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The brickwork state with fewer qubits in blind quantum computation

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Abstract

How to reduce the quantum cost of the client is a hot topic in blind quantum computation. However, making full use of the server’s quantum ability to reduce the client’s quantum cost is rarely acknowledged. In this research, we show that twice measurements on a three-qubits cluster state can be used to realize an indeterminate single-qubit gates by converting each qubit of the cluster state into another form in measurement-based quantum computation. In blind quantum computation, combining with this approach and adding different circuit-based quantum computation capabilities to the server, we present two models that the client can send fewer qubits to the server to implement a deterministic single-qubit gate. Finally, to keep the entire calculation process blind to the server, we substitute some bricks that realize a tensor of two single-qubit gates to the above models in the brickwork state. As a result, two types of the brickwork states with fewer particles were proposed, and we also give an upper bound for the number of replaced bricks. This reduces the quantum cost of the client.

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Acknowledgements

This work is supported by the National Natural Science Foundation of China (Grant No.11671284), Sichuan Science Foundation and Technology Program (Grant No.2020YFG0290).

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In fact, all of the authors’ contributions to this paper are important. The specific contributions are as follows. The first author plays a major role in the idea and writes the main content. The second author devotes to all the process of calculation and the language problem. The third author establishes the overall framework of this article.

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Correspondence to Ming-Qiang Bai.

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Appendix

Appendix

1.1 I: The concrete process from Eq.(1) to Eq.(3)

In this subsection, we will explain how Eq.(1) is evolved into Eq.(3). In the first place, we introduce a conclusion, from Ref. [19], that the result of measuring tangled state \(CZ_{12}|\psi \rangle _1|+\rangle _2\) based on \(M_1(\theta )\) is \(X^{s_1}HR_z(\theta )|\psi \rangle \), where \(\psi = \alpha |0\rangle + \beta |1\rangle \) and \(\alpha ,\beta \) satisfy \(|\alpha |^2+|\beta |^2 = 1\). To be specific, we can obtain that

$$\begin{aligned} \begin{aligned}&X^{s_1}HR_z(\theta )|\psi \rangle \\&\quad = M_1(\theta )CZ|\psi \rangle _1|+\rangle _2 \\&\quad = (\langle 0|+(-1)^{s_1}e^{i\theta }\langle 1|)_1 (\alpha |00\rangle +\alpha |01\rangle +\beta |10\rangle -\beta |11\rangle )_{12} \\&\quad = (\alpha +(-1)^{s_1}e^{i\theta }\beta )|0\rangle + (\alpha -(-1)^{s_1}e^{-i\theta }\beta )|1\rangle )_2 \end{aligned}\end{aligned}$$
(14)

For Eq.(1), we execute \(M_1(\delta _1)\) on qubit 1, then qubit 2 collapses into

$$\begin{aligned}&M_1^{\delta _1}CZ_{12} R_z(\theta _1)|\psi \rangle _1 |\theta _2\rangle \nonumber \\&\quad = (\langle 0|+(-1)^{s_1}e^{-i\delta _1}\langle 1|)_1(\alpha |00\rangle + e^{i\theta _2}\alpha |01\rangle + e^{i\theta _1}\beta |10\rangle - e^{i\theta _1}e^{i\theta _2}\beta |11\rangle )_{12} \nonumber \\&\quad = ((\alpha +(-1)^{s_1}e^{-i\delta _1}e^{i\theta _1}\beta )|0\rangle + e^{i\theta _2}(\alpha -(-1)^{s_1}e^{-i\delta _1}e^{i\theta _1}\beta )|1\rangle )_2\nonumber \\&\quad = R_z(\theta _2)[(\alpha +(-1)^{s_1}e^{-i\delta _1}e^{i\theta _1}\beta )|0\rangle + (\alpha -(-1)^{s_1}e^{-i\delta _1}e^{i\theta _1}\beta )|1\rangle ] \end{aligned}$$
(15)

from the result of Eq.(14), Eq.(15) can be evolved into

$$\begin{aligned} R_z(\theta _2)X^{s_1}HR_z(\theta _1-\delta _1)|\psi \rangle . \end{aligned}$$
(16)

