Quantum fast Poisson solver: the algorithm and complete and modular circuit design

Abstract

The Poisson equation has applications across many areas of physics and engineering, such as the dynamic process simulation of ocean current. Here we present a quantum algorithm for solving Poisson equation, as well as a complete and modular circuit design. The algorithm takes the HHL algorithm as the framework (where HHL is for solving linear equations). A more efficient way of implementing the controlled rotation, one of the crucial steps in HHL, is developed based on the arc cotangent function. The key point is that the inverse trigonometric function can be evaluated in a very simple recursive way by a binary expansion method. Quantum algorithms for solving square root and reciprocal functions are proposed based on the classical non-restoring method. These advances not only reduce the algorithm’s complexity, but more importantly make the circuit more complete and practical. We demonstrate our circuits on a quantum virtual computing system installed on the Sunway TaihuLight supercomputer. This is an important step toward practical applications of the present circuits as a fast Poisson solver in the near-term hybrid classical/quantum devices.

Appendices

Appendix 1

We take the maximum eigenvalue as the norm of the discretized matrix A. The norm of matrix A is $\lambda_{\hbox{max} } = 4N^{2} \sin^{2} \frac{(N - 1)\pi }{2N}$ as shown in Sect. 2 and that of matrix A⁻¹ is $\lambda^{\prime}_{\hbox{max} } = (4N^{2} \sin^{2} \frac{\pi }{2N})^{ - 1}$. So the condition number of matrix A, or the ratio between A’s largest and smallest eigenvalues [5], can be obtained as follows:

$$ \kappa = \left\| A \right\| \cdot \left\| {A^{ - 1} } \right\| = \lambda_{\hbox{max} } \cdot \lambda^{\prime}_{\hbox{max} } = \cot^{2} \frac{1}{2N}\pi . $$

(A.1)

The relationship between κ and N is apparently nonlinear. Assuming $x = {\pi \mathord{\left/ {\vphantom {\pi {2N}}} \right. \kern-0pt} {2N}} < 1$, the squared cotangent function satisfies the following equalities,

$$ \begin{aligned} \cot^{2} \frac{1}{2N}\pi & = \cot^{2} x = \left( {i\frac{{{\text{e}}^{ix} + {\text{e}}^{ - ix} }}{{{\text{e}}^{ix} - {\text{e}}^{ - ix} }}} \right)^{2} \\ & = \frac{1}{{x^{2} }} - 1 + o(x^{3} ) < \frac{{N^{2} }}{2} - 1 + o\left( {\left( {\frac{\pi }{2N}} \right)^{3} } \right), \\ \end{aligned} $$

(A.2)

where Taylor expansions of ${\text{e}}^{ix}$ and ${\text{e}}^{ - ix}$ are used. The condition number of matrix A is $\kappa = O(N^{2} )$. In addition, we have the relationship between N and the basic error of the solutions caused by the central-difference approximation, namely $N = \varepsilon^{ - \alpha }$. The α is a smoothness parameter depending on the smoothness of the solution function. (For example, when the solution function has uniformly bounded partial derivatives up to order four, α is 1/2 [26].) Therefore, the condition number of matrix A can be further expressed as $\kappa = O(N^{2} ) = O(\varepsilon^{ - 2\alpha } )$, which is independent of the dimension of matrix A. The additive preconditioner [47] would be used to reduce the κ.

Appendix 2

2.1 Non-restoring method for solving the square root function

The calculation procedure consists of the following five steps.

• 1st. Ignore the binary point of the binary number and expand the m-bits binary string to be 2m-bits by adding 0 on the right-hand side.

• 2nd. Divide the 2m-bits string into m parts in pairs from upper bit to lower.

• 3rd. Subtract 01 from the most left part. If the subtraction result is nonnegative, then the first bit of the solution is 1 and proceed to the next step. Otherwise, first bit of the solution is 0 and undo the subtraction operation.

