01-EM Models

1 Simplifying Maxwell Equations

Macroscopic Maxwell equations describes any EM waves, we thus need to start from these equations to build our model in a mixed dielectric media like photonic crystals.

Recall that Maxwell equations is

$$\begin{aligned} \nabla\cdot B=0\quad \nabla\times E+\frac{\partial B}{\partial t}=0\\ \nabla\cdot D=\rho \quad \nabla\times H-\frac{\partial D}{\partial t}=J \end{aligned}$$

The first condition we propose is that

The material is related to cartesian position vector $r$, and it’s independent of time, thus there are no free charges and currents, therefore

$$\rho=0\quad J=0\tag{1}$$

The next approximation is about the relation between $D$ and $E$, from Bloembergen, the component of $D$ and $E$ is related to a power series

$$D_i/\epsilon_0 = \sum_j \epsilon_{ij}E_j+\sum_{jk}\chi_{ijk}E_jE_k+O(E^3)$$

We give the following conditions

• Assume the field strengths are small, and in linear regime, the higher order terms are neglected.
• The material is macroscopic and isotropic, since it’s isotropic, then $\epsilon_{ij}=\epsilon(r)\delta_{ij}$, which means the material’s response is the same in all directions. This gives us $D_i/\epsilon_0=\epsilon(r)E_i$, which is $D = \epsilon_0\epsilon(r)E$
• Ignore any explicit frequency dependence of $\epsilon$
• The material is transparent so that $\epsilon(r)$ is real and positive

Combine these conditions, we have

$$D(r) = \epsilon_0\epsilon(r)E(r)\quad B(r)=\mu_0\mu(r)H(r) \tag{2}$$

and we introduce one more condition, which is that the most materials of interest have $\mu(r)\approx 1$, that gives us

$$B = \mu_0 H \tag{3}$$

In this case, we also have $n=\sqrt{\epsilon\mu}$

Combine (1), (2), (3), we have

$$\begin{aligned} \nabla\cdot H(r,t)=0\quad \nabla\times E(r,t)+\mu_0\frac{\partial H(r,t)}{\partial t}=0 \\ \nabla\cdot [\epsilon(r)E(r,t)]=0\quad \nabla\times H(r,t)-\epsilon_0\epsilon(r)\frac{\partial E(r,t)}{\partial t}=0 \end{aligned}\tag{4}$$

Since by Fourier analysis, we can build any solution by combining harmonic terms, we can reduce our situation to a simple sinusoidal field. That is, we use the trick of complex-valued field

$$H(r,t) = H(r)e^{-i\omega t}\quad E(r,t)=E(r)e^{-i\omega t}\tag{5}$$

We study this simple situation, where time dependence comes from the harmonic $e^{-i\omega t}$, other solutions can be obtained by combining these harmonic and according to Fourier, we can compose any well-behaved functions.

Now take (5) to (4), we have

$$\nabla\cdot H(r)=0\quad \nabla\cdot [\epsilon(r)E(r)]=0\tag{6}$$

This poses a condition that the EM waves are transverse, if we take (5) into the above equation, we will have something like $a\cdot k=0$, showing that the EM waves are perpendicular to the propagation direction.

Now back to the other two equations in (4), take (5) into (4) yields

$$\nabla\times E(r) - i\omega \mu_0H(r)=0\quad \nabla\times H(r)+i\omega \epsilon_0\epsilon(r)E(r)=0\tag{7}$$

Combining the above two equations, let’s divide the right equation with $\epsilon(r)$ and use the left equation to substitute $E(r)$, and since $c=1/\sqrt{\epsilon_0\mu_0}$, we have

$$\nabla\times \left(\frac{1}{\epsilon(r)}\nabla\times H(r)\right)=\left(\frac{\omega}{c}\right)^2H(r)\tag{8}$$

This is the master equation.

Now for any given $\epsilon(r)$, we can first solve (8) and use the right side equation in (7) to find $E(r)$ by

$$E(r)=\frac{i}{\omega\epsilon_0\epsilon(r)}\nabla\times H(r)$$

this ensures that $\nabla\cdot[\epsilon(r)E(r)]=0$ since divergence of a curl is always zero.

