Waves in composites and metamaterials/Bloch waves and the quasistatic limit

The content of these notes is based on the lectures by Prof. Graeme W. Milton (University of Utah) given in a course on metamaterials in Spring 2007.

Bloch Theorem

In the previous lecture we showed that Maxwell's equations at fixed frequency can be formulated in terms of the fields $𝐃$ and $𝐁$ as

$\text{(1)}𝜵\cdot 𝐃=0;𝜵\cdot 𝐁=0;i𝜵\times \left(\frac{𝐃}{ϵ}\right)=\omega 𝐁;-i𝜵\times \left(\frac{𝐁}{\mu }\right)=\omega 𝐃.$

Equations (1) suggest that we should look for solutions $𝐃$ and $𝐁$ in the space of divergence-free fields such that

$\text{(2)}ℒ\left[\begin{array}{c}𝐃\\ 𝐁\end{array}\right]=\omega \left[\begin{array}{c}𝐃\\ 𝐁\end{array}\right]\text{or}(ℒ-\omega 1)\left[\begin{array}{c}𝐃\\ 𝐁\end{array}\right]=0$

where the operator $ℒ$ is given by

$\text{(3)}ℒ:=\left[\begin{array}{cc}0& -i𝜵\times {\mu }^{-1}\\ i𝜵\times {ϵ}^{-1}& 0\end{array}\right].$

If the medium is such that the permittivity $ϵ(𝐱)$ and the permeability $\mu (𝐱)$ are periodic, i.e.,

$ϵ(𝐱)=ϵ(𝐱+𝐑);\mu (𝐱)=\mu (𝐱+𝐑)$

where $𝐑$ is a lattice vector (see Figure 1) then the operator $ℒ$ has the same periodicity as the medium.

Figure 1. Lattice vector in a periodic medium.

Also recall the translation operator ${𝒯}_R$ defined as

${𝒯}_R\left[\begin{array}{c}𝐃(𝐱)\\ 𝐁(𝐱)\end{array}\right]=\left[\begin{array}{c}𝐃(𝐱+𝐑)\\ 𝐁(𝐱+𝐑)\end{array}\right].$

Periodicity of the medium implies that ${𝒯}_R$ commutes with $ℒ$, i.e.,

${𝒯}_Rℒ=ℒ{𝒯}_R.$

The translation operator is unitary, i.e.,

${𝒯}_R{𝒯}_R^T={𝒯}_R{𝒯}_{-R}=1.$

This means that the adjoint operator ${𝒯}_R^T$ is equal to the inverse operator ${𝒯}_{-R}$.

The translation operator also commutes, i.e.,

${𝒯}_R{𝒯}_{R^{\prime }}={𝒯}_{R^{\prime }}{𝒯}_R={𝒯}_{R+R^{\prime }}.$

Also, since $ℒ$ and ${𝒯}_R$ commute, the operators $ℒ-\omega 1$ and ${𝒯}_R$ must also commute. This implies that

${𝒯}_R(ℒ-\omega 1)\left[\begin{array}{c}𝐃(𝐱)\\ 𝐁(𝐱)\end{array}\right]=0=(ℒ-\omega 1){𝒯}_R\left[\begin{array}{c}𝐃(𝐱)\\ 𝐁(𝐱)\end{array}\right]=(ℒ-\omega 1)\left[\begin{array}{c}𝐃(𝐱+𝐑)\\ 𝐁(𝐱+𝐑)\end{array}\right].$

Hence the eigenstates of $ℒ-\omega 1$ and the eigenstates of ${𝒯}_R$ lie in the same space. Therefore, any solution can be expressed in fields which are simultaneously eigenstates of all the ${𝒯}_R$, i.e., these eigenstates have the property

$\text{(4)}[{𝒯}_R-c(𝐑)1]\left[\begin{array}{c}𝐃(𝐱)\\ 𝐁(𝐱)\end{array}\right]=0.$

Since ${𝒯}_R{𝒯}_{R^{\prime }}={𝒯}_{R+R^{\prime }}$, we have

$c(𝐑)c({𝐑}^{\prime })=c(𝐑+{𝐑}^{\prime }).$

So it suffices to know $c(𝐑)$ when $𝐑={𝐚}_1,{𝐚}_2,{𝐚}_3$ where the ${𝐚}_i$'s are the primitive vectors of the lattice, i.e.,

$ϵ(𝐱+{𝐚}_j)=ϵ(𝐱);\mu (𝐱+{𝐚}_j)=\mu (𝐱).$

Let us assume that

$c({𝐚}_j)=e^{2\pi i{\alpha }_j}j=1,2,3$

for a suitable choice of ${\alpha }_j$.

