The effect of rashba spin–orbit interaction on optical far-infrared transition of tuned quantum dot/ring systems

Abstract

In the present work, optical properties of tuned quantum dot/ring systems with considering the effect of Rashba spin–orbit interaction are theoretically investigated. To this end, we describe our system by using an appropriate model potential to obtain the energy levels and wave functions analytically. The system is excited by a monochromatic electromagnetic field and intersubband transitions for the electrons are considered. Then, analytical expressions for optical absorption coefficients and refractive index changes are used. The effect of quantum dot radius and the Rashba coupling constant on the optical properties have been investigated. The results show that: (i) The effect of Rashba spin–orbit interaction on the peak values of the refractive index changes is negligible. (ii) The total absorption coefficient is enhanced and shift toward higher energies with considering the effect of Rashba spin–orbit interaction. (iii) The impact rate of spin–orbit interaction on the optical properties relates to the quantum dot radius, depth of confining potential and the Rashba coupling constant.

Appendix 1

Here, we will show the details of obtaining energy eigenvalues and eigenstates of the system. The Schrödinger equation is expressed by

$$H\phi =E\phi$$

(28)

where

$$H=-\frac{{\hslash }^{2}}{2{m}^{*}}\left(\frac{{\partial }^{2}}{\partial {r}^{2}}+\frac{1}{r}\frac{\partial }{\partial r}+\frac{1}{{r}^{2}}\frac{{\partial }^{2}}{\partial {\theta }^{2}}\right)+\frac{1}{2}{m}^{*}\left({\omega }_{0}^{2}+\frac{{\omega }_{c}^{2}}{4}\right){r}^{2}+\frac{{V}_{0}{r}^{2}}{{R}_{0}^{2}}-{V}_{0}-i\hslash \frac{{\omega }_{c}}{2}\frac{\partial }{\partial \theta }+\frac{{\hslash }^{2}}{2{{m}^{*}}^{2}}\frac{\xi }{{r}^{2}}$$

(29)

Here $\phi$ and $E$ are the 2D eigenstate and its eigenvalue, respectively. We assume that

$$\phi \left(r,\theta \right)=f\left(r\right)\frac{{e}^{im\theta }}{\sqrt{2\pi }}$$

(30)

where $m$ is the magnetic quantum number. Inserting Eq. (30) into Eq. (28) and remove the term of $\frac{{e}^{im\theta }}{\sqrt{2\pi }}$ in both sides of the equation, we have

$$\left[-\frac{{\hslash }^{2}}{2{m}^{*}}\left(\frac{{\partial }^{2}}{\partial {r}^{2}}+\frac{1}{r}\frac{\partial }{\partial r}-\frac{{m}^{2}}{{r}^{2}}\right)+\frac{1}{2}{m}^{*}{\Omega }^{2}{r}^{2}-{V}_{0}+\frac{m{\omega }_{c}\hslash }{2}+\frac{{\hslash }^{2}}{2{{m}^{*}}^{2}}\frac{\xi }{{r}^{2}}\right]f\left(r\right)=Ef\left(r\right)$$

(31)

To solve above equation, the following relation has been employed

$$f\left(r\right)={r}^{-\frac{1}{2}}g\left(r\right)$$

(32)

Inserting above equation into Eq. (31), we obtain

$$\frac{{d}^{2}g(r)}{d{r}^{2}}+\left[\frac{2{m}^{*}\eta }{{\hslash }^{2}}-\frac{{m}^{{*}^{2}}{\Omega }^{2}{r}^{2}}{{\hslash }^{2}}-\frac{l(l+1)}{{r}^{2}}\right]g\left(r\right)=0$$

(33)

where

$$\eta =E+{V}_{0}-\frac{m\hslash {\omega }_{c}}{2}, l\left(l+1\right)={m}^{2}+\xi -\frac{1}{4}$$

(34)

Now, we use the following variables

$${r}^{2}=\frac{\hslash }{{m}^{*}\Omega }x, {l}_{m}=\sqrt{{m}^{2}+\frac{\xi }{2}}+\frac{1}{4}$$

(35)

Substituting Eq. (35) into Eq. (33), we have

$$\frac{{d}^{2}g(z)}{d{z}^{2}}+\frac{1}{2z}\frac{dg(z)}{dz}-\left[\frac{{l}_{m}\left({l}_{m}-\frac{1}{2}\right)}{{z}^{2}}+\frac{1}{4}-\frac{\eta }{2\hslash\Omega z}\right]g\left(z\right)=0$$

(36)

Inserting the following relation into Eq. (36),

$$g\left(z\right)={z}^{{l}_{m}}{e}^{-\frac{z}{2}}h\left(z\right)$$

(37)

We obtain the below equation

$$z\frac{{d}^{2}h(z)}{d{z}^{2}}+\left[\left(2{l}_{m}+\frac{1}{2}\right)-z\right]\frac{dh(z)}{dz}-\left({l}_{m}+\frac{1}{4}-\frac{\eta }{2\hslash\Omega }\right)h\left(z\right)=0$$

(38)

It is clear that above equation is the Kummer’s differential equation if

$$\left(2{l}_{m}+\frac{1}{2}\right)=b, \left({l}_{m}+\frac{1}{4}-\frac{\eta }{2\hslash\Omega }\right)=a$$

(39)

To make $g\left(z\right)$ finite, we have $a=-n$. Therefore, we can obtain the energy levels as

$${E}_{mn}=\left(2n+\sqrt{{m}^{2}+\upxi }+1 \right){\hslash \Upomega }+\frac{m{\hslash }{\omega }_{c}}{2}-{V}_{0}$$

(40)

and the wave function as

$$\phi \left(r,\theta \right)=N{r}^{\left(2{l}_{m}-\frac{1}{2}\right)}{e}^{-\frac{{r}^{2}}{2{\Omega }_{1}^{2}}}{L}_{n}^{\left(2{l}_{m}-\frac{1}{2}\right)}\left(\frac{{r}^{2}}{{\Omega }_{1}^{2}}\right){e}^{im\theta }$$

(41)

where $N$ is the normalization constant and ${\Omega }_{1}=\sqrt{\frac{\hslash }{{m}^{*}\Omega }}$.