3. THE COMPLEX DIELECTRIC FUNCTION AND REFRACTIVE INDEX

Transcription

1 3. THE COMPLEX DIELECTRIC FUNCTION AND REFRACTIVE INDEX 3.1 Conductivity and dielectric constant In this this chapter I ll define the complex dielectric function as I ll use it for the remainder of the text. The first topic is the relaxation between conductivity (σ) and dielectric function (ɛ). For a moment, let me return to Eq. 2.4d, substituting into it Eqs and 2.17: H = 4π c σ 1E + ɛ 1 E c t. (3.1) As discussed on page 9, the Maxwell equations are only complex if complex fields are used with them. But how am I to think about σ 1 and ɛ 1? Are they real numbers or complex numbers? Well, if I use sines and cosines for the fields, they had better both be real numbers. If I use complex fields (as I plan to do) I can take one of two philosophical positions. The first position omits the subscripts and identifies σe as the free current density. Similarly ɛ E/ t contains the bound current density and Maxwell s displacement current. One can defend this distinction. For example, the conduction electrons in a metal are free carriers and their motion in response to an electric field is the free current. Because the carriers have inertia, their response will lag the field as frequency increases. This phase delay can be represented by having σ be complex, with the imaginary part being an inductive response that results from this inertia. Similarly, the remaining electrons in the solid are bound to their atoms, and their response is the polarization current. Inertial and dissipative effects will ensure that ɛ also is complex. I have 4 independent quantities (the real and the imaginary parts of these functions) but can identify them with the free and with the bound carriers. The second position asserts that the conductivity and dielectric function in Eq. 3.1 are both real quantities, just as they would be if real-valued fields were used. It goes on to note that the distinction between free carrier and bound carrier response is clear in the case of well understood and well characterized systems such as simple metals. However, for some new material, perhaps with no good theoretical understanding, it would not be possible to make such a clear distinction. It is better to reduce the number of materials constants to two. I ll take the second position. There are two independent quantities, the real conductivity σ 1 and the real dielectric function ɛ 1. The first leads to a current that is in phase with E (and hence is dissipative) and the second to a current that is 90 out of phase (and hence dispersive). I will combine them into a single complex dielectric function. 16

2 3.2 The complex dielectric function Using this approach, I can substitute our plane-wave fields into Eq. 3.1; indeed, I did that to derive Eq. 2.18d (the plane-wave-generated algebraic equation) which I copy here: q H = ω c andthenidefine a complex dielectric function ɛ as [ ɛ 1 + 4πi ] ω σ 1 E. ɛ ɛ 1 + 4πi ω σ 1. (3.2) With this definition, I arrive at the form of Maxwell s equations which is appropriate for local, nonmagnetic, linear, isotropic, and homogeneous media. I ll put them in a box to make them stand out. q E =0 q H =0 q E = ω c H q H = ω c ɛe (3.3a) (3.3b) (3.3c) (3.3d) These equations have a very agreeable symmetry. The materials properties enter only through the presence of the (complex function) ɛ in Eq. 3.3d. Equations 3.3a and 3.3b tell me that q is perpendicular to E and H, justastheywere in vacuum. Equation 3.3c tells me that H is perpendicular to q and E; the three vectors form a right handed set, with unit vectors {ˆq, ê, ĥ}. Then, with ˆq ê = ĥ, the vector part of Eq. 3.3d gives ˆq ĥ = ê as it should, telling nothing new about vector directions. I ll generally write complex quantities in the form Z = Z 1 +iz 2,withZ 1 therealpartand Z 2 the imaginary part. The dielectric function is one such complex quantity, ɛ = ɛ 1 + iɛ 2. From the definition, Eq. 3.2, I see that the imaginary part of the dielectric function is related to the real part of the conductivity, ɛ 2 = 4π ω σ 1. (3.4) Some authors use primes instead, Z = Z + iz. 17

