Dark states in an integrable $XY$ central spin model

Jaco van Tonder^∗, Jon Links School of Mathematics and Physics, The University of Queensland, 4072, Australia $^*$w.vantonder@student.uq.edu.au

Abstract

Eigenstates of central spin models in which the central spin is unentangled with the environment are known as dark states. They have recently been observed in a class of integrable $XX$ models. Here we find that dark states are present in $XY$ models, but only for particular configurations of the central spin magnetic field. We show this via an explicit construction of the Bethe states.

1 Introduction

Central spin models have gained renewed attention due to their possible applications in modern quantum technologies focused on quantum sensing and metrology [16]. This is due to their integrability allowing high-fidelity control of these systems at a mesoscopic scale, where the exponentially increasing size of the Hilbert space would in general be prohibitive [6]. The central spin allows the dynamics of the spin bath to be monitored, and for feedback to be used to steer the dynamics of the bath in a desired direction. There are several physical or engineered systems for which central spin models provide a theoretical model. These include nitrogen vacancy (NV) centres in diamonds [23], room temperature quantum memory storage [18, 21, 22], quantum batteries [11] and quantum dots [2, 19] – an active area of research which gained public attention through the 2023 Nobel prize for Chemistry [13].

On the theoretical side, these models have been known for their integrability since Gaudin’s seminal paper [8]. Integrability allows for the analytic solution of the eigenstates and eigenspectrum of the models through the Bethe Ansatz. This makes these models well suited for studies of their equilibrium and dynamical behaviour; in particular, tests of the Eigenstate Thermalisation Hypothesis (ETH) and investigation of many-body localisation [1, 12, 17, 22]. Integrability is also essential for perturbative solutions of physical models. For example, integrability provides a wavefunction Ansatz in variational eigensolvers to model strong electron correlation in quantum chemistry [7].

Integrability is also of great interest in mathematical physics. There is ongoing work on extending or modifying known integrable models to obtain new ones. This has recently led to establishing that the $XX$ model, which models certain resonant dipolar spin systems, is integrable [5, 21]. Integrability was subsequently extended to the $XY$ model [20] using conserved charges discussed in [14], and the same charges with self-interaction obtained by Skrypnyk [15]. These charges were diagonalised using several modifications of the standard algebraic Bethe Ansatz for arbitrary spin [15]. These expressions for the eigenvalues can also be obtained through the use of functional relations as described in [20].

Here we derive the $XY$ central spin model Hamiltonian and its charges through a limit of the charges for the $XYZ$ Gaudin model in [15]. We also derive a second integrable class of $XY$ central spin models, which are seen to simply be a different parametrisation of the original model. With this reparametrisation we obtain regularised eigenvalue and eigenstate results for the central spin model. We show that these reproduce the diagonalisation results in the literature [3, 21] in the appropriate limits. It is found that dark states, states in which the central spin is not entangled with the environment, are present for special configurations of the magnetic field $\vec{B}=(B^{x},B^{y},B^{z})$ . This is reminiscent of the $XX$ model for which these occur for an out-of-plane magnetic field, $B^{x}=B^{y}=0$ [5, 21].

In Sect. 2 we derive two examples of $XY$ central spin Hamiltonians and their conserved charges for spin- $1/2$ central spin. We then show how these are different parametrisations of the same integrable model. After performing a reparametrisation which regularises the eigenstate results of [15], we find the eigenstates for the $XY$ central spin model in Sect. 3. We show that for special magnetic field configurations dark states occur, in analogy to those that were seen in the out-of-plane $XX$ model [21]. Concluding remarks are given in Sect. 4. In the Appendix 5 we take the isotropic limit to confirm that these recover the results for the $XX$ model. This is supplemented with some heuristic arguments and numerical results about how the dark states emerge from the states with generic magnetic field configurations.

2 Derivation of integrable $XY$ central spin Hamiltonians

Consider a set of $L+1$ spin operators $\{S^{x}_{j},\,S_{j}^{y},\,S^{z}_{j}:j=0,1,\dots,L\}$ satisfying the standard canonical commutation relations

\displaystyle[S^{\theta}_{j},\,S^{\eta}_{k}]=i\delta_{jk}\sum_{\kappa\in\{x,y,z\}}\varepsilon^{\theta\eta\kappa}S^{\kappa}_{j},

(1)

where $i=\sqrt{-1}$ and $\varepsilon^{\theta\eta\kappa}$ is the Levi-Civita symbol. Introduce a set of distinct parameters $\{\beta\}\cup\{\epsilon_{j}:j=0,1,\dots,L\}$ with $\beta>0$ and such that $\epsilon_{j}-\beta>0$ for all $j=0,1,\dots,L$ . Also define $f_{j}^{\pm}=\sqrt{\epsilon_{j}\pm\beta}$ .

Integrable central spin models, with spin-1/2 central spin and arbitrary bath spins, are derived from the following $L+1$ conserved charges ¹¹1We use $\zeta$ instead of $B^{z}$ to prevent confusion with $B^{z}$ of the $XY$ Hamiltonian, Eq. (7) below.

	$\displaystyle\mathcal{Q}_{i}=\zeta S^{z}_{i}+\frac{B^{x}}{f_{i}^{+}}S^{x}_{i}+\frac{B^{y}}{f_{i}^{-}}S^{y}_{i}+\frac{f_{i}^{-}}{f_{i}^{+}}S^{x}_{i}S^{x}_{i}+\frac{f_{i}^{+}}{f_{i}^{-}}S^{y}_{i}S^{y}_{i}$
	$\displaystyle\quad\quad\quad\quad+2\sum_{j\neq i}^{L}\frac{1}{\epsilon_{i}-\epsilon_{j}}(f_{i}^{+}f_{j}^{-}S^{x}_{i}S^{x}_{j}+f_{i}^{-}f_{j}^{+}S^{y}_{i}S^{y}_{j}+f_{j}^{+}f_{j}^{-}S^{z}_{i}S^{z}_{j}),$		(2)
	$\displaystyle[\mathcal{Q}_{i},\mathcal{Q}_{j}]=0,\quad\forall{i,j}\in\{0,1,\dots,L\}.$

To do this, we take $\{S^{x}_{0},\,S_{0}^{y},\,S^{z}_{0}\}$ to be the spin- $1/2$ operators, identify the zero subscript with the central spin and the other subscripts with the bath spins. The zeroth charge $\mathcal{Q}_{0}$ will then become the central spin Hamiltonian and ${\mathcal{Q}}_{j}$ , $j=1,\dots,L$ its conserved charges. Noting that for spin-1/2 operators we have

\displaystyle\left(S^{x}_{0}\right)^{2}=\left(S^{y}_{0}\right)^{2}=\frac{1}{4}I,

it is convenient for the zeroth charge to subtract off the squared spin operator terms. After making the change of variables $\{\zeta\mapsto B^{z}/\sqrt{\epsilon_{0}}\}$ for the magnetic field, we obtain the first class of XY central spin models through the limit as $\epsilon_{0}\to\infty$ with the Hamiltonian being

	$\displaystyle H^{\rm I}$	$\displaystyle=\lim\limits_{\epsilon_{0}\to\infty}\left[\sqrt{\epsilon_{0}}\mathcal{Q}_{0}-\frac{\sqrt{\epsilon_{0}}}{4}\left(\frac{f^{-}_{0}}{f^{+}_{0}}+\frac{f^{+}_{0}}{f^{-}_{0}}\right)\right]$			(3)
		$\displaystyle=B^{x}S^{x}_{0}+B^{y}S^{y}_{0}+B^{z}S^{z}_{0}+2\sum_{j=1}^{L}(f_{j}^{-}S^{x}_{0}S^{x}_{j}+f_{j}^{+}S^{y}_{0}S^{y}_{j}),$			(3)

and the charges

$\displaystyle Q_{i}^{\rm I}$	$\displaystyle=\lim\limits_{\epsilon_{0}\to\infty}\mathcal{Q}_{i}$	(4)
	$\displaystyle=-2S_{0}^{z}S_{i}^{z}+\frac{B^{x}}{f_{i}^{+}}S_{i}^{x}+\frac{B^{y}}{f_{i}^{-}}S^{y}_{i}+\frac{f_{i}^{-}}{f_{i}^{+}}S_{i}^{x}S_{i}^{x}+\frac{f_{i}^{+}}{f_{i}^{-}}S_{i}^{y}S_{i}^{y}$
	$\displaystyle\quad+2\sum_{\stackrel{{\scriptstyle[}}{{j}}\neq i]{}{j=1}}^{L}\frac{1}{\epsilon_{i}-\epsilon_{j}}(f_{i}^{+}f_{j}^{-}S^{x}_{i}S^{x}_{j}+f_{i}^{-}f_{j}^{+}S^{y}_{i}S^{y}_{j}+f_{j}^{+}f_{j}^{-}S^{z}_{i}S^{z}_{j}),$

reproducing those introduced in [20].

For the second class of Hamiltonians we use the variables $\{\tilde{B^{x}},\tilde{\zeta},\tilde{B^{z}}\}$ for the magnetic field and $\{\phi_{j}\}$ instead of $\{\epsilon_{j}\}$ , letting $h^{\pm}_{j}=\sqrt{\phi_{j}\pm\beta}$ . We make a change of variables for the magnetic field, $\{\tilde{B^{x}}\mapsto-f_{0}^{+}\tilde{B^{x}},\tilde{\zeta}\mapsto-f_{0}^{-}\tilde{B^{y}},\tilde{B^{z}}\mapsto-\tilde{B^{z}}\}$ . Taking the limit as $\phi_{0}\to\beta$ leads to the Hamiltonian

