$\displaystyle(L_{j}^{1})^{n+2}=(2J)^{n+2}((9x_{n}+3)-(9x_{n}+2)\tau^{z}_{j-1}\sigma_{0}s_{y}$
$\displaystyle-(9x_{n}+2)\tau^{z}_{j+2}\sigma_{y}s_{0}+(9x_{n}+2)\tau^{z}_{j-1}\tau^{z}_{j+2}\sigma_{y}s_{y}.$
(84)

giving

$\displaystyle
x_{n+2}=9x_{n}+2\quad\Rightarrow\quad(x_{n+2}+\frac{1}{4})=3^{2}(x_{n}+\frac{1}{4})$
$\displaystyle\Rightarrow$ $\displaystyle
x_{n}+\frac{1}{4}=3^{n-2}(x_{2}+\frac{1}{4})\quad\Rightarrow\quad
x_{n}=\frac{3^{n}-1}{4}.$ (85)

Thus, for $1\leq m\in\mathbb{Z}$,

$\displaystyle
L_{j}^{2m}=\frac{(2J)^{2m}}{4}\left((3^{2m}+3)\sigma_{0}s_{0}\right.$
$\displaystyle\left.+(3^{2m}-1)(-\tau^{z}_{j-1}\sigma_{0}s_{y}-\tau^{z}_{j+2}\sigma_{y}s_{0}+\tau^{z}_{j-1}\tau^{z}_{j+2}\sigma_{y}s_{y})\right),$
(86) $\displaystyle
L_{j}^{2m-1}=\frac{(2J)^{2m-1}}{4}\left((3^{2m-1}-3)\sigma_{0}s_{0}\tau^{z}_{j-1}\tau^{z}_{j+2}\right.$
$\displaystyle\left.+(3^{2m-1}+1)(-\tau^{z}_{j+2}\sigma_{0}s_{y}-\tau^{z}_{j-1}\sigma_{y}s_{0}+\sigma_{y}s_{y})\right).$
(87)

From these results, one could obtain arbitrary iterated commutation between
$H_{2}$ and the two-spin terms $B_{j,j+1}$. For instance,

$\displaystyle[H_{2}^{(2m)},\tau^{x}_{j}\tau^{x}_{j+1}]=(1,0,0,0)L_{j}^{2m}B_{j,j+1}$
$\displaystyle=\frac{(2J)^{2m}}{4}\left((3^{2m}+3)\tau^{x}_{j}\tau^{x}_{j+1}+(3^{2m}-1)\right.$
$\displaystyle\quad\times\left.(\tau^{z}_{j-1}i\tau^{x}_{j}\tau^{y}_{j+1}+i\tau^{y}_{j}\tau^{x}_{j+1}\tau^{z}_{j+2}-\tau^{z}_{j-1}\tau^{y}_{j}\tau^{y}_{j+1}\tau^{z}_{j+2})\right)$
(88)
$\displaystyle[H_{2}^{(2m-1)},\tau^{x}_{j}\tau^{x}_{j+1}]=(1,0,0,0)L_{j}^{2m-1}B_{j,j+1}$
$\displaystyle=\frac{(2J)^{2m-1}}{4}\left((3^{2m-1}-3)\tau^{z}_{j-1}\tau^{x}_{j}\tau^{x}_{j+1}\tau^{z}_{j+2}+\right.$
$\displaystyle\left.(3^{2m-1}+1)(i\tau^{x}_{j}\tau^{y}_{j+1}\tau^{z}_{j+2}+\tau^{z}_{j-1}i\tau^{y}_{j}\tau^{x}_{j+1}-\tau^{y}_{j}\tau^{y}_{j+1})\right)$
(89)

where the vector $(1,0,0,0)$ is acting on the $4\times 4$ matrices
$\sigma_{0}s_{y}$ etc. in Eqs. (86) and (87). In sum, the commutation of
$H_{2}$ with two-spin terms would result in the generic form

$\displaystyle[H_{2}^{(n)},\tau^{\nu_{1}}_{j}\tau^{\nu_{2}}_{j+1}]=\sum_{\scriptsize\begin{array}[]{l}m_{1}m_{2}=0,1\\\
\mu_{1}\mu_{2}=x,y\end{array}}\beta^{(n)}_{m_{1}m_{2},\mu_{1}\mu_{2}}(\tau^{z}_{j-1})^{m_{1}}\tau^{\mu_{1}}_{j}\tau^{\mu_{2}}_{j+1}(\tau^{z}_{j+2})^{m_{2}},$
(92)

where $\beta^{(n)}_{m_{1}m_{2},\mu_{1}\mu_{2}}$’s are given in Eqs. (88) and
(89).

##  Appendix B Algebraic proof for scaling relations

###  B.1 IPR scaling: amplitudes of original cat scar components and DTC
amplitudes

Here we would focus on the first-order correction to the wave function and
obtain the leading order deviation of IPR from the unperturbed values. This is
intimately related to the DTC amplitudes. Specifically, for the fine-tuned
solutions in Eq. (5), the only non-degenerate scars take either the FM or AFM
configuration. We would denote both of them by $\\{s_{j}^{(\text{cat})}\\}$ as
analysis for the two configurations are the same,

$\displaystyle|\ell,\\{s_{j}^{(\text{cat})}\\}\rangle$
$\displaystyle=|\ell,\text{FM}\rangle=|\ell,\\{s_{j}=(+1)^{j}\\}\rangle$ or
$\displaystyle=|\ell,\text{AFM}\rangle=|\ell,\\{s_{j}=(-1)^{j}\\}\rangle.$
(93)

Under perturbation, the first-order perturbation gives

$\displaystyle|\tilde{\omega}_{\ell,\text{cat}}\rangle=\alpha_{0}\left(|\ell,\\{s_{j}^{(\text{cat})}\\}\rangle+i\lambda\sum_{\ell^{\prime},\\{s_{j}\\}^{\prime}\neq\ell,\\{s_{j}^{(\text{cat})}\\})}\frac{\langle\ell^{\prime},\\{s_{j}\\}^{\prime}|V_{1}|\ell,\\{s_{j}^{(\text{cat})}\\}\rangle}{e^{i\left(E(\ell,\\{s_{j}^{(\text{cat})}\\})-E(\ell^{\prime},\\{s_{j}\\}^{\prime})\right)}-1}|\ell^{\prime},\\{s_{j}\\}^{\prime}\rangle\right)+O(\lambda^{2}).$
(94)

Now, the generic form for $V_{1}$ in Eq. (71) means that only consecutive one
or two spins can be flipped. That means all
$|\ell^{\prime},\\{s_{j}\\}^{\prime}\rangle$ in Eq. (94) must differ from the
scar configurations $|\ell,\\{s_{j}^{(\text{cat})}\\}\rangle$ by two domain
walls. Recall $E(\ell,\\{s_{j}\\})=E(\ell,w)=\pi\ell-J(L-2w)$, with $w=0,L$
for FM and AFM configurations respectively. All quasienergy differences in the
denominator of Eq. (94) then give identical contributions