Then for any lattice vector

$𝐑=n_1{𝐚}_1+n_2{𝐚}_2+n_3{𝐚}_3$

we have

$\begin{array}{cc}c(𝐑)& =c(n_1{𝐚}_1+n_2{𝐚}_2+n_3{𝐚}_3)=c(n_1{𝐚}_1)c(n_2{𝐚}_2)c(n_3{𝐚}_3)\\ & =c\left(\sum _{i=1}^{n_1}{𝐚}_1\right)c\left(\sum _{i=1}^{n_2}{𝐚}_2\right)c\left(\sum _{i=1}^{n_3}{𝐚}_3\right)=[c({𝐚}_1){]}^{n_1}[c({𝐚}_2){]}^{n_2}[c({𝐚}_3){]}^{n_3}\\ & =e^{2\pi i{\alpha }_1{n}_1}{e}^{2\pi i{\alpha }_2{n}_2}{e}^{2\pi i{\alpha }_3{n}_3}=e^{2\pi i({\alpha }_1{n}_1+{\alpha }_2{n}_2+{\alpha }_3{n}_3)}\end{array}$

or,

$c(𝐑)=e^{2\pi i({\alpha }_1{n}_1+{\alpha }_2{n}_2+{\alpha }_3{n}_3)}.$

Define a vector

$𝐤:={\alpha }_1{𝐛}_1+{\alpha }_2{𝐛}_2+{\alpha }_3{𝐛}_3$

where the vectors ${𝐛}_i$ are the reciprocal lattice vectors satisfying

${𝐛}_i\cdot {𝐚}_j=2\pi {\delta }_{ij}.$

Then,

$𝐤\cdot 𝐑=({\alpha }_1{𝐛}_1+{\alpha }_2{𝐛}_2+{\alpha }_3{𝐛}_3)\cdot (n_1{𝐚}_1+n_2{𝐚}_2+n_3{𝐚}_3)={\alpha }_1{n}_{1}2\pi +{\alpha }_2{n}_{2}2\pi +{\alpha }_3{n}_{3}2\pi$

or,

$𝐤\cdot 𝐑=2\pi ({\alpha }_1{n}_1+{\alpha }_2{n}_2+{\alpha }_3{n}_3).$

Therefore, we have

$c(𝐑)=e^{i𝐤\cdot 𝐑}.$

Plugging this expression into (4), we get

$[{𝒯}_R-e^{i𝐤\cdot 𝐑}1]\left[\begin{array}{c}𝐃(𝐱)\\ 𝐁(𝐱)\end{array}\right]=0$

or,

$\text{(5)}\left[\begin{array}{c}𝐃(𝐱+𝐑)\\ 𝐁(𝐱+𝐑)\end{array}\right]=e^{i𝐤\cdot 𝐑}\left[\begin{array}{c}𝐃(𝐱)\\ 𝐁(𝐱)\end{array}\right].$

Equation (5) is called the Bloch condition.

In summary, the solutions to the electromagnetic equations in a periodic medium can be expressed in Bloch waves where each Bloch wave is a time harmonic solution to the electromagnetic equations which in addition satisfies the Bloch condition for all lattice vectors $𝐑$ and for some appropriate choice of $𝐤$.

Note that for any vector $𝐱$, the Bloch condition implies that

$e^{-i𝐤\cdot (𝐱+𝐑)}\left[\begin{array}{c}𝐃(𝐱+𝐑)\\ 𝐁(𝐱+𝐑)\end{array}\right]=e^{i𝐤\cdot 𝐑-i𝐤\cdot 𝐑-i𝐤\cdot 𝐱}\left[\begin{array}{c}𝐃(𝐱)\\ 𝐁(𝐱)\end{array}\right]=e^{-i𝐤\cdot 𝐱}\left[\begin{array}{c}𝐃(𝐱)\\ 𝐁(𝐱)\end{array}\right].$

Therefore the quantity

$e^{-i𝐤\cdot 𝐱}\left[\begin{array}{c}𝐃(𝐱)\\ 𝐁(𝐱)\end{array}\right]$

is periodic.

Quasistatic Limit

Let us now consider the solution of Maxwell's equation in periodic media in the quasistatic limit. Consider the periodic medium shown in Figure 2. The lattice spacing is $\eta$.

Figure 2. Periodic medium with $x$ and $y$ spaces.

