Hardware efficient decomposition of the Laplace operator and its application to the Helmholtz and the Poisson equation on quantum computer

Abstract

With the rapid advancement of quantum computers in the past few years, there is ongoing development of algorithms aimed at solving problems that are difficult to tackle with classical computers. A pertinent instance of this is the resolution of partial differential equations (PDEs), where a current trend involves the exploration of variational quantum algorithms (VQAs) tailored to efficiently function on the noisy intermediate-scale quantum (NISQ) devices. Recently, VQAs for solving the Poisson equation have been proposed, and these algorithms require highly entangled quantum states or specific types of qubit entanglement to compute the expectation value of the Laplace operator. Implementing such requirements on NISQ devices poses a significant challenge. To overcome this problem, we propose a new method for representing the Laplace operator in the finite difference formulation. Since the quantum circuits introduced for evaluating the expectation value of the Laplace operator through proposed method do not require processes that degrade the fidelity of computation, such as swap operations or generation of highly entangled states, they can be easily implemented on NISQ devices. In the regime of quantum supremacy (the number of qubits is approximately 50), our proposed approach necessitates approximately one-third fewer CNOT operations compared to conventional methods. To assess the effectiveness of the proposed method, we conduct computations for finding the eigenvalues of the Helmholtz equation and solving the Poisson equation on cloud-based quantum hardware. We calculate the fidelity of quantum states required for each method through quantum tomography and also estimate the fidelity in the quantum supremacy regime. We believe that the proposed method can be applied to other PDEs having the Laplace operator and greatly assists in solving them.

Appendices

Appendix A: Derivation of Hamiltonian decomposition

Here, we prove the Proposition 1 in the Section III A.

We prove

$${\text{P}}_{\text{n}}^{-1}\left({\text{I}}^{\otimes \text{n}-1}\otimes \text{X}\right){\text{P}}_{\text{n}}= \sum_{\text{i}=1}^{\text{n}-1}{\text{B}}_{\text{i}}\left(\text{n}\right)$$

(A1)

by using the mathematical induction.

For the case of n = 2, it is easy to show that Eq. (A1) holds by construction. By definition of operations,

$${\text{P}}_{2}= \left(\begin{array}{cc}\begin{array}{cc}0& 0\\ 1& 0\end{array}& \begin{array}{cc}0& 1\\ 0& 0\end{array}\\ \begin{array}{cc}0& 1\\ 0& 0\end{array}& \begin{array}{cc}0& 0\\ 1& 0\end{array}\end{array}\right),\text{ I}\otimes \text{X}=\left(\begin{array}{cc}\begin{array}{cc}0& 1\\ 1& 0\end{array}& \begin{array}{cc}0& 0\\ 0& 0\end{array}\\ \begin{array}{cc}0& 0\\ 0& 0\end{array}& \begin{array}{cc}0& 1\\ 1& 0\end{array}\end{array}\right),{\text{T}}_{1}\left(2\right)=\left(\begin{array}{cc}\begin{array}{cc}1& 0\\ 0& -1\end{array}& \begin{array}{cc}0& 0\\ 0& 0\end{array}\\ \begin{array}{cc}0& 0\\ 0& 0\end{array}& \begin{array}{cc}1& 0\\ 0& -1\end{array}\end{array}\right),\text{C}\left(1\right)=\frac{1}{\sqrt{2}} \left(\begin{array}{cc}\begin{array}{cc}1& 0\\ 1& 0\end{array}& \begin{array}{cc}0& 1\\ 0& -1\end{array}\\ \begin{array}{cc}0& 1\\ 0& -1\end{array}& \begin{array}{cc}1& 0\\ 1& 0\end{array}\end{array}\right),$$

Both LHS and RHS of Eq. (A1) are given by,

$$\left(\begin{array}{cc}\begin{array}{cc}0& 0\\ 0& 0\end{array}& \begin{array}{cc}0& 1\\ 1& 0\end{array}\\ \begin{array}{cc}0& 1\\ 1& 0\end{array}& \begin{array}{cc}0& 0\\ 0& 0\end{array}\end{array}\right).$$

Then, we will show that for every $\text{n}\ge 2$, if Eq. (A1) for n holds, then the Eq. (A1) for n + 1 also holds. For the sake of simplicity, we denote the LHS and the RHS of equation (A1) to $\text{L}(\text{n})$ and $\text{R}\left(\text{n}\right)$ , respectively. By the definition of the shift operation, the matrix element of the $\text{L}(\text{n})$ is given by,

$$\text{L}{\left(\text{n}\right)}_{\text{ij}}=\left\{\begin{array}{c}1, if \left(\text{i},\text{j}\right)=\left(2\text{k}, 2\text{k}+1\right)\, or \,\left(2\text{k}+1, 2\text{k}\right) for 1\le k\le {2}^{\text{n}-1}-1\\ 0, otherwise\end{array}\right..$$

(A2)

By equation (A2), $\text{L}\left(\text{n}+1\right)$ can be expressed by,

$$\text{L}\left(\text{n}+1\right)=\text{I}\otimes \text{L}\left(\text{n}\right)+\left(-\text{I}+\text{X}\right)\otimes \left({\text{X}}_{0}^{\otimes \text{n}}+{\text{X}}_{1}^{\otimes \text{n}}\right), $$

(A3)

where ${\text{X}}_{0}= \left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ and ${\text{X}}_{1}= \left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$. By the definition of ${\text{B}}_{\text{i}}\left(\text{n}\right),\text{ C}(\text{i})$ and ${\text{T}}_{\text{i}}(\text{n})$ in Eq. (27)–(30) in the main text, we can write the operations acting on the n + 1 qubits quantum state as that of on n qubits quantum state,

$$\text{C}\left(\text{i},\text{n}+1\right)=\text{I}\otimes \text{C}\left(\text{i},\text{ n}\right),$$

(A4)

$${\text{T}}_{\text{i}}\left(\text{n}+1\right)=\text{I}\otimes {\text{T}}_{\text{i}}\left(\text{n}\right)\text{ for }1\le \text{i}\le \text{n}-2,$$

(A5)

$${\text{T}}_{\text{n}-1}\left(\text{n}+1\right)=\text{I}\otimes {\text{T}}_{\text{n}-1}\left(\text{n}\right)-\text{I}\otimes \left({\text{I}}_{0}^{\otimes \text{n}-1}\otimes \text{Z}\right),$$

(A6)

$${\text{T}}_{\text{n}}\left(\text{n}+1\right)=\text{I}\otimes {\text{I}}_{0}^{\text{n}-1}\otimes \text{Z},$$

(A7)

where, $\text{C}\left(\text{i},\text{ n}\right)$ is the $\text{C}(\text{i})$ in the n qubits space. Hence,

$${\text{B}}_{\text{i}}\left(\text{n}+1\right)=\text{I}\otimes {\text{B}}_{\text{i}}\left(\text{n}\right)\text{ for }1\le \text{i}\le \text{n}-2,$$