$\displaystyle
E(\ell,\\{s_{j}^{\text{cat}}\\})-E(\ell^{\prime},\\{s_{j}\\}^{\prime})=\pi(\ell-\ell^{\prime})\mp
4J,$ (95)

where $\mp$ signs correspond to FM or AFM scars. Thus, the denominator can be
factored out from the summation. Meanwhile, for the generic form of $V_{1}$,

$\displaystyle V_{1}=\sum_{j=1}^{L}V_{1j},$ $\displaystyle
V_{1j}=\sum_{m_{1},m_{2}=0,1}\left(\sum_{\mu_{1}=x,y}\alpha_{m_{1}m_{2},\mu_{1}}(\tau^{z}_{j-1})^{m_{1}}\tau^{\mu_{1}}_{j}(\tau^{z}_{j+1})^{m_{2}}\right.$
$\displaystyle+\left.\sum_{\mu_{1}\mu_{2}=x,y}\beta_{m_{1}m_{2},\mu_{1}\mu_{2}}(\tau^{z}_{j-1})^{m_{1}}\tau^{\mu_{1}}_{j}\tau^{\mu_{2}}_{j+1}(\tau^{z}_{j+2})^{m_{2}}\right),$
(96)

translation invariance implies that

$\displaystyle\mathbb{T}_{x}V_{1}\mathbb{T}_{x}^{-1}=V_{1}\qquad\Rightarrow\qquad\mathbb{T}_{x}V_{1j}\mathbb{T}_{x}^{-1}=V_{1,j+1}.$
(97)

Also, recall that cat scars satisfy the projective translation symmetry ($\pm$
signs below are for FM and AFM configurations respectively)

$\displaystyle\mathbb{T}_{x}|\ell,\\{s_{j}^{(\text{cat})}\\}\rangle=(\pm
1)^{\ell}|\ell,\\{s_{j}^{(\text{cat})}\\}\rangle.$ (98)

Then, the matrix elements for each site $V_{1j}$ are identical to the same
terms two sites away,

$\displaystyle\sum_{\ell^{\prime},\\{s_{j}\\}^{\prime}}\langle\ell^{\prime},\\{s_{j}\\}^{\prime}|V_{1,j+2}|\ell,\\{s_{j}^{(\text{cat})}\\}\rangle|\ell^{\prime},\\{s_{j}\\}^{\prime}\rangle$
$\displaystyle=\sum_{\ell^{\prime},\\{s_{j}\\}^{\prime}}\langle\ell^{\prime},\\{s_{j}\\}^{\prime}|\mathbb{T}_{x}^{2}V_{1j}(\mathbb{T}_{x}^{-1})^{2}|\ell,\\{s_{j}^{(\text{cat})}\\}\rangle|\ell^{\prime},\\{s_{j}\\}^{\prime}\rangle$
$\displaystyle=\sum_{\ell^{\prime},\\{s_{j}\\}^{\prime}}\langle\ell^{\prime},\\{s_{j}\\}^{\prime}|\mathbb{T}_{x}^{2}V_{1j}|\ell,\\{s_{j}^{(\text{cat})}\\}\rangle|\ell^{\prime},\\{s_{j}\\}^{\prime}\rangle$
$\displaystyle=\sum_{\ell^{\prime},\\{s_{j}\\}^{\prime}}\langle\ell^{\prime},\\{s_{j}\\}^{\prime}|V_{1,j}|\ell,\\{s_{j}^{(\text{cat})}\\}\rangle\left(\mathbb{T}_{x}^{2}|\ell^{\prime},\\{s_{j}\\}^{\prime}\rangle\right),$

where in the last step we shift the dummy configuration
$\sum_{\\{s_{j}\\}^{\prime}}|\ell^{\prime},\\{s_{j}\\}^{\prime}\rangle\langle\ell^{\prime},\\{s_{j}\\}^{\prime}|=\sum_{\\{s_{j}\\}^{\prime}}\mathbb{T}_{x}^{2}|\ell^{\prime},\\{s_{j}\\}^{\prime}\rangle\langle\ell^{\prime},\\{s_{j}\\}^{\prime}|(\mathbb{T}_{x}^{-1})^{2}$.
That means Eq. (94) can be written as

$\displaystyle|\tilde{\omega}_{\ell,\text{cat}}\rangle=\alpha_{0}\left(|\ell,\\{s_{j}^{(\text{cat})}\\}\rangle+\lambda\sum_{\ell^{\prime},\\{s_{j}\\}^{\prime}}\frac{\langle\ell^{\prime},\\{s_{j}\\}^{\prime}|(V_{1,j=1}+V_{1,j=2})|\ell,\\{s_{j}^{(\text{cat})}\\}\rangle}{e^{i(\pi(\ell-\ell^{\prime})\mp
4J}-1}\sum_{m=0}^{L/2-1}\mathbb{T}_{x}^{2m}|\ell^{\prime},\\{s_{j}\\}^{\prime}\rangle\right)+O(\lambda^{2}).$
(99)

Denote the averaged strength of the first order perturbation as

$\displaystyle\bar{V}_{1}^{2}$
$\displaystyle=\frac{1}{2}\sum_{\ell^{\prime},\\{s_{j}\\}^{\prime}\neq\ell,\\{s_{j}^{(\text{cat})}\\}}\left|\frac{\langle\ell^{\prime},\\{s_{j}\\}^{\prime}|(V_{1,j=1}+V_{1,j=2})|\ell,\\{s_{j}^{(\text{cat})}\\}\rangle}{e^{i(\pi(\ell-\ell^{\prime})\mp
4J)}-1}\right|^{2}$
$\displaystyle=\frac{1}{8}\sum_{\ell^{\prime},\\{s_{j}\\}^{\prime}\neq\ell,\\{s_{j}^{(\text{cat})}\\}}\left|\langle\ell^{\prime},\\{s_{j}\\}^{\prime}|(V_{1,j=1}+V_{1,j=2})|\ell,\\{s_{j}^{(\text{cat})}\\}\rangle\right|^{2}$
$\displaystyle\qquad\qquad\times\csc^{2}\left(\frac{\pi(\ell-\ell^{\prime})\mp
4J}{2}\right),$ (100)

the normalization constant $\alpha_{0}$, related to amplitudes for the
original cat scar components $|\ell,\\{s_{j}^{(\text{cat})}\\}\rangle$ in the
unperturbed cat scars $|\tilde{\omega}_{\ell,\text{cat}}\rangle$, is rescaled
to

