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OPTOELECTRONIC AND PHOTONIC MATERIALS

Problem set 2

Note:1) Unless otherwise specified, in this entire homework the units will be CGS-Gaussian

2) At the plane boundary between two media the tangential component of the electric field, E, must remaincontinuous, and the tangential component of the auxiliary magnetic field, H, must remain continuous (in theabsence of free currents). Also, the normal component of the magnetic field, B, remains continuous as well as thenormal component of electric displacement field, D in the absence of free charges. With tangential we mean thecomponent parallel to the interface separating the two media, and with normal component we mean the componentperpendicular to the interface. These principles, that we have discussed briefly in class, are known as electromagneticboundary continuity conditions. They can be derived directly from Maxwell’s equation in integral form when theseare applied around the boundary between two media. For further reading on electromagnetic continuity boundaryconditions see ‘Introduction to Electrodynamics, Part II’, by D. J. Griffiths, Greek Edition (CUP) 1997, pp. 57-60(paragraph entitled Συνoριακǫς Συνθηκǫς). All EM problems in the following, which involve boundaries betweendifferent media can be solved by employing some or all of the EM boundary continuity equations, at each boundary.

3) You can use as granted findings/relationships of the previous homework set. However, please do specify thenumber of equation you have used (e.g. from eq. (2) Hwset1)

Problem 1

Reflection and refraction at plane boundary between media of different electric and magnetic properties

Consider an EM plane wave coming from a medium with an electric permittivity ǫ1 and magnetic permeability µ1,hitting the interface of a semi-infinite medium with electric permittivity ǫ2 and magnetic permeability µ2, with anangle θI with respect to the surface normal n (see figure).This is depicted in Fig. 1. The wave reflects at the boundary, with an angle θR with respect to the surface normaln, and refracts (enters into the second medium) with an angle θT with respect to the surface normal n. We havechosen a coordinate system where xy is the plane of the interface, and so z is the direction normal to the interface.We have chosen the coordinate system so that the incident wave vector, KI lies entirely on the xz plane. The xzplane, which represents the plane formed by the surface normal n and the incident wave vector kI is called plane ofincidence. Assume all real (no material losses) and positive permitivities and permeabilities.i) A requirement for the electromagnetic (EM) continuity conditions to hold for any point of the interface plane, isthat the phase of incident, reflected and transmitted EM waves must be equal at all points of the interface plane.This condition, requires

k||,I = k||,R = k||,T (1)

where the symbol || denotes the projections of all respective wave vectors onto the plane boundary (which for thechosen coordinate system of Fig. 1 are along the x direction).Condition (1) would be always valid when an EM wave crosses a boundary between two media.For its proof see: ‘Introduction to Electrodynamics, Part II’, by D. J. Griffiths, Greek Edition (CUP) 1997, pp.126-128 (paragraph entitled: Aνακλαση και Mǫταδoση για Πλαγια Πρoσπτωση).Note also, that the requirement of the electromagnetic continuity condition to hold for any time yields also that thefrequencies of the incident, reflected and refracted waves, are equal, i.e.,

ωI = ωR = ωT = ω (2)

Taking into account eq. (1), along with the relations between the magnitude of the wavevectors and frequencies inmedia 1 and 2, show Snell’s law, namely:

n1sinθI = n2sinθT (3)

where n1 =√

ǫ1µ1, n2 =√

ǫ2µ2 are the refractive indices in medium 1 and medium 2 respectively.

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FIG. 1: Reflection and refraction at plane boundary between the two media (ǫ1, µ1),(ǫ2, µ2). Incident, reflected and transmittedwave vectors are indicated( KI, KR, KT, respectively). Incident, reflected, refracted (transmitted) angles are also indicated(θI , θR, θT , respectively)

ii) Taking into account eq. (1), together with the relation between the magnitude of the wavevector, k, andfrequency ω for medium 1 show

θR = θI (4)

(where the above angles represent magnitude of angle)

iii) Assume that the electric field is perpendicular to the plane of incidence (xz plane). This type of polarization inknown in optics as TE polarization, or S-polarization from the german word senkrecht, which means perpendicular.Calculate the reflection and transmitted coefficients R and T respectively, defined as:

R =n · < SR >

n · < SI >(5)

and

T =n · < ST >

n · < SI >(6)

where the symbol <> refers to the time averaged quantities.Hints:- First show that from eq’s (5) and (6) it can be shown that for the TE polarization case:

R =E2

R

E2I

(7)

and

T =Z1

Z2

cosθT

cosθI

E2T

E2I

, (8)

where EI , ER, ET are the amplitudes (not the rms values) of the electric field of the incident, reflected andtransmitted plane wave. (Note: All amplitudes are also real valued).

Z1 =√

µ1

ǫ1, Z2 =

√

µ2

ǫ2are the impedances for medium 1 and medium 2.

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- Use plane wave expressions as in eqs. (11) or (12) of previous homework set for all waves incident, reflected and

transmitted, but express in terms of amplitude not rms value (i.e. without√

2). Do not get confused with the Rsubscripts in the fields in previous homework set; there it meant rms value, here it is amplitude of reflected wave.- Use the fact that the tangential component of the electric field (along y), and the tangential component of theauxiliary magnetic field (along x) are continuous in the boundary (z=0).- In the boundary continuity conditions at z=0 you need to enter only the amplitudes of the incident, reflected andtransmitted plane wave. You do not need to carry the spatial dependence on x, or the time dependence on t. Wediscussed that the requirement for the continuity conditions at the boundary to be valid ∀x and ∀t lead to eqs. (1)and (2), which as a result imply that the x and t dependence on both hand sides will be the same and so it can becanceled out- For second boundary continuity equation you will use the continuity of the tangential (along x) component of theauxiliary magnetic field. So, H should be expressed in terms of the tangential component of E (along y) in order toget two equations with two unknowns (ER and ET ), for which you can solve. (Use eq. (13) of previous homework).- Find ER in terms of EI , and ET in terms of EI . Eqs. (7) and (8), from which T and R will be calculated, haveonly the corresponding fractions, so the actual value of EI does not need to be known.

