Lecture 23 Notes: Layered Materials and Photonic Band Diagrams

Lecture 23 Layered Materials and Photonic Band Diagrams Today 1. Transfer matrix approaches to layered media 2. Periodic layered media 3. Bloch solutions and photonic band diagrams Questions you should be able to answer by the end of today’s lecture 1. How to derive the transfer matrix that describes the field transformation at interfaces? 2. Transfer matrix approach to solving periodic systems of dielectric materials. 3. Eigenvalue problem and its solutions 4. Photonic band diagrams and band gaps 1 Transferr Matrix approach: gen neral treatm ment of multtilayered op ptical materrials. Consider a 3--layered matterial with ann index of reefraction proofile:  n, x0  1 n(x)   n2 , 0  x  d  dx  n3 , As discussed in the previoous lecture, the z compoonent off the waveveector kz   does not chhange througghout th he problem ( consequencee of phase coontinuity). This leeads to a sim mple dependeence for the eevolution inn z diirection:   i  t z  EE x e  However for x and y directions the sccenario is moore omplicated a s we have noticed in thee previous co lecture when we deriived the refleection and trransmission coefficients for the single interface. Now we would d like to find d a general way to relate any pair of electric andd magnetic field amplituddes in any layerr to those in any other layer. a As usual we would break our eleectric (and co onsequently magnetic) fi field componnents into s- and p-polarizzations. Heree we will onlly consider s-polarized e lectric field for simpliciity (y-directiion for E and d x-z plane, perpendiculaar to waveveector for H). For s-pollarized field::  E1eik1 x x  E '1 eik1 x x , x  0  E  x    E2 eik2 x x  E '2 eik2 x x , 0  x  d  ik3 x  x d  ik xd  E '3 e 3 x   , x  d  E3e Where E1, E2 , E3 arre the amplittudes of com mponents of the electric field propaggating forwarrd and E '1, E '2 , E '3 aree the amplitu udes of the reflected com mponents proopagating baackward. If we werre to relate the amplitudes in the lay yer 1 to the laayer 3, we will have to consider propagation across tw wo interfaces 1-2 and 2-3 3 and the sim mple propaggation througgh the neous layer 2. homogen The effecct of propagaation through h a medium of index n2 and thickneess d is captuured by the propagation matrix:  iik2 x d e P2    0 0 eik2 x d     2 Which is just simple phase accumulation for a plane wave propagating distance d in a homogeneous medium of index n 2 . The interface matrices can be derived from the boundary conditions akin to the way we treated a single interface:   E1||  E1||  E1y  E '1y  E2 y  E '2 y H1||  H 2||  H1z  H '1z  H 2 z  H '2 z  n1 cos1E1y  n1 cos1E '1y  n2 cos 2 E2 y  n2 cos 2 E '2 y Then can rewrite the equations above in a matrix form: E1y  E '1y  E2 y  E '2 y n1 cos1E1y  n1 cos1E '1y  n2 cos 2 E2 y  n2 cos 2 E '2 y       E1y     E2 y 1 1        E '    n2 cos 2 n2 cos 2  E ' 2y   1y     E   E   E   E  1y   2y  1y   2y  1     D12  ; D  D21D1 D2 D1  E '1y   E '2 y   E '1y   E '2 y  12         Then the matrix D12 can be simply found as:  1 1   n1 cos1 n1 cos1   1 D12  D21D1   cos 2 n  2  n cos  n cos 2 1 1  2 2n2 cos 2  D12   n cos 2  n1 cos1  2  2n2 cos 2  1 n2 cos 2 1      1   n1 cos1      