Electrodynamics of Evanescent Wave in Negative Refractive Index Superlens

1. Introduction

Early in 1968, Veselago[1] predicted that a new type of artificial metamaterial, which possesses simultaneously negative permittivity and permeability, could function as a lens to focus electromagnetic waves. These research direction was promoted by Pendry’s work[2, 3] and other latter works [4, 5, 15, 16, 23, 22, 7, 8, 9, 10, 11, 12, 13, 14, 21, 20, 17, 18, 24, 19, 25, 6, 26]. They show that with such a metamaterial lens, not only the radiative waves but also the evanescent waves, can be collected at its image, so the lens could be a superlens which can break through or overcome the diffraction limit of the conventional imaging system. This beyond-limit property gives us a new window to design devices.

It is well-known that evanescent wave plays an important role in the beyond-limit property of the metamaterial superlens. Furthermore, evanescent waves become more and more important when the metamaterial devices enter sub-wavelength scales[28, 27]. Therefore, the quantitatively study of pure evanescent waves in the metamaterial superlens is very significant. However, the quantitatively effects of pure evanescent wave in the metamaterial superlens have not been intensively studied, since so far almost all studies were only interested in the image properties with global field[8, 9], which is the summation of radiative wave and evanescent wave. On the other hand, many theoretical works were performed to study the metamaterial superlens, employing either finite-difference-time domain (FDTD) simulations[16] or some approximate approaches[29, 30]. However, one cannot obtain the rigorous pure evanescent wave by these numerical methods, since the image field of the metamaterial superlens obtained by FDTD is global field, and other approximate approaches cannot be rigorous.

In reviewing these existing efforts, we feel desirable to develop a rigorous method that can be used to study quantitatively the transient phenomena of the evanescent wave in the image of the metamaterial superlens. In this paper, we will present a new method based on the Green’s function[7, 8] to serve this purpose. Our method can be successfully used to calculate the evanescent wave, as well as the radiative wave and the global field. The main idea of our method can be briefly illustrated as follows. Since the metamaterial superlens is a linear system, so all dynamical processes can be solved by sum of multi-frequency components. And each frequency component can be solved by sum of multi-wavevector components. So we can use Green’s function of multi-frequency components to obtain the strict numerical results. Therefore, our method based on the Green’s function is strict, and it is quite a universe method in any linear system, for example, it can be used to study the two dimensional (2D) and three dimensional (3D) metamaterial superlens.

The content of the chapter is organized as follows. We mainly focus on the details of our theory and method in Section 2. After that, in Section 3, we will calculate the field of the image of a NIR superlens by using our method, including radiative waves, evanescent waves, SEWs, and global field. The method will be confirmed by using FDTD simulations. In Section 4, we will present our study on the group delay of evanescent wave in the superlens by using our method. Finally, conclusions are presented in Section 5.

2. Theoretical method

In this section, we will focus on the theoretical details of our method. First, a time-dependent Green’s function will be introduced. Then, based on the Green’s function, the method to obtain evanescent waves as well as radiative waves will be presented.

2.1. A time-dependent Green’s function

First, we would like to introduce a very useful time-dependent Green’s function for the solution of inhomogeneous media. The time-dependent Green’s function can be applied to study the dynamical scattering processes[7, 22]. In the inhomogeneous media, the problem we study can be solved by Maxwell equations:

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E1

where $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ is the current density and $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$is the conductivity. We rewrite Eq.(1) as:

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E2

To solve Eq.(2), we introduce a dynamic Green’s function$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, which satisfies:

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E3

where $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ is a unit dyad. In this system, when $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$and $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$is given, $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$is just a function of $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ in time domain, so it yields

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E4

Therefore, Eq.(3) becomes:

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E5

If $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ is known, then the $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ field can be found as

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E6

By Fourier transforming Eq.(5) from time domain to frequency domain, we obtain

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E7

where $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$and $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$are the relative permittivity and the relative permeability of the dispersive material at a frequency$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, respectively, and $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ is the Green’s Function in the frequency domain, which satisfies

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E8

By Fourier transforming the electric field $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ in time domain to the frequency domain, we obtain$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, satisfying

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E9

Considering the general properties:

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E10

and

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E11

Substituting Eq.(8), Eq.(9), Eq(10) and Eq.(11) into Eq.(6), we can obtain

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E12

Eq.(12) exhibits the role of $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ that $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ is a propagator for the field in the frequency domain between $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$and$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$. Here we have supposed that $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$in the source region for simplicity. If $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ is the exciting source’s electric field in the frequency domain, i.e., $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, then Eq.(12) can be rewritten as:

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E13

where $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ is the spectrum of the exciting source.

