A Partially Random Trotter Algorithm for Quantum Hamiltonian Simulations

Abstract

Given the Hamiltonian, the evaluation of unitary operators has been at the heart of many quantum algorithms. Motivated by existing deterministic and random methods, we present a hybrid approach, where Hamiltonians with large amplitude are evaluated at each time step, while the remaining terms are evaluated at random. The bound for the mean square error is obtained, together with a concentration bound. The mean square error consists of a variance term and a bias term, arising, respectively, from the random sampling of the Hamiltonian terms and the operator-splitting error. Leveraging on the bias/variance trade-off, the error can be minimized by balancing the two. The concentration bound provides an estimate of the number of gates. The estimates are verified using numerical experiments on classical computers.

Appendices

Appendix A The Proof of Proposition 1

Proof

To estimate this one-step error, we separate it into two terms,

$$\left\{ \begin{aligned} \begin{aligned} {|{ \chi _1(\Delta t) }\rangle }=&\exp \big ( -{\text{i}} \Delta tH_1 - {\text{i}} \Delta t\delta H \big ) {|{\psi (0)}\rangle } - \exp \big ( -{\text{i}} \Delta tH_1 \big ) {|{\psi (0)}\rangle }, \\ {|{ \chi _2(\Delta t) }\rangle }=&\exp \big ( -{\text{i}} \Delta tH_0 \big ) \exp \big ( -{\text{i}} \Delta tH_1 \big ) {|{\psi (0)}\rangle } - \exp \big ( -{\text{i}} \Delta tH \big ) {|{\psi (0)}\rangle }. \end{aligned} \end{aligned} \right.$$

(A1)

As a result, we have the total one-step error

$$\begin{aligned} {|{\chi (\Delta t)}\rangle } = \exp \big ( -{\text{i}} \Delta tH_0 \big ) {|{\chi _1(\Delta t)}\rangle } + {|{\chi _2(\Delta t)}\rangle }. \end{aligned}$$

(A2)

Using the triangle inequality and the estimates from Theorems 1 and 3, we arrive at the stated estimate.

Appendix B The Proof of Theorem 3

Proof

We first extend the function $|\chi \rangle$ to $t\in [0,\Delta t]$,

$$\begin{aligned} {|{\chi (t)}\rangle } = \exp \big ( -{\text{i}} t H - {\text{i}} t \delta H \big ), {|{\psi (0)}\rangle } - \exp \big ( -{\text{i}} t H \big ) {|{ \psi (0)}\rangle }. \end{aligned}$$

It can be directly verified that ${|{\chi (t)}\rangle }$ satisfies the differential equation

$$\begin{aligned} \frac{\text{d}}{{\text{d}}t} {|{\chi (t)}\rangle } = -{\text{i}} (H+\delta H) {|{\chi (t)}\rangle } +{\text{i}} \delta H {|{\psi (t)}\rangle }. \end{aligned}$$

(B1)

Since $\delta H$ is Hermitian, we have

$$\begin{aligned} \frac{\text{d}}{{\text{d}}t} {\langle { \chi (t)}\rangle } = {\text{i}} {\langle { \chi (t)}|} \delta H {|{\psi (t)}\rangle } - {\text{i}} {\langle { \psi (t)}|} \delta H {|{\chi (t)}\rangle }=2 \text {Im} {\langle {\psi (t)}|} \delta H {|{\chi (t)}\rangle }. \end{aligned}$$

(B2)

Therefore, using the Cauchy-Schwarz inequality, one has

$$\begin{aligned} \frac{\text{d}}{{\text{d}}t} {\langle { \chi (t)}\rangle } \leqslant 2 {\langle { \chi (t)}\rangle }^{1/2} {\langle {\psi (t)}|} \delta H^2 {|{\psi (t)}\rangle }^{1/2}. \end{aligned}$$

By direct integration, or using the Ou-Iang inequality [42], this inequality leads to the desired bound (33).

