# The fractional Landau-Lifshitz-Gilbert equation
Supplementary Material

R.C. Verstraten1, T. Ludwig1, R.A. Duine1,2, C. Morais Smith1 1Institute for
Theoretical Physics, Utrecht University, Princetonplein 5, 3584CC Utrecht, The
Netherlands
2Department of Applied Physics, Eindhoven University of Technology, P.O. Box
513, 5600 MB Eindhoven, The Netherlands

###### Contents

  1. I Keldysh microscopic model
    1. I.1 Hamiltonian
    2. I.2 Keldysh partition function
    3. I.3 Quasi-classical equation of motion
    4. I.4 Generalized Landau-Lifshitz-Gilbert equation
  2. II Fractional derivative from non-Ohmic spectral function
    1. II.1 Calculating the effective Greens functions
    2. II.2 Ohmic spectral function
    3. II.3 Sub-Ohmic spectral function
    4. II.4 Super-Ohmic spectral function
    5. II.5 Comparison Ohmic versus non-Ohmic
  3. III FMR powerlaw derivation
    1. III.1 Ferromagnetic Resonance
    2. III.2 Resonance frequency and amplitude
    3. III.3 Calculating the FWHM linewidth
  4. IV Dimensional analysis

##  I Keldysh microscopic model

For pedagogical reasons we start with a microscopic derivation of the usual
LLG equation before going into the fractional one. In this section, we combine
spin coherent states with the Keldysh formalism [1, 2] to derive a stochastic
Langevin-like equation of motion of a (macro) spin [3].

###  I.1 Hamiltonian

In the main text, we introduced a spectral function $J(\omega)$ with a cutoff
frequency $\Omega_{c}$. This was originally done from the perspective that any
spectral function could be expanded to linear order; hence, the model would
only be valid up to some highest frequency. However, the cutoff is also
important for the model to be realistic, since any physical spectral function
should vanish as $\omega\to\infty$. In the main text, we stated that the same
results can be obtained by introducing a constant counter term in the
Hamiltonian. This is a term which exactly completes the square of the coupling
term and the harmonic potential of the bath and can be seen as a normalization
of the zero-energy level. If we instead start the model with this counter term
and drop the cutoff, we will get a Greens function
$\alpha_{\text{ct}}(\omega)$, which is precisely such that the original Greens
function can be written as
$\alpha(\omega)=\alpha(0)+\alpha_{\text{ct}}(\omega)$, i.e., the counter term
in the Hamiltonian removes the zero frequency contribution of the Greens
function. This $\alpha(\omega=0)$ generates a term in the equation of motion
that goes as
$\int_{0}^{\infty}d\epsilon\frac{J(\epsilon)}{\pi\epsilon}[\bm{S}(t)\times\bm{S}(t)]$.
Since the integral is finite, with a frequency cutoff in $J(\omega)$, the
entire term is zero due to the cross product. This means that the equation of
motion will be identical if we start either from the regular Hamiltonian with
a frequency cutoff, or with a counter term and no cutoff. Here, we choose to
show the method that includes a counter term, because then we do not need to
calculate terms which would have canceled either way.

The microscopic system that we describe is a large spin in an external
magnetic field, where the spin is linearly coupled to a bath of harmonic
oscillators in the same way as in Refs. [4, 5, 6, 7, 8, 9]. Therefore, our
Hamiltonian has the form of a system, coupling, bath, and counter term;
$H(t)=H_{s}+H_{c}+H_{b}+H_{ct}$, where

$\displaystyle H_{s}$ $\displaystyle=\bm{B}\cdot\hat{\bm{S}}-KS_{z}^{2},$
$\displaystyle H_{c}$
$\displaystyle=\sum_{\alpha}\gamma_{\alpha}\hat{\bm{S}}\cdot\hat{\bm{x}}_{\alpha},$
$\displaystyle H_{b}$
$\displaystyle=\sum_{\alpha}\frac{\hat{\bm{p}}_{\alpha}^{2}}{2m_{\alpha}}+\frac{m_{\alpha}\omega_{\alpha}^{2}}{2}\hat{\bm{x}}_{\alpha}^{2},$
$\displaystyle H_{ct}$
$\displaystyle=\sum_{\alpha}\frac{\gamma_{\alpha}^{2}}{2m_{\alpha}\omega_{\alpha}^{2}}\hat{\bm{S}}^{2}.$
(1)

Here, $\bm{B}$ is the (effective) magnetic field, $\hat{\bm{S}}$ is the spin,
$K$ is the $z$-axis anisotropy, $\gamma_{\alpha}$ is the coupling strength,
and $\alpha$ is the index over all harmonic oscillators which have position
$\hat{\bm{x}}_{\alpha}$, momentum $\hat{\bm{p}}_{\alpha}$, mass $m_{\alpha}$
and natural frequency $\omega_{\alpha}$. Notice that the counter term is
constant, since $S^{2}$ is a conserved quantity, and that we have indeed
completed the square, such that

$\displaystyle
H(t)=\bm{B}\cdot\hat{\bm{S}}-KS_{z}^{2}+\sum_{\alpha}\frac{\hat{\bm{p}}_{\alpha}^{2}}{2m_{\alpha}}+\left[\sqrt{\frac{m_{\alpha}\omega_{\alpha}^{2}}{2}}\hat{\bm{x}}_{\alpha}+\sqrt{\frac{\gamma_{\alpha}^{2}}{2m_{\alpha}\omega_{\alpha}^{2}}}\hat{\bm{S}}\right]^{2}.$
(2)

###  I.2 Keldysh partition function

We will use the Keldysh formalism to derive a quasi-classical equation of
motion. Since this is an out-of-equilibrium system, a common choice would be
to use the Lindblad formalism with a master equation [10]. However, Lindblad
can only describe Markovian systems, which will not be the case when we
introduce a non-Ohmic bath. In the Keldysh formalism, one starts with an
equilibrium density matrix in the far past (effectively infinite on the
relevant time scale). This then gets evolved with the time evolution operator
as usual. However, in contrast to ordinary path integrals, once the present
has been reached, one evolves back to the infinite past. Since there is
infinite time for evolution, we can reach out-of-equilibrium states
adiabatically. The benefit of integrating back to the infinite past is that we
begin and end with the same in-equilibrium system, which means equilibrium
techniques can be used, at the cost of having both the forward
($\mathcal{O}^{+}$) and backward ($\mathcal{O}^{-}$) quantities to take care
of. To reach useful results, one can apply a Keldysh rotation to the classical
($\mathcal{O}^{c}=(\mathcal{O}^{+}+\mathcal{O}^{-})/2$) and quantum
($\mathcal{O}^{q}=\mathcal{O}^{+}-\mathcal{O}^{-}$) components with the added
notation $\vec{\mathcal{O}}=\begin{pmatrix}\mathcal{O}^{c}\\\
\mathcal{O}^{q}\end{pmatrix}$. To derive a quasi-classical equation of motion,
the action can be expanded in all the quantum components, after which the
Euler-Lagrange equation for the quantum components provides the equation of
motion in terms of the classical components.

Figure 1: Figure extracted from Ref. [2]. The Keldysh contour starts at
$t=-\infty$, evolves forward to some time $t$, and then evolves backwards in
time to $t=-\infty$.

To begin, we write down the Keldysh partition function

$\displaystyle Z$
$\displaystyle=\operatorname{Tr}\left\\{T_{K}\exp\left[-i\oint_{K}dtH(t)\right]\rho_{0}\right\\},$
(3)

where $T_{K}$ is the Keldysh time ordering, $\rho_{0}$ is the density matrix
at $t=-\infty$, and the integral runs over the Keldysh contour, as shown in
FIG. 1. After discretizing the Keldysh time integral in the way of FIG. 1, we
can rewrite the trace as path-integrals over the spin coherent state $\lvert
g\rangle$ and the oscillators $\lvert\hat{\bm{x}}_{\alpha}\rangle$ and
$\lvert\hat{\bm{p}}_{\alpha}\rangle$. This yields

$Z=\int Dg\prod_{\alpha}\int D\hat{\bm{x}}_{\alpha}\int
D\hat{\bm{p}}_{\alpha}\;e^{iS[g,\\{\hat{\bm{x}}_{\alpha}\\},\\{\hat{\bm{p}}_{\alpha}\\}]},$
(4)

with the Keldysh action

$\displaystyle
S[g,\\{\hat{\bm{x}}_{\alpha}\\},\\{\hat{\bm{p}}_{\alpha}\\}]=\oint_{K}dt$
$\displaystyle\Bigg{[}(-i\partial_{t}\langle g\rvert)\lvert
g\rangle-\bm{B}\cdot\bm{S}_{g}+KS_{z,g}^{2}$
$\displaystyle+\left.\sum_{\alpha}\left(-\gamma_{\alpha}\bm{S}_{g}\cdot\hat{\bm{x}}_{\alpha}+\hat{\bm{p}}_{\alpha}\cdot\dot{\hat{\bm{x}}}_{\alpha}-\frac{\gamma_{\alpha}^{2}\bm{S}_{g}^{2}}{2m_{\alpha}\omega_{\alpha}^{2}}-\frac{\hat{\bm{p}}_{\alpha}^{2}}{2m_{\alpha}}-\frac{m_{\alpha}\omega_{\alpha}^{2}}{2}\hat{\bm{x}}_{\alpha}^{2}\right)\right],$
(5)

where we defined $\bm{S}_{g}=\langle g\rvert\hat{\bm{S}}\lvert g\rangle$.

The continuous path-integral seems to miss the boundary term
$\langle\hat{\bm{x}}_{1,\alpha},g_{1}\rvert\rho_{0}\lvert\hat{\bm{x}}_{2N,\alpha},g_{2N}\rangle\langle\hat{\bm{p}}_{2N,\alpha}|\hat{\bm{x}}_{2N,\alpha}\rangle$,
but it is included in the Keldysh contour, as it connects the beginning and
final contour time at $t=-\infty$; see Ref. [2].

Now, we will integrate out the bath degrees of freedom, beginning by
completing the square and performing the Gaussian integral over
$\hat{\bm{p}}_{\alpha}$. The Gaussian contribution in $\hat{\bm{p}}_{\alpha}$
will act as a constant prefactor, so it will drop out of any calculation of an
observable due to the normalization. Hence, we can effectively set it to one
to find

$\displaystyle\int
D\hat{\bm{p}}_{\alpha}\exp\left[-i\oint_{K}dt\left(\frac{\hat{\bm{p}}_{\alpha}^{2}}{2m_{\alpha}}-\hat{\bm{p}}_{\alpha}\cdot\dot{\hat{\bm{x}}}_{\alpha}\right)\right]$
$\displaystyle=\exp\left[i\oint_{K}dt\left(-\frac{m_{\alpha}}{2}\hat{\bm{x}}_{\alpha}\partial_{t}^{2}\hat{\bm{x}}_{\alpha}\right)\right],$
(6)

where we also did a partial integration in $\hat{\bm{x}}_{\alpha}$. Next we
will perform a similar approach for the positions, but it is useful to apply
the Keldysh rotation first. Note that we can directly rewrite the integral
over the Keldysh contour as a regular time integral over the quantum
components. However, one must still rewrite the contents of the integral in
terms of the quantum and classical parts of the variables, since the Keldysh
rotation does not immediately work for products. The action can first be
written as

$\displaystyle iS[g,\\{\hat{\bm{x}}_{\alpha}\\}]$ $\displaystyle=i\int
dt\left(\left[(-i\partial_{t}\langle g\rvert)\lvert
g\rangle\right]^{q}-\left[\bm{B}\cdot\bm{S}_{g}\right]^{q}+K\left[S_{z,g}^{2}\right]^{q}\right.$
$\displaystyle-\sum_{\alpha}\left.\left\\{\left[\gamma_{\alpha}\bm{S}_{g}\cdot\hat{\bm{x}}_{\alpha}\right]^{q}+\frac{\gamma_{\alpha}^{2}[\bm{S}_{g}^{2}]^{q}}{2m_{\alpha}\omega_{\alpha}^{2}}+\left[\frac{m_{\alpha}}{2}\hat{\bm{x}}_{\alpha}\left(\partial_{t}^{2}+\omega_{\alpha}^{2}\right)\hat{\bm{x}}_{\alpha}\right]^{q}\right\\}\right).$
(7)

We can then derive that

$\displaystyle-\left[\gamma_{\alpha}\bm{S}_{g}\cdot\hat{\bm{x}}_{\alpha}\right]^{q}$
$\displaystyle=-\gamma_{\alpha}\left[\bm{S}_{g}^{+}\cdot\hat{\bm{x}}_{\alpha}^{+}-\bm{S}_{g}^{-}\cdot\hat{\bm{x}}_{\alpha}^{-}\right]=-\gamma_{\alpha}\left[\left(\bm{S}_{g}^{c}+\frac{1}{2}\bm{S}_{g}^{q}\right)\cdot\left(\hat{\bm{x}}_{\alpha}^{c}+\frac{1}{2}\hat{\bm{x}}_{\alpha}^{q}\right)-\left(\bm{S}_{g}^{c}-\frac{1}{2}\bm{S}_{g}^{q}\right)\cdot\left(\hat{\bm{x}}_{\alpha}^{c}-\frac{1}{2}\hat{\bm{x}}_{\alpha}^{q}\right)\right]$
$\displaystyle=-\gamma_{\alpha}\left[\bm{S}_{g}^{c}\hat{\bm{x}}_{\alpha}^{q}+\bm{S}_{g}^{q}\hat{\bm{x}}_{\alpha}^{c}\right]=-\gamma_{\alpha}\left[\begin{pmatrix}\bm{S}_{g}^{c}&\bm{S}_{g}^{q}\end{pmatrix}\tau_{x}\begin{pmatrix}\hat{\bm{x}}_{\alpha}^{c}\\\
\hat{\bm{x}}_{\alpha}^{q}\end{pmatrix}\right],$ (8)

where we introduced $\tau_{x}=\begin{pmatrix}0&1\\\ 1&0\end{pmatrix}$ in the
Keldysh (classical, quantum) space represented by an upper index $c$ and $q$
respectively. Next, we want to derive a similar form for the part of the
action that is quadratic in $\hat{\bm{x}}_{\alpha}$. Since these are harmonic
oscillators in equilibrium, we can refer the reader to Ref. [2], noting that a
unit mass was used there, and conclude that

$\displaystyle\left[-\frac{m_{\alpha}}{2}\hat{\bm{x}}_{\alpha}\left(\partial_{t}^{2}+\omega_{\alpha}^{2}\right)\hat{\bm{x}}_{\alpha}\right]^{q}$
$\displaystyle=\begin{pmatrix}\hat{\bm{x}}_{\alpha}^{c}&\hat{\bm{x}}_{\alpha}^{q}\end{pmatrix}\begin{pmatrix}0&\left[G_{\alpha}^{-1}\right]^{A}\\\
\left[G_{\alpha}^{-1}\right]^{R}&\left[G_{\alpha}^{-1}\right]^{K}\end{pmatrix}\begin{pmatrix}\hat{\bm{x}}_{\alpha}^{c}\\\
\hat{\bm{x}}_{\alpha}^{q}\end{pmatrix},$ (9)

where the retarded and advanced Greens functions read

$[G^{-1}_{\alpha}]^{R/A}(t-t^{\prime})=\delta(t-t^{\prime})\frac{m_{\alpha}}{2}[(i\partial_{t}\pm
i0)^{2}-\omega_{\alpha}^{2}].$ (10)

The $\pm i0$ is introduced because we need an infinitesimal amount of
dissipation on the bath for it to remain in equilibrium and the sign is tied
to causality. This is because there is also an infinitesimal amount of energy
transfer from the macroscopic spin to each of the oscillators. This results in
an extra first-order derivative term, which is found by multiplying out the
square with $i0$. One might want to set these terms to zero immediately, but
as it turns out, these are very important limits, which shift away poles from
integrals that we need to compute later. Once that is done, the limits are no
longer important for the final result, and they may finally be put to zero.
Since the bath is in equilibrium, we can use the fluctuation dissipation
theorem to compute the Keldysh component using