1  n1 cos1      1 2 1 2   E   D  1y 1   E '1y   1 2n2 cos 2  1 2n2 cos 2   E   D  2y 2   E '2 y      1 1  n cos n cos 1 1 1  1       n2 cos 2  n1 cos1   2n2 cos 2   n2 cos 2  n1 cos1   2n2 cos 2  Then in order to get find the connection between the electric field amplitudes in layer 1 and layer 3 we need to do the following:  E   E  D23 P2 D12  1    3   E'   E'  3   1   3     Periodicc Medium We have previously considered electronic prroperties of p eriodic matterials. We have found thhat periodiciity led to the gaps in the electronic en nergy levelss. Let’s now consider a material that is periodic on the scale similar to th he wavelength of light. T hese materiials are calleed “photonicc crystals”. The index of refractio on is a perio odic function n:  n  2 n  x    n1 l 1 a  x  l 1 a  d2 la a  d1  x  la a n  x  la  n  x  d1  d2  a The soluttion in each medium hass the form: E  E  x  ei  t   z  Focus iniitially on thee x dependen nce of the so olutions:  ik x la ik xla E1l e 1 x    E '1l e 1 x    E  x   ik xla d1  ik xlad1   E '2l e 2 x   E2l e 2 x   la  d1  x  la l  1 a  x  l 1 a  d2 Recall th hat interface matrices D have the folllowing form m: 4  1 Dik  Dk1Di    nk cos k 1 nk cos k 1      1   ni cosi   n cos  n cos k i i  k   2nk cos k 1  ni cos i   nk cos k  ni cosi   2nk cos k  nk cos k  ni coos i   2nk cos k  nk cos k  ni coosi   2nk cos k  To get th he matrix mu ultiplication right it is necessary to caarefully conssider the forrm of the solution: 1 x  la  denotes the solution in thhe medium 1 at the interrface x  la (Left Hand Side = LHS) 1 HS) x  la  d1  denotes the solution in the mediuum 1 at the iinterface x  la  d1 (RH Consequently: E  x  xla1  E1l  E '1l E  x  xllad 1  E1l eikk1 x d1  E '1l eik1 x d1 1   ik1x d1  e   P1    0      0 eik1 x d1 From thee previous lecture we kno ow that relatting the fieldds across thee interface: E  x  xlad 2  E2l  E '2l 1  E 2ll D2   E '2 l    E   D1P1  1l     E '1l   E    2l   E '2l     E   D21D1P1  1l     E '1l     Now we are at the LH HS of the x  na  d1 intterface (i.e. iin medium 2), let’s propaagate the waave in medium 2 to the RHS of interfface: x  la  d1  d2   l 1 a . E  x  xlad 2  E2l  E '2l 1 E  x  x l1 a2  E2l eik2 x d2  E '2l eik2 x d2     ik2 x d2  e   P2    0   0 eik2 x d2    And oncee again we match the fieelds across th he interface t o extract E1l1 and E '1l1 : E  x  x l1 a2  E2l eikk2 x d2  E '2l eik2 x d2   E  x  x l1 a1  E1l1  E '1l1    E 1ll1 D1  E'  1l1     D P  E2l 2 2    E '2l    E1l1    E'   1l1     D 1D P  E2l 1 2 2   E '2l       Then puttting it all tog gether we fin nally get:  E  1l1  E'  1l1     D1D P D 1D P  E1l 1 2 2 2 1 1   E '1l   5     Now recaall the Bloch h theorem th hat we have used to get t he electroniic wavefuncttion for the periodic material. Wee can use thee same logicc for the optiically perioddic structure or a “photonnic crystal”. Accordin ng to Bloch theorem the solution to a periodicallly constraineed system haas to have a form:  E  1l1  E'  1l1     eiKa  E1l   E '1l       Combinin ng these two o equations leads us to th he followingg expression::  E 1l D D2 P2 D D1P1   E '1l  1 1 1 2   E   eiKa  1l     E '1l   E   M  1l  E '1l      E   eiKa  1l   E '1l       M  D11D2 P2 D21D1P1 The