The inverse Berreman problem Bill Lionheart and Chris Newton School - - PowerPoint PPT Presentation

Advances in the determination of liquid crystal director profiles

The inverse Berreman problem

Bill Lionheart and Chris Newton

School of Mathematics University of Manchester and HP Labs Bristol

Advances in the determination of liquid crystal director profiles – p. 1/13

Overview

• In stratified anisotropic media Maxwell’s equations can be replaced by Berreman’s ODE.
• We want to solve for the unknown permittivity tensor in Berreman using data from a range of

incidence angles.

• Is such data sufficient to uniquely determine the parameters?
• How ill-conditioned is the inverse problem and why?

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Berreman’s formulation

Berreman field vector X = (Ex, Hy, Ey, −Hx)T ∂ ∂z X = − iω c MX

(1)

where the matrix M is M = 2 6 6 6 6 6 6 4 − ǫ13

ǫ33 ξ

µ0c ǫ33−ξ2

ǫ33

− ǫ23

ǫ33 ξ

ǫoc “ ǫ11 − ǫ132

ǫ33

” − ǫ13

ǫ33 ξ

ǫ0c “ ǫ12 − ǫ13ǫ23

ǫ33

” µ0c ǫ0c “ ǫ12 − ǫ13ǫ23

ǫ33

” − ǫ23

ǫ33 ξ

ǫ0c „ ǫ22 − ǫ2

23

ǫ33 − ξ2

« 3 7 7 7 7 7 7 5

(2)

The final value problem for this linear ODE can be solved to give the linear relationship between ‘initial data’ X(0) and X(z) = P(z)X(0), where P(z) is a propagation matrix. We will write P(d) as simply P.

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Perturbation formula

The inverse problem is to determine ǫ from P for a range of incident angles ξ, a kind of generalized inverse spectral problem for an ODE. We consider a perturbation δM in the material properties from some initial guess. The Berreman matrix changes to M + δM. We need to know how the transmitted light depends on the perturbation in M to first order. The final data can be obtained by solving this ODE for its final value δX(d), or from the integral δX(d) =

d

Z G(d, z)δM(z)X(z)dz

(3)

where G is the Green function of the ODE.

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Perturbation formula, constant background

For the case of a constant background M we can solve explicitly, and while an over simplification it gives us some insight X(z) = − iω c exp( iω c Mz) @

z

Z exp(− iω c Mz′)h(z′)dz′ + X(0) 1 A

(4)

• r G(z, z′) = exp(− iω

c M(z − z′)) so that (3) becomes

δX(d) = iω c exp(− iω c Md)

d

Z exp( iω c Mz)δM(z) exp(−i ω c Mz)X(0) dz

(5)

as for non-lossy materials Mξ is real, we can consider (5) as a generalized Fourier transform of δM.

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Fourier transform of the unknowns – sort of!

Eigenvalues qi of M, distinct, real and q1 = −q2, q3 = −q4. U matrix of eigenvectors of M, Q = diag(qi). Equation (5) becomes δX(d) = (iω/c) exp((iω/c)Md)

d

Z U exp(−iQz)U −1δM(z)U exp(iQz)U −1X(0)dz

(6)

Set A = U−1δM(z)U. Note exp(−iQz)A exp(iQz) has elements ei(qi−qj)zAij. As we know all possible pairs of initial and final data X(0), X(d), we know the linear response T defined by δX(d) = TX(0) hence we know the matrix Y = −i(c/ω) exp(−(iω/c)Md)U −1TU =

d

Z exp(−iQz)A exp(iQz) dz with (7) Yij =

d

Z ei(qi−qj)zAijdz = d Aij(qi − qj)

(8)

where the b denotes the Fourier transform.

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Fourier transform of the unknowns, continued

The qi are functions of ξ and in general varying ξ over an interval will vary qi − qj over an interval, for i = j. This data over a variety of incident angles gives us some information about the deviation of the permittivity tensor from our initial constant assumption in terms of the deviation of the measurements from those we could calculate for the constant case.

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Orthorhombic example

We can best illustrate this by an example that is tractable analytically, the orthorhombic case ǫij = 0, i = j. M reduces to 2 6 6 6 6 6 4 µ0c −

ξ2 cǫ0ǫ33

ǫ0cǫ11 µ0c ǫ0c `ǫ22 − ξ2´ 3 7 7 7 7 7 5 The eigenvalues are q1 = √ǫ11 s 1 − ξ2 ǫ33 , q2 = −q1, q3 = √ǫ22 s 1 − ξ2 ǫ22 , q4 = −q3

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Orthorhombic example, δM

We consider a completely general perturbation δij from this orthorhombic case, and try to see if this can be detected by data from a range of ξ. The eigenvectors can be calculated explicitly, and of course are constant in z although they vary with ξ. We then calculate δM = 2 6 6 6 6 6 4 − ξ δ13

ǫ33 ξ2 δ33 cǫ0ǫ2

33

− ξ δ23

ǫ33

cǫ0δ11 − ξ δ1,3

ǫ33

cǫ0δ12 cǫ0δ12 − ξ δ23

ǫ33

cǫ0δ22 3 7 7 7 7 7 5

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Orthorhombic example, what can we find?

Look at coeff ofδij in U−1δMU. We will be able to obtain Fourier data from off-diagonal terms. coefficient of δ11 = q3 2cǫ0ǫ11 2 6 6 6 6 6 4 −1 1 −1 1 3 7 7 7 7 7 5 coefficient of δ12 = 1 2cǫ0 2 6 6 6 6 6 4 − 1

q3 1 q3

− 1

q3 1 q3

− q1

2ǫ11 q1 2ǫ11

− q1

2ǫ11 q1 2ǫ11

3 7 7 7 7 7 5

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Orthorhombic example, what can we find? cont

coefficient of δ13 = − ξ cǫ0ǫ33 2 6 6 6 6 6 4 1 1 3 7 7 7 7 7 5 coefficient of δ22 = cµ0 2q3 2 6 6 4 −1 1 −1 1 3 7 7 5

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Orthorhombic example, what can we find? cont

coefficient of δ23 = 2 6 6 6 6 6 4 −

ξ√ǫ11 2√ǫ33q1q3 ξ√ǫ11 2√ǫ33q1q3 ξ√ǫ11 2√ǫ33q1q3

−

ξ√ǫ11 2√ǫ33q1q3

−

ξ 2ǫ33

−

ξ 2ǫ33

−

ξ 2ǫ33

−

ξ 2ǫ33

3 7 7 7 7 7 5 coefficient of δ33 = ξǫ11 2ǫ33q1 2 6 6 6 6 6 4 −1 −1 1 1 3 7 7 7 7 7 5 We see that for all except δ13 there are non-trivial off diagonal components so that for data using incident light in this plane for an interval of incident angles the Fourier transforms of all other δij are determined for some interval of frequencies, hence by analytic continuation the perturbation is determined uniquely everywhere. Although of course their recovery in the presence of noise will be unstable. Note now that by rotating the plane of incidence through a right angle, the x and y axis swap roles and we can with the data for these two planes recover all the perturbed coefficients.

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Conclusions

• The linearization or Jacobian can be calculated by solving an ODE.
• Boundary data for a range of incident angles contains information about a depth dependent

permittivity tensor.

• The illposedness is similar to analytic continuation so is severe, but the problem is better

conditioned when the range of incident angles produces a wider separation of eigenvalues of the Berreman matrix.

• Sufficiency of data shown for perturbation to orthorhomic, where data is full optical transfer
• matrix. Gives some insight in to real problem.

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