# 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 ${\displaystyle \mathbf {D} }$ and ${\displaystyle \mathbf {B} }$ as [1]

${\displaystyle {\text{(1)}}\qquad {\boldsymbol {\nabla }}\cdot \mathbf {D} =0~;~~{\boldsymbol {\nabla }}\cdot \mathbf {B} =0~;~~i~{\boldsymbol {\nabla }}\times \left({\cfrac {\mathbf {D} }{\epsilon }}\right)=\omega ~\mathbf {B} ~;~~-i~{\boldsymbol {\nabla }}\times \left({\cfrac {\mathbf {B} }{\mu }}\right)=\omega ~\mathbf {D} ~.}$

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

${\displaystyle {\text{(2)}}\qquad {\mathcal {L}}{\begin{bmatrix}\mathbf {D} \\\mathbf {B} \end{bmatrix}}=\omega ~{\begin{bmatrix}\mathbf {D} \\\mathbf {B} \end{bmatrix}}\qquad {\text{or}}\qquad {({\mathcal {L}}-\omega ~{\boldsymbol {1}}){\begin{bmatrix}\mathbf {D} \\\mathbf {B} \end{bmatrix}}={\boldsymbol {0}}}}$

where the operator ${\displaystyle {\mathcal {L}}}$ is given by

${\displaystyle {\text{(3)}}\qquad {{\mathcal {L}}:={\begin{bmatrix}0&-i~{\boldsymbol {\nabla }}\times \mu ^{-1}\\i~{\boldsymbol {\nabla }}\times \epsilon ^{-1}&0\end{bmatrix}}~.}}$

If the medium is such that the permittivity ${\displaystyle \epsilon (\mathbf {x} )}$ and the permeability ${\displaystyle \mu (\mathbf {x} )}$ are periodic, i.e.,

${\displaystyle \epsilon (\mathbf {x} )=\epsilon (\mathbf {x} +\mathbf {R} )~;~~\mu (\mathbf {x} )=\mu (\mathbf {x} +\mathbf {R} )}$

where ${\displaystyle \mathbf {R} }$ is a lattice vector (see Figure 1) then the operator ${\displaystyle {\mathcal {L}}}$ has the same periodicity as the medium.

 Figure 1. Lattice vector in a periodic medium.

Also recall the translation operator ${\displaystyle {\mathcal {T}}_{R}}$ defined as

${\displaystyle {{\mathcal {T}}_{R}{\begin{bmatrix}\mathbf {D} (\mathbf {x} )\\\mathbf {B} (\mathbf {x} )\end{bmatrix}}={\begin{bmatrix}\mathbf {D} (\mathbf {x} +\mathbf {R} )\\\mathbf {B} (\mathbf {x} +\mathbf {R} )\end{bmatrix}}~.}}$

Periodicity of the medium implies that ${\displaystyle {\mathcal {T}}_{R}}$ commutes with ${\displaystyle {\mathcal {L}}}$, i.e.,

${\displaystyle {{\mathcal {T}}_{R}~{\mathcal {L}}={\mathcal {L}}~{\mathcal {T}}_{R}~.}}$

[2] The translation operator is unitary, i.e.,

${\displaystyle {{\mathcal {T}}_{R}~{\mathcal {T}}_{R}^{T}={\mathcal {T}}_{R}~{\mathcal {T}}_{-R}={\boldsymbol {1}}~.}}$

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

The translation operator also commutes, i.e.,

${\displaystyle {{\mathcal {T}}_{R}~{\mathcal {T}}_{R'}={\mathcal {T}}_{R'}~{\mathcal {T}}_{R}={\mathcal {T}}_{R+R'}~.}}$

[3] Also, since ${\displaystyle {\mathcal {L}}}$ and ${\displaystyle {\mathcal {T}}_{R}}$ commute, the operators ${\displaystyle {\mathcal {L}}-\omega {\boldsymbol {1}}}$ and ${\displaystyle {\mathcal {T}}_{R}}$ must also commute. This implies that

