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One way quantum computer

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There are some discrepancies between the results here and the paper. But the results are basically the same.

Rotation Gates

This part corresond to the Gates in paper 0301052v2.

$$S_{ab}=\frac{1+Z_a}{2}+\frac{1-Z_a}{2}Z_b=U_a+D_aZ_b$$

The \(u_i, d_i\) are defined by \(\ket{\psi_i}=u_i\ket{0_i}+d_i\ket{1_i}\). And we have \(Z_b\ket{+_b}=\ket{-_b}\)

$$\begin{aligned} \mathcal{R}&=\bra{\psi_1}\bra{\psi_2}\bra{\psi_3}\bra{\psi_4}S_{12}S_{23}S_{34}S_{45}\ket{+_2}\ket{+_3}\ket{+_4}\ket{+_5}\\ &=\left(\ket{+_5}\bra{\psi_4}U_4+\ket{-_5}\bra{\psi_4}D_4\right) \bra{\psi_1}\bra{\psi_2}\bra{\psi_3}S_{12}S_{23}S_{34}\ket{+_2}\ket{+_3}\ket{+_4}\\ &=\prod_{i=4}^1 \Big(\ket{+_{i+1}}\bra{0_i}u_i+\ket{-_{i+1}}\bra{1_i}d_i\Big)\\ &=\prod_{i=4}^1 {\frac{1}{\sqrt{2}}\begin{bmatrix}u_i & d_i\\-u_i & -d_i\end{bmatrix}}=\prod_{i=4}^1 H\begin{bmatrix}u_i & \\ & d_i\end{bmatrix}\\ &=\prod_{i=4}^1 H\mathcal{Z}_{\phi_i},\quad \mathcal{Z}_\phi=\exp(-\ii \phi Z/2)\\ &=(H\mathcal{Z}_\zeta H)\mathcal{Z}_\eta (H\mathcal{Z}_\xi H)\\ &=\mathcal{X}_\zeta\mathcal{Z}_\eta\mathcal{X}_\xi\end{aligned}$$

In the basis of \(Z_1\rightarrow Z_5\). If all measurement results \(\psi_i\) are positive for directions \((0, \xi,\eta,\zeta)\), respectively, we can verify that the rotation matrix \(\mathcal{R}\) is equivalent to \(\exp(-\ii \zeta X/2)\exp(-\ii \eta Z/2)\exp(-\ii \xi X/2)\). Hadamard gate is simply a special case.

CNOT Gates

This part corresponds to PRL.86.5188(Page 3, upper left corner), so we are using a different \(S\):

$$S_{ab}=1-\frac{(1+Z_{a})(1-Z_{b})}{2}=D_a+U_aZ_b=U_b-D_bZ_a$$

So

$$S_{ab}S_{bc}=\frac{Z_c-Z_a}{2}+\frac{Z_c+Z_a}{2}Z_b=U_bZ_c-D_bZ_a$$
$$\begin{aligned} \mathcal{C}&=\bra{\psi_1}\bra{\psi_2}S_{12}S_{23}S_{24}\ket{+_2}\ket{+_3}\\ &=\Big(\ket{+_{3}}\bra{1_2}d_2+\ket{-_3}\bra{0_2}u_2\Big)\bra{\psi_1}S_{12}S_{24}\ket{+_2}\\ &=\Big(\ket{+_{3}}\bra{1_2}d_2+\ket{-_3}\bra{0_2}u_2\Big)\bra{\psi_1}U_2Z_4-D_2Z_1\ket{+_2}\\ &=\Big(\pm_2\ket{+_{3}}\bra{1_2}+\ket{-_3}\bra{0_2}\Big)\Big(\ket{0_2}\bra{\psi_1}Z_4-\ket{1_2}\bra{\psi_1}Z_1\Big)/2\\ &=\Big( \mp_2\ket{+_{3}}\bra{\mp_1}+\ket{-_3}\bra{\pm_1}Z_4\Big)/2 \\\end{aligned}$$

If \(s_1=1, s_2=1\), then \(\mathcal{C}_n=\ket{-_{3}}\bra{-_1}Z_4+\ket{+_3}\bra{+_1}I_4\)

$$\mathcal{C}=I_4\otimes\begin{bmatrix} 1& 1\\ 1& 1 \end{bmatrix}+Z_4\otimes\begin{bmatrix} 1& -1\\ -1& 1 \end{bmatrix}=\begin{bmatrix} 1& &&\\ & 1&&\\ &&&1\\ &&1&\\ \end{bmatrix}$$

In the basis of \(Z_4\otimes Z_1\rightarrow Z_4\otimes Z_3\)

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