(A8)

$${\text{B}}_{\text{n}-1}\left(\text{n}+1\right)=\text{I}\otimes {\text{B}}_{\text{n}-1}\left(\text{n}\right)-\text{C}{\left(\text{n}-1,\text{n}+1\right)}^{-1}\left[\text{I}\otimes {\text{I}}_{0}^{\otimes \text{n}-1}\otimes \text{Z}\right]\text{C}\left(\text{n}-1,\text{ n}+1\right),$$

(A9)

$${\text{B}}_{\text{n}}\left(\text{n}+1\right)=\text{C}{\left(\text{n},\text{ n}+1\right)}^{-1}{\text{T}}_{\text{n}}\left(\text{n}+1\right)\text{C}\left(\text{n},\text{ n}+1\right).$$

(A10)

Using above relations, it can be shown that,

$$\text{R}\left(\text{n}+1\right)=\text{I}\otimes \text{R}\left(\text{n}\right)-\text{C}{\left(\text{n}-1,\text{ n}+1\right)}^{-1}\left[\text{I}\otimes {\text{I}}_{0}^{\otimes \text{n}-1}\otimes \text{Z}\right]\text{C}\left(\text{n}-1,\text{ n}+1\right)+\text{C}{\left(\text{n},\text{ n}+1\right)}^{-1}{\text{T}}_{\text{n}}\left(\text{n}+1\right)\text{C}\left(\text{n},\text{n}+1\right).$$

(A11)

By the definition of $\text{C}(\text{i},\text{n})$,

$$\text{C}\left(\text{n},\text{ n}+1\right)=\text{C}\left(\text{n}-1,\text{ n}+1\right)\text{CX}\left(0,\text{ n}\right),$$

(A12)

and the last term in the Eq. (A11) can be written by,

$$\begin{aligned}\text{C}{\left(\text{n},\text{ n}+1\right)}^{-1}{\text{T}}_{\text{n}}\left(\text{n}+1\right)\text{C}\left(\text{n},\text{ n}+1\right)=\text{CX}\left(0,\text{n}\right)\text{C}{\left(\text{n}-1,\text{n}+1\right)}^{-1}\\ \left[\text{I}\otimes {\text{I}}_{0}^{\otimes \text{n}-1}\otimes \text{Z}\right]\text{C}\left(\text{n}-1,\text{n}+1\right)\text{CX}\left(0,\text{ n}\right).\end{aligned}$$

(A13)

From $\text{C}\left(\text{n}-1,\text{ n}+1\right)=\text{H}(0)\prod_{\text{q}=1}^{\text{n}-1}\text{CX}(0,\text{ q})$,

$$\text{C}\left(\text{n}-1,\text{ n}+1\right)=\text{H}\left(0\right)\text{I}\otimes \left[{\text{I}}^{\otimes \text{n}-1}\otimes {\text{I}}_{0}+{\text{X}}^{\otimes \text{n}-1}\otimes {\text{I}}_{1}\right],$$

(A14)

where, ${\text{I}}_{0}= \left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$ and ${\text{I}}_{1}= \left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)$. If we insert the Eq. (A14) into the second term of Eq. (A11),

$$\text{C}{\left(\text{n}-1,\text{ n}+1\right)}^{-1}\left[\text{I}\otimes {\text{I}}_{0}^{\otimes \text{n}-1}\otimes \text{Z}\right]\text{C}\left(\text{n}-1,\text{ n}+1\right)=\text{I}\otimes \left\{\left({\text{I}}^{\otimes \text{n}-1}\otimes {\text{I}}_{0}+{\text{X}}^{\otimes \text{n}-1}\otimes {\text{I}}_{1}\right)\left[{\text{I}}^{\otimes \text{n}-1}\otimes {\text{I}}_{0}+{\text{X}}^{\otimes \text{n}-1}\otimes {\text{I}}_{1}\right]\left({\text{I}}^{\otimes \text{n}-1}\otimes {\text{I}}_{0}+{\text{X}}^{\otimes \text{n}-1}\otimes {\text{I}}_{1}\right)\right\}=\text{I}\otimes \left({\text{X}}_{0}^{\otimes \text{n}}+{\text{X}}_{1}^{\otimes \text{n}}\right).$$

(A15)

Similarly, if we insert Eq. (A14) into the third term of Eq. (A11) through Eq. (A13) using the matrix representation of CNOT operation, $\text{CX}\left(0,\text{ n}\right)= {\text{I}}^{\otimes \text{n}}\otimes {\text{I}}_{0}+\text{X}\otimes {\text{I}}^{\otimes \text{n}-1}\otimes {\text{I}}_{1}$,

$$\text{C}{\left(\text{n},\text{ n}+1\right)}^{-1}{\text{T}}_{\text{n}}\left(\text{n}+1\right)\text{C}\left(\text{n},\text{n}+1\right)=\text{X}\otimes \left({\text{X}}_{0}^{\otimes \text{n}}+{\text{X}}_{1}^{\otimes \text{n}}\right).$$

(A16)

Therefore, $\text{R}(\text{n}+1)$ in Eq. (A11) is written as,

$$\text{R}\left(\text{n}+1\right)=\text{I}\otimes \text{R}\left(\text{n}\right)+\left(-\text{I}+\text{X}\right)\otimes \left({\text{X}}_{0}^{\otimes \text{n}}+{\text{X}}_{1}^{\otimes \text{n}}\right).$$

(A17)

By comparing Eq. (A3) and Eq. (A17), we can check $\text{L}\left(\text{n}+1\right)=\text{R}(\text{n}+1)$, and we can conclude that Eq. (A1) holds for all $\text{n}\ge 2$. Similar to this argument, we can prove Eqs. (25) and (26) for ${\text{B}}_{\text{D}}\left(\text{n}\right)$ and ${\text{B}}_{\text{N}}\left(\text{n}\right)$ , respectively. □

Appendix B: Derivation of Hamiltonian decomposition

Here, we prove the Proposition 2 in the Sec. III A.

We prove.

$$\text{C}{\left(\text{i}\right)}^{-1}{\text{T}}_{\text{i}}\left(\text{n}\right)\text{C}\left(\text{i}\right)=\widetilde{\text{C}}{\left(\text{i}\right)}^{-1}{\text{T}}_{\text{i}}\left(\text{n}\right)\widetilde{\text{C}}\left(\text{i}\right),$$

(B1)

here,

$${\text{T}}_{\text{i}}\left(\text{n}\right)=\left\{\begin{array}{c}{\text{I}}^{\otimes \text{n}-1-\text{i}}\otimes {\text{I}}_{1}\otimes {\text{I}}_{0}^{\otimes \text{i}-1}\otimes Z, 1\le i\le n-2\\ I\otimes {\text{I}}_{0}^{\otimes \text{n}-2}\otimes Z, i=n-1\end{array}\right.,$$

(B2)

$$\text{C}\left(\text{i}\right)=\text{H}\left(0\right)\prod_{\text{q}=1}^{\text{i}}\text{CX}\left(0,\text{ q}\right),$$

(B3)

and,

$$\widetilde{\text{C}}\left(\text{i}\right)=\text{H}\left(0\right)\prod_{\text{q}=1}^{\text{i}}\text{CX}\left(\text{q}-1,\text{ q}\right).$$

(B4)

For the sake of convenience, we denote the LHS for i and n as LHS(i, n) and RHS for i and n as RHS(i, n).

