傅里叶采样实际上如何工作（并解决奇偶校验问题）？

我写的是Umesh Vazirani教授的傅立叶采样视频讲座的第一部分和第二部分。

我从一部分开始：

在Hadamard变换中：

| 0...0 ⟩ \to \sum_{{0, 1}^{n}} \frac{1}{2^{n / 2}} | x ⟩

$|0...0\rangle \to \sum_{\{0,1\}^n}\frac{1}{2^{n/2}}|x\rangle$

| u ⟩ = | u_{1} . . . u_{n} ⟩ \to \sum_{{0, 1}^{n}} \frac{(- 1)^{u . x}}{2^{n / 2}} | x ⟩ (where u . x = u_{1} x_{1} + u_{2} x_{2} + . . . + u_{n} x_{n})

$|u\rangle =|u_1...u_n\rangle \to \sum_{\{0,1\}^n}\frac{(-1)^{u.x}}{2^{n/2}}|x\rangle \quad \text{(where $u.x=u_1x_1+u_2x_2+...+u_nx_n$)}$

在傅立叶采样中：

| ψ ⟩ = \sum_{{0, 1}}^{n} α_{x} | x ⟩ \to \sum_{x} \hat{α_{x}} | x ⟩ = | \hat{ψ} ⟩

$|\psi\rangle=\sum_{\{0,1\}}^{n}\alpha_x|x\rangle \to \sum_{x}\hat{\alpha_x}|x\rangle=|\hat{\psi}\rangle$

$|\hat{\psi}\rangle$ $x$ $|\hat{\alpha_x}|^2$

第二部分：

奇偶问题：

$f:\{0,1\}^n\to\{0,1\}$ $f(x)=u.x$ $u_1x_1+u_2x_2+...+u_nx_n (\text{mod 2})$ $u\in\{0,1\}^{n}$ $u$ $f$

$u$

$\frac{1}{2^{n/2}}\sum_{x}(-1)^{f(x)}|x\rangle$
$u$

$u$

他们进一步建立了这样的量子门：

$|0\rangle$ $|-\rangle$ $- \oplus f(0...0)$ $\oplus$

quantum-gate fourier-sampling

— Sanchayan Dutta
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$\left| 0\right\rangle^{\otimes n}\left| -\right\rangle$ $H^{\otimes n}\otimes I$

(\sum_{x = {0, 1}^{n}} \frac{1}{2^{n / 2}} | x ⟩) | - ⟩ = \frac{1}{2^{n / 2}} {(| 0 ⟩ + | 1 ⟩)}^{\otimes n} | - ⟩ .

$\left(\sum_{x=\{0,1\}^n}\frac{1}{2^{n/2}}|x\rangle\right)\left|-\right\rangle = \frac{1}{2^{n/2}}\left(\left|0\right\rangle + \left|1\right\rangle\right)^{\otimes n}\left|-\right\rangle.$

U_{f}

$U_f$

U_{f} (\sum_{x = {0, 1}^{n}} \frac{1}{2^{n / 2}} | x ⟩) | - ⟩ = \sum_{x = {0, 1}^{n}} \frac{1}{2^{n / 2}} | x ⟩ | - \oplus f (x) ⟩ .

$U_f\left(\sum_{x=\{0,1\}^n}\frac{1}{2^{n/2}}|x\rangle\right)\left|-\right\rangle = \sum_{x=\{0,1\}^n}\frac{1}{2^{n/2}}|x\rangle\left|-\oplus f\left(x\right)\right\rangle.$

$\oplus$

(\sum_{x = {0, 1}^{n}} \frac{1}{2^{n / 2}} {(- 1)}^{f (x)} | x ⟩) | - ⟩ .

$\left(\sum_{x=\{0,1\}^n}\frac{1}{2^{n/2}}\left(-1\right)^{f\left(x\right)}\left|x\right\rangle\right)\left|-\right\rangle.$

U_{f} | x ⟩ (| 0 ⟩ - | 1 ⟩) = | x ⟩ | f (x) ⟩ - | 1 \oplus f (x) ⟩ = {(- 1)}^{f (x)} | x ⟩ (| 0 ⟩ - | 1 ⟩)

$U_f\left|x\right\rangle\left(\left|0\right\rangle - \left|1\right\rangle\right) = \left|x\right\rangle\left|f\left(x\right)\right\rangle - \left|1\oplus f\left(x\right)\right\rangle = \left(-1\right)^{f\left(x\right)}\left|x\right\rangle\left(\left|0\right\rangle - \left|1\right\rangle\right)$

$x$ $x = \prod_ix_i$

H | x_{i} ⟩ = \frac{1}{\sqrt{2}} (| 0 ⟩ + {(- 1)}^{x_{i}} | 1 ⟩) = \frac{1}{\sqrt{2}} \sum_{y = {0, 1}} {(- 1)}^{x_{i} . y} | y ⟩ .

$H\left|x_i\right\rangle = \frac{1}{\sqrt{2}}\left(\left|0\right\rangle + \left(-1\right)^{x_i}\left|1\right\rangle\right) = \frac{1}{\sqrt{2}}\sum_{y=\left\lbrace0, 1\right\rbrace}\left(-1\right)^{x_i.y}\left|y\right\rangle.$

H^{\otimes n} | x ⟩ = \frac{1}{2^{n / 2}} \sum_{y \in {0, 1}^{n}} {(- 1)}^{x . y} | y ⟩ .

$H^{\otimes n}\left| x\right\rangle = \frac{1}{2^{n/2}}\sum_{y\in\left\lbrace0, 1\right\rbrace^n}\left(-1\right)^{x.y}\left|y\right\rangle.$

\frac{1}{2^{n}} (\sum_{x, y = {0, 1}^{n}} {(- 1)}^{f (x) \oplus x . y} | y ⟩) | - ⟩ .

$\frac{1}{2^n}\left(\sum_{x, y=\{0,1\}^n}\left(-1\right)^{f\left(x\right) \oplus x.y}\left|y\right\rangle\right)\left|-\right\rangle.$

$f\left(x\right) = u.x = x.u$ $\left(-1\right)^{f\left(x\right) \oplus x.y} = \left(-1\right)^{x.\left(u\oplus y\right)}$ $x$ $\sum_x\left(-1\right)^{x.\left(u\oplus y\right)} = 0,\, \forall\, u\oplus y \neq 0$ $u\oplus y = 0$ $u=y$ $\left|u\right\rangle\left|-\right\rangle$ $u$

$\left|+\right\rangle^{\otimes n}$ $\left|u\right\rangle$

关键是，使用叠加，我们可以同时对所有量子位执行此操作，而不必像传统情况那样单独检查每个量子位。

— Mithrandir24601
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