Define

${ϵ}_{\eta }(𝐱):=ϵ\left(\frac{𝐱}{\eta }\right)=ϵ(𝐲);{\mu }_{\eta }(𝐱):=\mu \left(\frac{𝐱}{\eta }\right)=\mu (𝐲).$

These are periodic functions, i.e.,

$ϵ(𝐲+{𝐚}_i)=ϵ(𝐲);\mu (𝐲+{𝐚}_i)=\mu (𝐲)$

where ${𝐚}_i$ are the primitive lattice vectors. We may also write these periodicity conditions as

${ϵ}_{\eta }(𝐱+\eta {𝐚}_i)={ϵ}_{\eta }(𝐱);{\mu }_{\eta }(𝐱+\eta {𝐚}_i)={\mu }_{\eta }(𝐱).$

Similarly, define

${𝐃}_{\eta }(𝐱):=𝐃(𝐲);{𝐁}_{\eta }(𝐱):=𝐁(𝐲);{𝐄}_{\eta }(𝐱):=𝐄(𝐲);{𝐇}_{\eta }(𝐱):=𝐇(𝐲).$

Then Maxwell's equations can be written as

$\text{(6)}𝜵\cdot {𝐃}_{\eta }=0;𝜵\cdot {𝐁}_{\eta }=0;𝜵\times {𝐄}_{\eta }-i\omega {𝐁}_{\eta }=0;𝜵\times {𝐇}_{\eta }+i\omega {𝐃}_{\eta }=0.$

Let us look for Bloch wave solutions of the form

${𝐄}_{\eta }(𝐱)=e^{i𝐤\cdot 𝐱}{𝐞}_{\eta }(𝐱);{𝐃}_{\eta }(𝐱)=e^{i𝐤\cdot 𝐱}{𝐝}_{\eta }(𝐱);{𝐇}_{\eta }(𝐱)=e^{i𝐤\cdot 𝐱}{𝐡}_{\eta }(𝐱);{𝐁}_{\eta }(𝐱)=e^{i𝐤\cdot 𝐱}{𝐛}_{\eta }(𝐱)$

where ${𝐞}_{\eta },{𝐝}_{\eta },{𝐡}_{\eta },{𝐛}_{\eta }$ have the same periodicity as $ϵ$ and $\mu$, i.e.,

${𝐞}_{\eta }(𝐱+\eta {𝐚}_i)=𝐞(𝐱);{𝐝}_{\eta }(𝐱+\eta {𝐚}_i)=𝐝(𝐱);{𝐡}_{\eta }(𝐱+\eta {𝐚}_i)=𝐡(𝐱);{𝐛}_{\eta }(𝐱+\eta {𝐚}_i)=𝐛(𝐱).$

From the constitutive relations, we get

${𝐝}_{\eta }(𝐱)={ϵ}_{\eta }(𝐱){𝐞}_{\eta }(𝐱);{𝐛}_{\eta }(𝐱)={\mu }_{\eta }(𝐱){𝐡}_{\eta }(𝐱).$

Recall that, for periodic media, Maxwell's equations may be expressed as

$(ℒ-\omega 1)\left[\begin{array}{c}𝐃\\ 𝐁\end{array}\right]=0.$

Here $\omega$ is an eigenvalue of $ℒ$. However, $ℒ$ depends on $\omega$ via $ϵ(\omega )$ and $\mu (\omega )$. {\bf Bloch wave solutions do not exists unless $\omega$ takes one of a discrete set of values.}

Let these discrete values be

$\omega ={\omega }_{\eta }^j(𝐤)$

where the superscript $j$ labels the solution branches.

Let us see what the Bloch wave solutions reduce to as $\eta \to 0$. Following standard multiple scale analysis, let us assume that the periodic complex fields have the expansions

$\text{(7)}\begin{array}{cc}{𝐞}_{\eta }(𝐱)& ={𝐞}_0(𝐲)+\eta {𝐚}_1(𝐲)+{\eta }^2{𝐚}_2(𝐲)+\dots \\ {𝐝}_{\eta }(𝐱)& ={𝐝}_0(𝐲)+\eta {𝐝}_1(𝐲)+{\eta }^2{𝐝}_2(𝐲)+\dots \\ {𝐡}_{\eta }(𝐱)& ={𝐡}_0(𝐲)+\eta {𝐡}_1(𝐲)+{\eta }^2{𝐡}_2(𝐲)+\dots \\ {𝐛}_{\eta }(𝐱)& ={𝐛}_0(𝐲)+\eta {𝐛}_1(𝐲)+{\eta }^2{𝐛}_2(𝐲)+\dots \end{array}$

Let us also assume that the dependence of $\omega$ on $\eta$ and $𝐤$ has an expansion of the form

$\text{(8)}\omega ={\omega }_{\eta }^j(𝐤)={\omega }_0^j(𝐲)+\eta {\omega }_1^j(𝐲)+{\eta }^2{\omega }_2^j(𝐲)+\dots$

