SCIENCE CHINA Information Sciences, Volume 63 , Issue 9 : 192501(2020) https://doi.org/10.1007/s11432-019-2706-9

## Quantum algorithms of state estimators in classical control systems

• AcceptedOct 24, 2019
• PublishedAug 13, 2020
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### Abstract

In this paper, quantum algorithms are applied to the design of state estimators in classical control systems under the condition that quantum algorithms can be physically implemented. We demonstrate that the design of state estimators can be solved by quantum algorithms, which may achieve significant acceleration in comparison to traditional classical algorithms.The time complexity can be reduced from $O(n^6)$ to $O(qn)$ when the system matrix is sparse and the condition number $\kappa$ and the reciprocal of precision $\epsilon$ are small in size $O$(poly log$(n)$), where $n$ is the dimension of state $x(t)$ and $q$ is the dimension of input $u(t)$.Our research will provide an entire quantum scheme of constructing state estimators and can be regarded as an attempt to widen application scope of quantum computation.

### Acknowledgment

This work was supported by National Natural Science Foundation of China (Grant Nos. 61673389, 61273202).

### Supplement

Appendix

The impact of the inaccuracy of quantum measurement on the state estimator

The design goal of the state estimator is to make the state estimator's output $\hat{x}(t)$ as close as possible to the system state $x(t)$. Therefore, we make the steady state error $J=\lim_{t\rightarrow+\infty}\|\hat{T}(\hat{x}(t)-x(t))\|$ as a measure of the quality of the state observer constructed by matrices $\hat{T}$ and $\hat{L}$. Suppose there are $N$ measured quantum states, then the measurement error $\delta=O(\frac{1}{N})$.

After measuring quantum states $|T\rangle$ and $|L\rangle$, we get the normalized matrices $\hat{T_0}$ and $\hat{L_0}$. Because of the inaccuracy of quantum measurement, there is a deviation between the real values $T_0,~L_0$ and measured values $\hat{T_0},~\hat{L_0}$. Therefore, the coefficients $\hat{C_1}$ and $\hat{C_2}$ obtained based on $\hat{T_0},~\hat{L_0}$ also deviate from the true values $C_1$ and $C_2$, $$T=C_1T_0, \hat{T}=\hat{C_1}\hat{T_0}=\hat{C_1}\bigg(T_0+\Delta\bigg(\frac{1}{N}\bigg)\bigg), \tag{35}$$ $$L=C_2L_0, \hat{L}=\hat{C_2}\hat{L_0}=\hat{C_2}\bigg(L_0+\Delta\bigg(\frac{1}{N}\bigg)\bigg), \tag{36}$$ where $T,~L$ are the expected target matrices, $\hat{T},~\hat{L}$ are the corresponding matrices with errors and $\Delta(\frac{1}{N})$ is a set of matrices whose element $\delta_{ij}$ satisfies $\delta_{ij}=O(\frac{1}{N})$. The matrices $T_0$ and $L_0$ are normalized, so $C_1=\|T\|$ and $C_2=\|L\|$. Substituting $\hat{T}$ into (4), we get \begin{aligned} &\hat{C_1}\bigg(FT_0+F\Delta\bigg(\frac{1}{N}\bigg)-T_0A-\Delta\bigg(\frac{1}{N}\bigg)A\bigg)_{11}=(GC)_{11} \\ &\mapsto\hat{C_1}\bigg(\frac{1}{C_1}GC+F\Delta\bigg(\frac{1}{N}\bigg)-\Delta\bigg(\frac{1}{N}\bigg)A\bigg)_{11}=(GC)_{11} \\ &\mapsto\hat{C_1}\bigg(\frac{1}{C_1}V_{11}+\lambda(f_1)O\bigg(\frac{1}{N}\bigg)-\sum_ia_{i1}O\bigg(\frac{1}{N}\bigg)\bigg)=V_{11}, \end{aligned} \tag{37} where $V_{11}=(GC)_{11}$ is the first element of matrix $GC$, $a_{i1}$ is the non-zero element of the first column of $A$, the matrix $F$ is a diagonal matrix constructed based on eigenvalue $\lambda(f_i)~(\lambda(f_i)<0)$ and $A$ is a sparse matrix. According to (37), there is $$\hat{C_1}=\frac{V_{11}}{V_{11}+(\lambda(f_1)-\sum_ia_{i1})C_1O(\frac{1}{N})}C_1. \tag{38}$$ The value of $N$ is big enough, so $\hat{C_1}\approx~C_1=\|T\|$. Based on the equation $\hat{L}=\hat{T}B$, we can get \begin{aligned} &\hat{C_2}\|\hat{L_0}\|=\|C_1\bigg(T_0+\Delta\bigg(\frac{1}{N}\bigg)\bigg)B\|, \\ & \hat{C_2}=\|L+C_1\Delta\bigg(\frac{1}{N}\bigg)B\|\approx\|L\|. \end{aligned} \tag{39}