	$\displaystyle H^{\rm II}$	$\displaystyle=\lim\limits_{\epsilon_{0}\to\beta}\left[-\mathcal{Q}_{0}+\frac{1}{4}\left(\frac{h^{-}_{0}}{h^{+}_{0}}+\frac{h^{+}_{0}}{h^{-}_{0}}\right)\right]$			(5)
		$\displaystyle=\tilde{B^{x}}S^{x}_{0}+\tilde{B^{y}}S^{y}_{0}+\tilde{B^{z}}S^{z}_{0}+2\sum_{j=1}^{L}\left(\frac{\sqrt{2\beta}}{h_{j}^{-}}S^{x}_{0}S^{x}_{j}+\frac{h^{+}_{j}}{h^{-}_{j}}S^{z}_{0}S^{z}_{j}\right),$			(5)

and the charges

$\displaystyle Q^{\rm{II}}_{i}$	$\displaystyle=\lim\limits_{\epsilon_{0}\to\beta}\left[-\mathcal{Q}_{i}\right]$	(6)
	$\displaystyle=-2\frac{\sqrt{2\beta}}{h_{i}^{-}}S^{y}_{0}S^{y}_{i}+\tilde{B^{z}}S^{z}_{i}+\tilde{B^{x}}\frac{\sqrt{2\beta}}{h_{i}^{+}}S_{i}^{x}-\frac{h_{i}^{-}}{h_{i}^{+}}S_{i}^{x}S_{i}^{x}-\frac{h_{i}^{+}}{h_{i}^{-}}S_{i}^{y}S_{i}^{y}$
	$\displaystyle\quad-2\sum_{\stackrel{{\scriptstyle[}}{{j}}\neq i]{}{j=1}}^{L}\frac{1}{\phi_{i}-\phi_{j}}(h_{i}^{+}h_{j}^{-}S^{x}_{i}S^{x}_{j}+h_{i}^{-}h_{j}^{+}S^{y}_{i}S^{y}_{j}+h_{j}^{+}h_{j}^{-}S^{z}_{i}S^{z}_{j})$
	$\displaystyle=-2\frac{\sqrt{2\beta}}{h_{i}^{-}}S^{y}_{0}S^{y}_{i}+\tilde{B^{x}}S^{x}_{i}+\tilde{B^{y}}\frac{\sqrt{2\beta}}{h_{i}^{+}}S_{i}^{y}+\frac{2\beta}{h_{i}^{+}h_{i}^{-}}S_{i}^{y}S_{i}^{y}+\frac{h_{i}^{+}}{h_{i}^{-}}S_{i}^{z}S_{i}^{z}$
	$\displaystyle\quad-2\sum_{\stackrel{{\scriptstyle[}}{{j}}\neq i]{}{j=1}}^{L}\frac{1}{\phi_{i}-\phi_{j}}(h_{i}^{+}h_{j}^{-}S^{x}_{i}S^{x}_{j}+h_{i}^{-}h_{j}^{+}S^{y}_{i}S^{y}_{j}+h_{j}^{+}h_{j}^{-}S^{z}_{i}S^{z}_{j})$

where the last equality is obtained through a shift by $\mathbf{S}_{i}^{2}=(S^{x}_{i})^{2}+(S^{y}_{i})^{2}+(S^{z}_{i})^{2}$ . Note that for this class the limit as $\beta\to 0$ is an $X$ model (after making the transformation $S^{x}\mapsto-S^{z},\,S^{z}\mapsto S^{x}$ ) in contrast to the first class.

These two classes can be seen as different parametrisations of the parameters $\{\epsilon_{j}\}$ and the magnetic field components. The integrable $XY$ central spin Hamiltonian

\displaystyle H_{XY}=B^{x}S_{0}^{x}+B^{y}S_{0}^{y}+B^{z}S_{0}^{z}+2\sum_{j=1}^{L}(X_{j}S_{0}^{x}S_{j}^{x}+Y_{j}S_{0}^{y}S_{j}^{y})

(7)

was seen in [20] to be characterised by $Y_{j}^{2}-X_{j}^{2}=c$ , with the constant $c$ being free to be chosen through rescalings of the parameters $\{\epsilon_{j}\}$ and magnetic field components. The isotropic case where $c=0$ corresponds to the $XX$ model, as seen for example in [21]. In the above two cases $c$ is seen to be respectively $2\beta$ and $-1$ . The reparametrisation relating the two classes is

\displaystyle\phi_{j}=\beta+\frac{4\beta^{2}}{\epsilon_{j}-\beta}.

Specifically, the charges (2) in the variables $\{\phi_{j}\}$ and $\{\tilde{B^{x}},\tilde{\zeta},\tilde{B^{z}}\}$ , under this reparametrisation take the form (after making the rotation $S^{y}\to S^{z},\;S^{z}\to-S^{y}$ and adding $\mathbf{S}_{i}^{2}$ )

$\displaystyle\tilde{\mathcal{Q}}_{i}$	$\displaystyle=-\frac{f_{i}^{-}}{\sqrt{2\beta}}\bigg{\{}-\frac{f_{i}^{+}}{f_{i}^{-}}s_{i}(s_{i}+1)-\frac{\tilde{B^{y}}}{\sqrt{2\beta}}S^{z}_{i}-\frac{\tilde{B^{x}}}{f^{+}_{i}}S^{x}_{i}+\tilde{\zeta}\frac{\sqrt{2\beta}}{f^{-}_{i}}S^{y}_{i}+\frac{f_{i}^{-}}{f_{i}^{+}}S_{i}^{x}S_{i}^{x}$	(8)
	$\displaystyle\quad+\frac{f_{i}^{+}}{f_{i}^{-}}S_{i}^{y}S_{i}^{y}+2\sum_{j\neq i}^{L}\frac{1}{\epsilon_{i}-\epsilon_{j}}\left(f^{+}_{i}f^{-}_{j}S^{x}_{i}S^{x}_{j}+f^{-}_{i}f^{+}_{j}S^{y}_{i}S^{y}_{j}+f^{+}_{j}f^{-}_{j}S^{z}_{i}S^{z}_{j}\right)\bigg{\}}$
	$\displaystyle=-\frac{f_{i}^{-}}{\sqrt{2\beta}}\left(\mathcal{Q}_{i}-\frac{f_{i}^{+}}{f_{i}^{-}}s_{i}(s_{i}+1)\right)$

where

	$\displaystyle\mathcal{Q}_{i}=\zeta S^{z}_{i}+\frac{B^{x}}{f^{+}_{i}}S^{x}_{i}+\frac{B^{y}}{f^{-}_{i}}S^{y}_{i}+\frac{f_{i}^{-}}{f_{i}^{+}}S_{i}^{x}S_{i}^{x}+\frac{f_{i}^{+}}{f_{i}^{-}}S_{i}^{y}S_{i}^{y}$
	$\displaystyle\qquad+2\sum_{j\neq i}^{L}\frac{1}{\epsilon_{i}-\epsilon_{j}}\left(f^{+}_{i}f^{-}_{j}S^{x}_{i}S^{x}_{j}+f^{-}_{i}f^{+}_{j}S^{y}_{i}S^{y}_{j}+f^{+}_{j}f^{-}_{j}S^{z}_{i}S^{z}_{j}\right)$

with the redefined magnetic field components

\displaystyle\tilde{B^{x}}=-B^{x},\quad\tilde{\zeta}=-\zeta\sqrt{2\beta},\quad\tilde{B^{z}}=B^{y}/\sqrt{2\beta}.

Henceforth we will only consider the parametrisation given in the first class and take $H=H^{\rm I},\;Q_{j}=Q^{\rm I}_{j}$ . However, we will see that the second parametrisation will be useful for obtaining the eigenstates of the $XY$ model.

3 The eigenstates

The eigenstates of the charges $\mathcal{Q}_{j}$ , as well as their eigenvalues, have been found in [15]. Here we recall the algebraic setup and the results required in order to obtain the eigenstates and eigenvalues of $H$ and its conserved charges $Q_{j}$ .

3.1 Relations for the Algebraic Bethe Ansatz

Introduce the Lax algebra generators (respectively corresponding to $B(w)$ , $C(w)$ and $A(w)-D(w)$ in [15]) ²²2Again using $\tilde{\zeta}$ to prevent confusion with $\tilde{B^{y}}$ of the $XY$ model.

	$\displaystyle S^{-}(w)$	$\displaystyle=\frac{-\tilde{B^{x}}+i\tilde{\zeta}}{\sqrt{2\beta}}-i\tilde{B^{z}}+2\sum_{j=0}^{L}\frac{h_{j}^{+}h_{j}^{-}}{w-\phi_{j}}T_{j}^{12}-i\sum_{j=0}^{L}\left(T_{j}^{11}-T_{j}^{22}\right),$
	$\displaystyle S^{+}(w)$	$\displaystyle=\frac{-\tilde{B^{x}}+i\tilde{\zeta}}{\sqrt{2\beta}}+i\tilde{B^{z}}+2\sum_{j=0}^{L}\frac{h_{j}^{+}h_{j}^{-}}{w-\phi_{j}}T_{j}^{21}-i\sum_{j=0}^{L}\left(T_{j}^{11}-T_{j}^{22}\right),$
	$\displaystyle S^{z}(w)$	$\displaystyle=-\frac{2i\tilde{B^{x}}}{\sqrt{2\beta}}(w-\beta)-\frac{2\tilde{\zeta}}{\sqrt{2\beta}}(w+\beta)+2\sum_{j=0}^{L}\frac{w^{2}-\beta^{2}}{w-\phi_{j}}\left(T_{j}^{11}-T_{j}^{22}\right)$

where $h_{j}^{\pm}=\sqrt{\phi_{j}\pm\beta}$ and the operators $\{T_{k}^{ij}\}$ satisfy the $\mathfrak{gl}(2)$ commutation relations

\displaystyle[T_{a}^{ij},T_{b}^{kl}]=\delta_{ab}(\delta^{kj}T_{a}^{il}-\delta^{il}T_{a}^{kj});\quad\quad a,b=1,\dots,L.

The Lax algebra generators satisfy the relations

$\displaystyle{[}S^{\mp}(u),S^{\mp}(v){]}$	$\displaystyle=\pm 2i\left(S^{\mp}(u)-S^{\mp}(v)\right),$
$\displaystyle{[}S^{z}(u),S^{\mp}(v){]}$	$\displaystyle=\mp 4\frac{u^{2}-\beta^{2}}{u-v}\left(S^{\mp}(u)-S^{\mp}(v)\right),$
$\displaystyle{[}S^{z}(u),S^{z}(v){]}$	$\displaystyle=0,$
$\displaystyle{[}S^{+}(u),S^{-}(v){]}$	$\displaystyle=\frac{2}{u-v}\left(S^{z}(u)-S^{z}(v)\right)-2iS^{-}(v)-2iS^{+}(u).$	(9)

We also define operators shifted by a constant term, necessary to build the appropriate Bethe states,

\displaystyle S^{\mp}_{k}(w)=S^{\mp}(w)\mp(2k-1)iI,\quad k\in\mathbb{Z}

and with these define

\displaystyle\mathcal{S}^{\mp}(w_{1},w_{2},\dots,w_{M})=S^{\mp}_{1}(w_{1})S^{\mp}_{2}(w_{2})\cdots S^{\mp}_{M}(w_{M}).

