]

# Transforming spin chains with a continuous driving field

Hudaiba Soomro, Adam Zaman Chaudhry  [<EMAIL_ADDRESS>School of
Science and Engineering, Lahore University of Management Sciences (LUMS),
Opposite Sector U, D.H.A, Lahore 54792, Pakistan

###### Abstract

A continuous, sinusoidal control field is used to suitably transform quantum
spin chains. In particular, we are able to transform the quantum Ising chain
to the quantum XY model, and the XY model to the XYZ spin chain. Our applied
control field can also mitigate the effect of noise on the spin chain. We show
how these spin chain transformations can be useful for quantum state transfer
as well as entanglement generation.

††preprint: APS/123-QED

##  I Introduction

Quantum spin chains have served as prototypical models in explaining a host of
physical phenomena. For instance, spin chain models have been used to study
many body localization [1, 2], Anderson localization, quantum chaos [3], phase
transitions [4, 5], and quantum correlations [6]. Their theoretical prowess
has paved the way for their experimental realization in the form of optical
lattices [7], ion traps such as Paul traps [8, 9], solid state setups such as
nitrogen vacancy centers connected by a chain of implanted nitrogen impurities
[10, 11], as well as photonic systems [12]. Moreover, numerical simulations
also assist research directions, especially for complex spin chains which
unlike low dimensional spin chains (1D and 2D) are not exactly solvable [13].
For instance, Ref. [14] uses a transformation scheme to simulate the transport
of polarization beyond exactly solvable methods. Similarly, a numerical pulse
optimization technique and a unitary operator decomposition algorithm to
simulate unitary operators in spin chains. Furthermore, quantum simulators
allow the possibility of simulating spin chains using quantum circuits to
study phenomena such as phase transitions in the thermodynamic limit [15].

In the field of quantum information, quantum spin chains have yielded
extremely useful results, from accurate state transfer to generation and
transfer of entangled states [16, 17, 18, 19]. Experimental possibilities of
attaining perfect state transfer have included implementations such as hybrid
integrated quantum photonic circuits and superconducting qubits [20, 21].
Theoretically, there are various methods to achieve perfect state transfer in
spin chains [22, 23, 24, 25, 26]. One particularly useful method works by
modulating neighboring coupling strengths to achieve perfect state transfer in
the XY spin chain [27]. Similarly, entanglement improvement in quantum spin
system is another crucial site of work. While there are methods of improving
long-distance entanglement such as the use of quantum mediators - for example,
sets of antiferromagnetically coupled spin dimers [28] \- we focus on the use
of control field configurations [29, 30, 31]. The use of control fields has a
dual advantage. First, they can be used to perform useful Hamiltonian
transformations. For instance, in some cases, the use of oscillatory and
static control fields can effectively introduce additional spin-spin
interaction terms which can in turn improve entanglement generation [32, 33,
34, 35]. Similarly, a staggered field configuration has been shown to produce
perfect entanglement between any two spin sites of an XY spin chain [36].
Second, control fields can be used to dynamically decouple the system from the
environment so that the effect of noise on the spin chain is mitigated [37,
38, 39, 40, 41], thereby helping to protect the fragile quantum states from
decoherence. More so, the method of dynamical decoupling can lead to
additional benefits such as scaling of the squeezing performance in spin
chains [42].

Keeping in mind the usefulness of suitable control fields, we propose the use
of a continuous driving field that can effectively transform spin-spin
interactions. In particular, our work is inspired by the use of a continuous
driving field to transform a one-axis spin squeezing Hamiltonian can be
transformed to a two-axis spin squeezing Hamiltonian [43]. Using a similar
scheme, we show how we can effectively transform the quantum Ising spin chain
to the XY spin chain, and the XY spin chain to an engineered XYZ spin chain.
These spin chain transformations introduce additional interactions in the spin
chain Hamiltonian which can be controlled by tuning the control field
parameters. Suitably engineering the spin-spin interactions allows us to yield
perfect fidelity of state transfer for the quantum Ising spin chain system
with a control field as this transformation allows us to realize the isotropic
XY spin chain [21]. Similarly, by our control fields effectively transform the
XY spin chain such that significantly higher entanglement is generated between
the ends of the spin chain. Moreover, while we observe no significant
entanglement in the Ising spin chain, we are able to show that by taking into
account the next nearest-neighbor interactions and subjecting the spin chain
to a transverse control field, considerably greater entanglement can again be
generated between the ends of the spin chain. Apart from suitably engineering
the spin-spin interactions, the highly oscillatory control fields are also
useful in averaging out system-environment interaction terms which manifest as
local noise terms and lead to decoherence of the fragile quantum states [40].

