Interferometry of multi-level systems: rate-equation approach for a charge qu${ d }$it

Abstract

We theoretically describe a driven two-electron four-level double-quantum dot (DQD) tunnel coupled to a fermionic sea using the rate-equation formalism. This approach allows to find occupation probabilities of each DQD energy level in a relatively simple way, compared to other methods. Calculated dependencies are compared with the experimental results. The system under study is irradiated by a strong driving signal, and as a result, one can observe Landau–Zener–Stückelberg–Majorana (LZSM) interferometry patterns which are successfully described by the considered formalism. The system operation regime depends on the amplitude of the excitation signal and the energy detuning, so one can transfer the system to the necessary quantum state in the most efficient way by setting these parameters. Obtained results give insights about initializing, characterizing, and controlling the quantum system states.

Appendix A: Energy levels of a parallel double-quantum dot

In the current research, we study the DQD proposed in Ref. [30]; see Fig. 1a. The first step in the system analysis is finding of the DQD energy levels. In the general form, system electrostatic energy can be written by

$$\begin{aligned} E=\frac{1}{2}\overrightarrow{V}\cdot {\textbf{C}}\overrightarrow{V}=\frac{1}{2} \overrightarrow{V}\cdot \overrightarrow{Q}=\frac{1}{2}\overrightarrow{Q} \cdot {\textbf{C}}^{-1}\overrightarrow{Q}, \end{aligned}$$

(A1)

where $\overrightarrow{V}$ and $\overrightarrow{Q}$ are the vectors of voltages and charges, respectively, ${\textbf{C}}$ is the capacitance matrix. In Eq. (A1), we used $\overrightarrow{Q}=\textbf{C }\overrightarrow{V}$. Thus, to obtain the system energy levels, we need to find the vector of charges and the inverse capacitance matrix of the system.

The charges $Q_{1,2}$ in the quantum dots can be written as follows:

$$\begin{aligned} Q_{1}= \,& {} C_{\textrm{T1}}(V_{1}-V_{\textrm{TG}})+C_{\textrm{B1}}(V_{1}-V_{\textrm{BG}})+C_{\textrm{S}}(V_{1}-V_{\textrm{S}}) \nonumber \\{} & {} +C_{\textrm{D}}(V_{1}-V_{\textrm{D}})+C_{\textrm{M}}(V_{1}-V_{2}), \end{aligned}$$

(A2)

$$\begin{aligned} Q_{2}= \,& {} C_{\textrm{T2}}(V_{2}-V_{\textrm{TG}})+C_{\textrm{B2}}(V_{2}-V_{\textrm{BG}})+C_{\textrm{S}}(V_{2}-V_{\textrm{S}}) \nonumber \\{} & {} +C_{\textrm{D}}(V_{2}-V_{\textrm{D}})+C_{\textrm{M}}(V_{2}-V_{1}). \end{aligned}$$

(A3)

It is convenient to rewrite expressions (A2, A3) in a matrix form

$$\begin{aligned} \overrightarrow{Q}=\, & {} \begin{pmatrix} Q_{1}+C_{\textrm{T1}}V_{\textrm{TG}}+C_{\textrm{B1}}V_{\textrm{BG}}+C_{\textrm{S}}V_{\textrm{S}}+C_{\textrm{D}}V_{\textrm{D}} \\ Q_{2}+C_{\textrm{T2}}V_{\textrm{TG}}+C_{\textrm{B2}}V_{\textrm{BG}}+C_{\textrm{S}}V_{\textrm{S}}+C_{\textrm{D}}V_{\textrm{D}} \end{pmatrix} \nonumber \\=\, & {} \begin{pmatrix} C_{1}V_{1}-C_{\textrm{M}}V_{2} \\ C_{2}V_{2}-C_{\textrm{M}}V_{1} \end{pmatrix}, \end{aligned}$$

(A4)

where $C_{1}$ and $C_{2}$ are the capacitances, connected to the first and the second quantum dots, respectively

$$\begin{aligned} C_{1}= \,& {} C_{\textrm{T1}}+C_{\textrm{B1}}+C_{\textrm{D}}+C_{\textrm{S}}+C_{\textrm{M}}, \\ C_{2}= \,& {} C_{\textrm{T2}}+C_{\textrm{B2}}+C_{\textrm{D}}+C_{\textrm{S}}+C_{\textrm{M}}. \nonumber \end{aligned}$$

(A5)

Then, the capacitance matrix has the following form:

$$\begin{aligned} {\textbf{C}}= \begin{pmatrix} C_{1} &{} -C_{\textrm{M}} \\ -C_{\textrm{M}} &{} C_{2} \end{pmatrix}. \end{aligned}$$

(A6)

