where $A_{i_-,i_+}$ is the identity, if $i_-=i_+$ and a product of some operators $Q$, $\mathds{1}_{I_1}$, and $\mathds{1}_{I_2}$ otherwise. As all of the operators are projections, their operator norm can be bounded by $1$. As we are interested in the trace, we will bound the trace norm. To do so, it suffices to bound two operators in the Hilbert--Schmidt norm and all others in the operator norm. The operators we will bound in Hilbert--Schmidt norm are $\mathds{1}_{I_{j_0}} Q \mathds{1}_{I_{j_{i_-}}}$ and $ \mathds{1}_{I_{j_{i_+}}}Q \mathds{1}_{I_{j_0}}$. These operators are adjoint and hence have the same Hilbert--Schmidt norm. As $j_{i_-} \neq j_0 \neq j_{i_+}$, we know $\{j_0,j_{i_-}\}=\{j_0,j_{i_+}\}=\{1,2\}$. Thus, we are left to estimate $\lVert \mathds{1}_{I_1} Q \mathds{1}_{I_2} \rVert_2^2$. Since the operator $Q$ has integral kernel $Q(x,y)= k_\mu(x-y) = \frac{\sin(\sqrt{\mu}(x-y))}{\pi (x-y)}, x,y\in\R$, the square of the Hilbert--Schmidt norm can be easily calculated as the square of the integral of this kernel for $x \in I_1$ and $y \in I_2$. By translation invariance we may assume that $I_1=(0,\ell_1)$. By the definition of $d_1$, we know $I_2 \subset (\ell_1+d_1, \infty)$. Hence, using the estimate $|k_\mu(z)|\le C/(1+|z|)$ for some constant $C$ we get
\begin{align}
\big|\tr \eqref{C.3}\big|\, \le  \, \lVert \mathds{1}_{I_1} Q \mathds{1}_{I_2} \rVert_2^2 &\le C \int_{I_1} \mathrm{d}x\, \int_{I_2} \mathrm{d}y\, \frac 1 {(1+ y-x)^2} \\
& \le C \int_{0}^{\ell_1} \mathrm{d}x\, \int_{\ell_1+d_1}^\infty \mathrm{d}y\, \frac1{(1+ y-x)^2}\\
&= C \int_{0}^{\ell_1} \mathrm{d}x\,\frac 1 {1+d_1+\ell_1-x} = C \ln \left( \frac{ 1+ d_1+\ell_1}{ 1+d_1} \right)\,.
\end{align}
The number of such error terms is $2^m-1$. Thus, the error bound in $m$ is quite bad, but we only need to be good in $k$. The proof is now finished.
\end{proof}

Here is our second lemma on the mentioned improved asymptotic expansion for a single interval of Landau and Widom. This agrees with the improvement of Widom in [48]. As the paper of Landau and Widom [25] is freely accessible, but the later paper by Widom [48] is not\footnote{as of August 25, 2022}, we provide this different proof for the reader's convenience. We do not claim any originality.

\begin{lemma}\label{improved Landau-Widom}
Let $\Omega \subset \mathbb R$ be an open and bounded interval of length $\ell>0$ and let $\mu>0$. Then for any $m\in\N$ and $L >0$, we have with $\mathsf{I}(m)$ explained after \eqref{def: I},
\begin{align}\label{C.7}
\tr \big( \mathds{1}_{L\Omega} \mathds{1}[(-\mathrm{i}\nabla^\parallel)^2 \le \mu] \mathds{1}_{L\Omega}\big)^m &= \frac{\sqrt \mu}\pi L\ell +4 \,\mathsf{I}(m)\ln(1+L\ell) + O(1)\,,
\end{align}
where the order one error term is independent of $L$ and $\ell$ but depends on $\mu$.
\end{lemma}

\begin{proof}
The case $m=1$ is trivial, as the integral kernel is constant on the diagonal and only the volume term $\frac{\sqrt \mu}\pi L\ell$ appears. Thus, by linearity, it suffices to show the statement for a basis of the polynomials vanishing at $0$ and $1$.

As $\mu$ is fixed, the result depends only on $L\ell$, which can be small or large. If $L\ell\le1$ then the trace on the left-hand side of \eqref{C.7} is bounded uniformly for these $L,\ell$ by continuity as a function of $L\ell\in[0,1]$. The same is true for the first two terms on the right-hand side of \eqref{C.7} and hence the equality holds true with an $O(1)$ error term. In the following we will assume that $L\ell>1$.

