# Fractional Floquet theory
[ Chaos, Solitons & Fractals 168 (2023) 113196 ]

Alexander Iomin<EMAIL_ADDRESS>Department of Physics, Technion,
Haifa, 32000, Israel,
Max Planck Institute for the Physics of Complex Systems, 01187 Dresden,
Germany

###### Abstract

A fractional generalization of the Floquet theorem is suggested for fractional
Schrödinger equations (FTSE)s with the time-dependent periodic Hamiltonians.
The obtained result, called the fractional Floquet theorem (fFT), is
formulated in the form of the Mittag-Leffler function, which is considered as
the eigenfunction of the Caputo fractional derivative. The suggested formula
makes it possible to reduce the FTSE to the standard quantum mechanics with
the time-dependent Hamiltonian, where the standard Floquet theorem is valid.
Two examples related to quantum resonances are considered as well to support
the obtained result.

###### keywords:

Floquet theorem, Fractional Schrödinger equation, Caputo fractional
derivative, Mittag-Leffler function

##  1 Introduction

In this paper, we consider a Floquet theory for fractional time Schrödinger
equations. Separately, these two issues of both the Floquet theorem and
fractional quantum mechanics are well defined and well reviewed. The Floquet
theory is introduced to treat linear differential equations with time-periodic
coefficients [1] and it is widely used in description of driven quantum
systems such as the interaction of radiation with matter, quantum nonlinear
resonances, quantum chaos and so on [2, 3, 4, 5]. The Floquet theorem states
that if the time-dependent Hamiltonian $\hat{H}(t)=\hat{H}(t+T)$ is periodic
in time with the period $T$, then the wave function, as the solution to the
corresponding Schrödinger equation, has the form $\psi(t)=e^{-i\epsilon
t}u(t)$. Here $u(t+T)=u(t)$ is the periodic eigenfunction of a so called
Floquet operator $\hat{F}$,

$\hat{F}u(t)\equiv[\hat{H}(t)-i\hbar\partial_{t}]u(t)=\epsilon u(t),$ (1.1)

where $\hbar$ is an effective dimensionless Planck constant and $\epsilon$ is
the quasienergy spectrum [6, 7] (see also Ref. [3]). It is worth mentioning
that in many cases, Eq. (1.1) has no analytic solutions, like in quantum chaos
[4]. Moreover, the spectrum can have fractal band structures, such as the
Hofstadter butterfly [8, 9], see also an extended review [5]. This situation
is also reflected in the exploration of the Floquet engineering of quantum
materials [10, 11].

Fractional time Schödinger equations belong to the field of fractional quantum
mechanics. The latter has been introduced in Refs. [12, 13] and currently it
is a well established field of quantum mechanics [14, 15, 16, 17], which is
also supported by experimental evidences [18, 19, 20]. The fractional time
Schrödinger equation (FTSE) has been introduced by analogy with a fractional
diffusion equation [21], where the time derivative is replaced by the
fractional time derivative, namely
$i\hbar\partial_{t}\rightarrow(i\hbar)^{\alpha}\partial_{t}^{\alpha}$ with
$\alpha\in(0,2)$, where $\partial_{t}^{\alpha}$ is the Caputo fractional
derivative,

$\partial_{t}^{\alpha}\equiv\fourIdx{C}{t_{0}}{\alpha}{t}{D}f(t)=\frac{1}{\Gamma(\alpha)}\int_{t_{0}}^{t}\frac{d^{n}f(\tau)}{d\tau^{n}}(t-\tau)^{\alpha-n}d\tau,$
(1.2)

with $n-1<\alpha<n$. In the present study, we consider $\alpha<1,~{}(n=1)$ and
$t_{0}=0$. An alternative replacement of the time derivative,
$i\hbar\partial_{t}\rightarrow i(\hbar)^{\alpha}\partial_{t}^{\alpha}$ has
been suggested as well in Refs. [16, 22] with the main argument that such
“fractionalization” of the time derivative does not lead to any artificial
non-physical effects. It is worth noting that the latter approach to the
fractional time derivative in the Schrödinger equation is based on the
generalized Taylor series that has been proposed in Ref. [23], and relates to
a general memory in the system e.g., in the form of a fading memory [24]. An
exact example of the relation between a fractional diffusion equation and its
quantum counterpart is studied in Ref. [25].

So far, the extended studies of the FTSE are restricted by consideration of
conservative systems, where the Hamiltonian is time independent. The main
reason for this restriction is that the Caputo fractional derivative destroys
the periodicity of any periodic functions, see A. Another important obstacle
of the treatment of the FTSE with the time-dependent Hamiltonians is the
violation of the general Leibniz rule by fractional derivatives, see e.g.,
Ref. [26].

In this paper, the FTSEs with the time-periodic operators,
$\hat{H}(t+T)=\hat{H}(t)$ are considered. We also suggest the extension of the
Floquet theorem for the FTSE,

$\partial_{t}^{\alpha}\Psi(t)=\hat{H}(t)\Psi(t).$ (1.3)

We call this theorem by the “fractional Floquet theorem” (fFT). To apply this
theorem for the various physical systems, it is convenient to use
dimensionless variables and parameters111Following Refs. [21, 27], one
introduces the Planck length $L_{P}=\sqrt{\hbar G/c^{3}}$, time
$T_{P}=\sqrt{\hbar G/c^{5}}$, mass $M_{P}=\sqrt{\hbar c/G}$, and energy
$E_{P}=M_{P}c^{2}$, where $\hbar,~{}G,$ and $c$ are the Planck constant, the
gravitational constant and the speed of light, respectively. Therefore,
quantum mechanics of a particle with dimensionless mass $m/M_{P}\rightarrow m$
is described by the dimensionless units of the coordinates and time
$x/L_{P}\rightarrow x,~{}t/T_{P}\rightarrow t$, while the dimensionless Planck
constant is $\hbar/(E_{P}T_{P})\rightarrow 1$. We however keep the notation of
$\hbar$. Note, that the dimensionless frequency is
$\hbar\omega/E_{P}\rightarrow\omega$, which keeps $\omega t$ being the
dimensionless parameter as well. [21].