(10)

Now in order to describe the eigenstates of the charges $\{\mathcal{Q}_{j}\}$ we use the following representation of the $\mathfrak{gl}(2)$ operators

	$\displaystyle T_{k}^{11}$	$\displaystyle=-i\frac{h_{k}^{-}}{\sqrt{2\beta}}S_{k}^{x}-\frac{h_{k}^{+}}{\sqrt{2\beta}}S_{k}^{y},\quad T_{k}^{22}=i\frac{h_{k}^{-}}{\sqrt{2\beta}}S_{k}^{x}+\frac{h_{k}^{+}}{\sqrt{2\beta}}S_{k}^{y},$
	$\displaystyle T_{k}^{21}$	$\displaystyle=iS_{k}^{z}-\frac{h_{k}^{+}}{\sqrt{2\beta}}S_{k}^{x}+i\frac{h_{k}^{-}}{\sqrt{2\beta}}S_{k}^{y},\quad T_{k}^{12}=-iS_{k}^{z}-\frac{h_{k}^{+}}{\sqrt{2\beta}}S_{k}^{x}+i\frac{h_{k}^{-}}{\sqrt{2\beta}}S_{k}^{y}.$

Following the notation of [15] we also introduce the reference states $v^{\pm}$ defined as

\displaystyle v^{\pm}

\displaystyle=v_{0}^{\pm}\otimes v_{1}^{\pm}\otimes\cdots\otimes v_{L}^{\pm}

(11)

where the $v_{k}^{\pm}$ are respectively lowest and highest weight states satisfying

	$\displaystyle T_{k}^{11}v_{k}^{+}=-s_{k}v_{k}^{+},\;$	$\displaystyle T_{k}^{22}v_{k}^{+}=s_{k}v_{k}^{+},\;$	$\displaystyle T_{k}^{21}v_{k}^{+}=0,$
	$\displaystyle T_{k}^{11}v_{k}^{-}=s_{k}v_{k}^{-},\;$	$\displaystyle T_{k}^{22}v_{k}^{-}=-s_{k}v_{k}^{-},\;$	$\displaystyle T_{k}^{12}v_{k}^{-}=0$

for half non-negative integers $s_{k}$ , i.e. $s_{k}\in\{0,1/2,1,3/2,\dots\}$ . For the spin labelled by the 0 subscript, for which $s_{0}=1/2$ , we use the following notational convention

\ket{+}_{0}=v^{-}_{0},\quad\ket{-}_{0}=v^{+}_{0}.

For conciseness we will also use the notation

v_{B}^{\pm}=v_{1}^{\pm}\otimes\cdots\otimes v_{L}^{\pm}

when identifying spins 1 to $L$ as bath spins.

3.1.1 Reparametrised operators.

In the following we will also need the expressions of the operators in the reparametrisation introduced in Sect. 2

		$\displaystyle\epsilon_{j}=\beta+\frac{4\beta^{2}}{\phi_{j}-\beta},\quad u=\beta+\frac{4\beta^{2}}{w-\beta},$			(12)
		$\displaystyle B^{x}=-\tilde{B^{x}},\quad\zeta=-\tilde{\zeta}/\sqrt{2\beta},\quad B^{y}=-\tilde{B^{z}}\sqrt{2\beta}.$			(12)

These being (after also rescaling by $\sqrt{2\beta}$ )

	$\displaystyle S^{-}(u)$	$\displaystyle=B^{x}-iB^{y}-i\zeta\sqrt{2\beta}+2\sum_{j=0}^{L}\frac{f^{+}_{j}(u-\beta)}{\epsilon_{j}-u}T_{j}^{12}-i\sqrt{2\beta}\sum_{j=0}^{L}\left(T_{j}^{11}-T_{j}^{22}\right),$
	$\displaystyle S^{+}(u)$	$\displaystyle=B^{x}+iB^{y}-i\zeta\sqrt{2\beta}+2\sum_{j=0}^{L}\frac{f^{+}_{j}(u-\beta)}{\epsilon_{j}-u}T_{j}^{21}-i\sqrt{2\beta}\sum_{j=0}^{L}\left(T_{j}^{11}-T_{j}^{22}\right)$

with

	$\displaystyle T_{k}^{11}$	$\displaystyle=-i\frac{\sqrt{2\beta}}{f_{k}^{-}}S_{k}^{x}-\frac{f_{k}^{+}}{f_{k}^{-}}S_{k}^{z},\quad T_{k}^{22}=i\frac{\sqrt{2\beta}}{f_{k}^{-}}S_{k}^{x}+\frac{f_{k}^{+}}{f_{k}^{-}}S_{k}^{z},$
	$\displaystyle T_{k}^{21}$	$\displaystyle=-iS_{k}^{y}-\frac{f_{k}^{+}}{f_{k}^{-}}S_{k}^{x}+i\frac{\sqrt{2\beta}}{f_{k}^{-}}S_{k}^{z},\quad T_{k}^{12}=iS_{k}^{y}-\frac{f_{k}^{+}}{f_{k}^{-}}S_{k}^{x}+i\frac{\sqrt{2\beta}}{f_{k}^{-}}S_{k}^{z}.$

We define $\mathcal{S}^{\pm}(u_{1},\dots,u_{M})$ similarly to (10) using instead

\displaystyle S^{\mp}_{k}(u)=S^{\mp}(u)\mp(2k-1)i\sqrt{2\beta}I,\quad k\in\mathbb{Z}.

3.2 Eigenstates of $H_{XY}$

To fully capture the possible types of eigenstates of the central spin model we first give the construction of the states via the supersymmetry of the model, before obtaining the states for generic magnetic field configurations from the limits of Sect. 2.

3.2.1 Construction from supersymmetry.

Define the spin raising and lowering operators $S_{0}^{\pm}=S_{0}^{x}\pm iS_{0}^{y}$ , and supercharges $\mathcal{A}^{\pm}=S_{0}^{\mp}A^{\pm}$ , where

\displaystyle A^{\pm}=(B^{x}\pm iB^{y})I/2+\sum_{j=1}^{L}(f_{j}^{-}S^{x}_{j}\pm if_{j}^{+}S^{y}_{j}).

(13)

The supercharges satisfy $(\mathcal{A}^{\pm})^{2}=0$ and $S_{0}^{z}\mathcal{A}^{\pm}=-\mathcal{A}^{\pm}S_{0}^{z}$ . Recall from [20] that the Hamiltonian (3) can be written as

\displaystyle H

\displaystyle=B^{z}S_{0}^{z}+\mathcal{A}^{+}+\mathcal{A}^{-}.

(14)

With this one sees that $H^{2}$ is related to a supersymmetric Hamiltonian through

	$\displaystyle\left(H\right)^{2}$	$\displaystyle=(B^{z})^{2}I/4+\mathcal{A}^{+}\mathcal{A}^{-}+\mathcal{A}^{-}\mathcal{A}^{+}$
		$\displaystyle=((B^{x})^{2}+(B^{y})^{2}+(B^{z})^{2})I/4+\sum_{j=1}^{L}f_{j}^{+}f_{j}^{-}Q_{j}.$

Rewriting this in the eigenbasis of $S^{z}_{0}$

	$\displaystyle H^{2}$	$\displaystyle=(B^{z})^{2}/4+(I/2-S^{z}_{0})A^{+}A^{-}+(I/2+S^{z}_{0})A^{-}A^{+}$
		$\displaystyle=\pmatrix{(B^{z})^{2}/4+A^{-}A^{+}&0\cr 0&(B^{z})^{2}/4+A^{+}A^{-}}$
		$\displaystyle=\frac{\|\vec{B}\|^{2}}{4}I+\pmatrix{\sum_{j=1}^{L}f_{j}^{+}f_{j}^{-}\mathcal{Q}_{j}\|_{\zeta=-1}&0\cr 0&\sum_{j=1}^{L}f_{j}^{+}f_{j}^{-}\mathcal{Q}_{j}\|_{\zeta=+1}}$

shows that $H^{2}$ can be diagonalised with eigenstates of the form

\ket{\Psi_{\Theta}}=\ket{\Theta}_{0}\otimes\ket{\psi_{\Theta}}_{B},\quad\Theta\in\{+,-\}

where the subscript $0$ denotes a central spin state and $B$ a bath-spin state. Observe that the the states $\ket{\psi_{\Theta}}_{B}$ are respectively eigenstates of the charges $\mathcal{Q}_{j}|_{\zeta=-\Theta}$ . Due to the supersymmetry, the eigenstates of $H^{2}$ are either singlets or doublets. The singlets $\ket{\Psi_{\Theta}}$ ,

\mathcal{A}^{+}\ket{\Psi_{\Theta}}=\mathcal{A}^{-}\ket{\Psi_{\Theta}}=0,

are also eigenstates of $H$ . Whereas for the doublets $\ket{\Psi_{\Theta}}$ ,

\displaystyle\mathcal{A}^{+}\ket{\Psi_{+}}=\Lambda\ket{\Psi_{-}},\quad\mathcal{A}^{-}\ket{\Psi_{-}}=\Lambda\ket{\Psi_{+}},\quad\Lambda\in\mathbb{R}

we can diagonalise $H$ to obtain the eigenstates

	$\displaystyle\ket{E_{\Theta}}$	$\displaystyle=\Lambda\ket{\Psi_{+}}+\left(E_{\Theta}-B^{z}/2\right)\ket{\Psi_{-}}$
		$\displaystyle=\left(\mathcal{A}^{-}+\left(E_{\Theta}-B^{z}/2\right)\right)\ket{\Psi_{-}}$
		$\displaystyle=\left(A^{-}+\left(E_{\Theta}-B^{z}/2\right)S_{0}^{-}\right)\ket{+}_{0}\otimes\ket{\psi_{-}}_{B}$

where $E_{\Theta}=\Theta\sqrt{(B^{z})^{2}/4+\Lambda^{2}}$ .

From the above construction it is clear that we can obtain the eigenstates of the $XY$ model from those of $\mathcal{Q}_{j}|_{\zeta=-\Theta}$ or equivalently, due to simultaneous diagonalisability, respectively from $A^{-\Theta}A^{\Theta}$ . To do so we observe that with the inverted reparametrisation introduced in Sect. 2

\displaystyle\epsilon_{j}=\beta+\frac{4\beta^{2}}{\phi_{j}-\beta},\quad B^{x}=-\tilde{B^{x}},\quad\zeta=-\tilde{\zeta}/\sqrt{2\beta},\quad B^{y}=\tilde{B^{z}}\sqrt{2\beta}

we have expressions in terms of the Lax algebra generators as

\displaystyle A^{-\Theta}A^{\Theta}=(\beta/2)S^{-\Theta}_{1}(\beta)S^{\Theta}_{0}(\beta),

noting that $\tilde{\zeta}=\Theta\sqrt{2\beta}$ .

3.2.2 Dark states.

Here we will demonstrate how dark states can be identified from the Bethe state construction. Assuming the roots $\{w_{m}\}_{m=1}^{M}$ are distinct and not equal to $\beta$ we find using the algebraic relations (3.1)

	$\displaystyle S^{-\Theta}_{1}(\beta)S^{\Theta}_{0}(\beta)$	$\displaystyle\mathcal{S}^{-\Theta}(w_{1},\dots,w_{M})v^{\Theta}=\mathcal{S}^{-\Theta}(\beta,w_{1},\dots,w_{M})S^{\Theta}_{-M}(\beta)v^{\Theta}$
		$\displaystyle+\sum_{l=1}^{M}\mathcal{S}^{-\Theta}(w_{1},\dots,w_{l}\to\beta,\dots,w_{M})$
		$\displaystyle\times\left\{8I+2\Theta\frac{S^{z}(w_{l})-S^{z}(\beta)}{w_{l}-\beta}+8\sum_{m\neq l}^{M}\frac{w_{l}+\beta}{w_{l}-w_{m}}I\right\}v^{\Theta}.$

Noting that

\displaystyle S^{\Theta}_{-M}(\beta)v^{\Theta}=\Theta\left(\left(N_{s}+\frac{\sqrt{2\beta}+\Theta i\tilde{B^{x}}}{\sqrt{2\beta}}+\tilde{B^{z}}\right)i-(2M+1)i\right)v^{\Theta},

where

\displaystyle N_{s}=2\sum_{j=1}^{L}s_{j},

we see that the Bethe state will be an eigenstate with zero eigenvalue if ³³3See Sect. 5.1 of the Appendix for the full argument that these are the only magnetic field configurations admitting dark states.

