Tunneling of Conduction Band Electrons Driven by a Laser Field in a Double Quantum dot: An Open Systems Approach

Abstract

In this paper, we investigate tunneling of conduction band electrons in a system of an asymmetric double quantum dot which interacts with an environment. First we consider the case in which the system only interacts with the environment and demonstrate that as time goes to infinity they both reach an equilibrium, which is expected, and there is always a maximum and minimum for the populations of the states of the system. Then we investigate the case in which an external resonant optical pulse (a laser) is applied to the system interacting with the environment. However, in this case for different intensities we have different populations of the states in equilibrium and as the intensity of the laser gets stronger, the populations of the states in equilibrium approach the same constant.

Appendix A:

In the following we obtain the exact solutions of (11), (12) and (13). Applying Laplace transform to these equations we have

$$ s\rho_{00}(s)-\rho_{00}(t=0) = l\rho_{11}(s)-m\rho_{00}(s), $$

(30)

$$ s\rho_{11}(s)-\rho_{11}(t=0) = -(l+n)\rho_{11}(s)+m\rho_{00}(s)+n\rho_{22}(s), $$

(31)

$$ s\rho_{22}(s)-\rho_{22}(t=0) = n\rho_{11}(s)-n\rho_{22}(s), $$

(32)

$$ \rho_{00}(s)+\rho_{11}(s)+\rho_{22}(s) = \frac{1}{s}, $$

(33)

where the last equality comes from the completeness relation, t r ρ = 1. After doing some calculations we find that

$$ \rho_{00}(s) = \frac{A}{s}+\frac{B}{s-\lambda_{0}}+\frac{C}{s-\lambda_{1}}, $$

(34)

where

$$ \lambda_{0}=\frac{-(l+m+2n)-\sqrt{(l+m+2n)^{2}-4(ln+2mn)}}{2}, $$

(35)

$$ \lambda_{1}=\frac{-(l+m+2n)+\sqrt{(l+m+2n)^{2}-4(ln+2mn)}}{2}. $$

(36)

Now applying $\mathcal {L}^{-1}$ to ρ ₀₀(s) gives ρ ₀₀(t) as

$$ \rho_{00}(t) = A+B\exp(\lambda_{0}t)+C\exp(\lambda_{1}t), $$

(37)

where A, B and C are constants. A is the stationary value of ρ ₀₀(t) when time goes to infinity. Substituting ρ ₀₀(t) into (11) and t r ρ = 1, one can easily obtain ρ ₁₁(t) and ρ ₂₂(t) as

$$ \rho_{11}(t) = \frac{mA}{l}+\frac{(\lambda_{0}+m)B}{l}\exp(\lambda_{0}t)+\frac{(\lambda_{1}+m)C}{l}\exp(\lambda_{1}t), $$

(38)

$$\begin{array}{@{}rcl@{}} \rho_{22}(t)&=&1-(A+\frac{mA}{l})-\left(B+\frac{(\lambda_{0}+m)B}{l}\right)\exp(\lambda_{0}t) \\ &-&\left(C+\frac{(\lambda_{1}+m)C}{l}\right)\exp(\lambda_{1}t). \end{array} $$

(39)

The stationary values of ρ ₀₀(t), ρ ₁₁(t) and ρ ₂₂(t) can be obtained by setting $\dot {\rho }_{00}(t)$, $\dot {\rho }_{00}(t)$ and $\dot {\rho }_{00}(t)$ equal to zero. Therefore from (13) we have

$$ \rho_{11}(t) = \rho_{22}(t). $$

(40)

Substituting (38) and (39) into (40) one gets

$$ A=\frac{l}{2m+l}, $$

(41)

which is the stationary value of ρ ₀₀(t). From the initial condition, ρ ₀₀(t = 0) = 1 and ρ ₁₁(t = 0) = 0, we find that

$$ B=1-\frac{l\left(m+(2m+l)(\lambda_{0}-\lambda_{1})\right)-2m(\lambda_{0}+m)}{(2m+l)(\lambda_{0}-\lambda_{1})} $$

(42)

and

$$ C=\frac{m\left(l+2(\lambda_{0}+m)\right)}{(2m+l)(\lambda_{0}-\lambda_{1})}. $$

(43)

Substituting A into (38) and (39) the stationary values of both ρ ₁₁(t) and ρ ₂₂(t) are obtained to be m/2m + l. Now consider the effect of the presence of an external resonant optical pulse on the stationary values of ρ ₀₀(t), ρ ₁₁(t) and ρ ₂₂(t). Setting $\dot {\rho }_{00}(t)$, $\dot {\rho }_{11}(t)$, $\dot {\rho }_{22}(t)$, $\dot {\rho }_{12}(t)$ and $\dot {\rho }_{21}(t)$ equal to zero and using completeness relation, after some calculations, the stationary values of ρ ₀₀(t), ρ ₁₁(t) and ρ ₂₂(t) can be easily obtained,

$$ \rho_{00}(t\rightarrow\infty) = 1-\frac{2\left(m+2p^{2}\left(\frac{1}{m+n}+\frac{1}{l+m+n}\right)\right)}{2m+l+6p^{2}\left(\frac{1}{m+n}+\frac{1}{l+m+n}\right)}, $$

(44)

$$ \rho_{11}(t\rightarrow\infty) = \rho_{22}(t\rightarrow\infty) = \frac{m+2p^{2}\left(\frac{1}{m+n}+\frac{1}{l+m+n}\right)}{2m+l+6p^{2}\left(\frac{1}{m+n}+\frac{1}{l+m+n}\right)}. $$

(45)

As can be seen from (44) and (45) the Rabi frequency appears in stationary values and as it gets bigger these stationary values approach 1/3 and in the limit of p → ∞ the three stationary values become 1/3.