Let $\delta~T=\hat{T}-T,~\delta~L=\hat{L}-L$, then we may know $\delta~T\subseteq\Delta(\frac{1}{N})\|T\|,~\delta~L\subseteq\Delta(\frac{1}{N})\|L\|$. Let $e(t)=\hat{T}(\hat{x}(t)-x(t))$ and we obtain \begin{align}\dot{e}(t)&=\dot{z}(t)-\hat{T}\dot{x}(t) =Fe(t)+(F\hat{T}+GC-\hat{T}A)x(t)+(\hat{L}-\hat{T}B)u(t) \\ &=Fe(t)+(F\delta T-\delta TA)x(t)+(\delta L-\delta TB)u(t). \tag{40} \end{align} Set $\delta~U=F\delta~T-\delta~TA\subseteq\|U\|\Delta(\frac{1}{N})$ and $\delta~V=\delta~L-\delta~TB\subseteq\|V\|\Delta(\frac{1}{N})$, then we get $$\|U\|\leq(\lambda(f)+a)\|T\|, \|V\|\leq(\|L\|+nb\|T\|), \tag{41}$$ where $\lambda(f)=\text{max}(|\lambda(f_i)|)$, $a=\text{max}(|a_{ij}|)$ and $b=\text{max}(|b_{ij}|)$.

Therefore, we can obtain $$\dot{e}(t)=Fe(t)+\delta Ux(t)+\delta Vu(t). \tag{42}$$ Therefore, Eq. (42) can be transformed into $$\dot{e_i}(t)=\lambda(f_i)e_i(t)+\sum_{j=1}^n{\delta U_{ij}x_j(t)}+\sum_{j=1}^q{\delta V_{ij}u_j(t)}, \tag{43}$$ and $$e_i(t)=E_i{\rm e}^{\lambda(f_i)t}+{\rm e}^{\lambda(f_i)t}\int{{\rm e}^{-\lambda(f_i)t}\left(\sum_{j=1}^n{\delta U_{ij}x_j(t)}+\sum_{j=1}^q{\delta V_{ij}u_j(t)}\right)}\text{d}t. \tag{44}$$

Let $y_i(t)=\sum_{j=1}^n{\delta~U_{ij}x_j(t)}+\sum_{j=1}^q{\delta~V_{ij}u_j(t)}$, then we may infer \begin{align}|y_i(t)|&=\left|\sum_{j=1}^n{\delta U_{ij}x_j(t)}+\sum_{j=1}^q{\delta V_{ij}u_j(t)}\right| \\ &\leq \left| \sum_{j=1}^n{\delta U_{ij}x_j(t)}\right|+\left|\sum_{j=1}^q{\delta V_{ij}u_j(t)}\right| \\ &\leq nx_m\|U\|O\bigg(\frac{1}{N}\bigg)+qu_m\|V\|O\bigg(\frac{1}{N}\bigg), \tag{45} \end{align} where $x_m=\text{max}(|x_{j}(t)|)$ and $u_m=\text{max}(|u_{j}(t)|)$.