We express LHS(i,n) and RHS(i, n) as the tensor products of $\text{I}, {\text{ I}}_{0}, {\text{ I}}_{1},\text{ X},{\text{ X}}_{0}$ and ${\text{X}}_{1}$, and the check that results are the same.

During the proof, we use the following identities,

$$\text{CX}\left(\text{0,1}\right)=\text{I}\otimes {\text{I}}_{0}+\text{X}\otimes {\text{I}}_{1},$$

(B6)

$$\left(\text{I}{\otimes \text{I}}_{0}+\text{X}{\otimes \text{I}}_{1}\right)\left[{\text{I}}_{0}\otimes {\text{X}}_{0}\right]\left(\text{I}{\otimes \text{I}}_{0}+\text{X}{\otimes \text{I}}_{1}\right)={\text{X}}_{0}^{\otimes 2},$$

(B7)

$$\left(\text{I}{\otimes \text{I}}_{0}+\text{X}{\otimes \text{I}}_{1}\right)\left[{\text{I}}_{0}\otimes {\text{X}}_{1}\right]\left(\text{I}{\otimes \text{I}}_{0}+\text{X}{\otimes \text{I}}_{1}\right)={\text{X}}_{1}^{\otimes 2},$$

(B8)

$$\left(\text{I}{\otimes \text{I}}_{0}+\text{X}{\otimes \text{I}}_{1}\right)\left[{\text{I}}_{1}\otimes {\text{X}}_{0}\right]\left(\text{I}{\otimes \text{I}}_{0}+\text{X}{\otimes \text{I}}_{1}\right)={\text{X}}_{1}{\otimes \text{X}}_{0},$$

(B9)

$$\left(\text{I}{\otimes \text{I}}_{0}+\text{X}{\otimes \text{I}}_{1}\right)\left[{\text{I}}_{1}\otimes {\text{X}}_{1}\right]\left(\text{I}{\otimes \text{I}}_{0}+\text{X}{\otimes \text{I}}_{1}\right)={\text{X}}_{0}\otimes {\text{X}}_{1}.$$

(B10)

For $\text{i}=1$, LHS(1, n) can be expressed by,

$$\text{LHS}\left(1,\text{ n}\right)=\text{CX}\left(\text{0,1}\right)\left[{\text{I}}^{\otimes \text{n}-2}\otimes {\text{I}}_{1}\otimes \text{X}\right]\text{CX}\left(\text{0,1}\right)={\text{I}}^{\otimes \text{n}-2}\otimes \left({\text{X}}_{1}\otimes {\text{X}}_{0}+{\text{X}}_{0}\otimes {\text{X}}_{1}\right).$$

(B11)

For $2\le \text{i}\le \text{n}-2$, LHS(i, n) is given by,

$$\begin{aligned} \text{LHS}\left(\text{i},\text{n}\right)& =\text{CX}\left(0,\text{i}\right)\text{CX}\left(0,\text{i}-1\right)\dots \text{CX}\left(\text{0,1}\right)\\ &\quad \left[{\text{I}}^{\otimes \text{n}-1-\text{i}}\otimes {\text{I}}_{1}\otimes {\text{I}}_{0}^{\otimes \text{i}-1}\otimes \text{X}\right]\text{CX}\left(\text{0,1}\right)\dots \text{CX}\left(0,\text{i}-1\right)\text{CX}\left(0,\text{i}\right). \end{aligned}$$

(B5)

We can express $\text{CX}\left(\text{0,1}\right)[{\text{I}}^{\otimes \text{n}-1-\text{i}}\otimes {\text{I}}_{1}\otimes {\text{I}}_{0}^{\otimes \text{i}-1}\otimes \text{X}]\text{CX}\left(\text{0,1}\right)$ as

$$\text{CX}\left(\text{0,1}\right)\left[{\text{I}}^{\otimes \text{n}-1-\text{i}}\otimes {\text{I}}_{1}\otimes {\text{I}}_{0}^{\otimes \text{i}-1}\otimes \text{X}\right]\text{CX}\left(\text{0,1}\right)={\text{I}}^{\otimes \text{n}-1-\text{i}}\otimes {\text{I}}_{1}\otimes {\text{I}}_{0}^{\otimes \text{i}-2}\otimes \left({\text{X}}_{0}^{\otimes 2}+{\text{X}}_{1}^{\otimes 2}\right),$$

(B11)

and $\text{CX}(\text{0,2})\text{CX}\left(\text{0,1}\right)[{\text{I}}^{\otimes \text{n}-1-\text{i}}\otimes {\text{I}}_{1}\otimes {\text{I}}_{0}^{\otimes \text{i}-1}\otimes \text{X}]\text{CX}\left(\text{0,1}\right)\text{CX}(\text{0,2})$ as

$$\text{CX}(\text{0,2})\text{CX}\left(\text{0,1}\right)\left[{\text{I}}^{\otimes \text{n}-1-\text{i}}\otimes {\text{I}}_{1}\otimes {\text{I}}_{0}^{\otimes \text{i}-1}\otimes \text{X}\right]\text{CX}\left(\text{0,1}\right)\text{CX}(\text{0,2})={\text{CX}\left(\text{0,2}\right)[\text{I}}^{\otimes \text{n}-1-\text{i}}\otimes {\text{I}}_{1}\otimes {\text{I}}_{0}^{\otimes \text{i}-2}\otimes \left({\text{X}}_{0}^{2}+{\text{X}}_{1}^{2}\right)]\text{CX}(\text{0,2})={[\text{I}}^{\otimes \text{n}-1-\text{i}}\otimes {\text{I}}_{1}\otimes {\text{I}}_{0}^{\otimes \text{i}-3}\otimes \left({\text{X}}_{0}^{\otimes 3}+{\text{X}}_{1}^{\otimes 3}\right)].$$

(B12)

If we repeat above steps, for $2\le \text{i}\le \text{n}-3$, we can obtain

$$\text{LHS}\left(\text{i},\text{n}\right)={\text{I}}^{\otimes \text{n}-\text{i}-1}\otimes {\text{I}}_{1}\otimes \left({\text{X}}_{0}^{\otimes \text{i}}+{\text{X}}_{1}^{\otimes \text{i}}\right).$$

(B13)

