Lecture 5 - Electromagnetic Waves IV 19

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1 Lecture 5 - Electromagnetic Waves IV 9 5. Electromagnetic Waves IV 5.. Symmetry in EM In applications, we often have symmetry in the structures we are interested in. For example, the slab waveguide we discussed above has reflection symmetry with respect to the center axis of the waveguide. It also has continuous translational symmetry along the direction. In periodic structures, we have discrete translational symmetry. In photonic crystals, symmetry plays an important role of determining the dispersion relation. In this section, we briefly discuss some of the consequences due to symmetry. We will show that structures that exhibit the same symmetry have common properties in their solutions. From Maxwell s equations, we have: H = H ε( r ) (5.) Previously, we have studied the solution of this wave equation using its electric field counterpart. The reason of not using the electric field equation will be explained later. But now we would lie to study the relationship between the symmetry of the structure (described by a real function ε ( r )) and the solution H. We define the operator ˆΘ as follows. ˆ Θ = ε (5.) We can easily verify that ˆΘ is both a linear and a Hermitian operator. That means if H and H are both solutions of (5.) and α and β are constants, ˆΘ satisfies: Θ ˆ( αh ˆ ˆ + βh) = αθ H+ βθh drh Θ ˆ H = dr ΘˆH H (5.3) ( ) To verify the hermicity of ˆΘ, we carry out the integral by parts in (5.3) as follows 3. drh H ε = dr H H = dr H H ε ε ( ) ( ) = dr H H ε (5.4) 3 We use the following identity (integration by parts): is the surface of Ω. ˆ 3 3 F d r = F nds + Fd r Ω Γ Ω where Γ EECS 598- Nanophotonics and Nanoscale Fabrication Winter 6, P.C.Ku

2 Lecture 5 - Electromagnetic Waves IV where surface terms vanish because it is reasonable to assume the field vanishes at infinity. Because ˆΘ is a linear Hermitian operator, it has several properties:. ˆΘ has real eigenvalues --> ω is real.. Solutions to (5.) that correspond to different eigenvalues are orthogonal to each other. That is drh H =. You can verify that if instead of magnetic fields, you use the electric field counterpart of (5.), you get: D = ε D (5.5) But it can be shown that the operator (5.5) in the following discussions. is not Hermitian. Hence we will use (5.) instead of ε We discuss two types of symmetry operations that are often encountered in the nanophotonic problems. One is the translational symmetry and the other is the symmetry with respect to a fixed point, i.e. rotation, inversion, and reflection symmetry. We will find the similarities between the EM waves and the electrons in a solid state lattice Translational symmetry (Bloch s theorem) If the periodicity of the structure can be described by a set of unit vectors { ai, i =,,3}, we can expand the inverse of the dielectric function ε into a Fourier s series as follows. ( ) ( ) i ε r = ε e r (5.6) where represents a discrete set of vectors comprising of = lb+ mb + nb3 (l,m,n are integers and ai bj = πδ ij.) We say that b i s are the unit vectors for the reciprocal lattice. If ε ( r ) is real everywhere, ε ( ) = ε ( ). But the following discussions hold true even when ε ( r ) is complex. Now we want to solve for H in (5.) with ε ( r ) given by (5.6). We first expand H in the plane-wave basis as in (.5): ( ) ( ) i H r = dh r e (5.7) Substituting (5.6) and (5.7) into (5.), we get: i( + ) r i r dε ( ) H( ) e + ω µ dh ( ) e = (5.8) ε ( ) H ( ) + ω µ H ( ) = ( ) ( ) EECS 598- Nanophotonics and Nanoscale Fabrication Winter 6, P.C.Ku