Verify that your result can be expressed as:

R = (Z2cosθI − Z1cosθT

Z2cosθI + Z1cosθT)2 (9)

and

T =4Z1Z2cosθIcosθT

(Z2cosθI + Z1cosθT )2(10)

iv) After you derived these expressions show that:

R + T = 1, (11)

signifying conservation of energy.

v) Assume now that it is the magnetic field, H, that is perpendicular to the plane of incidence and not the electricfield E. This is called TH polarization or P-polarization (from parallel because the electric field is parallel to theplane of incidence). Derive again for this case starting from eqs. (5) and (6) the reflection and transmissioncoefficients. Show that you expressions can be written as:

R = (Z1cosθI − Z2cosθT

Z1cosθI + Z2cosθT)2 (12)

and

T =4Z1Z2cosθIcosθT

(Z1cosθI + Z2cosθT )2(13)

Note, that the R and T expressions for TH polarization are different from the corresponding ones for TEpolarization implying that reflection and transmission on an interface between two media will be different fordifferent polarizations of the incoming light.Hints- Work out (5) and (6) to obtain similar expressions to (7) and (8) but in terms of amplitudes of the auxiliarymagnetic field instead (use eq. (23) of previous homework).- Use the continuity of the tangential component of the auxiliary magnetic field, Hy, at the boundary (z=0).- Use the continuity of the tangential component of the electric field, Ex. Before that, express Ex in terms of Hy,using eq. (15) of previous homework.- Again you do not need in these continuity boundary conditions for z=0 to include the x and t dependence of thefields, for the same reason explained in problem (iii)

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vi) Assume normal incidence, i.e, θI = 0, which evidently means also θR = 0. Then from Snell’s law it also followsthat θT = 0.Simplify the expressions you have derived for R and T, for both cases of TE-(S-) and TH-(P-) polarization.- Show that these will be equal. Why is that? Comment. (On the physical reason behind it, not how it resultedfrom the math).- For the special case of non-magnetic media (µ1=µ2), show that the expressions for the reflection and transmissioncoefficients R and T, reduce to the following familiar result:

R = (n2 − n1

n2 + n1)2 (14)

T =4n1n2

(n2 + n1)2(15)

Equations (14) and (15) are known as Fresnel formulas for reflection and refraction.

vii) Brewster’s angleUsing the expressions for the reflection coefficient for both the TE and TH polarization cases, calculate the incidentangle θB for which the reflection coefficient vanishes, i.e. R=0. This angle is known as Brewster angle.

Hints- Give expressions for the cos of the Brewster angle θB .- Read 9.5.1 of Prof. Soukoulis notes.

Subsequently, show that for non-magnetic media i.e., µ1 = µ2 = 1:a) For the case of TE- polarization, there is no Brewster angle and the reflection coefficient vanishes only if ǫ1 = ǫ2.b) For the TH- polarization case the Brewster angle is given by the following simple expression:

tan θB =n2

n1(16)

viii) Total internal reflectionAssume the particular case where n1 > n2. Consider incident angles:

θI > arcsinn2

n1(17)

Verify, that for these angles then Snell’s law gives:

sin θT > 1 (18)

This means that there exist no real angle that can satisfy Snell’s law, or in other words there is no refracted(transmitted) beam. This phenomenon is known as total internal reflection (TIR).

Show that the component of the wave vector inside medium 2 along z, which is the surface normal (dotted redvertical line in Fig. 1) is purely imaginary and is given by the expression:

k2z = iω

c

√

n21 sin2 θI − n2

2 (19)

Eq. (12) implies that the wave inside medium 2 is evanescent. The TIR phenomenon is the basis of AttenuatedTotal Reflection (ATR) experimental set-ups used to produce evanescent waves necessary to excite surface waves onmetal surfaces or on particular structured dielectric surfaces (defect surfaces on bulk photonic crystal structures).

Hint: We also discussed this in class. Use the dispersion relation inside medium 2 along with eq. (1). (Note || is thex-component, as the incident beam lies completely on xz plane).

ix) Impedance matching

Use the simplified expressions derived for normal incidence in subquestion vi) to show that for that case when theimpedances of the two media are equal, i.e. when Z1 = Z2, there is no reflection. Such condition is generally knownas impedance matching.

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x)All angle matched media

Take for granted that the formulas you have derived for transmission and reflection, also hold for set’s of ǫ, µ thatare simultaneously negative. Are there material parameters for which the reflection coefficient becomes zero for allincident angles, besides the trivial solution ǫ1 = ǫ2 and µ1 = µ2?

Problem 2

FIG. 2: A dielectric slab of thickness d, permittivity ǫ2 and permeability µ2 is embedded in a medium with permittivity ǫ1 andpermeability µ1. Plane wave is incident from medium 1. Note, that the angle of the final exit beam, θT will be equal to theangle of the incident beam, θI .

Consider a slab of thickness d, with permittivity ǫ2 and permeability µ2 embedded in a medium with permittivity ǫ1and permeability µ1. Assume that a plane wave is incident from medium 1, with permittivity ǫ1 and permeabilityµ1).

Calculate the reflection and transmission coefficients defined as in (5) and (6) for both polarizations.

Hints- Assume n2 =

√ǫ2µ2 > n1 =

√ǫ1µ1

- Separate conceptually the space in three areas.Area 1: Is the area where the incoming plane wave comes from.Area 2: Is the area inside the slab.Area 3: Is the area after the slabThe three areas are separated be two interfaces. These are the xy plane at z=0 and the xy plane at z=d.

- An incoming plane wave comes from z = −∞ and encounters the first interface, partial refracts and reflects. Thepart of the plane wave that enter the slab will again partly reflect on z=d and the rest will get transmitted througharea 3. The part which will get reflected, will then travel toward the (-z) direction encounter the first interface atz=0 and again partly will reflect and then travel along the (+z) direction and partly refract through into area 1. Thetotal reflected beam in area 1, the total beams inside the slab, and the total beam in area 3, will be a summation ofall the contributions caused by the many contribution of the multi-refrections (see Fig. 9.2 of Prof. Soukoulis notes).

-One approach would be to consider the reflectivities, -ratio of reflected to incident amplitude-, and transmissivities,-ratio of transmitted to incident amplitudes-, of the single interface problem. Then with the aid of that, and takinginto account the phase of the electromagnetic wave when traveling through the slab, δ, the fields in area 1, area 2and area 3 can be written as sum expression in terms of r, t and δ. The infinite sums are actually geometrical

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sequences converging to values giving the fields in all areas and accordingly the transmission/reflection.We will not follow this approach in this homework set. This description was given just for you to understand theproblem better conceptually.