$\displaystyle
1=|\langle\tilde{\omega}_{\ell,\text{cat}}|\tilde{\omega}_{\ell,\text{cat}}\rangle|^{2}=\alpha_{0}^{2}(1+\lambda^{2}\bar{V}_{1}^{2}L)$
$\displaystyle\Rightarrow\qquad$
$\displaystyle\alpha_{0}^{2}=\frac{1}{1+\bar{V}_{1}^{2}\lambda^{2}L}=|\langle\ell,\\{s_{j}^{(\text{cat})}\\}|\tilde{\omega}_{\ell,\text{cat}}\rangle|^{2}.$
(101)

###  B.2 Scaling for amplitudes of other spin components: Fock space
localization

IPR scaling in the previous section characterizes the amplitude rescaling for
FM or AFM components
$|\ell,\\{s_{j}^{(\text{cat})}\\}\rangle\sim|\pm\\{s_{j}^{(\text{cat})}\\}\rangle$
in perturbed cat scars eigenstates $|\tilde{\omega}_{\ell,\text{cat}}\rangle$.
It is chiefly contributed by the first-order DW fluctuations on top of the
scar configurations $\pm\\{s_{j}^{(\text{cat})}\\}$. Here, we would further
consider all higher-order corrections to cat scar eigenstates and observe the
amplitudes for other spin configurations in
$|\tilde{\omega}_{\ell,\text{cat}}\rangle$. To quantitatively describe the
deviations of spin configurations away from the cat scar patterns, we recall
the Fock space distance $\delta s$ in Eq. (59) and further define the pairwise
Fock space distance $\Delta s$,

$\displaystyle\delta s(\\{s_{j}\\},\\{s_{j}^{(\text{cat})}\\})$
$\displaystyle=\frac{1}{2}\sum_{j=1}^{L}|s_{j}-s_{j}^{(\text{cat})}|,$
$\displaystyle\Delta s(\\{s_{j}\\},\pm\\{s_{j}^{(\text{cat})}\\})$
$\displaystyle=\frac{1}{2}\min\left(\sum_{j=1}^{L}|s_{j}-s_{j}^{(\text{cat})}|,\sum_{j=1}^{L}|s_{j}+s_{j}^{(\text{cat})}|\right).$
(102)

Here $\Delta s(\\{s_{j}\\},\pm\\{s_{j}^{(\text{cat})}\\})$ is abbreviated in
the main text as $\Delta s_{\text{cat}}(\\{s_{j}\\})$. An example for $L=4$ is
given in Fig. 13 (a) and (d). Intuitively, $\delta
s(\\{s_{j}\\},\\{s_{j}^{(\text{cat})}\\})$ characterizes that staring from
$+\\{s_{j}^{(\text{cat})}\\}$, how many spins are flipped in order to end up
with $\\{s_{j}\\}$. Similarly, $\Delta
s(\\{s_{j}\\},\pm\\{s_{j}^{(\text{cat})}\\})$ counts the minimal numbers of
spin flips to go from either of the cat scar configuration pair
$\pm\\{s_{j}^{(\text{cat})}\\}$ to $\\{s_{j}\\}$. Correspondingly, Fock space
localization means that certain eigenstates, i.e. the perturbed cat scars
$|\tilde{\omega}_{\ell,\text{cat}}\rangle$, exhibit exponential decay for the
Fock basis coefficients
$|\langle\\{s_{j}\\}|\tilde{\omega}_{\ell,\text{cat}}\rangle|$ with the
increase of $\Delta s$.

In the following, let us quantitatively analyze the scaling of Fock basis
amplitudes for cat scars including all perturbation orders. General
perturbation series for the corrected cat scar eigenstates reads

$\displaystyle|\tilde{\omega}_{\ell,\text{cat}}\rangle=\cdots
e^{i\lambda^{k}S_{k}}\cdots e^{i\lambda^{2}S_{2}}e^{i\lambda
S_{1}}|\ell,\\{s_{j}^{(\text{cat})}\\}\rangle,$ (103)

where the Hermitian matrices $S_{1},S_{2},\cdots,S_{k}$ would diagonalize the
perturbed Floquet operator up to the $\lambda^{k}$-th order,

$\displaystyle\langle\ell_{1},\\{s_{j}\\}_{1}|e^{-i\lambda
S_{1}}e^{-i\lambda^{2}S_{2}}\cdots
e^{-i\lambda^{k}S_{k}}\left(U_{0}e^{i\sum_{k=1}^{\infty}\lambda^{k}V_{k}}\right)e^{i\lambda^{k}S_{k}}\cdots$
$\displaystyle\qquad\times e^{i\lambda^{2}S_{2}}e^{i\lambda
S_{1}}|\ell_{2},\\{s_{j}\\}_{2}\rangle\propto\delta_{\ell_{1},\ell_{2}}\delta_{\\{s_{j}\\}_{1},\\{s_{j}\\}_{2}}+O(\lambda^{k+1}).$
(104)

For our purposes here, it will be most useful and sufficient to obtain the
formal operator solutions for $S_{k}$’s, and especially to prove their
associated selection rules. To gain some intuition for general forms, we check
the first order results, where off-diagonal elements for $S_{1}$ satisfy (i.e.
expand Eq. (104) up to $\lambda^{1}$ orders)

$\displaystyle 1-(1-i\lambda S_{1})(U_{0}(1+i\lambda V_{1}))(1+i\lambda
S_{1})=0$ $\displaystyle\Rightarrow\qquad$
$\displaystyle[S_{1},U_{0}]=U_{0}V_{1}\quad\Rightarrow
S_{1}-U_{0}^{\dagger}S_{1}U_{0}=-V_{1}.$ (105)

Then, formal solutions for Eq. (105) can be written as

$\displaystyle S_{1}=-\sum_{p=0}^{\infty}(U_{0}^{\dagger})^{p}V_{1}U_{0}^{p}.$
(106)

Using the solution forms in Eq. (106), we see that $S_{1}$ corresponds to
repeated gauge transformation by $U_{0}$ for the first-order perturbation
$V_{1}$. So according to the generalized selection rules in Eq. (72), $S_{1}$
inherits the selection rules of $V_{1}$ as

$\displaystyle\langle\\{s_{j}\\}|S_{1}|\\{\tilde{s}_{j}\\}^{\prime}\rangle\neq
0\qquad\Rightarrow\qquad\delta s(\\{s_{j}\\},\pm\\{\tilde{s}_{j}\\})\leqslant
n_{\text{op}},$ (107)

where $n_{\text{op}}$ for operator product powers in perturbation Hamiltonians
is given in Eq. (50).