${\displaystyle {\mathcal {T}}_{R}~({\mathcal {L}}-\omega ~{\boldsymbol {1}})~{\begin{bmatrix}\mathbf {D} (\mathbf {x} )\\\mathbf {B} (\mathbf {x} )\end{bmatrix}}={\boldsymbol {0}}=({\mathcal {L}}-\omega ~{\boldsymbol {1}})~{\mathcal {T}}_{R}~{\begin{bmatrix}\mathbf {D} (\mathbf {x} )\\\mathbf {B} (\mathbf {x} )\end{bmatrix}}=({\mathcal {L}}-\omega ~{\boldsymbol {1}})~~{\begin{bmatrix}\mathbf {D} (\mathbf {x} +\mathbf {R} )\\\mathbf {B} (\mathbf {x} +\mathbf {R} )\end{bmatrix}}~.}$

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

${\displaystyle {\text{(4)}}\qquad {[{\mathcal {T}}_{R}-c(\mathbf {R} )~{\boldsymbol {1}}]{\begin{bmatrix}\mathbf {D} (\mathbf {x} )\\\mathbf {B} (\mathbf {x} )\end{bmatrix}}={\boldsymbol {0}}~.}}$

Since ${\displaystyle {\mathcal {T}}_{R}~{\mathcal {T}}_{R'}={\mathcal {T}}_{R+R'}}$, we have

${\displaystyle c(\mathbf {R} )~c(\mathbf {R} ')=c(\mathbf {R} +\mathbf {R} ')~.}$

So it suffices to know ${\displaystyle c(\mathbf {R} )}$ when ${\displaystyle \mathbf {R} =\mathbf {a} _{1},\mathbf {a} _{2},\mathbf {a} _{3}}$ where the ${\displaystyle \mathbf {a} _{i}}$'s are the primitive vectors of the lattice, i.e.,

${\displaystyle \epsilon (\mathbf {x} +\mathbf {a} _{j})=\epsilon (\mathbf {x} )~;~~\mu (\mathbf {x} +\mathbf {a} _{j})=\mu (\mathbf {x} )~.}$

Let us assume that

${\displaystyle c(\mathbf {a} _{j})=e^{2\pi ~i~\alpha _{j}}\qquad j=1,2,3}$

for a suitable choice of ${\displaystyle \alpha _{j}}$.

Then for any lattice vector

${\displaystyle \mathbf {R} =n_{1}~\mathbf {a} _{1}+n_{2}~\mathbf {a} _{2}+n_{3}~\mathbf {a} _{3}}$

we have

{\displaystyle {\begin{aligned}c(\mathbf {R} )&=c(n_{1}~\mathbf {a} _{1}+n_{2}~\mathbf {a} _{2}+n_{3}~\mathbf {a} _{3})=c(n_{1}~\mathbf {a} _{1})~c(n_{2}~\mathbf {a} _{2})~c(n_{3}~\mathbf {a} _{3})\\&=c\left(\sum _{i=1}^{n_{1}}\mathbf {a} _{1}\right)~c\left(\sum _{i=1}^{n_{2}}\mathbf {a} _{2}\right)~c\left(\sum _{i=1}^{n_{3}}\mathbf {a} _{3}\right)=[c(\mathbf {a} _{1})]^{n_{1}}~[c(\mathbf {a} _{2})]^{n_{2}}~[c(\mathbf {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{aligned}}}

or,

${\displaystyle c(\mathbf {R} )=e^{2\pi i(\alpha _{1}~n_{1}+\alpha _{2}~n_{2}+\alpha _{3}~n_{3})}~.}$

Define a vector

${\displaystyle \mathbf {k} :=\alpha _{1}~\mathbf {b} _{1}+\alpha _{2}~\mathbf {b} _{2}+\alpha _{3}~\mathbf {b} _{3}}$

where the vectors ${\displaystyle \mathbf {b} _{i}}$ are the reciprocal lattice vectors satisfying

