ), e2 ( k), k ) so that (e1(k ), e2 (k))

= 0,

334

Stochastics, Control & Robotics

(e1(k ), k ) (e1(k ), e1 (k))

and

= (e2 ( k ), k ) = 0 = (e2 (k ), e2 (k) ) = 1.

1 (t , k )e ( k ) + δν 2 (t , k )e (k ) δ v (t , k) = δν 1 2

(b) ⇒

for some complex scalar fields δ v1 and δ v 2 . Substituting into (a) this expression,  after taking k × gives using  k × e ( k) = e1 (k ) and so k × e ( k ) = e ( k ),  2

1

2

(

)

∂  v1 (t , k ′ ) d 3 k ′ δ v (t , k ) + i ∫ k ′, V 0 (t , k − k ′) δ ∂t 1 e ( k ′ ) δ v1 (t , k ′ )V 02 (t , k − k ′ ) d 3 k ′ + i k − k ′, 

∫ ((

1

(

)

)

2 (t, k ′ ) + vk 2 δ + k − k ′,  e2 ( k ′ ) δv v1 (t , k) = α {T1 + T2 + T3 } When

T1 =

    ∫   k × ((δν(t , k ′) × B 0 (t , k′′ ) ) × B 0 (t , k − k′ − k′′) ) 2 d 3k ′d 3k ′′

=

∫ k × ( B 01(t , k − k ′ − k ′′)δν1(t , k ′)

)  (t , k − k ′ − k ′′), B  (t , k ′′ ) δ − (B ) v (t, k ′)

 02 (t , k − k ′ − k ′′), δ  (t , k ′ ) +B v 2 (t , k ′ ) B 0 0

3

0

1

2

3

d k ′d k ′′

=

∫ {( B 01 (t , k − k ′ − k ′′), δv1 (t , k ′)

}  (t , k − k ′ − k ′′), B  (t , k ′′) δ − (B ) v (t, k ′) d k ′d k ′′  , = (e ( k ), B )  (t, k ′ ) × B = ∫  k × ((V (t , k − k ′ − k ′′) × δ B )  (t, k′′))   02 (t , k − k ′ − k ′′), δ  (t , k ′′ ) +B v 2 (t , k ′ ) B 01

where

(e (k ), B ), B 1

0

0

02

T2

0

2

3

1

0

0

3

3

0

2

3

d k ′d k ′′

=

∫ (k × ( B 0 (t , k ′′), V 0 (t , k − k ′ − k ′′)) δ B (t , k ′)

(

)

 (t , k ′′), δ B  (t , k′) V 0 (t , k − k ′ − k′′ ) − B

)

2

d 3k ′d 3k ′′

Large deviations, Classical & Quantum General Relativity with GPS Applications

=

335

∫ {( B 0 (t , k ′′), V 0 (t , k − k ′ − k ′′)) δ B (t, k′)

(

}

)

 (t , k ′′), δ B  (t , k′) V 0 (t , k − k ′ − k′′ ) d 3k ′d 3k ′′ − B 0 and T3 =

∫ k × ((V 0 (t , k − k ′ − k ′′) × B 0 (t , k ′)) × δ B (t , k ′′)) 2 d 3k ′d 3k ′′

=

∫ {(V 0 (t , k − k ′ − k ′′), δ B1 (t , k ′′))B1 (t , k ′ )

(

)

}

 (t , k ′ ), δ B  (t , k ′′) V 01 (t , k − k ′ − k′′ ) − B 0 0 d 3k ′d 3k ′′

∂  δv 2 (t , k ) ∂t on the L.H.S. These form a pair of linear intergo-differential equations for  (t , k) . δ v(t , k ), m = 1, 2, δ B

Likewise by connecting the (1)th component, We get another involving

{

}

Consider the special can when V 0 and B 0 are constant vectors. Then the mhd equation (linearized) for δv(t. r ) for becomes ∇δp + v∆δ v (t , r ) δv,t (t , r ) + α (δ v (t , r ) × B 0 ) × B 0 ρ + (V 0 × δ B(t , r ) ) × B 0 + (V 0 × B 0 ) × δ B(t , r ))