$G^{K}_{\alpha}(\omega)=\left[G^{R}_{\alpha}(\omega)-G^{A}_{\alpha}(\omega)\right]\coth\left(\frac{\omega}{2T}\right).$

The $\hat{\bm{x}}$ dependent part of the action is now given by

$\displaystyle iS_{X}$ $\displaystyle=i\int
dt\left[-\gamma_{\alpha}\begin{pmatrix}\bm{S}_{g}^{c}&\bm{S}_{g}^{q}\end{pmatrix}\tau_{x}\begin{pmatrix}\hat{\bm{x}}_{\alpha}^{c}\\\
\hat{\bm{x}}_{\alpha}^{q}\end{pmatrix}+\begin{pmatrix}\hat{\bm{x}}_{\alpha}^{c}&\hat{\bm{x}}_{\alpha}^{q}\end{pmatrix}G_{\alpha}^{-1}\begin{pmatrix}\hat{\bm{x}}_{\alpha}^{c}\\\
\hat{\bm{x}}_{\alpha}^{q}\end{pmatrix}\right],$ (11)

which we can compute by completing the square to find

$iS_{X}=i\int
dt\left[-\frac{\gamma_{\alpha}^{2}}{4}\vec{S}^{T}_{g}\begin{pmatrix}0&G_{\alpha}^{A}\\\
G_{\alpha}^{R}&G_{\alpha}^{K}\end{pmatrix}\vec{S}_{g}\right].$ (12)

Before we write down the final effective action, we also have to rewrite the
quadratic part in $\bm{S}$ in a similar vector form, which is

$\displaystyle-\frac{\gamma_{\alpha}^{2}[\bm{S}_{g}^{2}]^{q}}{2m_{\alpha}\omega_{\alpha}^{2}}$
$\displaystyle=\frac{-\gamma_{\alpha}^{2}}{2m_{\alpha}\omega_{\alpha}^{2}}\begin{pmatrix}\bm{S}_{g}^{c}&\bm{S}_{g}^{q}\end{pmatrix}\tau_{x}\begin{pmatrix}\bm{S}_{g}^{c}\\\
\bm{S}_{g}^{q}\end{pmatrix}.$ (13)

Combining everything together, we find that the partition function of the
system is given by $Z=\int Dg\;e^{iS[g]}$, with the effective action

$\displaystyle iS[g]$ $\displaystyle=i\int
dt\left\\{\left[(-i\partial_{t}\langle g\rvert)\lvert
g\rangle-\bm{B}\cdot\bm{S}_{g}+KS_{z,g}^{2}\right]^{q}-\int
dt^{\prime}\vec{S}_{g}^{T}(t)\begin{pmatrix}0&\alpha^{A}\\\
\alpha^{R}&\alpha^{K}\end{pmatrix}_{(t-t^{\prime})}\vec{S}_{g}(t^{\prime})\right\\},$
(14)

where
$\alpha^{A/R}(t-t^{\prime})=\sum_{\alpha}\left(\frac{\gamma_{\alpha}^{2}}{4}G_{\alpha}^{A/R}(t-t^{\prime})+\frac{\gamma_{\alpha}^{2}}{2m_{\alpha}\omega_{\alpha}^{2}}\delta(t-t^{\prime})\right)$
and
$\alpha^{K}(t-t^{\prime})=\sum_{\alpha}\frac{\gamma_{\alpha}^{2}}{4}G_{\alpha}^{K}(t-t^{\prime})$.

###  I.3 Quasi-classical equation of motion

In the quasi-classical regime, we are interested in solutions where the
quantum components $(q)$ are small compared to the classical components $(c)$.
We can thus neglect terms of $\mathcal{O}[(q)^{3}]$, but we must be careful
with $(q)^{2}$. We can use a Hubbard-Stratonovich transformation to convert
$(q)^{2}$ terms into an expression with just $(q)$, but with a new field $\xi$
added to the path integral [3]. The action will then contain only terms of
linear order in $(q)$, which means the partition function has the form
$Z\sim\int DcDq\exp[if(c)q]=\int Dc\frac{1}{2\pi}\delta[f(c)]$. Hence, only
solutions that satisfy $f(c)=0$ contribute to the path integral. Within that
subset, we want to minimize the action.

In order to derive the equation of motion of the system, we must understand
the relation between $\lvert g\rangle$ and $\bm{S}_{g}=\langle
g\rvert\bm{S}\lvert g\rangle$. Using the Euler angle representation [1], we
can describe $\lvert g\rangle$ as

$\lvert g\rangle=g\lvert\uparrow\rangle=e^{-i\phi S_{z}}e^{-i\theta
S_{y}}e^{-i\psi S_{z}}\lvert\uparrow\rangle=e^{-i\phi S_{z}}e^{-i\theta
S_{y}}\lvert\uparrow\rangle e^{-i\psi S}$ (15)

and similarly

$\langle g\rvert=e^{i\psi S}\langle\uparrow\rvert e^{i\theta S_{y}}e^{i\phi
S_{z}}.$ (16)

Note that the $\psi$ angle is now independent of the quantum state
$\lvert\uparrow\rangle$, since this angle is describing the rotation of the
vector pointing in the spin direction, which is symmetric. Hence, this will
yield a gauge symmetry.

Using the Euler angle representation in the first terms of Eq. (14), we see
that

$\displaystyle(-i\partial_{t}\langle g\rvert)\lvert g\rangle$
$\displaystyle=\left(\dot{\psi}Se^{i\psi S}\langle\uparrow\rvert e^{i\theta
S_{y}}e^{i\phi S_{z}}+e^{i\psi
S}\langle\uparrow\rvert\dot{\theta}S_{y}e^{i\theta S_{y}}e^{i\phi
S_{z}}+e^{i\psi S}\langle\uparrow\rvert e^{i\theta
S_{y}}\dot{\phi}S_{z}e^{i\phi S_{z}}\right)e^{-i\phi S_{z}}e^{-i\theta
S_{y}}\lvert\uparrow\rangle e^{-i\psi S}$
$\displaystyle=\dot{\psi}S+\dot{\theta}\langle\uparrow\rvert
S_{y}\lvert\uparrow\rangle+\dot{\phi}\langle\uparrow\rvert e^{i\theta
S_{y}}S_{z}e^{-i\theta S_{y}}\lvert\uparrow\rangle.$ (17)

We note that $\langle\uparrow\rvert S_{y}\lvert\uparrow\rangle=0$, while the
last term includes a rotation of the spin up state by $\theta$ degrees in the
$y$ direction and then measures the $S_{z}$ component of that state, which is
$S\cos\theta$. Hence,

$(-i\partial_{t}\langle g\rvert)\lvert
g\rangle=\dot{\psi}S+\dot{\phi}S\cos\theta.$ (18)

We now define a new variable $\chi$ such that $\psi=\chi-\phi$, which results
in

$(-i\partial_{t}\langle g\rvert)\lvert
g\rangle=\dot{\chi}S-\dot{\phi}(1-\cos\theta)S.$ (19)

Making use of the Euler angle representation, we also see that

$\bm{S}_{g}=S\begin{pmatrix}\sin\theta\cos\phi\\\ \sin\theta\sin\phi\\\
\cos\theta\end{pmatrix}.$ (20)

We see that
$\bm{B}\cdot\bm{S}_{g}=S[B_{x}\sin\theta\cos\phi+B_{y}\sin\theta\sin\phi+B_{z}\cos\theta]$.
Similarly, $K\bm{S}_{z,g}^{2}=KS^{2}\cos^{2}\theta$. Now, we still have to
compute the quantum parts of these quantities. We first note that

$\displaystyle S_{g,x}^{q}/S$
$\displaystyle=[\sin\theta\cos\phi]_{q}=2\cos\theta_{c}\sin\frac{\theta_{q}}{2}\cos\phi_{c}\cos\frac{\phi_{q}}{2}-2\sin\theta_{c}\cos\frac{\theta_{q}}{2}\sin\phi_{c}\sin\frac{\phi_{q}}{2};$
$\displaystyle S_{g,y}^{q}/S$
$\displaystyle=[\sin\theta\sin\phi]_{q}=2\sin\theta_{c}\cos\frac{\theta_{q}}{2}\cos\phi_{c}\sin\frac{\phi_{q}}{2}+2\cos\theta_{c}\sin\frac{\theta_{q}}{2}\sin\phi_{c}\cos\frac{\phi_{q}}{2};$
$\displaystyle S_{g,z}^{q}/S$
$\displaystyle=[\cos\theta]_{q}=-2\sin\theta_{c}\sin\frac{\theta_{q}}{2};$
$\displaystyle[\cos^{2}\theta]_{q}$
$\displaystyle=-2\sin\theta_{c}\cos\theta_{c}\sin\theta_{q}.$ (21)

Next, we will choose a gauge for $\chi$ as in Ref. [11], which is

$\displaystyle\dot{\chi}_{c}$ $\displaystyle=\dot{\phi}_{c}(1-\cos\theta_{c})$
$\displaystyle\chi_{q}$ $\displaystyle=\phi_{q}(1-\cos\theta_{c}).$ (22)

Defining $p=1-\cos\theta$, we see that $\left[(-i\partial_{t}\langle
g\rvert)\lvert
g\rangle\right]_{q}=\left[\dot{\chi}S-\dot{\phi}pS\right]_{q}=S\left[\phi_{q}\dot{p}_{c}-\dot{\phi}_{c}p_{q}\right]$.
Now, $p_{q}=2\sin\theta_{c}\sin\frac{\theta_{q}}{2}$ and
$\dot{p}_{c}=\dot{\theta}_{c}\sin\theta_{c}\cos\frac{\theta_{q}}{2}+\frac{\dot{\theta}_{q}}{2}\cos\theta_{c}\sin\frac{\theta_{q}}{2},$
which leads to

$\displaystyle\left[(-i\partial_{t}\langle g\rvert)\lvert g\rangle\right]_{q}$
$\displaystyle=S\left[\phi_{q}\dot{p}_{c}-\dot{\phi}_{c}p_{q}\right]=S\left[\phi_{q}\dot{\theta}_{c}\sin\theta_{c}\cos\frac{\theta_{q}}{2}+\phi_{q}\frac{\dot{\theta}_{q}}{2}\cos\theta_{c}\sin\frac{\theta_{q}}{2}-2\dot{\phi}_{c}\sin\theta_{c}\sin\frac{\theta_{q}}{2}\right].$
(23)

Next, we want to express $\bm{B}\cdot\bm{S}_{g}^{q}$ in terms of Euler angles.
We see that

$\displaystyle\bm{B}\cdot\bm{S}_{g}^{q}$
$\displaystyle=S[B_{x}\sin\theta\cos\phi+B_{y}\sin\theta\sin\phi+B_{z}\cos\theta]^{q}$
$\displaystyle=2S\Big{[}B_{x}\left(\cos\theta_{c}\sin\frac{\theta_{q}}{2}\cos\phi_{c}\cos\frac{\phi_{q}}{2}-\sin\theta_{c}\cos\frac{\theta_{q}}{2}\sin\phi_{c}\sin\frac{\phi_{q}}{2}\right)$
$\displaystyle\qquad+B_{y}\left(\sin\theta_{c}\cos\frac{\theta_{q}}{2}\cos\phi_{c}\sin\frac{\phi_{q}}{2}+\cos\theta_{c}\sin\frac{\theta_{q}}{2}\sin\phi_{c}\cos\frac{\phi_{q}}{2}\right)$
$\displaystyle\qquad-B_{z}\sin\theta_{c}\sin\frac{\theta_{q}}{2}\Big{]},$ (24)

where we used the results from Eq. (21). Similarly, we have

$\displaystyle K\left[S_{z,g}^{2}\right]^{q}$
$\displaystyle=KS^{2}[\cos^{2}\theta]^{q}=-2KS^{2}\sin\theta_{c}\cos\theta_{c}\sin\theta_{q}.$
(25)

Combining these results, we conclude that

$\displaystyle\left[(-i\partial_{t}\langle g\rvert)\lvert
g\rangle-\bm{B}\cdot\bm{S}_{g}+KS_{z,g}^{2}\right]^{q}=S\Bigg{[}$
$\displaystyle\phi_{q}\dot{\theta}_{c}\sin\theta_{c}\cos\frac{\theta_{q}}{2}+\phi_{q}\frac{\dot{\theta}_{q}}{2}\cos\theta_{c}\sin\frac{\theta_{q}}{2}-2(-B_{z}+KS\cos\theta_{c}+\dot{\phi}_{c})\sin\theta_{c}\sin\frac{\theta_{q}}{2}$
$\displaystyle-2B_{x}\left(\cos\theta_{c}\sin\frac{\theta_{q}}{2}\cos\phi_{c}\cos\frac{\phi_{q}}{2}-\sin\theta_{c}\cos\frac{\theta_{q}}{2}\sin\phi_{c}\sin\frac{\phi_{q}}{2}\right)$
$\displaystyle-
B_{y}\left(\sin\theta_{c}\cos\frac{\theta_{q}}{2}\cos\phi_{c}\sin\frac{\phi_{q}}{2}+\cos\theta_{c}\sin\frac{\theta_{q}}{2}\sin\phi_{c}\cos\frac{\phi_{q}}{2}\right)\Bigg{]}.$
(26)

Remark that this expression only contains odd powers of $(q)$, so that we can
neglect all higher-order terms to get

$\displaystyle\Big{[}(-i\partial_{t}\langle g\rvert)\lvert g\rangle$
$\displaystyle-\bm{B}\cdot\bm{S}_{g}+K\bm{S}_{z,g}^{2}\Big{]}^{q}=S\Big{[}-\theta_{q}\sin\theta_{c}(-B_{z}+KS\cos\theta_{c}+\dot{\phi}_{c})$
$\displaystyle-\theta_{q}\cos\theta_{c}(B_{x}\cos\phi_{c}+B_{y}\sin\phi_{c})+\phi_{q}\sin\theta_{c}(\dot{\theta}_{c}+B_{x}\sin\phi_{c}-B_{y}\cos\phi_{c})\Big{]}.$
(27)

Now, we focus on the part of the action in Eq. (14) that comes from the bath,
given by

$iS_{b}[g]=-i\int dt\int
dt^{\prime}\vec{S}_{g}^{T}(t)\begin{pmatrix}0&\alpha^{A}\\\
\alpha^{R}&\alpha^{K}\end{pmatrix}_{(t-t^{\prime})}\vec{S}_{g}(t^{\prime}).$
(28)

Let us first consider what $\bm{S}_{g}^{q}$ and $\bm{S}_{g}^{c}$ are in terms
of $\phi$ and $\theta$. By performing some trigonometric operations on each of
the components, we find that

$\displaystyle\bm{S}_{g}^{c}$
$\displaystyle=S\begin{pmatrix}\sin\theta_{c}\cos\frac{\theta_{q}}{2}\cos\phi_{c}\cos\frac{\phi_{q}}{2}-\cos\theta_{c}\sin\frac{\theta_{q}}{2}\sin\phi_{c}\sin\frac{\phi_{q}}{2}\\\
\sin\theta_{c}\cos\frac{\theta_{q}}{2}\sin\phi_{c}\cos\frac{\phi_{q}}{2}+\cos\theta_{c}\sin\frac{\theta_{q}}{2}\cos\phi_{c}\sin\frac{\phi_{q}}{2}\\\
\cos\theta_{c}\cos\frac{\theta_{q}}{2}\end{pmatrix}$ (29)

and

$\displaystyle\bm{S}_{g}^{q}$
$\displaystyle=2S\begin{pmatrix}\cos\theta_{c}\sin\frac{\theta_{q}}{2}\cos\phi_{c}\cos\frac{\phi_{q}}{2}-\sin\theta_{c}\cos\frac{\theta_{q}}{2}\sin\phi_{c}\sin\frac{\phi_{q}}{2}\\\
\sin\theta_{c}\cos\frac{\theta_{q}}{2}\cos\phi_{c}\sin\frac{\phi_{q}}{2}+\cos\theta_{c}\sin\frac{\theta_{q}}{2}\sin\phi_{c}\cos\frac{\phi_{q}}{2}\\\
-\sin\theta_{c}\sin\frac{\theta_{q}}{2}\end{pmatrix}.$ (30)