3 3.3 The optical conductivity There are times when it convenient to work with a complex conductivity σ = σ 1 + iσ 2 rather than a complex dielectric function. There is no difference in the physics when I use σ rather that ɛ, just a change of perspective. When using the conductivity, I have to get the case where σ = 0 correct, keeping the vacuum displacement current of Eq. 2.9d. This I do by starting with Eq. 2.7d, writing D = E +4πP, usingj pol = P/ t = iωp. Then, as before, I write Ohm s law, j = σ 1 E andthendefinej pol = σ 2 E. With all this, I get q H = ω c E 4πi c σe = ω c [ 1+ 4πi ] ω σ E, (3.5) where σ σ 1 +iσ 2 is complex. Equation 3.5 preserves Maxwell s displacement current in the vacuum. By equating the right sides of Eqs. 3.3d and 3.5, I see that the relation between ɛ and σ is ɛ 1+ 4πi σ, (3.6) ω and then the complex conductivity is σ = iω (1 ɛ). (3.7) 4π Finally, I put ɛ = ɛ 1 + iɛ 2 and σ = σ 1 + iσ 2 into Eqs. 3.6 and 3.7 and equate real and imaginary parts to obtain Eq. 3.4 and σ 2 = ω 4π (1 ɛ 1). (3.8) Note that σ 2 is positive when ɛ 1 is negative, or at least less than than +1. I will show in Chapter 4 that metals and insulators both may have ɛ 1 negative over certain frequency ranges and positive over others. A negative σ 2 should not worry you. Recall that the reactance of inductors (iωl, positive imaginary) and capacitors (1/iωC, negative imaginary) have opposite signs, allowing resonant circuits to be constructed. When considering a specific type of material (metal, insulator, ionic solid, superconductor) it is sometimes better to work out the conductivity and other times better to derive the dielectric function. Equations 3.6 and 3.7 allow one to translate back and forth between them. The vacuum has σ = 0! In fact, the vacuum has ɛ =1+i0 andσ =0+i0. A negative σ 1 should worry you. 18

4 3.4 The complex refractive index With the directions removed, Eqs. 3.3c and 3.3d become qe = ω c H and qh = ω c ɛe with both E and H varying as e i(q r ωt). I ll solve the first for H and substitute into the second to find q = ω ɛ, (3.9) c thus motivating the definition of the complex refractive index, N = n + iκ ɛ, (3.10) so that q = Nω/c. The real part of N, n, is the refractive index and the imaginary part, κ, is the extinction coefficient. The electric field for plane waves becomes For definiteness, let ˆq = ˆx, soˆq r = x, and E = E 0 e i(q r ωt) = E 0 e i(n ω c ˆq r ωt). (3.11) E = E 0 e i(n ω c x ωt) e κ ω c x. (3.12) Recall that in vacuum, q = ω/c, so the wave vector (wavelength) in the material is larger (smaller) by a factor n. The wave decays with a decay length δ = c/ωκ. (In terms of the vacuum wavelength λ 0 =2πc/ω = c/f, the wavelength inside is λ = λ 0 /n and the decay length is δ = λ 0 /2πκ.) Thewavecrestsmoveatc in vacuum and at v = c/n in the medium. Equation 3.12 can then be rewritten as E = E 0 e 2πi( x λ ft) e x δ. (3.13) The observable (real part) of the electric field, E = E 0 cos [2π(x/λ ft)] e x/δ,isshown in Fig The wavelength (λ) and the damping length (δ) are both equal to 1 here. With directions removed, Eq. 3.3c is H =(cq/ω)e, and,then,withq = ωn/c, Iget H = NE = ɛe. (3.14) N and ɛ are both generally complex, so the phase of H is different from the phase of E. The refractive index is a counterexample to the Z = Z 1 +iz 2 convention. Note also that some (older) texts write N = n(1 + iκ). Furthermore, it is more common to use k than κ as the symbol for the extinction coefficient. I choose the latter because I want to use k as a wave vector. The word wavelength sometimes means the vacuum wavelength, c/f, and sometimes means the wavelength inside the medium, c/nf. If not specified or obvious from context, take it to mean the vacuum wavelength. Here, I take E 0 to be real. 19