##  II The Theoretical Model

###  II.1 Transforming the Ising spin chain

The Hamiltonian for the quantum Ising spin chain can be writted as (we use
dimensionless units throughout with $\hbar=1$)

$H=\sum_{i=1}^{N-1}J_{i}\sigma_{x}^{i}\sigma_{x}^{i+1},$ (1)

where $i$ labels the sites, and $J_{i}$ describe the nearest-neighbor coupling
strengths. In the presence of a sinusoidal time-dependent field in a
transverse direction, the Hamiltonian becomes

$H(t)=\sum_{i=1}^{N-1}J_{i}\sigma_{x}^{i}\sigma_{x}^{i+1}+\sum_{i=1}^{N}h(t)\sigma_{z}^{i},$
(2)

where $h(t)=g\cos(\omega t)$, with $g$ being the strength of the driving
field, and $\omega$ its frequency. Since the Hamiltonian at different times
does not commute, finding the unitary time-evolution operator is far from
straightforward. To make progress, it is best to transform to the frame of the
applied field, that is, we define [43]

$H_{I}(t)=U_{c}^{\dagger}(t)VU_{c}(t),$ (3)

where $V$ is the time-independent part of the Hamiltonian
($\sum_{i=1}^{N-1}J_{i}\sigma_{x}^{i}\sigma_{x}^{i+1}$) and $U_{c}(t)$ is
found as

$\begin{split}U_{c}(t)=&\exp[-i\int_{0}^{t}\sum_{i=1}^{N}g\cos(\omega\tau)\sigma_{z}^{i}d\tau]\\\
=&\exp[-i\sum_{i=1}^{N}g\frac{\sin(\omega
t)}{\omega}\sigma_{z}^{i}].\end{split}$ (4)

Defining $C(t)=\frac{\sin(\omega t)}{\omega}$ for simplicity of notation, we
get

$H_{I}(t)=e^{igC(t)\sum_{i=1}^{N}\sigma_{z}^{i}}\left(\sum_{i=1}^{N-1}J_{i}\sigma_{x}^{i}\sigma_{x}^{i+1}\right)e^{-igC(t)\sum_{i=1}^{N}\sigma_{z}^{i}}.$
(5)

Introducing the raising and lowering operators in the usual fashion, that is,
$\sigma_{\pm}^{j}=\sigma_{x}^{j}\pm i\sigma_{y}^{j}$, we note that

$e^{igC(t)\sum_{j=1}^{N}\sigma_{z}^{j}}\sigma_{\pm}^{j}e^{-igC(t)\sum_{j=1}^{N}\sigma_{z}^{j}}=\sigma_{\pm}^{j}e^{\pm
2igC(t)}.$ (6)

The Hamiltonian in Eq. (5) then transforms as

$\displaystyle H_{I}(t)$
$\displaystyle=\frac{1}{4}[\sum_{i=1}^{N-1}J_{i}(e^{4igC(t)}\sigma_{+}^{i}\sigma_{+}^{i+1}+e^{-4igC(t)}\sigma_{-}^{i}\sigma_{-}^{i+1})+$
(7) $\displaystyle
J_{i}(\sigma_{+}^{i}\sigma_{-}^{i+1}+\sigma_{-}^{i}\sigma_{+}^{i+1})]$

The Hamiltonian in the frame of the applied field has split into a time-
dependent part and a time-independent part. We now make use of the Jacobi-
Anger expansion [44], namely

$e^{\frac{i4g\sin(\omega
t)}{\omega}}=\sum_{n=-\infty}^{\infty}\mathcal{J}_{n}\left(\frac{\pm
4g}{\omega}\right)e^{in\omega t},$

where $\mathcal{J}_{n}$ is the $n^{\text{th}}$ Bessel function of the first
kind. Note that until now, our treatment has been exact. We now assume that
the applied field has a high frequency such that $\omega\gg J_{i}$. In such a
scenario, the higher order terms in the Jacobi-Anger expansion can be ignored,
and we can consider only the $n=0$ term. Defining
$A=\mathcal{J}_{0}(\pm\frac{4g}{\omega})$, we achieve the time-independent
effective Hamiltonian

$H_{I}=\sum_{i}^{N-1}\frac{J_{i}}{2}[(A+1)\sigma_{x}^{i}\sigma_{x}^{i+1}-(A-1)\sigma_{y}^{i}\sigma_{y}^{i+1}].$
(8)