By expanding in the quantum components of $\bm{S}_{g}^{c}$ and
$\bm{S}_{g}^{q}$, we see that

$\displaystyle\bm{S}_{g}^{c}=(q)^{0}+\mathcal{O}\left((q)^{2}\right),$
$\displaystyle\bm{S}_{g}^{q}=(q)^{1}+\mathcal{O}\left((q)^{3}\right).$

Since the action only contains terms with at least one $\bm{S}_{g}^{q}$, we
know that the only way to obtain a term of order $(q)^{2}$ is from
$(\bm{S}_{g}^{q})^{2}$. Hence, we may neglect all terms beyond linear $(q)$ in
$\bm{S}_{g}^{\;(c/q)}$ in the quasi-classical regime. This results in

$\displaystyle\bm{S}_{g}^{c}$
$\displaystyle=S\begin{pmatrix}\sin\theta_{c}\cos\phi_{c}\\\
\sin\theta_{c}\sin\phi_{c}\\\ \cos\theta_{c}\end{pmatrix},$ (31)
$\displaystyle\bm{S}_{g}^{q}$
$\displaystyle=S\begin{pmatrix}\theta_{q}\cos\theta_{c}\cos\phi_{c}-\phi_{q}\sin\theta_{c}\sin\phi_{c}\\\
\phi_{q}\sin\theta_{c}\cos\phi_{c}+\theta_{q}\cos\theta_{c}\sin\phi_{c}\\\
-\theta_{q}\sin\theta_{c}\end{pmatrix}.$ (32)

A useful remark for later is that this shows that

$\bm{S}_{g}^{q}=\theta_{q}\frac{\partial}{\partial\theta_{c}}\bm{S}_{g}^{c}+\phi_{q}\frac{\partial}{\partial\phi_{c}}\bm{S}_{g}^{c}.$
(33)

Going back to $iS_{b}[g]$, we can rewrite this as a convolution, in the sense
that

$iS_{b}[g]=-i\int
dt\;\left[\bm{S}_{g}^{c}(t)\cdot\left(\alpha^{A}*\bm{S}_{g}^{q}\right)(t)+\bm{S}_{g}^{q}(t)\cdot\left(\alpha^{R}*\bm{S}_{g}^{c}\right)(t)+\bm{S}_{g}^{q}(t)\cdot\left(\alpha^{K}*\bm{S}_{g}^{q}\right)(t)\right],$
(34)

where
$(f*g)(t)=\int_{-\infty}^{\infty}dt^{\prime}f(t-t^{\prime})g(t^{\prime})$. We
see that the first two terms contain precisely one quantum component, but the
last term has two quantum components. When writing down the Euler-Lagrange
equation of motion, it is important to realize that the convolution operation
will act as if it is a simple multiplication, since the convolution obeys

$\frac{d}{dx}(f(x)*g)(t)=\left(\frac{df}{dx}*g\right)(t).$ (35)

We now concentrate on the $(q)^{2}$ part of this action, for which we would
like to use a Hubbard-Stratonovich transformation in order to reduce this to
linear in $(q)$. Recall that a Hubbard-Stratonovich transformation is given by

$\exp\left[-\frac{a}{2}x^{2}\right]=\sqrt{\frac{1}{2\pi a}}\int
D\xi\exp\left[-\frac{\xi^{2}}{2a}-ix\xi\right].$ (36)

However, we see that our action does not contain any purely quadratic terms,
but rather a Greens functional shape as
$\bm{S}_{g}^{q}(t)\alpha^{K}(t-t^{\prime})\bm{S}_{g}^{q}(t^{\prime})$. Hence,
to use a Hubbard-Stratonovich like transformation, we must derive it from a
Greens function exponential, similarly to Ref. [3]. Assuming that this is
renormalizable and that $\alpha^{K}$ can be rewritten into a distribution, we
have

$\displaystyle 1$ $\displaystyle=\int D\xi\exp\left[-\frac{1}{2}\int dt\int
dt^{\prime}\bm{\xi}(t)[-2i\alpha^{K}]^{-1}(t-t^{\prime})\bm{\xi}(t^{\prime})\right]$
$\displaystyle=\int D\xi\exp\left[-\frac{1}{2}\int dt\int
dt^{\prime}\left(\bm{\xi}(t)-2\int
dt^{\prime\prime}\bm{S}_{g}^{q}(t^{\prime\prime})\alpha^{K}(t^{\prime\prime}-t)\right)[-2i\alpha^{K}]^{-1}(t-t^{\prime})\left(\bm{\xi}(t^{\prime})-2\int
dt^{\prime\prime\prime}\alpha^{K}(t^{\prime}-t^{\prime\prime\prime})\bm{S}_{g}^{q}(t^{\prime\prime\prime})\right)\right]$
$\displaystyle=\int D\xi\exp\left[-\frac{1}{2}\int dt\int
dt^{\prime}\bm{\xi}(t)[-2i\alpha^{K}]^{-1}(t-t^{\prime})\bm{\xi}(t^{\prime})\right.$
$\displaystyle\qquad-\left.i\bm{S}_{g}^{q}(t)\delta(t-t^{\prime})\bm{\xi}(t^{\prime})-i\bm{\xi}(t)\delta(t-t^{\prime})\bm{S}_{g}^{q}(t^{\prime})-2i\bm{S}_{g}^{q}(t)\alpha^{K}(t-t^{\prime})\bm{S}_{g}^{q}(t^{\prime})\right]$
$\displaystyle=\int D\xi\exp\left[-\frac{1}{2}\int dt\int
dt^{\prime}\bm{\xi}(t)[-2i\alpha^{K}]^{-1}(t-t^{\prime})\bm{\xi}(t^{\prime})-2i\bm{S}_{g}^{q}(t)\delta(t-t^{\prime})\bm{\xi}(t^{\prime})-2i\bm{S}_{g}^{q}(t)\alpha^{K}(t-t^{\prime})\bm{S}_{g}^{q}(t^{\prime})\right],$

where we used that $\int
dt^{\prime}\alpha^{K}(t-t^{\prime})[\alpha^{K}]^{-1}(t^{\prime}-t^{\prime\prime})=\delta(t-t^{\prime\prime})$
and that $2i\alpha^{K}$ is positive real. Therefore, we find that

$\displaystyle\exp\left[-i\int dt\int
dt^{\prime}\bm{S}_{g}^{q}(t)\alpha^{K}(t-t^{\prime})\bm{S}_{g}^{q}(t^{\prime})\right]$
$\displaystyle=\int D\xi\exp\left[-\frac{1}{2}\int dt\int
dt^{\prime}\bm{\xi}(t)[-2i\alpha^{K}]^{-1}(t-t^{\prime})\bm{\xi}(t^{\prime})\right]$
$\displaystyle\qquad\cdot\exp\left[i\int
dt\bm{S}_{g}^{q}(t)\bm{\xi}(t)\right].$ (37)

The double integral in the first exponential signifies the statistical
properties of $\bm{\xi}$. For instance, if $\alpha^{K}$ is delta-like, then
$\bm{\xi}$ would have Gaussian statistics (e.g. white noise), but in general
we will have time correlated noise defined by $\alpha^{K}$ [3], such that

$\langle\bm{\xi}(t)\bm{\xi}(t^{\prime})\rangle=-2i\alpha^{K}(t-t^{\prime}).$
(38)

Since there is no $g$ dependence in the double $\bm{\xi}$ exponential, we will
leave it out of $S[g]$ and only remember these statistics. Our partition
function is then given by

$Z=\int D\bm{\xi}\exp\left(iS_{n}\left[\bm{\xi}\right]\right)\int
Dg\exp\left(iS_{sc}\left[g,\bm{\xi}\right]\right),$ (39)

where the noise action is given by

$iS_{n}\left[\bm{\xi}\right]=-\frac{1}{2}\int dt\int
dt^{\prime}\bm{\xi}(t)[-2i\alpha^{K}]^{-1}(t-t^{\prime})\bm{\xi}(t^{\prime})$
(40)

and the semi-classical action is given by

$\displaystyle iS_{sc}\left[g,\bm{\xi}\right]=i\int dt$
$\displaystyle\;S\Big{[}-\theta_{q}\sin\theta_{c}(-B_{z}+KS\cos\theta_{c}+\dot{\phi}_{c})-\theta_{q}\cos\theta_{c}(B_{x}\cos\phi_{c}+B_{y}\sin\phi_{c})$
$\displaystyle+\phi_{q}\sin\theta_{c}(\dot{\theta}_{c}+B_{x}\sin\phi_{c}-B_{y}\cos\phi_{c})\Big{]}+i\int
dt\;\left[\bm{\xi}(t)\bm{S}_{g}^{q}(t)\right]$ $\displaystyle-i\int
dt\;\left[\bm{S}_{g}^{c}(t)\cdot\left(\alpha^{A}*\bm{S}_{g}^{q}\right)(t)+\bm{S}_{g}^{q}(t)\cdot\left(\alpha^{R}*\bm{S}_{g}^{c}\right)(t)\right],$
(41)

where $\bm{S}_{g}^{c}(t)$ and $\bm{S}_{g}^{q}(t)$ include only up to first-
order corrections in quantum components. Assuming that $\alpha^{A/R}$ can be
written in terms of distributions, we can define the distribution
$\operatorname{\alpha_{\text{diss}}}(t)=-\alpha^{R}(t)-\alpha^{A}(-t)$ and
rewrite the semi-classical action as

$\displaystyle iS_{sc}\left[g,\bm{\xi}\right]$ $\displaystyle=i\int
dt\;S\Big{[}-\theta_{q}\sin\theta_{c}(-B_{z}+KS\cos\theta_{c}+\dot{\phi}_{c})-\theta_{q}\cos\theta_{c}(B_{x}\cos\phi_{c}+B_{y}\sin\phi_{c})$
$\displaystyle+\phi_{q}\sin\theta_{c}(\dot{\theta}_{c}+B_{x}\sin\phi_{c}-B_{y}\cos\phi_{c})\Big{]}+i\int
dt\;\left[\left(\operatorname{\alpha_{\text{diss}}}*\bm{S}_{g}^{c}\right)(t)+\bm{\xi}(t)\right]\bm{S}_{g}^{q}(t).$
(42)

Recall that, using the Euler angles, we have $\int Dg=\int D\theta
D\phi\sin(\theta)$. Technically, the factor of $\sin(\theta)$ would end up in
the action. However, since one could define $\rho=\cos(\theta)$ as a new
variable in order to avoid this, we know that this term is not relevant to the
physics. Hence, we can disregard it.

Since all terms in $iS_{sc}\left[g,\bm{\xi}\right]$ are either linear in
$\theta_{q}$ or $\phi_{q}$, we find two Euler-Lagrange equations of the form

$\frac{\delta\mathcal{L}_{sc}}{\delta\theta_{q}}=0\quad\text{and}\quad\frac{\delta\mathcal{L}_{sc}}{\delta\phi_{q}}=0.$
(43)

Remembering Eq. (33), we see that
$\frac{\delta\bm{S}_{g}^{q}(t)}{\delta\theta_{q}}=\frac{\delta\bm{S}_{g}^{c}(t)}{\delta\theta_{c}}$
and
$\frac{\delta\bm{S}_{g}^{q}(t)}{\delta\phi_{q}}=\frac{\delta\bm{S}_{g}^{c}(t)}{\delta\phi_{c}}$.
Hence, the e.o.m. can be rearranged to yield

$\displaystyle\dot{\phi}_{c}$
$\displaystyle=\frac{1}{S\sin\theta_{c}}\left[-\bm{B}(S_{z}^{c})+\left(\operatorname{\alpha_{\text{diss}}}*\bm{S}_{g}^{c}\right)(t)+\bm{\xi}(t)\right]\cdot\frac{\delta\bm{S}_{g}^{c}(t)}{\delta\theta_{c}}$
(44)

and

$\displaystyle\dot{\theta}_{c}$
$\displaystyle=-\frac{1}{S\sin\theta_{c}}\left[-\bm{B}(S_{z}^{c})+\left(\operatorname{\alpha_{\text{diss}}}*\bm{S}_{g}^{c}\right)(t)+\bm{\xi}(t)\right]\cdot\frac{\delta\bm{S}_{g}^{c}(t)}{\delta\phi_{c}},$
(45)

where $\bm{B}(S_{z}^{c})=\begin{pmatrix}B_{x}\\\ B_{y}\\\
B_{z}-KS_{z}^{c}\end{pmatrix}$.

###  I.4 Generalized Landau-Lifshitz-Gilbert equation

We want to show that the equations found by the microscopic model are in fact
precisely of the LLG form. For this, we will have to start from the LLG
equation, and introduce the same two Euler angles $\theta$ and $\phi$ for the
spin, and show that this gives rise to the same set of equations as previously
deduced.

We begin with the generalized LLG equation

$\dot{\bm{S}}(t)=\bm{S}(t)\times\left[-\bm{B}(S_{z})+\left(\operatorname{\alpha_{\text{diss}}}*\bm{S}\right)(t)+\bm{\xi}(t)\right],$
(46)

where $\operatorname{\alpha_{\text{diss}}}(t)=-\alpha^{R}(t)-\alpha^{A}(-t)$,
$\langle\bm{\xi}(t)\bm{\xi}(t^{\prime})\rangle=-2i\alpha^{K}(t-t^{\prime})$
and $\bm{B}(S_{z})=(B_{x},B_{y},B_{z}-KS_{z})^{T}$. Since the velocity of
$\bm{S}$ is always perpendicular to $\bm{S}$, we know that the magnitude of
$\bm{S}$ is constant. Hence, we can go to spherical coordinates, such that

$\bm{S}=S\begin{pmatrix}\sin\theta\cos\phi\\\ \sin\theta\sin\phi\\\
\cos\theta\end{pmatrix}.$ (47)

Inserting this into the LLG equation, we firstly see that

$\displaystyle\dot{\bm{S}}$
$\displaystyle=\dot{\theta}\frac{\partial\bm{S}}{\partial\theta}+\dot{\phi}\frac{\partial\bm{S}}{\partial\phi}=\dot{\theta}S\begin{pmatrix}\cos\theta\cos\phi\\\
\cos\theta\sin\phi\\\
-\sin\theta\end{pmatrix}+\dot{\phi}S\begin{pmatrix}-\sin\theta\sin\phi\\\
\sin\theta\cos\phi\\\ 0\end{pmatrix}.$

Now, we notice that the RHS of the LLG equation can, without loss of
generality, be written as $\bm{S}(t)\times\bm{r}$ with $\bm{r}=(x,y,z)^{T}$.
Working this out explicitly, we find that the LLG equation
$\dot{\bm{S}}=\bm{S}\times\bm{r}$ becomes

$S\begin{pmatrix}\dot{\theta}\cos\theta\cos\phi-\dot{\phi}\sin\theta\sin\phi\\\
\dot{\theta}\cos\theta\sin\phi+\dot{\phi}\sin\theta\cos\phi\\\
-\dot{\theta}\sin\theta\end{pmatrix}=S\begin{pmatrix}z\sin\theta\sin\phi-y\cos\theta\\\
x\cos\theta-z\sin\theta\cos\phi\\\
y\sin\theta\cos\phi-x\sin\theta\sin\phi\end{pmatrix}.$ (48)