Now, we obtain the generic form for $S_{k}$ using deductions. Suppose
$S_{k-1}$ already diagonalize Eq. (104) up to the $\lambda^{k-1}$-th order,
and $\\{S_{q}|q=1,2,\dots,k-1\\}$ satisfy the selection rule that
$\langle\\{s_{j}\\}_{1}|S_{q}|\\{s_{j}\\}_{2}\rangle\neq 0\Rightarrow\delta
s(\\{s_{j}\\}_{1},\\{s_{j}\\}_{2})\leqslant n_{\text{op}}q$. Note the
expansion

$\displaystyle e^{i\sum_{k=1}^{\infty}\lambda^{k}V_{k}}$
$\displaystyle=\sum_{\alpha=0}^{\infty}\frac{i^{\alpha}}{\alpha!}(\sum_{k=1}^{\infty}\lambda^{k}V_{k})^{\alpha}$
$\displaystyle=\sum_{\alpha=0}^{\infty}\frac{i^{\alpha}}{\alpha!}\sum_{\\{k_{1},k_{2},\dots,k_{\alpha}\geqslant
1\\}}\lambda^{\sum_{j=1}^{\alpha}k_{j}}V_{k_{1}}V_{k_{2}}\dots V_{k_{\alpha}}$
(108)

Then, for the $\lambda^{k}$-th order, we have for non-diagonal elements

$\displaystyle 0=i[U_{0},S_{k}]+iU_{0}V_{k}$
$\displaystyle+\sum_{\alpha=0}^{k-1}\sum_{\scriptsize\begin{array}[]{c}\\{m_{j},n_{j},=0,1,\dots,k-1,\\\
k_{j}=1,2,\dots,k-1|\\\
\sum_{j=1}^{k-1}j(m_{j}+n_{j})+\sum_{j=1}^{\alpha}k_{j}=k\\}\end{array}}\frac{i^{\alpha+\sum_{j=1}^{k-1}(m_{j}-n_{j})}}{\alpha!\prod_{j=1}^{k}m_{j}!n_{j}!}$
(112) $\displaystyle\times\left(S_{1}^{n_{1}}S_{2}^{n_{2}}\cdots
S_{k-1}^{n_{k-1}}\right)U_{0}\left(V_{k_{1}}V_{k_{2}}\dots
V_{k_{\alpha}}\right)\left(S_{k-1}^{m_{k-1}}\cdots
S_{2}^{m_{2}}S_{1}^{m_{1}}\right)$ $\displaystyle\Rightarrow$ $\displaystyle
S_{k}-U_{0}^{\dagger}S_{k}U_{0}$
$\displaystyle=-V_{k}+i\sum_{\alpha=1}^{k-1}\sum_{\scriptsize\begin{array}[]{c}\\{m_{j},n_{j}=0,1,\dots,k-1,\\\
k_{j}=1,2,\dots,k-1|\\\
\sum_{j=1}^{k-1}j(m_{j}+n_{j})+\sum_{j=1}^{\alpha}k_{j}=k\\}\end{array}}\frac{i^{\alpha+\sum_{j=1}^{k-1}(m_{j}-n_{j})}}{\alpha!\prod_{j=1}^{k}m_{j}!n_{j}!}$
(116) $\displaystyle\times
U_{0}^{\dagger}\left(S_{1}^{n_{1}}S_{2}^{n_{2}}\cdots
S_{k-1}^{n_{k-1}}\right)U_{0}\left(V_{k_{1}}V_{k_{2}}\dots
V_{k_{\alpha}}\right)\left(S_{k-1}^{m_{k-1}}\cdots
S_{2}^{m_{2}}S_{1}^{m_{1}}\right).$

Thus, the formal solutions can be similarly written as

$\displaystyle S_{k}=-\sum_{p=0}^{\infty}(U_{0}^{\dagger})^{p}V_{k}U_{0}^{p}$
$\displaystyle+i\sum_{p=0}^{\infty}(U_{0}^{\dagger})^{p}(\sum_{\alpha=1}^{k-1}\sum_{\scriptsize\begin{array}[]{c}\\{m_{j},n_{j}=0,1,\dots,k-1,\\\
k_{j}=1,2,\dots,k-1|\\\
\sum_{j=1}^{k-1}j(m_{j}+n_{j})+\sum_{j=1}^{\alpha}k_{j}=k\\}\end{array}}\frac{i^{\alpha+\sum_{j=1}^{k-1}(m_{j}-n_{j})}}{\alpha!\prod_{j=1}^{k}m_{j}!n_{j}!}$
(120) $\displaystyle\times
U_{0}^{\dagger}\left(S_{1}^{n_{1}}S_{2}^{n_{2}}\cdots
S_{k-1}^{n_{k-1}}\right)U_{0}\left(V_{k_{1}}V_{k_{2}}\dots
V_{k_{\alpha}}\right)$ $\displaystyle\times\left(S_{k-1}^{m_{k-1}}\cdots
S_{2}^{m_{2}}S_{1}^{m_{1}}\right))U_{0}^{p}.$ (121)

The first line in Eq. (121) takes the same form as Eq. (106), which is a gauge
transformation of $V_{k}$ with $U_{0}$. Thus, it inherits the selection rules
of $V_{k}$ of relating configurations at most $\delta s\leqslant
n_{\text{op}}k$ spins apart. Also, operators $V_{1},\dots,V_{k-1}$ and
$S_{1},\dots,S_{k-1}$, already satisfies the selection rules according to the
assumptions of deduction. The second line corresponds to gauge transformations
of these operators. Thus, they also satisfy the selection rules of flipping at
most $\delta
s\leqslant\sum_{j=1}^{k-1}(n_{\text{op}}j)(m_{j}+n_{j})+\sum_{j=1}^{\alpha}(n_{\text{op}}p_{j})q_{j}=n_{\text{op}}k$
spins. In sum, the operator $S_{k}$ satisfy the selection rule

$\displaystyle\langle\\{s_{j}\\}|S_{k}|\\{\tilde{s}_{j}\\}^{\prime}\rangle\neq
0\quad\Rightarrow\quad\delta s(\\{s_{j}\\},\\{\tilde{s}_{j}\\})\leqslant
n_{\text{op}}k.$ (122)

That completes the deductive proof for arbitrary orders $k$.