${\displaystyle \mathbf {b} _{i}\cdot \mathbf {a} _{j}=2\pi ~\delta _{ij}~.}$

Then,

${\displaystyle \mathbf {k} \cdot \mathbf {R} =(\alpha _{1}~\mathbf {b} _{1}+\alpha _{2}~\mathbf {b} _{2}+\alpha _{3}~\mathbf {b} _{3})\cdot (n_{1}~\mathbf {a} _{1}+n_{2}~\mathbf {a} _{2}+n_{3}~\mathbf {a} _{3})=\alpha _{1}~n_{1}~2~\pi +\alpha _{2}~n_{2}~2~\pi +\alpha _{3}~n_{3}~2~\pi }$

or,

${\displaystyle \mathbf {k} \cdot \mathbf {R} =2\pi (\alpha _{1}~n_{1}+\alpha _{2}~n_{2}+\alpha _{3}~n_{3})~.}$

Therefore, we have

${\displaystyle c(\mathbf {R} )=e^{i\mathbf {k} \cdot \mathbf {R} }~.}$

Plugging this expression into (4), we get

${\displaystyle [{\mathcal {T}}_{R}-e^{i\mathbf {k} \cdot \mathbf {R} }~{\boldsymbol {1}}]{\begin{bmatrix}\mathbf {D} (\mathbf {x} )\\\mathbf {B} (\mathbf {x} )\end{bmatrix}}={\boldsymbol {0}}}$

or,

${\displaystyle {\text{(5)}}\qquad {{\begin{bmatrix}\mathbf {D} (\mathbf {x} +\mathbf {R} )\\\mathbf {B} (\mathbf {x} +\mathbf {R} )\end{bmatrix}}=e^{i\mathbf {k} \cdot \mathbf {R} }~{\begin{bmatrix}\mathbf {D} (\mathbf {x} )\\\mathbf {B} (\mathbf {x} )\end{bmatrix}}~.}}$

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 ${\displaystyle \mathbf {R} }$ and for some appropriate choice of ${\displaystyle \mathbf {k} }$.

Note that for any vector ${\displaystyle \mathbf {x} }$, the Bloch condition implies that

${\displaystyle e^{-i\mathbf {k} \cdot (\mathbf {x} +\mathbf {R} )}{\begin{bmatrix}\mathbf {D} (\mathbf {x} +\mathbf {R} )\\\mathbf {B} (\mathbf {x} +\mathbf {R} )\end{bmatrix}}=e^{i\mathbf {k} \cdot \mathbf {R} -i\mathbf {k} \cdot \mathbf {R} -i\mathbf {k} \cdot \mathbf {x} }~{\begin{bmatrix}\mathbf {D} (\mathbf {x} )\\\mathbf {B} (\mathbf {x} )\end{bmatrix}}=e^{-i\mathbf {k} \cdot \mathbf {x} }~{\begin{bmatrix}\mathbf {D} (\mathbf {x} )\\\mathbf {B} (\mathbf {x} )\end{bmatrix}}~.}$

Therefore the quantity

${\displaystyle e^{-i\mathbf {k} \cdot \mathbf {x} }~{\begin{bmatrix}\mathbf {D} (\mathbf {x} )\\\mathbf {B} (\mathbf {x} )\end{bmatrix}}}$

is periodic.

## Quasistatic Limit

Let us now consider the solution of Maxwell's equation in periodic media in the quasistatic limit. [4] Consider the periodic medium shown in Figure 2. The lattice spacing is ${\displaystyle \eta }$.

 Figure 2. Periodic medium with ${\displaystyle x}$ and ${\displaystyle y}$ spaces.