5 Fig Damped sinusoidal wave. Squaring Eq gives N 2 = n 2 κ 2 +2inκ = ɛ = ɛ 1 + iɛ 2 so that it is easy to see that ɛ 1 = n 2 κ 2 (3.15) and ɛ 2 =2nκ. (3.16) It takes a bit more labor to calculate things the other way, but after this labor I get where ɛ = ɛ ɛ2 2. ɛ + ɛ1 n = 2 ɛ ɛ1 κ = 2 (3.17a) (3.17b) 3.5 Energy density, Poynting vector, and intensity Electromagnetic waves carry energy and can deposit that energy as heat in absorbing materials. So here I will work out energy aspects of electromagnetic waves in linear, isotropic, and homogeneous materials. I ll start with a quick derivation of Poynting s theorem. I note I know this because if I stand in the sun or near a fire, I become warm through radiative heat transfer. Indeed, humans have known of radiative heat transfer at least as early as the discovery of fire, and physicists investigated this process for centuries, culminating in the blackbody theory of Planck and the birth of the quantum theory. 20

6 that this theorem will be derived from Maxwell s equations, so that it adds no real new physics, only some additional insight. I start by taking the dot (or inner or scalar) product of H with Eq. 2.4c and E with Eq. 2.4d. After doing so, I subtract the results, obtaining H E E H = 1 c H B t 4π c E j 1 c E D t. (3.18) Now, I use the vector identity (E H) =H E E H. Moreover, B = μh and D = ɛ 1 E so that quantities on the right hand side can be manipulated to give H B t H = μh t = μ H 2 2 t and E D t = ɛ 1E E t = ɛ 1 E 2 2 t. Therefore, with a modest rearrangement of the right hand side, I get (E H) = 1 2c (ɛ 1 E 2 Almost done! I define the Poynting vector and the electromagnetic energy density t ) + μ H2 4π t c j E. S c E H, (3.19) 4π u 1 8π (ɛ 1E 2 + μh 2 )= 1 (D E + B H), (3.20) 8π and then S = u t j E (3.21) is Poynting s theorem. The meaning of Poynting s theorem is most easily seen by integrating Eq in some volume V and using the divergence theorem to convert the volume integral on the left hand side to an integral of S ˆn over the surface of V.ThenS ˆn represents the energy flowing into or out of V, u/ t is the change of energy density within V,andj E is the Joule heating (I 2 R heating) due to currents in the lossy medium within V. Energy is conserved, which I find reassuring. When I backed up to Eqs. 2.4c and 2.4d, I reverted to the state before I defined the complex dielectric function, so I must use D = ɛ 1 E and j = σ 1 E. Moreover, I ll take μ as real. Were it not, I d need a term like the j E term to account for the dissipation done by the magnetic field. 21

7 My next worry is how to handle the complex fields that I have been using within Maxwell s equations. Whenever I am thinking about energy, I better make sure when I insert e i(qx ωt) that I end up with pure real quantities when done. I of course will get pure real quantities if I write my fields as sines and cosines but there are good reasons already invoked to eschew sines and cosines and instead to use complex exponentials. Moreover, it is generally the case that I am interested in time-averaged energy flows and not the instantaneous energy flow. This consideration is true even when I have the ability to measure the time-instantaneous energy flux. Let me write the manifestly real fields (E+E )/2and(H+H )/2, with E and H complex functions of space and time, specifically the electric and magnetic fields of Eqs. 2.5 and 2.6. Moreover, as in the footnote on that page, I ll write the constant describing the direction, amplitude, and phase (at r =0andt = 0) of the electric field as E 0 = êe 0 e iφ 0. The electric and magnetic fields (and their complex conjugates) are solutions to Maxwell s equations. I look back at my solutions to see that E and H are orthogonal and that ê ĥ = ˆq. Moreover H = NE where N is the complex refractive index. I use these fields in Eq and S = ˆq c ( E + E )( H + H ), 4π 2 2 = ˆq c 16π (EH + E H + EH + E H ), = ˆq c [ (N + N)EE + NE 2 + N (E ) 2]. 16π As always N = n + iκ, sothatn + N =2n. The complex electric field has an amplitude E and a phase φ, sothate = E e iφ. If the wave is traveling along the x axis, the time and space dependent phase φ is of the form φ = nωx/c ωt + φ 0 while the amplitude is E = E 0 e ωκx/c. This amplitude is a function of x in absorbing media but is not a function of time. With these, I get S = ˆq c 8π [n + n cos 2φ κ sin 2φ] E 2. (3.22) S has a time-independent and a time-dependent component. The time-dependence is embedded in φ; S oscillates at twice the frequency of the electric field. Being mostly interested in the time-independent part, I calculate the average S = 1 T T Many authors argue at this point in the Poynting discussion that the reason to be interested in the time-averaged flux rather than the instantaneous flux is that it is impossible to follow the instantaneous flux. This statement is of course a canard. If I have a laser oscillating at 300 THz, I can mix it on a photodiode with a local-oscillator laser oscillating at 300 THz + 10 MHz and look at the 10 MHz beat note to determine the amplitude and phase of the primary laser. I can shine light at GHz frequencies on a fast photodiode and see the output on an oscilloscope. It is interesting to set up a photodiode looking up at 50 Hz or 60 Hz fluorescent lights illuminating an office or lab and observe that the intensity follows I cos 2 (2πft) =(1+cos4πft)/2. Of course my eyes cannot quite follow this 100 or 120 Hz flicker but electronics can S dt