For $\text{i}=\text{n}-2$,

$$\text{LHS}\left(\text{n}-2,\text{n}\right)=\text{CX}\left(0,\text{ n}-2\right)\text{LHS}\left(\text{n}-3,\text{n}\right)\text{CX}\left(0,\text{ n}-2\right)=\text{I}\otimes \left({\text{X}}_{1}{\otimes \text{X}}_{0}^{\otimes \text{n}-2}+{\text{X}}_{0}\otimes {\text{X}}_{1}^{\otimes \text{n}-2}\right).$$

(B14)

Similarly for $\text{i}=\text{n}-1$,

$$\begin{aligned} \text{LHS}\left(\text{n}-1,\text{n}\right) & =\text{C}{\left(\text{n}\right)}^{-1}\left[\text{I}\otimes {\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{Z}\right]\text{C}\left(\text{n}\right)\\ & =\text{CX}\left(0,\text{n}-1\right)\left[\text{I}\otimes {(\text{X}}_{0}^{\otimes \text{n}-1}+{\text{X}}_{1}^{\otimes \text{n}-1}) \right]\text{CX}\left(0,\text{n}-1\right)\\ & =\text{X}\otimes {(\text{X}}_{0}^{\otimes \text{n}-1}+{\text{X}}_{1}^{\otimes \text{n}-1}). \end{aligned}$$

(B15)

We repeat same procedure for RHS(i, n). For $\text{i}=1$, $\text{C}\left(\text{i}\right)=\widetilde{\text{C}}(\text{i})$, so $\text{LHS}\left(1,\text{n}\right)=\text{RHS}(1,\text{n})$. For $2\le \text{i}\le \text{n}-2$, RHS(i, n) is given by,

$$\text{RHS}\left(\text{i},\text{n}\right)=\text{CX}\left(\text{i}-1,\text{ i}\right)\text{CX}\left(\text{i}-2,\text{i}-1\right)\dots \text{CX}\left(\text{0,1}\right)\left[{\text{I}}^{\otimes \text{n}-1-\text{i}}\otimes {\text{I}}_{1}\otimes {\text{I}}_{0}^{\otimes \text{i}-1}\otimes \text{X}\right]\text{CX}\left(\text{0,1}\right)\dots \text{CX}\left(\text{i}-2,\text{i}-1\right)\text{CX}\left(\text{i}-1,\text{i}\right).$$

(B16)

We can express $\text{CX}\left(\text{1,2}\right)\text{CX}\left(\text{0,1}\right)\left[{\text{I}}^{\otimes \text{n}-1-\text{i}}\otimes {\text{I}}_{1}\otimes {\text{I}}_{0}^{\otimes \text{i}-1}\otimes \text{X}\right]\text{CX}\left(\text{0,1}\right)\text{CX}\left(\text{1,2}\right)$ as,

$$\text{CX}(\text{1,2})\text{CX}\left(\text{0,1}\right)\left[{\text{I}}^{\otimes \text{n}-1-\text{i}}\otimes {\text{I}}_{1}\otimes {\text{I}}_{0}^{\otimes \text{i}-1}\otimes \text{X}\right]\text{CX}\left(\text{0,1}\right)\text{CX}(\text{1,2})={\text{CX}\left(\text{1,2}\right)[\text{I}}^{\otimes \text{n}-1-\text{i}}\otimes {\text{I}}_{1}\otimes {\text{I}}_{0}^{\otimes \text{i}-2}\otimes \left({\text{X}}_{0}^{2}+{\text{X}}_{1}^{2}\right)]\text{CX}(\text{1,2})={\text{I}}^{\otimes \text{n}-1-\text{i}}\otimes {\text{I}}_{1}\otimes {\text{I}}_{0}^{\text{i}-3}\otimes [\text{CX}\left(\text{1,2}\right)\left({\text{I}}_{0}\otimes {\text{X}}_{0}\right)\text{CX}\left(\text{1,2}\right)\otimes {\text{X}}_{0}+\text{CX}\left(\text{1,2}\right)\left({\text{I}}_{0}\otimes {\text{X}}_{1}\right)\text{CX}\left(\text{1,2}\right)\otimes {\text{X}}_{1}]={[\text{I}}^{\otimes \text{n}-1-\text{i}}\otimes {\text{I}}_{1}\otimes {\text{I}}_{0}^{\otimes \text{i}-3}\otimes \left({\text{X}}_{0}^{\otimes 3}+{\text{X}}_{1}^{\otimes 3}\right)].$$

(B17)

If we repeat above steps, for $2\le \text{i}\le \text{n}-3$, we can obtain

$$\text{RHS}\left(\text{i},\text{n}\right)={\text{I}}^{\otimes \text{n}-\text{i}-1}\otimes {\text{I}}_{1}\otimes \left({\text{X}}_{0}^{\otimes \text{i}}+{\text{X}}_{1}^{\otimes \text{i}}\right).$$

(B18)

Eq. (B18) is same as Eq. (B13). This is because, when applying the CNOT(0, m) operation (in calculation of LHS) or CNOT(m-1,m) operation (in the calculation of RHS) into the tensor products, the matrices corresponding to the controlled qubit for each case are the same. Using this property, it is easy to check that $\text{LHS}\left(\text{i},\text{n}\right)=\text{RHS}(\text{i},\text{n})$ for $\text{n}-3\le \text{i}\le \text{n}-2$. □

Appendix C: Derivation of Hamiltonian matrix for $|\mathbf{u}\rangle$

Here, we prove the Proposition 2 in the Sec. III B. For the simplicity, we denote the quantum state $\left|\uppsi \right.\rangle = \frac{1}{\sqrt{2}}(|0\rangle \otimes \left|\upphi \right.\rangle +{\left(-1\right)}^{\text{P}}|1\rangle \otimes \text{J}\left(\frac{\text{N}}{2}\right)|\upphi \rangle )$ as,

$$\left|\uppsi \right.\rangle =\frac{1}{\sqrt{2}}\left(\genfrac{}{}{}{}{|\upphi \rangle }{{\left(-1\right)}^{\text{P}}\text{J}|\upphi \rangle }\right),$$

(B1)

where, $\text{J}=\text{J}\left(\frac{\text{N}}{2}\right)$. The energy function $\langle\uppsi \left|\text{A}\right|\uppsi \rangle$ for $\text{A}=\left(\begin{array}{cc}{\text{A}}_{00}& {\text{A}}_{01}\\ {\text{A}}_{10}& {\text{A}}_{11}\end{array}\right)$ is expressed by,

$$\begin{aligned}&\langle \uppsi \left|\text{A}\right|\uppsi \rangle = \frac{1}{2}\left(\begin{array}{cc}\langle\upphi |& {\left(-1\right)}^{\text{P}}\langle\upphi |\text{J}\end{array}\right)\left(\begin{array}{cc}{\text{A}}_{00}& {\text{A}}_{01}\\ {\text{A}}_{10}& {\text{A}}_{11}\end{array}\right)\left(\genfrac{}{}{0pt}{}{\left|\upphi \right.\rangle }{{\left(-1\right)}^{\text{P}}\text{J}\left|\upphi \right.\rangle }\right)= \frac{1}{2}\left(\langle \upphi \left|{\text{A}}_{00}\right|\upphi \rangle \right.\\ &\left. +{\left(-1\right)}^{\text{P}}\langle \upphi\left|{\text{A}}_{01}\text{J}\right|\upphi \rangle+{\left(-1\right)}^{\text{P}}\langle \upphi \left|{\text{JA}}_{10}\right|\upphi \rangle +\langle \upphi \left|{\text{JA}}_{11}\text{J}\right|\upphi \rangle \right),\end{aligned}$$