We note that the equation corresponding to the $z$ component can be written as

$\dot{\theta}=-\frac{1}{\sin\theta}\bm{r}\cdot\begin{pmatrix}-\sin\theta\sin\phi\\\
\sin\theta\cos\phi\\\
0\end{pmatrix}=-\frac{1}{S\sin\theta}\bm{r}\cdot\frac{\partial\bm{S}}{\partial\phi}.$
(49)

Now, we add up the $\hat{x}$ and $\hat{y}$ equations, such that the
$\dot{\theta}$ cancels (i.e. $-\hat{x}\sin\phi+\hat{y}\cos\phi$). This yields

$\displaystyle\dot{\phi}\sin\theta(\sin^{2}\phi+\cos^{2}\phi)$
$\displaystyle=-z\sin\theta(\sin^{2}\phi+\cos^{2}\phi)+y\cos\theta\sin\phi+x\cos\theta\cos\phi,$
which simplifies to $\displaystyle\dot{\phi}$
$\displaystyle=\frac{1}{\sin\theta}\bm{r}\cdot\begin{pmatrix}\cos\theta\cos\phi\\\
\cos\theta\sin\phi\\\
-\sin\theta\end{pmatrix}=\frac{1}{S\sin\theta}\bm{r}\cdot\frac{\partial\bm{S}}{\partial\theta}.$
(50)

By inserting
$\bm{r}=-\bm{B}(S_{z})+\left(\operatorname{\alpha_{\text{diss}}}*\bm{S}\right)(t)+\bm{\xi}(t)$,
we see that this is identical to the equations derived from the microscopic
model

$\displaystyle\dot{\phi}_{c}$
$\displaystyle=\frac{1}{S\sin\theta_{c}}\left[-\bm{B}(S_{z})+\left(\operatorname{\alpha_{\text{diss}}}*\bm{S}_{g}^{c}\right)(t)+\bm{\xi}(t)\right]\cdot\frac{\delta\bm{S}_{g}^{c}(t)}{\delta\theta_{c}};$
(51) $\displaystyle\dot{\theta}_{c}$
$\displaystyle=-\frac{1}{S\sin\theta_{c}}\left[-\bm{B}(S_{z})+\left(\operatorname{\alpha_{\text{diss}}}*\bm{S}_{g}^{c}\right)(t)+\bm{\xi}(t)\right]\cdot\frac{\delta\bm{S}_{g}^{c}(t)}{\delta\phi_{c}}.$
(52)

Therefore, we may conclude that our microscopic model is described by the
generalized LLG equation.

For the fractional LLG equation, we are in particular interested in the case
where $\operatorname{\alpha_{\text{diss}}}*\bm{S}=\alpha_{s}D_{t}^{s}\bm{S}$,
where $D_{t}^{s}$ is a fractional derivative. For instance, assuming $0<s<1$,
the Liouville fractional derivative is given by

$D_{t}^{s}f(t)=\frac{1}{\Gamma(1-s)}\int_{-\infty}^{t}(t-t^{\prime})^{-s}f^{\prime}(t^{\prime})dt^{\prime}.$
(53)

So, if
$\operatorname{\alpha_{\text{diss}}}=\frac{\alpha_{s}\Theta(t)}{\Gamma(1-s)}t^{-s}\partial_{t}$,
then, because of the convolution with $\bm{S}$, we would find a fractional LLG
equation, whereas
$\operatorname{\alpha_{\text{diss}}}=\alpha_{1}\delta(t)\partial_{t}$ would
give the regular LLG equation.

##  II Fractional derivative from non-Ohmic spectral function

Here, we will compute the type of dissipation which comes from the spectral
function. We will first derive the spectral function from microscopic
quantities to see how it ends up in the Greens function. Then, we will
calculate the dissipation for three different cases.

###  II.1 Calculating the effective Greens functions

We recall that

$\alpha^{A/R}(t-t^{\prime})=\sum_{\alpha}\left(\frac{\gamma_{\alpha}^{2}}{4}G_{\alpha}^{A/R}(t-t^{\prime})+\delta(t-t^{\prime})\frac{\gamma_{\alpha}^{2}}{2m_{\alpha}\omega_{\alpha}^{2}}\right),$

where

$[G^{-1}_{\alpha}]^{R/A}(t-t^{\prime})=\delta(t-t^{\prime})\frac{m_{\alpha}}{2}[(i\partial_{t}\pm
i0)^{2}-\omega_{\alpha}^{2}].$

By the fluctuation dissipation theorem, we also have

$\alpha^{K}(\omega)=\left[\alpha^{R}(\omega)-\alpha^{A}(\omega)\right]\coth\left(\frac{\omega}{2T}\right).$

We are interested in finding closed forms for $\alpha^{R/A/K}(t-t^{\prime})$.
Using the relation

$\int
dt^{\prime}G^{-1}(t-t^{\prime})G(t^{\prime}-t^{\prime\prime})=\delta(t-t^{\prime\prime}),$
(54)

we note that

$\frac{m_{\alpha}}{2}[(i\partial_{t}\pm
i0)^{2}-\omega_{\alpha}^{2}]G^{R/A}_{\alpha}(t-t^{\prime\prime})=\delta(t-t^{\prime\prime}).$
(55)

The Fourier transform111We use the convention
$f(\omega)=\int_{-\infty}^{\infty}dt\;e^{i\omega t}f(t)$ and
$f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}d\omega\;e^{-i\omega t}f(\omega)$,
with $\delta(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}d\omega\;e^{i\omega t}$.
yields

$G^{R/A}_{\alpha}(\omega)=\frac{1}{\frac{m_{\alpha}}{2}\left[(\omega\pm
i0)^{2}-\omega_{\alpha}^{2}\right]}.$ (56)

We therefore find that

$\displaystyle\alpha^{R/A}(\omega)$
$\displaystyle=\sum_{\alpha}\left(\frac{\gamma_{\alpha}^{2}}{4}G_{\alpha}^{A/R}(\omega)+\frac{\gamma_{\alpha}^{2}}{2m_{\alpha}\omega_{\alpha}^{2}}\right)=\sum_{\alpha}\frac{\gamma_{\alpha}^{2}}{2m_{\alpha}}\left[\frac{1}{(\omega\pm
i0)^{2}-\omega_{\alpha}^{2}}+\frac{1}{\omega_{\alpha}^{2}}\right]=\sum_{\alpha}\frac{\gamma_{\alpha}^{2}}{2m_{\alpha}\omega_{\alpha}^{2}}\frac{\omega^{2}}{(\omega\pm
i0)^{2}-\omega_{\alpha}^{2}}.$ (57)

The spectral function is given by the imaginary part of the Fourier transform
of the dynamical susceptibility

$\chi(\omega)=\frac{\delta}{\delta\bm{S}(\omega)}\sum_{\alpha}\gamma_{\alpha}\hat{\bm{x}}_{\alpha}(\omega).$
(58)

The Hamiltonian e.o.m. for the bath can be found by going back to our starting
Hamiltonian

$\displaystyle H$
$\displaystyle=\bm{B}\cdot\bm{S}+\sum_{\alpha}\gamma_{\alpha}\bm{S}\cdot\hat{\bm{x}}_{\alpha}+\sum_{\alpha}\frac{\hat{\bm{p}}_{\alpha}^{2}}{2m_{\alpha}}+\frac{m_{\alpha}\omega_{\alpha}^{2}}{2}\hat{\bm{x}}_{\alpha}^{2}+\sum_{\alpha}\frac{\gamma_{\alpha}^{2}}{2m_{\alpha}\omega_{\alpha}^{2}}\bm{S}^{2}.$

The e.o.m. reads

$\dot{\hat{\bm{x}}}_{\alpha}=\frac{\hat{\bm{p}}_{\alpha}}{m_{\alpha}}\quad\text{and}\quad\dot{\hat{\bm{p}}}_{\alpha}=-\gamma_{\alpha}\bm{S}-m_{\alpha}\omega_{\alpha}^{2}\hat{\bm{x}}_{\alpha}.$
(59)

Combining both equations and taking the Fourier transform, we find

$\hat{\bm{x}}_{\alpha}(\omega)=\frac{\gamma_{\alpha}}{m_{\alpha}[(\omega+i0)^{2}-\omega_{\alpha}^{2}]}\bm{S}(\omega),$
(60)

where we have taken an infinitesimal amount of dissipation $+i0$ on the
oscillators into account. This leads to

$\chi(\omega)=\sum_{\alpha}\frac{\gamma_{\alpha}^{2}}{m_{\alpha}[(\omega+i0)^{2}-\omega_{\alpha}^{2}]}=\sum_{\alpha}\frac{\gamma_{\alpha}^{2}}{2m_{\alpha}\omega_{\alpha}}\left(\frac{1}{\omega+i0-\omega_{\alpha}}-\frac{1}{\omega+i0+\omega_{\alpha}}\right).$
(61)

We remark that

$\operatorname{Im}\frac{1}{x+i0}=-\pi\delta(x),$ (62)

which leads to

$J(\omega)=\operatorname{Im}\chi(\omega)=-\sum_{\alpha}\frac{\pi\gamma_{\alpha}^{2}}{2m_{\alpha}\omega_{\alpha}}\left[\delta(\omega-\omega_{\alpha})-\delta(\omega+\omega_{\alpha})\right]=-\frac{\pi}{2}\sum_{\alpha}\frac{\gamma_{\alpha}^{2}}{m_{\alpha}\omega_{\alpha}}\delta(\omega-\omega_{\alpha}),$
(63)

where we used that all oscillator frequencies are positive. We can identify
the spectral function in $\alpha^{R/A}$ as

$\alpha^{R/A}(\omega)=\sum_{\alpha}\frac{\gamma_{\alpha}^{2}}{2m_{\alpha}\omega_{\alpha}^{2}}\frac{\omega^{2}}{(\omega\pm
i0)^{2}-\omega_{\alpha}^{2}}=-\int_{0}^{\infty}\frac{d\varepsilon}{\pi}\frac{\omega^{2}\varepsilon^{-1}J(\varepsilon)}{(\omega\pm
i0)^{2}-\varepsilon^{2}}.$ (64)

Now, we will assume a particular shape for $J(\varepsilon)$. This can be
either Ohmic or non-Ohmic, but in general we may assume a power-law behavior
as some $J(\varepsilon)=\alpha_{s}\varepsilon^{s}$.

###  II.2 Ohmic spectral function

Beginning with the Ohmic case $J(\varepsilon)=\alpha_{1}\varepsilon$, we see
that

$\displaystyle 2i\text{ Im }\alpha^{R/A}(\omega)$
$\displaystyle=\alpha^{R/A}(\omega)-\left[\alpha^{R/A}\right]^{*}(\omega)$
$\displaystyle=-\alpha_{1}\int_{0}^{\infty}\frac{d\varepsilon}{\pi}\left[\frac{\omega^{2}}{(\omega\pm
i0)^{2}-\varepsilon^{2}}-\frac{\omega^{2}}{(\omega\mp
i0)^{2}-\varepsilon^{2}}\right]$
$\displaystyle=-\alpha_{1}\int_{-\infty}^{\infty}\frac{d\varepsilon}{2\pi}\left[\frac{\omega^{2}}{(\omega\pm
i0)^{2}-\varepsilon^{2}}-\frac{\omega^{2}}{(\omega\mp
i0)^{2}-\varepsilon^{2}}\right]$
$\displaystyle=-\alpha_{1}\omega^{2}\int_{-\infty}^{\infty}\frac{d\varepsilon}{2\pi}\left[\frac{1}{(\omega\pm
i0+\varepsilon)(\omega\pm i0-\varepsilon)}-\frac{1}{(\omega\mp
i0+\varepsilon)(\omega\mp i0-\varepsilon)}\right]$ $\displaystyle=\pm
4i0\omega^{3}\alpha_{1}\int_{-\infty}^{\infty}\frac{d\varepsilon}{2\pi}\frac{1}{(\omega\pm
i0+\varepsilon)(\omega\pm i0-\varepsilon)(\omega\mp i0+\varepsilon)(\omega\mp
i0-\varepsilon)},$ (65)

which has four poles at $\varepsilon=\pm_{1}(\omega\pm_{2}i0)$. Since the
integral scales as $<1/|\varepsilon|$, we can add an infinite radius half
circle to complete a complex contour integral. Notice from symmetry that we
will always have one of each of the four poles. We can thus drop the $\pm$
signs inside since this only changes the notation order in the fraction. We
thus find that

$\displaystyle 2i\text{ Im }\alpha^{R/A}(\omega)$ $\displaystyle=\pm
4i0\omega^{3}\alpha_{1}\int_{-\infty}^{\infty}\frac{d\varepsilon}{2\pi}\frac{1}{(\varepsilon+\omega+i0)(\varepsilon-\omega-i0)(\varepsilon+\omega-i0)(\varepsilon-\omega+i0)}.$
(66)

Completing the contour along the top, we find poles at
$\varepsilon=\pm\omega+i0$, which yields

$\displaystyle 2i\text{ Im }\alpha^{R/A}(\omega)=\mp 0\cdot
4\omega^{3}\alpha_{1}$
$\displaystyle\left[\frac{1}{(\omega+i0+\omega+i0)(\omega+i0+\omega-i0)(\omega+i0-\omega+i0)}\right.$
$\displaystyle+\left.\frac{1}{(-\omega+i0+\omega+i0)(-\omega+i0-\omega-i0)(-\omega+i0-\omega+i0)}\right]$
$\displaystyle=\mp 0\cdot
4\omega^{3}\alpha_{1}\left[\frac{1}{2(\omega+i0)(2\omega)(2i0)}+\frac{1}{(2i0)(-2\omega)2(-\omega+i0)}\right]$
$\displaystyle=\pm\frac{i}{2}\omega^{3}\alpha_{1}\left[\frac{1}{\omega^{2}+i0\omega}+\frac{1}{\omega^{2}-i0\omega}\right]$
$\displaystyle=\pm\frac{i}{2}\omega^{3}\alpha_{1}\left[\frac{\omega^{2}-i0\omega+\omega^{2}+i0\omega}{\omega^{4}}\right]$
$\displaystyle=\pm i\omega\alpha_{1}.$ (67)

Hence, $\text{ Im }\alpha^{R/A}(\omega)=\pm\alpha_{1}\omega/2$. Similarly,

$\displaystyle 2\text{ Re }\alpha^{R/A}(\omega)$
$\displaystyle=\alpha^{R/A}(\omega)+\left[\alpha^{R/A}\right]^{*}(\omega)$
$\displaystyle=-\alpha_{1}\int_{0}^{\infty}\frac{d\varepsilon}{\pi}\left[\frac{\omega^{2}}{(\omega\pm
i0)^{2}-\varepsilon^{2}}+\frac{\omega^{2}}{(\omega\mp
i0)^{2}-\varepsilon^{2}}\right]$
$\displaystyle=-\alpha_{1}\int_{-\infty}^{\infty}\frac{d\varepsilon}{2\pi}\left[\frac{\omega^{2}}{(\omega\pm
i0)^{2}-\varepsilon^{2}}+\frac{\omega^{2}}{(\omega\mp
i0)^{2}-\varepsilon^{2}}\right]$
$\displaystyle=-\alpha_{1}\int_{-\infty}^{\infty}\frac{d\varepsilon}{2\pi}\left[\frac{\omega^{2}}{(\omega\pm
i0+\varepsilon)(\omega\pm i0-\varepsilon)}+\frac{\omega^{2}}{(\omega\mp
i0+\varepsilon)(\omega\mp i0-\varepsilon)}\right]$
$\displaystyle=\alpha_{1}\int_{-\infty}^{\infty}\frac{d\varepsilon}{\pi}\frac{\omega^{2}(\varepsilon^{2}-\omega^{2})}{(\varepsilon+\omega+i0)(\varepsilon-\omega-i0)(\varepsilon+\omega-i0)(\varepsilon-\omega+i0)}.$
(68)

