Supporting online material for: A Chiral Route to Negative Refraction

Transcription

1 Supporting online material for: A Chiral Route to Negative Refraction J. B. Pendry Department of Physics, The Blacett Laboratory, Imperial College London, London SW7AZ, UK page 1

2 Let us suppose, Dispersion in a Resonant Chiral Medium Dx EE EH Ex D y EE E EH y D z EE EH E z = Bx HE HH H x B y HE HH H y B HE z HH H z where both E and H lie along x-axis. In an isotropic medium we could imagine a wave propagating with directed along the z-axis and with E and H in the x-y plane: z y Hx Dx H = z xhy = Dy () y x H z D z z y Ex Bx E = z xey = By (3) y x E z B z or in 6-matrix form, z y Ex Dx z x E y D y y x E z D z = z y Hx Bx z x H y B y y x Hz Bz EE EH Ex EE EH E y EE EH Ez = HE HH H x HE HH H y HE HH H z Let us assume that the wave vector lies in the xz-plane so that y = then, = sin θ, =, = cosθ (5) x y z (1) (4) page

3 If we assume helical polarisation: E cosθ E cosθ ie ie E sinθ E sinθ F = = (6) cosθ cosθ H H sinθ sinθ Ecosθ Ecosθ ie ie Esinθ Esinθ F = = (7) cosθ cosθ H H sinθ sinθ then we have for the polarisation, z y z x y x F z y z x y x z y E cosθ z x ie y x E sinθ = z y cosθ z x H y x sinθ EE EH E cosθ EE EH ie EE EH E sinθ = HE HH cos θ (8) HE HH H HE HH sinθ page 3

4 Because the system is presumed isotropic we can solve for EH,, assuming θ = in which case the above reduce to, i E EE EH E i = HE HH (9) or, 1 1 EE EH i E 1 1 i = HE HH 1 1 EE EH 1 E i = (1) HE HH 1 1 EH EE E E i 1 1 = HH HE and for the - polarisation, 1 1 EE EH i E 1 1 = HE i HH 1 1 EE EH 1 E i = (11) HE HH 1 1 EH EH E E i 1 1 = HH HE Solving (1) for the ve frequency, = det = 1 1 EH i EE ( EH i ) HH EE HH HE i where we have exploited the fact that we now from what follows that, hence, 1 1 HE EH = (13) ± EH HH EE i = ± i (14) (1) ± = i EH ± HH EE which we express in terms of a refractive index, (15) page 4

5 n = = ieh HHEE c We have adopted the notation that ± is the wave vector corresponding to the ve polarised wave with ± sign to the group velocity. i.e., = (17) pol; vg The meaning of positive polarisation is that the projection of the polarisation on the z-axis is positive. We have assumed that is the solution with ve group velocity. Solving (11) for the ve frequency, = det = 1 1 EH i EE ( EH i ) HH EE HH HE i where we have exploited the fact that we now that, hence, 1 1 HE EH = (19) EH HH EE i ± = ± i () ± = i ± EH HH EE which we express in terms of a refractive index, n = = ieh HHEE c We have adopted the notation that ± is the wave vector corresponding to the ve polarised wave with ± sign to the group velocity. The meaning of negative polarisation is that the projection of the polarisation on the z-axis is positive. We have assumed that is the solution with ve group velocity. The impedance can also be retrieved, 1 EE 1 EE 1 EH 1 EH E Z = = = H i i ε µ Z (16) (18) (1) () (3) First consider a non-resonant chiral medium with a response, EE EH 1 = HE HH (4) page 5

6 and over the range of frequencies of interest we shall assume that is roughly constant. This tensor gives rise to normal optical activity with a different refractive index for left and right have polarised light as shown below. Figure 1. Dispersion curves for the wave vector in a normal chiral medium. We have 1 assumed that EH is a pure negative imaginary number. Note that ± always defines a positive group velocity. Next consider a second medium comprising a set of electrical resonators dispersed uniformly throughout space such that the response is given by, where, ε ( ) = 1 ( ) ( ) P ( ) α ε = 1 = 1 (6) This might be typical of a set of Mie-resonating spheres, for example. (5) page 6

7 Figure. Dispersion curves for the wave vector in a non-chiral medium filled with a set of resonators, frequency. We now tae the chiral medium and fill it with the same set of resonators of the second medium to create a resonant chiral medium where, P EE EH 3 = HE HH From which, ( ) 1 1 HH EH 3 = and from above, hence, ± ( P ( ) EE ) HH EH HE HE P ( ) EE = i EH ± HH EE (7) (8) (15) page 7