\displaystyle N_{s}/2+\frac{i\Theta\tilde{B^{x}}/2}{\sqrt{2\beta}}+\tilde{B^{z}}/2=M\in\{0,1,\dots,N_{s}\}

and the “unwanted terms” vanish. This corresponds to the roots $\{w_{m}\}$ satisfying

\displaystyle 4(w_{l}-\beta)-\Theta\left(z_{\Theta}(w_{l})-z_{\Theta}(\beta)\right)+4\sum_{m\neq l}^{M}\frac{w_{l}^{2}-\beta^{2}}{w_{l}-w_{m}}=0,

where $z_{\Theta}(w)$ are the eigenvalues of $S^{z}(w)$ for the reference states $v^{\Theta}$ .

Due to these being zero eigenvalue states of $A^{-\Theta}A^{\Theta}$ , they correspond to the singlet states. Hence, making the reparametrisation (12), the states

\displaystyle\left.\ket{\Theta}_{0}\otimes\mathcal{S}^{-\Theta}(u_{1},\dots,u_{M})\,v^{\Theta}_{B}\right|_{\zeta=-\Theta}

are eigenstates of $H$ with energy $E=\Theta B^{z}/2$ if the roots satisfy the Bethe Ansatz equations

\displaystyle 2\beta+\Theta iB^{x}\sqrt{2\beta}+2\sum_{j=1}^{L}\frac{(u_{l}+\beta)(\epsilon_{j}-\beta)}{u_{l}-\epsilon_{j}}s_{j}-2\sum_{m\neq l}^{M}\frac{(u_{l}+\beta)(u_{m}-\beta)}{u_{l}-u_{m}}=0

and the magnetic field components satisfy

\displaystyle\frac{-iB^{x}/2+\Theta B^{y}/2}{\sqrt{2\beta}}\in\{-N_{s}/2,-N_{s}/2+1,\dots,N_{s}/2\}.

(17)

3.2.3 Bright states

Due to $A^{\Theta}\propto S^{\Theta}(\beta)$ we see that the doublets are Bethe states with one of the roots equal to $\beta$ . In this case one finds that

	$\displaystyle S^{-\Theta}_{1}(\beta)S^{\Theta}_{0}(\beta)\mathcal{S}^{-\Theta}(\beta,w_{1},\dots,w_{M-1})v^{\Theta}$
	$\displaystyle=\mathcal{S}^{-\Theta}(\beta,\beta,w_{1},\dots,w_{M-1})S^{\Theta}_{-M}(\beta)v^{\Theta}$
	$\displaystyle\quad+\mathcal{S}^{-\Theta}(\beta,w_{1},\dots,w_{M-1})\left\{8I+2\Theta\dot{S^{z}}(\beta)+\sum_{m=1}^{M-1}\frac{16\beta}{\beta-w_{m}}\right\}v^{\Theta}$
	$\displaystyle\quad+\sum_{l=1}^{M-1}\mathcal{S}^{-\Theta}(\beta,w_{1},\dots,w_{l}\to\beta,\dots,w_{M-1})$
	$\displaystyle\quad\times\left\{8I+2\Theta\frac{S^{z}(w_{l})-S^{z}(\beta)}{w_{l}-\beta}+8\frac{w_{l}+\beta}{w_{l}-\beta}+8\sum_{m\neq l}^{M-1}\frac{w_{l}+\beta}{w_{l}-w_{m}}\right\}v^{\Theta}$

where

\dot{S^{z}}(\beta)=\frac{\partial S^{z}(w)}{\partial w}\bigg{|}_{w=\beta}.

We again see that this will be an eigenstate, with eigenvalue

\displaystyle 8+2\Theta z_{\Theta}^{\prime}(\beta)+\sum_{m=1}^{M-1}\frac{16\beta}{\beta-w_{m}},

\displaystyle N_{s}/2+\frac{i\tilde{B^{x}}/2}{\sqrt{2\beta}}+\Theta\tilde{B^{z}}/2=M\in\{0,1,\dots,N_{s}\}

and the roots $\{w_{m}\}$ satisfy

\displaystyle 8+2\Theta\frac{z_{\Theta}(w_{l})-z_{\Theta}(\beta)}{w_{l}-\beta}+8\frac{w_{l}+\beta}{w_{l}-\beta}+8\sum_{m\neq l}^{M-1}\frac{w_{l}+\beta}{w_{l}-w_{m}}=0.

Hence, making the reparametrisation (12), the states

\displaystyle\left.\left(A^{-}+\left(E_{\Theta}-B^{z}/2\right)S_{0}^{-}\right)\ket{+}_{0}\otimes A^{+}\mathcal{S}^{+}(u_{1},\dots,u_{M-1})v_{B}^{-}\right|_{\zeta=-1}

(18)

are eigenstates of $H$ with energy

\displaystyle E_{\Theta}^{2}=(B^{z})^{2}/4+2\beta-iB^{x}\sqrt{2\beta}+2\sum_{j=1}^{L}(\epsilon_{j}-\beta)s_{j}-2\sum_{m=1}^{M-1}(u_{m}-\beta)

if the roots satisfy the Bethe Ansatz equations

2u_{l}+4\beta-iB^{x}\sqrt{2\beta}+2\sum_{j=1}^{L}\frac{(\epsilon_{j}-\beta)(u_{l}+\beta)}{u_{l}-\epsilon_{j}}s_{j}-2\sum_{m\neq l}^{M-1}\frac{(u_{l}+\beta)(u_{m}-\beta)}{u_{l}-u_{m}}=0

and the magnetic field components satisfy

\displaystyle\frac{-iB^{x}/2-B^{y}/2}{\sqrt{2\beta}}\in\{-N_{s}/2,-N_{s}/2+1,\dots,N_{s}/2\}.

These states (18) are referred to as bright states [21] due to the central spin being entangled with the bath, in contrast to the dark states.

Before we conclude this section, we note that in the reparametrisation

\displaystyle A^{-\Theta}=\Theta i\sqrt{2\beta}+\lim_{u\to\infty}S^{-\Theta}(u)|_{\zeta=\Theta}=\lim_{u\to\infty}S_{1}^{-\Theta}(u)|_{\zeta={-\Theta}},

(19)

i.e. a root equal to $\beta$ corresponds to a root divergent in the reparametrisation.

3.2.4 Generic eigenstates of the charges $\mathcal{Q}_{j}$ .

To describe the eigenstates of the $XY$ model for generic magnetic field configurations we will use the corresponding results of [15] for the charges $\{\mathcal{Q}_{j}\}$ .

As was shown in [15], the charges

	$\displaystyle\tilde{\mathcal{Q}_{j}}$	$\displaystyle=\tilde{B^{z}}S_{j}^{z}+\frac{\tilde{B^{x}}}{h_{j}^{+}}S_{j}^{x}+\frac{\tilde{\zeta}}{h_{j}^{-}}S_{j}^{y}+\frac{h_{j}^{-}}{h_{j}^{+}}S_{j}^{x}S_{j}^{x}+\frac{h_{j}^{+}}{h_{j}^{-}}S_{j}^{y}S_{j}^{y}$
		$\displaystyle\quad+2\sum_{k\neq j}^{L}\frac{1}{\phi_{j}-\phi_{k}}(h_{j}^{+}h_{k}^{-}S_{j}^{x}S_{k}^{x}+h_{j}^{-}h_{k}^{+}S_{j}^{y}S_{k}^{y}+h_{k}^{+}h_{k}^{-}S_{j}^{z}S_{k}^{z}),$

have the eigenstates constructed respectively from lowering or raising operators

\displaystyle\ket{w_{1},\dots,w_{M}}_{-\Theta}

\displaystyle=\mathcal{S}^{-\Theta}(w_{1},\dots,w_{M})v^{\Theta},

(20)

with respective eigenvalues

	$\displaystyle\mathfrak{q}^{-\Theta}_{j}$	$\displaystyle=2\frac{\phi_{j}}{h^{+}_{j}h^{-}_{j}}s_{j}^{2}+\Theta\left(\frac{i\tilde{B^{x}}}{\sqrt{2\beta}}\frac{h^{-}_{j}}{h^{+}_{j}}+\frac{\tilde{\zeta}}{\sqrt{2\beta}}\frac{h^{+}_{j}}{h^{-}_{j}}\right)s_{j}$			(21)
		$\displaystyle\quad-2\sum_{m=1}^{M}\frac{h^{+}_{j}h^{-}_{j}}{\phi_{j}-w_{m}}s_{j}+2\sum_{k\neq j}^{L}\frac{h^{+}_{j}h^{-}_{j}}{\phi_{j}-\phi_{k}}s_{j}s_{k}.$			(21)

The Bethe roots $\{{w_{m}}\}_{m=1}^{M}$ , with $M=\sum_{j=1}^{L}2s_{j}$ , are the solutions to the set of Bethe equations

	$\displaystyle 2w_{l}-\Theta\frac{i\tilde{B^{x}}(w_{l}-\beta)+\tilde{\zeta}(w_{l}+\beta)}{\sqrt{2\beta}}+2\sum_{j=0}^{L}\frac{w_{l}^{2}-\beta^{2}}{\phi_{j}-w_{l}}s_{j}+2\sum_{m\neq l}^{M}\frac{w_{l}^{2}-\beta^{2}}{w_{l}-w_{m}}$
	$\displaystyle\quad=\frac{1}{4}\left(\left[M+1-\Theta\bigg{(}\frac{i\tilde{B^{x}}+\tilde{\zeta}}{\sqrt{2\beta}}\bigg{)}\right]^{2}-(B^{z})^{2}\right)\frac{\prod_{j=0}^{L}(w_{l}-\phi_{j})^{2s_{j}}}{\prod_{m\neq l}^{M}(w_{l}-w_{m})}.$		(22)

As we will mainly be working with the variables of the reparametrisation (12), using also the notation of Sect. 3.1.1, we also give their corresponding expressions. Here the Bethe roots $\{u_{m}\}$ satisfy the modified Bethe equations