• 4th. Expand the subtraction result of the last step by combining it with the next 2-bit part, and combine the solution obtained in the last step with 01. Then subtract the second number from the first number. If the subtraction result is nonnegative, the next bit of the solution is 1 and proceed to the next step. Otherwise, the next bit is 0 and undo the subtraction operation.

• 5th. Repeat step 4m − 1 times and the m-bits of the solution are obtained.

We take the square root of 01.00₂ as the example to illustrate the calculation procedure as shown in Fig. 15.

Fig. 15

Demo case of calculating square root of 01.00₂ using our revised non-restoring square root method

2.2 Non-restoring method for solving the reciprocal function

The calculation procedure consists of the following three steps.

• 1st. The highest bit of dividend 1 is the sign bit. Subtract the divisor from the dividend. If the sign bit of the result is 0, the first bit of reciprocal is 1, otherwise the first bit is 0. Then the result shifts left one bit except the sign bit with zero padding to the end to obtain the second dividend.

• 2nd. If the sign bit of the dividend of the last step is 0, subtract the divisor. If the sign bit is 1, add the divisor. If the sign bit of the result is 0, the next bit of reciprocal is 1, otherwise the next bit is 0. Then the result shifts left one bit except the sign bit with zero padding to the end to obtain the next dividend.

• 3rd. Repeat step 2m − 1 times and the m-bits of reciprocal of the divisor are obtained.

The reciprocal of 1000₂ is taken as the example as described in Fig. 16.

Fig. 16

Demo case of calculating reciprocal of 1000₂ using the non-restoring method

Appendix 3

3.1 Error accumulation in the iteration process of EVC module

As shown in Fig. 6, the EVC module is first initialized with q qubits to perform the intermediate iterative calculations and finally truncated to m qubits to store the approximated eigenvalues. There exists truncation error in each iteration which will be accumulated to the final result. So appropriate q should be selected to guarantee that all the m-bits of the output are exact.

According to the characteristics of the iterative formula as shown in Eq. (12), the maximum error accumulation occurs at the points of |cos(1/2 ± 1/2ⁿ)π|. We take one of them, namely ${j \mathord{\left/ {\vphantom {j N}} \right. \kern-0pt} N} = (0.0\mathop {\underline{11 \cdots 1} }\limits_{n - 1} )_{2}$, to evaluate the upper bound of the truncation error. Firstly, Eq. (12) can be expressed as

$$ y(x) = \left\{ {\begin{array}{*{20}c} {\sqrt {(1 + x)/2} ,\,\,if \, \,\,v_{i} = v_{i - 1} \;\;} \\ {\sqrt {(1 - x)/2} ,\,\,if\,\,\, \, v_{i} \ne v_{i - 1} } \\ \end{array} } \right.,x \in [0,1],y \in [0,1]. $$

(C.1.1)

The derivative of the first iteration function is decreasing and belongs to the interval of $[{1 \mathord{\left/ {\vphantom {1 4}} \right. \kern-0pt} 4},{{\sqrt 2 } \mathord{\left/ {\vphantom {{\sqrt 2 } 4}} \right. \kern-0pt} 4}]$, while the second is also decreasing and belongs to the interval of $[{{ - \sqrt 2 } \mathord{\left/ {\vphantom {{ - \sqrt 2 } 4}} \right. \kern-0pt} 4}, - \infty )$. The iteration process is as follows:

$$ \left\{ \begin{aligned} &a_{1} = y(a_{0} ) + 0 \hfill \\ &\hat{a}_{2} = y(a_{1} ) + \varepsilon_{2} \hfill \\ &\hat{a}_{3} = y(\hat{a}_{2} ) + \varepsilon_{3} \hfill \\ &\cdots \hfill \\ &\hat{a}_{n} = y(\hat{a}_{n - 1} ) + \varepsilon_{n} \hfill \\ \end{aligned} \right., $$