Define

${\displaystyle \epsilon _{\eta }(\mathbf {x} ):=\epsilon \left({\cfrac {\mathbf {x} }{\eta }}\right)=\epsilon (\mathbf {y} )~;~~\mu _{\eta }(\mathbf {x} ):=\mu \left({\cfrac {\mathbf {x} }{\eta }}\right)=\mu (\mathbf {y} )~.}$

These are periodic functions, i.e.,

${\displaystyle \epsilon (\mathbf {y} +\mathbf {a} _{i})=\epsilon (\mathbf {y} )~;~~\mu (\mathbf {y} +\mathbf {a} _{i})=\mu (\mathbf {y} )}$

where ${\displaystyle \mathbf {a} _{i}}$ are the primitive lattice vectors. We may also write these periodicity conditions as

${\displaystyle \epsilon _{\eta }(\mathbf {x} +\eta \mathbf {a} _{i})=\epsilon _{\eta }(\mathbf {x} )~;~~\mu _{\eta }(\mathbf {x} +\eta \mathbf {a} _{i})=\mu _{\eta }(\mathbf {x} )~.}$

Similarly, define

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

Then Maxwell's equations can be written as

${\displaystyle {\text{(6)}}\qquad {\boldsymbol {\nabla }}\cdot \mathbf {D} _{\eta }=0~;~~{\boldsymbol {\nabla }}\cdot \mathbf {B} _{\eta }=0~;~~{\boldsymbol {\nabla }}\times \mathbf {E} _{\eta }-i\omega ~\mathbf {B} _{\eta }={\boldsymbol {0}}~;~~{\boldsymbol {\nabla }}\times \mathbf {H} _{\eta }+i\omega ~\mathbf {D} _{\eta }={\boldsymbol {0}}~.}$

Let us look for Bloch wave solutions of the form

${\displaystyle \mathbf {E} _{\eta }(\mathbf {x} )=e^{i\mathbf {k} \cdot \mathbf {x} }~\mathbf {e} _{\eta }(\mathbf {x} )~;~~\mathbf {D} _{\eta }(\mathbf {x} )=e^{i\mathbf {k} \cdot \mathbf {x} }~\mathbf {d} _{\eta }(\mathbf {x} )~;~~\mathbf {H} _{\eta }(\mathbf {x} )=e^{i\mathbf {k} \cdot \mathbf {x} }~\mathbf {h} _{\eta }(\mathbf {x} )~;~~\mathbf {B} _{\eta }(\mathbf {x} )=e^{i\mathbf {k} \cdot \mathbf {x} }~\mathbf {b} _{\eta }(\mathbf {x} )}$

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

${\displaystyle \mathbf {e} _{\eta }(\mathbf {x} +\eta ~\mathbf {a} _{i})=\mathbf {e} (\mathbf {x} )~;~~\mathbf {d} _{\eta }(\mathbf {x} +\eta ~\mathbf {a} _{i})=\mathbf {d} (\mathbf {x} )~;~~\mathbf {h} _{\eta }(\mathbf {x} +\eta ~\mathbf {a} _{i})=\mathbf {h} (\mathbf {x} )~;~~\mathbf {b} _{\eta }(\mathbf {x} +\eta ~\mathbf {a} _{i})=\mathbf {b} (\mathbf {x} )~.}$

From the constitutive relations, we get

${\displaystyle \mathbf {d} _{\eta }(\mathbf {x} )=\epsilon _{\eta }(\mathbf {x} )~\mathbf {e} _{\eta }(\mathbf {x} )~;~~\mathbf {b} _{\eta }(\mathbf {x} )=\mu _{\eta }(\mathbf {x} )~\mathbf {h} _{\eta }(\mathbf {x} )~.}$

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

${\displaystyle ({\mathcal {L}}-\omega ~{\boldsymbol {1}}){\begin{bmatrix}\mathbf {D} \\\mathbf {B} \end{bmatrix}}={\boldsymbol {0}}~.}$

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

Let these discrete values be

${\displaystyle \omega =\omega _{\eta }^{j}(\mathbf {k} )}$

where the superscript ${\displaystyle j}$ labels the solution branches.