Appendix C Proof of Theorem 5

Proof

Following standard procedure for analyzing the accumulation of the error in time [17], we can derive an error equation, given by

$$\begin{aligned} |e_{n+1}\rangle = \exp \big ( -{\text{i}} \Delta tH_0 \big ) \exp \big ( -{\text{i}} \Delta tH_1 - {\text{i}} \Delta t\delta H_n \big ) |e_n\rangle + | \chi _n \rangle , \end{aligned}$$

(C1)

where

$$\begin{aligned} {|{\chi _n}\rangle } = \exp \big ( -{\text{i}} \Delta tH \big ) |\psi _{n} \rangle - \exp \big ( -{\text{i}} \Delta tH_0 \big ) \exp \big ( -{\text{i}} \Delta tH_1 - {\text{i}} \Delta t \delta H_n \big ) |\psi _n\rangle \end{aligned}$$

(C2)

can be interpreted as a one-step truncation error.

Estimating the $L^2$-error is quite straightforward, since,

$$\begin{aligned} \Vert e_{n+1} \Vert \leqslant \Vert e_{n} \Vert + \Vert \chi _{n} \Vert \Rightarrow \Vert e_n \Vert \leqslant \sum _{j=0}^{n-1} \Vert \chi _{n} \Vert .\end{aligned}$$

On the other hand, to connect to the MSE, one can start with (C1). Using the Cauchy-Schwarz inequality, one finds that

$$\begin{aligned} \langle e_{n+1} |e_{n+1}\rangle \leqslant \langle e_{n} |e_{n}\rangle + 2 \langle e_{n} |e_{n}\rangle ^{\frac{1}{2}} \langle \chi _n | \chi _n \rangle ^{\frac{1}{2}} + \langle \chi _n | \chi _n \rangle . \end{aligned}$$

(C3)

Using the inequality, $2 \langle e_{n} |e_{n}\rangle ^{\frac{1}{2}} \leqslant 1 + \langle e_{n} |e_{n}\rangle ,$ we arrive at

$$\begin{aligned} \langle e_{n+1} |e_{n+1}\rangle \leqslant \langle e_{n} |e_{n}\rangle (1 + \langle \chi _n | \chi _n \rangle ^{\frac{1}{2}}) + \langle \chi _n | \chi _n \rangle . \end{aligned}$$

(C4)

A bound can be found using the following discrete Gronwall’s inequality [42][Theorem 2.1.3]:

$$\begin{aligned} u_{n+1} \leqslant (1+ g_n) u_n + f_n, \forall n \geqslant 0 \Rightarrow u_n \leqslant \sum _{m=0}^{n-1} f_m \prod _{j=m+1}^{n-1} (1+g_j). \end{aligned}$$

(C5)

As a result, we have

$$\begin{aligned} \langle e_{n+1} |e_{n+1}\rangle \leqslant \displaystyle \sum _{k=0}^n \langle \chi _k | \chi _k \rangle \exp \bigg ( \sum _{k=0}^n \langle \chi _k | \chi _k \rangle ^{\frac{1}{2}} \bigg ). \end{aligned}$$

(C6)

This can be combined with the one-step error bounds (51) (applied to the exponential) and (52).

Appendix D The Proof of Proposition 2

Proof

Recall that ${|{\psi (t)}\rangle }:= \exp \big ( -{\text{i}} t H \big ) {|{\psi (0)}\rangle }$ is the exact solution. We will split the error as follows:

$$\begin{aligned} \begin{aligned} \left\| {|{\psi _1}\rangle } - {\mathbb {E}}[{|{\phi _1}\rangle } ] \right\|& \leqslant \left\| {|{\psi (\Delta t)}\rangle } - \exp \big ( -{\text{i}} \Delta tH_0\big ) \exp \big ( - {\text{i}} \Delta tH_1 \big ) {|{\psi _0}\rangle } \right\| \\& \quad + \left\| \exp \big ( -{\text{i}} \Delta tH_0\big ) \exp \big ( - {\text{i}} \Delta tH_1 \big ) {|{\psi _0}\rangle } - {\mathbb {E}}[{|{\phi (\Delta t)}\rangle } ] \right\| . \end{aligned} \end{aligned}$$

(D1)

The first term can be estimated by (45) from Theorem 4. For the second term, it is enough to consider the difference, denoted by ${|{\theta (t)}\rangle },$ of the following wave functions:

$$\begin{aligned} {|{\theta (t)}\rangle } = \exp \big ( -{\text{i}} t H_1 \big ) |\psi _0\rangle - {\mathbb {E}}\left[ \exp \big ( -{\text{i}} t H_1 - {\text{i}} t \delta H \big )\right] |\psi _0\rangle . \end{aligned}$$