(V 0 , ∇) δ v (t , r ) = −

The moncomeater equation (linearized) is ∇(δν(t , r )) = 0 and finally the linearized Maxwell eqn. are ∂ ∇ × δE = − δB , ∂t ∇ × δB = − µσ(δ E + δν × B 0 + V 0 × δB) + µεδE,t This latter equation gives on taking ∇ × and using the previous one along with ∇ .δB = 0, ∇2 δB − µσδB,t − µεδB,tt + µσδ v × B 0 + µσV 0 × δB = 0 In the spatial frequency domain, there read

(V 0 , k ) δv (t, k ) + δv ,t (t , k ) −

k  v (t , k ) δ p (t , k ) − vk 2 δ ρ

336

Stochastics, Control & Robotics

((

)

(

)

 (t, k ) × B + α δ v (t , k ) × B 0 × B 0 + V 0 × δ B 0

)

 (t , k) , + (V 0 × B 0 ) × δ B

(a)

(k , δv (t, k )) = 0

(b)

 (t , k ) − µσδB  , (t , k ) − µεδ B  , (t , k) − k 2δ B t tt  + µσV × δ B  (t , k ) = 0 v (t , k ) × B + µσδ 0 0

(g)

 (t , k ) , we get Assuming a time dependence of ejwt for δ v (t , k ) and δ B ( k , V ) + jw + vk 2 − α B 0 BT0 + αβo2  δ v 0  

(

)

T T T  = α (V 0 , B 0 ) I − V 0 B 0 + B0 V 0 − V 0 B 0 δ B

− (g) becomes

k  δp = 0 ρ

(a)

 − µσΩ δ  k 2 + jwµσ − µtw2 + µσΩv  δ B B0 v = 0 0 

(g′)

(

T ∈R 3 When for any vector X = X1, X 2 , X 3x

 0  WX = X 3   −X2

− X3 0 X1

X2  − X1   0 

This matrix has the defining property Ω x ξ = X × ξ ∀ ξ ∈R 3 (α ′ ) (β), (γ ′ ) can be cast in matrix form (block) as

(

)

 ( k , V ) + jw + vk 2 + αB 2 I − α B BT 0 0 0 0 3  T  k   −µσΩB0 

(

α (V 0 , B 0 ) I 3 − 2V 0 BT0 + B 0 V T0

(k

OT 2

)

+ jwµσ − µεw2 I + µσΩ 3

V0

)

k p  0   0   

−

 δ v    =0 δ B    δ p 

Large deviations, Classical & Quantum General Relativity with GPS Applications

337

For a non-zero solution to exist, we require that the determinant of the above 7 × 7 matrix vanish. This gives us for given B 0 , V 0 a dispersion relation between w and k . Now we analyze the special-relativistic mhd equation (3a) and (3b). Substituting for

(( ρ + p ) v ) ν

1ν

from (3b) into (3a) gives

k   ,α ν ρα ν v ν v α p ′ α − F ρσ Fρσ vα − kF,1ρα ν Fα vρ − k2 F Fα vν vρ  2   k αβ ,µ F Fαβ + kF,νµα Fαν + k2 F µα Fαν vν = 0 (4) 2 We note from (4) that when there is no em field, i.e. Fmn = 0, we get the standard special relativistic fluid dynamical equation without viscosity. µ − p ,µ + + (ρ + p) v ν v,ν

µ (ρ + p)v ν v,ν − p,µ + v µ vα p α = 0

(5)

and the man conservation equation (3b) becomes

((ρ + p ) v ) µ

,µ

− vµ p,µ = 0

(6)

Suppose we combine (5) and (6) with the equation of state p = p (ρ).

(7)

These are to be combined with

ηµν v µ v ν = 1

(8)

vµ v µ = 1.

or

(6) can be exprenal as

(ρv ) µ

µ + pv,µ =0

,µ

( )

Let v r

3 r =1

be the spatial component of vm. We note that (8) is the same as 3

2

v 02 − ∑ v r = 1 r =1

(

g (u) = 1 − u 2

Define

ur =

when Then (10) implies

)

−1/ 2

d τr dx r r 0 = = v v dζ dτ

v 02 − v 02u 2 = 1 or

(9)

(

2 v0 = 1 − u

)

−1/ 2

r r = γ (u ), v = γu1 1 ≤ r ≤ 3 .