8 ± Similarly, ± ( P P ( ) ( ) ( P i P ( ) EE HH EH HE EH HH EE EH ± ( EE ) HH EE HH EH = = i ± = i ± EH EE HH ( ( ) ( P ( ) ( P P EE HH i EH P EE HH i EH = i ± EH EE HH = EE HH i ± EH ( P P ( ) ( ) ( P i P ( ) EE HH EH HE EH HH EE EH ± ( EE ) HH EE HH EH = = i ± = i ± EH EE HH ( ( ) ( P ( ) ( P P EE HH i EH P EE HH i EH = i ± EH EE HH = EE HH i ± EH These equations are plotted schematically in the following figure. Note that the range of frequencies over which we see negative refraction depends strongly on EH. (9) (3) page 8

9 Figure 3. Dispersion curves for the wave vector in a chiral medium filled with a set of resonators, frequency. page 9

10 A Resonant Chiral Medium (perfect conductors) model: an array of cylinders each with a helical winding of metal foil Figure 4. Each cylinder is a Swiss roll structure wound in a helical fashion. Each layer of foil is separated from the next by a distance d, and the total thicness of foil is N layers. This is a right handed spiral. Figure 5. The helix is wound from a sheet of foil, one period of which is defined in the figure above. Figure 6. The Swiss rolls are arranged on a square lattice, side a. page 1

11 The important point is that there is a gap which prevents current from flowing around any one ring. However there is a considerable capacitance between the two rings which enables current to flow, Figure 7. The Swiss rolls viewed down the z-axis where there chiral structure is not apparent. Define three dimensionless quantities, 3 π r a πr A= ( N 1, ) B=, D= 1 dc c a to give, 4π ( 1 A) tan ( θ ) ( 1 AD) tan ( θ ) B = µ BD 1 1 HH 1 HE i r tanθ 1 iπr tanθ = = 1 AD tan BD 1 AD tan BD ( ) ( θ) ε µ λ ( ) ( θ) (31) (3) (33) 1 1 BDε EE = BD 1 EH ( 1 AD) tan ( θ ) tan ( θ) ( ) ( θ) tan ( θ) ( ) ( θ) ir 1 iπr = = 1 AD tan BD 1 AD tan BD ε µ λ where λ is the free space wavelength of the radiation. For a strong chiral effect we require that all four quantities are of the same order of magnitude which effectively requires that, tan ( ) r (34) (35) θ π λ (36) However we also require the dimensions of our structure to be small: r << λ (37) which inevitably leads to small values of θ, the winding angle. page 11

12 In the figures below we show the dispersion and impedance calculated for various parameters. 4 AN = 1.1 RAD = 5.D-3 D = 1.D-4 A =.D- THETA = 5.*(PI/18.) wave vector (m -1 ) frequency (GHz) Figure 8. Dispersion in a chiral medium for the parameters in equation (38). Imaginary component in blue; the light line in vacuo is shown in red AN = 1.1 RAD = 5.D-3 D = 1.D-4 A =.D- THETA = 5.*(PI/18.) impedance/z frequency (GHz) Figure 9. Impedance relative to free space of a chiral medium for the parameters in equation (38). Imaginary component in blue; the light line in vacuo is shown in red. a =. 1 m 4 d = 1. 1 m 3 r = 5. 1 m N = 1 θ= 5 (38) page 1

13 3 AN = 1.1 RAD = 5.D-3 D = 1.D-4 A =.D- THETA = 3.*(PI/18.) wave vector (m -1 ) frequency (GHz) Figure 1. Dispersion in a chiral medium for the parameters in equation (39). Imaginary component in blue; the light line in vacuo is shown in red. Only one half of the bands is shown in this figure...15 AN = 1.1 RAD = 5.D-3 D = 1.D-4 A =.D- THETA = 3.*(PI/18.) impedance/z frequency (GHz) Figure 11. Impedance relative to free space of a chiral medium for the parameters in equation (39). Imaginary component in blue; the light line in vacuo is shown in red. a =. 1 m 4 d = 1. 1 m 3 r = 5. 1 m N = 1 θ= 3 (39) page 13

14 1 AN = 1.1 RAD = 5.D-3 D = 1.D-4 A =.D- THETA = 1.*(PI/18.) wave vector (m -1 ) frequency (GHz) Figure 1. Dispersion in a chiral medium for the parameters in equation (4). Imaginary component in blue; the light line in vacuo is shown in red. Only one half of the bands is shown in this figure. 5 4 AN = 1.1 RAD = 5.D-3 D = 1.D-4 A =.D- THETA = 1.*(PI/18.) impedance/z frequency (GHz) Figure 13. Impedance relative to free space of a chiral medium for the parameters in equation (4). Imaginary component in blue; the light line in vacuo is shown in red. a =. 1 m 4 d = 1. 1 m 3 r = 5. 1 m N = 1 θ= 1 (4) page 14