Finally, the perturbed cat scar eigenstate can be sorted by orders of
$\lambda^{k}$ as

$\displaystyle|\tilde{\omega}_{\ell,\text{cat}}\rangle$ $\displaystyle=\cdots
e^{i\lambda^{k}S_{k}}\cdots e^{i\lambda^{2}S_{2}}e^{i\lambda
S_{1}}|\ell,\\{s_{j}^{(\text{cat})}\\}\rangle$
$\displaystyle=|\ell,\\{s_{j}^{(\text{cat})}\\}\rangle+\sum_{k=1}^{\infty}\lambda^{k}\sum_{\scriptsize\begin{array}[]{l}\\{m_{j}=0,1,\dots,k|\\\
\sum_{j=1}^{k}jm_{j}=k\\}\end{array}}\frac{i^{\sum_{j=1}^{k}m_{j}}}{\prod_{j=1}^{k}m_{j}!}$
(125) $\displaystyle\qquad\times S_{k}^{m_{k}}\cdots
S_{2}^{m_{2}}S_{1}^{m_{1}}|\ell,\\{s_{j}^{(\text{cat})}\\}\rangle.$ (126)

Using the selection rules for $S_{k}$ in Eq. (122), the maximal number of
spins that can be flipped by the $\lambda^{k}$ terms
$\leqslant\sum_{j=1}^{k}n_{\text{op}}j\times m_{j}=n_{\text{op}}k$. Now, note
that in the unperturbed eigenstate
$|\ell,\\{s_{j}^{(\text{cat})}\\}\rangle=(1/\sqrt{2})(|\\{s_{j}^{(\text{cat})}\\}\rangle+(-1)^{\ell}|-\\{s_{j}^{(\text{cat})}\\}\rangle)$,
there are pairwise Fock product states $|\pm\\{s_{j}^{(\text{cat})}\\}\rangle$
to start with. Thus, Fock states $|\\{s_{j}\\}\rangle$ with $\Delta
s(\\{s_{j}\\},\pm\\{s_{j}^{(\text{cat})}\\})$ spins flipped with respect to
cat configurations $|\pm\\{s_{j}^{(\text{cat})}\\}\rangle$ will only show up
above perturbation orders $\lambda^{k\geqslant\Delta
s(\\{s_{j}\\},\pm\\{s_{j}^{(\text{cat})}\\})/n_{\text{op}}}$, with pairwise
Fock space distance $\Delta s$ in Eq. (102). Then, we have the scaling
relation characterizing Fock space localization

$\displaystyle|\langle\\{s_{j}\\}|\tilde{\omega}_{\ell,\text{cat}}\rangle|^{2}=O(\lambda^{\Delta
s(\\{s_{j}\\},\pm\\{s_{j}^{(\text{cat})}\\})/\xi}),\qquad\xi\leqslant
n_{\text{op}}.$ (127)

Recall that $\lambda\lesssim 0.1$ in the DTC regime, so indeed the cat scar
eigenstates $|\tilde{\omega}_{\ell,\text{cat}}\rangle$ is exponentially
centered at the Fock space around $|\pm\\{s_{j}^{(\text{cat})}\\}\rangle$ as
the coefficients for other spin configurations decay exponentially
$\lambda^{\Delta s(\\{s_{j}\\},\pm\\{s_{j}^{(\text{cat})}\\})/\xi}=e^{-\Delta
s(\\{s_{j}\\},\pm\\{s_{j}^{(\text{cat})}\\})|\ln(1/\lambda)|/\xi}$ with the
increase of spin differences between $\\{s_{j}\\}$ and
$\pm\\{s_{j}^{(\text{cat})}\\}$.

###  B.3 Spectral gap scaling: exponential growth of DTC lifetime with the
increase of $L$

In the previous two sections, we have proved that for the perturbed cat scars
$|\tilde{\omega}_{\ell,\text{cat}=\text{FM/AFM}}\rangle$, the amplitudes for
original FM or AFM components experience a minor reduction to
$\alpha_{0}^{2}=|\langle\ell,\\{s_{j}^{(\text{cat})}\\}|\tilde{\omega}_{\ell,\text{cat}}\rangle|^{2}=1/(1+\bar{V}_{1}^{2}\lambda^{2}L)$
by the first-order domain wall fluctuations. Other spin configurations show
exponential Fock space localization. Thus, in intermediate scale systems
$L\lesssim 1/\lambda^{2}$, an FM (or AFM) initial state will chiefly overlap
with two cat scars
$|\tilde{\omega}_{\ell=0,\text{FM}}\rangle,|\tilde{\omega}_{\ell=1,\text{FM}}\rangle$
(or $|\tilde{\omega}_{\ell=0,\text{AFM}}\rangle$,
$|\tilde{\omega}_{\ell=1,\text{AFM}}\rangle$). In this section, we would
further prove that the spectral gap for the scars of a certain configuration,
i.e. $\tilde{\omega}_{1,\text{FM}}-\tilde{\omega}_{0,\text{FM}}$ or
$\tilde{\omega}_{1,\text{AFM}}-\tilde{\omega}_{0,\text{AFM}}$, approaches the
unperturbed value $\pi$ with exponential accuracy as the system size $L$
increases. In particular, we would explicitly demonstrate the origin of
exponentially small gap deviation from $\pi$ in finite-size systems, and give
a bound on the scaling exponents.

Let us first gain some intuitions for the general proof by checking the first-
order quasienergy correction,

$\displaystyle\omega_{\ell,\text{cat}}^{(1)}$
$\displaystyle=\langle\ell,\\{s_{j}^{(\text{cat})}\\}|V_{1}|\ell,\\{s_{j}^{(\text{cat})}\\}\rangle.$
(128)

Note that $V_{1}$ can flip up to $n_{\text{op}}$ spins, i.e. for perturbations
up to two-spin terms like $\phi\tau^{x}_{j}\tau^{x}_{j+1}$ we have
$n_{\text{op}}=2$. Then, for system sizes $L>4$, matrix elements like

$\displaystyle\langle\ell,\text{FM}|V_{1}|\ell,\text{FM}\rangle$
$\displaystyle={\color[rgb]{0,.5,0}\frac{1}{2}(\langle\uparrow\uparrow\uparrow\uparrow\cdots|V_{1}|\uparrow\uparrow\uparrow\uparrow\cdots\rangle+\langle\downarrow\downarrow\downarrow\downarrow\cdots|V_{1}|\downarrow\downarrow\downarrow\downarrow\cdots\rangle)}$
$\displaystyle+{\color[rgb]{1,0,0}\frac{(-1)^{\ell}}{2}(\langle\uparrow\uparrow\uparrow\uparrow\dots|V_{1}|\downarrow\downarrow\downarrow\downarrow\dots\rangle+\langle\downarrow\downarrow\downarrow\downarrow\cdots|V_{1}|\uparrow\uparrow\uparrow\uparrow\cdots\rangle)},$
$\displaystyle\langle\ell,\text{AFM}|V_{1}|\ell,\text{AFM}\rangle$
$\displaystyle={\color[rgb]{0,.5,0}\frac{1}{2}(\langle\uparrow\downarrow\uparrow\downarrow\dots|V_{1}|\uparrow\downarrow\uparrow\downarrow\dots\rangle+\langle\downarrow\uparrow\downarrow\uparrow\dots|V_{1}|\downarrow\uparrow\downarrow\uparrow\dots\rangle)}$
$\displaystyle+{\color[rgb]{1,0,0}\frac{(-1)^{\ell}}{2}(\langle\uparrow\downarrow\uparrow\downarrow\dots|V_{1}|\downarrow\uparrow\downarrow\uparrow\dots\rangle+\langle\downarrow\uparrow\downarrow\uparrow\dots|V_{1}|\uparrow\downarrow\uparrow\downarrow\dots\rangle)},$
(129)

shows vanishing cross terms (denoted by red), because those red terms require
simultaneously flipping $L$ spins, which violates the selection rules. Thus,
these matrix elements are completely independent of spectral pairing quantum
numbers $\ell$, and we see that it is again the selection rules enforcing the
identical quasienergy correction for pairwise scars,