Since the integral scales as $1/|\varepsilon|$, we can freely add the infinite
circular contour along the top. Applying the residue theorem, we find

$\displaystyle 2\text{ Re }\alpha^{R/A}(\omega)$
$\displaystyle=2i\alpha_{1}\left[\frac{\omega^{2}((\omega+i0)^{2}-\omega^{2})}{(\omega+i0+\omega+i0)(\omega+i0+\omega-i0)(\omega+i0-\omega+i0)}\right.$
$\displaystyle\quad\left.+\frac{\omega^{2}((-\omega+i0)^{2}-\omega^{2})}{(-\omega+i0+\omega+i0)(-\omega+i0-\omega-i0)(-\omega+i0-\omega+i0)}\right]$
$\displaystyle=2\alpha_{1}i\left[\frac{\omega^{2}(2\omega
i0)}{2(\omega+i0)(2\omega)(2i0)}+\frac{\omega^{2}(-2\omega
i0)}{(2i0)(-2\omega)2(-\omega+i0)}\right]$
$\displaystyle=\frac{\alpha_{1}}{2}i\left[\frac{\omega^{2}}{\omega+i0}-\frac{\omega^{2}}{\omega-i0}\right]$
$\displaystyle=\frac{\alpha_{1}}{2}i\left[\frac{\omega^{2}(-2i0)}{\omega^{2}}\right]$
$\displaystyle=\alpha_{1}0,$ (69)

which means that $\text{ Re }\alpha^{R/A}(\omega)=\frac{\alpha_{1}}{2}\cdot
0=0$. Hence,

$\alpha^{R/A}(\omega)=\pm\frac{\alpha_{1}i\omega}{2}$ (70)

and since
$\operatorname{\alpha_{\text{diss}}}(t)=-\alpha^{R}(t)-\alpha^{A}(-t)$ we have

$\operatorname{\alpha_{\text{diss}}}(\omega)=-\alpha^{R}(\omega)-\alpha^{A}(-\omega)=-\alpha_{1}i\omega.$
(71)

In the LLG equation, we thus find

$\displaystyle\left(\operatorname{\alpha_{\text{diss}}}*\bm{S}\right)(t)$
$\displaystyle=\frac{1}{2\pi}\int d\omega e^{-i\omega
t}\left(\operatorname{\alpha_{\text{diss}}}*\bm{S}\right)(\omega)$
$\displaystyle=\frac{1}{2\pi}\int d\omega e^{-i\omega
t}\operatorname{\alpha_{\text{diss}}}(\omega)\bm{S}(\omega)$
$\displaystyle=\frac{1}{2\pi}\int d\omega e^{-i\omega
t}(-\alpha_{1}i\omega)\bm{S}(\omega)$
$\displaystyle=\alpha_{1}\frac{\partial}{\partial t}\frac{1}{2\pi}\int d\omega
e^{-i\omega t}\bm{S}(\omega)$ $\displaystyle=\alpha_{1}\dot{\bm{S}}(t).$ (72)

Furthermore, we have that

$\displaystyle\alpha^{K}(\omega)$
$\displaystyle=[\alpha^{R}(\omega)-\alpha^{A}(\omega)]\coth\left(\frac{\omega}{2T}\right)$
$\displaystyle=i\alpha_{1}\omega\coth\left(\frac{\omega}{2T}\right).$ (73)

We have two regimes from the cotangent which crossover around $\omega\approx
2T$. If the temperature is large enough that we can approximate
$\coth\left(\frac{\omega}{2T}\right)\approx\frac{2T}{\omega}$, then we have
$\alpha^{K}(\omega)\approx 2i\alpha_{1}T$. Therefore, we find

$\alpha^{K}(t)=2i\alpha_{1}T\frac{1}{2\pi}\int_{-\infty}^{\infty}d\omega\;e^{-i\omega
t}=2i\alpha_{1}T\delta(t),$ (74)

and thus

$\langle\bm{\xi}_{m}(t)\bm{\xi}_{n}(t^{\prime})\rangle=-2i\delta_{m,n}\alpha^{K}(t-t^{\prime})=4\alpha_{1}T\delta_{m,n}\delta(t-t^{\prime}).$
(75)

We see that Eq. (72) and Eq. (75) combine to give the regular LLG equation
with first-order dissipation and white noise fluctuation:

$\dot{\bm{S}}(t)=\bm{S}(t)\times\left[-\bm{B}+\alpha_{1}\dot{\bm{S}}(t)+\bm{\xi}(t)\right].$
(76)

###  II.3 Sub-Ohmic spectral function

We now consider the case where

$J(\varepsilon)=\alpha_{s}\sin\left(\frac{\pi s}{2}\right)\varepsilon^{s},$
(77)

with $0<s<1$. In this case,

$\alpha^{R/A}(\omega)=-\alpha_{s}\sin\left(\frac{\pi
s}{2}\right)\int_{0}^{\infty}\frac{d\varepsilon}{\pi}\frac{\omega^{2}\varepsilon^{s-1}}{(\omega\pm
i0)^{2}-\varepsilon^{2}}.$ (78)

Considering the LLG equation, the relevant dissipation term is
$(\operatorname{\alpha_{\text{diss}}}*\bm{S})(t)$. In terms of the Fourier
transform of $\operatorname{\alpha_{\text{diss}}}(t)$, we find that

$\displaystyle(\operatorname{\alpha_{\text{diss}}}*\bm{S})(t)$
$\displaystyle=\int_{-\infty}^{\infty}dt^{\prime}\operatorname{\alpha_{\text{diss}}}(t-t^{\prime})\bm{S}(t^{\prime})$
$\displaystyle=-\int_{-\infty}^{\infty}dt^{\prime}[\alpha^{R}(t-t^{\prime})+\alpha^{A}(t^{\prime}-t)]\bm{S}(t^{\prime})$
$\displaystyle=-\frac{1}{2\pi}\int_{-\infty}^{\infty}dt^{\prime}\int_{-\infty}^{\infty}d\omega\left[e^{-i\omega(t-t^{\prime})}\alpha^{R}(\omega)+e^{i\omega(t-t^{\prime})}\alpha^{A}(\omega)\right]\bm{S}(t^{\prime})$
$\displaystyle=-\frac{1}{2\pi}\int_{-\infty}^{\infty}dt^{\prime}\int_{-\infty}^{\infty}d\omega\;e^{-i\omega(t-t^{\prime})}\left[\alpha^{R}(\omega)+\alpha^{A}(-\omega)\right]\bm{S}(t^{\prime})$
$\displaystyle=\alpha_{s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{2\pi^{2}}\int_{-\infty}^{\infty}dt^{\prime}\int_{-\infty}^{\infty}d\omega\int_{0}^{\infty}d\varepsilon\;e^{-i\omega(t-t^{\prime})}\left[\frac{\omega^{2}\varepsilon^{s-1}}{(\omega+i0)^{2}-\varepsilon^{2}}+\frac{(-\omega)^{2}\varepsilon^{s-1}}{(-\omega-i0)^{2}-\varepsilon^{2}}\right]\bm{S}(t^{\prime})$
$\displaystyle=-\alpha_{s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi^{2}}\int_{-\infty}^{\infty}dt^{\prime}\int_{0}^{\infty}d\varepsilon\int_{-\infty}^{\infty}d\omega\left[e^{-i\omega(t-t^{\prime})}\frac{\omega^{2}\varepsilon^{s-1}}{\varepsilon^{2}-(\omega+i0)^{2}}\bm{S}(t^{\prime})\right].$
(79)

Notice that we have two poles at $\omega=\pm\varepsilon-i0$, below the real
axis. If $t-t^{\prime}<0$, then the exponential will go to zero as
$\omega\to+i\infty$. Hence we could close the $\omega$ integration with a
complex contour as an infinite half-circle along the top, and get zero from
the Cauchy theorem. If $t-t^{\prime}>0$, however, we see that the exponential
goes to zero when $\omega\to-i\infty$. Hence, we can close the $\omega$
integration along the bottom. Thus, using the residue theorem (reversing the
integration direction), we find

$\displaystyle(\operatorname{\alpha_{\text{diss}}}*\bm{S})(t)$
$\displaystyle=-\alpha_{s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi^{2}}\int_{-\infty}^{\infty}dt^{\prime}\int_{0}^{\infty}d\varepsilon\;2\pi
i\Theta(t-t^{\prime})$
$\displaystyle\qquad\Big{[}e^{-i(\varepsilon-i0)(t-t^{\prime})}\frac{(\varepsilon-i0)^{2}\varepsilon^{s-1}}{(\varepsilon-i0+i0+\varepsilon)}+e^{-i(-\varepsilon-i0)(t-t^{\prime})}\frac{(-\varepsilon-i0)^{2}\varepsilon^{s-1}}{(-\varepsilon-i0+i0-\varepsilon)}\Big{]}\bm{S}(t^{\prime})$
$\displaystyle=-i\alpha_{s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi}\int_{-\infty}^{t}dt^{\prime}\int_{0}^{\infty}d\varepsilon\left[e^{-i\varepsilon(t-t^{\prime})}\frac{\varepsilon^{s+1}}{\varepsilon}+e^{i\varepsilon(t-t^{\prime})}\frac{\varepsilon^{s+1}}{-\varepsilon}\right]\bm{S}(t^{\prime})$
$\displaystyle=-i\alpha_{s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi}\int_{-\infty}^{t}dt^{\prime}\int_{0}^{\infty}d\varepsilon\left[e^{-i\varepsilon(t-t^{\prime})}-e^{i\varepsilon(t-t^{\prime})}\right]\varepsilon^{s}\bm{S}(t^{\prime})$
$\displaystyle=-2\alpha_{s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi}\int_{-\infty}^{t}dt^{\prime}\int_{0}^{\infty}d\varepsilon\sin[\varepsilon(t-t^{\prime})]\varepsilon^{s}\bm{S}(t^{\prime})$
$\displaystyle=-2\alpha_{s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi}\int_{0}^{\infty}d\varepsilon\left\\{\left[\varepsilon^{s-1}\cos[\varepsilon(t-t^{\prime})]\bm{S}(t^{\prime})\right]_{t^{\prime}=t_{0}}^{t^{\prime}=t}-\int_{t_{0}}^{t}dt^{\prime}\cos[\varepsilon(t-t^{\prime})]\varepsilon^{s-1}\dot{\bm{S}}(t^{\prime})\right\\}$
$\displaystyle=-2\alpha_{s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi}\int_{0}^{\infty}d\varepsilon\left\\{\varepsilon^{s-1}\bm{S}(t)-\varepsilon^{s-1}\cos[\varepsilon(t-t_{0})]\bm{S}(t_{0})-\int_{t_{0}}^{t}dt^{\prime}\cos[\varepsilon(t-t^{\prime})]\varepsilon^{s-1}\dot{\bm{S}}(t^{\prime})\right\\}.$
(80)

The first term vanishes because of the cross product with $\bm{S}(t)$ in the
LLG equation. The second term is where we had to be careful. Here, we should
realize that the $-\infty$ is physically only indicating that it is a time
very far in the past. So, to avoid unphysical infinities, we introduced a
finite initial time $t_{0}$ and we will take $t_{0}\to-\infty$ later. For
this, we need to introduce some fractional derivative notation. We define the
Riemann-Liouville (RL) and Caputo (C) derivatives of order $s$, with an
integer $n$ such that $n\leq s<n+1$, as

$\displaystyle\prescript{RL}{t_{0}}{D}_{t}^{s}f(t)$
$\displaystyle=\frac{d^{n}}{dt^{n}}\frac{1}{\Gamma(n-s)}\int_{t_{0}}^{t}dt^{\prime}(t-t^{\prime})^{n-1-s}f(t^{\prime}),$
(81) $\displaystyle\prescript{C}{t_{0}}{D}_{t}^{s}f(t)$
$\displaystyle=\frac{1}{\Gamma(n-s)}\int_{t_{0}}^{t}dt^{\prime}(t-t^{\prime})^{n-1-s}f^{(n)}(t^{\prime}),$
(82)

where we reserve the simpler $D_{t}^{s}$ notation for the Liouville derivative
that was used in the main text.

Now, rescaling $\varepsilon\to\varepsilon/(t-t_{0})$ and
$\varepsilon\to\varepsilon/(t-t^{\prime})$ in the second and third terms of
Eq. (80) respectively, we have

$\displaystyle(\operatorname{\alpha_{\text{diss}}}*\bm{S})(t)$
$\displaystyle=2\alpha_{s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi}\int_{0}^{\infty}d\varepsilon\cos(\varepsilon)\varepsilon^{s-1}\left[(t-t_{0})^{-s}\bm{S}(t_{0})+\int_{t_{0}}^{t}dt^{\prime}(t-t^{\prime})^{-s}\dot{\bm{S}}(t^{\prime})\right]$
$\displaystyle=2\alpha_{s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi}\left[\cos\left(\frac{\pi
s}{2}\right)\Gamma(s)\right]\left[(t-t_{0})^{-s}\bm{S}(t_{0})+\int_{t_{0}}^{t}dt^{\prime}(t-t^{\prime})^{-s}\dot{\bm{S}}(t^{\prime})\right]$
$\displaystyle=2\alpha_{s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi}\cos\left(\frac{\pi
s}{2}\right)\Gamma(s)\left[(t-t_{0})^{-s}\bm{S}(t_{0})+\Gamma(1-s)\prescript{C}{t_{0}}{D}_{t}^{s}\bm{S}(t)\right]$
$\displaystyle=2\alpha_{s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi}\cos\left(\frac{\pi
s}{2}\right)\Gamma(s)\Gamma(1-s)\left[\frac{(t-t_{0})^{-s}}{\Gamma(1-s)}\bm{S}(t_{0})+\prescript{C}{t_{0}}{D}_{t}^{s}\bm{S}(t)\right]$
$\displaystyle=2\alpha_{s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi}\cos\left(\frac{\pi s}{2}\right)\frac{\pi}{\sin(\pi
s)}\prescript{RL}{t_{0}}{D}_{t}^{s}\bm{S}(t)$
$\displaystyle=\alpha_{s}\prescript{RL}{t_{0}}{D}_{t}^{s}\bm{S}(t)=\alpha_{s}D_{t}^{s}\bm{S}(t),$
(83)

where we used several identities from Sec. 6 in the Sup. Mat. of Ref. [12] and
in the last line we sent $t_{0}\to-\infty$.