The structure of the paper is as follows. In Sec. 2 we formulate the fFT with
its proof. Then the result on the fFT is presented in the form of
subordination to the Floquet theorem, in Sec. 3. In Sec 4 we consider both the
Floquet theorem and the fFT in Fourier - Laplace space. We also consider two
examples to show how the fFT can be applied for the corresponding FTSEs. The
first example, considered in Sec. 5, is devoted to a toy model, where the
unperturbed spectrum is not affected by time. In the second example we
consider a quantum particle in a time-dependent potential and describe it in
the framework of the FTSE in Sec. 6. The latter situation is also related to
quantum nonlinear resonances [28, 29, 30]. In Conclusion, we summarize the
obtained result on the fFT. Appendix plays a dual role. First, it is an
original result on fractional differentiation of periodic functions. At the
same time, it is also a brief overview of fractional calculus.

##  2 Fractional Floquet theorem

We formulate the fractional Floquet theorem in the form of the following
proposition.

###### Proposition 1

Let us consider the FTSE (1.3)

$\partial_{t}^{\alpha}\Psi(t)=\hat{H}(t)\Psi(t)$

with the time-dependent periodic Hamiltonian $\hat{H}(t+T)=\hat{H}(t)$. The
solution to the FTSE (1.3) can be presented as follows

$\Psi(t)=\sum_{n}E_{\alpha}\left[i\hbar^{1-\alpha}(\omega
n-\epsilon)t^{\alpha}\right]C_{n},$ (2.1)

where $E_{\alpha.1}\equiv E_{\alpha}(z)$ is the one-parameter Mittag -Leffler
function [31], $\omega=\frac{2\pi}{T}$, and $C_{n}$ are the coefficients of
the Fourier series of $u(t)$, defined for the standard Floquet theorem in Eq.
(1.1).

###### Proof 1

Important part of the consideration is the eigenvalue problem for the Floquet
operator in Eq. (1.1), which after the Fourier expansion, can be presented as
follows

$\sum_{n}\left[\hat{H}\left(\frac{i}{\omega}\frac{d}{dn}\right)C_{n}+\hbar(\omega
n-\epsilon)C_{n}\right]e^{i\omega nt}=0,$ (2.2)

see Eq. (4.10) and Comments 4.2. Due to the completeness of the Fourier basis
$e^{i\omega nt}$, one obtains the eigenvalue equation

$\hat{H}\left(\frac{i}{\omega}\frac{d}{dn}\right)C_{n}=\hbar(\epsilon-\omega
n)C_{n},~{}\forall n.$

Then we take into account that the Mittag-Leffler function is the
eigenfunction of the Caputo fractional derivative,

$\partial_{t}^{\alpha}E_{\alpha}(iat^{\alpha})=iaE_{\alpha}(iat^{\alpha}),$

where $a=\hbar^{1-\alpha}(\omega n-\epsilon)$, see Comment 3-3. Thus, the
straightforward substitution of the fFT (4.17) in the FTSE (1.3) yields

$\sum_{n}\left[\hbar(-\omega
n-\epsilon)C_{n}+\hat{H}(t)C_{n}\right]E_{\alpha}\left[i\hbar^{1-\alpha}(\omega
n-\epsilon)t^{\alpha}\right]=0,$ (2.3)

see Comment 4.2-2. The expression in square brackets is just the expression in
Eq. (2.2) that reduces the FTSE to the standard Floquet theory in Eq. (1.1).

This proves the Proposition 1 for the fFT.

##  3 Subordination to the Floquet theorem

Equation (2.1) can be presented in a subordination form, where the Floquet
theorem subordinates the fFT. Taking into account the Laplace image of the
Mittag-Leffler function, we have the chain of transformations as follows [16]

$E_{\alpha}(iat^{\alpha})=\frac{1}{2\pi
i}\int_{-i\infty}^{i\infty}\frac{s^{\alpha-1}e^{st}ds}{s^{\alpha}-ia}=\mathcal{L}^{-1}\left[\frac{s^{\alpha-1}e^{st}ds}{s^{\alpha}-ia}\right]\\\
=\mathcal{L}^{-1}\left[\int_{-\infty}^{\infty}\frac{s^{\alpha-1}e^{st}ds}{s^{\alpha}-iz}\delta(z-a)dz\right]=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-ia\xi}\left[\int_{-\infty}^{\infty}e^{iz\xi}E_{\alpha}(izt^{\alpha})\right]d\xi\\\
=\int_{-\infty}^{\infty}\mathcal{K}(\xi,t)\exp\left[i\xi(\omega
n-\epsilon)\right]d\xi,$ (3.1)

where the Fourier transformation of the Mittag-Leffler function,

$\mathcal{K}(\xi,t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-iz\xi}E_{\alpha}(-izt^{\alpha})dz$
(3.2)

can be reduced to the superposition of the odd and even functions of $z$ and
$\xi$. Thus, we obtain

$E_{\alpha}(-izt^{\alpha})=\sum_{n=0}^{\infty}\frac{(-z^{2}t^{2\alpha})^{n}}{\Gamma(n2\alpha+1)}-izt^{\alpha}\sum_{n=0}^{\infty}\frac{(-z^{2}t^{2\alpha})^{n}}{\Gamma(n2\alpha+\alpha+1)}\\\
=E_{2\alpha}(-z^{2}t^{2\alpha})-izt^{\alpha}E_{2\alpha,1+\alpha}(-z^{2}t^{2\alpha}).$
(3.3)

The first term here is the even function of $z$, which stands for the cosine-
Fourier transformation and correspondingly leads to the even function of
$\xi$. The second term is the odd function of $z$ and stands for the sine-
Fourier transform that leads to the odd function of $\xi$. Therefore,
integration with respect to $\xi\in(-\infty,\infty)$ is reduced to the
integration with respect to $\xi\in(0,\infty)$. Correspondingly, $\xi$ is
treated as a time parameter. Eventually, taking an analogy with a continuous
time random walk (CTRW) theory [32], we suggest that $\mathcal{K}(\xi,t)$ is
the subordinator [33, 34], which subordinates the fractional evolution on the
time scale $t$ (physical time) to the unitary evolution on the time scale
$\xi$ (operational time). In other words, the fFT is subordinated to the
Floquet theorem:

$\Psi(t)=\sum_{n}\int_{0}^{\infty}\mathcal{K}(\xi,t)e^{-i\epsilon\xi}e^{i\omega
n\xi}C_{n}d\xi=\int_{0}^{\infty}\mathcal{K}(\xi,t)e^{-i\epsilon\xi}u(\xi).$
(3.4)