3 Lecture 5 - Electromagnetic Waves IV (5.8) is a set of linear eigenvalue equations with H ( ) lined by the reciprocal lattice. Therefore the solution must only be a superposition of these lined components, that is: H r H e v r e i( ) r i r ( ) = ( ) ( ) (5.9) where the Bloch s function v = H ( ) e i r has the same periodicity as ε ( r ) as we can easily verify as follows. v ( r + la + ma + na ) = H ( ) e 3 = H ( ) e e i ( r + la+ ma+ na3) i r πi i r = H ( ) e = v (5.) On the other hand because of the periodicity in the reciprocal lattice, we only need to consider in the first Brillouin one which is the unit lattice of the reciprocal lattice. As an example in a D periodic structure with a period Λ along the direction, the unit vector a = Λˆ and the unit vector in the reciprocal space is b = π / Λˆ. The dielectric function can be expanded as: i( π / Λ) i( π / Λ) ( ) ( e e ) The TM solution ( H = yh ˆ ) can be expanded as follows. ε = ε κ + κ + κ + (5.) π m H H e i( π m/ Λ) ( ) = ( ) m (5.) m= Λ Substituting (5.) and (5.) into (5.) and neglecting all the higher order terms in (5.), we get 4 : We have for m=: πm π( m+ ) π( m ) ε Hm = κ Hm + κ Hm+ + κ Hm Λ Λ Λ (5.3) ( ) π π ε κ H = κ H+ κ + H (5.4) Λ Λ For m=: π 4 π ε κ H = κ H + κ H (5.5) Λ Λ When π / Λ, H represents a counterpropagating wave along the direction. Let = ( π / Λ )( + δ ) with δ. In this case the dominating terms in (5.4) and (5.5) are H and H. We have: ω µε κ δ ε κ δ κκ δ δ Λ Λ Λ 4 π π π ( + ) ( ) = ( + ) ( ) (5.6) 4 [ π ] H ( ) yˆ = i( m/ Λ ) H ( ) yˆ EECS 598- Nanophotonics and Nanoscale Fabrication Winter 6, P.C.Ku

4 Lecture 5 - Electromagnetic Waves IV If ε ( r ) as: is real, κ =. If we only eep terms up to κ δ, we can derive the dispersion relation from (5.6) π κ ± κ± κ / κ ω = c κ ± κ δ Λ ( κ ± κ) (5.7) 5... Symmetry with respect to a fixed point Let ˆT be a symmetry operator on ε ( r ) with respect to a fixed point, that is ˆ Tε = ε. We now that if Tˆ, ˆ Tˆ ˆ ˆTˆ Θ = Θ Θ =, ˆΘ and ˆT have a simultaneous set of eigenfunctions. This is indeed the case for any ˆT that is a combination of rotation, inversion, and reflection symmetry on ε ( r ). ˆT can be written in a matrix form. For example, a rotation operation along the -axis by an angle ϕ can be written as follows. cosϕ sinϕ R = sinϕ cosϕ When ˆT is acting on a vector field F : ˆ TF = RF( R r ) We will discuss more details later when we discuss photonic crystals. (5.8) (5.9) 5.. Scaling in EM We now that EM waves can be classified by their wavelengths. In this section, we study the scaling behavior of EM waves. We as ourselves the following question. If we design a device structure that wors for infrared waves, what do we do to mae it to wor for visible lights? In general, if H is a solution to: r r H = H (5.) ε If we scale the structure such that the new dielectric function is identical to the original dielectric function in (5.) but with its magnitude and linear dimension scaled as follows. ε ' = mε ( sr) (5.) where m and s are two scaling factors. For example, if ε ( r ) is a period function with a period of a. The new dielectric function will have the dielectric constant m times bigger and with a period of a/s. If we mae the coordinate transformation in (5.) as r' = r / s and use r' = s r, (5.) becomes: r' r' H ( sr ') = H ( sr ') (5.) s ε( sr') s Using (5.) in (5.), we get: EECS 598- Nanophotonics and Nanoscale Fabrication Winter 6, P.C.Ku

5 Lecture 5 - Electromagnetic Waves IV 3 m r' r' H( sr') = H( sr') s ε '( r') s (5.3) ωs r' r' H ( sr ') = µ H ( sr ') ε '( r ') m We see that with the new dielectric function defined by (5.), H ( sr ') is a solution. That is if H is a solution for a dielectric function ε ( r ), H ( sr) is a solution to a scaled problem with a dielectric function given by mε ( sr) but with a frequency at ω s / m. For example, if we design an antireflection coating using a periodic structure at a certain frequency ω, we can reduce the period of the structure by two times and the same device will wor at a frequency ω. EECS 598- Nanophotonics and Nanoscale Fabrication Winter 6, P.C.Ku