(C.1.2)

where ε_i represents the truncation error of each iteration step. The error of the first step is always zero. For ${j \mathord{\left/ {\vphantom {j N}} \right. \kern-0pt} N} = (0.011 \cdots 1)_{2}$, the next n − 2 steps satisfy v_i= v_i − 1, so the function of $\sqrt {(1 + x)/2}$ is calculated; the last step calculates $\sqrt {(1 - x)/2}$. Using the fact that the derivative of $\sqrt {(1 + x)/2}$ is no larger than ${{\sqrt 2 } \mathord{\left/ {\vphantom {{\sqrt 2 } 4}} \right. \kern-0pt} 4}$, the error accumulation can be expressed as

$$ \begin{aligned} |a_{n} - \hat{a}_{n} | & = |y(a_{n - 1} ) - y(\hat{a}_{n - 1} ) - \varepsilon_{n} | \\ & < \delta |a_{n - 1} - \hat{a}_{n - 1} | + |\varepsilon_{n} | \\ & < \frac{\sqrt 2 }{4}\delta |a_{n - 2} - \hat{a}_{n - 2} | + \delta |\varepsilon_{n - 1} | + |\varepsilon_{n} | \\ & \cdots \, \\ & < \delta \left( {\left( {\frac{\sqrt 2 }{4}} \right)^{n - 3} |\varepsilon_{2} | + \cdots + \frac{\sqrt 2 }{4}|\varepsilon_{n - 2} | + |\varepsilon_{n - 1} |} \right) + |\varepsilon_{n} | \\ & < [4\delta + 1]2^{ - q} , \\ \end{aligned} $$

(C.1.3)

where δ is the derivative of $\sqrt {(1 - x)/2}$ satisfying $\delta = |y^{\prime}_{n} | \in [{{\sqrt 2 } \mathord{\left/ {\vphantom {{\sqrt 2 } 4}} \right. \kern-0pt} 4},\infty )$.

Now we need to evaluate the upper bound of δ. First we have the relation of cosx < 1 − x²/4 when $x \in (0,1)$. Then the output of the (n − 1)th iteration satisfies

$$ \hat{a}_{n - 1} < a_{n - 1} = \cos \frac{\pi }{{2^{n - 1} }} < 1 - \frac{1}{4}\left( {\frac{\pi }{{2^{n - 1} }}} \right)^{2} < 1 - 2^{ - 2n + 3} . $$

(C.1.4)

So the absolute value of the derivative at the nth iteration could be calculated,

$$ |y^{\prime}_{n} | = \frac{1}{\sqrt 2 }\frac{1}{{\sqrt {1 - \hat{a}_{n - 1} } }} < \frac{1}{\sqrt 2 }\frac{1}{{\sqrt {2^{ - 2n + 3} } }} = \frac{1}{{\sqrt {2^{ - 2n + 4} } }} = 2^{n - 2} . $$

(C.1.5)

Substitute the result in Eq. (C.1.3) and we obtain

$$ |a_{n} - \hat{a}_{n} | < [4 \cdot 2^{n - 2} + 1]2^{ - q} = (2^{n} + 1)2^{ - q} . $$

(C.1.6)

This is the error for |cos(jπ/N)|. We finally truncate the result of 2N²(1 − cos(jπ/N)) into m qubits, and m = 2n + 2+f (note that the n here represents the same parameter as that in, say Eq. (C.1.2)), where the 2n + 2 bits hold the integer part and the f bits hold the fractional part. In order to guarantee that all the m-bits of the output are totally exact, the q should satisfy the following inequality,

$$ q \ge m + n. $$

(C.1.7)

Without loss of generality, we would assume that f = 2n + 2, so in such case q = 5m/4 is adequate to make the high m-bits output being exact in theory.