(B2)

and, consequently leading to the Hamiltonian for $|\upphi \rangle$, ${\text{A}}^{\upphi }$,

$${\text{A}}^{\upphi }= \frac{1}{2}\left({\text{A}}_{00}+{\left(-1\right)}^{\text{P}}{\text{A}}_{01}\text{J}+{\left(-1\right)}^{\text{P}}{\text{JA}}_{10}+{\text{JA}}_{11}\text{J}\right),$$

(B3)

First, we derive the ${\text{A}}^{\upphi }$ for the Hamiltonian with the periodic boundary condition, ${\text{A}}_{\text{periodic}}\left(\text{n}\right)={\text{A}}_{1}+{\text{A}}_{2}+{\text{A}}_{3}= 2{\text{I}}^{\otimes \text{n}}-{\text{I}}^{\otimes \text{n}-1}\otimes \text{X}-{\text{B}}_{\text{P}}$(n) in Eq. (14) of main text. The first term of the Hamiltonian, ${\text{A}}_{1}= 2{\text{I}}^{\otimes \text{n}}$, is partitioned into $\left(\begin{array}{cc}2{\text{I}}^{\otimes \text{n}-1}& 0\\ 0& 2{\text{I}}^{\otimes \text{n}-1}\end{array}\right)$, and it is trivial to show that

$${\text{A}}_{1}^{\upphi }= \frac{1}{2}\left(2{\text{I}}^{\otimes \text{n}-1}+2{\text{JI}}^{\otimes \text{n}-1}\text{J}\right).$$

${\text{A}}_{2}= {\text{I}}^{\otimes \text{n}-1}\otimes \text{X}$ is partitioned into $\left(\begin{array}{cc}{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}& 0\\ 0& {\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\end{array}\right)$, and

$${\text{A}}_{2}^{\upphi }= \frac{1}{2}({\text{I}}^{\otimes \text{n}-2}\otimes \text{X}+\text{J}{[\text{I}}^{\otimes \text{n}-2}\otimes \text{X}]\text{J}).$$

The shift operation ${\text{P}}_{\text{n}}\in {\mathbb{R}}^{{2}^{\text{n}}\times {2}^{\text{n}}}$ in Eq. (20) of main text is partitioned into $\left(\begin{array}{cc}{\text{Q}}_{\text{n}-1}& {\text{R}}_{\text{n}-1} \\ {\text{R}}_{\text{n}-1} & {\text{Q}}_{\text{n}-1}\end{array}\right)$, where ${\text{Q}}_{\text{n}-1}\in {\mathbb{R}}^{{2}^{\text{n}-1} \times {2}^{\text{n}-1}}$ and ${\text{R}}_{\text{n}-1}\in {\mathbb{R}}^{{2}^{\text{n}-1} \times {2}^{\text{n}-1}}$ is given by,

$${\text{Q}}_{\text{n}-1}= \left(\begin{array}{cc}\begin{array}{ccc}0& 0& \dots \\ 1& 0& \dots \\ 0& 1& \dots \end{array}& \begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\\ \begin{array}{ccc}\dots & \dots & \dots \\ 0& 0& \dots \\ 0& 0& \dots \end{array}& \begin{array}{ccc}\dots & \dots & \dots \\ 1& 0& 0\\ 0& 1& 0\end{array}\end{array}\right), {\text{R}}_{\text{n}-1}= \left(\begin{array}{cc}\begin{array}{ccc}0& 0& \dots \\ 0& 0& \dots \\ 0& 0& \dots \end{array}& \begin{array}{ccc}0& 0& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\\ \begin{array}{ccc}\dots & \dots & \dots \\ 0& 0& \dots \\ 0& 0& \dots \end{array}& \begin{array}{ccc}\dots & \dots & \dots \\ 0& 0& 0\\ 0& 0& 0\end{array}\end{array}\right),$$

respectively. Therefore, ${\text{A}}_{3}={\text{P}}_{\text{n}}^{-1}\left({\text{I}}^{\otimes \text{n}-1}\otimes \text{X}\right){\text{P}}_{\text{n}}$ can be partitioned into

$${\text{A}}_{3}= \left(\begin{array}{cc}{\text{Q}}_{\text{n}-1}^{\text{T}}& {\text{R}}_{\text{n}-1}^{\text{T}}\\ {\text{R}}_{\text{n}-1}^{\text{T}}& {\text{Q}}_{\text{n}-1}^{\text{T}}\end{array}\right)\left(\begin{array}{cc}{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}& 0\\ 0& {\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\end{array}\right)\left(\begin{array}{cc}{\text{Q}}_{\text{n}-1}& {\text{R}}_{\text{n}-1} \\ {\text{R}}_{\text{n}-1} & {\text{Q}}_{\text{n}-1}\end{array}\right)= \left(\begin{array}{cc}{\text{Q}}_{\text{n}-1}^{\text{T}}[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}]{\text{Q}}_{\text{n}-1}+{\text{R}}_{\text{n}-1}^{\text{T}}[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}]{\text{R}}_{\text{n}-1}& {\text{Q}}_{\text{n}-1}^{\text{T}}[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}]{\text{R}}_{\text{n}-1}+{\text{R}}_{\text{n}-1}^{\text{T}}[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}]{\text{Q}}_{\text{n}-1}\\ {\text{Q}}_{\text{n}-1}^{\text{T}}[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}]{\text{R}}_{\text{n}-1}+{\text{R}}_{\text{n}-1}^{\text{T}}[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}]{\text{Q}}_{\text{n}-1}& {\text{Q}}_{\text{n}-1}^{\text{T}}[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}]{\text{Q}}_{\text{n}-1}+{\text{R}}_{\text{n}-1}^{\text{T}}[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}]{\text{R}}_{\text{n}-1}\end{array}\right).$$

By using ${\text{Q}}_{\text{n}-1}={\text{P}}_{\text{n}-1}-{\text{R}}_{\text{n}-1}$, the (0,0) component of ${\text{A}}_{3}$ is given by,

$${\text{A}}_{3}\left(\text{0,0}\right)=\left({\text{P}}_{\text{n}-1}^{\text{T}}-{\text{R}}_{\text{n}-1}^{\text{T}}\right)\left[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\right]\left({\text{P}}_{\text{n}-1}-{\text{R}}_{\text{n}-1}\right)+{\text{R}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{R}}_{\text{n}-1}= {\text{P}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{P}}_{\text{n}-1}-{\text{R}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{P}}_{\text{n}-1}-{\text{P}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{R}}_{\text{n}-1}+{2\text{R}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{R}}_{\text{n}-1}.$$