Form now on we use the same notation as in [25], where $c$ takes the role of $L$. The last equation in the proof of their Theorem 1, where they still carry the order one error term is \cite[(18)]{LW}. Afterwards they allow for a larger $o(\ln(c))$ error term and here we take a different route.

They consider the polynomials $(t(1-t))^n$ and $t(t(1-t))^n$ for $n \in \N$, which span all polynomials that vanish at $0$ and $1$. We proceed with these polynomials instead of $t^m$ as in the statement of our lemma. Their equation \cite[(18)]{LW} states
\be
\tr A_c [A_c(I-A_c)]^n =2 \,\tr K_c^n + O(1) = \frac 12 \tr [A_c(I- A_c)]^n + O(1)\,, \label{LW18}
\ee
where $A_c= P(0,c) Q(0,1)P(0,c)$, which is unitarily equivalent to $\mathds{1}_{[0,c]} \mathds{1}\big(-\Delta \le \frac 1 4 \big) \mathds{1}_{[0,c]}$ in our notation, and $K_c=P(1,c) Q(-\infty,0) P(-\infty, 0) Q( -\infty,0) P(1,c)$, as stated below \cite[(17)]{LW}. They state below \cite[(18)]{LW} that the integral kernel of the operator $K_c$ on $\Lp^2([1,c])$ is given for $1 \le x,y \le c$ by
\begin{align}
f(x,y) \coloneqq \frac 1 {4 \pi^2} \int_0^\infty \frac {\mathrm{d}u} {(u+x)(u+y)} = \frac 1 {4\pi^2}\begin{cases} \frac { \ln(x)- \ln(y)} {x-y} & \mbox{ if }x \neq y\\ \frac 1 x & \mbox{ if } x=y
\end{cases}\,.
\end{align}
Let $K$ be the operator on $\Lp^2(\mathbb R^+)$ with integral kernel $f(x,y)$ for $0 <  x,y < \infty$. Thus, $K_c=P(1,c) K P(1,c)$ and $K= Q(-\infty,0) P(-\infty, 0) Q( -\infty,0)$.
Hence, we can conclude
\be
\tr K_c^n = \int_{[1,c]^n} \mathrm{d}x\, \prod_{i=1}^n f_1(x_i, x_{i+1})\,,
\ee
with the convention $x_{n+1}=x_1$. We denote the integrand $f_n(x) \coloneqq \prod_{i=1}^n f(x_i, x_{i+1})$. It satisfies for $\lambda>0$ the homogeneity property $f_n(\lambda x)= \lambda^{-n} f_n(x)$, which indicates that we should use spherical coordinates to calculate the integral. The problem is, however, that the integration domain does not look particularly nice in spherical coordinates. Thus, we would like to change the integration domain without changing the integral too much.

The first thing to observe is that as $\ln$ is increasing, $f_n(x) \ge 0$ holds for any $n \in \N, x \in (\R^+)^n$. For any (Borel) measurable $X \subset (\R^+)^n$, we define
\be
\iota(X) \coloneqq \int_{X} \mathrm{d}x \, f_n(x)= \int_X \mathrm{d}x\,  \prod_{i=1}^n f(x_i, x_{i+1})\,.
\ee
As the integrand is non-negative, $\iota$ is a measure. We also observe that $\iota$ is invariant under the cyclic shift $(x_1, x_2, \dots, x_n) \mapsto (x_2, x_3, \dots, x_n,x_1)$. Assuming $n >1$, for $i=1, \dots, n$, we consider the set
\be
U_i \coloneqq \big\{ x \in (\mathbb R^+)^n  : x_i \le 1 \le x_{i+1}\big\}\,.
\ee
We observe $\iota(U_i)=\iota(U_1)$ by the cyclic shift property. We see
\be
\iota(U_1)= \tr P(0,1)K P( 1, \infty) (K P(0, \infty))^{n-2} K P(0,1)\,.
\ee
Since $P(0,1)K P(1,\infty)= P(0,1) Q(-\infty,0)P(-\infty, 0) Q(-\infty,0) P(1,\infty)$ is the operator $R$ \cite[(9)]{LW} with appropriately chosen intervals $J,M,K,N,L$ we see that this is trace class by \cite[Lemma, (L2)]{LW}.
By the homogeneity of $f_n$, we even have $\iota( c U_1)= c^{n-n} \iota(U_1)=\iota(U_1)$.