-The description in the above paragraph makes it clear then why the fields in the three area’s can be written as:

TE polarization case

Area 1

Ey(x, z) = [EI exp(ikxx) exp(ik1zz) + ER exp(ikxx) exp(−ik1zz)] exp(−iωt) (20)

Area 2

Ey(x, z) = [CE exp(ikxx) exp(ik2zz) + DE exp(ikxx) exp(−ik2zz)] exp(−iωt) (21)

Area 3

Ey(x, z) = [ET exp(ikxx) exp(ik1z(z − d))] exp(−iωt) (22)

TH polarization case

Area 1

Hy(x, z) = [HI exp(ikxx) exp(ik1zz) + HR exp(ikxx) exp(−ik1zz)] exp(−iωt) (23)

Area 2

Hy(x, z) = [CH exp(ikxx) exp(ik2zz) + DH exp(ikxx) exp(−ik2zz)] exp(−iωt) (24)

Area 3

Hy(x, z) = [HT exp(ikxx) exp(ik1z(z − d))] exp(−iωt) (25)

In the above expressions:

o We have assumed that xz is the plane of incidence, just like in problem 1, hence ky = 0.

o Accordingly, and since the slab has infinite extent in y, there is translational symmetry along y, therefore all fieldswill not depend on y.

o The component of the wave vector along x, kx, is the parallel component of the wave vector, and for the samereason as in problem 1 it will be the same in all areas. Therefore, there is no index 1,2 next to kx in the aboveexpressions. From the incident beam since kI = n1ω/c we can deduce that:

kx = n1ω

csin θI (26)

So, for the same reason as in problem 1 wherever boundary continuity equations are taken, the term exp(ikxx) and

exp(−iωt) can be ignored, because they will cancel out from both hand sides anyhow.o The amplitudes of the reflected wave, ER or HR, the transmitted wave, ET or HT , the downwards wave inside theslab, CE or CH , and the upwards wave inside the slab, DE or DH , are really a sum of many components resultingfrom the multireflection and can be evaluated in this manner as we have described in the previous paragraph.However, as we said, here we are not going to follow this approach for the evaluation of these amplitudes. Instead,we are going to consider here all these aforementioned amplitudes as unknows. You will consider the amplitude ofthe incoming wave, EI or HI , as a known quantity. In other words all other amplitudes must be expressed in termsof EI or HI . You will apply two of the electromagnetic boundary continuity equations for both boundaries, at z=0and z=d. These will give you a system of 4 equations, with 4 unknowns ER, ET , CE ,DE (TE-polarization) orHR,HT , CH ,DH (TH-polarization) which is solvable.

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o Note, that there is no plane wave upward contribution in area 3 since there can not possibly be a reflection comingfrom −∞.

- For the case of TE-polarization, first you take the continuity of the tangential component of the electric field, Ey,at both boundaries. Then you take the continuity of the tangential component of the auxiliary magnetic field, Hx,at both boundaries. You need to express Hx in terms of Ey before you apply the continuity conditions in eachboundary. Use Farday’s law from Maxwell’s equation in it’s time-independent (stationary) form, which will give you:

∂Ey

∂z=

−iωµ

cHx (27)

- For the case of TH-polarization, first you take the continuity of the tangential component of the auxiliary

magnetic field, Hy, at both boundaries. Then you take the continuity of the tangential component of the electricfield, Ex, at both boundaries. You need to express Ex in terms of Hy before applying the continuity condition at acertain boundary. For that you will need Amperes law from Maxwell’s equations in its time-independent form,which will give you:

∂Hy

∂z=

iωǫ

cEx (28)

- The final requested expressions for R and T will contain the z component of the wavevector in media 1 and 2, i.e.

k1z, k2z. Calculate these and show that these are equal to:

k1z =

√

ǫ1µ1ω2

c2− k2

x = kI cos θI , (29)

and

k2z =

√

ǫ2µ2ω2

c2− k2

x, (30)

respectively.

- In this case, < ST > in eq. (6) will be calculated in the area after the second interface of the finite width slab(designated as area 3). In this problem ER and ET could be complex. Show that application of Eqs. (5) and (6) forthis case lead to:For TE polarization

R = |ER

EI|2 (31)

and

T = |ET

EI|2 (32)

For TH polarization

R = |HR

HI|2 (33)

and

T = |HT

HI|2 (34)

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Now, after following the steps hinted in all comments above, work out your math and show that the results for the

transmission and reflection coefficients can be simplified for both polarizations in the following two elegantexpressions:The transmission coefficient, T is:

TTE,TH =1

1 + 14 (χTE,TH − 1

χT E,T H )2 sin2 k2zd(35)

and the reflection coefficient R is:

RTE,TH =

14 (χTE,TH − 1

χT E,T H )2 sin2 k2zd

1 + 14 (χTE,TH − 1

χT E,T H )2 sin2 k2zd(36)

The subscripts TE,TH denote the respective results for TE, TH polarization. In the above expressions we have that:

χTE =

k2z

µ2

k1z

µ1

(37)

χTH =k2z

ǫ2k1z

ǫ1

(38)

Note, that it is easily seen from the above equations that T+R=1 for both TE and TH polarizations, signifyingconservation of energy.

Problem 3

Assume now that the block of dielectric seen in Fig. 2 has refractive index n2 =√

ǫ2µ2 < n1 =√

ǫ1µ1. Also assumethat the incoming angle θI is above the TIR critical angle [case of problem 1 viii)], and the plane of incidence to bethe xz plane, exactly as in the case of problems 1 and 2 (Fig. 2). Also assume TE polarization, which implies thatelectric field lies along y, and that all fields have translation symmetry along y. Also, in this case we will have anincoming, reflected and outgoing plane wave outside the slab propagating along the +z, -z and +z directionsrespectively, so the applicable expressions for area 1 and are 3 of problem 2 still apply [Eqs. (20) and (22)].However, note that in this case due to the TIR condition, k2z inside the slab will be purely imaginary as we havealso seen in problem 1v(iii) [Eq. (19)]. Set in Eq. (21) k2z = ik

′

2z, with k′

2z being a purely real quantity given by:

k′

2z =ω

c

√

n21 sin2 θI − n2

2 (39)

Then you will see that the electric field inside the slab will be a superposition of two counterprogating evanescentwaves, given by:

E2y(x, z) = [CE exp(−k′

2zz) + DE exp(k′

2zz)] exp(ikxx) exp(−iωt) (40)

where, kx = n1ωc sin θI designates the parallel component of the wave vector conserved throughout all interfaces [see

eq. (1)].

i) Calculate the coefficients CE , DE , as a function of the amplitude of the incoming plane wave EI .

Hint: Repeat the same procedure, using the electromagnetic continuity boundary conditions at z=0 and z=d forthe fields given by Eqs. (20), (22) and (40)Hint for shortcut: Use the expressions CE , DE you have calculated in problem 2 and substitute k2z in ALL placeswhere it enters (including the χ-parameters), with ik

′

2z.

ii) After calculating CE and DE give the expression for the x- and z- components of the magnetic field inside theslab. For that use from Maxwell’s equation, the time-independent (stationary) expression of Faradays law, i.e use:

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∇× E =iωµ

cH (41)

iii) Now the transfer of energy across the slab is represented by the z-component of the time averaged Poyntingvector inside the slab. The latter will be given by the following expression:

< Sz >= Re(E × H∗)z = Re(−EyH∗x) (42)

You will find that this quantity is non-zero. This means unlike the case of a single evanescent wave which cannottransfer any energy (we saw that in problem 5(d) of problem set 1), two counterprogating evanescent waves dotransfer energy.