	$\displaystyle u_{l}+3\beta+\Theta\left(iB^{x}\sqrt{2\beta}+\zeta(u_{l}+\beta)\right)$
	$\displaystyle+2\sum_{j=0}^{L}\frac{(\epsilon_{j}-\beta)(u_{l}+\beta)}{u_{l}-\epsilon_{j}}s_{j}-2\sum_{m\neq l}^{M}\frac{(u_{m}-\beta)(u_{l}+\beta)}{u_{l}-u_{m}}$
	$\displaystyle=\frac{1}{4}\left(\left[(M+1)\sqrt{2\beta}+\Theta(iB^{x}+\zeta\sqrt{2\beta})\right]^{2}-(B^{y})^{2}\right)$
	$\displaystyle\quad\times\prod_{j=0}^{L}\left(\frac{\epsilon_{j}-u_{l}}{\epsilon_{j}-\beta}\right)^{2s_{j}}\prod_{m\neq l}^{M}\frac{u_{m}-\beta}{u_{m}-u_{l}},$		(23)

while the eigenvalues $\mathfrak{q}^{-\Theta}_{j}$ of the charges $\mathcal{Q}_{j}$ , using the result of (8), are

	$\displaystyle\mathfrak{q}^{-\Theta}_{j}$	$\displaystyle=\frac{f_{j}^{+}}{f_{j}^{-}}s_{j}(s_{j}+1)-\frac{\epsilon_{j}+3\beta}{f_{j}^{+}f_{j}^{-}}s_{j}^{2}+\Theta\left(iB^{x}\frac{\sqrt{2\beta}}{f_{j}^{+}f_{j}^{-}}+\zeta\frac{f_{j}^{+}}{f_{j}^{-}}\right)s_{j}$			(24)
		$\displaystyle\quad+2\frac{f_{j}^{+}}{f_{j}^{-}}\sum_{m=1}^{M}\frac{u_{m}-\beta}{u_{m}-\epsilon_{j}}s_{j}+2\frac{f_{j}^{+}}{f_{j}^{-}}\sum_{k\neq j}^{L}\frac{\epsilon_{k}-\beta}{\epsilon_{j}-\epsilon_{k}}s_{j}s_{k}.$			(24)

Before moving onto the eigenstates of the central spin model, we note that for generic model parameters the states constructed by either applying the lowering operators to the highest weight state $v^{+}$ , or the raising operators to the lowest weight state $v^{-}$ , constitute a complete set of eigenstates. This follows from quadratic identities of the charges $\mathcal{Q}_{j}$ similar to those seen in [20], and extended to the eigenvalues. From these the Bethe Ansatz equations are seen to be the consistency equations for the parametrisation in terms of the roots $\{u_{m}\}$ . Completeness is then argued similarly as in [9, 10] by assuming regularisability and the one-dimensionality of the simultaneous eigenspace for generic parameters. We also observe that there is a map between the states obtained via the raising and lowering operators through the symmetry of the charges, $\{S^{z}\to-S^{z},\;i\to-i,\;\zeta\to-\zeta\}$ .

3.3 Generic eigenstates of the $XY$ model

We now proceed to obtain the eigenstates of the $XY$ model with the limits defined in Sect. 2. Recall from (3) that we obtain the Hamiltonian by taking the limit of $\mathcal{Q}_{0}$ as $\epsilon_{0}\to\infty$ after applying a rescaling and a constant shift. Hence the eigenstates are found from the eigenstates of the charge $\mathcal{Q}_{0}$ in the above limit. As we saw in Sect. 3.2.3, the Bethe roots $\{u_{m}\}$ can diverge, so we analyse the behaviour as two cases.

Finite roots in the limit.

Assuming the roots stay finite in the limit, the raising operators become

	$\displaystyle\tilde{S}^{-\Theta}(u)$	$\displaystyle=B^{x}-\Theta iB^{y}+2i\sqrt{2\beta}S_{0}^{z}+2\sum_{j=1}^{L}\frac{f_{j}^{+}(u-\beta)}{\epsilon_{j}-u}\left(\pm iS_{j}^{y}-\frac{f_{j}^{+}}{f_{j}^{-}}S_{j}^{x}+i\frac{\sqrt{2\beta}}{f_{j}^{-}}S_{j}^{z}\right)$
		$\displaystyle\quad+2\sqrt{2\beta}\sum_{j=1}^{L}\left(-\frac{\sqrt{2\beta}}{f_{j}^{-}}S_{j}^{x}+i\frac{f_{j}^{+}}{f_{j}^{-}}S_{j}^{z}\right)$
		$\displaystyle=S^{-\Theta}(u)\|_{\zeta=-2S^{z}_{0}}.$

Hence the eigenstates are dark states

\displaystyle\ket{u_{1},\dots,u_{M}}_{\Theta}

\displaystyle=\left.\ket{\Theta}_{0}\otimes\mathcal{S}^{-\Theta}(u_{1},u_{2},\dots,u_{M})v_{B}^{\Theta}\right|_{\zeta=-\Theta}.

However, from Sect. 3.2.2 we saw that these will in general only occur when (17) holds. So in this case the Bethe Ansatz would have to be modified to that seen in the same section. We also remark that the dark states can be seen to come from entangled states in the limit as the magnetic field configuration satisfy (17).⁴⁴4See also Sect. 5.2.1 in the Appendix for a heuristic discussion of this.

One of the roots diverges.

Assuming that $u_{M}\to\infty$ we have

\displaystyle\lim_{\epsilon_{0},u_{M}\to\infty}S^{-\Theta}(u_{M})

\displaystyle=B^{x}-\Theta B^{y}i-2\kappa S_{0}^{-\Theta}+2i\sqrt{2\beta}S_{0}^{z}+2\sum_{j=1}^{L}\left(f_{j}^{-}S_{j}^{x}-\Theta if_{j}^{+}S_{j}^{y}\right)

where

\kappa=\lim_{\epsilon_{0},u_{M}\to\infty}\left[\frac{u_{M}}{\sqrt{\epsilon_{0}}}\right].

The eigenstates become

	$\displaystyle\ket{\kappa;u_{1},\dots,u_{M-1}}_{\Theta}$	$\displaystyle=\left[B^{x}-\Theta iB^{y}+2\sum_{j=1}^{L}\left(f_{j}^{-}S_{j}^{x}-\Theta if_{j}^{+}S_{j}^{y}\right)-2\kappa S_{0}^{-\Theta}\right]\ket{\Theta}_{0}$
		$\displaystyle\quad\otimes\left(S^{-\Theta}(u_{1})-\Theta 3i\sqrt{2\beta}I\right)\times\cdots$
		$\displaystyle\quad\times\left.\left(S^{-\Theta}(u_{M-1})-\Theta(2M-1)i\sqrt{2\beta}I\right)v_{B}^{\Theta}\right\|_{\zeta=-\Theta}$
		$\displaystyle=\left.2\left[A^{-\Theta}-\kappa S_{0}^{-\Theta}\right]\ket{\Theta}_{0}\otimes\mathcal{S}^{-\Theta}(u_{1},\dots,u_{M-1})v_{B}^{\Theta}\right\|_{\zeta=\Theta},$

using the expressions (13) for $A^{\Theta}$ in the last line. Here the roots satisfy

	$\displaystyle 2u_{l}+4\beta+\Theta iB^{x}\sqrt{2\beta}+2\sum_{j=1}^{L}\frac{(\epsilon_{j}-\beta)(u_{l}+\beta)}{u_{l}-\epsilon_{j}}s_{j}-2\sum_{m\neq l}^{M-1}\frac{(u_{m}-\beta)(u_{l}+\beta)}{u_{l}-u_{m}}$
	$\displaystyle=\frac{1}{4}\left((M+1)\sqrt{2\beta}-\Theta(-iB^{x}+B^{y})\right)\left((M+1)\sqrt{2\beta}-\Theta(-iB^{x}-B^{y})\right)$
	$\displaystyle\quad\times\prod_{j=1}^{L}\left(\frac{\epsilon_{j}-u_{l}}{\epsilon_{j}-\beta}\right)^{2s_{j}}\prod_{m\neq l}^{M-1}\frac{u_{m}-\beta}{u_{m}-u_{l}}$

and $\kappa$ satisfies

	$\displaystyle\kappa^{2}-\Theta\left(iB^{x}\sqrt{2\beta}+B^{z}\kappa\right)-2\beta-2\sum_{j=1}^{L}(\epsilon_{j}-\beta)s_{j}+2\sum_{m=1}^{M-1}(u_{m}-\beta)$
	$\displaystyle=-\frac{1}{4}\left(\left[(M+1)\sqrt{2\beta}+\Theta iB^{x}\right]^{2}-(B^{y})^{2}\right)\frac{\prod_{m=1}^{M-1}(u_{m}-\beta)}{\prod_{j=1}^{L}(\epsilon_{j}-\beta)^{2s_{j}}}.$

The energy is

\displaystyle E

\displaystyle=\Theta B^{z}/2-\kappa

(26)

and the eigenvalues $\{q^{-\Theta}_{i}\}$ of the charges $\{Q_{i}\}$ are

	$\displaystyle q^{-\Theta}_{i}=$	$\displaystyle-\frac{2\beta}{f_{i}^{+}f_{i}^{-}}s_{i}^{2}+\left(\Theta iB^{x}\frac{\sqrt{2\beta}}{f_{i}^{+}f_{i}^{-}}+2\frac{f_{i}^{+}}{f_{i}^{-}}\right)s_{i}$
		$\displaystyle+2\frac{f_{i}^{+}}{f_{i}^{-}}\sum_{m=1}^{M}\frac{u_{m}-\beta}{u_{m}-\epsilon_{i}}s_{i}+2\frac{f_{i}^{+}}{f_{i}^{-}}\sum_{j\neq i}^{L}\frac{\epsilon_{j}-\beta}{\epsilon_{i}-\epsilon_{j}}s_{i}s_{j}.$

Due to the quadratic equation for $\kappa$ this gives two sets of eigenstates with different energy. However, recalling the completeness argument of Sect. 3.2.4, the states with the same energy are the same. So a state constructed from lowering operators with $\kappa=k$ corresponds to a state constructed with raising operators with $\kappa=k-B^{z}$ (and vice-versa a “raised” state with $\kappa=k$ corresponds to a “lowered” state with $\kappa=k+B^{z}$ ).

We note that, since these are the eigenstates of the charges $\{Q_{i}\}_{i=1,\dots,L}$ , projecting onto $\ket{-\Theta}_{0}$ for the central spin, the bath states

\left.S^{-\Theta}(u_{1},\dots,u_{M-1})v_{B}^{\Theta}\right|_{\zeta=\Theta}

are eigenstates of the charges $\{\mathcal{Q}_{i}\}_{i=1,\dots,L}$ with $\zeta=\Theta$ . We see that this is consistent with the Bethe Ansatz equations. To see that the eigenstates are also consistent with the construction via supersymmetry we rewrite them as

	$\displaystyle\left.\ket{\kappa(E);u_{1},\dots,u_{M-1}}_{\Theta}\propto\ket{\Theta}_{0}\otimes A^{-\Theta}\mathcal{S}^{-\Theta}(u_{1},\dots,u_{M-1})v_{B}^{\Theta}\right\|_{\zeta=\Theta}$
	$\displaystyle\quad+\left.(E-\Theta B^{z}/2)\ket{-\Theta}_{0}\otimes\mathcal{S}^{-\Theta}(u_{1},\dots,u_{M-1})v_{B}^{\Theta}\right\|_{\zeta=\Theta}$		(27)

using expression (26) for the energy. Then by (19) we observe that

\displaystyle\left.A^{-\Theta}\mathcal{S}^{-\Theta}(u_{1},\dots,u_{M-1})v_{B}^{\Theta}\right|_{\zeta=\Theta}

\displaystyle=\lim_{u\to\infty}\left.\mathcal{S}^{-\Theta}(u,u_{1},\dots,u_{M-1})v_{B}^{\Theta}\right|_{\zeta=-\Theta}.