So the field $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ can be obtained by the inverse Fourier transformation:

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E14

where $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ is the working frequency of the exciting source.

Figure 1.

The shematic of the three-layer inhomogeneous medium.

Now the remaining problem is to solve Eq(7) to get$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$. With the exciting source $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ (whose spectrum is$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$), after obtain$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, then we can calculate $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ directly by Eq.(12).

As a typical example, the time-dependent Green’s function for the three-layer inhomogeneous media is presented, which is shown in Fig.1. The inhomogeneous media of the system can be described by:

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E15

$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$is related to the transmission coefficient and/or reflection coefficient which are dependent on the boundary conditions of each layer. Here, we only consider the Green’s function in the region that$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$. Under these conditions, $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$has been obtained in Ref.[7]. $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$is different in different spatial dimensions. For three-dimension, we can rewrite $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ as follows:

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E16

where $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ and $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ are the usual zeroth-order Bessel function and second-order Bessel function, respectively, and$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$are the dispersion relation in the region $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ and the region $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ respectively. Here, $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$and $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ are the transmission coefficients for TE wave ($\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$polarized) and TM wave ($\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$polarized), respectively, which are respectively given by

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E17

and

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E18

where $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$and$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$.

$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$is the conductivity in source region, i.e.,$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$. $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$has been assumed to be$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$. So Eq(16) becomes:

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E19

For the 2D case, the wave vector$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, we have

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E20

And for the 3D case, we have

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E21

After $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ is obtained, we can obtain the field in the frequency domain in the region $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$by Eq.(12). And then by the inverse Fourier transformation, the field in time domain can be obtained.

2.2. The Green’s function for radiative waves and evanescent waves

Now, we will apply a time-dependent Green’s function for a radiative wave and an evanescent wave. This Green’s function can be directly developed from the Green’s function introduced in the Sec.2.1. The schematic model is shown in Fig.1. As we know, the plane solution wave for the electric field in vacuum is of the form $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$exp($\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$), where $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ and $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ are wave numbers along the $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$plane and $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$directions respectively, and they satisfy the dispersion relation as follows:$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, where $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$is the light velocity in the vacuum. In the case of$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$is real, corresponding to the radiative waves along the $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$direction. While if$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$is imaginary, corresponding to the evanescent waves along the $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$direction. Similarly, in the slab region, the real or imaginary $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ corresponds to the radiative wave or the evanescent wave along $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$direction, respectively.

For simplicity, we first consider the 2D system, in which$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$. The global field of image region is the superposition of radiative waves and evanescent waves. We can rewrite Eq.(20),Eq.(13) and Eq(14) as follows:

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E22

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E23

and

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E24

where $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ is the integral range. The integral range is of great significance, since different integral range corresponds to different wave. For example, the integral range [$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$,$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$] is for global field, and the range [$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$,$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$] is for radiative wave. Obviously, for linear system, the integral range can be chosen arbitrarily.

From Eq.(22) to Eq.(24), we can directly calculate the radiative wave and evanescent wave. In the case of radiative wave ($\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$), the integral range is [$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$,$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$], so the radiative wave Green’s function $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ satisfies

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E25

for radiative waves.

In the case of evanescent wave ($\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$), the integral range is$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, so the evanescent wave Green’s function $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ satisfies:

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E26

for evanescent waves.