Next, we express qubit 2 as \(\alpha _2|0\rangle +\beta _2|1\rangle \) and carry out \(CZ_{23}\) between qubit 2 and 3. Then the total state can be shown as

$$\begin{aligned}&CZ_{23}(\alpha _2|0\rangle +\beta _2|1\rangle )_2(|0\rangle +e^{i\theta _3}|1\rangle )_3\nonumber \\&\qquad = (\alpha _2|00\rangle + e^{i\theta _3}\alpha _2|01\rangle + \beta _2|10\rangle - e^{i\theta _3}\beta _2|11\rangle )_{23} \end{aligned}$$
(17)

after that we measure qubit 2 based on \(M_2(\delta _2)\), which is similar to the process of transmitting Eq.(1) into Eq.(16), the qubit 3 from Eq.(17) collapses into

$$\begin{aligned} \begin{aligned}&M_2(\delta _2)CZ_{23}(\alpha _2|0\rangle + \beta _2|1\rangle )_2(|0\rangle + e^{i\theta _3}|1\rangle )_3 \\&\quad =(\langle 0|+(-1)^{s_2}e^{-i\delta _2}\langle 1|)_2(\alpha _2|00\rangle + e^{i\theta _3}\alpha _2|01\rangle + \beta _2|10\rangle - e^{i\theta _3}\beta _2|11\rangle )_{23} \\&\quad = [(\alpha _2+(-1)^{s_2}e^{-i\delta _2}\beta _2)|0\rangle + e^{i\theta _3}(\alpha _2-(-1)^{s_2}e^{-i\delta _2}\beta _2)|1\rangle ]_3 \\&\quad = R_z(\theta _3)[(\alpha _2+(-1)^{s_2}e^{-i\delta _2}\beta _2)|0\rangle + (\alpha _2-(-1)^{s_2}e^{-i\delta _2}\beta _2)|1\rangle ] \end{aligned}\end{aligned}$$
(18)

from Eq.(14), Eq.(18) can evolve into

$$\begin{aligned} R_z(\theta _3)X^{s_2}HR_z(-\delta _2)(\alpha _2|0\rangle + \beta _2|1\rangle ). \end{aligned}$$
(19)

Then, substituting Eq.(16) for \(\alpha _2|0\rangle +\beta _2|1\rangle \) and utilizing \(R_z(\delta )X^s = X^sR_z((-1)^s\delta )\) and \(R_z(\delta )Z^s = Z^sR_z(\delta ) \) for adjusting order, the qubit 3 will be expressed as Eq.(3) ultimately.

1.2 II: The concrete process from Eq.(4) to Eq.(6)

In this subsection, we make clear the detail of process from Eq.(4) to Eq.(6). After replacing \(\delta _i\;(i=1,2)\) in Eq.(4) for \(\theta _i+\phi _i+r_i\pi \), we get that

$$\begin{aligned} \begin{aligned}&X^{s_2}Z^{s_1}R_z((-1)^{s_2}\theta _3)R_x((-1)^{s_1}(\theta _2-\delta _2))R_z(\theta _1-\delta _1)|\psi \rangle \\&\quad = X^{s_2}Z^{s_1}R_z((-1)^{s_2}\theta _3)R_x((-1)^{s_1}(-\phi _2-r_2\pi ))R_z(-\phi _1-r_1\pi )|\psi \rangle \\&\quad = X^{s_2}Z^{s_1}R_z((-1)^{s_2}\theta _3)HR_z((-1)^{s_1}(-\phi _2))R_Z((-1)^{s_1}\\&\qquad (-r_2\pi ))HR_z(-\phi _1)R_z(-r_1\pi )|\psi \rangle \end{aligned}\end{aligned}$$
(20)

on the other hand

$$\begin{aligned} R_z((-1)^sr\pi ) = \left( \begin{array}{cc} 1 &{} 0 \\ 0 &{} e^{i(-1)^sr\pi } \\ \end{array} \right) =Z^{r} \end{aligned}$$

where \(s,r \in \{0,1\}\). So Eq.(20) can be transmitted to

$$\begin{aligned} \begin{aligned}&X^{s_2}Z^{s_1}R_z((-1)^{s_2}\theta _3)HR_z((-1)^{s_1}(-\phi _2))Z^{r_2}HR_z(-\phi _1)Z^{r_1}|\psi \rangle \\&\quad =X^{s_2+r_2}Z^{s_1+r_1}R_z((-1)^{s_2+r_2}\theta _3)R_x((-1)^{s_1+r_1}(-\phi _2))R_z(-\phi _1)|\psi \rangle \end{aligned}\end{aligned}$$
(21)

when a global phase factor is ignored.

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Yang, Z., Bai, MQ. & Mo, ZW. The brickwork state with fewer qubits in blind quantum computation. Quantum Inf Process 21, 125 (2022). https://doi.org/10.1007/s11128-022-03473-1

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