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I'm reading this paper and have a question about the computation to go from (3) to (4). Define the state $\vert\psi\rangle = \sum_b\alpha_b\vert b\rangle$.

In (3), we have the state

$$\sum_{b \in\{0,1\}} \alpha_b|b\rangle\left|x_{b, y}\right\rangle$$

where $b$ is a bit, $y$ is a bit string and $x_{b,y}$ is also a bit string determined by $b$ and $y$. If one applies the Hadamard to both registers, one obtains (according to (4))

$$\frac{1}{\sqrt{|\mathcal{X}|}} \sum_{d \in \mathcal{X}} X^{d \cdot\left(x_{0, y} \oplus x_{1, y}\right)} H|\psi\rangle \otimes Z^{x_{0, y}}|d\rangle$$

Can someone show exactly what happened here? My solution attempt is as follows. Using $H\vert x\rangle = \sum_y (-1)^{y.x}\vert y\rangle$, we get

$$H\otimes H\sum_{b \in\{0,1\}} \alpha_b|b\rangle\left|x_{b, y}\right\rangle = \sum_{d}\alpha_0\frac{1}{\sqrt{2}}(\vert 0 \rangle + \vert 1\rangle)(-1)^{x_{0,y}.d}\vert d \rangle + \alpha_1\frac{1}{\sqrt{2}}(\vert 0 \rangle - \vert 1\rangle)(-1)^{x_{1,y}.d}\vert d \rangle$$

I'm not sure how to proceed to obtain (4) although I see that you gather the $\vert d\rangle$ terms and manipulate the phase somehow.

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This is a little fiddly to get right the first time. Let's start by rewriting eq (3) as $$ \sum_{b\in\{0,1\}}\alpha_b|b\rangle\otimes X^{x_{b,y}}|0\rangle. $$ Now, it probably helps to think of this $X^{b,y}$ operation as two steps: always apply $X^{x_{0,y}}$ and then, if the first qubit is in state $b=1$, apply $X^{x_{0,y}\oplus x_{1,y}}$.

Now you can apply the Hadamard transform. We have the relation $$ HX^p=Z^pH $$ First, let's just apply the Hadamard to the second register $$ I\otimes H\sum_{b\in\{0,1\}}\alpha_b|b\rangle\otimes X^{x_{b,y}}|0\rangle=\sum_{b\in\{0,1\}}\alpha_b|b\rangle\otimes Z^{x_{b,y}}\sum_d|d\rangle. $$ Let's think about that $Z^{x_{b,y}}$ operation again. Always apply $Z^{x_{0,y}}$ and then apply a phase $Z^{x_{0,y}\oplus x_{1,y}}$ if $b=1$. So that second part wants to introduce a minus sign if $b=1$ and $d\cdot (x_{0,y}\oplus x_{1,y})=1\text{ mod }2$. We can achieve this effect not by acting on the second system, but by acting on the first using $Z^{d\cdot (x_{0,y}\oplus x_{1,y})}$. Thus, after $I\otimes H$, we have $$ \sum_d\sum_{b\in\{0,1\}}\alpha_bZ^{d\cdot (x_{0,y}\oplus x_{1,y})}|b\rangle\otimes Z^{x_{0,y}}|d\rangle $$ In doing so, nothing depends on the $b$ index any more, so we can substitute back the original state $$ \sum_dZ^{d\cdot (x_{0,y}\oplus x_{1,y})}|\psi\rangle\otimes Z^{x_{0,y}}|d\rangle. $$ Finally, we can apply the first Hadamard, $H\otimes I$, to give $$ \longrightarrow \sum_dX^{d\cdot (x_{0,y}\oplus x_{1,y})}H|\psi\rangle\otimes Z^{x_{0,y}}|d\rangle. $$

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  • $\begingroup$ That's a nice trick to remove the $b$ dependence from the phase! $\endgroup$
    – JRT
    Jan 22 at 12:26

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