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

{\displaystyle {\text{(7)}}\qquad {\begin{aligned}\mathbf {e} _{\eta }(\mathbf {x} )&=\mathbf {e} _{0}(\mathbf {y} )+\eta ~\mathbf {a} _{1}(\mathbf {y} )+\eta ^{2}~\mathbf {a} _{2}(\mathbf {y} )+\dots \\\mathbf {d} _{\eta }(\mathbf {x} )&=\mathbf {d} _{0}(\mathbf {y} )+\eta ~\mathbf {d} _{1}(\mathbf {y} )+\eta ^{2}~\mathbf {d} _{2}(\mathbf {y} )+\dots \\\mathbf {h} _{\eta }(\mathbf {x} )&=\mathbf {h} _{0}(\mathbf {y} )+\eta ~\mathbf {h} _{1}(\mathbf {y} )+\eta ^{2}~\mathbf {h} _{2}(\mathbf {y} )+\dots \\\mathbf {b} _{\eta }(\mathbf {x} )&=\mathbf {b} _{0}(\mathbf {y} )+\eta ~\mathbf {b} _{1}(\mathbf {y} )+\eta ^{2}~\mathbf {b} _{2}(\mathbf {y} )+\dots \end{aligned}}}

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

${\displaystyle {\text{(8)}}\qquad \omega =\omega _{\eta }^{j}(\mathbf {k} )=\omega _{0}^{j}(\mathbf {y} )+\eta ~\omega _{1}^{j}(\mathbf {y} )+\eta ^{2}~\omega _{2}^{j}(\mathbf {y} )+\dots }$

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

{\displaystyle {\text{(9)}}\qquad {\begin{aligned}{\boldsymbol {\nabla }}\cdot \left(e^{i\eta ~\mathbf {k} \cdot \mathbf {y} }~[\mathbf {d} _{0}(\mathbf {y} )+\eta ~\mathbf {d} _{1}(\mathbf {y} )+\dots ]\right)&=0\\{\boldsymbol {\nabla }}\cdot \left(e^{i\eta ~\mathbf {k} \cdot \mathbf {y} }~[\mathbf {b} _{0}(\mathbf {y} )+\eta ~\mathbf {b} _{1}(\mathbf {y} )+\dots ]\right)&=0\\{\boldsymbol {\nabla }}\times \left(e^{i\eta ~\mathbf {k} \cdot \mathbf {y} }~[\mathbf {e} _{0}(\mathbf {y} )+\eta ~\mathbf {e} _{1}(\mathbf {y} )+\dots ]\right)-i[\omega _{0}^{j}+\eta ~\omega _{1}^{j}+\dots ][\mathbf {b} _{0}(\mathbf {y} )+\eta ~\mathbf {b} _{1}(\mathbf {y} )+\dots ]&={\boldsymbol {0}}\\{\boldsymbol {\nabla }}\times \left(e^{i\eta ~\mathbf {k} \cdot \mathbf {y} }~[\mathbf {h} _{0}(\mathbf {y} )+\eta ~\mathbf {h} _{1}(\mathbf {y} )+\dots ]\right)+i[\omega _{0}^{j}+\eta ~\omega _{1}^{j}+\dots ][\mathbf {d} _{0}(\mathbf {y} )+\eta ~\mathbf {d} _{1}(\mathbf {y} )+\dots ]&={\boldsymbol {0}}~.\end{aligned}}}

Define

${\displaystyle {\boldsymbol {\nabla }}_{y}:=\left({\frac {\partial }{\partial y_{1}}},{\frac {\partial }{\partial y_{2}}},{\frac {\partial }{\partial y_{3}}}\right)~.}$

Then, for a vector field ${\displaystyle \mathbf {v} (\mathbf {y} )}$, using the chain rule we get