Let us assume for simplicity that $V_{\textrm{S}}=V_{\textrm{D}}=0$. Then, putting expressions for the vector of charges Eq. (A4) and for the inverse capacitance matrix Eq. (A6), for the DQD electrostatic energy, we obtain

$$\begin{aligned} E= \,& {} \frac{1}{C_{1}C_{2}-C_{\textrm{M}}^{2}}\left[ \frac{1}{2}C_{1}Q_{2}^{2}+\frac{1}{ 2}C_{2}Q_{1}^{2}+C_{\textrm{M}}Q_{1}Q_{2}\right] \nonumber \\{} & {} + \frac{V_{\textrm{TG}}}{C_{1}C_{2}-C_{\textrm{M}}^{2}}\left[ C_{\textrm{T1}}\left( C_{\textrm{M}}Q_{2}+C_{2}Q_{1}\right) +C_{\textrm{T2}}\left( C_{1}Q_{2}+C_{\textrm{M}}Q_{1}\right) \right] \nonumber \\{} & {} +\frac{V_{\textrm{BG}}}{C_{1}C_{2}-C_{\textrm{M}}^{2}}\left[ C_{\textrm{B1}}\left( C_{\textrm{M}}Q_{2}+C_{2}Q_{1}\right) +C_{\textrm{B2}}\left( C_{1}Q_{2}+C_{\textrm{M}}Q_{1}\right) \right] \nonumber \\{} & {} + \frac{V_{\textrm{TG}}^{2}}{C_{1}C_{2}-C_{\textrm{M}}^{2}}\left[ \frac{1}{2}C_{1}C_{\textrm{T2}}^{2}+ \frac{1}{2}C_{2}C_{\textrm{T1}}^{2}+ C_{\textrm{T1}}C_{\textrm{T2}}C_{\textrm{M}}\right] \nonumber \\{} & {} +\frac{V_{\textrm{BG}}^{2}}{C_{1}C_{2}-C_{\textrm{M}}^{2}}\left[ \frac{1}{2}C_{1}C_{\textrm{T2}}^{2}+ \frac{1}{2}C_{2}C_{\textrm{T1}}^{2}+C_{\textrm{T1}}C_{\textrm{T2}}C_{\textrm{M}}\right] . \end{aligned}$$

(A7)

To simplify Eq. (A7), let us introduce new values, $N_{i}=-\frac{Q_{i}}{\left| e\right| }$, for the number of electrons in the ith quantum dot, and reduced top-gate $n_{t}$ and back-gate $n_{b}$ voltages

$$\begin{aligned} n_{t}=\, & {} \frac{C_{\textrm{T1}}V_{\textrm{TG}}}{\left| e\right| }=\frac{1}{1+a}\frac{ C_{\textrm{T2}}V_{\textrm{TG}}}{\left| e\right| }, \end{aligned}$$

(A8)

$$\begin{aligned} n_{b}= \,& {} \frac{C_{\textrm{B1}}V_{\textrm{BG}}}{\left| e\right| }=\frac{1}{1-a}\frac{ C_{\textrm{B2}}V_{\textrm{BG}}}{\left| e\right| }, \end{aligned}$$

(A9)

where a is an asymmetry factor in the gate couplings. Assuming $C_{1}=C_{2}=mC_{\textrm{M}}$ and rewriting the quantum dot charges as $Q_{1(2)}=-\left| e\right| N_{1(2)}$, then for the energy of the quantum dot, we have

$$\begin{aligned} \frac{E_{N_1,N_2}}{E_{\textrm{C}}}=\, & {} \frac{1}{2}N_{1}^{2}+\frac{1}{2}N_{2}^{2}+\frac{N_{1}N_{2} }{m}\nonumber \\{} & {} -n_{t}\left[ N_{1}+\frac{N_{2}}{m}+(1+a)\left( N_{2}+\frac{N_{1}}{m} \right) \right] \nonumber \\{} & {} -n_{b}\left[ N_{1}+\frac{N_{2}}{m}+(1-a)\left( N_{2}+\frac{N_{1}}{m} \right) \right] \nonumber \\{} & {} + n_{t}^{2}\left[ \frac{1}{2}+\frac{1}{2}(1+a)^{2}+\frac{1+a}{m}\right] \nonumber \\{} & {} +n_{b}^{2}\left[ \frac{1}{2}+\frac{1}{2}(1-a)^{2}+\frac{1-a}{m}\right] \nonumber \\{} & {} +n_{t}n_{b}\left[ 2(1+\frac{1}{m})-a^{2}\right] , \end{aligned}$$

(A10)

where we defined $E_{\textrm{C}}=E_{\mathrm{C_1}}=E_{\mathrm{C_2}}=mE_{\mathrm{C_M}}=e^{2}\frac{C_{1}}{{C_{1}^{2}-C_{\textrm{M}}^{2}}}$. We plot the energy-level diagram in Fig. 1b for the following parameters: $a=0.1$, $n_{b}=0.25$, $m=10$, and $N_{1},~N_{2}$ are equal to 0 or 1.