Localization of two-particle quantum walk on glued-tree and its application in generating Bell states

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.

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:

$$\begin{aligned} M = |\varPhi \rangle \langle \varPhi |+|\varPhi ^-\rangle \langle \varPhi ^-|+|\varPsi \rangle \langle \varPsi |+|\varPsi ^-\rangle \langle \varPsi ^-| \end{aligned}$$

(19)

where $|\varPhi \rangle$ and $|\varPsi \rangle$ are defined in Eq. (9), and

$$\begin{aligned} \begin{array}{rl} |\varPhi ^-\rangle =&{}\frac{1}{\sqrt{2}}(|1,2\rangle _1|1,2\rangle _2-|1,3\rangle _1|1,3\rangle _2) \\ |\varPsi ^-\rangle =&{}\frac{1}{\sqrt{2}}(|1,2\rangle _1|1,3\rangle _2-|1,3\rangle _1|1,2\rangle _2) \end{array} \end{aligned}$$

(20)

This hypothesis can be proved as follows:

Proof

Form Eqs. (9) and (20), we can verify that:

$$\begin{aligned} \begin{array}{c} |\varPhi \rangle \langle \varPhi |+|\varPhi ^-\rangle \langle \varPhi ^-|= (|1,2\rangle _1|1,2\rangle _2)(_1\langle 1,2|_2\langle 1,2|)+(|1,3\rangle _1|1,3\rangle _2)(_1\langle 1,3|_2\langle 1,3|) \\ |\varPsi \rangle \langle \varPsi |+|\varPsi ^-\rangle \langle \varPsi ^-|= (|1,2\rangle _1|1,3\rangle _2)(_1\langle 1,2|_2\langle 1,3|)+(|1,3\rangle _1|1,2\rangle _2)(_1\langle 1,3|_2\langle 1,2|) \end{array} \end{aligned}$$

(21)

And we have the conclusion as follows:

$$\begin{aligned} \begin{array}{rl} M= &{} |1\rangle _{s_1}\langle 1|\otimes I_{c_1}\otimes |1\rangle _{s_2}\langle 1|\otimes I_{c_2} \\ = &{} |1\rangle _{s_1}\langle 1|\otimes \sum \nolimits _{i=1}^{N}{|i\rangle _{c_1}\langle i|}\otimes |1\rangle _{s_2}\langle 1|\otimes \sum \limits _{j=1}^{N}|j\rangle _{c_1}\langle j|\\ \end{array} \end{aligned}$$

(22)

Because the walkers only move along the edges of the graph, so the Hilbert space where the walkers evolve is actually

$$\begin{aligned} H' = \hbox {span}\left\{ |j,k\rangle _1|i,l\rangle _2: j,i=\{1,2,\ldots ,N\},(j,k)\in E\ \hbox {and}\ (i,l)\in E\right\} \end{aligned}$$

(23)

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:

$$\begin{aligned} M= & {} |1\rangle _{s_1}\langle 1|\otimes \sum \nolimits _{i=1}^{N}{|i\rangle _{c_1}\langle i|}\otimes |1\rangle _{s_2}\langle 1|\otimes \sum \nolimits _{j=1}^{N}|j\rangle _{c_1}\langle j|\nonumber \\= & {} |1\rangle _{s_1}\langle 1|\otimes \sum \nolimits _{i=2}^{3}{|i\rangle _{c_1}\langle i|}\otimes |1\rangle _{s_2}\langle 1|\otimes \sum \nolimits _{j=2}^{3}|j\rangle _{c_1}\langle j|\nonumber \\= & {} |1\rangle _{s_1}\langle 1|\otimes |2\rangle _{c_1}\langle 2|\otimes |1\rangle _{s_2}\langle 1|\otimes |2\rangle _{c_2}\langle 2|\nonumber \\&+|1\rangle _{s_1}\langle 1|\otimes |2\rangle _{c_1}\langle 2|\otimes |1\rangle _{s_2}\langle 1|\otimes |3\rangle _{c_2}\langle 3|\nonumber \\&+|1\rangle _{s_1}\langle 1|\otimes |3\rangle _{c_1}\langle 3|\otimes |1\rangle _{s_2}\langle 1|\otimes |2\rangle _{c_2}\langle 2|\nonumber \\&+|1\rangle _{s_1}\langle 1|\otimes |3\rangle _{c_1}\langle 3|\otimes |1\rangle _{s_2}\langle 1|\otimes |3\rangle _{c_2}\langle 3|\nonumber \\= & {} |\varPhi \rangle \langle \varPhi |+|\varPhi ^-\rangle \langle \varPhi ^-|+|\varPsi \rangle \langle \varPsi |+|\varPsi ^-\rangle \langle \varPsi ^-| \end{aligned}$$

(24)