Plugging (8) and (7) into (6) gives

$\text{(9)}\begin{array}{cc}𝜵\cdot (e^{i\eta 𝐤\cdot 𝐲}[{𝐝}_0(𝐲)+\eta {𝐝}_1(𝐲)+\dots ])& =0\\ 𝜵\cdot (e^{i\eta 𝐤\cdot 𝐲}[{𝐛}_0(𝐲)+\eta {𝐛}_1(𝐲)+\dots ])& =0\\ 𝜵\times (e^{i\eta 𝐤\cdot 𝐲}[{𝐞}_0(𝐲)+\eta {𝐞}_1(𝐲)+\dots ])-i[{\omega }_0^j+\eta {\omega }_1^j+\dots ][{𝐛}_0(𝐲)+\eta {𝐛}_1(𝐲)+\dots ]& =0\\ 𝜵\times (e^{i\eta 𝐤\cdot 𝐲}[{𝐡}_0(𝐲)+\eta {𝐡}_1(𝐲)+\dots ])+i[{\omega }_0^j+\eta {\omega }_1^j+\dots ][{𝐝}_0(𝐲)+\eta {𝐝}_1(𝐲)+\dots ]& =0.\end{array}$

Define

${𝜵}_y:=(\frac{\partial }{\partial y_1},\frac{\partial }{\partial y_2},\frac{\partial }{\partial y_3}).$

Then, for a vector field $𝐯(𝐲)$, using the chain rule we get

$\text{(10)}𝜵\cdot 𝐯(𝐲)=\frac{1}{\eta }{𝜵}_y\cdot 𝐯(𝐲);𝜵\times 𝐯(𝐲)=\frac{1}{\eta }{𝜵}_y\times 𝐯(𝐲).$

Using definitions (10) in (9) and collecting terms of order $1/\eta$ gives

$\text{(11)}\begin{array}{c}{𝜵}_y\cdot {𝐝}_0(𝐲)=0\\ {𝜵}_y\cdot {𝐛}_0(𝐲)=0\\ {𝜵}_y\times {𝐞}_0(𝐲)=0\\ {𝜵}_y\times {𝐡}_0(𝐲)=0\end{array}$

These are the solutions in the quasistatic limit. Also, from the constitutive equations

${𝐝}_0(𝐲)=ϵ(𝐲){𝐞}_0(𝐲);{𝐛}_0(𝐲)=\mu (𝐲){𝐡}_0(𝐲).$

Similarly, collecting terms of order 1 from the expanded Maxwell's equations (9) we get

$\text{(12)}\begin{array}{c}i𝐤\cdot {𝐝}_0(𝐲)+{𝜵}_y\cdot {𝐝}_1(𝐲)=0\\ i𝐤\cdot {𝐛}_0(𝐲)+{𝜵}_y\cdot {𝐛}_1(𝐲)=0\\ i𝐤\times {𝐞}_0(𝐲)+{𝜵}_y\times {𝐞}_1(𝐲)-i{\omega }_0^j{𝐛}_0(𝐲)=0\\ i𝐤\times {𝐡}_0(𝐲)+{𝜵}_y\times {𝐡}_1(𝐲)+i{\omega }_0^j{𝐝}_0(𝐲)=0.\end{array}$

Since ${𝐝}_1(𝐲),{𝐛}_1(𝐲),{𝐞}_1(𝐲),{𝐡}_1(𝐲)$ are periodic, this implies that

$\text{(13)}\begin{array}{c}⟨{𝜵}_y\cdot {𝐝}_1(𝐲)⟩=0\\ ⟨{𝜵}_y\cdot {𝐛}_1(𝐲)⟩=0\\ ⟨{𝜵}_y\times {𝐞}_1(𝐲)⟩=0\\ ⟨{𝜵}_y\times {𝐡}_1(𝐲)⟩=0.\end{array}$

where $⟨\bullet ⟩$ is the volume average over the unit cell. So a necessary condition that equations (12) have a solution is that

$\text{(14)}\begin{array}{c}i𝐤\cdot ⟨{𝐝}_0(𝐲)⟩=0\\ i𝐤\cdot ⟨{𝐛}_0(𝐲)⟩=0\\ i𝐤\times ⟨{𝐞}_0(𝐲)⟩-i{\omega }_0^j⟨{𝐛}_0(𝐲)⟩=0\\ i𝐤\times ⟨{𝐡}_0(𝐲)⟩+i{\omega }_0^j⟨{𝐝}_0(𝐲)⟩=0.\end{array}$

Note that the second pair of (14) implies the first pair.

References

• N. W. Ashcroft and N. D. Mermin. Solid State Physics. Saunders, New York, 1976.
• G. W. Milton. Theory of Composites. Cambridge University Press, New York, 2002.