$\displaystyle\omega_{0,\text{cat}}=\omega_{1,\text{cat}}=\langle\ell,\\{s_{j}^{(\text{cat})}\\}|V_{1}|\ell,\\{s_{j}^{(\text{cat})}\\}\rangle$
$\displaystyle={\color[rgb]{0,.5,0}\frac{1}{2}\sum_{m=0,1}\langle(-1)^{m}\\{s_{j}^{(\text{cat})}\\}|V_{1}|(-1)^{m}\\{s_{j}^{(\text{cat})}\\}\rangle.}$
(130)

Based on the first-order solutions, we now prove the fixed spectral gap for
higher-order quasienergy corrections $\tilde{\omega}_{\ell,\text{cat}}$ via
deductions. Assuming that $\omega_{\ell,\text{cat}}^{1\leqslant q\leqslant
k-1}$ up to the $\lambda^{k-1}$-th order are all independent of spectral
pairing quantum numbers $\ell$, let us consider the $\lambda^{k}$-th order
corrections. Generically, with the corrected eigenstates
$|\tilde{\omega}_{\ell,\text{cat}}\rangle=\cdots e^{i\lambda^{k}S_{k}}\dots
e^{i\lambda S_{1}}|\ell,\\{s_{j}^{(\text{cat})}\\}\rangle$, the corresponding
corrected quasienergy reads

$\displaystyle e^{i\tilde{\omega}_{\ell,\text{cat}}}$
$\displaystyle=\langle\ell,\\{s_{j}^{(\text{cat})}\\}|e^{-i\lambda
S_{1}}e^{-i\lambda^{2}S_{2}}\cdots
e^{-i\lambda^{k}S_{k}}\cdots\left(U_{0}e^{i\sum_{k=1}^{\infty}\lambda^{k}V_{k}}\right)\cdots$
$\displaystyle\times e^{i\lambda^{k}S_{k}}\cdots
e^{i\lambda^{2}S_{2}}e^{i\lambda
S_{1}}|\ell,\\{s_{j}^{(\text{cat})}\\}\rangle$
$\displaystyle=e^{i\left(E(\ell,\\{s_{j}^{(\text{cat})}\\})+\delta\omega_{\ell,\text{cat}}\right)},\qquad\delta\omega_{\ell,\text{cat}}=\sum_{k=1}^{\infty}\lambda^{k}\omega_{\ell,\text{cat}}^{(k)},$
(131)

where $E(\ell,\\{s_{j}^{(\text{cat})}\\})$ is the cat scar quasienergy at
$\lambda=0$. Using
$\langle\ell,\\{s_{j}^{(\text{cat})}\\}|U_{0}^{\dagger}=e^{-iE(\ell,\\{s_{j}^{(\text{cat})}\\})}\langle\ell,\\{s_{j}^{(\text{cat})}\\}|$,
we have

$\displaystyle
e^{i\delta\omega_{\ell,\text{cat}}}=e^{i\left(\tilde{\omega}_{\ell,\text{cat}}{\color[rgb]{1,0,0}-E(\ell,\\{s_{j}^{(\text{cat})}\\})}\right)}$
$\displaystyle=\langle\ell,\\{s_{j}^{(\text{cat})}\\}|{\color[rgb]{1,0,0}U_{0}^{\dagger}}e^{-i\lambda
S_{1}}e^{-i\lambda^{2}S_{2}}\cdots
e^{-i\lambda^{k}S_{k}}\cdots\left(U_{0}e^{i\sum_{k=1}^{\infty}\lambda^{k}V_{k}}\right)\cdots$
$\displaystyle\qquad\times e^{i\lambda^{k}S_{k}}\cdots
e^{i\lambda^{2}S_{2}}e^{i\lambda
S_{1}}|\ell,\\{s_{j}^{(\text{cat})}\\}\rangle,$ (132)

and therefore a power counting gives that the $\lambda^{k}$-th order terms
read

$\displaystyle i\omega_{\ell,\text{cat}}^{(k)}$
$\displaystyle=-\sum_{\alpha=2}^{k}\frac{i^{\alpha}}{\alpha!}\sum_{\\{1\leqslant
k_{j}\leqslant
k-1|\sum_{j=1}^{\alpha}k_{j}=k\\}}\omega_{\ell,\text{cat}}^{(k_{1})}\dots\omega_{\ell,\text{cat}}^{(k_{\alpha})}$
$\displaystyle+\sum_{\alpha=0}^{k}\sum_{\scriptsize\begin{array}[]{c}\\{m_{j},n_{j}=0,1,\dots,k,\\\
k_{j}=1,2,\dots,k|\\\
\sum_{j=1}^{k}j(m_{j}+n_{j})+\sum_{j=1}^{\alpha}k_{j}=k\\}\end{array}}\frac{i^{\alpha+\sum_{j=1}^{k}(m_{j}-n_{j})}}{\alpha!\prod_{j=1}^{k}m_{j}!n_{j}!}$
(136)
$\displaystyle\times\langle\ell,\\{s_{j}^{(\text{cat})}\\}|F_{\\{n_{j},m_{j},k_{j}\\}}^{(k,\alpha)}|\ell,\\{s_{j}^{(\text{cat})}\\}\rangle,$
$\displaystyle F_{\\{n_{j},m_{j},k_{j}\\}}^{(k,\alpha)}$
$\displaystyle=U_{0}^{\dagger}\left(S_{1}^{n_{1}}S_{2}^{n_{2}}\cdots
S_{k}^{n_{k}}\right)U_{0}\left(V_{k_{1}}V_{k_{2}}\dots V_{k_{\alpha}}\right)$
$\displaystyle\qquad\times\left(S_{k}^{m_{k}}\cdots
S_{2}^{m_{2}}S_{1}^{m_{1}}\right).$ (137)