For the noise correlation, we have to compute the Keldysh component. This is

$\displaystyle\alpha^{K}(t)$
$\displaystyle=\frac{1}{2\pi}\int_{-\infty}^{\infty}d\omega\;e^{-i\omega
t}\alpha^{K}(\omega)$
$\displaystyle=\frac{1}{2\pi}\int_{-\infty}^{\infty}d\omega\;e^{-i\omega
t}[\alpha^{R}(\omega)-\alpha^{A}(\omega)]\coth\left(\frac{\omega}{2T}\right)$
$\displaystyle=\alpha_{s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{2\pi^{2}}\int_{0}^{\infty}d\varepsilon\int_{-\infty}^{\infty}d\omega\;e^{-i\omega
t}\left[\frac{\omega^{2}\varepsilon^{s-1}}{\varepsilon^{2}-(\omega+i0)^{2}}-\frac{\omega^{2}\varepsilon^{s-1}}{\varepsilon^{2}-(\omega-i0)^{2}}\right]\coth\left(\frac{\omega}{2T}\right)$
(now send $\omega\to-\omega$ in the advanced part)
$\displaystyle=\alpha_{s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{2\pi^{2}}\int_{0}^{\infty}d\varepsilon\int_{-\infty}^{\infty}d\omega\;\left[\frac{e^{-i\omega
t}\omega^{2}\varepsilon^{s-1}}{\varepsilon^{2}-(\omega+i0)^{2}}+\frac{e^{i\omega
t}\omega^{2}\varepsilon^{s-1}}{\varepsilon^{2}-(-\omega-i0)^{2}}\right]\coth\left(\frac{\omega}{2T}\right)$
$\displaystyle=\alpha_{s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi^{2}}\int_{0}^{\infty}d\varepsilon\int_{-\infty}^{\infty}d\omega\;\cos(\omega
t)\frac{\omega^{2}\varepsilon^{s-1}}{\varepsilon^{2}-(\omega+i0)^{2}}\coth\left(\frac{\omega}{2T}\right).$
(84)

Now, we send $\omega\to\omega/t$ and $\varepsilon\to\varepsilon/t$, which
yields

$\displaystyle\alpha^{K}(t)$
$\displaystyle=\alpha_{s}t^{-1-s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi^{2}}\int_{0}^{\infty}d\varepsilon\int_{-\infty}^{\infty}d\omega\;\cos(\omega)\frac{\omega^{2}\varepsilon^{s-1}}{\varepsilon^{2}-(\omega+i0)^{2}}\coth\left(\frac{\omega}{2tT}\right).$
(85)

Taking the high temperature limit, we get

$\displaystyle\alpha^{K}(t)$
$\displaystyle=\alpha_{s}t^{-1-s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi^{2}}\int_{0}^{\infty}d\varepsilon\int_{-\infty}^{\infty}d\omega\;\cos(\omega)\frac{\omega^{2}\varepsilon^{s-1}}{\varepsilon^{2}-(\omega+i0)^{2}}\frac{2tT}{\omega}$
$\displaystyle=2T\alpha_{s}t^{-s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi^{2}}\int_{0}^{\infty}d\varepsilon\int_{-\infty}^{\infty}d\omega\;\cos(\omega)\frac{\omega\varepsilon^{s-1}}{\varepsilon^{2}-(\omega+i0)^{2}}$
$\displaystyle=-2T\alpha_{s}t^{-s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi^{2}}\int_{0}^{\infty}d\varepsilon\int_{-\infty}^{\infty}d\omega\;\frac{\cos(\omega)\omega\varepsilon^{s-1}}{(\omega+i0-\varepsilon)(\omega+i0+\varepsilon)}.$
(86)

Now, we want to close the integral over $\omega$ with an infinite half-circle.
For this, we need fast enough convergence of the integrand to zero. Splitting
the cosine into two exponential parts
$\cos(\omega)=(e^{i\omega}+e^{-i\omega})/2$, we see that the first term goes
to zero when $\omega\to i\infty$ and that the second term goes to zero when
$\omega\to-i\infty$. Since we have poles at $\omega=\pm\varepsilon-i0$, the
integral along the top half-plane vanishes. The integral along the bottom is
then computed as

$\displaystyle\alpha^{K}(t)$
$\displaystyle=-T\alpha_{s}t^{-s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi^{2}}\int_{0}^{\infty}d\varepsilon\int_{-\infty}^{\infty}d\omega\;\frac{e^{-i\omega}\omega\varepsilon^{s-1}}{(\omega+i0-\varepsilon)(\omega+i0+\varepsilon)}$
$\displaystyle=-T\alpha_{s}t^{-s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi^{2}}\int_{0}^{\infty}d\varepsilon\;-2\pi
i\left[\frac{e^{-i(\varepsilon-i0)}(\varepsilon-i0)\varepsilon^{s-1}}{\varepsilon-i0+i0+\varepsilon}+\frac{e^{-i(-\varepsilon-i0)}(-\varepsilon-i0)\varepsilon^{s-1}}{-\varepsilon-i0+i0-\varepsilon}\right]$
$\displaystyle=2iT\alpha_{s}t^{-s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi}\int_{0}^{\infty}d\varepsilon\;\left[\frac{e^{-i\varepsilon}\varepsilon^{s}}{2\varepsilon}+\frac{e^{i\varepsilon}\varepsilon^{s}}{2\varepsilon}\right]$
$\displaystyle=2iT\alpha_{s}t^{-s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi}\int_{0}^{\infty}d\varepsilon\;\cos(\varepsilon)\varepsilon^{s-1}$
$\displaystyle=2iT\alpha_{s}t^{-s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi}\cos\left(\frac{\pi s}{2}\right)\Gamma(s)$
$\displaystyle=iT\alpha_{s}t^{-s}\frac{\sin\left(\pi s\right)}{\pi}\Gamma(s)$
$\displaystyle=iT\alpha_{s}\frac{t^{-s}}{\Gamma(1-s)}.$ (87)

We therefore find that

$\langle\bm{\xi}_{m}(t)\bm{\xi}_{n}(t^{\prime})\rangle=-2i\delta_{m,n}\alpha^{K}(t-t^{\prime})=2\alpha_{s}T\delta_{m,n}\frac{(t-t^{\prime})^{-s}}{\Gamma(1-s)},$
(88)

where we assumed that $t\geq t^{\prime}$. Therefore, we have now found the
fractional LLG equation

$\dot{\bm{S}}(t)=\bm{S}(t)\times\left[-\bm{B}+\alpha_{s}D_{t}^{s}\bm{S}(t)+\bm{\xi}(t)\right].$
(89)

###  II.4 Super-Ohmic spectral function

We now consider the case where

$J(\varepsilon)=\alpha_{s}\sin\left(\frac{\pi s}{2}\right)\varepsilon^{s},$
(90)

with $1<s<2$. In this case, everything is equivalent to the sub-Ohmic case, up
to Eq. (80), where we wanted to rewrite the dissipation into a fractional
derivative. We had to introduce a finite initial time $t_{0}$, which lead to

$\displaystyle(\operatorname{\alpha_{\text{diss}}}*\bm{S})(t)$
$\displaystyle=-2\alpha_{s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi}\int_{-\infty}^{t}dt^{\prime}\int_{0}^{\infty}d\varepsilon\sin[\varepsilon(t-t^{\prime})]\varepsilon^{s}\bm{S}(t^{\prime})$
$\displaystyle=-2\alpha_{s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi}\int_{0}^{\infty}d\varepsilon\left\\{\left[\varepsilon^{s-1}\cos[\varepsilon(t-t^{\prime})]\bm{S}(t^{\prime})\right]_{t^{\prime}=t_{0}}^{t^{\prime}=t}-\int_{t_{0}}^{t}dt^{\prime}\cos[\varepsilon(t-t^{\prime})]\varepsilon^{s-1}\dot{\bm{S}}(t^{\prime})\right\\}$
$\displaystyle=-2\alpha_{s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi}\int_{0}^{\infty}d\varepsilon\left\\{\varepsilon^{s-1}\bm{S}(t)-\varepsilon^{s-1}\cos[\varepsilon(t-t_{0})]\bm{S}(t_{0})-\int_{t_{0}}^{t}dt^{\prime}\cos[\varepsilon(t-t^{\prime})]\varepsilon^{s-1}\dot{\bm{S}}(t^{\prime})\right\\}.$
(91)

The first term vanishes because of the cross product with $\bm{S}(t)$ in the
LLG equation. However, the second term is more problematic compared to the
sub-Ohmic case, since the identity used to rewrite it only holds for $s<1$. We
solve this by writing it as a time derivative

$\varepsilon^{s-1}\cos[\varepsilon(t-t_{0})]\bm{S}(t_{0})=\frac{d}{dt}\varepsilon^{s-2}\sin[\varepsilon(t-t_{0})]\bm{S}(t_{0}),$
(92)

and then switching the ordering of the derivative and integral. Performing
also one more partial integration in $t^{\prime}$, we get

$\displaystyle(\operatorname{\alpha_{\text{diss}}}*\bm{S})(t)$
$\displaystyle=2\alpha_{s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi}\int_{0}^{\infty}d\varepsilon\left\\{\frac{d}{dt}\varepsilon^{s-2}\sin[\varepsilon(t-t_{0})]\bm{S}(t_{0})+\int_{t_{0}}^{t}dt^{\prime}\cos[\varepsilon(t-t^{\prime})]\varepsilon^{s-1}\dot{\bm{S}}(t^{\prime})\right\\}$
$\displaystyle=2\alpha_{s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi}\int_{0}^{\infty}d\varepsilon\left(\frac{d}{dt}\varepsilon^{s-2}\sin[\varepsilon(t-t_{0})]\bm{S}(t_{0})+\left[\varepsilon^{s-2}\sin[\varepsilon(t-t^{\prime})]\dot{\bm{S}}(t^{\prime})\right]_{t^{\prime}=t_{0}}^{t^{\prime}=t}-\int_{t_{0}}^{t}dt^{\prime}\left\\{-\sin[\varepsilon(t-t^{\prime})]\varepsilon^{s-2}\ddot{\bm{S}}(t^{\prime})\right\\}\right)$
$\displaystyle=2\alpha_{s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi}\int_{0}^{\infty}d\varepsilon\left\\{\frac{d}{dt}\varepsilon^{s-2}\sin[\varepsilon(t-t_{0})]\bm{S}(t_{0})+\varepsilon^{s-2}\sin[\varepsilon(t-t_{0})]\dot{\bm{S}}(t_{0})+\int_{t_{0}}^{t}dt^{\prime}\sin[\varepsilon(t-t^{\prime})]\varepsilon^{s-2}\ddot{\bm{S}}(t^{\prime})\right\\}.$
(93)

Now, rescaling $\varepsilon\to\varepsilon/(t-t_{0})$ and
$\varepsilon\to\varepsilon/(t-t^{\prime})$ respectively, we have

$\displaystyle(\operatorname{\alpha_{\text{diss}}}*\bm{S})(t)$
$\displaystyle=2\alpha_{s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi}\int_{0}^{\infty}d\varepsilon\sin(\varepsilon)\varepsilon^{s-2}\left[\frac{d}{dt}(t-t_{0})^{1-s}\bm{S}(t_{0})+(t-t_{0})^{1-s}\dot{\bm{S}}(t_{0})+\int_{t_{0}}^{t}dt^{\prime}(t-t^{\prime})^{1-s}\ddot{\bm{S}}(t^{\prime})\right]$
$\displaystyle=2\alpha_{s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi}\left[\sin\left(\frac{\pi(s-1)}{2}\right)\Gamma(s-1)\right]\left[(1-s)(t-t_{0})^{-s}\bm{S}(t_{0})+(t-t_{0})^{1-s}\dot{\bm{S}}(t_{0})+\int_{t_{0}}^{t}dt^{\prime}(t-t^{\prime})^{1-s}\ddot{\bm{S}}(t^{\prime})\right]$
$\displaystyle=-2\alpha_{s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi}\cos\left(\frac{\pi
s}{2}\right)\Gamma(s-1)\left[(1-s)(t-t_{0})^{-s}\bm{S}(t_{0})+(t-t_{0})^{1-s}\dot{\bm{S}}(t_{0})+\Gamma(2-s)\prescript{C}{t_{0}}{D}_{t}^{s}\bm{S}(t)\right]$
$\displaystyle=-2\alpha_{s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi}\cos\left(\frac{\pi
s}{2}\right)\Gamma(s-1)\Gamma(1-(s-1))\left[\frac{(t-t_{0})^{-s}}{\Gamma(1-s)}\bm{S}(t_{0})+\frac{(t-t_{0})^{1-s}}{\Gamma(2-s)}\dot{\bm{S}}(t_{0})+\prescript{C}{t_{0}}{D}_{t}^{s}\bm{S}(t)\right]$
$\displaystyle=-\alpha_{s}\frac{\sin\left(\pi
s\right)}{\pi}\frac{\pi}{\sin[\pi(s-1)]}\prescript{RL}{t_{0}}{D}_{t}^{s}\bm{S}(t)$
$\displaystyle=\alpha_{s}D_{t}^{s}\bm{S}(t),$ (94)

where we used several identities from Sec. 6 in the Sup. Mat. of Ref. [12],
Ref. [13, p.893], and in the last line we sent $t_{0}\to-\infty$.

For the noise correlation, we have to compute the Keldysh component. This is

$\displaystyle\alpha^{K}(t)$
$\displaystyle=\frac{1}{2\pi}\int_{-\infty}^{\infty}d\omega\;e^{-i\omega
t}\alpha^{K}(\omega)$
$\displaystyle=\frac{1}{2\pi}\int_{-\infty}^{\infty}d\omega\;e^{-i\omega
t}[\alpha^{R}(\omega)-\alpha^{A}(\omega)]\coth\left(\frac{\omega}{2T}\right)$
$\displaystyle=\alpha_{s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{2\pi^{2}}\int_{0}^{\infty}d\varepsilon\int_{-\infty}^{\infty}d\omega\;e^{-i\omega
t}\left[\frac{\omega^{2}\varepsilon^{s-1}}{\varepsilon^{2}-(\omega+i0)^{2}}-\frac{\omega^{2}\varepsilon^{s-1}}{\varepsilon^{2}-(\omega-i0)^{2}}\right]\coth\left(\frac{\omega}{2T}\right).$
(95)

Now, we send $\omega\to-\omega$ in the advanced part

$\displaystyle\alpha^{K}(t)$
$\displaystyle=\alpha_{s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{2\pi^{2}}\int_{0}^{\infty}d\varepsilon\int_{-\infty}^{\infty}d\omega\;\left[\frac{e^{-i\omega
t}\omega^{2}\varepsilon^{s-1}}{\varepsilon^{2}-(\omega+i0)^{2}}+\frac{e^{i\omega
t}\omega^{2}\varepsilon^{s-1}}{\varepsilon^{2}-(-\omega-i0)^{2}}\right]\coth\left(\frac{\omega}{2T}\right)$
$\displaystyle=\alpha_{s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi^{2}}\int_{0}^{\infty}d\varepsilon\int_{-\infty}^{\infty}d\omega\;\cos(\omega
t)\frac{\omega^{2}\varepsilon^{s-1}}{\varepsilon^{2}-(\omega+i0)^{2}}\coth\left(\frac{\omega}{2T}\right).$
(96)

Then, we insert $\cos(\omega t)=\frac{d}{dt}\frac{\sin(\omega t)}{\omega}$ and
we send $\omega\to\omega/t$ and $\varepsilon\to\varepsilon/t$, which yields

$\displaystyle\alpha^{K}(t)$
$\displaystyle=\frac{d}{dt}\alpha_{s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi^{2}}\int_{0}^{\infty}d\varepsilon\int_{-\infty}^{\infty}d\omega\;\sin(\omega
t)\frac{\omega\varepsilon^{s-1}}{\varepsilon^{2}-(\omega+i0)^{2}}\coth\left(\frac{\omega}{2T}\right)$
$\displaystyle=\frac{d}{dt}\alpha_{s}t^{-s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi^{2}}\int_{0}^{\infty}d\varepsilon\int_{-\infty}^{\infty}d\omega\;\sin(\omega)\frac{\omega\varepsilon^{s-1}}{\varepsilon^{2}-(\omega+i0)^{2}}\coth\left(\frac{\omega}{2tT}\right).$
(97)

Taking the high-temperature limit, we get

$\displaystyle\alpha^{K}(t)$
$\displaystyle=\frac{d}{dt}\alpha_{s}t^{-s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi^{2}}\int_{0}^{\infty}d\varepsilon\int_{-\infty}^{\infty}d\omega\;\sin(\omega)\frac{\omega\varepsilon^{s-1}}{\varepsilon^{2}-(\omega+i0)^{2}}\frac{2tT}{\omega}$
$\displaystyle=\frac{d}{dt}2T\alpha_{s}t^{1-s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi^{2}}\int_{0}^{\infty}d\varepsilon\int_{-\infty}^{\infty}d\omega\;\sin(\omega)\frac{\varepsilon^{s-1}}{\varepsilon^{2}-(\omega+i0)^{2}}$
$\displaystyle=-2T\alpha_{s}(1-s)t^{-s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi^{2}}\int_{0}^{\infty}d\varepsilon\int_{-\infty}^{\infty}d\omega\;\frac{\sin(\omega)\varepsilon^{s-1}}{(\omega+i0-\varepsilon)(\omega+i0+\varepsilon)}.$
(98)