Work out your math and simplify your expressions to show that < Sz >is given by the following expression:

< Sz >=

k1z

µ1

cω

1 + 14 ( 1

χT E′ + χTE′)2 sinh2 k′

2zd(43)

with

χTE′

=

k′

2z

µ2

k1z

µ1

(44)

iv) Comment how the derived Poynting vector < Sz >, given by Eq. (43), behaves with increasing slab width, d.

What happens at d → ∞?

Problem 4

Transfer matrix formalism for transmission calculations

An elegant way in transmission calculations involves expressing the relations between transmitted and incomingfields in matrix form. We analyze here this method only for the TE polarization case. We take z to be the directionof the surface normal, xz the plane of incidence and so y would be the direction of the electric field, exactly as wehad for Problem 2 (see Fig. 2). The total field in area 1 (before encountering the slab) would be:

Ey(x, z) = [EI exp(ik1zz) + ER exp(−ik1zz)] exp[i(kxx − ωt)] (45)

The total field in area 3 (after encountering the slab) would be:

Ey(x, z) = ET exp(ik1z(z − d)) exp(i(kxx − ωt) (46)

with kx being conserved through all interfaces (parallel to the interface wavevector component).Another schematic with only the xz plane is shown in Fig. 3. The contributions from the +z and -z propagatingfield is given at each location (ignoring the x,t dependence which is the same for all fields).

To find all these fields ER, F,G,H,K and ET in terms of incident field EI , one should apply again the continuityboundary conditions at both interfaces. However, by considering only a single boundary and assuming that therefracted waves can have both +z and -z contributions for generality, -and which is the case when a secondboundary exists- (for proof see Prof. Soukoulis notes of chapter 9) one can show that:

(

EI

ER

)

= M12

(

FG

)

(M1)

where

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FIG. 3: The xz plane of the slab under study with the contributions of the +z and -z propagating fields indicated at certainlocations. White space represent a medium with material parameters ǫ1 and µ1, while the green slab represents a medium withmaterial parameters ǫ2, µ2. It is assumed that medium 2 extends infinitely in x- and y-direction.

M12 = 12

(

1 + χE12 1 − χE

12

1 − χE12 1 + χE

12

)

(M2)

with

χE12 =

kz2

µ2

kz1

µ1

(47)

I must alert you at this point, that the formalism developed in chapter 9 and 10 of Prof. Soukoulis notes, are a bitdifferent from the one we follow in this homework. In both cases, fields that come from the left and are travelingtoward the right of Fig. 3 are considered. In the formalism of the notes, the fields in the right are related through amatrix with the fields on the left. In the formalism of this homework, the fields in the left are related through amatrix with the fields on the right. Both formalism are equivalent and will lead to the same results and both areused in the photonics literature. Each time you must check which formalism is the one adopted, since the form ofthe corresponding matrix expressions does depend on the particular formalism adopted. To convince yourselvesapply continuity boundary conditions at a single boundary assuming for generality that there are also refractedfields along the (-z). Then, you can see if one wants to express the field relation in a matrix form [eq. (M1)], thematrix M12 will indeed be given by eq. (M2). The superscripts 12 designate that this matrix will relate incomingand refracted fields, for a EM plane wave traveling in the +z direction from medium 1, impinging on the surface ofmedium 2. Again, you see that matrix given in (M2) is different from the corresponding matrix in the formalism ofprof. Soukoulis notes, [eq. 9.22]. Evidently, from (M1) it follows that:

(

FG

)

= (M12)−1

(

EI

ER

)

(M3)

where (M12)−1 denotes the inverse of matrix M12.We use different symbols from the ones that were used in the notes of Prof. Soukoulis. To make the correspondenceEI ↔ E+

1 , ER ↔ E−1 , F ↔ E+

2 , G ↔ E−2 . Thus, by comparison it follows that (M12)−1 must be equal with the

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matrix, M (s) given in eq. (9.22) of Prof. Soukoulis notes. For practice and to convince yourselves you can checkthat indeed this equality holds.This parenthesis for comparison between the two formalisms was to avoid confusion when you read the notes Prof.Soukoulis gave you. From this point thereon, we will focus on the formalism denoted in this homework designatedby matrix equations (M1) and (M2).Now, it is evident that the fields on the left and the right on the second interface of the slab by analogy will be alsorelated by the following matrix equation:

(

HK

)

= M21

(

ET

0

)

(M4)

where

M21 = 12

(

1 + χE21 1 − χE

21

1 − χE21 1 + χE

21

)

(M5)

with

χE21 =

kz1

µ1

kz2

µ2

, (48)

where M21 substitutes the matrix M12 because in this interface we have refraction from medium 2 to medium 1.The bottom element of the right column matrix is zero because there are no waves reflecting at z = ∞.Now it is very easy to relate the fields F, G and H, K. Since no other boundary intervenes between them, they willbe different only by the phase the EM wave acquires while propagating through medium 2. Thus since F, Hrepresent the fields propagating in the +z directions at z=0 and z=d respectively we would have H = exp(ik2zd)F ,so F = exp(−ik2zd)H. Conversely, since G and K represent the fields propagating in the -z directions at z=0 andz=d respectively we would have K = exp(−ik2zd)G, so G = exp(+ik2zd)K. Accordingly, if we expressing therelation between these fields in matrix form we will get:

(

FG

)

= P 2(d)

(

HK

)

(M6)

P 2(d) =

(

exp(−ik2zd) 00 exp(ik2zd)

)

(M7).

P 2(d) is called the propagation matrix, and it is defined for propagation from left to right through the slab ofmedium 2 (hence the superscript 2) as depicted in Fig. 3, for a distance d (hence the d in parenthesis).Again do not get confused with the expressions in the notes Prof. Soukoulis has given you, they are not supposed tobe the same due to the different formalism.Now, it is easy to see that:

(

EI

ER

)

= M12

(

FG

)

= M12 P 2(d)

(

HK

)

= M12 P 2(d)M21

(

ET

0

)

(M8)

Equation (M8) implies that the fields in the left and right of the slab depicted in Fig. 3 are related through thefollowing matrix relation:

(

EI

ER

)

= Mslab

(

ET

0

)

(M9),

12

where Mslab is called the transfer matrix through the material slab. Evidently by comparison with eq. (M8),

Mslab = M12 P 2(d)M21 (49)

Take a closer look at eq. (49). You see that the transfer matrix through the entire slab equals to the transfer matrixfor waves impinging medium 2 from medium 1, times the propagation matrix through the slab, times the transfermatrix for waves coming from medium 2 impinging medium 1.