Since this effective Hamiltonian is time-independent, we can find the
corresponding time-evolution operator $U_{I}(t)=e^{-iH_{I}t}$. The full time-
evolution operator is then given by $U(t)=U_{c}(t)U_{I}(t)$. However, the
time-evolution operator $U_{c}(t)$ is periodic with period $2\pi/\omega$. As
such, at times equal to integer multiples of $2\pi/\omega$, the time evolution
generated by the full Hamiltonian is simply given by $U_{I}(t)$, provided of
course that $\omega\gg J_{i}$. In particular, starting from the quantum Ising
model, it is clear that we have effectively transformed to the XY spin chain
via the applied control field. The anisotropy for this effectively spin chain
can be changed by tweaking the control field’s parameters. For instance,
taking $A=0$ yields the fully isotropic XY spin chain - this entails requiring
that $\frac{4g}{\omega}$ equals the roots of the zeroeth Bessel function. On
the other extreme, if we choose $A=1$, we recover the initial Ising spin
chain. It should be noted that there is a restriction on the range of values
$A$ can take, namely $-0.4\leq A\leq 1$; this simply follows from the
properties of the Bessel function.

###  II.2 Transforming the XY spin chain

The transformation that we have discussed can analogously be extended to
transforming the XY spin chain. We begin with the Hamiltonian

$\displaystyle H(t)$
$\displaystyle=\sum_{i}^{N-1}[J_{x}^{i}\sigma_{x}^{i}\sigma_{x}^{i+1}+J_{y}^{i}\sigma_{y}^{i}\sigma_{y}^{i+1}]+$
(9) $\displaystyle\sum_{i=1}^{N}[g\cos({\omega t})\sigma_{z}^{i}].$

As before, we transform to the interaction picture where the Hamiltonian
transforms as $H_{I}(t)=U_{c}^{\dagger}(t)VU_{c}(t)$ where $V$ is the time-
independent part of the Hamiltonian in Eq. (9). As in Eq. (4), we find
$U_{c}(t)$ is given by $U_{c}(t)=\exp[-i\sum_{i=1}^{N}gC(t)\sigma_{z}^{i}]$,
with $C(t)=\frac{\sin(\omega t)}{\omega}$ as defined previously. We then get

$\displaystyle H_{I}$
$\displaystyle=\sum_{i}^{N-1}[\frac{1}{2}\\{(J_{x}^{i}(A+1)-J_{y}^{i}(A-1))\sigma_{x}^{i}\sigma_{x}^{i+1}+$
(10)
$\displaystyle(J_{y}^{i}(A+1)-J_{x}^{i}(A-1))\sigma_{y}^{i}\sigma_{y}^{i+1}\\}],$

where, as before, $A=\mathcal{J}_{0}(\pm\frac{4g}{\omega})$. We see that if we
choose $J_{x}^{i}=J_{y}^{i}$, we get back the fully isotropic XY spin chain.
Our chosen control field is unable to yield additional interaction terms. To
get these, let us consider the more interesting case where we apply the field
along one of the two axes of the Hamiltonian. The term along the same axis
does not undergo any transformation while the term along the transverse axis
transforms yielding an interaction term along the third axis. As such, we now
apply a control field to the XY spin chain along the $y$ direction, that is,
we consider

$\displaystyle H(t)$
$\displaystyle=\sum_{i}^{N-1}[J_{x}^{i}\sigma_{x}^{i}\sigma_{x}^{i+1}+J_{y}^{i}\sigma_{y}^{i}\sigma_{y}^{i+1}]+$
(11) $\displaystyle\sum_{i=1}^{N}[g\cos({\omega t})\sigma_{y}^{i}]$

Carrying out an analogous procedure as done before, but now with
$U_{c}(t)=\exp[-i\sum_{i=1}^{N}gC(t)\sigma_{y}^{i}]$, we get

$\displaystyle H_{I}$
$\displaystyle=\sum_{i}^{N-1}[\frac{1}{2}\\{J_{x}^{i}(A+1)\sigma_{x}^{i}\sigma_{x}^{i+1}-J_{x}^{i}(A-1)\sigma_{z}^{i}\sigma_{z}^{i+1}$
(12) $\displaystyle+2J_{y}^{i}\sigma_{y}^{i}\sigma_{y}^{i+1}\\}].$

Taking $A=0$, $J_{i}^{x}=2$, and $J_{i}^{y}=0$ yields the XXX spin chain given
by