##  4 Discussion: Floquet theorem in Fourier-Laplace space

We start the discussion with the case of $\alpha=1$, when
$E_{\alpha}\left[i\hbar^{1-\alpha}(\omega
n-\epsilon)t^{\alpha}\right]=e^{i(\omega n-\epsilon)t}$. Then Eq. (2.1)
reduces to the standard Floquet theorem, which states that the solution to the
Schrödinger equation

$i\hbar\partial_{t}\psi(t)=\hat{H}\psi(t),\quad\quad\hat{H}(t+T)=\hat{H}(t)$
(4.1)

is

$\psi(t)=\sum_{n}e^{i(\omega n-\epsilon)t}C_{n}=e^{-i\epsilon t}u(t),$ (4.2)

where $u(t)$ is the periodic function, which together with the Hamiltonian can
be presented in the form of the the Fourier expansion

$\displaystyle u(t)=\sum_{n=-\infty}^{\infty}C_{n}e^{in\omega
t},\quad\quad\omega=\frac{2\pi}{T},$ (4.3a)
$\displaystyle\hat{H}(t)=\sum_{m=-\infty}^{\infty}\hat{h}(m)e^{im\omega t}.$
(4.3b)

The Laplace transform of Eqs. (4.2) and (4.3a) yields

$\mathcal{L}\left[\psi(t)\right]=\tilde{\psi}(s)=\sum_{n}C_{n}\left(s+i\epsilon-i\omega
n\right)^{-1},$ (4.4)

and the Laplace transform of the Schrödinger Eq. (4.1) yields

$i\hbar s\sum_{n}\frac{C_{n}}{s+i\epsilon-i\omega
n}=\sum_{n}\left(\sum_{m}\frac{\hat{h}(m)C_{n-m}}{s+i\epsilon-i\omega
n}+i\hbar C_{n}\right),$ (4.5)

where we take into account that the initial condition is
$\Psi_{0}=\psi_{0}=\psi(t=0)=\sum_{n}C_{n}$, according to the Floquet theorem
(4.2) and Eq. (4.3a).

Considering $C_{n}$ as a “lattice amplitude”, we have

$C_{n}=\sum_{q}U_{q}e^{iqn},$ (4.6)

and substituting the series of Eq. (4.6) in Eq. (4.5), we obtain,

$\sum_{q}\sum_{n}\left[\frac{1}{s+i\epsilon-i\omega n}\left(i\hbar
s-\sum_{m}\hat{h}(m)e^{-iqm}\right)-i\hbar\right]U_{q}e^{iqn}=0.$ (4.7)

According to Eq. (4.3b), summation with respect to $m$ yields
$\sum_{m}\hat{h}(m)e^{-iqm}=\hat{H}(-q/\omega)$. Then taking into account Eq.
(4.6) and the fact that

$\hat{H}\left(\frac{i}{\omega}\frac{d}{dn}\right)e^{iqn}=\hat{H}\left(-\frac{q}{\omega}\right)e^{iqn},$
(4.8)

we obtain Eq. (4.7) as follows

$\sum_{n}\frac{1}{s+i\epsilon-i\omega
n}\left[\hat{H}\left(\frac{i}{\omega}\frac{d}{dn}\right)C_{n}-(\hbar\epsilon-\hbar\omega
n)C_{n}\right]=0.$ (4.9)

Therefore for any values of $\epsilon$ and $n$, the coefficients $C_{n}$ are
eigenfunctions of the Hermitian operator
$\hat{H}\left(\frac{i}{\omega}\frac{d}{dn}\right)$, namely

$\hat{H}\left(\frac{i}{\omega}\frac{d}{dn}\right)C_{n}=(\hbar\epsilon-\hbar\omega
n)C_{n}.$ (4.10)

Note that the Laplace argument $s$ is not restricted as well, since performing
the Laplace inversion, the denominator in Eq. (4.9) becomes just the
exponential function, $e^{-it(\epsilon-\omega n)}$.

###  4.1 Floquet theorem for the FTSE

Let us return to the FTSE (1.3), and consider its solution in the form

$\Psi(t)=\sum_{n}\Lambda_{n}(t)C_{n}e^{i\omega nt},$ (4.11)

where $\Lambda_{n}(t)$ is an unknown function of time. The Laplace
transformations of both the FTSE (1.3) and the solution (4.11) are

$\displaystyle
i\hbar^{\alpha}s^{\alpha}\tilde{\Psi}(s)=\sum_{m}\hat{h}(m)\sum_{n}C_{n}\mathcal{L}\left[\Lambda_{n}(t)e^{it\omega(n+m)}\right]+i\hbar^{\alpha}s^{\alpha-1}\Psi_{0}$
$\displaystyle=\sum_{m}\sum_{n}\hat{h}(m)C_{n}\tilde{\Lambda}(s-i\omega
n-i\omega m)+i\hbar^{\alpha}s^{\alpha-1}\Psi_{0},$ (4.12a)
$\displaystyle\tilde{\Psi}(s)=\sum_{n}\tilde{\Lambda}(s-i\omega
n)C_{n},\quad\quad\Psi_{0}=\Lambda(0)\sum_{n}C_{n}.$ (4.12b)

Performing the shift of indexes $\sum_{n}\tilde{\Lambda}(s-i\omega n-i\omega
m)C_{n}=\sum_{n}\tilde{\Lambda}(s-i\omega n)C_{n-m}$, we obtain the equation
for $\tilde{\Lambda}(s-i\omega n)$ from Eqs. (4.12a) and (4.12b) as follows

$\sum_{n}\left[\tilde{\Lambda}(s-i\omega
n)\left(i\hbar^{\alpha}s^{\alpha}C_{n}-\sum_{m}\hat{h}(m)C_{n-m}\right)-i\hbar^{\alpha}s^{\alpha-1}\Lambda(0)C_{n}\right]=0.$
(4.13)

One should bear in mind that the coefficients $C_{n}$ and the Hamiltonian
$\hat{H}(t)$ are the same as in the standard Floquet theory. Therefore,
performing the expansion (4.6) for the lattice amplitudes $C_{n}$ and then
performing summation with respect to $m$, we obtain
$\sum_{m}\hat{h}(m)e^{-iqm}=\hat{H}(-q/\omega)$. Then taking into account Eq.
(4.8) and the eigenvalue Eq. (4.10), we obtain Eq. (4.13) as follows