The lower-order corrections $\omega_{\ell,\text{cat}}^{q\leqslant k-1}$ are
already independent of $\ell$ according to the assumptions of deductions.
Further, for $F_{\\{n_{j},m_{j},k_{j}\\}}^{(k,\alpha)}$, one could apply the
selection rules Eq. (68) for $V_{k_{j}}$, Eq. (122) for $S_{1}\dots S_{k}$, as
well as Eq. (72) for operators under gauge transformation by
$U_{0}^{\dagger}\dots U_{0}$. Altogether,
$F_{\\{n_{j},m_{j},k_{j}\\}}^{(k,\alpha)}$ could flip as many as
$\sum_{j=1}^{n_{k}}(n_{\text{op}}j)(n_{j}+m_{j})+\sum_{j=1}^{\alpha}n_{\text{op}}k_{j}=n_{\text{op}}k$
spins, namely,

$\displaystyle\langle\\{s_{j}\\}_{1}|F_{\\{n_{j},m_{j},k_{j}\\}}^{(k,\alpha)}|\\{\tilde{s}_{j}\\}_{2}\rangle\neq
0\quad\Rightarrow\quad\delta
s(\\{s_{j}\\}_{1},\\{\tilde{s}_{j}\\}_{2})\leqslant n_{\text{op}}k.$ (138)

Thus, for a system of size $L$, up to perturbation order $k<L/n_{\text{op}}$,
the cross terms denoted by red color below vanish,

$\displaystyle\langle\ell,\\{s_{j}^{(\text{cat})}\\}|F_{\\{n_{j},m_{j},k_{j}\\}}^{(k,\alpha)}|\ell,\\{s_{j}^{(\text{cat})}\\}\rangle$
$\displaystyle={\color[rgb]{0,.5,0}\frac{1}{2}\left(\langle\\{s_{j}^{(\text{cat})}\\}|F_{\\{n_{j},m_{j},k_{j}\\}}^{(k,\alpha)}|\\{s_{j}^{(\text{cat})}\\}\rangle+\langle-\\{s_{j}^{(\text{cat})}\\}|F_{\\{n_{j},m_{j},k_{j}\\}}^{(k,\alpha)}|-\\{s_{j}^{(\text{cat})}\\}\rangle\right)}$
$\displaystyle+{\color[rgb]{1,0,0}\frac{(-1)^{\ell}}{2}\left(\langle\\{s_{j}^{(\text{cat})}\\}|F_{\\{n_{j},m_{j},k_{j}\\}}^{(k,\alpha)}|-\\{s_{j}^{(\text{cat})}\\}\rangle+\langle-\\{s_{j}^{(\text{cat})}\\}|F_{\\{n_{j},m_{j},k_{j}\\}}^{(k,\alpha)}|\\{s_{j}^{(\text{cat})}\\}\rangle\right)}.$
(139)

This is again because the cross terms require flipping all $L$ spins in order
to related $\pm\\{s_{j}^{(\text{cat})}\\}$, which violates the selection rules
for $F_{\\{n_{j},m_{j},k_{j}\\}}^{(k,\alpha)}$ when $k<L/n_{\text{op}}$. Then,
we have proved that up to the order $\lambda^{k<L/n_{\text{op}}}$, quasienergy
corrections $\omega_{0,\text{cat}}^{(k)}=\omega_{1,\text{cat}}^{(k)}$ at each
perturbation order. On the other hand, for $k\geqslant L/n_{\text{op}}$,
operators $F_{\\{n_{j},m_{j},k_{j}\\}}^{(k,\alpha)}$ can indeed flip all $L$
spins in the system, so the red terms in Eq. (139) no longer vanish such that
$\omega_{\ell,\text{cat}}^{(k\geqslant L/n_{\text{op}})}$ start to depend on
the quantum numbers $\ell$. In sum, the quasienergy difference for pairwise
cat scars $\tilde{\omega}_{0,\text{cat}},\tilde{\omega}_{1,\text{cat}}$
approaches the unperturbed value $\pi$ with exponential accuracy

$\displaystyle\exp\left(i\left(\tilde{\omega}_{1,\text{cat}}-\tilde{\omega}_{0,\text{cat}}\right)\right)$
$\displaystyle=\exp\left(i\left(E(1,\\{s_{j}^{(\text{cat})}\\})-E(0,\\{s_{j}^{(\text{cat})}\\})\right)+i\left(\delta\omega_{1,\text{cat}}-\delta\omega_{0,\text{cat}}\right)\right)$
$\displaystyle=\exp\left(i\pi+i\sum_{k=1}^{\infty}\lambda^{k}\left(\omega_{1,\text{cat}}^{(k)}-\omega_{0,\text{cat}}^{(k)}\right)\right)$
$\displaystyle=\exp\left(i\pi+i\sum_{k=L/n_{\text{op}}}^{\infty}\lambda^{k}\left(\omega_{1,\text{cat}}^{(k)}-\omega_{0,\text{cat}}^{(k)}\right)\right)$
$\displaystyle=\exp\left(i(\pi+O(\lambda^{k\geq L/n_{\text{op}}}))\right)$
(140)

Thus, we could introduce the exponent $\nu$ and write the spectral gap
deviations for finite size systems as

$\displaystyle\tilde{\omega}_{1,\text{cat}}-\tilde{\omega}_{0,\text{cat}}=\pi+(\delta\omega_{1,\text{cat}}-\delta\omega_{0,\text{cat}})=\pi+O(\lambda^{L/\nu}),$
$\displaystyle\nu\leqslant n_{\text{op}}.$ (141)

##  Appendix C Derivation of effective Hamiltonian for strongly interacting
cases

For the model in Eq. (35), with strong interactions $J\sim 1$, we need to
include all higher-order terms $J^{n}$ using the BCHD formula

$\displaystyle
e^{A}e^{B}=e^{\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sum_{p_{j}+q_{j}>0,p_{j},q_{j}\geqslant
0}\frac{[A^{(p_{1})},[B^{(q_{1})},[\dots,[A^{(p_{n})},B^{(q_{n})}]\dots]}{(\sum_{j=1}^{n}(p_{j}+q_{j}))\prod_{j=1}^{n}p_{j}!q_{j}!}}.$
(142)

Here we keep terms up to perturbation strength $\lambda^{1}$, while include
all powers of $J$. Then,