$\displaystyle H_{I}$
$\displaystyle=\sum_{i}^{N-1}[\sigma_{x}^{i}\sigma_{x}^{i+1}+\sigma_{y}^{i}\sigma_{y}^{i+1}+\sigma_{z}^{i}\sigma_{z}^{i+1}].$
(13)

Another useful case is obtained if we choose $J_{x}^{i}=J_{y}^{i}=J_{i}$ and
$A=0$ in Eq. (12). We then obtain

$\displaystyle H_{I}$
$\displaystyle=\sum_{i}^{N-1}[\frac{J_{i}}{2}\\{\sigma_{x}^{i}\sigma_{x}^{i+1}+\sigma_{z}^{i}\sigma_{z}^{i+1}+2\sigma_{y}^{i}\sigma_{y}^{i+1}\\}],$
(14)

which is essentially a rotated version of the usual XXZ spin chain.

We can also recast our results for the XY spin chain in terms of an anisotropy
parameter $\gamma$. Let us rewrite our spin chain Hamiltonian in the presence
of the control field as

$\displaystyle H(t)$
$\displaystyle=\sum_{i}^{N-1}J_{i}[(\gamma+1)\sigma_{x}^{i}\sigma_{x}^{i+1}+(1-\gamma)\sigma_{y}^{i}\sigma_{y}^{i+1}]$
(15) $\displaystyle+\sum_{i=1}^{N}[g\cos({\omega t})\sigma_{y}^{i}]$

We correspondingly have that

$\displaystyle H_{I}$
$\displaystyle=\sum_{i}^{N-1}\frac{J_{i}}{2}[(\gamma+1)(A+1)\sigma_{x}^{i}\sigma_{x}^{i+1}-$
(16)
$\displaystyle(\gamma+1)(A-1)\sigma_{z}^{i}\sigma_{z}^{i+1}+2(1-\gamma)\sigma_{y}^{i}\sigma_{y}^{i+1}].$

###  II.3 Introducing Noise

An experimentally realizable model of any quantum system must take into
account noise introduced due to the environment. Assuming that each spin
locally interacts with the environment, the system-environment interaction can
be modelled as follows:

$H_{SB}=\sum_{i=1}^{N}(B_{x}^{i}\sigma_{x}^{i}+B_{y}^{i}\sigma_{y}^{i}+B_{z}^{i}\sigma_{z}^{i}).$
(17)

In Eq. (17), $B_{m}^{i}$ represent the randomly fluctuating noise terms; these
could even be environment operators. In order to eliminate these system-
environment interaction terms, we use the method of dynamically decoupling the
open quantum system by means of a control field [38]. Since our control field
acts along one of the axes, it is able to cancel out noise terms along two of
the three general degrees of freedom [39]. This is because $U_{c}(t)$ is a
rotation operator and causes the noise terms that are perpendicular to it to
rotate such that they average out. However, the noise term which is placed
along the same axis as $U_{c}(t)$ remains undisturbed. Examining this in more
detail, we note that in order for the periodic control field to decouple the
noise terms, we need [37, 38, 39, 40, 41]

$\int_{0}^{t_{c}}U_{c}^{\dagger}(t)H_{SB}U_{c}(t)dt=0.$ (18)

For the control field in the $z$ direction, we have
$U_{c}(t)=\exp(-i\sum_{i=1}^{N}g\frac{\sin(\omega t)}{\omega}\sigma_{z}^{i})$.
We then consider

$\displaystyle\int_{0}^{\frac{2\pi}{\omega}}e^{i\sum_{i=1}^{N}g\frac{\sin(\omega
t)}{\omega}\sigma_{z}^{i}}\sum_{i=1}^{N}B_{x}^{i}\sigma_{x}^{i}e^{-i\sum_{i=1}^{N}g\frac{\sin(\omega
t)}{\omega}\sigma_{z}^{i}}$
$\displaystyle+e^{i\sum_{i=1}^{N}g\frac{\sin(\omega
t)}{\omega}\sigma_{z}^{i}}\sum_{i=1}^{N}B_{y}^{i}\sigma_{y}^{i}e^{-i\sum_{i=1}^{N}g\frac{\sin(\omega
t)}{\omega}\sigma_{z}^{i}}dt.$

To solve this, we note that

$\displaystyle
e^{iC(t)\sum_{i=1}^{N}\sigma_{x}^{i}}\sigma_{y}^{j}e^{-iC(t)\sum_{i=1}^{N}\sigma_{x}^{i}}=$
$\displaystyle\sigma_{y}^{j}\cos(2C(t))-\sigma_{x}^{j}\sin(2C(t)).$

It is straightforward to see that integrating over $\sin{(2C(t))}$ yields zero
while integrating over $\cos{(2C(t))}$ yields the zeroth Bessel functions of
the first kind being divided by $\omega^{2}$. As the zeroth Bessel functions
are bounded by $1$, increasing $\omega$ leads to the integral approaching
zero. Having analytically satisfied the decoupling condition, we will model,
for the purposes of the ensuing numerical simulations, the noise as Ornstein-
Uhlenbeck processes keeping the mean ($\mu=0$), standard deviation
($\sigma=0.5$), and the correlation time ($\tau=0.005$) [45].