And for the global field, the global field Green’s function $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ satisfies:

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E27

Additionally, we can also focus our observation on the subdivided evanescent wave (SEW), with a certain integral range$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$. The SEW, with a certain integral range, can also be obtained from Eq.(22) to Eq.(24) such that

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E28

As an typical example, a SEW with an integral range [$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$], whose Green’s function can be obtained easily from Eq.(28) as:

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E29

In this way, we can obtain the Green’s function for the SEW with any integral range. Obviously, evanescent wave could be regarded as the superposition of SEWs. Therefore, we have

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E30

From Eq.(28), Eq.(26), Eq.(25), and Eq.(27), one can obtain the SEW Green’s function$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, the evanescent wave Green’s function$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, the radiative wave Green’s function$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, and the global field Green’s function$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, respectively. Substituting them to Eq.(23) and Eq.(24) respectively, we can obtain the field of the SEW, the evanescent wave, the radiative wave, and the global field, respectively.

For the 3D system, obviously, the methods to get Green’s function for the SEW, the evanescent wave, the radiative wave, and the global field are respectively very similar with the above discussion, i.e., just replace $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ by $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ and let $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ in Eq.(28), Eq.(26), Eq.(25), and Eq.(27), respectively.

3. Electromagnetic waves in the image of the superlens

In this section, we will discuss the image’s field of a 2D metamaterial superlens, which is shown in Fig.2. The thickness of the metamaterial slab is$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, which is placed at the $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$-$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ between $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$and$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$. The source is set in the object plane at ($\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$). Obviously, the image will be formed in the image plane at$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$. The source is the quasi-monochromatic random source with the field expressed as $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$=$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$), where $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ is a slow-varying random function, $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$is the central frequency of our random source (the details of the random source can be seen in Ref.[29]). The exciting source $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ is a quasi-monochromatic field with the central frequency$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, whose electric field $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ and frequency spectrum $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ are shown in Fig.3(a) and Fig.3(b), respectively. In this paper, only the TM modes are investigated (the TM modes have the electric field perpendicular to the two-dimensional plane of our model).

The inhomogeneous media of the metamaterial superlens system are described by:

Figure 2.

The schematic diagram of our model.

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E31

The negative relative permittivity $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ and the negative relative permeability $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ of metamaterial are phenomenologically introduced by the Lorenz model. The negative relative permittivity $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ and the negative relative permeability $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ are satisfied as follows:

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E32

where $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$and $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$are the resonant frequency and the resonant line-width of the “resonators" in the metamaterial respectively, and $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ is the plasma frequency. At$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, we have$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$.

Figure 3.

a) Electric field of the source. (b) Spectrum of the source (top) and the image obtained by using our method (bottom) in units of$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$. (c) Electric field of the image vs time. The global field calculated by using our method (top) and by using FDTD (bottom). (d) The evanescent wave of the image calculated by using our method. (e) Two typical SEWs of the image.

In order to excite the evanescent wave strong enough in the image of the metamaterial superlens, the distance $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$between the source and the superlens should be small enough. Here we choose$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, where $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ is the wavelength corresponding to the central frequency$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$.

For this metamaterial superlens system, it is very easy to obtain$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, and $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ from Eq.(26) to Eq.(28), respectively. Let $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$and$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, and thus the Green’s functions for the image will be obtained. After that, we can obtain the global field, the evanescent wave and the SEW of the image via:

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E33

respectively, where

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E34

The numerical results calculated by our method are shown in Fig.3. Fig.3(c)(up, the blue one), (d) and (e) show the global field, the evanescent wave, and two typical SEWs respectively. The integral $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ range of the two SEWs are [$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$] (shown in Fig.3(e)(up, the black one)), and [$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$] (shown in Fig.3(e)(down, the blue one)), respectively.

In order to convince our method, FDTD simulation is also applied to calculate the field of the image, which is shown in Fig.3(c) (down, the green one). Comparison with the results calculated by our method and FDTD shown in Fig.3(c), we can see they coincide with each other very well. In addition, we also calculate the frequency sepctrum of the image by our method, as shown in Fig.3(b) (down, the red one). Comparing the spectra of source and image, we can find they are very close to each other. This result also agrees with the Ref.[29]. Therefore, our method is convincible, which can be used to obtain the pure evanescent waves, the SEWs, and the global field effectively.