This simple elegant expression applies also in any system however complex provided only planar interfaces, infinitelyextending in xy are present. When you are called to write a transfer matrix in a general more complex system (asan example, you will do that in Problem 6), you will start walking through the structure from left to right. Everytime you have a boundary between medium a and b you will write a matrix Mab, and everytime you propagatethrough a medium, e.g. (a), for a certain distance distance da, you will write the corresponding propagation matrix,P a(da). The resulting transfer matrix through the entire structure will be the product of the matrices you will havewritten down. Hence, however complex the system maybe you can easily obtain the entire transfer matrix. Thenfrom the transfer matrix you can find the relation between the incoming and the outgoing fields, easily. Once youstart having more boundaries, the methodology applied for problem 2 becomes too tedious, while the transfermatrix will be always a product of known matrices which can be also evaluated numerically. Note that if we want towrite the field relations in terms of the individual matrix elements of the transfer matrix we would have that,

(

EI

ER

)

= Mslab

(

ET

0

)

=

(

Mslab,11 Mslab,12

Mslab,21 Mslab,22

)(

ET

0

)

=

(

Mslab,11ET

Mslab,21ET

)

(M10)

So, from (M10) we would have

t =ET

EI=

1

Mslab,11(50)

and

r =ER

EI=

Mslab,21

Mslab,11(51)

The above equations in conjunction with eq. (31) and (32) suggest that the transmission coefficient T is

T =1

|Mslab,11|2(52)

and the reflection coefficient R is:

R =|Mslab,21|2|Mslab,11|2

(53)

So, it is now clear that the calculation of the transfer matrix through the slab Mslab is sufficient for the calculationof the transmission and reflection coefficients, T and R respectively. Note, this holds true for any complex layered(infinite in the xy plane) structures as well. In such case the transfer matrix through the entire structure will becalculated.

i) Give the expressions for all matrix elements of transfer matrix Mslab.bf Hint Use eq. (49) along with eqs. M2, M5 and M7.

ii) Show that the matrix elements that you have calculated obey the following relations:

13

Mslab,11 = M∗slab,22, (54)

and

Mslab,21 = M∗slab,12, (55)

where the ∗ denotes the complex conjugate value.

Note Of course this is trivially shown once you know the matrix elements that you have calculated in i). I would liketo stress that eqs. (54) and (55) are universal and hold for any transfer matrix through any layered structuredembedded in a uniform medium. Eqs. (54) and (55) can also be shown without using the knowledge for the values ofthe matrix elements of Mslab, by requiring that the scattering problems of EM waves impinging the same slab fromleft to right and right to left are equivalent.

iii) Show that the transfer matrix through the slab is unimodular, i.e., det(Mslab) = 1.Of course this is shown trivially from the result in i). First perform the math using the matrix elements youcalculated in i). This would be also a check that your math for the transfer matrix calculation are indeed correct.Then try to show the unimodular property of the transfer matrix Mslab without using the knowledge for the matrixelements of Mslab, but relying solely on physical principles (Use principle of energy conservation). Thus again, theunimodular property of the transfer matrix is universal for any transfer matrix through any layered structuredembedded in a uniform medium.

iv) Use eqs. (52) and (53) to calculate the transmission and reflection coefficients from the transfer matrix. Workout your result to compare it with Eqs. (35) and (36) of problem 2. Evidently, it must be identical.

Problem 5

Fabry-Perot modes

For the case of TE polarization, show by using eq. (35) that for non-magnetic media (µ1 = µ2 = 1) the transmissioncan be expressed in the following form:

T =1

1 + 14 [k2z

k1z− k2z

k1z]2 sin2 k2zd

(56)

i) Show that transmission maxima occur when k2zd = mπ, where m=0,1,2,3 etc.

ii) Manipulate the above equation, in order to find the frequency location ω of the Fabry-Perot maxima in terms ofthe incident angle θI . (Use Snell’s law and assume n2 > n1.)

Problem 6

Transmission through periodic multilayer stacks, i.e. 1D photonic crystals

The usefulness of the transfer matrix method, seen in problem 4 for one slab becomes evident when morecomplicated structures are considered. Assume a periodic multilayer stack structure like the one seen in Fig. 4.Namely, we have alternating stacks of medium 2, with permittivity ǫ2, permeability µ2 and thickness d2, with stacksof medium 1, with permittivity ǫ1, permeability µ1 and thickness d1. The entire multilayer system is embedded in amedium, identical to the one of the multilayer constituents and in particular medium 1. The building block unit ofthe multilayer stack, magnified in the bottom of Fig. 4, repeats itself periodically with repetition period equal toa = d1 + d2. In other words, this multilayer stack is essentially a one-dimensional (1D) photonic crystal. The entiremultilayer system consists of N total building block units. Assume for simplicity that you have normal incidence, i.e.that θI = 0, and polarization is as seen in Fig. 4. It is easy to see to that in such a case both TE and THpolarizations are degenerate, due to the uniformity in the xy plane. [See also question in Problem 1v).] The TE caseis drawn here, and you can use the relations derived for problem 4 but for θI = 0.

In problem 4, we saw the transfer matrix through the entire slab, which relates the field componets (along +z and -z)on the left and right of the slab seen in Fig. 3, can be written as a product of matrices as indicated by eq. (49). Thisis a very elegant formalism because it allows the construction of the transfer matrix in more complicated systems.

14

FIG. 4: The multilayer stack system (1D photonic crystal) under study.

So, repeating the findings of problem 4, to construct the transfer matrix relating the fields to the left and right ofany intervening structure you follow the following rule. You walk conceptually from the left (where the fields arecoming from) to the right (where the fields will exit). Every time you encounter an interface you factor a matrixMab where a is the medium on the left and b is the medium on the right of the interface, with Mab, given by thefollowing expression:

Mab = 12

(

1 + µa

µb

kbz

kaz1 − µa

µb

kbz

kaz

1 − µa

µb

kbz

kaz1 + µa

µb

kbz

kaz

)

(M11)

where indices a and b represent either medium 1, or 2. Every time you walk a distance da within a medium, a, youfactor a propagation matrix P a(da) which corresponds to medium a for distance da, with

P a(da) =

(

exp(−ikazda) 00 exp(+ikazda)

)

(M12)

with kaz being the z-component of the wave vector inside medium, a. Then the entire transfer matrix will essentiallybe a product of Mab and P a(da). (Note that in problem 6 we concentrate on the special case of normal incidence, asindicated in Fig. 4, i.e. θI = 0, and kx = 0). Then:

k1z = n1ω

c(57)

and

k2z = n2ω

c(58)

where n1, n2, are the refractive indices in medium 1 and medium 2 respectively.

i) Consider the fields on the left and right of the ith building block, as seen in Fig. 4, where the superscript +,denotes the amplitudes for plane wave contributions propagating in the +z direction. Conversely the superscript -,

15

denotes the amplitudes for the plane wave contributions propagating in the -z direction. Using the logic in thepreceding paragraph write the transfer matrix for the unit building block Munit, relating the fields on the left andright if the ith building block, i.e.:

(

Ei,+

Ei,−

)

= Munit

(

Ei+1,+

Ei+1,−

)

(M13)

In other words explain that the transfer matrix can be written as:

Munit = M12P 2(d2)M21P 1(d1) (59)

ii) Express explicitly Munit in terms of the kz wavevectors, the ǫ and µ and d parameters.

iii) After ii) show that:

Munit,11 = M∗unit,22 (60)

Munit,12 = M∗unit,21 (61)

where Munit,ij represents the Matrix element of Munit, on the ith row and jth column.