(10)

338

Stochastics, Control & Robotics

So (9) can be exprinced as and (5) as

( γρ),0 + ( γρu r ),r + p ( γ ,0 + (γu r ), r ) = 0

(

(10)

)

γ (ρ + p) γ ,0 + γ (ρ + p)u rs ( γu r ), s − p r + γu r γp s 0 + γu s p, s = 0 or

( )

γγ ,0 + γu s γu r

,s

+

2 r

γ u p ,0 p, r + = 0 ( p + ρ) ( p + ρ)

(11)

+ γ 2u r u s p, s / ( p + ρ) Perturbed version of (10) and (11). r ρ = ρ0 + δρ(t , r ), u = U 0r + δ u r (t , r ) p = p(ρ0 ) + p ′(ρ0 )δP(t , r ) r The unperturbed versions of (10) and (11) are consistent with p(ρ0 ), U 0 , ρ0 as constants. We look at the perturbed versions of (10) and (11):

δγ ,0 ρ0 + γ 0δρ,0 +ρ0U 0r δγ ,r + γ 0U 0r δρ,r + γ 0ρ0 δu,rr + p0 (δγ ,0 + U 0r δγ ,r + γ 0 δu,rr ) = 0

(12) 

7 Quantum Signal Processing [1] Quantum Scattering of a Rigid Body:

where

 ∂2  m ∂2 ∂2 + k 2  ψ s ( X , Y , θ)  2+ 2+ 2 I ∂θ ∂Y   ∂X 2m = 2 χψ i ( X , Y , θ) £ 2mE m k2 = = k X2 + kY2 + n 2 . 2 I 

By the Green's function for the Helmholtz equation, this has the solution Ys(X, Y, q) =

−m

∑ 2π2

r∈Z

1

∫

2    I   2 2 2 (θ − θ ′ + 2r π )   exp  ik  ( X − X ′) + (Y − Y ′) +   m    

X ′ ,Y ′ ∈R , θ ′ ∈[ 0, 2 π ]

1

2   2 2 2  I  ( X − X ′) + (Y − Y ′) +  (θ − θ′ + 2rπ)   m   

× (χψ i )( X ′, Y ′, θ′ )dX ′dY ′ ×

I d θ′ m



[2] Optimal Control of a Field: The filed j: Rn → Rp j = (j1, .., jp) satisfies the field equation

Lj(X) = J(X) where L is a linear partial differential operator.

340

Stochastics, Control & Robotics

J: Rn → Rq is a source (input) signal field. We wish to select J so that the cost functional

∫ C (ϕk ( X ), ϕk ,m ( X ), J ( X )) d

n

X

is a minimum. Incorporating the equations as constraints using a Lagrange multiplier field L: Rn → Rq (Note L: C∞(Rn, Rp) → C∞(Rn, Rq)) the functional to be minimized is S[j, J, L] =

∫ C (ϕk , ϕk ,m , J ) d

δϕ S = 0 ⇒ k

n

X − ∫ ( Λ, Lϕ − J )d n X

∂C ∂ ∂C −∑ − ( L * Λ )k = 0 ∂ϕ k m ∂X m ∂ϕ k ,m

∂C + Λk = 0 ∂J k dLS = 0 ⇒ Lj – J = 0

δ Jk S = 0 ⇒ Eliminating L gives

∂C ∂ ∂C ∂C   −∑ +  L*  = 0,  ∂ϕ k m ∂X m ∂ϕ k ,m ∂J  k

Lj – J = 0 

[3] Example from Quantum Mechanics: Calculate {f(t), 0 ≤ t ≤ T} so that T

∫

2

ψ d (t ) − ψ (t ) dt is a minimum

0

where y(.) satisfies 1 ∆ψ(t ) + f (t )V ψ(t ) = iy′(t) 2m Here y(t) = y(t, r) and yd(t) = yd(t, r), −

||yd(t) – y(t)||2 =

∫ ψ d (t , r ) − ψ (t , r )

(1) 2 3

d r.

and V = V(r) (multiplication operator). Incorporating the eqns. of motion (Schrödinger's equation) using the complex Lagrange multiplier L(t, r), we have to minimize S{y, f, L] =

∫ ψ d (t , r ) − ψ (t , r )

2

dtd 3r

 1  ∆ψ (t , r ) − f (t )V ( r )ψ(t , r ) )) dtd 3r + ∫ Re  Λ (t , r )  iψ ,t (t , r ) +   2m

Quantum Signal Processing

341

y, ψ , f, L, Λ are independent functions. δψ S = 0 i 1 ∆Λ(t , r ) ⇒ − (ψ d (t , r) − ψ (t , r ) ) + Λ ,t (t , r ) + 2 4m − ⇒

1 f (t )V ( r )Λ(t , r ) = 0 2 dfS = 0

∫ Re (Λ(t , r )V ( r )ψ (t , r ))d

3

r

(2)

=0

(3)

Equations (1), (2) and (3) are to be solved for y, L and f. Note that f(t) is real. 