According to (44), it follows: \begin{align}\lim_{t\rightarrow+\infty}|e_i(t)|&\leq\lim_{t\rightarrow+\infty}|{\rm e}^{\lambda(f_i)t}\int{{\rm e}^{-\lambda(f_i)t}y_i(t)}\text{d}t| \\ &\leq-\frac{nx_m}{\lambda(f_i)}\|U\|O\bigg(\frac{1}{N}\bigg)-\frac{qu_m}{\lambda(f_i)}\|V\|O\bigg(\frac{1}{N}\bigg) \\ &\leq-\frac{nx_m(\lambda(f)+a)}{\lambda(f_i)}\|T\|O\bigg(\frac{1}{N}\bigg)-\frac{qu_m}{\lambda(f_i)}(\|L\|+nb\|T\|)O\bigg(\frac{1}{N}\bigg), \tag{46} \end{align} and the steady state error of state estimator $J=\lim_{t\rightarrow+\infty}\|\hat{T}(\hat{x(t)}-x(t))\|$ satisfies \begin{align}J&=\lim_{t\rightarrow+\infty}\|e(t)\| \\ &\leq\sqrt{\sum_{i=1}^{n}\bigg(\frac{nx_m(\lambda(f)+a)}{\lambda(f_i)}\|T\|O\bigg(\frac{1}{N}\bigg)+\frac{qu_m}{\lambda(f_i)}(\|L\|+nb\|T\|)O\bigg(\frac{1}{N}\bigg)\bigg)^2}. \tag{47} \end{align} When all eigenvalues of the system matrix $A$ are negative, the original linear system $\Sigma_1$ is stable, $\lim_{t\rightarrow+\infty}x(t)=x$ and \begin{align}J&\leq\sqrt{\sum_{i=1}^{n}\bigg(\frac{nx_m(\lambda(f)+a)}{\lambda(f_i)}\|T\|O\bigg(\frac{1}{N}\bigg)+\frac{qu_m}{\lambda(f_i)}(\|L\|+nb\|T\|)O\bigg(\frac{1}{N}\bigg)\bigg)^2} \\ &\approx O\bigg(\frac{\sqrt{n}n^2q}{N}\bigg). \tag{48} \end{align} With the increase of the number $N$ of measured quantum states, the steady state error $J$ of state estimator decreases. We can simulate the error in Figure A1.

In addition, when the system matrix $A$ has positive eigenvalues, the $\Sigma_1$ is an unstable system. With the increase of $x(t)$, the steady state error $J$ increases. For the unstable system $\Sigma_1$, the inaccuracy of quantum measurement will seriously affect the state estimator.

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• Figure 1

The system diagram about state estimator. $\Sigma_1$ is the original linear system, $\Sigma_2$ is the state estimator to be designed, and state feedback $K$ is the coefficient matrix.

• Figure 2

The flow chart of quantum algorithms of constructing state estimators. Information of the matrices $T$ and $L$ is stored in quantum registers.

• Figure 3

Quantum circuit for phase estimation. There are two registers in the circuit, one for storing state $|v\rangle$ and the other named M for storing the eigenvalues. Initialize M to $|0\rangle^{\bigotimes~m}$. And the bottom half of the picture is a Fourier transform. The $x_i$ is the representation of qubits of eigenvalue. Accuracy is reserved to four decimal places.

• Figure 4

Quantum circuit for the quantum state transition. Convert the output of the HHL algorithm to the input of the quantum algorithms of matrix multiplication. The position information of the quantum state $|x\rangle$ is stored in the register G, which controls the number of rows of the quantum state $|T\rangle$ in the register N. Where X is the NOT gate.

• Table 1

Table 1The time complexity comparison between classical algorithms and our scheme

 Time complexity Our scheme when Our scheme when Step of classical quantum states can be quantum states cannot algorithms efficiently prepared be efficiently prepared Select $F$ and $G$ 0 0 0 Preparing states $|\nu\rangle$ and $|B\rangle$ $/$ $O(\log(n)/\epsilon^2)$[10] $O(n^2)$[21] $H|r\rangle=|\nu\rangle$ $O(n^6)$ $O(\kappa_1^2\log(n)/\epsilon)$[6,8] $O(\kappa_1^2\log(n)/\epsilon)$[6,8] $|L\rangle=|TB\rangle$ $O(n^2q)$ $O(\kappa_2\sqrt{n}/\epsilon+nq)$[10] $O(\kappa_2\sqrt{n}/\epsilon+nq)$[10] Sum $O(n^6)$ $O(\kappa_2\sqrt{n}/\epsilon+nq)$ $O(\kappa_2\sqrt{n}/\epsilon+n^2)$

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