$\displaystyle
e^{-i\sum_{j}J\tau^{z}_{j}\tau^{z}_{j+1}}e^{i\lambda\sum_{j}(\cos(h)\tau^{x}_{j}-\sin(h)\tau^{y}_{j})}$
$\displaystyle=e^{i\sum_{j}(-J\tau^{z}_{j}\tau^{z}_{j+1}+\lambda(\cos(h)\tau^{x}_{j}-\sin(h)\tau^{y}_{j}))+i\lambda
H^{\prime}},$ (143)

where

$\displaystyle H^{\prime}$
$\displaystyle=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sum_{p_{1}\dots
p_{n}=0}^{\infty}\frac{(-iJ)^{p_{1}+p_{2}+\dots+p_{n}}}{(1+\sum_{j=1}^{n}p_{j})\prod_{j=1}^{n}p_{j}!}$
$\displaystyle\quad\times[(\tau^{z}_{j}\tau^{z}_{j+1})^{(p_{1}+p_{2}+\dots+p_{n})},\cos(h)\tau^{x}_{j}-\sin(h)\tau^{y}_{j}]$
$\displaystyle=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sum_{r=1}^{\infty}\frac{(-iJ)^{r}}{r+1}[(\tau^{z}_{j}\tau^{z}_{j+1})^{(r)},\cos(h)\tau^{x}_{j}-\sin(h)\tau^{y}_{j}]$
$\displaystyle\quad\times\sum_{p_{1}+\dots+p_{n}=r}\frac{1}{p_{1}!p_{2}!\dots
p_{n}!}$
$\displaystyle=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sum_{r=1}^{\infty}\frac{(-iJ)^{r}}{r+1}[(\tau^{z}_{j}\tau^{z}_{j+1})^{(r)},\cos(h)\tau^{x}_{j}-\sin(h)\tau^{y}_{j}]\frac{n^{r}}{r!}$
$\displaystyle=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sum_{r=1}^{\infty}\frac{(-iJn)^{r}}{(r+1)!}[(\tau^{z}_{j}\tau^{z}_{j+1})^{(r)},\cos(h)\tau^{x}_{j}-\sin(h)\tau^{y}_{j}]$
(144)

Now, use

$\displaystyle[(\sum_{j}\tau^{z}_{j}\tau^{z}_{j+1})^{(1)},\sum_{k}\cos(h)\tau^{x}_{k}-\sin(h)\tau^{y}_{k}]$
$\displaystyle=\sum_{j}i(\cos(h)\tau^{y}_{j}+\sin(h)\tau^{x}_{j})(\tau^{z}_{j-1}+\tau^{z}_{j+1}),$
$\displaystyle[(\sum_{j}\tau^{z}_{j}\tau^{z}_{j+1})^{(2)},\sum_{k}\cos(h)\tau^{x}_{k}-\sin(h)\tau^{y}_{k}]$
$\displaystyle=\sum_{j}(\cos(h)\tau^{x}_{j}-\sin(h)\tau^{y}_{j})2(1+\tau^{z}_{j-1}\tau^{z}_{j+1}),$
$\displaystyle[(\sum_{j}\tau^{z}_{j}\tau^{z}_{j+1})^{(3)},\sum_{k}\cos(h)\tau^{x}_{k}-\sin(h)\tau^{y}_{k}]$
$\displaystyle=\sum_{j}i(\cos(h)\tau^{y}_{j}+\sin(h)\tau^{x}_{j})4(\tau^{z}_{j-1}+\tau^{z}_{j+1}),$
(145)

which gives

$\displaystyle[(\sum_{j}\tau^{z}_{j}\tau^{z}_{j+1})^{(2m-1)},\sum_{k}\cos(h)\tau^{x}-\sin(h)\tau^{y}_{k}]$
$\displaystyle=\sum_{j}i(\cos(h)\tau^{y}_{j}+\sin(h)\tau^{x}_{j})2^{2(m-1)}(\tau^{z}_{j-1}+\tau^{z}_{j+1}),$
$\displaystyle[(\sum_{j}\tau^{z}_{j}\tau^{z}_{j+1})^{(2m)},\sum_{k}\cos(h)\tau^{x}-\sin(h)\tau^{y}_{k}]$
$\displaystyle=\sum_{j}(\cos(h)\tau^{x}_{j}-\sin(h)\tau^{x}_{j})2^{2m-1}(1+\tau^{z}_{j-1}\tau^{z}_{j+1}),$
(146)

Thus,

$\displaystyle H^{\prime}$
$\displaystyle=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}$
$\displaystyle\times\sum_{m=1}^{\infty}\left(\frac{(-iJn)^{2m-1}}{(2m)!}2^{2m-2}i(\cos(h)\tau^{y}_{j}+\sin(h)\tau^{x}_{j})(\tau^{z}_{j-1}+\tau^{z}_{j+1})\right.$
$\displaystyle\qquad+\left.\frac{(-iJn)^{2m}}{(2m+1)!}2^{2m-1}(\cos(h)\tau^{x}_{j}-\sin(h)\tau^{y}_{j})(1+\tau^{z}_{j-1}\tau^{z}_{j+1})\right)$
$\displaystyle=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\left(\frac{1}{-4iJn}(-1+\cos(2Jn))\right.$
$\displaystyle\qquad\times\left.i(\cos(h)\tau^{y}_{j}+\sin(h)\tau^{x}_{j})(\tau^{z}_{j-1}+\tau^{z}_{j+1})\right.$
$\displaystyle\qquad+\frac{1}{-4iJn}(2iJn-i\sin(2Jn))$
$\displaystyle\qquad\times\left.(\cos(h)\tau^{x}_{j}-\sin(h)\tau^{y}_{j})(1+\tau^{z}_{j-1}\tau^{z}_{j+1})\right)$
$\displaystyle=\sum_{n=1}^{\infty}\frac{(-1)^{n}}{4Jn^{2}}\left((-1+\cos(2Jn))(\cos(h)\tau^{y}_{j}+\sin(h)\tau^{x}_{j})(\tau^{z}_{j-1}+\tau^{z}_{j+1})\right.$
$\displaystyle\qquad\left.+(2Jn-\sin(2Jn))(\cos(h)\tau^{x}_{j}-\sin(h)\tau^{y}_{j})(1+\tau^{z}_{j-1}\tau^{z}_{j+1})\right)$
$\displaystyle=\frac{1}{4J}\left(f(J)(\cos(h)\tau^{y}_{j}+\sin(h)\tau^{x}_{j})(\tau^{z}_{j-1}+\tau^{z}_{j+1})\right.$
$\displaystyle\quad\left.+g(J)(\cos(h)\tau^{x}_{j}-\sin(h)\tau^{y}_{j})(1+\tau^{z}_{j-1}\tau^{z}_{j+1})\right)$
(147)

where

$\displaystyle f(J)$
$\displaystyle=\frac{1}{4J}\left(\frac{\pi^{2}}{12}+\frac{\text{Li}_{2}(-e^{-2iJ})+\text{Li}_{2}(-e^{2iJ})}{2}\right),$
$\displaystyle g(J)$
$\displaystyle=\frac{1}{4J}\left(-2J\ln(2)+\frac{\text{Li}_{2}(-e^{-2iJ})-\text{Li}_{2}(-e^{2iJ})}{2i}\right)$
(148)

and $\text{Li}_{s}(z)$ is the polylogarithm function

$\displaystyle\text{Li}_{s}(z)=\sum_{k=1}^{\infty}\frac{z^{k}}{k^{s}}.$ (149)

Thus, we recover the results in Eq. (42).

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