$\sum_{n}\sum_{q}\left\\{\tilde{\Lambda}(s-i\omega
n)\left[i\hbar^{\alpha}s^{\alpha}-\hat{H}(-q/\omega)\right]-i\hbar^{\alpha}s^{\alpha-1}\Lambda(0)\right\\}U_{q}e^{iqn}\\\
=\sum_{n}\left\\{\tilde{\Lambda}(s-i\omega
n)\left[i\hbar^{\alpha}s^{\alpha}-\hat{H}\left(\frac{i}{\omega}\frac{d}{dn}\right)\right]-i\hbar^{\alpha}s^{\alpha-1}\Lambda(0)\right\\}C_{n}\\\
=\sum_{n}\left\\{\tilde{\Lambda}(s-i\omega
n)\left[i\hbar^{\alpha}s^{\alpha}-\hbar(\epsilon-\omega
n)\right]-i\hbar^{\alpha}s^{\alpha-1}\Lambda(0)\right\\}C_{n}=0.$ (4.14)

Since the coefficients $C_{n}$ are the functions in question in the Floquet
theory, it is reasonable to suppose that the Laplace image is

$\tilde{\Lambda}(s-i\omega
n)=\frac{s^{\alpha-1}\Lambda(0)}{s^{\alpha}+i\hbar^{1-\alpha}(\epsilon-\omega
n)},$ (4.15)

which is the Laplace image of the Mittag-Leffler function [31]. Performing the
Laplace inversion and setting $\Lambda(0)=1$, we obtain the solution in the
form of the one parameter Mittag-Leffler function,

$\Lambda_{n}(t)e^{i\omega nt}=E_{\alpha}\left[i\hbar^{1-\alpha}(\omega
n-\epsilon)t^{\alpha}\right].$ (4.16)

Taking into account the definition (4.11), we arrived at the wave function as
follows

$\Psi(t)=\sum_{n}\Lambda_{n}(t)e^{i\omega
nt}C_{n}=\sum_{n}E_{\alpha}\left[i\hbar^{1-\alpha}(\omega
n-\epsilon)t^{\alpha}\right]C_{n}.$ (4.17)

We stress that it has been proven in Sec. 2 that the wave function in the form
of (4.17) is the fFT.

###  4.2 Comments

  * 1

Evolution operator

The evolution of the wave function of the FTSE (1.3) is according to the
evolution operator $\hat{U}(t)$ (do not confuse it with the band amplitude
$U_{q}$), such that $\Psi(t)=\hat{U}(t)\Psi_{0}$. However, the explicit
operator form of the evolution operator is unknown, since the fractional
Caputo derivative destroys the periodicity of the wave function, see A.
Therefore, the Fourier expansion in Eqs. (4.3a) and (4.11) is formal, such
that the coefficient $C_{n}$ can be both an operator valued function and a
function of coordinate space, depending on the explicit operator form of the
Hamiltonian $\hat{H}(t)$. The former case is considered as an example in Sec.
5, while the latter case is an example of Sec. 6.

  * 2

Lattice amplitude

Important part of the analysis is relates to the consideration of the
amplitude $C_{n}$ as a “lattice amplitude” in the form of the expansion (4.6),
namely $C_{n}=\sum_{q}U_{q}e^{iqn}$. Then taking into account the expansion
(4.3b) and performing summation with respect to $m$, one obtains that the time
dependence in the Hamiltonian can be replaced as follows
$\sum_{m}\hat{h}(m)e^{-iqm}=\hat{H}(-q/\omega)$. The latter leads to
expression (4.8), which is

$\hat{H}\left(\frac{i}{\omega}\frac{d}{dn}\right)e^{iqn}=\hat{H}\left(-\frac{q}{\omega}\right)e^{iqn}.$

Therefore, taking into account this chain of transformations (that always can
be done), one can made the formal replacement

$\hat{H}(t)C_{n}\equiv\hat{H}(\omega
t)C_{n}\rightarrow\hat{H}\left(i\frac{d}{dn}\right)C_{n}.$

  * 3

Mittag-Leffler function

Note, that the Mittag-Leffler function is the eigenfunction of the Caputo
fractional derivative. Indeed, taking into account Eq. (A.7), we obtain

$\partial_{t}^{\alpha}E_{\alpha}(iat^{\alpha})=\sum_{k=0}^{\infty}\frac{(ia)^{k}}{\Gamma(k\alpha+1)}\partial_{t}^{\alpha}t^{k\alpha}\\\
=\sum_{k=1}^{\infty}\frac{(ia)^{k}}{\Gamma(k\alpha+1)}\cdot\frac{\Gamma(k\alpha+1)}{\Gamma(k\alpha-\alpha+1)}t^{(k-1)\alpha}\\\
=ia\sum_{k=0}^{\infty}\frac{(ia)^{k}t^{k\alpha}}{\Gamma(k\alpha+1)}=iaE_{\alpha}(iat^{\alpha}),$

where $a=\hbar^{1-\alpha}(\omega n-\epsilon)$.

##  5 Example 1: Toy model

In this section, we consider a toy Hamiltonian presented in the form

$\hat{H}(t)=\hat{H}_{0}\cos(\omega t),$ (5.1)

where $\hat{H}_{0}$ can be an arbitrary time-independent Hamiltonian, which
can describe a variety of conservative systems. The standard Schrödinger
equation for the toy model (5.1) can be easily solved, and the evolution of
the initial wave function $\psi_{0}$ is

$\psi(t)=\exp\left(-\frac{i}{\hbar}\int_{0}^{t}\hat{H}(t^{\prime})dt^{\prime}\right)\psi_{0}=\exp\left[-i\frac{\hat{H}_{0}}{\hbar\omega}\sin(\omega
t)\right]\psi_{0}.$ (5.2)

The initial wave function $\psi_{0}$ can be e.g., the eigenfunction of
$\hat{H_{0}}$, namely $\hat{H}_{0}\psi_{0}=e\psi_{0}$. Then
$\psi(t)=\exp[-i(e/\hbar\omega)\sin(\omega t)]$. Since for $t=kT$ ($k$ is
integer), $\psi(kT)=\psi_{0}$, correspondingly the quasienergy is zero,
$\epsilon=0$ and $\psi(t)=u(t)$. The Fourier transformation yields