3.2 The error in controlled rotation

According to Eq. (13), the probability amplitude of the quantum state after controlled rotation should be

$$ \sin \theta_{j} = \frac{1}{{\sqrt {1 + \cot^{2} \theta_{j} } }} = \frac{1}{{\hat{\lambda }_{j} }},\theta_{j} \in (0,\pi /2). $$

(C.2.1)

Substituting Eq. (14) in Eq. (C.2.1), then it turns to

$$ \sin \theta^{\prime}_{j} = \frac{1}{{\sqrt {1 + \hat{\lambda }_{j}^{2} } }},\theta^{\prime}_{j} \in (0,\pi /2). $$

(C.2.2)

Based on Eq. (14), we know that ${1 \mathord{\left/ {\vphantom {1 {\tilde{\lambda }_{j} }}} \right. \kern-0pt} {\tilde{\lambda }_{j} }} = {1 \mathord{\left/ {\vphantom {1 {\sqrt {1 + \hat{\lambda }_{j}^{2} } }}} \right. \kern-0pt} {\sqrt {1 + \hat{\lambda }_{j}^{2} } }}$. So the error is

$$ \begin{aligned} \varepsilon_{2} (0) & = |\sin \theta^{\prime}_{j} - \sin \theta_{j} |= \left| {\frac{1}{{\sqrt {1 + \hat{\lambda }_{j}^{2} } }} - \frac{1}{{\hat{\lambda }_{j} }}} \right| \\ & \le \left| {\frac{1}{{\sqrt {1 + \hat{\lambda }_{\hbox{min} }^{2} } }} - \frac{1}{{\hat{\lambda }_{\hbox{min} } }}} \right|{ = }\left| {\frac{1}{{\sqrt {1 + 8^{2} } }} - \frac{1}{8}} \right| < 2^{ - 10} . \\ \end{aligned} $$

(C.2.3)

We can reduce error by amplifying the approximated eigenvalues, i.e., shift the binary numeral of $\hat{\lambda }_{j}$ left, say, i bits. Now the error turns to

$$ \begin{aligned} \varepsilon_{2} (i) & = \left| {\frac{1}{{\tilde{\lambda }_{j} }} - \frac{1}{{\hat{\lambda }_{j} }}} \right| = \left| {\frac{1}{{\sqrt {2^{ - 2i} + \hat{\lambda }_{j}^{2} } }} - \frac{1}{{\hat{\lambda }_{j} }}} \right| \le \left| {\frac{1}{{\sqrt {2^{ - 2i} + 8^{2} } }} - \frac{1}{8}} \right| \\ & = 2^{ - 3} \left( {2^{0} - \frac{1}{{\sqrt {2^{0} + 2^{ - 2i - 6} } }}} \right)\mathop < \limits_{NR} 2^{ - 3} \left( {2^{0} - \frac{1}{{2^{0} + 2^{ - 2i - 7} }}} \right) \\ & \mathop = \limits_{NR} 2^{ - 3} \left( {2^{0} - \sum\nolimits_{j = 1}^{i} {2^{ - 2j - 7} } } \right) = 2^{ - 2i - 10} , \\ \end{aligned} $$

(C.2.4)

where NR means the operation is calculated by non-restoring method. So the upper bound of error $\varepsilon_{2} (i)$ reduces exponentially with i. Shifting left one bit makes the error reduce about 2² times. And the state after controlled rotation turns to $\sqrt {1 - \left( {{{C^{\prime}} \mathord{\left/ {\vphantom {{C^{\prime}} {2^{i} \tilde{\lambda }_{j} }}} \right. \kern-0pt} {2^{i} \tilde{\lambda }_{j} }}} \right)^{2} } |0\rangle + {{C^{\prime}} \mathord{\left/ {\vphantom {{C^{\prime}} {2^{i} \tilde{\lambda }_{j} }}} \right. \kern-0pt} {2^{i} \tilde{\lambda }_{j} }} |1\rangle$, where $C^{\prime}$ represents the normalizing constant. Factor ${1 \mathord{\left/ {\vphantom {1 {2^{i} }}} \right. \kern-0pt} {2^{i} }}$ will be contained in the normalizing constant.