Abstract
Studies on two-particle quantum walks show that the spatial interaction between walkers will dynamically generate complex entanglement. However, those entanglement states are usually on a large state space and their evolutions are complex. It makes the entanglement states generated by quantum walk difficult to be applied directly in many applications of quantum information, such as quantum teleportation and quantum cryptography. In this paper, we firstly analyse a localization phenomena of two-particle quantum walk and then introduce how to use it to generate a Bell state. We will show that one special superposition component of the walkers’ state is localized on the root vertex if a certain interaction exists between walkers. This localization is interesting because it is contrary to our knowledge that quantum walk spreads faster than its classical counterpart. More interestingly, the localized component is a Bell state in the coin space of two walkers. By this method, we can obtain a Bell state easily from the quantum walk with spatial interaction by a local measurement, which is required in many applications. Through simulations, we verify that this method is able to generate the Bell state \(\frac{1}{\sqrt{2}}(|A \rangle _1|A\rangle _2 \pm |B\rangle _1|B\rangle _2)\) in the coin space of two walkers with fidelity greater than \(99.99999\,\%\) in theory, and we have at least a \(50\,\%\) probability to obtain the expected Bell state after a proper local measurement.
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This work was supported by the National Natural Science Foundation of China (NSFC) No. 61402506 and the Open Fund from HPCL No. 20150101.
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Appendix
Appendix
In Sect. 4, in order to obtain Eq. (15) we conclude that \(M(SC^I)^t|\varPhi \rangle = |\varPhi \rangle \) [Eq. (13)] and \(M(SC^I)^t|\varPsi \rangle = |\varPsi \rangle \langle \varPsi |(SC^I)^t|\varPsi \rangle \) (Eq. (14)) when t is an even number. Here we give the proof of these conclusions. Firstly, we prove a hypothesis that the measurement operator M in Eq. (12) can be expressed as:
where \(|\varPhi \rangle \) and \(|\varPsi \rangle \) are defined in Eq. (9), and
This hypothesis can be proved as follows:
Proof
Form Eqs. (9) and (20), we can verify that:
And we have the conclusion as follows:
Because the walkers only move along the edges of the graph, so the Hilbert space where the walkers evolve is actually
where E is the edges set of the graph. So, for the glued-tree, a measurement operator \(|1\rangle _{s_1}\langle 1|\otimes |i\rangle _{c_1}\langle i|\otimes |1\rangle _{s_2}\langle 1|\otimes |j\rangle _{c_1}\langle j|\) is meaningful only when \((1,i)\in E\) and \((1,j)\in E\). So the measurement operator in Eq. (22) on glued-tree can be expressed as:
\(\square \)
Then the conclusion that “\(M(SC^I)^t|\varPhi \rangle = |\varPhi \rangle \) when t is an even number” can be proved as follows:
Proof
From Eq. (11), we have \((SC^I)^t|\varPhi \rangle = |\varPhi \rangle \) when t is an even number. So
\(\square \)
The conclusion that “\(M(SC^I)^t|\varPsi \rangle = |\varPsi \rangle \langle \varPsi |(SC^I)^t|\varPsi \rangle \)” can be proved as follows:
Proof
From Eq. (19) we have
-
1. From Eq. (11) we have:
$$\begin{aligned} (SC^I)^t|\varPhi \rangle =|\varPhi \rangle \ \ \text{ when } \text{ t } \text{ is } \text{ an } \text{ even } \text{ number } \end{aligned}$$(27)So we can conclude that
$$\begin{aligned} \langle \varPhi |(SC^I)^t|\varPsi \rangle = \langle \varPhi |(C^{I\dag }S^\dag )^t(SC^I)^t|\varPsi \rangle = \langle \varPhi |\varPsi \rangle = 0 \end{aligned}$$(28)when t is an even number
-
2. In the same way, we can verify that:
$$\begin{aligned} (SC^I)^t|\varPhi ^-\rangle =|\varPhi ^-\rangle \ \ \text{ when } \text{ t } \text{ is } \text{ an } \text{ even } \text{ number } \end{aligned}$$(29)So we can conclude that
$$\begin{aligned} \langle \varPhi ^-|(SC^I)^t|\varPsi \rangle = \langle \varPhi ^-|(C^{I\dag }S^\dag )^t(SC^I)^t|\varPsi \rangle = \langle \varPhi ^-|\varPsi \rangle = 0 \end{aligned}$$(30) -
3. Because t is an even number, we suppose that \(t'=t/2\). Then we have:
$$\begin{aligned} \begin{array}{rl} \langle \varPsi ^-|(SC^I)^t|\varPsi \rangle = &{} \langle \varPsi ^-|(SC^I)^{t'}(SC^I)^{t'}|\varPsi \rangle \\ = &{} \langle \varPsi ^-|(S^\dag C^{I\dag })^{t'}(SC^I)^{t'}|\varPsi \rangle \ \ \text{ For } C^{I\dag }=C^I \text{ and } S^{I\dag }=S^I\\ = &{} \langle \varPsi ^-|(C^{I\dag }S^\dag )^{t'}C^{I\dag }(SC^I)^{t'}|\varPsi \rangle \ \ \text{ For } C^I|\varPsi ^-\rangle = |\varPsi ^-\rangle \end{array} \end{aligned}$$(31)
We suppose \(\frac{1}{\sqrt{2}}|1,2\rangle _1|1,3\rangle _2=|a\rangle \) and \(\frac{1}{\sqrt{2}}|1,3\rangle _1|1,2\rangle _2=|b\rangle \). Then we have \(|\varPsi \rangle = |a\rangle +|b\rangle \) and \(|\varPsi ^-\rangle = |a\rangle -|b\rangle \).
where \(|a'\rangle = (SC^I)^{t'}|a\rangle \) and \(|b'\rangle = (SC^I)^{t'}|b\rangle \). From Eqs. (31) and (32), we have
Because of the symmetry of glued-tree (see Fig. 1), we can verify that \(\langle a'|C^{I\dag }|a'\rangle = \langle b'|C^{I\dag }|b'\rangle \). So we conclude that
From Eq. (24), we have
Then from Eq. (28), we have
From Eq. (30), we have
And from Eq. (34), we have
So finally, we can conclude that
\(\square \)
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Wang, H., Wu, J., He, H. et al. Localization of two-particle quantum walk on glued-tree and its application in generating Bell states. Quantum Inf Process 15, 3619–3635 (2016). https://doi.org/10.1007/s11128-016-1414-9
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DOI: https://doi.org/10.1007/s11128-016-1414-9