8 where T is either the period of the wave or a time large compared to many periods. The oscillating terms average to zero, leaving only the constant-in-time part, S = ˆq c 8π n E 2 = ˆqI, (3.23) where I is the wave intensity, the energy flux/unit area. Equation 3.23 gives the energy flow in isotropic, homogeneous, linear media. One upshot of this discussion is that the usual definition of the time-averaged Poynting vector for harmonically-varying fields, 2 S = c 8π E H (3.24) is only correct for non absorbing media. It does give the direction and take care of the factor e i(qx ωt) in both E and H because the effect of the complex conjugation is to eliminate this term. But H = NE and N is a complex function. If however I decide that the real part is the observable, Eq agrees with Eq Returning to Eq for the field, I have at last the simple result for the Poynting vector as a function of distance traveled in the solid: S(x) = ˆq c 8π n E(0) 2 e 2 ω c κx, where E(0) is the value of the electric field at x = 0. I ll simplify by writing the intensity as where I 0 is the intensity in the y-z plane at x =0and I = I 0 e 2 ω c κx I 0 e αx (3.25) α =2 ω c κ (3.26) is the absorption coefficient (with units cm 1 ) of the material. It is understood that I follow the ray traced by the light wave from x = 0 in calculating the intensity. Equation 3.25 is known as Lambert s Law, Beer s Law, or the Beer-Lambert Law. Note that because ɛ 2 =2nκ, one can also write α = ω ɛ 2 c n = 4πσ 1 nc. (3.27) Eq is exact but is often most useful when ɛ 2 ɛ 1 so that n ɛ 1. Another useful result is that α =2/δ with δ the penetration depth of the field (the skin depth ). If ɛ 2 =0andɛ 1 > 0,thenn = ɛ 1 and κ = 0. Because κ governs the damping of the wave, a material with positive ɛ 1 and ɛ 2 = 0 is nonabsorbing; the electromagnetic wave propagates to infinity without loss of amplitude. If ɛ 1 < 0andɛ 2 =0,thenκ = ɛ 1 and n = 0. The electric field is damped as e κ ω c x with no oscillatory part at all. (The wavelength is infinite.) Beer s Law is conventionally written in terms of the concentration c of a molecule in solution, and breaks the absorption coefficient into two parts: α = cε where ε is a molecular constant called the molar extinction coefficient. It is not immediately obvious whether this damping in the negative ɛ 1 plus ɛ 2 = 0 case is accompanied by energy absorption. I need to consider energy flow across an interface to determine that, in fact, there is no absorption. I can argue here that the current j is 90 out of phase with the electric field E, sothetime average dissipation, j E, is zero. 23