$\psi(t)=\sum_{n=-\infty}^{\infty}J_{-n}\left(\frac{\hat{H}_{0}}{\hbar\omega}\right)e^{i\omega
nt},$ (5.3)

where $C_{n}=(-1)^{n}J_{n}(z)\equiv
J_{-n}\left(\frac{\hat{H}_{0}}{\hbar\omega}\right)$ is the Bessel function of
the first kind, which also satisfies the eigenvalue Eq. (4.10). The latter
reads

$z\cos\left(\frac{d}{dn}\right)J_{-n}(z)=-nJ_{-n}(z),$ (5.4)

which is just the recurrence relation of the Bessel functions [35]:

$zJ_{n+1}(z)+zJ_{n-1}(z)=2nJ_{n}(z).$

###  5.1 Fractional toy model

The fractional generalization of the toy model is formulated in the framework
of the FTSE as follows

$i\hbar^{\alpha}\partial_{t}^{\alpha}\Psi(t)=\hat{H}_{0}\cos(\omega t)\Psi(t)$
(5.5)

where the initial condition is $\Psi_{0}=\psi_{0}$. According to the fFT, we
look for the solution to the FTSE (5.5) as follows

$\Psi(t)=\sum_{n}E_{\alpha}\left(i\hbar^{1-\alpha}\omega
nt^{\alpha}\right)(-1)^{n}J_{n}\left(\frac{\hat{H}_{0}}{\hbar\omega}\right)\psi_{0},$
(5.6)

which for $\alpha=1$ reduces to Eq. (5.2).

Let us substitute this fFT solution into the FTSE (5.4). Taking into account
the Comment 3-3, we obtain

$\sum_{n}(-\hbar\omega n)E_{\alpha}\left(i\hbar^{1-\alpha}\omega
nt^{\alpha}\right)(-1)^{n}J_{n}\left(\frac{\hat{H}_{0}}{\hbar\omega}\right)\psi_{0}\\\
=\hat{H}_{0}\cos(\omega t)\sum_{n}E_{\alpha}\left(i\hbar^{1-\alpha}\omega
nt^{\alpha}\right)(-1)^{n}J_{n}\left(\frac{\hat{H}_{0}}{\hbar\omega}\right)\psi_{0}.$
(5.7)

Performing the chain of transformations according to the Comment 2-2, we
obtain the rhs of Eq. (5.7) as follows

$\hat{H}_{0}\sum_{n}E_{\alpha}\left(i\hbar^{1-\alpha}\omega
nt^{\alpha}\right)\cdot\left[\cos\left(\frac{d}{dn}\right)(-1)^{n}J_{n}\left(\frac{\hat{H}_{0}}{\hbar\omega}\right)\right]\psi_{0}.$

Then from Eq. (5.4), we get the identity for Eq. (5.7).

##  6 Example 2: Fractional quantum dynamics of a particle in time-dependent
potential

In this section, we consider another, more realistic example, where the
coefficients $C_{n}=C_{n}(x)$ are defined in the coordinate space $x\in R$. It
is a quantum particle in a nonlinear time-dependent potential $V(x)\cos(\omega
t)$, which can be described e.g., by the Schrödinger equation

$i\hbar\partial_{t}\psi(x,t)=\hat{H}(t)\psi(x,t)\equiv\left[-\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}}+V(x)\cos(\omega
t)\right]\psi(x,t),$ (6.1)

where $m$ is the mass of a quantum particle affected by the time-dependent
potential $V(x)\cos(\omega t)$, where $V(x)$ is a nonlinear function of $x$.
Such systems can describe a quantum nonlinear resonance [28, 29]. We
concentrate our attention on the Floquet theory of the Schrödinger Eq. (6.1)
and shall not discuss the physics of the quantum resonance, which is well
studied and well reviewed in literature [36, 37, 38]. Therefore, according to
the Floquet theorem, the eigenvalue problem for the Floquet operator reads

$\hat{H}(t)u(x,t)-i\hbar\partial_{t}u(x,t)=\hbar\epsilon u(x,t),$ (6.2)

where $u(x,t)=\sum_{n}C_{n}(x)e^{i\omega nt}$. This yields the equation for
the coefficients $C_{n}(x)$ as follows

$\hat{K}C_{n}(x)\equiv\\\
-\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}}C_{n}(x)+\frac{1}{2}V(x)\left[C_{n+1}(x)+C_{n-1}(x)\right]+\hbar(\omega
n-\epsilon)C_{n}(x)=0.$ (6.3)

After diagonalization, when $C_{n}(x)=\sum_{q}U_{q}(x)e^{iqn}$, Eq. (6.3)
reads

$-\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}}U_{q}(x)+V(x)\cos(qn)U_{q}(x)+\hbar\omega
nU_{q}(x)=\hbar\epsilon U_{q}(x).$ (6.4)

Depending on the explicit form of the potential $V(x)$, this equation can be
treated either analytically, or numerically [36, 37, 38]. We however, do not
concern with the issue, as admitted above.

Equation (6.3) is the skeleton equation for the consideration of the
fractional quantum nonlinear resonance in the framework of the FTSE (6.1),
which reads now

$i\hbar\partial_{t}^{\alpha}\Psi(x,t)=\hat{H}(t)\Psi(x,t)\equiv\left[-\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}}+V(x)\cos(\omega
t)\right]\Psi(x,t),$ (6.5)

To obtain Eq. (6.3), let us substitute expression (2.1) into FTSE (6.5).
Following the Comment 3-2, we obtain

$\sum_{n}\hbar(\epsilon-\omega
n)E_{\alpha}\left[i\hbar^{1-\alpha}(\epsilon-\omega
n)t^{\alpha}\right]C_{n}(x)\\\
=\sum_{n}E_{\alpha}\left[i\hbar^{1-\alpha}(\epsilon-\omega
n)t^{\alpha}\right]\left[-\frac{\hbar^{2}}{2m}\frac{d^{2}C_{n}(x)}{dx^{2}}+V(x)\cos(\omega
t)C_{n}\right].$ (6.6)

Following the Comment 2, we have $\cos(\omega
t)C_{n}\rightarrow\frac{1}{2}(C_{n+1}+C_{n-1})$. Therefore, in the Laplace
space, the real and imaginary parts of Eq. (6.6) are

$\displaystyle\sum_{n}\frac{s^{2\alpha-1}}{s^{2\alpha}+\hbar^{2-2\alpha}(\epsilon-\omega
n)^{2}}\left[\hat{K}C_{n}(x)\right]=0,$ (6.7a) $\displaystyle
i\hbar^{1-\alpha}\sum_{n}\frac{(\epsilon-\omega
n)}{s^{2\alpha}+\hbar^{2-2\alpha}(\epsilon-\omega
n)^{2}}\left[\hat{K}C_{n}(x)\right]=0.$ (6.7b)

Then we arrive at Eq. (6.3): $\hat{K}C_{n}(x)=0$.