4. Group delay time of SEWs and its impacts on the temporal coherence

4.1. Group delay time of SEWs

From Figs.3(c)-(e), we can find that the profile of evanescent wave and that of SEWs look like that of radiative wave with a group delay time$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$. So the field evanescent wave$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, as well as that of SEWs$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, can be written as an expression such as$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, where $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ is parameter function of$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$. In order to quantitatively study the delay time$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, we introduce a function $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$which satisfies:

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E35

where $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ and $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ are the field of radiative wave and the evanescent wave respectively, $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$is an independent variable with the time dimension. Since the profile of the radiative wave and the evanescent wave are very similar, obviously, the function $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$will get the maximal value when$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$. Therefore, the delay time can be defined quantitatively as follows:

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E36

Here $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ means the inverse function.

Similarly, we can also study the group delay time of the SEWs. We rewrite Eq.(34) as follows:

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E37

where $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ is the field of the SEW with a certain integral $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ range. So we can calculate the delay time of the SEWs from Eq.(36) and Eq.(35).

Figure 4.

a) The group delay time $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ of SEWs. (b) $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$of SEWs.

In our numerical experiment, in order to calculate the group delay time of the SEWs, we choose 70 SEWs with integral $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ range as [$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$,$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$], where $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$=$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, respectively. The delay time of the 70 SEWs is shown in Fig.4(a). In this figure, we can see that the SEW with larger integral variable $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ will have a larger delay time. Obviously, the function $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ is a continuous-monotone increasing function, when the integral range $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ is infinitesimal ($\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$0). The result can be obtained by using a polynomial fitting, as shown in Fig.4(a). Therefore, from Fig.4(a), we can find that the SEWs with larger integral variable $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ corresponds to a larger delay time, which means the group velocity of SEWs with larger integral variable is smaller in the superlens system.

Since the field profile of SEWs looks like that of a radiative wave, so we can write the field of the SEW with the integral range $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ as follows:

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E38

where$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$is a slowly-verifying function of$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, and $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$is an exponentially-decreasing function of $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ and a general-function of$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$. Obviously, when $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$(or$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$) becomes larger, $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$will trends to become smaller. Then the field of evanescent wave $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ can be obtained by:

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E39

The physical meaning of $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$and its impacts will be discussed in the following.

4.2. Impacts of the group delay of SEWs on temporal coherence gain in the image of the superlens

One of the most interesting impacts of the group delay of SEWs is related to the first-order temporal coherence gain (CG). Here, we would like to discuss the CG caused by the SEWs in the image of the superlens. In our previous work [29, 30], we have investigated a prominent CG of the image by the radiative waves even when the frequency-filtering effects are very weak. Then, a natural question is what about the role of the evanescent waves play in the CG of a superlens? In this section, we will show that not only the radiative waves but also the evanescent waves, and the SEWs that can be responsible for the CG. Furthermore, we will show that the total CG in the image of a superlens is the weighted averaged of evanescent-wave coherence gain (ECG), radiative-wave coherence gain (RCG), radiative-wave and evanescent-wave coherence gain (RECG).

First of all, let’s consider the contributions of the evanescent waves on the CG. For this, we calculate the normalized first-order temporal coherence $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ of the superlens with the random source $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ exciting, which are shown in Fig.5. Here, the normalized first-order temporal coherence function $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ is defined by

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E40

where $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ is a coherence function, which is defined by

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E41

or

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E42

here $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ means the statistic average (ensemble average) and $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$is time delay. From Fig.5, we can see that the temporal coherence of image of evanescent wave, radiative wave and global field, are all obviously better than that of source. Comparing with the temporal coherence of source and image, there are three kinds of coherence gain as follows. The first one is the radiative-wave coherence gain (RCG, this mechanism has been discussed in our previous work[29, 30]), which is determined by the radiative waves. The second one is the evanescent-wave coherence gain (ECG), which is determined by the evanescent waves. The last one is the global field coherence gain (GCG), which is from the global field with the co-effect of RCG and ECG.