##  III Numerical Results

In this section, we present simulation results that help illustrate the
usefulness of the spin chain transformations from the point of view of quantum
information processing tasks. We present these results for the following
effectively transformed time-independent spin chain systems: (i) the isotropic
XY spin chain, (ii) the XXX spin chain, (iii) the rotated XXZ spin chain, and
(iv) the XY spin chain with next nearest neighbor interactions. For the first
case, we present the fidelity of state transfer as well entanglement
generation while for the rest of the cases, we present the concurrence of
entanglement generation between the ends of the spin chain, thereby showing
that one can effectively modulate the interaction couplings to enhance
entanglement generation.

###  III.1 The isotropic XY spin chain

####  III.1.1 Quantum State Transfer

We first consider the isotropic XY spin chain which we get as the effective
spin chain after applying a transverse field to the Ising spin chain,

$H_{I}(t)=\sum_{i}^{N-1}\frac{J_{i}}{2}[\sigma_{x}^{i}\sigma_{x}^{i+1}+\sigma_{y}^{i}\sigma_{y}^{i+1}].$
(19)

This requires that we set $A=0$ in Eq. (8). From the definition of
$A=\mathcal{J}_{0}(\frac{4g}{\omega})$, we observe that $A=0$ requires that
($\frac{4g}{\omega}$) equals the roots of the zeroeth Bessel function. We use
the first of these zeroeth roots in our simulations
($g=\frac{\omega}{4(2.4048)}$) for the corresponding time-dependent spin chain
system described by Eq. (2). Moreover, it is important to note that in
deriving the result in Eq. (19), we assumed that $\omega\gg J_{i}$. As such,
we use a rapidly oscillating field throughout the simulations. Having decided
the control field’s parameters, we initialize the spin chain sites in the
state
$\ket{\Psi(0)}=(\alpha\ket{0}_{z}+\beta\ket{1}_{z})\otimes\ket{1}_{z}\ldots\otimes\ket{1}_{z}$.
For quantum state transfer, after some time $\tau$, the state should evolve to
$\ket{\Psi(\tau)}=\ket{1}_{z}\otimes\ket{1}_{z}\ldots\otimes(\alpha\ket{0}_{z}+\beta
e^{i\phi}\ket{1}_{z})$, where $\phi$ is known and can be corrected at the end
of the state transfer [32]. We study the state transfer
$\ket{\Psi_{i}}=\ket{0}_{z}\otimes\ket{1}_{z}\ldots\otimes\ket{1}_{z}\rightarrow\ket{\Psi_{f}}=\ket{1}_{z}\otimes\ket{1}_{z}\ldots\otimes\ket{0}_{z}$;
by showing that we can effectively perform this state transfer, we can show
that we can perform state transfer for an arbitrary superposition state. To
quantitatively measure the state transfer, we find the fidelity,
$F(t)=\bra{\psi_{f}}\rho(t)\ket{\psi_{f}}$, where $\rho(t)$ is the spin
density matrix as a function of time.

For an isotropic XY spin chain, it has been shown that perfect fidelity of
state transfer can be achieved by engineering the coupling interaction $J_{i}$
as $\sqrt{i(N-i)}$ [27]. As such, we select the nearest neighbor coupling
terms as $J_{i}=\sqrt{i(N-i)}$ and simulate the fidelity plots corresponding
to Eq. (19) and its time-dependent equivalent. These are described the dashed
magenta curve and the smooth black curve respectively for $N=7$ in Fig. (1)
which are shown to overlap. In Fig. (1), the blue dotted curve is for the
fidelity of the Ising spin chain in the absence of any fields showing the
relative advantage of using the Ising spin chain with the control field.
Moreover, we have investigated that the fidelity for higher $N$ values further
falls for the Ising spin chain alone while it remains practically perfect in
the presence of the control field.

