2. The model

The system is presented in Figure 1. The system consists of an infinite nanowire (in the longitudinal x-direction), whose width (in the transverse y-direction) is w, a point defect, whose distance from the boundary of the nanowire is wε, and an external capacitor, whose voltage and charge can be controlled by a power source.

### Figure 1.

Model schematic. The capacitor creates the transverse electric field, and the point defect simulates the single atom.

Single-Atom Field-Effect Transistor DOI: http://dx.doi.org/10.5772/intechopen.81526

Mathematically, the model can be described by the 2D stationary Schrödinger equation

$$\frac{\partial^2 \Psi}{\partial \mathbf{x}^2} + \frac{\partial^2 \Psi}{\partial \mathbf{y}^2} + (E - U(\mathbf{y}) + F\mathbf{y})\Psi = -D(\mathbf{r} - \mathbf{r\_0})\Psi \tag{1}$$

in which normalized units (where Planck constant is ℏ ¼ 1 and the electron's mass is m ¼ 1=2) were used. In this equation

$$U(\boldsymbol{\nu}) = \begin{cases} \mathbf{0} & \mathbf{0} \prec \boldsymbol{\mathcal{Y}} \prec \boldsymbol{w} \\ \infty & \text{else} \end{cases} \tag{2}$$

is the boundaries' potential, which confines the dynamics to the wire's geometry. F is the electric field.

The point defect, which models the single atom, is presented by the asymmetric impurity D function (IDF) [19–21]

$$D(\mathbf{r}) \equiv \lim\_{\rho \to 0} \frac{2\sqrt{\pi}\delta(y)\exp\left(-\mathbf{x}^2/\rho^2\right)}{\rho\ln\left(\rho\_0/\rho\right)}\tag{3}$$

which is located at r<sup>0</sup> ¼ ^yε, and ρ<sup>0</sup> is related to the impurity's bound eigenenergy by

$$E\_0 = -\frac{16\exp\left(-\gamma\right)}{\rho\_0^2} \cong -\frac{8.98}{\rho\_0^2} \tag{4}$$

where γ ffi 0:577 is Euler constant [22]. On the other hand, the relation between ρ<sup>0</sup> and the physical properties of a real physical impurity (which has a finite radius a and a finite local potential V0) is

$$\rho\_0 = 2a \exp\left(\frac{2}{V\_0 a^2} + \frac{\chi}{2}\right). \tag{5}$$

The homogeneous eigenstates solutions of the wire, i.e., solutions without the point defect, are

$$
\Psi\_{n,E}(\mathbf{x}, \boldsymbol{\uprho}) = \exp\left(i k\_n \boldsymbol{\uprho}\right) \chi\_n(\boldsymbol{\uprho}) \tag{6}
$$

where χnð Þy are the eigenstates of the one-dimensional differential equation

$$\frac{\partial^2 \chi\_n(\mathbf{y})}{\partial \mathbf{y}^2} + (E\_n - U(\mathbf{y}) + F\mathbf{y})\chi\_n(\mathbf{y}) = \mathbf{0} \tag{7}$$

with the corresponding eigenvalues

$$k\_n^2 = E - E\_n.\tag{8}$$

These eigenstates can be written using the Airy functions Ai and Bi [22] and the normalized parameter

$$\xi \equiv \mathcal{y}F^{1/3} + \left(E - k\_n^2\right)/F^{2/3} = F^{1/3}\left\{\mathcal{y} + \left(E - k\_n^2\right)/F\right\},$$

as

$$\chi\_n(\xi) = N\{Ai(\xi)Bi(\xi\_0) - Ai(\xi\_0)Bi(\xi)\}\tag{9}$$

where <sup>N</sup> is the normalization constant, <sup>ξ</sup><sup>0</sup> � <sup>F</sup>1=<sup>3</sup> fEn=Fg, where the eigenvalues En are determined by the infinite solutions of the transcendental equation.

<sup>χ</sup>nðξwÞ ¼ 0, when <sup>ξ</sup><sup>w</sup> � <sup>F</sup>1=<sup>3</sup> fw þ En=Fg.