matrrix elements are given by y the followiing expressioon:  M 11 M   M 21  M12   M 22    k  1 k M11  eiik1 x d1 cosk2 x d2  i  1x  2 x  sin k2 x d2  2  k2 x k1x    1 k k  M12  eikk1x d1 i  1x  2 x  sin k2 x d2 2  k2 x k1x  1 k k  M 21  eeik1 x d1 i  1x  2 x  sin k2 x d2 2  k2 x k1x    k  1 k M 22  eikk1x d1 cosk2 x d2  i  1x  2 x  sin k2 x d2  2  k2 x k1x     n cos1  n cos 2 k1x  k1 cos1  1 , k2 x  k2 cos 2  2 , k1 sin1  k2 sin  2   c0 c0 k1x   n1 c0 2 2 n   n  1 sin 1   1    2 , k2 x   2    2  c0   c0  2 Each onee of the matrrix elements depends on ω and β. 6 Photonic band structures. The matrix equation above is simply an eigenvalue problem. For a 2x2 matrix the eigenvalues are given by: eiKa  1 1 2  M11  M 22    M11  M 22   M12 M 21 2 4 This equation defines the dispersion relations for the Bloch wavenumber K and  and  .  M  M 22  eiKa  eiKa 1 1  cos Ka   M11  M 22   K  ,    cos1  11    2 2 a 2 Here we have taken into account that matrix M has a determinant of unity (easy to check yourself), which means both eiKa and eiKa are the eigenvalues. The eigenvectors then are:  E  10   E '10   M 12    iKa   e  M 11     Now we can distinguish between two qualitative regimes: M 11  M 22  1  K is real, waves are propagating 2 M 11  M 22 m  1  defines the band edges  cosKa=1  K  2 a M 11  M 22  1  K is complex, waves are evanescent (decaying in amplitude) 2 Then the general solution:  E1lK  x  eiKla  E10 e ik1 x  xla  E '10 e ik1x  xla e iKla Recall that: k1x   n1 c0 2 2 n   n  1 sin 1   1    2 , k2 x   2    2  c0   c0  2 Lets first consider the case of   0 which corresponds to normal incidence (incidence angle of 0). Then: k1x  k1   n1 c0 , k2 x  k2   n2 c0 Then the dispersion relation  vs. K can be expressed as: 7 cos Ka   1   M 11  M 22 1 ik1d1  1 k k  1 k k  cos k2 d 2  i  2  1  sin k2 d 2  e ik1d1 ccos k2 d 2  i 2  1  sin k2 d 2   e 2 2 2  k1 k2  2  k1 k2   2    1n n  cos Ka  cos k1d1 cos k2 d 2   2  1  sin k1d1 sin k2 d 2 2  n1 n2  Finally we find: coss Ka  cos  n1d1 c0 cos  n2 d2 c0 1  n n  n d n d   2  1  sin 1 1 sin 2 2 2  n1 n2  c0 c0 Figure removed due to copyright restrictions. Bloch waves corresponding to the A and B solutions for frequency at the edge of the Brillouin zone: Unknown source. On the riight picture A and B are the Bloch waves correspponding to thhe A and B solutions forr  2 g frequency y at the edgee of the Brilllouin zone  K   . Herre g  .  a 2 When   0 , i.e. the wave hits th he periodic medium at ann angle otheer than zero, we can findd plot the band diagram for     : Figure removed due to copyright restrictions. TM and TE polarization: Unknown source. 8 Reflection from a dielectric mirror. Consider a situation where N periods are assembled into a stack this arrangement is also called a dielectric or distributed Bragg reflector (DBR) and is used extensively in applications ranging from laser cavities to telecommunication filters.  E  0  E' 0       M 11 M 12   M M 22 21       M 21 2 R  rN  M 21 2 N  E  N   E 'N     2  sin Ka     sin NKa  2 This means we can achieve nearly perfect reflection from a material with a large number of layers: M 21 2 R  rN  M 21 2 2  sin Ka     sin NKa  M 21   Ka0 2 M 21 2 2  Ka     NKa  9 2 M 21  M 21 2 2 1   N N  2    M 21 M 21 2 2 1