Figure 1: The smooth black curve shows the fidelity of state transfer for the
Ising spin chain in the presence of a transverse field for $N=7$. The dashed
magenta curve shows the fidelity in the time-effective XY spin chain while the
dotted blue curve is for fidelity in the Ising spin chain without any fields.
Here, we have set the neighboring coupling strengths as $J_{i}=\sqrt{i(N-i)}$.

Having verified the behaviors of the time-dependent as well as the
corresponding time-independent spin chain, we now introduce noise into the
system to see how closely the behavior of our noisy time-dependent spin chain
system matches with the effective noiseless case. As discussed before, we
consider the noise terms to be given by

$H_{SB}=\sum_{i=1}^{N}(B_{x}^{i}\sigma_{x}^{i}+B_{y}^{i}\sigma_{y}^{i}),$

since these noise terms can be removed with our control field. We regard the
noise terms $B_{m}^{i}$ as independent random variables that are found by
solving the Ornstein-Uhlenbeck equation

$dB_{m}=-\frac{(B_{m}-\mu)}{\tau}dt+\sigma\sqrt{\frac{2}{\tau}}dW,$

where $m=x,y$, while $\mu$ is the mean, $\sigma$ is the standard deviation,
$\tau$ is the correlation time, and $W$ is the standard Weiner process [45].
Here, $\mu$ and $\sigma$ have the same dimensions as the $B_{m}$. For all the
simulations dealing with noise in this paper, we have kept $\sigma=0.5$,
$\mu=0$, and $\tau=0.005$ while sampling the noise over $20$ trials.

As explained before, we expect that as a result of the decoupling condition
being met, the noise simulations should match closely with the ones without
noise. As such, the fidelity simulation for the Ising spin system with the
control field in the $z$ direction for $N=7$ is shown in Fig. (2) by the solid
black curve while the dashed magenta curve reveals the effect of noise.
Furthermore, we also plot the XY spin chain (no control fields) in the
presence of the same noisy environment and find that the fidelity for the XY
spin chain and noise dips more rapidly as compared to the Ising spin chain
with the transverse field in the presence of noise.

Figure 2: The smooth black curve here shows the fidelity of state transfer for
an Ising spin chain in the presence of a transverse field for $N=7$. The
dashed magenta curve is for the Ising spin chain with both the field and the
noise while the dotted blue curve is for the XY spin chain in the presence of
noise. Here, we have set the neighboring coupling strengths as
$J_{i}=\sqrt{i(N-i)}$.

As a digression, let us consider the case of the isotropic XY spin chain upon
which is incident a control field in the $z$ direction. Earlier, in Eq. (10),
we had shown that this does not transform the spin chain. Let us now
demonstrate this by simulating the fidelity of state transfer in such a
system. Setting $J_{x}^{i}=J_{y}^{i}=J_{i}=\sqrt{i(N-i)}$, we plot the
fidelity of state transfer for the XY spin chain in the presence of the
control field (the black curve) in Fig. (3); the dashed magenta curve shows
the effect of noise on the same system. As is the case for the XY spin chain
demonstrated in Fig. (2), the fidelity remains practically perfect, hence
verifying that the transformation has not affected the spin chain. Moreover,
the control field cancels out the effect of noise as we observe the magenta
curve to closely follow the the smooth black curve. Moreover, the dotted blue
curve in Fig. (3) is for just the XY spin chain in the presence of noise (that
is, no control fields) which we observe dips faster than the magenta curve.

Figure 3: The smooth black curve shows the fidelity of state transfer of the
XY spin chain in the presence of a control field along the $z$ direction for
$N=7$. The dashed magenta curve accounts for the XY spin chain with the field
in the $z$ direction in the presence of noise while the dotted blue curve is
for the initial XY spin chain with just noise. Here, as usual, we have set
$J_{i}=\sqrt{i(N-i)}$

####  III.1.2 Quantifying Entanglement Generation

In this section, we turn to quantify the generation of entanglement between
two spin sites in a spin chain by finding the concurrence between the two
spins. We do so for an initially bit flipped state
$\ket{\psi_{i}}=\ket{0}_{z}\otimes\ket{1}_{z}\ldots\otimes\ket{1}_{z}$. First,
in the time evolved spin density matrix, we take the trace over all the spins
other than the two being considered. After finding this density matrix,
$\rho^{\prime}(t)$, we find the concurrence by first finding

$R=\sqrt{\sqrt{\rho^{\prime}}\widetilde{\rho^{\prime}}\sqrt{\rho^{\prime}}},$

where

$\widetilde{\rho^{\prime}}=(\sigma_{y}\otimes\sigma_{y})\widetilde{\rho^{\prime}}^{*}(\sigma_{y}\otimes\sigma_{y}).$

The concurrence, $C(t)$ is then given by

$C(\rho)=\textrm{max}(0,\lambda_{1}-\lambda_{2}-\lambda_{3}-\lambda_{4}),$

where each of the $\lambda_{i}$ above are eigenvalues of $R$ in descending
order [32].