Also check, as in problem 4, that indeed Munit is unimodular, i.e., show that det|Munit| = 1.

iv) Find the two eigenvalues Λ1,2, of the Munit matrix. Show that these are given by:

Λ1,2 = Re(Munit,11) ±√

(Re(Munit,11))2 − 1 (62)

where√

here denotes the generilzed complex square root since the quantity Re(Munit,11))2 − 1 can take both

positive and negative values. In the above equation we have that,

Re(Munit,11) = cos (k1zd1) cos (k2zd2) −1

2(χE

12 + χE21) sin (k1zd1) sin (k2zd2), (63)

where χE12 and χE

21 are given be eqs. (47) and (48) respectively.

v) We have seen in class, just like the wave function for electrons in a crystal, the EM fields in a periodic mediumalso obey Bloch’s theorem.This means:

(

Ei,+

Ei,−

)

= exp(−iqa)

(

Ei+1,+

Ei+1,−

)

(M14)

which implies, exp(−iqa), is an eigenvalue of the matrix Munit, and q is the Bloch wave vector, restricted in the firstBrillouin zone (−π

a < q < πa ). Now, notice the two eigenvalues in problem iv) Λ12. Bloch’s theorem implies that

exp(−iqa) = Λ12. This means that Λ12 must be complex or Λ212 = 1.

-When (Re(Munit,11))2 − 1 > 0 then Λ1,2 is real and not equal to ±1. This implies that no real q exists which

satisfies Bloch’s theorem, i.e. no modes are allowed to propagate inside the multilayer (bandgap).-When (Re(Munit,11))

2 − 1 < 0 then Λ1,2 is complex, implying real q exist which satisfy Bloch’s theorem (bandregion). The complex eigenvalues will be given by:

Λ12 = exp(−iqa) = Re(Munit,11) ± i√

(1 − Re(Munit,11))2, (64)

where here the√

is the real square root, since the restriction in the inequality guarantees 1 − (Re(Munit,11))2 to be

positive. Note the two roots are one complex conjugate of each other, which means that if for a certain frequency ω

16

we get a Bloch momentum q, as a solution from eq. (64), a Bloch momentum of (-q) will be also a solution. In otherwords, propagation of an EM wave inside the multilayer will be equivalent in both directions.

- In the limiting case where Re(Munit,11))2 = 1, then Λ12 = exp(iqa) = ±1. Real q solutions exist in such a case and

represent the band edge before the onset of the forbidden propagation region.

Use the fact that exp(iqa) and exp(−iqa) are both eigenvalues to show for a certain frequency ω, that the Blochmomentum, q, of the multilayer system, is given in term of the material and structural parameters of the systemfrom the following relation:

cos(qa) = cos (n1ω

cd1) cos (n2

ω

cd2) −

1

2(Z2

Z1+

Z1

Z2) sin (n1

ω

cd1) sin (n2

ω

cd2), (65)

where Z1 =√

ǫ1µ1

, Z2 =√

ǫ2µ2

, represent the impedances in medium 1 and medium 2 respectively.

Note: The above simplified relation is valid only under normal incindence (θI = 0)!

vi) Write now the transfer matrix for the entire multilayer stack according to the recipe described in the beginningof the problem. Assume the total multilayer stack comprises of N unit cells. Use Chebyshev’s identity to make thematrix multiplication.

Chebyshev’s identity

The N th power of a unimodular matrix V:

V =

(

A BC D

)

(M15)

can be simplified according to the following equation:

(

A BC D

)N

=

(

AUN−1 − UN−2 BUN−1

CUN−1 DUN−1 − UN−2

)

(M16)

where,

UN =sin[(N + 1)QW ]

sin(QW )(66)

with

QW = arccosA + D

2(67)

Hint Use the knowledge that cos(qa) = 12Tr[Munit], obtained in subquestion v) to express the final total transfer

matrix in terms of the Bloch momentum q.

vii) After the calculations in subquestion vi) use the knowledge that T = |ET

EI|2 = 1/|Mtotal,11|2, where Mtotal is the

transfer matrix through the entire multilayer stack consisting of N unit building blocks to derive the followingsimple expression for the transmission through the entire 1D multilayer system [The simplified expression of T, interms of Mtotal,11 is due to the fact that the entry and exit medium is the same (medium 1)]. Thus, you will get:

T =1

1 + |runit/tunit|2 sin2(Nqa)sin2(qa)

, (68)

where runit and tunit refer to the values corresponding through transmission through one building block only, i.e.:

runit

tunit= Munit,21 =

i

2(Z2

Z1− Z1

Z2) exp(−i

n1ω

cd1) sin(

n2ω

cd2) (69)

Problem 7

17

Use the result obtained in 6vii) [i.e, eqs. (68), along with eq. (69) and eqs. (65)] to calculate the transmissionthrough a multilayer stack in the arrangement of Fig. 4 for the following specific parameters: ǫ1 = 1, d1 = 310nm,ǫ2 = 4, and d2 = 155nm. Assume non-magnetic media case i.e. µ1 = µ2 = 1. Plot the transmission vs. thedimensionless frequency ωa/c, with a = d1 + d2, for a total number of unit cells N=2, N=4 and N=16 respectively(use linear scale for the y-axis) (Again, a is the period, thus a=310 nm+ 155 nm= 465 nm).For the particular case for N=16 plot the transmission T versus the photon energy in (eV) with logarithmic scalingfor the transmission axis.Note: Unit conversion: 1eV ∼ 1.5191015rad · sec−1 (conversion to cyclic frequency ω).

Compare your plot, with the one shown in lecture 5 of notes in page 20 (same case).

Problem 8

Band structure of 1D periodic photonic crystal

The multilayer stack seen in Fig. 4, is in fact a 1D periodic photonic crystal with period a. We derived in class forthe case of a photonic crystal with three-dimensional periodicity an eigenvalue equation, the eigenvalues of whichprovide the dispersion relation, i.e. the band structure, for EM wave propagation inside the periodic medium. Wewill do here the same for the case of the 1D PC. Follow through the instructions and verify all steps as presented.Assume non magnetic media, i.e. µ1=µ2=1.