##  7 Conclusion

In the paper we concern with the fractional Schrödinger equations (FTSE) for
systems, which are described by the time-dependent periodic Hamiltonians. An
analog of the well known Floquet theorem is suggested to solve this class of
the FTSEs, and this new result is called the fractional Floquet theorem (fFT).
The main result can be formulated in the form of the following generalization
of the Floquet theorem for the wave function. Namely, considering the standard
Floquet theorem as follows

$\psi(t)=e^{-i\epsilon t}u(t)=\sum_{n}C_{n}e^{-i\epsilon t+i\omega nt},$ (7.1)

we generalize the exponential by the one-parameter Mittag-Leffler function
$E_{\alpha}\left[i\hbar^{1-\alpha}(\omega n-\epsilon)t^{\alpha}\right]$, which
yields the fFT (2.1) as follows

$\Psi(t)=\sum_{n}C_{n}E_{\alpha}\left[i\hbar^{1-\alpha}(\omega
n-\epsilon)t^{\alpha}\right],$ (7.2)

which for $\alpha=1$ reduces to the Floquet theorem (7.1). The fFT has been
verified by the straightforward substitution of Eq. (2.1) in the FTSE that
yields the identity. Two examples supporting the obtained result have been
considered as well.

In conclusion, it should be admitted that by means of the Mittag-Leffler
function the fFT reduces the FTSE (1.3) with the time dependent Hamiltonian to
the standard Floquet theorem consideration. In this case, the quasienergy
spectrum of the FTSE exists and is determined by the standard Floquet theorem.

## Acknowledgments

It is my pleasure to acknowledge the hospitality at the Max Planck Institute
for the Physics of Complex Systems, Dresden, where a part of the work has been
done.

##  Appendix A Fractional differentiation of periodic functions

In this notes, we consider fractional calculus as an example of fractional
differentiation of periodic functions. Recently, this issue has attracted some
attention in the form of non-existence of periodic solutions in fractional-
order dynamical systems [39]. It has been shown in the framework of the Mellin
transform consideration that fractional differentiation of periodic functions
destroys their periodicity. The result was obtained for the Caputo, Riemann-
Liouville and Grünwald-Letnikov definitions of fractional-order derivatives.
We suggest an alternative approach, following the fundamental work of Ref.
[26], and the Fourier and the Laplace transformations are the main machinery
of the analysis.

Any periodic function $g(t)$ of the period $T$, such that $g(t+T)=g(t)$, can
be presented in the form of its Fourier series

$g(t)=\sum_{l=-\infty}^{\infty}g_{l}e^{i\bar{l}t}\,,\quad
g_{l}=\frac{1}{T}\int_{0}^{T}g(t)e^{-i\bar{l}t}dt\,,\quad\bar{l}=2\pi l/T\,.$
(A.1)

Therefore, $g(t)$ is determined by its Fourier image $g_{l}$ if the Fourier
series (A.1) converges. In particular, the $n$-th derivative is a periodic
function, determined by $\bar{l}^{n}g_{l}$ as the convergence of
$g^{(n)}(t)=\tfrac{d^{n}g(t)}{dt^{n}}$ in Eq. (A.1). Therefore, to define
periodicity or non-periodicity of the fractional integro-differentiation of
$g(t)$ one considers the exponential $\exp[2\pi ilt/T]$ only.

###  A.1 Riemann-Liouville fractional integral

Note that the Caputo and Riemann-Liouville fractional derivatives are
regularization of the fractional derivative

$\fourIdx{}{a+}{\mu}{t}{D}f(t)=\frac{1}{\Gamma(\mu)}\int_{a}^{t}(t-\tau)^{-\mu-1}f(\tau)\,d\tau,$
(A.2)

where we take $a=0$. Therefore, taking $n-1<\mu<n$ and $\nu=n-\mu$, and
dropping out the integer part of the derivative from the consideration, we
arrive at the Riemann-Liouville fractional integration of the exponential
$f(t)=e^{i\bar{l}\tau}$,

$\fourIdx{}{0}{-\nu}{t}{D}f(t)\equiv
I_{0+}^{\nu}f(t)=\frac{1}{\Gamma(\nu)}\int_{0}^{t}(t-\tau)^{\nu-1}e^{i\bar{l}\tau}\,d\tau.$
(A.3)

To treat this integral, we use the Laplace transform of the convolution
integral, which we present it in the form of the identity

$I_{0+}^{\nu}e^{i\bar{l}t}=\mathcal{L}^{-1}\left\\{\mathcal{L}\left[I_{0+}^{\nu}e^{i\bar{l}t}\right]\right\\}=\frac{1}{2\pi
i}\int_{C}\frac{s^{\nu}e^{st}}{s-i\bar{l}}ds=t^{-\nu}E_{1,1-\nu}\left(i\bar{l}t\right)\,,$
(A.4)

where the last line is the integral representation of the two parameter
Mittag-Leffler function $E_{\alpha,\beta}(z)$, [31]. Taking into account that
$1-\nu=\\{\mu\\}$ is the fractional part of $\mu$, and presenting the Mittag-
Leffler function as a superposition of the real and imaginary parts, we obtain

$E_{1,\\{\mu\\}}\left(i\bar{l}t\right)=\sum_{k=0}^{\infty}\left[\frac{\left(i\bar{l}t\right)^{2k}}{\Gamma(2k+\\{\mu\\})}+\frac{\left(i\bar{l}t\right)^{2k+1}}{\Gamma(2k+1+\\{\mu\\})}\right]\\\
=E_{2,\\{\mu\\}}\left(-\bar{l}^{2}t^{2}\right)+i\bar{l}tE_{2,1+\\{\mu\\}}\left(-\bar{l}^{2}t^{2}\right)\,.$
(A.5)

Eventually, we show that the Riemann-Liouville fractional integration destroys
the periodicity of periodic functions. Therefore, both the Caputo and Riemann-
Liouville fractional derivatives act in the same way.