We first consider the concurrence between the first and second spins of the
Ising spin chain in the presence of a transverse control field as described by
Eq. (4). Results are shown in Fig. (4). The smooth black curve is for the
Ising spin chain in the presence of the transverse field while the dashed blue
curve is for the time-effective XY spin chain. The magenta curve shows the
effect of noise which we find to match closely, indicating that the effect of
the noise is largely mitigated by our control field. The dotted red curve is
for the Ising spin chain in the absence of any fields, and we see that this is
significantly lower. These results demonstrate the usefulness of using our
simple control field to engineer the spin chain Hamiltonian so as to generate
greater entanglement.

Figure 4: The smooth black curve here shows the concurrence between the first
and second spin of an Ising spin chain in the presence of a transverse field
for $N=7$ while the time effective XY spin chain is shown by the dashed blue
curve. The dot-dashed magenta curve shows the effect of noise while the dotted
red curve is for the Ising spin chain in the absence of fields. Here, we have
set the neighboring coupling strengths as $J_{i}=1$.

###  III.2 The XXX spin chain

Earlier, in Sec. (II.2), we saw that we can arrive at the XXX spin chain by
choosing $J_{i}^{x}=2$, $J_{i}^{y}=1$, and $A=0$ in Eq. (12) which we obtained
by applying a control field along the $y$ direction to the XY spin chain. The
choice of setting $A=0$ again implies that $\mathcal{J}(\frac{4g}{\omega})$
should be set to zero and that $\frac{4g}{\omega}$ equals the zeroth roots of
the Bessel function. Once again, we set $g=\frac{\omega}{4(2.4048)}$.

We now examine the generation of end to end entanglement in the XXX spin chain
for our initially bit flipped state. Figures (5) and (6) present the
concurrence of entanglement generation between the ends of the XXX spin chain
for $N=5$ and $N=8$ respectively. We see that the time-dependent system
closely matches with the our effective time-independent treatment since the
solid black curve and dashed blue curve coincide. The system is also able to
effectively cancel out noise terms as is shown by the dot-dashed magenta
curves. At the same time, we see that the concurrence of the initial XY spin
chain without any external fields is considerably less - this is demonstrated
by the dotted red curve in Figures (5) and (6). Once again, these results
demonstrate the usefulness of applying our simple control field.

Figure 5: The smooth black curve shows the concurrence of entanglement
generation between the ends of the XY spin chain in the presence of the
control field along the $y$ axis such that the neighboring coupling strength
are set as $J_{i}^{x}=2$ and $J_{i}^{y}=1$ for $N=5$, the dashed blue curve
shows the time effective XXX spin chain, the dot-dashed magenta curve shows
the effect of noise on the time-dependent system, while the red dotted curve
is for the system in the absence of any fields. Figure 6: Same as Fig. (5)
except for $N=8$.

###  III.3 The rotated XXZ spin chain

We now present results for the generation of entanglement between two ends of
a rotated XXZ spin chain as found earlier in Eq. (14). We recall that we
applied the control field along the $y$ axis to the XY spin chain and chose
our parameters such that $J_{x}^{i}=J_{y}^{i}=J_{i}=1$ and set $A=0$ in Eq.
(12). For this case too, we initialize the state as the bit-flipped state;
however, since we have the rotated XXZ spin chain, we now consider the initial
state to be
$\ket{\Psi_{i}}=\ket{0}_{y}\otimes\ket{1}_{y}\ldots\otimes\ket{1}_{y}$.
Moreover, as the control field is along the $y$ axis this time, it can only
decouple noise terms along the $x$ and $z$ directions. Hence, we to show the
usefulness of our control field in eliminating noise, we now consider
$H_{SB}=\sum_{i=1}^{N}(B_{x}^{i}\sigma_{x}^{i}+B_{z}^{i}\sigma_{z}^{i})$. We
have found significant entanglement generation between the ends of a rotated
XXZ spin chain by setting the neighboring interactions equal. This is
demonstrated in Figures (7) and (8) for $N=3$ and $N=7$ respectively where the
smooth black curve is for the XY spin chain in the presence of a control field
along the $y$ axis while the dashed blue curve is for the time-effective
rotated XXZ spin chain. These practically coincide. The magenta curve
demonstrating the effect of noise for the time-dependent system shows that the
control field significantly cancels out the noise terms. More so, the dot-
dashed red curve in Fig. (8) shows the nearly non-existent concurrence in the
XY spin chain without the field, which we have found this to be the case for
other values of $N$ too. We have investigated and found the same qualitative
behavior for increasing values of $N$.