Each term in the above equation is simplified by,

$${\text{R}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{P}}_{\text{n}-1}={\text{D}}_{\text{A}},$$

$${\text{P}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{R}}_{\text{n}-1}={\text{D}}_{\text{A}}^{\text{T}},$$

$${\text{R}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{R}}_{\text{n}-1}=0,$$

where,

$${\text{D}}_{\text{A}}= \left(\begin{array}{cc}\begin{array}{ccc}0& 0& \dots \\ 0& 0& \dots \\ 0& 0& \dots \end{array}& \begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\\ \begin{array}{ccc}\dots & \dots & \dots \\ 0& 0& \dots \\ 1& 0& \dots \end{array}& \begin{array}{ccc}\dots & \dots & \dots \\ 0& 0& 0\\ 0& 0& 0\end{array}\end{array}\right).$$

The above calculation can be repeated for other terms in (B3),

$${\text{A}}_{3}\left(\text{0,1}\right)\text{J}= {\text{Q}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{R}}_{\text{n}-1}\text{J}+{\text{R}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{Q}}_{\text{n}-1}\text{J}=\left({\text{P}}_{\text{n}-1}^{\text{T}}-{\text{R}}_{\text{n}-1}^{\text{T}}\right)\left[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{R}}_{\text{n}-1}\text{J}+{\text{R}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\right]\left({\text{P}}_{\text{n}-1}-{\text{R}}_{\text{n}-1}\right)\text{J}.$$

Each term in the above equation is simplified by,

$${\text{P}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{R}}_{\text{n}-1}\text{J}={\text{D}}_{3},$$

$${\text{R}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{R}}_{\text{n}-1}\text{J}=0,$$

$${\text{R}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{P}}_{\text{n}-1}\text{J}={\text{D}}_{2}-{\text{D}}_{3}.$$

${\text{JA}}_{3}\left(1, 0\right)={\left({\text{A}}_{3}\left(0, 1\right)\text{J}\right)}^{\text{T}}$. ${\text{JA}}_{3}(\text{1,1})\text{J}$ is given by,

$${\text{JA}}_{3}\left(\text{1,1}\right)\text{J}={\text{JQ}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{Q}}_{\text{n}-1}\text{J}+{\text{JR}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{R}}_{\text{n}-1}\text{J}=\text{J}\left({\text{P}}_{\text{n}-1}^{\text{T}}-{\text{R}}_{\text{n}-1}^{\text{T}}\right)\left[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\right]\left({\text{P}}_{\text{n}-1}-{\text{R}}_{\text{n}-1}\right)\text{J}+ {\text{JR}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{R}}_{\text{n}-1}\text{J}$$

Each term in the above equation is simplified by,

$${\text{JR}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{P}}_{\text{n}-1}\text{J}={\text{D}}_{\text{B}},$$

$${\text{JP}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{R}}_{\text{n}-1}\text{J}={\text{D}}_{\text{B}}^{\text{T}},$$

$${\text{JR}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{R}}_{\text{n}-1}\text{J}=0,$$

where,

$${\text{D}}_{\text{B}}= \left(\begin{array}{cc}\begin{array}{ccc}0& 0& \dots \\ 0& 0& \dots \\ 0& 0& \dots \end{array}& \begin{array}{ccc}0& 0& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\\ \begin{array}{ccc}\dots & \dots & \dots \\ 0& 0& \dots \\ 0& 0& \dots \end{array}& \begin{array}{ccc}\dots & \dots & \dots \\ 0& 0& 0\\ 0& 0& 0\end{array}\end{array}\right).$$

By above calculations, ${\text{A}}_{3}^{\upphi }$ is given by,

$${\text{A}}_{3}^{\upphi }= \frac{1}{2}\left({\text{P}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{P}}_{\text{n}-1}-{\text{D}}_{1}+{\left(-1\right)}^{\text{P}}{(\text{D}}_{2}+{\text{D}}_{2}^{\text{T}})+{\text{JP}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{P}}_{\text{n}-1}\text{J}-{\text{D}}_{1}\right)= \frac{1}{2}\left({\text{P}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{P}}_{\text{n}-1}+ {\text{JP}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{P}}_{\text{n}-1}\text{J}\right)-{\text{D}}_{1}+{\left(-1\right)}^{\text{P}}{\text{D}}_{2}$$

In summary of the above calculation results,

$${\text{A}}_{\text{periodic}}^{\upphi }={\text{A}}_{1}^{\upphi }+{\text{A}}_{2}^{\upphi }+{\text{A}}_{3}^{\upphi }= \frac{1}{2}\left(2{\text{I}}^{\otimes \text{n}-1}+2{\text{JI}}^{\otimes \text{n}-1}\text{J}-{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}-\text{J}{[\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\right]\text{J}- {\text{P}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{P}}_{\text{n}-1}- {\text{JP}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{P}}_{\text{n}-1}\text{J})+{\text{D}}_{1}-{\left(-1\right)}^{\text{P}}{\text{D}}_{2}= \frac{1}{2}\left(2{\text{I}}^{\otimes \text{n}-1}- {\text{I}}^{\otimes \text{n}-2}\otimes \text{X}- {\text{P}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{P}}_{\text{n}-1}\right)+ \frac{1}{2}\text{J}\left(2{\text{I}}^{\otimes \text{n}-1}- {\text{I}}^{\otimes \text{n}-2}\otimes \text{X}- {\text{P}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{P}}_{\text{n}-1}\right)\text{J}+{\text{D}}_{1}-{\left(-1\right)}^{\text{P}}{\text{D}}_{2}= \frac{1}{2}{\text{A}}_{\text{periodic}}^{\text{u}}\left(\text{n}-1\right)+\text{ J}(\frac{1}{2}{\text{A}}_{\text{periodic}}^{\text{u}}\left(\text{n}-1\right))\text{J}+{\text{D}}_{1}-{\left(-1\right)}^{\text{P}}{\text{D}}_{2},$$

which is the Eq. (30) in the main text.