Now, we want to close the integral over $\omega$ with an infinite half-circle.
For this, we need fast enough convergence of the integrand to zero. Splitting
the cosine into two exponential parts
$\sin(\omega)=(e^{i\omega}-e^{-i\omega})/2i$, we see that the first term goes
to zero when $\omega\to i\infty$ and that the second term goes to zero when
$\omega\to-i\infty$. Since we have poles at $\omega=\pm\varepsilon-i0$, the
integral along the top half-plane vanishes. The integral along the bottom is
then computed as

$\displaystyle\alpha^{K}(t)$
$\displaystyle=-iT\alpha_{s}(1-s)t^{-s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi^{2}}\int_{0}^{\infty}d\varepsilon\int_{-\infty}^{\infty}d\omega\;\frac{e^{-i\omega}\varepsilon^{s-1}}{(\omega+i0-\varepsilon)(\omega+i0+\varepsilon)}$
$\displaystyle=-iT\alpha_{s}(1-s)t^{-s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi^{2}}\int_{0}^{\infty}d\varepsilon\;-2\pi
i\left[\frac{e^{-i(\varepsilon-i0)}\varepsilon^{s-1}}{\varepsilon-i0+i0+\varepsilon}+\frac{e^{-i(-\varepsilon-i0)}\varepsilon^{s-1}}{-\varepsilon-i0+i0-\varepsilon}\right]$
$\displaystyle=-2T\alpha_{s}(1-s)t^{-s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi}\int_{0}^{\infty}d\varepsilon\;\left[\frac{e^{-i\varepsilon}\varepsilon^{s-2}}{2}-\frac{e^{i\varepsilon}\varepsilon^{s-2}}{2}\right]$
$\displaystyle=i2T\alpha_{s}(1-s)t^{-s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi}\int_{0}^{\infty}d\varepsilon\;\sin(\varepsilon)\varepsilon^{s-2}$
$\displaystyle=-i2T\alpha_{s}(1-s)t^{-s}\frac{\sin\left(\frac{\pi
s}{2}\right)}{\pi}\cos\left(\frac{\pi s}{2}\right)\Gamma(s-1)$
$\displaystyle=iT\alpha_{s}t^{-s}\frac{\sin\left(\pi s\right)}{\pi}\Gamma(s)$
$\displaystyle=iT\alpha_{s}\frac{t^{-s}}{\Gamma(1-s)}.$ (99)

We therefore find the same expression as in the sub-Ohmic case, which leads to

$\langle\bm{\xi}_{m}(t)\bm{\xi}_{n}(t^{\prime})\rangle=-2i\delta_{m,n}\alpha^{K}(t-t^{\prime})=2\alpha_{s}T\delta_{m,n}\frac{(t-t^{\prime})^{-s}}{\Gamma(1-s)}.$
(100)

Therefore, we have now also found the fractional LLG equation in the super-
Ohmic case.

###  II.5 Comparison Ohmic versus non-Ohmic

In the Ohmic case, we started with $J(\omega)=\alpha_{1}\omega$ and ended up
with a $\alpha_{1}\dot{\bm{S}}(t)$ friction term, and noise correlation
$4\alpha_{1}k_{B}T\delta(t-t^{\prime})$. On the other hand, in the non-Ohmic
case, we started with $J(\varepsilon)=\alpha_{s}\sin\left(\frac{\pi
s}{2}\right)\varepsilon^{s}$ and ended up with a friction
$\alpha_{s}D_{t}^{s}\bm{S}(t)$ and noise correlation
$2\alpha_{s}k_{B}T\frac{(t-t^{\prime})^{-s}}{\Gamma(1-s)}$. Although it is
clear that both the non-Ohmic $J(\omega)$ and the friction term will go to the
Ohmic case when $s\to 1$, the noise is less straightforward. We can see that,
as $s\to 1$, the Gamma function will blow up, hence sending the correlation to
zero, with the exception of $t=t^{\prime}$. In this case, the numerator blows
up even before taking the limit of $s\to 1$. Hence, we can expect this
function to behave as a delta function. To find the correct prefactor, we
integrate the distribution with a test-function $f(t^{\prime})=1$ to find

$\displaystyle\lim_{s\to
1}\int_{-\infty}^{\infty}d(t-t^{\prime})\frac{(t-t^{\prime})^{-s}}{\Gamma(1-s)}\cdot
1$ $\displaystyle=\lim_{s\to
1}\left[\frac{(t-t^{\prime})^{1-s}}{(1-s)\Gamma(1-s)}\right]_{(t-t^{\prime})=-\infty}^{\infty}$
$\displaystyle=\lim_{s\to
1}\left[\frac{(t-t^{\prime})^{1-s}}{\Gamma(2-s)}\right]_{(t-t^{\prime})=-\infty}^{\infty}$
$\displaystyle=\left[\frac{(t-t^{\prime})}{|t-t^{\prime}|\Gamma(1)}\right]_{(t-t^{\prime})=-\infty}^{\infty}$
$\displaystyle=2.$

Hence, we see that the limit of the noise correlation becomes
$4\alpha_{1}k_{B}T\delta(t-t^{\prime})$, which is precisely as in the Ohmic
case.

##  III FMR powerlaw derivation

Here, we calculate the response of this spin-bath system to a rotating
magnetic field. We will find the steady state solutions and calculate several
quantities that experiments could measure.

###  III.1 Ferromagnetic Resonance

We will study the effects of a rotating magnetic field on the fractional LLG
(FLLG) equation

$\dot{\bm{S}}(t)=\bm{S}(t)\times\big{\\{}-\bm{B}_{\text{eff}}[t,\bm{S}(t)]+\alpha_{s}D_{t}^{s}\bm{S}(t)+\bm{\xi}(t)\big{\\}}.$
(101)

For simplicity, we assume that the temperature of the bath is low compared to
the energy of the external fields, such that the thermal noise $\bm{\xi}(t)$
may be neglected. We apply a rotating magnetic field

$\bm{B}_{\text{eff}}[t,\bm{S}(t)]=\begin{pmatrix}\Omega\cos(\omega_{d}t)\\\
\Omega\sin(\omega_{d}t)\\\ B_{0}-KS_{z}(t)\end{pmatrix}$ (102)

and use spherical coordinates

$\bm{S}=S\begin{pmatrix}\sin\theta\cos\phi\\\ \sin\theta\sin\phi\\\
\cos\theta\end{pmatrix}.$ (103)

We will assume a small $\theta$ approximation, where the ground state is in
the positive $z$ direction, i.e. $0<\Omega\ll B_{0}-KS$ and
$\alpha_{s}S\ll(B_{0}-KS)^{1-s}$. As shown in Section I.4, we may rewrite the
FLLG equation of this form in spherical coordinates as

$\displaystyle\dot{\phi}$
$\displaystyle=\frac{1}{\sin\theta}\left[-\bm{B}_{\text{eff}}[t,\bm{S}(t)]+\alpha_{s}D_{t}^{s}\bm{S}(t)\right]\cdot\begin{pmatrix}\cos\theta\cos\phi\\\
\cos\theta\sin\phi\\\ -\sin\theta\end{pmatrix};$ (104)
$\displaystyle\dot{\theta}$
$\displaystyle=\left[-\bm{B}_{\text{eff}}[t,\bm{S}(t)]+\alpha_{s}D_{t}^{s}\bm{S}(t)\right]\cdot\begin{pmatrix}\sin\phi\\\
-\cos\phi\\\ 0\end{pmatrix}.$ (105)

In the rotating frame, where $\bm{B}(t)$ is constant, we could expect the
system to go to a steady state after some time. Hence, we introduce a new
coordinate such, that $\phi=\omega_{d}t-\varphi$. We may then set
$\dot{\varphi}=\dot{\theta}=0$ to find the steady state in the rotating frame,
where

$\displaystyle\omega_{d}\sin\theta$
$\displaystyle=\left[-\bm{B}_{\text{eff}}[t,\bm{S}(t)]+\alpha_{s}D_{t}^{s}\bm{S}(t)\right]\cdot\begin{pmatrix}\cos\theta\cos(\omega_{d}t-\varphi)\\\
\cos\theta\sin(\omega_{d}t-\varphi)\\\ -\sin\theta\end{pmatrix};$ (106)
$\displaystyle 0$
$\displaystyle=\left[-\bm{B}_{\text{eff}}[t,\bm{S}(t)]+\alpha_{s}D_{t}^{s}\bm{S}(t)\right]\cdot\begin{pmatrix}\sin(\omega_{d}t-\varphi)\\\
-\cos(\omega_{d}t-\varphi)\\\ 0\end{pmatrix}.$ (107)

We note that $\bm{S}(t)$ is now only time dependent in the rotating-frame
term, which means that we can explicitly calculate the fractional derivative.
The Liouville derivative works well with Fourier transforms, hence the
fractional derivative of a trigonometric function is given by

$D_{t}^{s}\sin(\omega t)=|\omega|^{s}\sin\left(\omega
t+\text{sign}(\omega)\frac{\pi s}{2}\right),$ (108)

and similarly for a cosine. We remark that the Liouville derivative of a
constant can only be described by setting the initial time to some finite
$t_{0}$, in which case it becomes zero222Since physically the infinite time
only models a long time in the past, we will apply it as such. Formally, there
are some restrictions on the functions that the Liouville derivative can be
applied to. These include restrictions such as its integral over the whole
domain being finite. Since this is not the case for a non-zero constant on an
infinite interval, we have to regularize the lower integral boundary $-\infty$
as a finite $t_{0}$.. Combining this with the steady state expression for
$\bm{S}$, we find

$\alpha_{s}D_{t}^{s}\bm{S}(t)=\alpha_{s}D_{t}^{s}S\begin{pmatrix}\sin\theta\cos(\omega_{d}t-\varphi)\\\
\sin\theta\sin(\omega_{d}t-\varphi)\\\
\cos\theta\end{pmatrix}=\alpha_{s}S\sin\theta|\omega_{d}|^{s}\begin{pmatrix}\cos\left(\omega_{d}t-\varphi+\text{sign}(\omega_{d})\frac{\pi
s}{2}\right)\\\ \sin\left(\omega_{d}t-\varphi+\text{sign}(\omega_{d})\frac{\pi
s}{2}\right)\\\ 0\end{pmatrix}.$ (109)

Hence, we find that

$\displaystyle\omega_{d}\tan\theta$
$\displaystyle=\left[-\bm{B}_{\text{eff}}[t,\bm{S}(t)]+\alpha_{s}D_{t}^{s}\bm{S}(t)\right]\cdot\begin{pmatrix}\cos(\omega_{d}t-\varphi)\\\
\sin(\omega_{d}t-\varphi)\\\ -\tan\theta\end{pmatrix}$
$\displaystyle=\left[-\begin{pmatrix}\Omega\cos(\omega_{d}t)\\\
\Omega\sin(\omega_{d}t)\\\
B_{0}-KS\cos\theta\end{pmatrix}+\alpha_{s}S\sin\theta|\omega_{d}|^{s}\begin{pmatrix}\cos\left(\omega_{d}t-\varphi+\text{sign}(\omega_{d})\frac{\pi
s}{2}\right)\\\ \sin\left(\omega_{d}t-\varphi+\text{sign}(\omega_{d})\frac{\pi
s}{2}\right)\\\
0\end{pmatrix}\right]\cdot\begin{pmatrix}\cos(\omega_{d}t-\varphi)\\\
\sin(\omega_{d}t-\varphi)\\\ -\tan\theta\end{pmatrix}$
$\displaystyle=-\Omega\cos\varphi+B_{0}\tan\theta-
KS\sin\theta+\alpha_{s}S\sin\theta|\omega_{d}|^{s}\cos\left(\text{sign}(\omega_{d})\frac{\pi
s}{2}\right)$ (110)

and

$\displaystyle 0$
$\displaystyle=\left[-\bm{B}_{\text{eff}}[t,\bm{S}(t)]+\alpha_{s}D_{t}^{s}\bm{S}(t)\right]\cdot\begin{pmatrix}\sin(\omega_{d}t-\varphi)\\\
-\cos(\omega_{d}t-\varphi)\\\ 0\end{pmatrix}$
$\displaystyle=\left[-\begin{pmatrix}\Omega\cos(\omega_{d}t)\\\
\Omega\sin(\omega_{d}t)\\\
B_{0}-KS\cos\theta\end{pmatrix}+\alpha_{s}S\sin\theta|\omega_{d}|^{s}\begin{pmatrix}\cos\left(\omega_{d}t-\varphi+\text{sign}(\omega_{d})\frac{\pi
s}{2}\right)\\\ \sin\left(\omega_{d}t-\varphi+\text{sign}(\omega_{d})\frac{\pi
s}{2}\right)\\\
0\end{pmatrix}\right]\cdot\begin{pmatrix}\sin(\omega_{d}t-\varphi)\\\
-\cos(\omega_{d}t-\varphi)\\\ 0\end{pmatrix}$
$\displaystyle=\Omega\sin\varphi-\alpha_{s}S\sin\theta|\omega_{d}|^{s}\sin\left(\text{sign}(\omega_{d})\frac{\pi
s}{2}\right).$ (111)

With some rearranging, we have

$\varphi=\arcsin\left[\frac{\alpha_{s}S}{\Omega}|\omega_{d}|^{s}\sin\left(\text{sign}(\omega_{d})\frac{\pi
s}{2}\right)\sin\theta\right]$ (112)

and

$\displaystyle\Omega\cos\varphi$
$\displaystyle=\Omega\cos\arcsin\left[\frac{\alpha_{s}S}{\Omega}|\omega_{d}|^{s}\sin\left(\text{sign}(\omega_{d})\frac{\pi
s}{2}\right)\sin\theta\right]$
$\displaystyle=\sqrt{\Omega^{2}-\left[\alpha_{s}S|\omega_{d}|^{s}\sin\left(\text{sign}(\omega_{d})\frac{\pi
s}{2}\right)\sin\theta\right]^{2}}$
$\displaystyle=(B_{0}-KS\cos\theta-\omega_{d})\tan\theta+\alpha_{s}S\sin\theta|\omega_{d}|^{s}\cos\left(\text{sign}(\omega_{d})\frac{\pi
s}{2}\right).$ (113)

Squaring this, we get

$\displaystyle\Omega^{2}$
$\displaystyle=(B_{0}-KS\cos\theta-\omega_{d})^{2}\tan^{2}\theta+\left(\alpha_{s}S|\omega_{d}|^{s}\right)^{2}\sin^{2}\theta+2\alpha_{s}S|\omega_{d}|^{s}(B_{0}-KS\cos\theta-\omega_{d})\cos\left(\frac{\pi
s}{2}\right)\frac{\sin^{2}\theta}{\cos\theta},$

and multiplying by $\cos^{2}\theta=1-\sin^{2}\theta$, we have

$\displaystyle(1-\sin^{2}\theta)\Omega^{2}$
$\displaystyle=(B_{0}-KS\sqrt{1-\sin^{2}\theta}-\omega_{d})^{2}\sin^{2}\theta+\left[\left(\alpha_{s}S|\omega_{d}|^{s}\right)^{2}-2\alpha_{s}S|\omega_{d}|^{s}KS\cos\left(\frac{\pi
s}{2}\right)\right]\sin^{2}\theta(1-\sin^{2}\theta)$
$\displaystyle\quad+2\alpha_{s}S|\omega_{d}|^{s}(B_{0}-\omega_{d})\cos\left(\frac{\pi
s}{2}\right)\sin^{2}\theta\sqrt{1-\sin^{2}\theta}.$ (114)

We could go further and make this into (effectively) a 4th order equation for
$\sin^{2}\theta$. However, since we are in a small $\theta$ limit, we will
solve this equation up to first order in $\sin^{2}\theta$, which yields

$\displaystyle\Omega^{2}$
$\displaystyle=\sin^{2}\theta\Big{\\{}\Omega^{2}(B_{0}-KS-\omega_{d})^{2}+\left[\left(\alpha_{s}S|\omega_{d}|^{s}\right)^{2}-2\alpha_{s}S|\omega_{d}|^{s}KS\cos\left(\frac{\pi
s}{2}\right)\right]$
$\displaystyle\quad+2\alpha_{s}S|\omega_{d}|^{s}(B_{0}-\omega_{d})\cos\left(\frac{\pi
s}{2}\right)\Big{\\}}.$ (115)

Hence, we find that

$\displaystyle\sin^{2}\theta$
$\displaystyle=\frac{\Omega^{2}}{\Omega^{2}+(B_{0}-KS-\omega_{d})^{2}+\left(\alpha_{s}S|\omega_{d}|^{s}\right)^{2}+2\alpha_{s}S|\omega_{d}|^{s}(B_{0}-KS-\omega_{d})\cos\left(\frac{\pi
s}{2}\right)}$
$\displaystyle=\frac{\Omega^{2}}{(B_{0}-KS-\omega_{d})^{2}+\left(\alpha_{s}S|\omega_{d}|^{s}\right)^{2}+2\alpha_{s}S|\omega_{d}|^{s}(B_{0}-KS-\omega_{d})\cos\left(\frac{\pi
s}{2}\right)}+\mathcal{O}(\Omega^{4}).$ (116)

Note that the $\mathcal{O}(\Omega^{4})$ should formally be dimensionless, but
we explain what we mean with terms being small in Section IV, as this is more
subtle with fractional dimensions.