Figure 7: The smooth black shows the concurrence between the first and last
spins of an XY spin chain in the presence of the control field along the $y$
axis for $N=3$. The dashed blue curve shows the effective rotated XXZ spin
chain while the magenta dot-dashed curve shows the effect of noise. Here, we
have taken $J_{x}^{i}=J_{y}^{i}=1$. Figure 8: Same as Fig. (7) except for
$N=8$ and $J_{i}=2$. The dot-dashed red curve tracing the horizontal axis is
for the XY spin chain without any fields.

###  III.4 Ising spin chain with nearest and next nearest neighbor
interactions

So far, we have modeled spin chains taking into account only the nearest
neighbor interactions. A more realistic model however would take into account
interactions beyond the nearest neighbor. As such, we take into account the
next nearest neighbor interactions and modify the Ising spin chain Hamiltonian
in the presence of a transverse sinusoidal field as

$H(t)=\sum_{i=1}^{N-1}J_{i}\sigma_{x}^{i}\sigma_{x}^{i+1}+\sum_{i=1}^{N-2}\frac{L_{i}}{N}\sigma_{x}^{i}\sigma_{x}^{i+2}\sum_{i=1}^{N}h(t)\sigma_{z}^{i},$
(20)

where, as before, $h(t)=g\cos(\omega t)$. Again, we transform Eq. (20) to the
interaction picture and find that it yields the time-effective Hamiltonian

$\displaystyle
H_{I}=\sum_{i}^{N-1}\frac{J_{i}}{2}[(A+1)\sigma_{x}^{i}\sigma_{x}^{i+1}-(A-1)\sigma_{y}^{i}\sigma_{y}^{i+1}]+$
$\displaystyle\sum_{i}^{N-2}\frac{L_{i}}{2N}[(A+1)\sigma_{x}^{i}\sigma_{x}^{i+2}-(A-1)\sigma_{y}^{i}\sigma_{y}^{i+2}].$
(21)

The result derived in Eq. (21) is analogous to the transformation of the Ising
spin chain with only nearest neighbor interactions in the presence of a
transverse field as shown earlier in Eq. (8) - we have transformed the Ising
spin chain with nearest and next nearest neighbor couplings to the completely
anisotropic XY spin chain with nearest and next nearest neighbor couplings. If
we choose $A=0$ in Eq. (21), we find that this yields the completely isotropic
XY spin chain with both nearest and next nearest neighbor interactions. Let us
then present results for the concurrence of the entanglement generated between
the ends of the XY spin chain with both nearest and next nearest neighbor
interactions. As usual, this time-independent spin chain corresponds to the
time dependent system in Eq. (20) with $g=\frac{\omega}{4(2.4048)}$ for the
time-dependent system. Results for the concurrence of end to end entanglement
generation for $N=9$ are presented in Fig. (9). Here the smooth black curve
shows the concurrence for the time-independent effective spin chain in Eq.
(21) while the dashed blue curve tracing the $x$ axis is for the Ising spin
chain with nearest and next nearest without any fields. Once again, the
drastic advantage of our control field is evident.

Figure 9: The smooth black curve shows the concurrence for the entanglement
generated between the ends the Ising spin chain with next nearest neighbor
interactions in the presence of the control field along the $z$ direction for
$N=9$ while the dashed blue curve is for the system without any fields. Here,
we have set $J_{i}=L_{i}=1.$

##  IV Conclusion

In summary, we have applied a continuous driving field to spin chains to
effectively generate additional spin-spin interactions. These interactions can
be tuned; in particular, interaction strengths depend on Bessel functions of
the first kind, and can be physically changed by varying the strength of the
driving field. In particular, we have considered how we can obtain the
isotropic XY spin chain from the quantum Ising model, as well as the XXX and
XXZ spin chains from the XY model. We have also briefly extended our results
to include next-nearest neighbor interactions. Moreover, our control field has
been shown to also mitigate the effects of noise on the spin chain. The
particular cases we have analyzed yield interesting results from the point of
view of quantum information processing tasks. While the isotropic XY spin
chain yields perfect state transfer, the other time effective spin chains help
us generate significantly high entanglement as compared to the case where no
control field is applied.

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