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Again, I am new to quantum computing and have a CS background, so apologies if this seems like an obvious question or if I seem unclear. $\newcommand{\braket}[1]{\langle #1 \rangle}\newcommand{\bra}[1]{\langle #1 |}\newcommand{\ket}[1]{| #1 \rangle}$

Let $U_l = \text{exp}(i\frac{\pi}{4}l(z)X)$ where $X$ is the pauli $\sigma_x$ gate and $l(z)$ is either $+1$ or $-1$

The paper that I've been reading claims that $$U_l^{\dagger}YU_l = \cos(\frac{\pi}{2}l(z))Y + \sin(\frac{\pi}{2}l(z))Z$$ where $Y$ is the pauli $\sigma_{y}$ gate and $Z$ is $\sigma_z$

Finally, they conclude that $$\bra{z,0}U_l^{\dagger}YU_l\ket{z,0} = \sin(\frac{\pi}{2}l(z)) = l(z)$$ , where $z = z_1,...,z_n$

When I try this myself, I don't get the same result and I am not able to spot my mistake. Here's my approach:

$U_l = \text{exp}(i\frac{\pi}{4}l(z)X) = I + e^{i\frac{\pi}{4}l(z)}X$ where $I$ is the identity matrix

Thus, I have that $$U_l^{\dagger}Y = Y + ie^{-i\frac{\pi}{4}l(z)}Z$$ and therefore $$U_l^{\dagger}YU_l = -ie^{i\frac{\pi}{4}l(z)}Z + ie^{-i\frac{\pi}{4}l(z)}Z$$ which, when we apply Euler's formula, reduces to: $$2\sin(\frac{\pi}{4}l(z))Z$$ which doesn't seem to be correct.

Could you please help me out? I've been stuck on it for a couple of days now.

Where did you get that $U_l = I + e^{...}X$? – Bobak Hashemi – 2020-03-09T02:29:41.497

This is Farhi’s paper on qnn right ?! – Enrique Segura – 2020-03-09T05:44:02.923

@Enrique Segura, yes this is indeed the paper on qnn – Skyris – 2020-03-09T07:24:01.383