Whereas, we see from (3.2.1) that

\displaystyle\left.\mathcal{S}^{-\Theta}(u_{1},\dots,u_{M-1})v_{B}^{\Theta}\right|_{\zeta=\Theta}

is an eigenstate of $A^{\Theta}A^{-\Theta}$ with eigenvalue $E^{2}-(B^{z})^{2}/4$ , so by the previous relation

(E^{2}-(B^{z})^{2}/4)\left.\mathcal{S}^{-\Theta}(u_{1},\dots,u_{M-1})v_{B}^{\Theta}\right|_{\zeta=\Theta}=\left.\mathcal{A}^{\Theta}\ket{\Theta}_{0}\otimes\mathcal{S}^{-\Theta}(\infty,u_{1},\dots,u_{M-1})v_{B}^{\Theta}\right|_{\zeta=-\Theta}.

Hence we find that

\ket{\kappa(E);u_{1},\dots,u_{M-1}}_{\Theta}\propto\left((E+\Theta B^{z}/2)+\mathcal{A}^{\Theta}\right)\ket{\Theta}_{0}\otimes\left.\mathcal{S}^{-\Theta}(\infty,u_{1},\dots,u_{M-1})v_{B}^{\Theta}\right|_{\zeta=-\Theta},

which matches (3.2.1).

4 Conclusion

We have derived the $XY$ central spin model Hamiltonian $H_{XY}$ , (7), and charges given in [20] as a limit of the $XYZ$ model charges with self-interaction $\mathcal{Q}_{i}$ obtained by Skrypnyk [15]. This allowed the diagonalisation results found in [15] for the charges $\mathcal{Q}_{i}$ via a modified Algebraic Bethe Ansatz to be used to diagonalise $H_{XY}$ . However, in order for these results to connect with those found for the $XX$ model we regularised the charges $\mathcal{Q}_{i}$ via a reparametrisation. Lastly, we showed that the dark states of the $XX$ model without an in-plane magnetic field [5, 21] can be seen as a special case of dark states of the $XY$ model for special configurations of the magnetic field.

These results open the way for finding form factors and correlation functions for the $XY$ central spin model as well as probing the dynamics for large systems. The characterisation of the magnetic field configurations supporting dark states and their emergence via special limits of the parameter $\kappa$ can provide experimental criteria for preparing these states with high-fidelity and determining how far a bright state is from being a product state.

Another extension of this work would be to the case of a spin-1 central spin. Here it has been shown that the $XX$ central spin model is integrable for an out-of-plane magnetic field [17] and more recently for an arbitrarily oriented magnetic field [4]. Integrability for the $XY$ model can also be shown through a minor modification of the charges found in [4]. In the latter work it was numerically demonstrated that dark states can emerge at asymptotically large coupling, just as seen in the central spin-1/2 model for an arbitrarily oriented magnetic field [5]. This raises the possibility of the existence of these states for certain special magnetic field configurations of the $XX$ and $XY$ spin-1 central spin models. However, determining whether these dark states actually exist is complicated by the fact that the relationship between the model and the known class of Richardson-Gaudin models is not yet understood.

The authors thank Pieter Claeys and Taras Skrypnyk for helpful comments that enabled progress on this research problem. The authors also acknowledge the traditional owners of the land on which The University of Queensland at St. Lucia operates, the Turrbal and Jagera people. This work was supported by the Australian Research Council through Discovery Project DP200101339.

5 Appendix

5.1 Characterisation of magnetic field configurations admitting dark states

In this section we supplement Subsect. 3.2.1 – to argue that the eigenstates given by the singlet states are the only dark states, i.e. product states of the central spin and bath spins – and Subsec. 3.2.2, to characterise when they occur.

Firstly, with regards to the singlets we note that for the eigenstates

\ket{E}=\Lambda\ket{-}_{0}\otimes\ket{\psi_{-}}_{B}+\left(E+B^{z}/2\right)\ket{+}_{0}\otimes\ket{\psi_{+}}_{B}

to be dark states requires $(E+B^{z}/2)\ket{\psi_{+}}_{B}\propto\ket{\psi_{-}}_{B}.$ If we assume that neither of these is zero this implies that $\ket{\psi_{\Theta}}_{B},\,\Theta\in\{+,-\}$ are eigenstates of both charges $\mathcal{Q}_{j}|_{\zeta=\mp 1}$ since $\ket{\psi_{\Theta}}_{B}$ are respectively eigenstates of the charges $\mathcal{Q}_{j}|_{\zeta=-\Theta}$ due to

A^{+}\mathcal{Q}_{j}|_{\zeta=-1}=\mathcal{Q}_{j}|_{\zeta=+1}A^{+}.

Hence they are also eigenstates of $\mathcal{Q}_{j}|_{\zeta=+1}-\mathcal{Q}_{j}|_{\zeta=-1}=2S_{j}^{z}$ for all $j=1,\dots,L$ . This determines $\ket{\psi_{\Theta}}_{B},\,\Theta\in\{+,-\}$ and would lead to a contradiction if we took e.g. $B^{x}\gg 1$ since this holds for all $B^{x}$ . This also means that dark states can only occur if $\Lambda\ket{\psi_{-}}_{B}=0$ or $(E+B^{z}/2)\ket{\psi_{+}}_{B}=0$ in which case the central spin state is respectively $\ket{\Theta}_{0}$ . Furthermore note that by the supersymmetry this means (as these are originally doublets) that $\ket{\Theta}_{0}\otimes\ket{\psi_{\Theta}}_{B}$ is a singlet.

Next we will show that these dark states only occur when

\displaystyle\frac{-iB^{x}/2\pm B^{y}/2}{\sqrt{2\beta}}

\displaystyle=\frac{N_{s}}{2}-\sum_{j=1}^{N}n_{j}\in\{-N_{s}/2,\dots,N_{s}/2\}

(28)

where $N_{s}=\sum_{j=1}^{L}2s_{j}$ and $n_{j}\in\{0,1,\dots,2s_{j}\}$ . Define $\mathfrak{f}_{j}^{\pm}:=(f_{j}^{-}\pm f_{j}^{+})/2$ , so that we can write the local operators of $A^{\pm}$ in terms of $S_{j}^{\pm}$ as

\displaystyle f^{-}_{j}S_{j}^{x}\pm if_{j}^{+}S_{j}^{y}

\displaystyle=\mathfrak{f}_{j}^{\pm}S_{j}^{+}+\mathfrak{f}_{j}^{\mp}S_{j}^{-}.

If we now define operators

	$\displaystyle E_{j}^{\pm}$	$\displaystyle=-\frac{\sqrt{\mathfrak{f}_{j}^{\pm}}}{2\sqrt{\mathfrak{f}_{j}^{\mp}}}S_{j}^{+}+S_{j}^{z}+\frac{\sqrt{\mathfrak{f}_{j}^{\mp}}}{2\sqrt{\mathfrak{f}_{j}^{\pm}}}S_{j}^{-},\;H_{j}^{\pm}=\frac{\sqrt{\mathfrak{f}_{j}^{\pm}}}{\sqrt{\mathfrak{f}_{j}^{\mp}}}S_{j}^{+}+\frac{\sqrt{\mathfrak{f}_{j}^{\mp}}}{\sqrt{\mathfrak{f}_{j}^{\pm}}}S_{j}^{-},$
	$\displaystyle F_{j}^{\pm}$	$\displaystyle=\frac{\sqrt{\mathfrak{f}_{j}^{\pm}}}{2\sqrt{\mathfrak{f}_{j}^{\mp}}}S_{j}^{+}+S_{j}^{z}-\frac{\sqrt{\mathfrak{f}_{j}^{\mp}}}{2\sqrt{\mathfrak{f}_{j}^{\pm}}}S_{j}^{-},$

then it can be checked that these respectively satisfy the $\mathfrak{sl}(2)$ relations

\displaystyle[H_{j}^{\pm},E_{j}^{\pm}]=2E_{j}^{\pm},\quad[E_{j}^{\pm},F_{j}^{\pm}]=2H_{j}^{\pm},\quad[H_{j}^{\pm},F_{j}^{\pm}]=-2F_{j}^{\pm}.

Noting that

	$\displaystyle A^{\pm}$	$\displaystyle=(B^{x}\pm iB^{y})I/2+\sum_{j=1}^{N}\sqrt{\mathfrak{f}^{+}_{j}\mathfrak{f}^{-}_{j}}\,H_{j}^{\pm}$
		$\displaystyle=(B^{x}\pm iB^{y})I/2+i\sum_{j=1}^{N}\sqrt{\beta/2}\,H_{j}^{\pm}$

we can respectively diagonalise $A^{\pm}$ with the eigenvectors

\displaystyle\ket{n_{1},\dots,n_{N}}^{\pm}=\prod_{j=1}^{N}(F_{j}^{\pm})^{n_{j}}\chi^{\pm},

where the highest weight vector is defined as

	$\displaystyle\chi^{\pm}=\chi_{1}^{\pm}\otimes\cdots\chi_{N}^{\pm},$
	$\displaystyle E_{j}^{\pm}\chi_{j}^{\pm}=0,\;H_{j}^{\pm}\chi_{j}^{\pm}=2s_{j}\chi_{j}^{\pm};\;-2s_{j}\leq n_{j}\leq 2s_{j},$

to find that it has respective eigenvalues

\displaystyle(B^{x}\pm iB^{y})/2+i\sqrt{2\beta}\sum_{j=1}^{N}n_{j}-\frac{N_{s}}{2}i\sqrt{2\beta},\quad-2s_{j}\leq n_{j}\leq 2s_{j}.

We see that these are zero (i.e. the corresponding states are dark states) when (28) holds.

For the following section we briefly mention the $XX$ model case. Here the dark states are zero eigenvalue eigenstates of

A^{\pm}=(B^{x}\pm iB^{y})I/2+\sum_{j=1}^{L}\sqrt{\epsilon_{j}}S_{j}^{\pm}.

From this we see that $A^{\pm}$ is strictly upper or lower triangular if and only if $B^{x}=B^{y}=0$ , and so dark states can only occur if $B^{x}=B^{y}=0$ .