In a similar manner, we can derive ${\text{A}}^{\upphi }$ for the Hamiltonian with the Dirichlet boundary condition, ${\text{A}}_{\text{Dirichlet}}(\text{n})={\text{A}}_{\text{periodic}}+{\text{B}}_{\text{D}}(\text{n})$, by performing the aforementioned calculations again for ${\text{B}}_{\text{D}}(\text{n})$. ${\text{B}}_{\text{D}}\left(\text{n}\right)={\text{P}}_{\text{n}}^{-1}\left({\text{I}}_{0}^{\otimes \text{n}-1}\otimes \text{X}\right){\text{P}}_{\text{n}}$ can be partitioned into

$${\text{B}}_{\text{D}}\left(\text{n}\right)= \left(\begin{array}{cc}{\text{Q}}_{\text{n}-1}^{\text{T}}& {\text{R}}_{\text{n}-1}^{\text{T}}\\ {\text{R}}_{\text{n}-1}^{\text{T}}& {\text{Q}}_{\text{n}-1}^{\text{T}}\end{array}\right)\left(\begin{array}{cc}{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{X}& 0\\ 0& 0\end{array}\right)\left(\begin{array}{cc}{\text{Q}}_{\text{n}-1}& {\text{R}}_{\text{n}-1} \\ {\text{R}}_{\text{n}-1} & {\text{Q}}_{\text{n}-1}\end{array}\right)= \left(\begin{array}{cc}{\text{Q}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{Q}}_{\text{n}-1}& {\text{Q}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{R}}_{\text{n}-1}\\ {\text{R}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{Q}}_{\text{n}-1}& {\text{R}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{R}}_{\text{n}-1}\end{array}\right).$$

By using ${\text{Q}}_{\text{n}-1}={\text{P}}_{\text{n}-1}-{\text{R}}_{\text{n}-1}$, the (0,0) component of ${\text{B}}_{\text{D}}(\text{n})$ is given by,

$${\text{B}}_{\text{D}}\left(\text{n}\right)\left(0, 0\right)=\left({\text{P}}_{\text{n}-1}^{\text{T}}-{\text{R}}_{\text{n}-1}^{\text{T}}\right)\left[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{X}\right]\left({\text{P}}_{\text{n}-1}-{\text{R}}_{\text{n}-1}\right)= {\text{P}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{P}}_{\text{n}-1}-{\text{R}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{P}}_{\text{n}-1}-{\text{P}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{R}}_{\text{n}-1}+{\text{R}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{R}}_{\text{n}-1}.$$

Each term in the above equation is simplified by,

$${\text{R}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{P}}_{\text{n}-1}={\text{D}}_{\text{A}},$$

$${\text{P}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{R}}_{\text{n}-1}={\text{D}}_{\text{A}}^{\text{T}},$$

$${\text{R}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{R}}_{\text{n}-1}=0.$$

${\text{B}}_{\text{D}}\left(\text{n}\right)\left(0, 1\right)\text{J}$ is given by,

$${\text{B}}_{\text{D}}\left(\text{n}\right)\left(\text{0,1}\right)\text{J}= {\text{Q}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{R}}_{\text{n}-1}\text{J}=\left({\text{P}}_{\text{n}-1}^{\text{T}}-{\text{R}}_{\text{n}-1}^{\text{T}}\right)\left[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{R}}_{\text{n}-1}\text{J}$$

Each term in the above equation is simplified by,

$${\text{P}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{R}}_{\text{n}-1}\text{J}={\text{D}}_{3},$$

$${\text{R}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{R}}_{\text{n}-1}\text{J}=0.$$

${\text{JB}}_{\text{D}}\left(\text{n}\right)\left(1, 0\right)={\left({\text{B}}_{\text{D}}\left(\text{n}\right)\left(0, 1\right)\text{J}\right)}^{\text{T}}$. ${\text{JB}}_{\text{D}}\left(\text{n}\right)(\text{1,1})\text{J}$ is given by,

$${\text{JB}}_{\text{D}}\left(\text{n}\right)\left(\text{1,1}\right)\text{J}={\text{JR}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{X}\right]{\text{R}}_{\text{n}-1}\text{J}=0.$$

By above calculations, ${\text{B}}_{\text{D}}^{\upphi }(\text{n})$ is given by,

$$ {\text{B}}_{{\text{D}}}^{\upphi } \left( {\text{n}} \right) = \frac{1}{2}\left( {{\text{P}}_{{{\text{n}} - 1}}^{{\text{T}}} \left[ {{\text{I}}_{0}^{{ \otimes {\text{n}} - 2}} \otimes {\text{X}}} \right]{\text{P}}_{{{\text{n}} - 1}} - {\text{D}}_{{\text{A}}} - {\text{D}}_{{\text{A}}}^{{\text{T}}} + \left( { - 1} \right)^{{\text{P}}} {\text{D}}_{3} + \left( { - 1} \right)^{{\text{P}}} {\text{D}}_{3}^{{\text{T}}} } \right) = \frac{1}{2}\left( {{\text{P}}_{{{\text{n}} - 1}}^{{\text{T}}} \left[ {{\text{I}}_{0}^{{ \otimes {\text{n}} - 2}} \otimes {\text{X}}} \right]{\text{P}}_{{{\text{n}} - 1}} - {\text{D}}_{1} + \left( { - 1} \right)^{{\text{P}}} 2{\text{D}}_{3} } \right) = \frac{1}{2}\left( {{\text{P}}_{{{\text{n}} - 1}}^{{\text{T}}} \left[ {{\text{I}}_{0}^{{ \otimes {\text{n}} - 2}} \otimes {\text{X}}} \right]{\text{P}}_{{{\text{n}} - 1}} - {\text{D}}_{1} } \right) + \left( { - 1} \right)^{{\text{P}}} {\text{D}}_{3} = \frac{1}{2}{\text{B}}_{{\text{D}}} \left( {{\text{n}} - 1} \right) - \frac{1}{2}{\text{D}}_{1} + \left( { - 1} \right)^{{\text{P}}} {\text{D}}_{3} . $$

In summary of the above calculation results,

$${\text{A}}_{\text{Dirichlet}}^{\upphi }={\text{A}}_{\text{periodic}}^{\upphi }+{\text{B}}_{\text{D}}^{\upphi }\left(\text{n}\right)={\text{A}}_{\text{periodic}}^{\upphi }+ \frac{1}{2}{\text{B}}_{\text{D}}\left(\text{n}-1\right)-\frac{1}{2}{\text{D}}_{1}+{\left(-1\right)}^{\text{P}}{\text{D}}_{3},$$

which is Eq. (31) in the main text.

