# 对于Grover算法为何有效，是否有外行的解释？

27

Answers:

20

$$U|a⟩=−|a⟩,U|b⟩=|b⟩U|a⟩=−|a⟩,U|b⟩=|b⟩U | a \rangle = - | a \rangle, \,\,\,\,\,\,\,\,\,\,\,\,\, U | b \rangle = | b \rangle$$

$$D:∑jαj|bj⟩↦∑j(2μ−αj)|bj⟩D:∑jαj|bj⟩↦∑j(2μ−αj)|bj⟩D: \,\,\,\, \sum_j \alpha_j \, | b_j \rangle \,\,\,\,\,\, \mapsto \,\,\,\,\,\, \sum_j (2\mu \, - \, \alpha_j) \, | b_j \rangle$$

4

• 我们可以生产的状态下与该'标记'状态非零重叠：。$|\psi ⟩$$|\psi\rangle$$|x⟩$$|x\rangle$$⟨x|\psi ⟩\ne 0$$\langle x|\psi\rangle\neq 0$
• 我们可以实现操作${U}_{1}=-\left(\mathbb{I}-2|\psi ⟩⟨\psi |\right)$$U_1=-(\mathbb{I}-2|\psi\rangle\langle\psi|)$
• 我们可以实现操作。${U}_{2}=\mathbb{I}-2|x⟩⟨x|$$U_2=\mathbb{I}-2|x\rangle\langle x|$

${U}_{2}$$U_2$${U}_{1}$$U_1$$2\theta$$2\theta$$|x⟩$$|x\rangle$

$|\psi ⟩$$|\psi\rangle$$\theta$$\theta$$|x⟩$$|x\rangle$$x$$x$

$|x⟩$$|x\rangle$$|\psi ⟩$$|\psi\rangle$$p\ll 1$$p\ll 1$$O\left(1/p\right)$$O(1/p)$$\sqrt{p}=\mathrm{sin}\theta \approx \theta$$\sqrt{p}=\sin\theta\approx\theta$$\theta$$\theta$$r$$r$$\mathrm{sin}\left(\left(2r+1\right)\theta \right)\approx 1$$\sin((2r+1)\theta)\approx 1$$r\approx \frac{\pi }{2\theta }\approx \frac{\pi }{2\sqrt{p}}$$r\approx \frac{\pi}{2\theta}\approx \frac{\pi}{2\sqrt{p}}$

3

$1/\sqrt{N}$$1/\sqrt{N}$$\sqrt{N}$$\sqrt{N}$$1$$1$

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