Continuous dynamical decoupling of spin chains: Inducing two-qubit interactions to generate perfect entanglement

Efficient control over entanglement in spin chains is useful for quantum information processing applications. In this paper, we propose the use of a combination of two different configurations of strong static and oscillating fields to control and generate near-perfect entanglement between any two spins in a spin chain, even in the presence of noise. This is made possible by the fact that our control fields not only decouple the spin chain from its environment but also selectively modify the spin-spin interactions. By suitably tuning these spin-spin interactions via the control fields, we show that the quantum state of any two spins in the spin chain can be made to be a Bell state. We illustrate our results for various spin chains, such as the $\mathit{XY}$ model, the XYZ model, and the Ising spin chain.

Figure 1

A tree diagram is a useful way to keep track of the evolution of the spin chain. Each state vector in this diagram should be appended with $|\mathbf{0}\rangle$, meaning that the other spins are in the state $|0\rangle$. As we continue evolving the spin chain using the two-qubit Hamiltonian, the state of the chain becomes a linear combination of a larger number of different state vectors. The complete state just after each interaction is given by the sum of the states in each column. For example, the sum of the four terms in the third column represents the state of the system when spins 1 and 3 have interacted.

Figure 2

A plot of $D\left(t\right)$ against time, where $D\left(t\right)=|{\gamma }_1\left(t\right){\gamma }_2\left(t\right)|$. Note that we are using dimensionless units throughout with $\hslash =1$.

Figure 3

A plot of $D\left(t\right)$ against time, where now $D\left(t\right)=|{\gamma }_1\left(t\right){\eta }_1\left(t\right)|$.

Figure 4

The diagram shows the general scheme we use to entangle the ends of a spin chain with $N=5$. In the complete Hamiltonian picture, a white circle at site $i$ represents a control field Hamiltonian at site $i$ described by the constants $n_x\mathrm{and}{n}_y$, and a black circle at site $i$ represents a control field Hamiltonian at site $i$ described by the constants $m_x\mathrm{and}{m}_y$. Notice that at each step, the two neighboring spins that are interacting have the same control field Hamiltonian at their sites, while the control field Hamiltonian for the rest of the chain is staggered. In the effective Hamiltonian picture, the two gray circles in each step represent the two-qubit interaction ${\overline{H}}_i$ [see Eq. (15)] that is caused by the same control field Hamiltonians at those sites. The white circles represent no interaction, which is caused by staggered fields at those sites.

Figure 5

Step 1. Spins 1 and 2 in an $\mathit{XY}$ chain ($N=5$) interact for ${\tau }_1$ ($\approx 1.4$) to become perfectly entangled, after which the interaction is effectively turned off. The concurrence between spins 1 and 2 is given by $C\left(t\right)$. The solid blue curve is the concurrence using the effective Hamiltonian, the dotted-dashed gray curve is the concurrence using the complete Hamiltonian, and the dashed black curve is the concurrence without the control fields. We use dimensionless units throughout with $\hslash =1$, and we have set ${\zeta }_{j,k}=1$.

Figure 6

Step 2. Spins 2 and 3 interact for ${\tau }_2$ ($\approx 11.3$), after which spins 1 and 3 are perfectly entangled, as shown by their concurrence $C\left(t\right)$. Notice that $C\left(t\right)=0$ for the duration ${\tau }_1$ since that is the duration for which spins 1 and 2 were interacting while all other spins were decoupled from each other.

Figure 7

Step 3. Spins 3 and 4 interact for ${\tau }_3$ ($\approx 11.3$), after which spins 1 and 4 are perfectly entangled, as shown by their concurrence $C\left(t\right)$. $C\left(t\right)$ begins to increase after $({\tau }_1+{\tau }_2)$ since that is the duration during which the first two steps of the scheme take place.

Figure 8

Step 4. Spins 4 and 5 interact for ${\tau }_4$ ($\approx 11.3$), after which spins 1 and 5 are perfectly entangled, as shown by their concurrence $C\left(t\right)$. We have checked that when $C\left(t\right)$ is approximately one, the state of spins 1 and 5 is very close to $\frac{1}{\sqrt{2}}(|00\rangle +i|11\rangle )$, which is a perfectly entangled state.

Figure 9

$C\left(t\right)$ is the concurrence between spins 1 and 10 in an $\mathit{XY}$ chain ($N=10,{\zeta }_{jk}=1$). From $t=0$ to $t\approx 80$, consecutive pairs of neighboring spins interact. After that, spins 9 and 10 are made to effectively interact until spins 1 and 10 are perfectly entangled.

Figure 10

$C\left(t\right)$ is the concurrence between spins 1 and 10 for the XXX model with $N=10$. From $t=0$ to $t\approx 5.9$, consecutive pairs of neighboring spins interact. From $t\approx 5.9$ onwards spins 9 and 10 are made to interact until spins 1 and 10 are perfectly entangled. As always, we are using dimensionless units with $\hslash =1$, and we have set ${\zeta }_{j1}={\zeta }_{j2}={\zeta }_{j3}=1$.

Figure 11

$C\left(t\right)$ is the concurrence between spins 1 and 10 for the quantum Ising model with $N=10$ following our scheme.

Figure 12

A plot of $D\left(t\right)$ against time, where $D\left(t\right)=|{\mu }_1\left(t\right){\mu }_2\left(t\right)|$.

Figure 13

A plot of $D\left(t\right)$ against time, where $D\left(t\right)=|{\mu }_1\left(t\right){\nu }_1\left(t\right)|$.

Figure 14

Step 1. Spins 1 and 2 in an anisotropic XYZ chain ($N=5,{\zeta }_{j1}=1.1,{\zeta }_{j2}=1,{\zeta }_{j3}=1$) interact for ${\tau }_1$ ($\approx 15.7$).

Figure 15

Step 2. Spins 2 and 3 interact for ${\tau }_2$ ($\approx 0.77$). $C\left(t\right)$ is the concurrence between spins 1 and 3.

Figure 16

Step 3. Spins 3 and 4 interact for ${\tau }_3$ ($\approx 0.77$). $C\left(t\right)$ is the concurrence between spins 1 and 4.

Figure 17

Step 4. Spins 4 and 5 interact for ${\tau }_4$ ($\approx 0.77$). $C\left(t\right)$ is the concurrence between spins 1 and 5. At the point when the interaction stops, $C\left(t\right)=0.997$ rounded off to three decimal places.