(D2)

Let ${|{\phi (t)}\rangle }= \big ( -{\text{i}} t H_1 - {\text{i}} t \delta H \big ) |\psi _0\rangle$, and notice that

$$\begin{aligned} \displaystyle \frac{\text{d}}{{\text{d}}t} {\mathbb {E}}[{|{\phi (\Delta t)}\rangle } ] = {\mathbb {E}}[ -{\text{i}} (H_1+\delta H) {|{\phi (\Delta t)}\rangle } ]. \end{aligned}$$

Therefore, the error term term ${|{\theta (t)}\rangle }$ satisfies the equation

$$\begin{aligned} \frac{\text{d}}{{\text{d}}t} {|{ \theta (t)}\rangle } = - {\text{i}} H_1 {|{\theta (t)}\rangle } - {\text{i}} {\mathbb {E}} [ \delta H {|{\phi (t)}\rangle }]. \end{aligned}$$

(D3)

Using the variation-of-constant formula, and using the fact that ${\mathbb {E}} [ \delta H]=0,$ we can write

$$\begin{aligned} \begin{aligned} {|{\theta (\Delta t)}\rangle } =&-{\text{i}} \int _0^{\Delta t} U_1(\Delta t-t) {\mathbb {E}} \left[ \delta H {|{\phi (t)}\rangle } \right] {\text{d}}t\\ =&-{\text{i}} \int _0^{\Delta t} U_1(\Delta t-t) {\mathbb {E}} \left[ \delta H ( {|{\phi (t)}\rangle } - {|{\phi (0)}\rangle } ) \right] {\text{d}}t\\ =&- \int _0^{\Delta t} U_1(\Delta t-t) {\mathbb {E}} \left[ \delta H (H_1 + \delta H) \int _0^t {|{\phi (\tau )}\rangle } {\text{d}}\tau \right] {\text{d}}t. \end{aligned} \end{aligned}$$

Here $U_1(t)=\exp ( -{\text{i}} t H_1).$ In light of the fact that $\Vert {|{\phi }\rangle }\Vert =1,$ one can take the norms and arrive at

$$\begin{aligned} \Vert {|{\chi (\Delta t)}\rangle } \Vert \leqslant \int _0^{\Delta t} \int _0^t {\mathbb {E}} \left[ \Vert \delta H (H_1 + \delta H)\Vert \right] ^{\frac{1}{2}} {\text{d}}\tau {\text{d}}t,\end{aligned}$$

which yields the second term in the estimate (59).

This estimate can be extended to time $t=n\Delta t$ [9] using the fact that for a unitary matrix V, $\Vert {\mathbb {E}}[V] \Vert \leqslant 1.$ More specifically, we can split the error as follows:

$$\begin{aligned} \begin{aligned} \left\| {|{\psi _n}\rangle } - {\mathbb {E}}[ {|{\phi _n}\rangle }] \right\|&\leqslant \left\| U(\Delta t) {|{\psi _{n-1}}\rangle } - {\mathbb {E}} [V_{n-1}] {|{\psi _{n-1}}\rangle } \right\| + \left\| {\mathbb {E}} [V_{n-1}] {|{\psi _{n-1}}\rangle } - {\mathbb {E}} [V_{n-1}] {\mathbb {E}}[ {|{\phi _{n-1}}\rangle } ] \right\| \\&\leqslant \left\| U(\Delta t) {|{\psi _{n-1}}\rangle } - {\mathbb {E}} [V_{n-1}] {|{\psi _{n-1}}\rangle } \right\| + \left\| {|{\psi _{n-1}}\rangle } - {\mathbb {E}}[ {|{\phi _{n-1}}\rangle } ] \right\| \\&\leqslant \cdots \\ \Longrightarrow \left\| {|{\psi _n}\rangle } - {\mathbb {E}}[ {|{\phi _n}\rangle }] \right\|&\leqslant \frac{t \Delta t}{2} \Vert [H_0,H_1]\Vert + \frac{t \Delta t}{2} {\mathbb {E}}\left[ \Vert (H_1+\delta H) \delta H^2 (H_1+\delta H) \Vert \right] ^{\frac{1}{2}}. \end{aligned} \end{aligned}$$