###  A.2 Fox $H$-function

Let us present exponential in the form of the Fox $H$-function,

$e^{-z}=H_{0,1}^{1,0}\left[z\Big{|}\binom{}{(0,1)}\right].$

Thus the exponential can be presented in the form of the Mellin-Barnes
integral [40]

$e^{i\bar{l}t}=\frac{1}{2\pi i}\int_{C}\Gamma(\xi)(-i\bar{l}t)^{-\xi}d\xi$
(A.6)

Making the variable change $\xi\rightarrow-\xi$, we obtain the singularities
of the gamma functions $\Gamma(-\xi)$ at $\Re{\xi}\geq 0$. (In this case, Eq.
(A.6) corresponds to the Meijer $G$-function [31].) Now let us consider the
Caputo fractional derivative $\partial^{\\{\mu\\}}_{t}$ of the power law
function $t^{\xi}$, which yields

$\partial^{\\{\mu\\}}_{t}(-i\bar{l}t)^{\xi}=(-i\bar{l})^{\xi}\frac{\Gamma(\xi+1)}{\Gamma(1-\\{\mu\\})}\int_{0}^{t}\tau^{\xi-1}(t-\tau)^{-\\{\mu\\}}d\tau\\\
=(-i\bar{l})^{\xi}\frac{\Gamma(\xi+1)}{\Gamma(\xi-\\{\mu\\}+1)}t^{\xi-\\{\mu\\}}\,,$
(A.7)

where $\\{\mu\\}=\mu+1-n$. Combining this result with Eq. (A.6), then
performing the variable change $\xi\rightarrow-\xi$ with corresponding
deformation of the countour, we obtain the Mellin-Barnes integral for the
Mittag-Leffler function

$\partial^{\\{\mu\\}}_{t}e^{i\bar{l}t}=\frac{t^{-\\{\mu\\}}}{2\pi
i}\int_{C}\frac{\Gamma(\xi)\Gamma(1-\xi)}{\Gamma(1-\mu-\xi)}(-i\bar{l}t)^{-\xi}d\xi=t^{-\\{\mu\\}}E_{1,1-\\{\mu\\}}\left(i\bar{l}t\right)$
(A.8)

###  A.3 Riesz fractional derivatives

In this section we treat the periodic function $g(x)=g(x+L)$, where $x\in R$
in the framework of the symmetric Riesz-Feller integration. Let us consider
the symmetric Riesz-Feller derivative, which can be defined in Fourier space

$\mathcal{F}\left[\fourIdx{\textrm{RF}}{0}{\mu}{x}{D}f(x)\right](k)=-|k|^{\mu}\mathcal{F}\left[f(x)\right](k).$
(A.9)

Its integral representation reads

$\fourIdx{\textrm{RF}}{0}{\mu}{x}{D}f(x)=\frac{\Gamma(1+\mu)}{\pi}\sin\frac{\mu\pi}{2}\int_{0+}^{\infty}\frac{f(x+\xi)-2f(x)+f(x-\xi)}{\xi^{1+\mu}}d\xi\\\
=\frac{\mu}{2\Gamma(1-\mu)\cos\frac{\mu\pi}{2}}\int_{0}^{\infty}\frac{f(x+\xi)-2f(x)+f(x-\xi)}{\xi^{1+\mu}}d\xi$
(A.10)

Setting $f(x)=e^{i\bar{l}x}$ with $\bar{l}=2\pi l/L$, integration in Eq.
(A.10) yields

$\fourIdx{\textrm{RF}}{0}{\mu}{x}{D}f(x)=\frac{\mu
e^{i\bar{l}x}}{2\Gamma(1-\mu)\cos\frac{\mu\pi}{2}}\int_{0}^{\infty}\frac{\sin^{2}(\bar{l}\xi)}{\xi^{1+\mu}}d\xi\\\
=-\frac{2e^{i\bar{l}x}}{\Gamma(1-\mu)\cos\frac{\mu\pi}{2}}\int_{0}^{\infty}\sin^{2}(\bar{l}\xi)d(\xi^{-\mu})=A(\mu)e^{i\bar{l}x},\quad
0<\mu<2\,,$ (A.11)

where $A(\mu)=(2\bar{l})^{\mu}$. Therefore, the symmetric Riesz-Feller
differentiation of periodic functions does not destroy the periodicity, namely
$\fourIdx{\textrm{RF}}{0}{\mu}{x}{D}g(x)$ is the periodic function with the
same period $L$ of $g(x)$ for $0<\mu<2$.

###  A.4 Riesz fractional derivative in Fourier space

Let us consider left $(+)$ and right $(-)$ fractional integrals

$\displaystyle(I_{+}^{\mu}f)(x)=\frac{1}{\Gamma(\mu)}\int_{-\infty}^{x}(x-y)^{\mu-1}f(y)dy,$
(A.12a)
$\displaystyle(I_{-}^{\mu}f)(x)=\frac{1}{\Gamma(\mu)}\int^{\infty}_{x}(y-x)^{\mu-1}f(y)dy.$
(A.12b)

Correspondingly the R-L fractional derivatives are [26](§ 5.1)

$\displaystyle(D_{+}^{\mu}f)(x)=\frac{1}{\Gamma(n-\mu)}\frac{d^{n}}{dx^{n}}\int_{-\infty}^{x}(x-y)^{n-\mu-1}f(y)dy,$
(A.13a)
$\displaystyle(D_{-}^{\mu}f)(x)=\frac{(-1)^{n}}{\Gamma(n-\mu)}\frac{d^{n}}{dx^{n}}\int^{\infty}_{x}(y-x)^{n-\mu-1}f(y)dy,$
(A.13b)

where $n-1<\mu<n$. Then the Fourier transform of Eq. (A.13) reads [26](§ 7.1)

$\mathcal{F}\left[(D_{\pm}^{\mu}f)(x)\right](k)=(\mp ik)^{\mu}\tilde{f}(k)\,.$
(A.14)

Introducing the Riesz fractional derivative, we consider superposition of the
Fourier images in Eq. (A.14), which is

$\mathcal{F}\left[(D_{-}^{\mu}f)(x)\right](k)+\mathcal{F}\left[(D_{+}^{\mu}f)(x)\right](k)\\\
=\left[(ik)^{\mu}+(-ik)^{\mu}\right]\tilde{f}(k)=2|k|^{\mu}\cos\frac{\pi\mu}{2}\tilde{f}(k)\,.$
(A.15)

Applying this result to the periodic function $f(x)=e^{i\bar{l}x}$, we obtain
$\tilde{f}(k)=2\pi\delta(k-\bar{l})$ that ensures the fractional operator
keeps the periodicity of $g(x)$, which is immediate by the Fourier inversion.