$\square$

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

$$\begin{aligned} M(SC^I)^t|\varPhi \rangle= & {} \left( |\varPhi \rangle \langle \varPhi |+|\varPhi ^-\rangle \langle \varPhi ^-|+|\varPsi \rangle \langle \varPsi |+|\varPsi ^-\rangle \langle \varPsi ^-|\right) |\varPhi \rangle \nonumber \\= & {} |\varPhi \rangle \ \ \ \ \text{ when } \text{ t } \text{ is } \text{ an } \text{ even } \text{ number } \end{aligned}$$

(25)

$\square$

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

$$\begin{aligned} M(SC^I)^t|\varPsi \rangle = \left( |\varPhi \rangle \langle \varPhi |+|\varPhi ^-\rangle \langle \varPhi ^-|+|\varPsi \rangle \langle \varPsi |+|\varPsi ^-\rangle \langle \varPsi ^-|\right) (SC^I)^t|\varPsi \rangle \end{aligned}$$

(26)

• 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 $|\varPsi ^-\rangle = |a\rangle -|b\rangle$.

$$\begin{aligned} \begin{array}{l} (SC^I)^{t'}|\varPsi \rangle = (SC^I)^{t'}|a\rangle +(SC^I)^{t'}|b\rangle =|a'\rangle +|b'\rangle \\ (SC^I)^{t'}|\varPsi ^-\rangle = (SC^I)^{t'}|a\rangle -(SC^I)^{t'}|b\rangle = |a'\rangle -|b'\rangle \end{array} \end{aligned}$$

(32)

where $|a'\rangle = (SC^I)^{t'}|a\rangle$ and $|b'\rangle = (SC^I)^{t'}|b\rangle$. From Eqs. (31) and (32), we have

$$\begin{aligned} \begin{array}{rl} \langle \varPsi ^-|(SC^I)^t|\varPsi \rangle = &{}(\langle a'|- \langle b'|)C^{I\dag }(|a'\rangle +|b'\rangle )\\ = &{} \langle a'|C^{I\dag }|a'\rangle + \langle a'|C^I|b'\rangle -\langle b'|C^{I\dag }|a'\rangle \\ &{} - \langle b'|C^{I\dag }|b'\rangle \ \ \text{ For } C^I=C^{I\dag } \\ = &{} \langle a'|C^{I\dag }|a'\rangle - \langle b'|C^{I\dag }|b'\rangle \ \ \ \ \text{ For } \langle a'|C^I|b'\rangle = \langle b'|C^{I\dag }|a'\rangle \end{array} \end{aligned}$$

(33)

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

$$\begin{aligned} \langle \varPsi ^-|(SC^I)^t|\varPsi \rangle =0 \end{aligned}$$

(34)

From Eq. (24), we have

$$\begin{aligned} \begin{array}{rl} M(SC^I)^t|\varPsi \rangle = &{} \left( |\varPhi \rangle \langle \varPhi |+|\varPhi ^-\rangle \langle \varPhi ^-|+|\varPsi \rangle \langle \varPsi |+|\varPsi ^-\rangle \langle \varPsi ^-|\right) (SC^I)^t|\varPsi \rangle \\ = &{} |\varPhi \rangle \langle \varPhi |(SC^I)^t|\varPsi \rangle +|\varPhi ^-\rangle \langle \varPhi ^-|(SC^I)^t|\varPsi \rangle \\ &{} +|\varPsi \rangle \langle \varPsi |(SC^I)^t|\varPsi \rangle +|\varPsi ^-\rangle \langle \varPsi ^-|(SC^I)^t|\varPsi \rangle \end{array} \end{aligned}$$

(35)

Then from Eq. (28), we have

$$\begin{aligned} |\varPhi \rangle \langle \varPhi |(SC^I)^t|\varPsi \rangle = 0\ \ \text{ when } \text{ t } \text{ is } \text{ an } \text{ even } \text{ number } \end{aligned}$$

(36)

From Eq. (30), we have

$$\begin{aligned} |\varPhi ^-\rangle \langle \varPhi ^-|(SC^I)^t|\varPsi \rangle = 0\ \ \text{ when } \text{ t } \text{ is } \text{ an } \text{ even } \text{ number } \end{aligned}$$

(37)

And from Eq. (34), we have

$$\begin{aligned} |\varPsi ^-\rangle \langle \varPsi ^-|(SC^I)^t|\varPsi \rangle = 0\ \ \text{ when } \text{ t } \text{ is } \text{ an } \text{ even } \text{ number } \end{aligned}$$

(38)

So finally, we can conclude that

$$\begin{aligned} \begin{array}{rl} M(SC^I)^t|\varPsi \rangle = &{} \left( |\varPhi \rangle \langle \varPhi |+|\varPhi ^-\rangle \langle \varPhi ^-|+|\varPsi \rangle \langle \varPsi |+|\varPsi ^-\rangle \langle \varPsi ^-|\right) (SC^I)^t|\varPsi \rangle \\ = &{} |\varPsi \rangle \langle \varPsi |(SC^I)^t|\varPsi \rangle \ \ \text{ when } \text{ t } \text{ is } \text{ an } \text{ even } \text{ number } \end{array} \end{aligned}$$

(39)

$\square$