9 3.6 Normal-incidence reflectance I ll discuss reflection and transmission of a sample (with a front and a back surface) in Chapter 7. For the moment I want to consider the simpler case of light incident normally on an interface between two semi-infinite media, one with complex dielectric function ɛ a (refractive index N a ), occupied by the incoming and reflected fields, and a second with complex dielectric function ɛ b (refractive index N b ), occupied by the transmitted field. See Fig Fig Field vectors for incoming, reflected, and transmitted waves at an interface between two media. Letmeorienttheˆq vector of the incoming field along the ˆx axis and its E field along the ŷ axis, so that the H field is along ẑ. The interface is in the y-z plane at x =0. The incident fields are E i = E 0 ŷe i(qx ωt) and H i = N a E 0 ẑe i(qx ωt), where I ve used Eq to write the magnetic field amplitude. The reflected field is traveling along the x axis, so q r = qx or ˆq = ˆx. Its amplitude is proportional to E 0 and to the amplitude reflection coefficient r. E r = re 0 ŷe i( qx ωt) and H r = N a re 0 ẑe i( qx ωt). Reversing ˆq means that one of ê and ĥ (but not both!) must reverse also. I chose ĥ. Both fields have q = ωn a /c. 24

10 The transmitted field is in the second medium (b). Hence, it has q = ωn b /c. Italsois linear in E 0 and in the amplitude transmission coefficient t. E t = te 0 ŷe i(qx ωt) and H t = tn b E 0 ẑe i(qx ωt). All fields are parallel to the interface, so the boundary conditions are that the tangential components of the total fields are continuous across the interface so lets dot with the appropriate unit vector, cancel E 0, which appears in every field, and evaluate the boundary conditions at the interface, x = 0. I can evaluate the boundary conditions at any convenient time, zero being the most convenient. The boundary conditions produce two equations: E is continuous (E i + E r = E t ): 1+r = t. (3.28) H is continuous (H i + H r = H t ): N a (1 r) =N b t. (3.29) Substitute t from Eq into Eq and solve for r to find: This clearly gives the intuitive answer when N a = N b. Now that I know r, itiseasytofindt using Eq. 3.28: r = N a N b N a + N b. (3.30) t = 2N a N a + N b. (3.31) For the moment, I ll specialize to the case where the incident and reflected waves are traveling in vacuum, so N a = 1. I can let N b N and write Eq as r = 1 N 1+N. (3.32) It seems that for N>1 and mostly real (a common case) the amplitude reflection coefficient is negative. This fact is usually expressed by saying the reflection from less-dense to moredense media has a 180 phase shift. The condition on H is actually that the fields are discontinuous by the surface current density 4πK/c but for the moment let me assume no surface charges or currents. 25

11 In most experiments, one measures the intensity of incident and reflected light. The normal-incidence reflectance is proportional to E r 2,i.e.,R = rr.withn = n + iκ, the reflectance is R = 1 N 2 1+N = (n 1)2 + κ 2 (3.33) (n +1) 2 + κ 2. The range of R is 0 1. To see how R depends on the refractive index, let me first look at the case where the material is weakly absorbing, so that κ can be neglected, and R (n 1) 2 /(n +1) 2. Then, if n is not large (say in the range of 1.1 to 1.9, common values for glass and other materials in the visible), R is in the range to 0.1. It requires n = 6 to raise R to 0.5 or so. R =1onlywhenn is very very large If κ is large, n can be close to unity and the reflectance still will be large. In the case when n 1, R approaches 1 independent of κ. 3.7 What if my solid is magnetic? If I keep the requirements of linear, isotropic, and homogeneous materials, but allow there to be an induced magnetization M, I will have a non-zero magnetic susceptibility χ m.the magnetization in a linear material is proportional to the magnetizing field H: M = χ m H. Then, using B = H +4πM, IseethatB = μh (rather than B = H). In a microscopic picture, the magnetic dipoles within the material are rearranged by the external field. Materials with μ>1are paramagnetic (aluminum, platinum, MnO 2 ); those with with μ<1 are diamagnetic (bismuth, silver, water, benzene). Typical deviations from one are small. IhavetomodifyEq.3.3whenM 0. UsingB = μh, Eq. 3.3c becomes q E = ω μh. (3.34) c Having the permeability in Eq. 3.3c increases the symmetry; now two equations have materials properties in them. Note that μ is complex, with μ = μ 1 + iμ 2. Detectors typically can respond to amplitudes at low frequencies (up to GHz) but not at optical frequencies (THz to PHz and more). Instead they respond to the intensity or power in the optical beam. That case occurs in metals. In the extreme limit of materials with n =0andκ finite, R =1. Anyvalue for κ suffices here. This result is the proof for the assertion in the footnote on page 23 that there is no absorption when ɛ 2 =0. The conduction electrons of a metal exhibit Pauli paramagnetism, a definite quantum mechanical effect. The core electrons or filled shells exhibit core diamagnetism, another quantum effect. The sum of these terms determines whether the material is overall paramagnetic or diamagnetic. 26