Figure 5.

The first temporal coherence of image and source. The dashed-dotted one is the $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ of the source. Others are for the image, i.e., the red one, the black one (and also the green one), and the blue one are the $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ of the evanescent waves, the global field, and the radiative waves, respectively.

To show how the ECG is produced from the interference of the SEWs with different group delay time, we assume there are only two SEWs, such as $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ and $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ with the range $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ and $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ respectively, which correspond to the delay time $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ and$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, respectively. The two SEWs can be expressed by Eq.(37). We assume the two SEWs have an integral range very close to each other, i.e., $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, and so we have $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$and both of them are close to a certain constant$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$.

From Eq.(38), we have the evanescent wave field:

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E43

then the temporal coherence of the evanescent wave in the image is given by

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E44

The first two terms are the same as the coherence function of the radiative wave, so they do not contribute to ECG. The last two terms are from the interference between SEWs, they can be very large at the condition$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$. This condition can always be satisfied between SEWs since $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ is a continuous variable. So the relative delay time $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ of SEWs are responsible for ECG.

Therefore, ECG can always exist in the superlens when two conditions are satisfied: (1)$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$; and (2)$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$. Unfortunately, the two conditions could not always be satisfied at the same time. When $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$is large, which also means the integral variable $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ and $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ are far from each other, so the value of $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ will be very large, and thus the former condition could not be satisfied. The condition that $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$is the direct reflection that only the interference of those SEWs with close integral variable $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ can produce the ECG.

Here the integral variable “$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$close to $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$" means when $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ is given, for any $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ satisfying$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, the condition $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$is always satisfied, where $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$is a threshold value with small positive value near zero(for example $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$=0.01). For two SEWs with the integral range $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ and $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ respectively, we can obtain the relative delay time$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$. As the discussion above, the relative delay time $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$is responsible to the ECG. When $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ is close to$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$is a monotonic increasing function of $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ in the range$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, which gives a threshold value $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ as:

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E45

$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$shows the upper-limit of the effective coherent relative delay time that only the $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$is the effective responsible to ECG. As $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$increasing, when$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, which means $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ (i.e. $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$and $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ is not close to each other), the difference between $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$and $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$becomes greater, and so their interference becomes weaker and their contribution to the coherence gain will decrease rapidly. While when$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, which means the integral variable $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ and $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ are very far from each other, then the SEW with the much larger integral variable one is too weak to have any effective interference, so their contribution to the coherence gain trends to be zero.

Therefore, the coherence gain from the interference of SEWs is limited by$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$. The SEW with larger $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ corresponds larger coherence gain. In order to study the temperas coherence gain of SEWs, we introduce a parameter function as:$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, which is shown in Fig.4(b). The physical meaning of $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ is very clear, which gives a relation

$$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$$E46

when$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, we can get$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$. Thus, $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$is proportional to$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$. From Fig.4(b) we can see that $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ is an increasing function of$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, so the SEWs with larger integral variable $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ correspond larger $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ and stronger ECG, and so the field of the SEWs with larger integral various $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ will have better temporal-coherence$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$.

Figure 6.

$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$of SEW$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, SEW$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, and SEW$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, comparing with $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ of evanescent wave.

To convince it, three SEWs (SWE$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$,SWE$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$,SWE$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$)with the different integral $\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ range [$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$], [$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$] and [$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$] respectively are chosen to calculate the normalized first-order temporal-coherence function$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$, which are shown in Fig.6. In this figure, we can see the temporal-coherence of SWE$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ with the largest integral variable is the best (black), the temporal coherence of SWE$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ with the second largest integral variable(red) is the second best, and the temporal coherence of SWE$\overrightarrow{J}(r,t)=\sigma (r,t)\overrightarrow{E}(r,t)$ with the smallest integral variable is the worst, which is expected.

5. Conclusion

In conclusion, based on the Green’s function, we have numerically and theoretically obtained the evanescent wave, as well as the SEWs, separating from the global field. This study could help us to investigate the effect of an evanescent wave on a metamaterial superlens directly and give us a new way to design new devices.