5.2 Eigenstates of the $XX$ central spin model

Due to the reparametrised expressions for the eigenstates and eigenvalues of the $XYZ$ Gaudin model charges all being non-singular in the limit $\beta\to 0$ , we can rederive the diagonalisation results for the $XX$ model. This will also give us an idea of how the dark states emerge in the limit of the special magnetic field configurations of Subsect. 3.2.2.

Taking the limit as $\beta\to 0$ , the eigenstates of the charges

	$\displaystyle\mathcal{Q}_{i}=\zeta S^{z}_{i}+\frac{B^{x}}{\sqrt{\epsilon_{i}}}S^{x}_{i}+\frac{B^{y}}{\sqrt{\epsilon_{i}}}S^{y}_{i}+S_{i}^{x}S_{i}^{x}+S_{i}^{y}S_{i}^{y}$
	$\displaystyle\qquad+2\sum_{j\neq i}^{L}\frac{1}{\epsilon_{i}-\epsilon_{j}}\left(\sqrt{\epsilon_{i}}\sqrt{\epsilon_{j}}S^{x}_{i}S^{x}_{j}+\sqrt{\epsilon_{i}}\sqrt{\epsilon_{j}}S^{y}_{i}S^{y}_{j}+\epsilon_{j}S^{z}_{i}S^{z}_{j}\right)$

become

\displaystyle\ket{u_{1},\dots,u_{M}}_{-\Theta}

\displaystyle=S^{-\Theta}(u_{1})S^{-\Theta}(u_{2})\cdots S^{-\Theta}(u_{M})v^{\Theta},

with $\Theta\in\{+,-\}$ . Here

\displaystyle S^{-\Theta}(u)

\displaystyle=B^{x}-\Theta iB^{y}+2\sum_{j=1}^{L}\frac{u\sqrt{\epsilon_{j}}}{u-\epsilon_{j}}S_{j}^{-\Theta}

and the reference states $v^{\Theta}=v_{0}^{\Theta}\otimes v_{1}^{\Theta}\otimes\cdots\otimes v_{L}^{\Theta}$ now satisfy

	$\displaystyle S_{k}^{z}v_{k}^{+}=s_{k}v_{k},\;$	$\displaystyle S_{k}^{+}v_{k}^{+}=0,$
	$\displaystyle S_{k}^{z}v_{k}^{-}=-s_{k}v_{k},\;$	$\displaystyle S_{k}^{-}v_{k}^{-}=0,$

or in other words

v^{\Theta}=\ket{\Theta s_{1},\dots,\Theta s_{L}}.

The Bethe roots in turn now satisfy

	$\displaystyle 1+\Theta\zeta+$	$\displaystyle 2\sum_{j=1}^{L}\frac{\epsilon_{j}}{u_{l}-\epsilon_{j}}s_{j}-2\sum_{m\neq l}^{M}\frac{u_{m}}{u_{l}-u_{m}}$			(29)
		$\displaystyle=-\frac{(B^{x})^{2}+(B^{y})^{2}}{4}\frac{\prod_{j=1}^{L}\left(u_{l}^{-1}-\epsilon_{j}^{-1}\right)^{2s_{j}}}{\prod_{m\neq l}^{M}(u_{l}^{-1}-u_{m}^{-1})}$			(29)

with $M=\sum_{j=1}^{L}2s_{j}$ , while the eigenvalues $\{\mathfrak{q}^{-\Theta}_{j}\}$ of the charges $\{\mathcal{Q}_{j}\}$ are

\displaystyle\mathfrak{q}^{-\Theta}_{i}

\displaystyle=\left(1+\Theta\zeta\right)s_{i}+2\sum_{m=1}^{M}\frac{u_{m}}{u_{m}-\epsilon_{i}}s_{i}+2\sum_{j\neq i}^{L}\frac{\epsilon_{j}}{\epsilon_{i}-\epsilon_{j}}s_{i}s_{j}.

To obtain the eigenstates of the $XX$ central spin model we can simply use the results for the $XY$ model in the limit $\beta\to 0$ . For a magnetic field with a non-zero in-plane component the eigenstates are entangled states of the central spin and bath, the bright states of [21], taking the form

	$\displaystyle\ket{\kappa;u_{1},\dots,u_{M-1}}_{\Theta}=$	$\displaystyle\left[B^{x}-\Theta iB^{y}+2\sum_{j=1}^{L}\sqrt{\epsilon_{j}}S_{j}^{-\Theta}-2\kappa S_{0}^{-\Theta}\right]\ket{\Theta}_{0}$
		$\displaystyle\otimes\left.\mathcal{S}^{-\Theta}(u_{1},u_{2},\dots,u_{M-1})\ket{\Theta s_{1},\dots,\Theta s_{L}}\right\|_{\zeta=\Theta}$

with the $M-1=\sum_{j=1}^{L}2s_{j}$ roots satisfying the Bethe Ansatz equations

\displaystyle 2+2\sum_{j=1}^{L}\frac{\epsilon_{j}}{u_{l}-\epsilon_{j}}s_{j}-2\sum_{m\neq l}^{M-1}\frac{u_{m}}{u_{l}-u_{m}}

\displaystyle=-\frac{(B^{x})^{2}+(B^{y})^{2}}{4}\frac{\prod_{j=1}^{L}(u_{l}^{-1}-\epsilon_{j}^{-1})^{2s_{j}}}{\prod_{m\neq l}^{M-1}(u_{l}^{-1}-u_{m}^{-1})}

and the equation for $\kappa$ coming from the Bethe Ansatz equation for $u_{M}$ ,

\displaystyle\kappa^{2}-\Theta B^{z}\kappa-2\sum_{j=1}^{L}\epsilon_{j}s_{j}+2\sum_{m=1}^{M-1}u_{m}

\displaystyle=-\frac{(B^{x})^{2}+(B^{y})^{2}}{4}\frac{\prod_{m=1}^{M-1}u_{m}}{\prod_{j=1}^{L}(\epsilon_{j})^{2s_{j}}}.

(30)

The energy is

\displaystyle E

\displaystyle=\Theta B^{z}/2-\kappa

while the eigenvalues $\{q_{i}\}$ of the charges $\{Q_{i}\}$ are

\displaystyle q_{i}

\displaystyle=2s_{i}-2\sum_{m=1}^{M-1}\frac{u_{m}}{\epsilon_{i}-u_{m}}s_{i}+2\sum_{j\neq i}^{L}\frac{\epsilon_{j}}{\epsilon_{i}-\epsilon_{j}}s_{i}s_{j}.

Observe here that if $\kappa$ is a solution to the first equation of $(\ref{eq: kappa XX})$ then $\kappa^{\prime}=\kappa-B^{z}$ is a solution to the second (respectively, if $\kappa$ is a solution to the second equation of $(\ref{eq: kappa XX})$ then $\kappa^{\prime}=\kappa+B^{z}$ is a solution to the first equation), due to the Bethe equations for the roots being the same for both. These have identical energy and charge eigenvalues. Therefore, based on the argument of the correspondence between eigenstates constructed using raising or lowering operators, they are proportional to the same eigenstate. We note these expressions match those of [3] and [21], where in the former $s_{j}=1/2$ and in the latter $B^{x}=B^{y}=0$ .

5.2.1 Recovering the dark states for $B^{x}=B^{y}=0$ .

As argued in Sect. 5.1, in the case of a non-zero in-plane magnetic field, $(B^{x})^{2}+(B^{y})^{2}>0$ , there are no dark states. We find numerically (see also Sect. 5.3) that the dark states of the $XX$ model with no in-plane field are recovered in the limit as $(B^{x})^{2}+(B^{y})^{2}\to 0$ via $\kappa\to\Theta B^{z}$ or $\kappa\to 0$ for some of the eigenstates. For simplicity we will only discuss the case where for all the bath spins $s_{j}=1/2$ .

In the case where $\kappa\to\Theta B^{z}$ , the states become

\ket{\psi_{\Theta}}=2\left[\sum_{j=1}^{L}\sqrt{\epsilon_{j}}S_{j}^{-\Theta}+B^{z}S_{0}^{-\Theta}\right]\ket{\Theta}_{0}\otimes\left.\mathcal{S}^{-\Theta}(u_{1},u_{2},\dots,u_{\mathfrak{M}-1})\ket{\Theta s_{1},\dots,\Theta s_{L}}\right|_{\zeta=\Theta}

where, from (29), the roots satisfy

\displaystyle\underbrace{1+\Theta\left(\Theta\right)}_{=2}+2\sum_{j=1}^{L}\frac{\epsilon_{j}}{u_{l}-\epsilon_{j}}s_{j}-2\sum_{m\neq l}^{\mathfrak{M}-1}\frac{u_{m}}{u_{l}-u_{m}}=0,

with $\mathfrak{M}\leq M$ due to some of the roots becoming zero and their corresponding raising and lowering operators being asymptotically $S^{-\Theta}(u)\propto c(B^{x},B^{y})I$ . This follows by noting that the Bethe states will become eigenstates of the total $S^{z}$ operator, as the raising and lowering operators are proportional to products of total spin raising or lowering operators in the zero in-plane field limit, and since the $XX$ model in this limit has a global $U(1)$ symmetry.

Using the correspondence coming from the symmetry of the charges, $\{S^{z}\to-S^{z},\;i\to-i,\;\zeta\to-\zeta\}$ , this can be written as

\ket{\psi_{\Theta}}\propto 2\left[\sum_{j=1}^{L}\sqrt{\epsilon_{j}}S_{j}^{-\Theta}+B^{z}S_{0}^{-\Theta}\right]\ket{\Theta}_{0}\otimes\left.\mathcal{S}^{\Theta}(\tilde{u}_{1},\tilde{u}_{2},\dots,\tilde{u}_{\mathcal{M}-1})\ket{-\Theta s_{1},\dots,-\Theta s_{L}}\right|_{\zeta=-\Theta}

with the roots $\{\tilde{u}_{m}\}_{m=1}^{\mathcal{M}-1}$ satisfying

\displaystyle\underbrace{1+\Theta\left(-\Theta\right)}_{=0}+2\sum_{j=1}^{L}\frac{\epsilon_{j}}{\tilde{u}_{l}-\epsilon_{j}}s_{j}-2\sum_{m\neq l}^{\mathcal{M}-1}\frac{\tilde{u}_{m}}{\tilde{u}_{l}-\tilde{u}_{m}}=0.

Note that the bath component of the state now satisfies (see for example [21])

\displaystyle\underbrace{\left(\sum_{j=1}^{L}\sqrt{\epsilon_{j}}S_{j}^{-\Theta}\right)}_{A^{-\Theta}}\left.\mathcal{S}^{\Theta}(\tilde{u}_{1},\tilde{u}_{2},\dots,\tilde{u}_{\mathcal{M}-1})\ket{-\Theta s_{1},\dots,-\Theta s_{L}}\right|_{\zeta=-\Theta}=0,

giving the dark state

\displaystyle\ket{\psi_{\Theta}}

\displaystyle\propto\ket{-\Theta}_{0}\otimes\left.\mathcal{S}^{\Theta}(\tilde{u}_{1},\tilde{u}_{2},\dots,\tilde{u}_{\mathcal{M}-1})\ket{-\Theta s_{1},\dots,-\Theta s_{L}}\right|_{\zeta=-\Theta}.