Step 1

Use the two stationary Maxwell’s equations for harmonic fields varying like exp(−iωt), (i.e. Use stationaryFaraday’s law and Amprere’s law), for normal incidence and a polarization for the electric field as seen Fig. 4 (orFig.5). Then show that the governing equation for the electric field lying in the y-axis, becomes:

d2Ey

dz2= −ω2

c2ǫ(z)Ey, (70)

where ǫ(z) is the periodic dielectric function along z, and within the unit cell extending from -a/2 to a/2, it is givenby:

ǫ(z) =

ǫ1 if −a/2 ≤ z < (−a + d1)/2ǫ2 if (−a + d1)/2 ≤ z < d2/2 , (M17)ǫ1 if d2/2 ≤ z < a/2

Note: The zero of the coordinate system has been offset for the unit cell for the band structure problem. For clarity,the unit cell with the off-setted coordinate system is seen in Fig. 5.

Step2

The electric field must satisfy Bloch’s theorem. In other words:

Ey(z + a) = exp(iqa)Ey(z) (71)

This means that the electric field can be written:

Ey(z) = exp(iqz)u(z) (72)

where u(z + a) = u(z) is a periodic function in z. A direct consequence of this periodicity is that it’s integral fourierdecomposition reduces to a discrete sum given by (see class notes of lecture 4):

u(z) =∑

Gn

un exp(iGnz), (73)

with Gn = 2nπa .

18

FIG. 5: The unit cell of the 1D photonic crystal.

Use eq. (72) and (73) to show that the proper sum representation for the Electric field would be:

Ey(z) =∑

n

un exp[i(2nπ

a+ q)z] (74)

Step3

The dielectric function is also periodic and so can be written as:

ǫ(z) =∑

n

ǫn exp(i2nπ

az) (75)

Since ǫ(z) is known and given by eq. (M17), the coefficients ǫn can be evaluated by applying the inverse Fouriertransform, i.e.

ǫn =1

a

∫ a/2

−a/2

ǫ(z) exp(−i2nπ

az)dz (76)

Show that the coefficients ǫn, will be given by the following simple formula:

ǫn = ǫ1δn,0 + f(ǫ2 − ǫ1)sin(nπf)

nπf, (77)

where δn,0 is Kroeneckers delta function, and f, the filling ratio of medium 2 inside the unit cell (i.e. f = d2/a).

Hint use eq. (76) for the calculation, first with n = 0 then with n 6= 0. Merge the two obtained results into one singleformula seen in eq. 77. Use the fact that sin x

x with x → ∞ goes to 1.

Step4

Insert (74), (75) into (70) and show that:

(q +2πn

a)2un =

ω2

c2

∑

n′

ǫn−n′un′ (78)

Hint: After inserting (74), (75) into (70), notice that exp(iqz) cancels out from both hand-sides.

Then multiply both hand sides with exp(i2πn′′

/az) and use the following useful formula

19

1

a

∫ a/2

−a/2

exp(i2πn

az)exp(−i

2πn′

az)dz = δn,n′ (79)

where δn,n′ is Kroenecker’s delta function.Some additional index (subscript) manipulations/transformations is needed to bring the equation in the desiredform shown in eq. (78).

Step5

Eq. (79) is a generalized eigenvalue problem of the form:

AU = λBU (80)

Note that both eigenvalues λ = ω2

c2 and eigenvectors U depend on q and will be different for different q. Hence, theeigenvalues will give the dispersion relation (band structure) and the eigenvectors the expansion coefficients un fromwhich the electric field can be calculated. Normally, the eigenvalue problem represented in eq. (80) is solvednumerically, where a certain number of terms are kept in the sum (truncation). The more terms taken the smallerthe truncation errors in the resulting band structure.

Assume here for simplicity that it is sufficient to take three terms in the sum, those corresponding for n=0, n=+1

and n=-1. Express the simplified 3X3 eigenvalue problem in such a case, and find the eigenvalues ω2

c2 express as afunction of q, and the structural and material parameters of the multilayer.

Step6

Assume the same structural and material parameters as in problem 7. Using the approximation made in step 5 plotall three dimensionless eigenvalues ωa/c as a function of q, with q ranging from −π/a to π/a. Do you find a bandgap with limits agreeing with the gap seen in the transmission spectrum of Fig. 7? How well is the agreement.Please comment.

Problem 9

Band structure of 1D periodic crystal; method 2

i) Eqs. (65) in problem 6 is essentially the band structure for the 1D photonic crystal (input Bloch wave vector qand obtain frequency solutions). Use the same material and structural parameters as in problem 7 and plot ωa/cversus q from the equation derived in problem 6.

ii) Compare the result with that obtained from problem 8 step 6 for the three first bands. (Plot together first bandof one method with first band of the other method, second band of one method with second band of the other method,etc.) How good is the agreement? Is the quality of the agreement same or different for the different bands.Comment.

iii) Set parameters all equal as in problem 7, except for ǫ2. Set here ǫ2 = 2.0 and plot again the three first bands asin ii) with the two different methods. Is the agreement same, better or worse than in the case ii). If you see anydifference in the quality of agreement can you explain it?

Problem 10

Surface states at the interface between two media

[Reading chapter 11 of Prof. Soukoulis notes]

In this problem, we will investigate for the existence of surface wave at the planar boundary, (along the xy plane),between two media with material parameters ǫ1, µ1 and ǫ2, µ2 respectively as seen in Fig. 5. We assume non-lossymaterial, i.e., all material parameters are purely real. In Fig. 5 we show the xz plane, which again like in theprevious problem will be assumed to be the plane of incidence (i.e. ky = 0). Again kx will be considered to remaincontinuous across the interface and yields the phase velocity (ω/kx) of the surface wave. However kz in this problemwill not be real, but will be purely imaginary both in area 1 and area 2. This is because we will investigate for caseswhere propagation is prohibited in the z-direction in both medium 1 and medium 2, which would result inpropagation only along the x-direction and hence a surface wave. We will take medium 1 to have ǫ1 > 0 and µ1 > 0.However medium 2 can have:

20

FIG. 6: Surface wave propagating laterally at the interface between the two media with material parameters ǫ1, µ1 and ǫ2, µ2,respectively. Away from the interface propagation is forbidden, so energy and thus fields will decay exponentially.

a) both ǫ2 > 0 and µ2 > 0.b) Only ǫ2 < 0, i.e. it could be a metal.c) Only µ2 < 0, i.e. it could be a magnetic material.d) Both ǫ2 < 0 and µ2 < 0, i.e., could be a negative index metamaterial.