Next, we introduce the set
\[ V \coloneqq \big\{x \in (\mathbb R^+)^n : \sqrt n \le \|x \| \le \sqrt{n} \, c\big\}\,,
\]
which looks very nice in spherical coordinates. For $n=1$, we just have $V=[1,c]$. For $n>1$, we observe the chain
\be
[1,c]^n \subset V \subset [1,c]^n \cup \bigcup_{i=1}^n ( U_i \cup c U_i ) \,.
\ee
The first inclusion is trivial. We call $x_1, \dots, x_n$ the coordinates of $x$. If $x$ has both a coordinate above $\lambda$ and one below $\lambda$, then it has to be in the set $\bigcup _{i=1} ^n \lambda U_i$. Any $x \in V$ has at least one coordinate above $1$ and a coordinate below $c$. Thus, if $x \not \in [1,c]^n$, it has to have a coordinate above and below $1$ or a coordinate above and below $c$, which proves the second inclusion. These inclusions and the subbadditivity and monotonicity of $\iota$ imply that there is a constant $C_n$ (depending on $n$ but not on $c$) such that
\begin{align}
&\iota([1,c]^n) \le \iota (V) \le \iota([1,c]^n) + C_n \qquad
\implies \quad & \iota([1,c]^n)= \iota(V) + O(1)\,. \label{iota V+O1}
\end{align}
This holds with $O(1)$ replaced by $0$ for $n=1$. Finally, we introduce $W \coloneqq \big\{ x \in (\R^+)^n : \| x \| = \sqrt n\big\}$ with Hausdorff measure $\mathcal H^{n-1}$ and observe $V= [1,c] W = \big\{\lambda x : 1\le\lambda\le c, x\in W\big\}$. Now, we are just left to calculate
\begin{align}
\iota(V) &= \int _V \mathrm{d}x\, f_n(x) = \int_{1}^c \mathrm{d}r\,r^{n-1}  \int_{W}\mathrm{d}\mathcal H^{n-1}(x)\, f_n(r x)
\\
& = \int_{1}^c \mathrm{d}r\, r^{n-1}  \int_{W} \mathrm{d}\mathcal H^{n-1}(x) \, r^{-n} f_n(x)  = \int_{1}^c \mathrm{d}r\,r^{-1} \int_{W} \mathrm{d}\mathcal H^{n-1}(x) \, f_n(x)
\\
&= \ln(c) \int_{W} \mathrm{d}\mathcal H^{n-1}(x) \,f_n(x) = \tilde{C}(n) \ln(c)\,,
\end{align}
where $\tilde{C}(n)$ is the result of the surface integral. We did a change to spherical coordinates in the second step. As the integrand is positive, $\tilde{C}(n)\in (0,\infty]$ is well-defined. By \eqref{iota V+O1}, we conclude (for fixed $n$ and as $c\to\infty$)
\begin{align}
\tr K_c^n = \tilde{C}(n) \ln(c) + O(1)\,.
\end{align}
From \cite[(19)]{LW}, we know that $\tilde{C}(n)= \frac 1 {4\pi^2} \int_0^1 \mathrm{d}t\, (t(1-t))^{n-1}$. In conjunction with \eqref{LW18}, we get the improved error term $O(1)$ with the same leading term for any polynomial, which vanishes at $0$ and $1$.

To get the claim of our lemma, we just have to replace $c$ by $2\sqrt \mu L\ell$ and then use $\ln(2\sqrt \mu L\ell) = \ln(2)+ \frac 1 2 \ln(\mu) +\ln(L\ell)=\ln(1+L\ell)+ O(1)$, which relies on $L\ell \ge 1$.
\end{proof}

Now we are in position to present and prove the main result in this section. The dependency of our error term on $\Omega$ is not just $O(1)$ as in [48] but explicit in terms of the number, lengths, and distances of the constituent intervals of $\O$. Sobolev in \cite[Chapter 8]{Sob:AMS} has a similar error term, which however, does not seem to suffice for our purposes.

\begin{cl} \label{1D final asym} We assume the same conditions on the set $\Omega$ as in \autoref{1d multiple interval error lem}, $\mu>0$ and $m\in\N$. Then, with $\mathsf{I}(m)$ explained after \eqref{def: I}, we have for any $L \ge 1$,
\begin{align}
\tr \left( \mathds{1}_{L\Omega} \mathds{1}[(-\mathrm{i}\nabla^\parallel)^2 \le \mu] \mathds{1}_{L\Omega}\right)^m &= \frac{\sqrt \mu}\pi L\lvert \Omega \rvert +4k \,\mathsf{I}(m)\ln(1+L)
\\
&+ O\Big(k+\lvert \ln(\ell_k)\rvert+\sum_{j=1}^{k-1} \lvert \ln(\ell_j) \rvert +  \lvert \ln (d_j) \rvert \Big) \,.\label{1d multiple interval error term eq}
\end{align}
\end{cl}