As explained in Sect. 5.2, the states with $\kappa^{\prime}=\kappa+B^{z}$ constructed with lowering operators correspond to states with $\kappa$ constructed with raising operators (and vice versa for $\kappa^{\prime}=\kappa-B^{z}$ and $\kappa$ ). So the states constructed with raising operators for which $\kappa\to 0$ are the same dark states as those constructed with lowering operators with the behaviour $\kappa\to+B^{z}$ (and vice versa for $\kappa\to 0$ and $\kappa\to-B^{z}$ ) that we just found. However, for completeness we will sketch here the behaviour in this limit. For conciseness let $\rho_{-\Theta}=(B^{x}-\Theta iB^{y})/2$ so that the state asymptotically becomes

	$\displaystyle\ket{\psi_{\Theta}}\sim\ket{\Theta}_{0}\otimes$	$\displaystyle\left(\rho_{-\Theta}+\sum_{j=1}^{L}\sqrt{\epsilon_{j}}S_{j}^{-\Theta}\right)\left(\rho_{-\Theta}+\sum_{j=1}^{L}\frac{u_{1}\sqrt{\epsilon_{j}}}{u_{1}-\epsilon_{j}}S_{j}^{\Theta}\right)\times$
		$\displaystyle\cdots\left(\rho_{-\Theta}+\sum_{j=1}^{L}\frac{u_{\mathfrak{M}-1}\sqrt{\epsilon_{j}}}{u_{\mathfrak{M}-1}-\epsilon_{j}}S_{j}^{\Theta}\right)\ket{\Theta s_{1},\dots,\Theta s_{L}}$

where again some of the roots $\{u_{i}\}_{i=1}^{M-1}$ go to zero. Expanding this out, the state has the form (for $s_{i}=1/2$ )

	$\displaystyle\ket{\psi_{\Theta}}\sim\ket{\Theta}_{0}\otimes$	$\displaystyle\left(\rho_{-\Theta}^{\mathfrak{M}}+\rho_{-\Theta}^{\mathfrak{M}-1}\sum_{i_{1}}c_{i_{1}}S_{i_{1}}^{\mp}+\cdots+\rho_{-\Theta}^{\mathfrak{M}-K}\sum_{i_{1}<\cdots<i_{K}}c_{i_{1}\cdots i_{K}}S_{i_{1}}^{-\Theta}\cdots S_{i_{K}}^{-\Theta}\right.$
		$\displaystyle\left.+\cdots+\sum_{i_{1}<\cdots<i_{\mathfrak{M}}}^{L}c_{i_{1}\cdots i_{\mathfrak{M}}}S_{i_{1}}^{-\Theta}\cdots S_{i_{\mathfrak{M}}}^{-\Theta}\right)\ket{\Theta s_{1},\dots,\Theta s_{L}}$

where

c_{i_{1}\cdots i_{K}}=\sum_{j_{1}<\cdots<j_{K-1}}\sum_{\sigma\in S_{K}}\epsilon_{i_{\sigma(1)}}\frac{u_{j_{1}}\sqrt{\epsilon_{i_{\sigma(2)}}}}{u_{j_{1}}-\epsilon_{i_{\sigma(2)}}}\cdots\frac{u_{j_{K-1}}\sqrt{\epsilon_{i_{\sigma(K)}}}}{u_{j_{K}-1}-\epsilon_{i_{\sigma(K)}}}.

For a dark state with $\mathcal{M}$ roots, all the $c_{i_{1}\cdots i_{K}}$ for $K>\mathcal{M}$ are zero while

c_{i_{1}\cdots i_{\mathcal{M}}}=\sum_{\sigma\in S_{\mathcal{M}}}\left(\frac{\tilde{u}_{1}\sqrt{\epsilon_{i_{\sigma(1)}}}}{\tilde{u}_{1}-\epsilon_{i_{\sigma(1)}}}\cdots\frac{\tilde{u}_{\mathcal{M}}\sqrt{\epsilon_{i_{\sigma(\mathcal{M})}}}}{\tilde{u}_{\mathcal{M}}-\epsilon_{i_{\sigma(\mathcal{M})}}}\right).

The state in the limit then becomes equal to the dark state

\displaystyle\ket{\psi_{\Theta}}\propto\ket{\Theta}_{0}\otimes\left(\sum_{j=1}^{L}\frac{\tilde{u}_{1}\sqrt{\epsilon_{j}}}{\tilde{u}_{1}-\epsilon_{j}}S_{j}^{\Theta}\right)\cdots\left(\sum_{j=1}^{L}\frac{\tilde{u}_{\mathcal{M}}\sqrt{\epsilon_{j}}}{\tilde{u}_{\mathcal{M}}-\epsilon_{j}}S_{j}^{\Theta}\right)\ket{\Theta s_{1},\dots,\Theta s_{L}}

with $\mathcal{M}\leq\mathfrak{M}$ and where the roots satisfy

2\sum_{j=1}^{L}\frac{\epsilon_{j}}{\tilde{u}_{l}-\epsilon_{j}}s_{j}-2\sum_{m\neq l}^{\mathcal{M}}\frac{\tilde{u}_{m}}{\tilde{u}_{l}-\tilde{u}_{m}}=0.

5.3 Emergence of dark states

In the following we show numerical examples of dark states emerging due to $\kappa\to 0,\Theta B^{z}$ when condition (28) is met. We set the model parameters as $L=3,\beta=0.1,B^{z}=1$ , ⁵⁵5This $\underline{\epsilon}$ was generated in Mathematica with SeedRandom 1324 using RandomReal in the range 0.4 to 1.3.

\displaystyle\underline{\epsilon}=\{0.5869853530170386,1.270553831408777,1.2426482643150194\},

and only look at the case where all spins are $s_{i}=1/2$ . Then for different $B^{y}$ we change $B^{x}\to 0$ so that we approach conditions where dark states emerge. As we use states constructed from $v^{+}$ with the lowering operators $S^{-}$ , we should observe for dark states that $\kappa\to 0,+B^{z}$ . We label states by the eigenvalue $q_{1}$ of $Q_{1}$ .

5.3.1 Dark states for $B^{y}=L\sqrt{2\beta}$ .

In this case there are only two dark states in the limit, both with the same eigenvalues for the charges but with different energies.

$B^{x}/2$	$\kappa_{1}$	$\kappa_{2}$	$q_{1}$
$10^{0}$	$-5.72938\times 10^{-2}$	$1+5.72938\times 10^{-2}$	$-2.57221$
$10^{-1}$	$-6.70952\times 10^{-5}$	$1+6.70952\times 10^{-5}$	$-2.14310$
$10^{-2}$	$-6.33204\times 10^{-7}$	$1+6.33204\times 10^{-7}$	$-2.13838$
$10^{-3}$	$-6.32830\times 10^{-9}$	$1+6.32830\times 10^{-9}$	$-2.13833$
$10^{-4}$	$-6.32829\times 10^{-11}$	$1+6.32829\times 10^{-11}$	$-2.13833$

5.3.2 Dark states for $B^{y}=\left(L-2\right)\sqrt{2\beta}$ .

In this case there are six dark states in the limit.

$B^{x}/2$	$\kappa_{1}$	$\kappa_{2}$	$q_{1}$
$10^{-1}$	$-1.03242\times 10^{-5}$	$1+1.03242\times 10^{-5}$	$-1.90365$
$10^{-2}$	$-9.33203\times 10^{-8}$	$1+9.33203\times 10^{-8}$	$-1.89852$
$10^{-3}$	$-9.32225\times 10^{-10}$	$1+9.32225\times 10^{-10}$	$-1.89846$
$10^{-4}$	$-9.30916\times 10^{-12}$	$1+9.30916\times 10^{-12}$	$-1.89846$

$B^{x}/2$	$\kappa_{1}$	$\kappa_{2}$	$q_{1}$
$10^{-1}$	$-2.97399\times 10^{-3}$	$1+2.97399\times 10^{-5}$	$-9.88279\times 10^{-2}$
$10^{-2}$	$-2.91182\times 10^{-5}$	$1+2.91182\times 10^{-5}$	$-8.68202\times 10^{-2}$
$10^{-3}$	$-2.91118\times 10^{-7}$	$1+2.91118\times 10^{-7}$	$-8.66989\times 10^{-2}$
$10^{-4}$	$-2.91117\times 10^{-9}$	$1+2.91117\times 10^{-9}$	$-8.66977\times 10^{-2}$

$B^{x}/2$	$\kappa_{1}$	$\kappa_{2}$	$q_{1}$
$10^{-1}$	$-1.07577\times 10^{-3}$	$1+1.07577\times 10^{-3}$	$3.78262$
$10^{-2}$	$-1.03663\times 10^{-5}$	$1+1.03663\times 10^{-5}$	$3.77776$
$10^{-3}$	$-1.03624\times 10^{-7}$	$1+1.03624\times 10^{-7}$	$3.77771$
$10^{-4}$	$-1.03623\times 10^{9}$	$1+1.03623\times 10^{-9}$	$3.77771$

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Dark states in an integrable X​YXY central spin model

Abstract

1 Introduction

2 Derivation of integrable X​YXY central spin Hamiltonians

3 The eigenstates

3.1 Relations for the Algebraic Bethe Ansatz

3.1.1 Reparametrised operators.

3.2 Eigenstates of HX​YH_{XY}

3.2.1 Construction from supersymmetry.

3.2.2 Dark states.

3.2.3 Bright states

3.2.4 Generic eigenstates of the charges 𝒬j\mathcal{Q}_{j}.

3.3 Generic eigenstates of the X​YXY model

Finite roots in the limit.

One of the roots diverges.

4 Conclusion

5 Appendix

5.1 Characterisation of magnetic field configurations admitting dark states

5.2 Eigenstates of the X​XXX central spin model

5.2.1 Recovering the dark states for Bx=By=0B^{x}=B^{y}=0.

5.3 Emergence of dark states

5.3.1 Dark states for By=L​2​βB^{y}=L\sqrt{2\beta}.

5.3.2 Dark states for By=(L−2)​2​βB^{y}=\left(L-2\right)\sqrt{2\beta}.

References

References

Dark states in an integrable $XY$ central spin model

2 Derivation of integrable $XY$ central spin Hamiltonians

3.2 Eigenstates of $H_{XY}$

3.2.4 Generic eigenstates of the charges $\mathcal{Q}_{j}$ .

3.3 Generic eigenstates of the $XY$ model

5.2 Eigenstates of the $XX$ central spin model

5.2.1 Recovering the dark states for $B^{x}=B^{y}=0$ .

5.3.1 Dark states for $B^{y}=L\sqrt{2\beta}$ .

5.3.2 Dark states for $B^{y}=\left(L-2\right)\sqrt{2\beta}$ .