In the case of a weak electric field, the eigenstates can be written to a first order in the electric field as a superposition of the free (zero electric field) eigenstates

$$\Phi\_m(y) = \sqrt{\frac{2}{w}} \sin \left( m \pi y / w \right) \tag{10}$$

namely,

$$\chi\_m(\mathbf{y}) = \phi\_m(\mathbf{y}) + F\nu^3 \sum\_q \phi\_q(\mathbf{y}) \frac{(-1)^{m+q} - 1}{2} \frac{mq}{(m+q)^3 (m-q)^3} \frac{8}{\pi^4} \tag{11}$$

with the corresponding eigenenergies (again to the first order in F)

$$E\_n \cong \left(n\pi/w\right)^2 + \frac{1}{2}Fw.\tag{12}$$

Clearly, in the absence of the point defect (the atom), there is no coupling between the transverse direction and the longitudinal one, i.e., the capacitor cannot affect the longitudinal conductance.

There is an exception, of course, if the capacitor occupies a finite region in space, in which case the electric field does create a coupling between the modes. But in the regime of a weak electric field, this coupling amq is also very weak

$$a\_{mq} = Fw^3 \frac{(-1)^{m+q} - 1}{2} \frac{mq}{\left(m+q\right)^3 \left(m-q\right)^3 \pi^4} \,\mathrm{s}.\tag{13}$$

Even to adjacent modes (where most of the energy is transferred), the coupling is very weak

$$a\_{m,m\pm1} = \mp F \nu^3 \frac{m(m\pm1)}{(2m\pm1)^3} \frac{8}{\pi^4} \tag{14}$$

For example, the coupling between the first and the second modes is as small as <sup>a</sup>1, <sup>2</sup> ffi <sup>0</sup>:0061Fw<sup>3</sup> .

However, the presence of the point defect breaks the Cartesian symmetry and increases the coupling between the modes.

The general solution, which takes the point defect into account, is

$$\Psi(\mathbf{r}) = \Psi\_{\rm inc}(\mathbf{r}) - \frac{G^+(\mathbf{r}, \mathbf{r}\_0)\Psi\_{\rm inc}(\mathbf{r}\_0)}{1 + \int d\mathbf{r}' G^+(\mathbf{r}', \mathbf{r}\_0)D(\mathbf{r}' - \mathbf{r}\_0)} \left[d\mathbf{r}'D(\mathbf{r}' - \mathbf{r}\_0)\tag{15}\right]$$

where ψincð Þr is the incoming wavefunction, which can be chosen as one of the eigenmodes χnð Þy , which in the case of a weak electric field can be approximated by Eq. (11) (or Eq. (10), i.e., by ϕmð Þy ); Gþðr; r0Þ is the outgoing 2D Green function, i.e., Gþðr; r0Þ is the solution of the partial differential equation

$$-\nabla^2 G^+(\mathbf{r}, \mathbf{r}\_0) + [U(\mathbf{y}) - E - F\mathbf{y}]G^+(\mathbf{r}, \mathbf{r}\_0) = -\delta(\mathbf{r} - \mathbf{r}\_0) \tag{16}$$

which can be written in terms of the 1D eigenstates χmð Þy as

$$G(\mathbf{r}, \mathbf{r}') = \sum\_{n=1}^{\infty} \frac{\chi\_n(\mathbf{y}) \chi\_n^\*(\mathbf{y}')}{2i\sqrt{E - E\_n}} \exp\left(i\sqrt{E - E\_n}|\mathbf{x} - \mathbf{x}'|\right). \tag{17}$$

The scattered solution is therefore

$$\Psi(\mathbf{r}) = \exp\left(ik\_n \mathbf{x}\right) \chi\_n(\boldsymbol{y}) - \sum\_{m=1}^{\infty} A\_{n,m} \chi\_m(\boldsymbol{y}) \exp\left(ik\_m |\mathbf{x}|\right) \tag{18}$$

with the coefficients

$$A\_{n,m} \equiv \lim\_{\rho \to 0} \frac{\frac{\chi\_{\mathfrak{a}}(\epsilon)\chi\_{m}^{\*}(\epsilon)}{2ik\_{m}}}{\frac{\ln \left(\rho\_{0}/\rho\right)}{2\pi} + \sum\_{q=1}^{\infty} \frac{\chi\_{q}(\epsilon)\chi\_{q}^{\*}(\epsilon)}{2ik\_{q}} \exp\left(-\left|E - E\_{q}\right|\rho^{2}/4\right)}.\tag{19}$$

The transmitted solution (x > 0) is thus

$$\Psi(\mathbf{r}) = \sum\_{m=1}^{\infty} \exp\left(ik\_m \mathbf{x}\right) \chi\_m(\mathbf{y}) \{\delta\_{n,m} - A\_{n,m}\}.\tag{20}$$

Eq. (18) is a generic solution; however, there are two types of energies, for which the solution reveals a universal pattern.