First requirement to have surface states is that the propagation is forbidden in the z-direction in both media. Thismeans that the general field expressions would be:

A) CASE OF TE-POLARIZATIONarea 1:

Ey(x, z) = AE exp(ikxx) exp(k′

1zz) exp(−iωt) (81)

Hx(x, z) = AEic

ωµ1k

′

1z exp(ikxx) exp(k′

1zz) exp(−iωt) (82)

area 2:

Ey(x, z) = BE exp(ikxx) exp(−k′

2zz) exp(−iωt) (83)

Hx(x, z) = −BEic

ωµ2k

′

2z exp(ikxx) exp(−k′

2zz) exp(−iωt) (84)

In the above expressions in both areas the expressions for the x-component of the magnetic field are derived with

the use of the stationary form of Faraday’s law which leads to Hx = icωµ

∂Ey

∂z .

We will discuss the values of k′

1z and k′

2z at a later point. For now we assume, that both quantities are real, andpositive so that eqs. (81)-(84) manifest decaying waves from the interface at z=0. The continuity of the parallelcomponent of the wave vector is already taken in the above expressions by assuming the same kx in area 1 and area2.

B) CASE OF TH-POLARIZATIONarea 1:

Hy(x, z) = AH exp(ikxx) exp(k′

1zz) exp(−iωt) (85)

Ex(x, z) = −AHic

ωǫ1k

′

1z exp(ikxx) exp(k′

1zz) exp(−iωt) (86)

21

area 2:

Hy(x, z) = BH exp(ikxx) exp(−k′

2zz) exp(−iωt) (87)

Ex(x, z) = BHic

ωǫ2k

′

2z exp(ikxx) exp(−k′

2zz) exp(−iωt) (88)

In the above expressions in both areas the expressions for the x-component of the electric field is derived with the

use of the stationary form of Amperes’s law which leads to Ex = − icωǫ

∂Hy

∂z .

i) Applying continuity for the tangential to the interface components of the electric field, and the auxiliary magneticfield show that the generalized conditions to have a surface wave are:For TE-polarization

k′

1z

µ1+

k′

2z

µ2= 0 (89)

For TH-polarization

k′

1z

ǫ1+

k′

2z

ǫ2= 0 (90)

ii) Look at these conditions carefully. You know that the decaying profile away from the interface mandates k′

1z and

k′

2z to be positive. So, the only way for eqs. (89) to be satisfied is that if µ2 < 0. Likewise, the only way for eq. (90)to be satisfied is that if ǫ2 < 0. (We assumed earlier that ǫ1 and µ1 are both positive).Convince yourselves that this means that:ifa) both ǫ2 > 0 and µ2 > 0.No surface waves under any conditions.

b) Only ǫ2 < 0, i.e. it could be a metal.Surface waves possible for TH-polarization, where condition (90) is satisfied

c) Only µ2 < 0, i.e. it could be a magnetic material.

Surface waves possible for TE-polarization, where condition (89) is satisfied

d) Both ǫ2 < 0 and µ2 < 0, i.e., could be a negative index metamaterial.Surface waves possible for both polarizations. For each respective polarization the respective conditions [(89) or (90)]should be satisfied

iii) Now, we know from the wave equation in a certain medium that for a plane wave propagating inside the mediumon the x-z plane we get:

k2x + k2

z = ǫµω2

c2, (91)

which is the dispersion relation of EM wave inside the medium. When we have a decaying profile in the z directionthen kz will be imaginary and the plane wave exp(ikzz) becomes a decaying exponential exp(−k

′

zz), with kz = ikz′,

and kz′ a real positive quantity. Thus, since we have decaying exponentials in the z-direction away from the

interface the proper form of the dispersion relations of EM waves inside the two media would be:

k2x − (k

′

1z)2 = ǫ1µ1

ω2

c2, (92)

and

k2x − (k

′

2z)2 = ǫ2µ2

ω2

c2, (93)

From eqs. (92) it follows that in order to have a decaying wave along z [i.e., k1z′

real and positive, so (k′

1z)2 must be

positive (positive real root will be chosen)] in medium 1 the following condition must be met:

kx >√

ǫ1µ1ω

c, (94)

22

Now, for the surface waves of case b) or c), we have the condition ǫ2µ2 < 0, which automatically makes (k2z′

)2 > 0

always, ensuring the reality of k2z′

.However, for the case d) with ǫ2µ2 > 0 in order to have (k2z

′

)2 > 0, we must additionally impose:

kx >√

ǫ2µ2ω

c(95)

Note, that the surface wave is characterized by the frequency ω and the lateral wavevector kx. The relation

between the two is called dispersion relation for the surface mode.

To recapa) If ǫ2 > 0 and µ2 > 0.No surface waves under any conditions

b) If ǫ2 < 0 and µ2 > 0.

Surface wave for TH-polarization given by eq. (90), with additional requirement (94) to be satisfied. Then k′

1z and

k′

2z are obtained by eqs. (92) and (93) respectively.

c) If ǫ2 > 0 and µ2 < 0.

Surface wave for TE-polarization given by eq. (89), with additional requirement (94) to be satisfied. Then k′

1z and

k′

2z are obtained by eqs. (92) and (93) respectively.d) If ǫ2 < 0 and µ2 < 0.Surface wave for both polarizations possible.Surface wave for TE-polarization from eqs. (89), with additional requirements (94) and (95) to be satisfied. Then

k′

1z and k′

2z are obtained by eqs. (92) and (93) respectively.Surface wave for TH-polarization from eqs. (90), with additional requirements (94) and (95) to be satisfied. Then

k′

1z and k′

2z are obtained by eqs. (92) and (93) respectively.

iv) Investigate a specific case of surface plasmon on metal surface

Assume:ǫ1 = 1µ1=1

ǫ2 = 1 − ω2

p

ω2

µ2 = 1where, we have assumed a lossless metal.- Use eqs. (90) along with condition (94), and eqs. (92), (93) to obtain the surface plasmon dispersion relationω(kx). You will find that it is easier to express the wave vector kx in terms of the frequency ω rather than theopposite. Show that eq. (90) yields the following relation for the surface plasmon dispersion.

k̃ = ω̃

√

ω̃2 − 1

2ω̃2 − 1(96)

where both k̃ and ω̃ are dimensionless quantities, relating to the wave vector and frequency through the followingrelations:

k̃ =kxωp

c

(97)

and

ω̃ =ω

ωp(98)

Show that condition (94) expressed in the dimensionless quantities reduces to:

k̃ > ω̃ (99)

One, must not forget the condition for ǫ2 to be negative, which poses the additional restriction of ω̃ < 1. Then from(96) we get that in order for kx to be real (propagating surface wave), also ω̃ < 1√

2. So finally, keeping the most

strict of the two conditions we must have ω̃ < 1√2.

23

- Show also that if ω̃ < 1√2, eqs. (96) also guarantees the satisfaction of condition (99).

Then, show graphically eq. (96) (Plot k̃ vs. ω̃ and inverse the axis, to have k̃ as the x-axis). You do not need furtherinformation for material parameters because the units are dimensionless. Vary ω̃ between 0 and 1√

2, in small steps.