许多论文断言汉密尔顿模拟是BQP完全的（例如，汉密尔顿模拟几乎所有参数都具有最佳依赖关系，而汉密顿模拟则是通过量化进行的）。

不难发现，汉密尔顿模拟是BQP难的，因为任何量子算法都可以简化为汉密尔顿模拟，但是BQP中的汉密尔顿模拟又如何呢？

即，BQP中的汉密尔顿模拟决策问题到底是什么？在汉密尔顿方程的什么条件下？

— 组组组
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有很多不同的变体，特别是关于哈密顿量的条件。例如，尝试找到最简单的哈密顿主义者类别（这仍然是BQP完成的）是一件小游戏。

该语句大致如下：let 为（标准化）产品状态，是从一些特定类别的哈密顿（例如，仅最近邻耦合的上一维点阵组成可观察到的包括一个体运营商，使得的张量积，和是时间。鉴于承诺， $|\psi\rangle$ $H$ $\hat O$ $\|\hat O\|\leq 1$ $t$ $\langle\psi|e^{iHt}\hat O e^{-iHt}|\psi\rangle$ 大于或小于 $\frac12+a$ 代表一些（例如 $\frac12-a$ $a$ ），确定是哪种情况。 $a=\frac16$

更多细节

哈密顿模拟是BQP难点

基本构造（最初由于费因曼，在这里进行了一些调整）基本上显示了如何设计实现任何量子计算（包括任何BQP完整计算）的哈密顿量。您要测量的可观测值只是特定输出量子位上的，两个测量结果分别对应“是”和“否”。 $Z$

你可能会认为汉密尔顿的最简单的排序是要考虑的一个计算连续unitaries 作用于量子比特，从状态开始。然后，您可以引入一个额外的量子位，并指定哈密顿量 $N-1$ $U_n$ $M$ $|0\rangle^{\otimes M}$ $N$

H = \frac{2}{N} \sum_{n = 1}^{N - 1} \sqrt{n (N - n)} (| 10 ⟩ ⟨ 01 |_{n, n + 1} \otimes U + | 01 ⟩ ⟨ 10 |_{n, n + 1} \otimes U^{†}) .

$H=\frac{2}{N}\sum_{n=1}^{N-1}\sqrt{n(N-n)}\left(|10\rangle\langle 01|_{n,n+1}\otimes U+|01\rangle\langle 10|_{n,n+1}\otimes U^\dagger\right).$ If you prepare your initial state as

| 1 ⟩ | 0 ⟩^{\otimes (N - 1)} | 0 ⟩^{\otimes M}

$|1\rangle|0\rangle^{\otimes(N-1)}|0\rangle^{\otimes M}$ then after a time

N π / 4

$N\pi/4$ , it will be in a state

| 0 ⟩^{\otimes (N - 1)} | 1 ⟩ | Φ ⟩

$|0\rangle^{\otimes (N-1)}|1\rangle|\Phi\rangle$ where

| Φ ⟩

$|\Phi\rangle$ is the output of the desired computation. The funny coupling strengths that I've used here, the

\sqrt{n (N - n)}

$\sqrt{n(N-n)}$ are chosen specifically to give deterministic evolution, and are related to the concept of perfect state transfer. Usually you'll see results stated with equal couplings, but probabilistic evolution.

To see how this works, you define a set of states

| ψ_{n} ⟩ = | 0 ⟩^{\otimes (n - 1)} | 1 ⟩ | 0 ⟩^{\otimes N - n} \otimes (U_{n - 1} U_{n - 2} \dots U_{1} | 0 ⟩^{\otimes M}) .

$|\psi_n\rangle=|0\rangle^{\otimes(n-1)}|1\rangle|0\rangle^{\otimes{N-n}}\otimes\left(U_{n-1}U_{n-2}\ldots U_1|0\rangle^{\otimes M}\right).$ The action of the Hamiltonian is then

H | ψ_{n} ⟩ = \frac{2}{N} \sqrt{(n - 1) (N + 1 - n)} | ψ_{n - 1} ⟩ + \frac{2}{N} \sqrt{n (N - n)} | ψ_{n + 1} ⟩,

$H|\psi_n\rangle=\frac2N\sqrt{(n-1)(N+1-n)}|\psi_{n-1}\rangle+\frac2N\sqrt{n(N-n)}|\psi_{n+1}\rangle,$ which proves that the evolution is restricted to an

N \times N

$N\times N$ subspace which is represented by a tridiagonal matrix (which is the specific thing studied in perfect state transfer).

Of course, this Hamiltonian doesn't have any particularly nice properties - it is highly non-local, for example. There are many tricks that can be played to simplify the Hamiltonian to being, for example, one-dimensional. It can even be translationally invariant if you want, at the cost of having to prepare a more complex initial product state (at that point, the computation is no longer encoded in the Hamiltonian, which is universal, but is encoded in the input state). See here, for example.

Hamiltonian Simulation

The evolution of any Hamiltonian which is local on some lattice, acting on an initial product state, for a time that is no more than polynomial in the system size, can be simulated by a quantum computer, and any efficiently implementable measurement can be applied to estimate an observable. In this sense, you can see that Hamiltonian simulation is no harder than a quantum computation, the counter-point to the previous statement that quantum computation is no harder than Hamiltonian simulation.

There are many ways to do this (and there have been some recent papers that show significant improvements in error scaling for certain classes of Hamiltonian). Hre's quite a simple one. Take the Hamiltonian $H$ that you want to simulate. Split it up into different parts, $H_i$ , each of which commutes. For example, on a nearest-neighbour Hamiltonian on some graph, you don't need more pieces than the maximum degree of the graph. You then Trotterize the evolution, writing the approximation

e^{i H t} \approx {(e^{- i H_{1} δ t} e^{- i H_{2} δ t} \dots e^{- i H_{n} δ t})}^{t / δ t}

$e^{iHt}\approx \left(e^{-iH_1\delta t}e^{-iH_2\delta t}\ldots e^{-iH_n\delta t}\right)^{t/\delta t}$ So, you just have to construct a circuit that implements terms like

e^{- i H_{1} δ t}

$e^{-iH_1\delta t}$ , which is composed of commuting terms

H_{1} = \sum_{n} h_{n}

$H_1=\sum_nh_n$ , each of which acts only on a small number of qubits.

e^{- i H_{1} δ t} = \prod_{n} e^{- i h_{n} δ t}

$e^{-iH_1\delta t}=\prod_{n}e^{-ih_n\delta t}$ Since this is just a unitary on a small number of terms, a universal quantum computer can implement it.

— DaftWullie
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哈密​​顿模拟是BQP完全的

更多细节

哈密​​顿模拟是BQP难点

Hamiltonian Simulation

哈密顿模拟是BQP完全的

哈密顿模拟是BQP难点