[4] Example from Quantum Field Theory: The field f (t, r) = f (X) is coupled to an external source J(t, r) = J(X) via the interaction action ∫ J ( X )φ( X )d 4 X . The Lagrangian density is thus

(

£ J φ, φ,*µ

)=

The propagator is 〈 Vac |T{f(X) f(Y)}|Vac 〉 =

1 1 ∂ µ φ∂ µ φ − m 2φ2 + J φ 2 2

∫ exp (i∫ £ J (φ(ξ), φ, µ(ξ), ξ) d φ( X )φ(Y )

∏

4

ξ

)

dφ( η)

η∈R 4

=

(

) )

−δ 2 exp i ∫ £ J φ (ξ ) , φ,µ (ξ ) , ξ df ξ δJ ( X )δJ (Y ) ∫

∏

(

dφ( η)

η∈R 4

We wish to make this as close to a given function K(X, Y) on R4 × R4 as possible. The propagator 〈Vac|T{f(X). f(Y)}|Vac〉 controls the scattering probability amplitude between times tX and tY when X = (tX, rX), Y = (tY, rY). Thus, we must select the function {J ( X )} X ∈R 4 so that

∫

2

Vac T {φ( X )φ(Y )} Vac − K ( X , Y ) d 4 Xd 4Y

is a minimum. The optimal minimizing equations for the current density J(.) are (J is real and f is also real)

342

Stochastics, Control & Robotics

∫

{

δ Vac T {φ( X ).φ(Y )} Vac δJ ( Z )

( Vac T {φ( X )φ(Y )} Vac

)

− K ( X , Y ) d 4 Xd 4Y = 0

or equivalently,   δ3  ∫  δJ ( X )δJ (Y )δJ (Z ) ∫ exp ( −iS J (φ)) ∏ d φ(ξ) ξ     δ2 . exp(iS J (φ))∏ d φ(ξ ′ ) − K ( X , Y )  d 4 Xd 4Y = 0 ∫  δJ ( X )δJ (Y )   ξ′



[5] Quantum Signal Processing (Contd.): n

d ×d Typical subspace: Let ρn ∈C where r∈Cd×d is a density matrix.

Let p(x) ≥ 0

r=

n

be a density matrix. Assume that rn = r⊗n

∑ p ( x) x > < x

x∈A

A = {1, 2, .., d}.

∑ p( x) = 1, {(x): x∈A} an ONB for Cd.

x∈A

rn = r⊗n =

Then

∑

x1 ,.., xn ∈A

p( x1 )...p( xn ) x1...xn > < x1...xn .

H(r) = –Tr(r log r) ≡ − ∑ p( x)log p ( x) .

Let

x∈A

Tnd

Let

n   n 1 x .., x ∈ A , log p ( xi ) + H (ρ) < δ  = ( 1 ∑ n) n i=1  

By Chebrshev's inequality P(Tnd) ≡

∑

( x1 ,.., xn )∈Tδn

where

Var (log p(x)) =

p( x1 )...p( xn ) ≥ 1 −

∑ p( x) (log p( x))

2

+ H (ρ)2

x∈A

=

∑ p( x) (log p( x) + H (ρ))

x∈A

Thus, lim P(Tδn ) = 1, ∀ d > 0.

n→∞

2

Var (log p( x)) nδ 2

Quantum Signal Processing

343

Now for (x1, .., xn)∈Td, – d– H(r) ≤

1 n ∑ log p( xi ) ≤ d – H(r) n i=1

and so exp(– h(d + H(r))) ≤ p(x1) ...p(xn) ≤ exp(n(d – H(p))) 1 ≥ P(Tdn) ≥ # T(dn). exp(– n(d + H(r)))