12 Continuing with the algebra which gave Eqs. 3.9 and 3.10 I find q = ω c ɛμ (3.35) and N = ɛμ. (3.36) The algebra which gave Eq leads to ɛ H = μ E. Finally, the normal incidence amplitude reflectivity (Eq. 3.30) becomes r = ɛa μ a ɛa μ a + ɛb μ b ɛb μ b (3.37) and the reflectance (intensity) is R = rr. 3.8 Negative index materials The refractive index, Eq. 3.36, is the square root of the product of the dielectric function and the permeability. It is interesting to ask if there is ever a case where I should use the negative square root? The answer to the question is yes To simplify the discussion of negative refractive indices, let me restrict myself to materials where ɛ and μ are pure real numbers, and allow them to be either positive or negative. In this case, the material is nonabsorbing. Start by writing N 2 = ɛμ. In the usual case of transparent materials, ɛ>0andμ>0. Then I have a pure real and positive refractive index, N = n = ɛμ. If one or the other of ɛ and μ is negative and the other one positive, then I have a pure imaginary refractive index, N = iκ. Now let me consider the case where both ɛ and μ are negative. Their product is positive so their square root is real. I ll start my discussion with Eqs. 2.7a 2.7d. I ll rewrite them using constitutive relations D = ɛe and B = μh. Itakej = 0 because j = σ 1 E =(ɛ 2 /4πω)E and ɛ 2 is zero for real ɛ. Then, ɛq E =0 μq H =0 q E = μ ω c H (3.38a) (3.38b) (3.38c) q H = ɛ ω c E (3.38d) The first two show as always that q is orthogonal to E and H. The third shows that E and H are orthogonal. (The fourth is redundant as to direction; it also shows H and E are orthogonal.) I have not yet shown evidence that negative dielectric functions or permeability exit. In Chapter 4 I will discuss classical models of the Drude metal and the Lorentz oscillator. In both, there are ranges of frequencies where the dielectric function is negative. Oscillator models for the permeability also give frequencies where it is negative. 27

13 To be specific, I can without loss of generality orient once again E along ŷ and H along ẑ. Then, according to Eq. 3.23, the Poynting vector S is along ˆx. Indeed, ê ĥ = ŝ, so that S, E, andh form a right-handed set. Energy flows along the positive x axis. Now, both μ and ɛ are negative, so that the directional parts of Eqs. 3.38c and 3.38d read ˆq ê = ĥ and ˆq ĥ =+ê. These equations both show that q is oriented along ˆx. Thus q, E, andh form a left-handed set. For this reason the term left handed materials is sometimes used. Wave motion is along the negative x axis. I can achieve this by writing Eq (with κ =0) E = E 0 e i(n ω c x ωt), and defining n = ɛμ to be a negative number. Thus I ve shown that I must chose the negative square root when both ɛ and μ are negative. Finding a material with both ɛ and μ negative at the same frequency is not easy. Moreover, many applications want the magnitudes not to be too large. Therefore, studies in this area have relied on synthetic materials, usually called metamaterials. A common example is a periodic array of circuit boards with unit cells made up of conducting straight wires for the negative ɛ elements and metallic split-ring resonators (looking like a c inside a reversed C) for the negative μ elements. 14 The materials are however far from non absorbing, requiring both ɛ and μ to be complex. The theory can handle this, but at some cost in complexity. 9,10,12 The normal incidence amplitude refection is still given by Eq More interesting things happen for oblique incidence. Snell s law (Appendix E) for refraction across an interface separating an ordinary material with refractive index n o > 0 and a negative index material with refractive index n n < 0readsn o sin θ o = n n sin θ n, appearing identical to the ordinary Snell s law. However, the negative n n then requires a negative θ n, and the refracted ray is on the same side of the interface normal as is the incident ray, rather than crossing over it as happens in the usual case. 28