In a similar manner, we can derive ${\text{A}}^{\upphi }$ for the Hamiltonian with the Neumann boundary condition, ${\text{A}}_{\text{Neumann}}\left(\text{n}\right)={\text{A}}_{\text{periodic}}+{\text{B}}_{\text{D}}\left(\text{n}\right)-{\text{B}}_{\text{N}}(\text{n})$, by performing the aforementioned calculations again for ${\text{B}}_{\text{N}}\left(\text{n}\right)$. ${\text{B}}_{\text{N}}\left(\text{n}\right)={\text{P}}_{\text{n}}^{-1}\left({\text{I}}_{0}^{\otimes \text{n}-1}\otimes \text{I}\right){\text{P}}_{\text{n}}$ can be partitioned into

$${\text{B}}_{\text{N}}(\text{n})= \left(\begin{array}{cc}{\text{Q}}_{\text{n}-1}^{\text{T}}& {\text{R}}_{\text{n}-1}^{\text{T}}\\ {\text{R}}_{\text{n}-1}^{\text{T}}& {\text{Q}}_{\text{n}-1}^{\text{T}}\end{array}\right)\left(\begin{array}{cc}{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{I}& 0\\ 0& 0\end{array}\right)\left(\begin{array}{cc}{\text{Q}}_{\text{n}-1}& {\text{R}}_{\text{n}-1} \\ {\text{R}}_{\text{n}-1} & {\text{Q}}_{\text{n}-1}\end{array}\right)= \left(\begin{array}{cc}{\text{Q}}_{\text{n}-1}^{\text{T}}[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{I}]{\text{Q}}_{\text{n}-1}& {\text{Q}}_{\text{n}-1}^{\text{T}}[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{I}]{\text{R}}_{\text{n}-1}\\ {\text{R}}_{\text{n}-1}^{\text{T}}[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{I}]{\text{Q}}_{\text{n}-1}& {\text{R}}_{\text{n}-1}^{\text{T}}[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{I}]{\text{R}}_{\text{n}-1}\end{array}\right)$$

By using ${\text{Q}}_{\text{n}-1}={\text{P}}_{\text{n}-1}-{\text{R}}_{\text{n}-1}$, the (0,0) component of ${\text{B}}_{\text{N}}(\text{n})$ is given by,

$${\text{B}}_{\text{N}}\left(\text{n}\right)\left(0, 0\right)=\left({\text{P}}_{\text{n}-1}^{\text{T}}-{\text{R}}_{\text{n}-1}^{\text{T}}\right)\left[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{I}\right]\left({\text{P}}_{\text{n}-1}-{\text{R}}_{\text{n}-1}\right)= {\text{P}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{I}\right]{\text{P}}_{\text{n}-1}-{\text{R}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{I}\right]{\text{P}}_{\text{n}-1}-{\text{P}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{I}\right]{\text{R}}_{\text{n}-1}+{\text{R}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{I}\right]{\text{R}}_{\text{n}-1}.$$

Each term in the above equation is simplified by,

$${\text{R}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{I}\right]{\text{P}}_{\text{n}-1}={\text{D}}_{3}-{\text{D}}_{4},$$

$${\text{P}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{I}\right]{\text{R}}_{\text{n}-1}={\text{D}}_{3}^{\text{T}}-{\text{D}}_{4}^{\text{T}}={\text{D}}_{3}-{\text{D}}_{4},$$

$${\text{R}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{I}\right]{\text{R}}_{\text{n}-1}={\text{D}}_{3}-{\text{D}}_{4}.$$

${\text{B}}_{\text{N}}\left(\text{n}\right)\left(0, 1\right)\text{J}$ is given by,

$${\text{B}}_{\text{N}}\left(\text{n}\right)\left(\text{0,1}\right)\text{J}= {\text{Q}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{I}\right]{\text{R}}_{\text{n}-1}\text{J}=\left({\text{P}}_{\text{n}-1}^{\text{T}}-{\text{R}}_{\text{n}-1}^{\text{T}}\right)\left[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{I}\right]{\text{R}}_{\text{n}-1}\text{J}$$

Each term in the above equation is simplified by,

$${\text{P}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{I}\right]{\text{R}}_{\text{n}-1}\text{J}={\text{D}}_{\text{A}},$$

$${\text{R}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{I}\right]{\text{R}}_{\text{n}-1}\text{J}={\text{D}}_{\text{A}}.$$

${\text{JB}}_{\text{N}}\left(\text{n}\right)\left(1, 0\right)={\left({\text{B}}_{\text{N}}\left(\text{n}\right)\left(0, 1\right)\text{J}\right)}^{\text{T}}$. ${\text{JB}}_{\text{N}}\left(\text{n}\right)(\text{1,1})\text{J}$ is given by,

$${\text{JB}}_{\text{N}}\left(\text{n}\right)\left(\text{1,1}\right)\text{J}={\text{JR}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{I}\right]{\text{R}}_{\text{n}-1}\text{J}={\text{D}}_{3}.$$

By above calculations, ${\text{B}}_{\text{N}}^{\upphi }(\text{n})$ is given by,

$${\text{B}}_{\text{N}}^{\upphi }\left(\text{n}\right)=\frac{1}{2}\left({\text{P}}_{\text{n}-1}^{\text{T}}\left[{\text{I}}_{0}^{\otimes \text{n}-2}\otimes \text{I}\right]{\text{P}}_{\text{n}-1}+{\text{D}}_{4}+{\left(-1\right)}^{\text{P}}2{\text{D}}_{1}\right)= \frac{1}{2}{\text{B}}_{\text{N}}^{\upphi }\left(\text{n}-1\right)+\frac{1}{2}{\text{D}}_{4}$$

In summary of the above calculation results,

$${\text{A}}_{\text{Neumann}}^{\upphi }={\text{A}}_{\text{periodic}}^{\upphi }+{\text{B}}_{\text{D}}^{\upphi }\left(\text{n}\right)-{\text{B}}_{\text{N}}^{\upphi }\left(\text{n}\right)={\text{A}}_{\text{periodic}}^{\upphi }+ \frac{1}{2}{\text{B}}_{\text{D}}\left(\text{n}-1\right)-\frac{1}{2}{\text{D}}_{1}+{\left(-1\right)}^{\text{P}}{\text{D}}_{3}-\frac{1}{2}{\text{B}}_{\text{N}}\left(\text{n}-1\right)-\frac{1}{2}{\text{D}}_{4},$$

which is Eq. (32) in the main text. □

Appendix D: Expectation value calculation using different machine

Here, we present the results of the expectation value calculation for the ground state and the excited states of the Helmholtz equation using various hardware from IBM Quantum systems. Figures 

Fig. 15

Energy expectation values for the ground state and excited states of the Helmholtz equation using various hardware are shown for 8-node FDM. The height of the bar represents the mean, and the error bar indicates the standard deviation of repeated calculations. The dashed horizontal lines show the exact value obtained by classical algorithm

15,

Fig. 16

Energy expectation values for the ground state and excited states of the Helmholtz equation using various hardware are shown for 16-node FDM. The height of the bar represents the mean, and the error bar indicates the standard deviation of repeated calculations. The dashed horizontal lines show the exact value obtained by classical algorithm

16 and

Fig. 17

Energy expectation values for the ground state and excited states of the Helmholtz equation using various hardware are shown for 32-node FDM. The height of the bar represents the mean, and the error bar indicates the standard deviation of repeated calculations. The dashed horizontal lines show the exact value obtained by classical algorithm

17 are result of the 8, 16, and 32-node FDM, respectively, and correspond to Fig. 4 in the main text.

Appendix E: Qubit map and calibration data of ibmq_mumbai

Figure 18 shows the qubit map of the ibmq_mumbai. The calibration data of ibmq_mumbai is tabulated in Table 

Fig. 18

Qubit map of ibmq_mumbai

Table 2 Calibration data of ibmq_mumbai

2.