Thus

# (Tdn) ≤ exp(n(d + H(r)))

or Also, 1 −

Var (log p( x)) nδ 2

≤ P(Tdn) ≤ exp (n(d – H(r))) # (Tdn)

Thus, v   1 − 2  exp(n(H(r) – d)) nδ ≤ # (Tdn) ≤ exp(n(d + H(r)))

(1)

v = Var(log p(x)).

where

Choose any d > 0 and encode the elements of Tdn (n-length sequences) into k length binary sequences where 2k ≤ exp(n(d+H(r))). This encoding can be performed in a 1 – 1 way. The original product source consisted of dn sequences from the alphabet A. Encode all the elements of Tdnc ≡ An \ Tc into a single k length binary sequence.

The compression achieved in this system of encoding is smaller than dn → exp(n(d + H(r))) or in terms of bits n log2 d → n(d + H(r))log2 e

i.e. the bits rate reduction is by the fraction

(δ + H (ρ)) (δ + H (ρ)) log 2 e = log e d log 2 d The error of decoding probability is P(Tdnc) = 1 – P(Tdn) ≤

v

nδ 2 By choosing n sufficiently large, the decoding error probability can be made (δ + H (ρ)) arbitrarily small while retaining the bit compression rate as ≤ . log e d

344

Stochastics, Control & Robotics

Thus choosing d sufficiently small, we can achieve a compression rate arbitrarily H (ρ) clone to and making n large enough the corresponding error probability log 2 d can be made arbitrarily small. This is the Shannon nóiselen coding theorem. 

[6] em Field Theory Based Image Processing:  ( w, r ) E is the electric field incident on the object H0(w, r) describes the filter i  ( w, r ) . The structure of the object at r. The e-field signal at this pixel is H(w, r) E i

surface magnetic current density at r (r∈S, S is the object surface) is

 υ( w, r ) MS(w, r) = H 0 ( w, r ) n ( r ) × E

where n ( r ) is the unit normal to S at r. The electric vector potented at r is then

  jw r − r′ M S ( w, r ′) exp  −   ε c F(w, r) = dS( r ′ ) 4π ∫S r − r′ ε = 4π ∫S

 υ( w, r ′ ) exp  − jw r − r ′  H 0 ( w, r ′ ) n ( r ′ ) × E   c dS ( r ′ ) r − r′

{H0(w, r′), r′∈S} describes the object pattern and {E(w, r), r∈R3\S} the recorded image. The above equation tells us how to recover the function {H0(w, r′), r′∈S} from em  i ( w, r ) in the incident wave is given by field measurements. The magnetic field H i  ∇E i ( w, r ) wµ and the surface electric current density is then  i ( w, r ) = H

JS(w, r) = −

(

j   ( w, r ) n( r ) × ∇ × E i wµ

)

Note: Assuming  ( w, r ) = E i

∫ Ei (w, m ) exp  − j c (m , r ) d Ω(m )

 satisfies the wave equation so that E i  2 w2    ∇ + 2  E i ( w, r ) = 0 c we get



w



Quantum Signal Processing

345

(

)

 i ( w, r ) = 1 m  × E ( w, m  ) exp  − jw ( m  , r ) d Ω(m ) H i ∫  c  µc JS(w, r) = −

and thus

{

(

1  × E ( w, m ) n ( r ) × m i µc ∫

)}

 w   ) exp  − j (m , r ) d Ω(m   c Problem from Revuz and Yor "Continuous Martingales and Brownian Motion" P.185 Zt = Bt + f(t)∈R2. t

∫

Zs

1 n

−1

df ( s ) =

nt

−1

∫ Zsn 1

{

E Zs

−1 n

}

 s df    n

( )

 = E  Bs + f s n  n

−1 

 

 s = E Z0 + f s n  n

( )

Z0 =

where

{

E Zs

Thus

−1 n

}

≤

Bs

n

n K s

{

m∈R 2

t

{

Thus, E Z s ∫ 1 n

−1

df ( s )

}

  

∼ N (0, I2 ) .

s n

K = sup E m + G

where

−1 

nt

≤ K n∫ 1

−1

} < ∞ (G ∼ N (0, I2 ))

1  s df    n s

( ) 

 df s n ≤ K n sup   ds  1≤ s ≤ nt

nt

 ∫1

1 ds s

( )