2 Eigenvalue Problem

As we can see from (8), we can define operator $\hat \Theta$ such that

$$\hat\Theta H(r) :=\nabla\times \left(\frac{1}{\epsilon(r)}\nabla \times H(r)\right)\tag{9}$$

Then we have

$$\hat \Theta H(r)=\left(\frac{\omega}{c}\right)^2 H(r)$$

The eigenvectors $H(r)$ are spatial patterns of the harmonic modes, and the eigenvalues are proportional to the squared frequencies of those modes.

The operator is linear, so $\alpha H_1(r)+\beta H_2(r)$ is a solution. Note that a simple scaling, like $\alpha=2,\beta=0$ is considered as two same modes.

Here, we define inner product of two vector fields as

$$(F,G):=\int d^3 r F^*(r)\cdot G(r)$$

we have the following properties:

• $(F,G)=(G,F)^*$
• $(F,F)$ is always real and non-negative

For the second property, we can achieve this by normalizing fields.

$$F(r) = \frac{F'(r)}{\sqrt{(F',F')}}$$

Since by definition, a simple scaling results in the same field, we then have $(F,F)=1$

We then define an operator is Hermitian if $(F,\hat \Theta G)=(\hat\Theta F, G)$, this is true for operator $\hat \Theta$, it can be proved by using integration by parts.

Another property is that the eigenvalues of $\hat \Theta$ must be real numbers, we can show that

$$(H,\hat\Theta H)=(\omega^2/c^2)(H,H)\quad (H,\hat\Theta H)^*=(\omega^2/c^2)(H,H)$$

and taking properties $(H,\hat\Theta H)=(\hat\Theta H, H)$ together with $(H,\hat \Theta H)=(\hat \Theta H, H)^*$, we have

$$(H,\hat\Theta H)^*=(\omega^2/c^2)(H,H)=(\hat\Theta H, H)=(\omega^2/c^2)(H,H)$$

Hence $(\omega^2/c^2)^*=(\omega^2/c^2)$, $\omega^2$ is real. We can also show that $\omega^2$ is always non-negative for $\epsilon>0$, the operator $\hat\Theta$ is said to be positive semi-definite.

In addition, the Hermiticity of $\hat \Theta$ makes two modes $H_1$ and $H_2$ to have an inner product of zero.

$$\omega_1^2(H_2,H_1) = c^2(H_2,\hat\Theta H_1) = c^2(\hat\Theta H_2,H_1)=\omega_2^2(H_2,H_1)$$

this gives us $(\omega_1^2-\omega_2^2)(H_2,H_1)=0$, so $(H_1,H_2)=0$ for any non-degenerate situation.

3 EM Energy and Variational Principle

We can show that, the smallest eigenvalue, let us denote as $\omega_0^2/c^2$, it is corresponding to find the field pattern that minimizes the functional

$$U_f:=\frac{(H,\hat\Theta H)}{(H,H)}$$

the above functional is called Rayleigh quotient, and it’s the Rayleigh quotient for $\hat \Theta$, and by variational principle, the minimum value of the Rayleigh quotient over all non-zero $H$ is the ​smallest eigenvalue​ $\omega_0^2/c^2$, and hence the field is in harmonic mode with frequency $\omega_0$.

Recall that, by definition,

$$(F, \hat\Theta G) = \int F^*\cdot [\nabla\times (\frac{1}{\epsilon}\nabla\times G)]d^3 r$$

Using the integral by part trick: $\int F\cdot (\nabla\times G)=\int (\nabla\times F)\cdot G$, we have

$$(F, \hat\Theta G) = \int (\nabla \times F^*)\cdot \frac{1}{\epsilon}(\nabla\times G)d^3 r$$

Then, we can calculate

$$U_f(H) = \frac{(\nabla\times E,\nabla\times E)}{(E,\epsilon(r)E)} = \frac{\int d^3r|\nabla\times E(r)|^2}{\int d^3 r \epsilon(r)|E(r)|^2}$$

In order to minimize $U_f$, we need to maximize the denominator and minimize the numerator, thus $E(r)$ must concentrate, or have more effective value around high dielectric constant $\epsilon$ region(denominator), and also we must minimize the amount of spatial oscillations(numerator).