${\displaystyle {\text{(10)}}\qquad {\boldsymbol {\nabla }}\cdot \mathbf {v} (\mathbf {y} )={\cfrac {1}{\eta }}~{\boldsymbol {\nabla }}_{y}\cdot \mathbf {v} (\mathbf {y} )~;~~{\boldsymbol {\nabla }}\times \mathbf {v} (\mathbf {y} )={\cfrac {1}{\eta }}~{\boldsymbol {\nabla }}_{y}\times {\mathbf {v} (\mathbf {y} )}~.}$

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

{\displaystyle {\text{(11)}}\qquad {\begin{aligned}{\boldsymbol {\nabla }}_{y}\cdot \mathbf {d} _{0}(\mathbf {y} )=0\\{\boldsymbol {\nabla }}_{y}\cdot \mathbf {b} _{0}(\mathbf {y} )=0\\{\boldsymbol {\nabla }}_{y}\times {\mathbf {e} _{0}(\mathbf {y} )}={\boldsymbol {0}}\\{\boldsymbol {\nabla }}_{y}\times {\mathbf {h} _{0}(\mathbf {y} )}={\boldsymbol {0}}\end{aligned}}}

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

${\displaystyle \mathbf {d} _{0}(\mathbf {y} )=\epsilon (\mathbf {y} )~\mathbf {e} _{0}(\mathbf {y} )~;~~\mathbf {b} _{0}(\mathbf {y} )=\mu (\mathbf {y} )~\mathbf {h} _{0}(\mathbf {y} )~.}$

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

{\displaystyle {\text{(12)}}\qquad {\begin{aligned}i~\mathbf {k} \cdot \mathbf {d} _{0}(\mathbf {y} )+{\boldsymbol {\nabla }}_{y}\cdot \mathbf {d} _{1}(\mathbf {y} )=0\\i~\mathbf {k} \cdot \mathbf {b} _{0}(\mathbf {y} )+{\boldsymbol {\nabla }}_{y}\cdot \mathbf {b} _{1}(\mathbf {y} )=0\\i~\mathbf {k} \times \mathbf {e} _{0}(\mathbf {y} )+{\boldsymbol {\nabla }}_{y}\times {\mathbf {e} _{1}(\mathbf {y} )}-i~\omega _{0}^{j}~\mathbf {b} _{0}(\mathbf {y} )={\boldsymbol {0}}\\i~\mathbf {k} \times \mathbf {h} _{0}(\mathbf {y} )+{\boldsymbol {\nabla }}_{y}\times {\mathbf {h} _{1}(\mathbf {y} )}+i~\omega _{0}^{j}~\mathbf {d} _{0}(\mathbf {y} )={\boldsymbol {0}}~.\end{aligned}}}

Since ${\displaystyle \mathbf {d} _{1}(\mathbf {y} ),\mathbf {b} _{1}(\mathbf {y} ),\mathbf {e} _{1}(\mathbf {y} ),\mathbf {h} _{1}(\mathbf {y} )}$ are periodic, this implies that

{\displaystyle {\text{(13)}}\qquad {\begin{aligned}\langle {\boldsymbol {\nabla }}_{y}\cdot \mathbf {d} _{1}(\mathbf {y} )\rangle =0\\\langle {\boldsymbol {\nabla }}_{y}\cdot \mathbf {b} _{1}(\mathbf {y} )\rangle =0\\\langle {\boldsymbol {\nabla }}_{y}\times {\mathbf {e} _{1}(\mathbf {y} )}\rangle ={\boldsymbol {0}}\\\langle {\boldsymbol {\nabla }}_{y}\times {\mathbf {h} _{1}(\mathbf {y} )}\rangle ={\boldsymbol {0}}~.\end{aligned}}}

where ${\displaystyle \langle \bullet \rangle }$ is the volume average over the unit cell. So a necessary condition that equations (12) have a solution is that