( )  ( ) 

 s s 1  df ( s )  2 nt  df n 2K 1 df n sup  = s | = n d s  ds  1  ds n1 ≤ s ≤t  n n

346

Stochastics, Control & Robotics

 df ( s )  1  = 2K  t −  sup  ds   n 1 n

≤ s ≤t

→  df ( s )  2K t sup   t Pf dT {U * (T ) ( I ⊗ Yin (t ))U (T )}

= dU *(T ) ( I ⊗ Yin (t ))U (T ) + U * (T ) (I ⊗ Yin (t )) dU (T ) + dU * (T )( I ⊗ Yin (t )) dU (T ) = 0, T > t Since

dU (T) =

∑ Pk dWk (T ) k



Quantum Signal Processing

475

(

Where Pk is a system operator, i.e., Pk ∈ L( L2 ( h )) and Wk (T ) ∈L2 Γs ( L2 (0, T )) So

)

dT {U * (T ) ( I ⊗Yin (t ))U (t )} =

∑ U * (T ) {Pj* ⊗ Yin (t )}U (t )dWk* (T ) k

+ ∑ U * (T ) {Pk ⊗ Yin (t )}U (t )dWk (T ) k

+

∑ U * (T ) {Pk* Pm ⊗ Yin (t )}U (t)dWk* (T ) dWm (T )

k ,m

and since dT (U* (T) U(T)) = 0, we have

∑ ( Pk*dWk* (T ) + Pk dWk (T )) + ∑ Pk*Pm dWk* (T ) dWm (T )

=0

k, m

k

Note: dWk (T), dW*k (T)

Commute with U (T), U* (t), Yin (t) 2 Now, put X ∈ L ( h ) (system observable) and X (t) = U * (t )( X ⊗ I )U (t), t ≥ 0

Then for t ≥ s ,

[X (t) = Yout (s)] = [U * (t )( X ⊗ I ) U (t),U * (t ) (I ⊗ Yin ( s )) U (t)] = U * (t )[ X ⊗ I , I ⊗ Yin ( s )]U (t ) = 0

i.e.

Hence {Yout (t ), t ≥ 0} is a non-demolition measurement for { X (t), t ≥ 0} in the sense of Belavkin, Now, dYout (t ) = dYin (t ) + dU * (t ) dYin (t )U (t) + U * (t ) dYin (t ) dU (t) =

∑ λ j dA j + λ j d A j + ∑ Ckj d Λ kj

(

)

+ U *  ∑ L(j1)*dA*j + L(j2)*dA j + ∑ Skj*d Λ kj   j   ∑ λ r dAr + λ r dAr*  + C.C. U    r

Problem: (a) Find

Tr {( A + ε B)log ( A + ε B)} − Tr ( A log A)

Up to O (e) when A, B > 0.

476

Stochastics, Control & Robotics

(b) Find Tr { A (log ( A + ε B ) − log A)} Up to O(e) where A, B > 0. Remark: The first formula can be used in finding the rate of change of the entropy (Von-Neumann) of a quantum system whose density matrix satisfies the Sudarshan-Lindblad equation. dρt 1 = − i [ H , ρt ] − {L * Lρt + ρt L * L − 2 Lρt L*} dt 2 and the second can be used to find the relative entropy of rt per unit time. 

[32] Problem on Rigid Body Motion: Consider the two link Robot with each link being a 3-D top. We have seen that the group of motions is defines by   R2 R1

G =   0

Show that if we define

Then

 R1 − R2 R1  : R1, R2 ∈SO(3)  R1  

  R S − R : R, S ∈SO(3) =   S   0   R S − R S 

T (R, S) =  0

 R2

T (R2, S2).T (R1, S1) =  0

S2 − R2   R1 S1 − R1  S1  S2   0

 R2 R1 S2 S1 − R2 R1  =   = T ( R2 R1, S2 S1 ) S2 S1  0  

[33] Problem from Revuz and Yor: lim

x→∞

x + a x  n t   − δ  Bs −   ds δ  Bs −   ∫ 0  2   n  n  = γ (aLt ), t ≥ 0

in distribution, i.e.

= lim

x→∞

n  Lt 2 

( x+ a) n

 − Ltx / n  = γ (aLt ), t ≥ 0 

In distribution, where g is a Brownian motion. 