gates.md

title	date	lang	summary
One way quantum computer	2017-02-18	en	The one-way or measurement based quantum computer (MBQC) is a method of quantum computing that first prepares an entangled resource state, usually a cluster state or graph state, then performs single qubit measurements on it. It is "one-way" because the resource state is destroyed by the measurements.

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

$\def\bra#1{\mathinner{\langle{#1}|}} \def\ket#1{\mathinner{|{#1}\rangle}} \def\braket#1{\mathinner{\langle{#1}\rangle}} \newcommand{\ii}{\mathrm{i}} \newcommand{\tr}{\mathrm{tr}}$

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 $Z_4\otimes Z_1\rightarrow Z_4\otimes Z_3$