\begin{proof}
For the case of a single interval, we use \autoref{improved Landau-Widom}
\begin{align}
\tr \left( \mathds{1}_{L I_j} Q \mathds{1}_{L I_j}\right)^m& =  \frac {\sqrt \mu} \pi L \ell_j +4\,\mathsf{I}(m) \ln(1+L \ell_j ) + O(1) \\
&=  \frac {\sqrt \mu} \pi L \ell_j +4\,\mathsf{I}(m) \Big[\ln(1+L)+\ln\Big(\frac{L}{1+L}\Big)+\ln\Big(\frac 1 L + \ell_j\Big)\Big] + O(1) \,. \label{C10}
\end{align}
As $L \ge 1$, we have
\begin{align}
\left \lvert \ln\left(\frac{L}{1+L}\right) + \ln\left(\frac 1 L + \ell_j\right) \right\rvert < 3+\lvert \ln(\ell_j)\rvert\,. \label{C11}
\end{align}
Next, we observe\footnote{If $ab\le 1$ then $\ln(1+ab)\le ab\le 1$ and \eqref{C.13} holds. If $ab\ge1$ then we distinguish between the case that both $a\ge1$ and $b\ge1$ and the case where one of them is smaller than $1$. In the first case \eqref{C.13} is equivalent to $\ln(1+ab)-\ln(ab) = \ln(1+1/(ab))\le1$ which holds because $\ln(1+1/(ab))\le 1/(ab)\le1$. In the remaining case we may assume $a\le1$ and $b\ge1$ (but still $ab\ge1$). Then $ab\le b/a$ and $\ln(1+ab) = \int_1^{ab} \mathrm{d}x/x \le \int_1^{b/a} \mathrm{d}x/x =\ln(b/a)$.} that for any $a>0$ and $b>0$, we have
\be\label{C.13}
\ln(1+ab) < \lvert \ln(a) \rvert + \lvert \ln(b) \rvert + 1\,.
\ee
Now, we only need to rewrite the error term from \autoref{1d multiple interval error lem} in the form we claim in this corollary. Thus, we estimate
\begin{align}
\ln \left(1+ \frac {L\ell_j}{1+Ld_j}\right) = &\ln\left( 1+ \frac{\ell_j}{\frac 1 L + d_j } \right) \le \ln(1+\frac{\ell_j}{d_j})  <   \lvert \ln(\ell_j) \rvert + \lvert \ln(d_j) \rvert +1\,. \label{C15}
\end{align}
Once we sum the error term estimate in \eqref{C15} for $i=1, \dots, k-1$ and the one in \eqref{C11} for $i=1, \dots, k$, we arrive at the claimed error estimate in \eqref{1d multiple interval error term eq}. We also see that the sum of the main terms in \eqref{C10} for $i=1,\dots, k$ agrees with the main term in \eqref{1d multiple interval error term eq}. This finishes the proof.
\end{proof}

\section{A technical lemma on decaying functions}

This section contains a technical lemma that was useful to construct the sequence $(a_i)_{i\in\N}$ and the region $\L$ in the proof of \autoref{thm error term}.

\begin{lemma}\label{technical function existence}
Let $f \colon \R^+ \to \R^+$ be bounded and satisfy $\lim_{L \to \infty} f(L)=0$. Then there is a convex, non-increasing function $\mathrm{Env}(f) \colon \R^+ \to \R^+$ satisfying $\mathrm{Env}(f)(0)=1, \lim_{L \to \infty} \mathrm{Env}(f)(L)= \lim_{L \to \infty} f(L)/\mathrm{Env}(f)(L) =0$ and $\mathrm{Env}(f)(L) \ge C/\sqrt{\ln(2+L)}$ for some $C>0$.
\end{lemma}