###  III.2 Resonance frequency and amplitude

Since we are studying ferromagnetic resonance, we want to find the driving
frequency for which we get the largest response from the magnetic system.
Since we know that the resonance for an Ohmic system is at
$\omega_{d}=B_{0}-KS$, we will expand the formula around this point to find
the new maximum. To compute the resonance frequency $\omega_{\text{res}}$, we
thus first assume that $\omega_{\text{res}}\approx(B_{0}-KS)(1+y)$ with $y$
small, such that $|\omega_{\text{res}}|^{s}\approx(B_{0}-KS)^{s}(1+sy)$ (for
$B_{0}-KS>0$). This results in

$\displaystyle\sin^{2}\theta$
$\displaystyle=\frac{\Omega^{2}}{(B_{0}-KS-\omega_{d})^{2}+\left(\alpha_{s}S|\omega_{d}|^{s}\right)^{2}+2\alpha_{s}S|\omega_{d}|^{s}(B_{0}-KS-\omega_{d})\cos\left(\frac{\pi
s}{2}\right)}$
$\displaystyle\approx\frac{\Omega^{2}}{(B_{0}-KS)^{2}y^{2}+\left(\alpha_{s}S\right)^{2}(B_{0}-KS)^{2s}(1+2sy)-2\alpha_{s}S(B_{0}-KS)^{s+1}(1+sy)y\cos\left(\frac{\pi
s}{2}\right)}.$ (117)

Now, we put the derivative with respect to $y$ equal to zero, to get

$(B_{0}-KS)^{2}y+s\left(\alpha_{s}S\right)^{2}(B_{0}-KS)^{2s}-\alpha_{s}S(B_{0}-KS)^{s+1}(1+2sy)\cos\left(\frac{\pi
s}{2}\right)=0.$ (118)

Hence, we find that

$\displaystyle y$
$\displaystyle=\frac{-s\left(\alpha_{s}S\right)^{2}(B_{0}-KS)^{2s}+\alpha_{s}S(B_{0}-KS)^{s+1}\cos\left(\frac{\pi
s}{2}\right)}{(B_{0}-KS)^{2}-2s\alpha_{s}S(B_{0}-KS)^{s+1}\cos\left(\frac{\pi
s}{2}\right)}$ $\displaystyle=\alpha_{s}S(B_{0}-KS)^{s-1}\cos\left(\frac{\pi
s}{2}\right)+\mathcal{O}(\alpha_{s}S)^{2},$ (119)

which results in

$\omega_{\text{res}}\approx(B_{0}-KS)\left[1+\alpha_{s}S(B_{0}-KS)^{s-1}\cos\left(\frac{\pi
s}{2}\right)\right]=(B_{0}-KS)+\alpha_{s}S(B_{0}-KS)^{s}\cos\left(\frac{\pi
s}{2}\right).$ (120)

We see that the resonance frequency gets shifted by a small amount, depending
on $s$, which scales non-linearly. Inserting this result into Eq. (116), we
can now also find an approximation for the amplitude at resonance:

$\displaystyle\sin^{2}\theta_{\text{res}}$
$\displaystyle=\frac{\Omega^{2}}{(B_{0}-KS-\omega_{\text{res}})^{2}+\left(\alpha_{s}S|\omega_{\text{res}}|^{s}\right)^{2}+2\alpha_{s}S|\omega_{\text{res}}|^{s}(B_{0}-KS-\omega_{\text{res}})\cos\left(\frac{\pi
s}{2}\right)}$
$\displaystyle\approx\Omega^{2}\Big{\\{}\left[\alpha_{s}S(B_{0}-KS)^{s}\cos\left(\frac{\pi
s}{2}\right)\right]^{2}+\left(\alpha_{s}S\left|(B_{0}-KS)+\alpha_{s}S(B_{0}-KS)^{s}\cos\left(\frac{\pi
s}{2}\right)\right|^{s}\right)^{2}$
$\displaystyle\quad-2\alpha_{s}S\left|(B_{0}-KS)+\alpha_{s}S(B_{0}-KS)^{s}\cos\left(\frac{\pi
s}{2}\right)\right|^{s}\alpha_{s}S(B_{0}-KS)^{s}\cos^{2}\left(\frac{\pi
s}{2}\right)\Big{\\}}^{-1}$
$\displaystyle=\frac{\Omega^{2}}{\left[\alpha_{s}S(B_{0}-KS)^{s}\right]^{2}}\Big{[}\cos^{2}\left(\frac{\pi
s}{2}\right)+\left|1+\alpha_{s}S(B_{0}-KS)^{s-1}\cos\left(\frac{\pi
s}{2}\right)\right|^{2s}$
$\displaystyle\quad-2\left|1+\alpha_{s}S(B_{0}-KS)^{s-1}\cos\left(\frac{\pi
s}{2}\right)\right|^{s}\cos^{2}\left(\frac{\pi s}{2}\right)\Big{]}^{-1}$
$\displaystyle=\frac{\Omega^{2}}{\left[\alpha_{s}S(B_{0}-KS)^{s}\right]^{2}}\Big{[}1-\cos^{2}\left(\frac{\pi
s}{2}\right)+\mathcal{O}(\alpha_{s}S)\Big{]}^{-1}$
$\displaystyle\approx\frac{\Omega^{2}}{\left[\alpha_{s}S(B_{0}-KS)^{s}\sin\left(\frac{\pi
s}{2}\right)\right]^{2}}.$ (121)

Since the sine function decreases as $s$ moves away from one, we see that the
amplitude actually increases for non-Ohmic environments.

###  III.3 Calculating the FWHM linewidth

Next, we are interested not only in the location of the resonance, but also
how sensitive the resonance is to the driving frequency. One way to describe
this is by using the Full Width at Half Maximum measure. This provides a well-
defined line width independently of the shape of the peak. It is found by
measuring the width of the peak at half the height of its maximum. This can be
measured in the laboratories, but it can also be computed. Since our function
of interest is of the form
$\sin^{2}\theta(\omega_{d})=\Omega^{2}/g(\omega_{d})$, it makes sense to
approximate the inverse function instead of the regular one. To this end, we
will translate the FWHM measurement to the inverse function, and then Taylor
expand $g(\omega_{d})$ near resonance as a parabola to solve for the new
condition of this inverse function. Notice that from Eq. (116), we have

$g(\omega_{d})=(B_{0}-KS-\omega_{d})^{2}+\left(\alpha_{s}S|\omega_{d}|^{s}\right)^{2}+2\alpha_{s}S|\omega_{d}|^{s}(B_{0}-KS-\omega_{d})\cos\left(\frac{\pi
s}{2}\right).$ (122)

The FWHM condition is

$\frac{\Omega^{2}}{g(\omega_{d})}=\sin^{2}\theta(\omega_{d})=\frac{\sin^{2}\theta(\omega_{\text{res}})}{2}=\frac{\Omega^{2}}{2g(\omega_{\text{res}})},$
(123)

hence we must solve for $2g(\omega_{\text{res}})=g(\omega_{d})$. To this end,
let us assume that $\omega_{d}=\omega_{\text{res}}+y$ and expand
$g(\omega_{d})$ in $y$. We will use that

$|a+y|^{n}\approx a^{n}+na^{n-1}y+\frac{1}{2}n(n-1)a^{n-2}y^{2}$

for small $y$ and $a>0$. Then,

$\displaystyle g(\omega_{\text{res}}+y)$
$\displaystyle=(B_{0}-KS-\omega_{\text{res}}-y)^{2}+\left(\alpha_{s}S|\omega_{\text{res}}+y|^{s}\right)^{2}+2\alpha_{s}S|\omega_{\text{res}}+y|^{s}(B_{0}-KS-\omega_{\text{res}}-y)\cos\left(\frac{\pi
s}{2}\right)$
$\displaystyle\approx(B_{0}-KS-\omega_{\text{res}})^{2}+\left(\alpha_{s}S\omega_{\text{res}}^{s}\right)^{2}+2\alpha_{s}S\omega_{\text{res}}^{s}(B_{0}-KS-\omega_{\text{res}})\cos\left(\frac{\pi
s}{2}\right)$
$\displaystyle+y\left(-2(B_{0}-KS-\omega_{\text{res}})+2s(\alpha_{s}S)^{2}\omega_{\text{res}}^{2s-1}-2\alpha_{s}S\cos\left(\frac{\pi
s}{2}\right)\left\\{\omega_{\text{res}}^{s}+s\omega_{\text{res}}^{s-1}[\omega_{\text{res}}-(B_{0}-KS)]\right\\}\right)$
$\displaystyle+y^{2}\left[1+s(2s-1)(\alpha_{s}S)^{2}\omega_{\text{res}}^{2s-2}-2s\alpha_{s}S\omega_{\text{res}}^{s-1}\cos\left(\frac{\pi
s}{2}\right)+s(s-1)\alpha_{s}S\omega_{\text{res}}^{s-2}(B_{0}-KS-\omega_{\text{res}})\cos\left(\frac{\pi
s}{2}\right)\right]$
$\displaystyle=g(\omega_{\text{res}})+y\Bigg{(}2\alpha_{s}S(B_{0}-KS)^{s}\cos\left(\frac{\pi
s}{2}\right)+2s(\alpha_{s}S)^{2}(B_{0}-KS)^{2s-1}\left[1+\alpha_{s}S(B_{0}-KS)^{s-1}\cos\left(\frac{\pi
s}{2}\right)\right]^{2s-1}$
$\displaystyle\qquad-2\alpha_{s}S(B_{0}-KS)^{s}\cos\left(\frac{\pi
s}{2}\right)\Bigg{\\{}\left[1+\alpha_{s}S(B_{0}-KS)^{s-1}\cos\left(\frac{\pi
s}{2}\right)\right]^{s}$
$\displaystyle\qquad\qquad+s\alpha_{s}S\cos\left(\frac{\pi
s}{2}\right)(B_{0}-KS)^{s-1}\left[1+\alpha_{s}S(B_{0}-KS)^{s-1}\cos\left(\frac{\pi
s}{2}\right)\right]^{s-1}\Bigg{\\}}\Bigg{)}$
$\displaystyle+y^{2}\Bigg{\\{}1+s(2s-1)(\alpha_{s}S)^{2}(B_{0}-KS)^{2s-2}\left[1+\alpha_{s}S(B_{0}-KS)^{s-1}\cos\left(\frac{\pi
s}{2}\right)\right]^{2s-2}$
$\displaystyle\qquad-2s\alpha_{s}S(B_{0}-KS)^{s-1}\cos\left(\frac{\pi
s}{2}\right)\left[1+\alpha_{s}S(B_{0}-KS)^{s-1}\cos\left(\frac{\pi
s}{2}\right)\right]^{s-1}$
$\displaystyle\qquad-s(s-1)(\alpha_{s}S)^{2}(B_{0}-KS)^{2s-2}\cos^{2}\left(\frac{\pi
s}{2}\right)\left[1+\alpha_{s}S(B_{0}-KS)^{s-1}\cos\left(\frac{\pi
s}{2}\right)\right]^{s-2}\Bigg{\\}}$ $\displaystyle\approx
g(\omega_{\text{res}})+y\left\\{2s(\alpha_{s}S)^{2}(B_{0}-KS)^{2s-1}\left[1-2\cos^{2}\left(\frac{\pi
s}{2}\right)\right]\right\\}$
$\displaystyle+y^{2}\Big{\\{}1-2s(\alpha_{s}S)(B_{0}-KS)^{s-1}\cos\left(\frac{\pi
s}{2}\right)+(\alpha_{s}S)^{2}(B_{0}-KS)^{2s-2}\left[s(2s-1)-3s(s-1)\cos^{2}\left(\frac{\pi
s}{2}\right)\right]\Big{\\}}+\mathcal{O}(\alpha_{s}S)^{3}.$ (124)

Now, we set
$2g(\omega_{\text{res}})=g(\omega_{\text{res}}+y)=g(\omega_{\text{res}})+by+ay^{2}$,
and remark that
$g(\omega_{\text{res}})=(\alpha_{s}S)^{2}(B_{0}-KS)^{2s}\sin^{2}\left(\frac{\pi
s}{2}\right)$, in order to find that

$\displaystyle y$
$\displaystyle=\frac{-b\pm\sqrt{b^{2}+4ag(\omega_{\text{res}})}}{2a}\Longrightarrow\Delta_{FWHM}=\frac{\sqrt{b^{2}+4ag(\omega_{\text{res}})}}{a}.$
(125)

Hence, we find that the lowest-order contribution to the linewidth is given by

$\displaystyle\Delta_{FWHM}$
$\displaystyle\approx\frac{\sqrt{4(\alpha_{s}S)^{2}(B_{0}-KS)^{2s}\sin^{2}\left(\frac{\pi
s}{2}\right)+\mathcal{O}(\alpha_{s}S)^{3}}}{1+\mathcal{O}(\alpha_{s}S)}$
$\displaystyle=2(\alpha_{s}S)(B_{0}-KS)^{s}\sin\left(\frac{\pi
s}{2}\right)+\mathcal{O}(\alpha_{s}S)^{2}.$ (126)

##  IV Dimensional analysis

The fractional derivative in the LLG equation has an impact on the dimensions
of quantities. We can firstly see that in the chosen units, we have
$[B_{0}-KS]=[\omega_{d}]=\text{time}^{-1}$. Assuming $S$ to be dimensionless,
then $[\omega_{d}]=[\alpha_{s}D_{t}^{s}\bm{S}]=[\alpha_{s}][\omega_{d}]^{s}$,
hence $[\alpha_{s}]=[\omega_{d}]^{1-s}$. We can now start to understand what
we mean when we say that certain quantities are small, since this has to be
relative to something else. For instance, when we say $\alpha_{s}S$ is small,
we understand this as $\alpha_{s}S\ll(B_{0}-KS)^{1-s}$. For $\Omega$ it is
simpler, since there is no fractional derivative acting with it. Hence, for
$\Omega$ small we simply mean $\Omega\ll B_{0}-KS$. We can now also define
some dimensionless variables, such as
$\alpha_{s}^{\prime}=\alpha_{s}S(B_{0}-KS)^{s-1}$ and
$\Omega^{\prime}=\Omega/(B_{0}-KS)$. We have used these variables in the
figures to show the general behavior of the quantities.

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