By performing the Fourier inversion of Eq. (A.15), the Riesz fractional
derivative $(-\Delta)^{\frac{\mu}{2}}$ can be defined. In the one dimensional
case it is

$(-\partial^{2}_{x})^{\frac{\mu}{2}}f(x)\equiv\partial^{\mu}_{|x|}f(x)=\frac{1}{2\cos\frac{\pi\mu}{2}}\left[D_{+}^{\mu}+D_{-}^{\mu}\right]\\\
=\int_{-\infty}^{\infty}dyf(y)\frac{1}{2\pi}\int_{-\infty}^{\infty}|k|^{\mu}e^{-ik(x-y)}dk\,,$
(A.16)

where $\mathcal{F}^{-1}[|k|^{\mu}](x-y)$ is the kernel. Let us obtain it for
$\mu\in(0\,,2)$. Thus we have

$\frac{1}{2\pi}\int_{-\infty}^{\infty}|k|^{\mu}e^{-ik(x-y)}dk=\frac{1}{\pi}\int_{0}^{\infty}k^{\mu}\cos[k(x-y)]dk\\\
=\frac{1}{\pi}\left(-\frac{d^{2}}{dx^{2}}\right)\int_{0}^{\infty}k^{\mu-2}\cos[k(x-y)]dk\\\
=\left(-\frac{d^{2}}{dx^{2}}\right)\frac{(x-y)^{1-\mu}}{2\Gamma(2-\mu)\cos\left[\frac{(2-\mu)\pi}{2}\right]}=\frac{(x-y)^{-\mu-1}}{2\Gamma(-\mu)\cos\frac{\mu\pi}{2}}\,.$
(A.17)

Eventually, we obtain

$\partial^{\mu}_{|x|}f(x)=\frac{1}{2\Gamma(-\mu)\cos\frac{\mu\pi}{2}}\int_{-\infty}^{\infty}|x-y|^{-\mu-1}f(y)dy\\\
=\frac{1}{2\Gamma(-\mu)\cos\frac{\mu\pi}{2}}\int_{-\infty}^{\infty}f(x-y)|y|^{-\mu-1}\,,$
(A.18)

which for $\mu>0$ diverges and demands regularisation. As shown here, one of
the regularization for $\mu\in(0\,,2)$ is

$\partial^{\mu}_{|x|}f(x)=\frac{1}{2\Gamma(-\mu)\cos\frac{\mu\pi}{2}}\int_{0+}^{\infty}y^{-\mu-1}\left[f(x+y)-2f(x)+f(x-y)\right]dy\,.$
(A.19)

Taking into account that $\Gamma(1-\mu)=-\mu\Gamma(-\mu)$, Eq. (A.19) can be
rewritten in the form of the Grünwald-Letnikov-Riesz fractional derivative
[26]

$f^{\mu)}(x)=\frac{\mu}{2\Gamma(1-\mu)\cos\frac{\mu\pi}{2}}\int_{0+}^{\infty}\frac{2f(x)-f(x+y)-f(x-y)}{y^{\mu+1}}dy\,.$
(A.20)

###  A.5 Grünwald-Letnikov derivative

The Grünwald-Letnikov fractional derivative relates to the differences of a
fractional order $\Delta_{h}^{\mu}f(x)$ of a function $f(x)$ defined on the
$x$ axis: $x\in R$. It can be presented as a finite difference with a step $h$
of order $\mu$ of the function $f(x)$

$\Delta_{h}^{\mu}f(x)=\sum_{k=0}^{\infty}(-1)^{k}\binom{\mu}{k}f(x-kh)\,,\quad\mu>0\,.$
(A.21)

Then the function

$f^{\mu)}_{\pm}(x)=\lim_{h\to+0}\frac{(\Delta_{\pm
h}f)(x)}{h^{\mu}}\,,\quad\mu>0$ (A.22)

is referred as the Grünwald-Letnikov fractional derivative [26]. It has been
rigorously proven that the expression (A.22) coincides with the Marchaud
derivative with the same domain of definition [26](§ 20.1), and it reads

$f^{\mu)}_{\pm}(x)=D_{\pm}^{\mu}f(x)=\lim_{\epsilon\to
0}\frac{\mu}{\Gamma(1-\mu)}\int_{\epsilon}^{\infty}\frac{f(x)-f(x\mp
y)}{y^{\mu+1}}dy\,,\quad 0<\mu<1\,,$ (A.23)

and its symmetric form is

$f^{\mu)}(x)=\frac{1}{2\cos\frac{\mu\pi}{2}}\lim_{h\to
0}\frac{(\Delta_{+h}f)(x)+\Delta_{-h}f)(x)}{h^{\mu}}\\\
=\frac{\mu}{2\Gamma(1-\mu)\cos\frac{\mu\pi}{2}}\int_{0+}^{\infty}\frac{2f(x)-f(x+y)-f(x-y)}{y^{\mu+1}}dy\,,$
(A.24)

which coincides with Eq. (A.20). As it follows from Eqs. (A.10) and (A.11),
the fractional operator (A.24) keeps $g(x)$ being periodic.

It is worth noting that the Marchaud derivatives $f^{\mu)}_{\pm}(x)$ in Eq.
(A.23) are in the regularization form for $0<\mu<1$, while their symmetric
combination in Eq. (A.24) is regularized for $0<\mu<2$. Both the form do not
destroy the periodicity of the exponential $e^{i\bar{l}x}$.

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