{\displaystyle {\text{(14)}}\qquad {\begin{aligned}i~\mathbf {k} \cdot \langle \mathbf {d} _{0}(\mathbf {y} )\rangle =0\\i~\mathbf {k} \cdot \langle \mathbf {b} _{0}(\mathbf {y} )\rangle =0\\i~\mathbf {k} \times \langle \mathbf {e} _{0}(\mathbf {y} )\rangle -i~\omega _{0}^{j}~\langle \mathbf {b} _{0}(\mathbf {y} )\rangle ={\boldsymbol {0}}\\i~\mathbf {k} \times \langle \mathbf {h} _{0}(\mathbf {y} )\rangle +i~\omega _{0}^{j}~\langle \mathbf {d} _{0}(\mathbf {y} )\rangle ={\boldsymbol {0}}~.\end{aligned}}}

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

## Footnotes

1. The following discussion is based on Ashcroft76 (p. 133-139).
2. We can see that the two operators commute by working out the operations. Thus,
${\displaystyle {\mathcal {T}}_{R}~{\mathcal {L}}~{\begin{bmatrix}\mathbf {D} \\\mathbf {B} \end{bmatrix}}={\mathcal {T}}_{R}~{\begin{bmatrix}-i~{\boldsymbol {\nabla }}\times [\mu ^{-1}(\mathbf {x} )~\mathbf {B} (\mathbf {x} )]\\i~{\boldsymbol {\nabla }}\times [\epsilon ^{-1}(\mathbf {x} )~\mathbf {D} (\mathbf {x} )]\end{bmatrix}}={\begin{bmatrix}-i~{\boldsymbol {\nabla }}\times [\mu ^{-1}(\mathbf {x} +\mathbf {R} )~\mathbf {B} (\mathbf {x} +\mathbf {R} )]\\i~{\boldsymbol {\nabla }}\times [\epsilon ^{-1}(\mathbf {x} +\mathbf {R} )~\mathbf {D} (\mathbf {x} +\mathbf {R} )]\end{bmatrix}}={\mathcal {L}}~{\begin{bmatrix}\mathbf {D} (\mathbf {x} +\mathbf {R} )\\\mathbf {B} (\mathbf {x} +\mathbf {R} )\end{bmatrix}}={\mathcal {L}}~{\mathcal {T}}_{R}~{\begin{bmatrix}\mathbf {D} \\\mathbf {B} \end{bmatrix}}~.}$
3. We can see that the translation operator commutes by working out the operations. Thus,
${\displaystyle {\mathcal {T}}_{R}~{\mathcal {T}}_{R'}~{\begin{bmatrix}\mathbf {D} (\mathbf {x} )\\\mathbf {B} (\mathbf {x} )\end{bmatrix}}={\mathcal {T}}_{R}~{\begin{bmatrix}\mathbf {D} (\mathbf {x} +\mathbf {R} ')\\\mathbf {B} (\mathbf {x} +\mathbf {R} ')\end{bmatrix}}={\begin{bmatrix}\mathbf {D} (\mathbf {x} +\mathbf {R} '+\mathbf {R} )\\\mathbf {B} (\mathbf {x} +\mathbf {R} '+\mathbf {R} )\end{bmatrix}}={\mathcal {T}}_{R+R'}~{\begin{bmatrix}\mathbf {D} (\mathbf {x} )\\\mathbf {B} (\mathbf {x} )\end{bmatrix}}={\mathcal {T}}_{R'}~{\begin{bmatrix}\mathbf {D} (\mathbf {x} +\mathbf {R} )\\\mathbf {B} (\mathbf {x} +\mathbf {R} )\end{bmatrix}}={\mathcal {T}}_{R'}~{\mathcal {T}}_{R}~{\begin{bmatrix}\mathbf {D} (\mathbf {x} )\\\mathbf {B} (\mathbf {x} )\end{bmatrix}}~.}$
4. The following discussion is based on Milton02

## 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.