\begin{proof}
The conditions on $\mathrm{Env}(f)$ only get worse if we increase $f$. Hence, we can replace $f$ by
\be \hat f(s) \coloneqq  \sup_{t >s} f(t)\,.
\ee
This is non-increasing. To achieve the $f(L)/\mathrm{Env}(f)(L)$ condition we consider $\sqrt{\hat f}$. However, we still need to make sure that $\mathrm{Env}(f)$ is convex. For this reason, we need to consider the lower convex envelope $ \breve {\sqrt {\hat f}}$. It is given by the supremum over all convex functions below $\sqrt {\hat f}$. Another way to think of it is that the area above the graph of $\breve{ \sqrt {\hat f}}$ is the convex hull of the area above $\sqrt {\hat f}$. Finally, we define
\be \mathrm{Env}(f)(t) \coloneqq N \left( \breve{ \sqrt {\hat f}} \left( \frac t 2\right) + \frac 1 {\sqrt{ \ln(2+t)}} \right)\,,\quad t\ge0\,,\ee
where $N$ is a normalization constant to be chosen below. As the lower convex envelope and $\frac 1 {\sqrt{ \ln(2+t)}}$ are convex, so is $\mathrm{Env}(f)$.  As the lower convex envelope lies below the function, we have $\mathrm{Env}(f)(t) \le N\Big(\sqrt{\hat f (t/2)} + \frac 1 {\sqrt{ \ln(2+t)}}\Big) \to 0$ as $t \to \infty$. As $\mathrm{Env}(f)$ is convex and $\lim_{L \to \infty}\mathrm{Env}(f)(L)=0$, $\mathrm{Env}(f)$ is non-increasing.  The condition $\mathrm{Env}(f)(L) \ge C/\sqrt{\ln(2+L)}$
is trivially satisfied and implies $\mathrm{Env}(f)(0)>0$ and hence allows us to choose $N$ such that $\mathrm{Env}(f)(0)=1$. We are only left with the condition $\lim_{L \to \infty} f(L)/\mathrm{Env}(f)(L) =0$. To show this, it is sufficient to prove
\be \mathrm{Env}(f)(L) \ge C \sqrt{f(L)}\,. \ee
By the definition of the convex envelope, for any $t \ge0$, there are $0 \le t_1 \le t < t_2$ such that
\be
\mathrm{Env}(f)(t)/N=\frac{t_2-t}{t_2-t_1} \sqrt {\hat f}\left( \frac {t_1} 2 \right) + \frac{t-t_1}{t_2-t_1} \sqrt {\hat f}\left( \frac {t_2} 2 \right) + \frac 1 {\sqrt{ \ln(2+t)}}\,.
\ee
If $t_2 \le 2t$, as $\hat f$ is non-increasing, we get $\sqrt {\hat f}\left( \frac {t_j} 2 \right)  \ge \sqrt {\hat f}(t) $ for $j=1,2$ and thus are finished. If $t_2 > 2t$, then, as $t_1 \ge0$, we have
$\frac{t_2-t}{t_2-t_1}  > \frac 12 $. Thus, it suffices to bound the first summand from below, which we already did.
\end{proof}

\end{appendix}

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\end{thebibliography}

\end{document}


# Observation of the quantum boomerang effect

Roshan Sajjad  Jeremy L. Tanlimco  Hector Mas  Alec Cao  Eber Nolasco-Martinez
Ethan Q. Simmons  Department of Physics, University of California, Santa
Barbara, California 93106, USA  Flávio L. N. Santos  Departamento de Física
Teórica e Experimental, Universidade Federal do Rio Grande do Norte, 59072-970
Natal, Rio Grande do Norte, Brazil  Patrizia Vignolo  Université Côte d’Azur,
CNRS, Institut de Physique de Nice, 06560 Valbonne, France  Tommaso Macrì
Departamento de Física Teórica e Experimental, Universidade Federal do Rio
Grande do Norte, 59072-970 Natal, Rio Grande do Norte, Brazil  David M. Weld
<EMAIL_ADDRESS>Department of Physics, University of California, Santa Barbara,
California 93106, USA

###### Abstract

A particle in an Anderson-localized system, if launched in any direction,
should on average return to its starting point and stay there. Despite the
central role played by Anderson localization in the modern understanding of
condensed matter, this “quantum boomerang” effect, an essential feature of the
localized state, was only recently theoretically predicted and has not
previously been observed. We report the experimental observation of the
quantum boomerang effect. Using a degenerate gas and a phase-shifted pair of
optical lattices, we probe the role of time reversal symmetry breaking,
Floquet gauge, and initial state symmetry in supporting or disrupting the
boomerang effect. Highlighting the key role of localization, we observe that
as stochastic kicking destroys dynamical localization, the quantum boomerang
effect also disappears. Measured dynamics are in agreement with analytical and
numerical predictions. These results showcase a unique experimental probe of
the underlying quantum nature of Anderson localized matter.

Anderson localization is a quintessential effect of disorder in a variety of
physical systems [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]. Disordered lattice
models enjoy a robust theoretical proof of localization and remain a hotbed
for discovery even today. While transport in disordered media has been studied
for more than 50 years, a fundamental _dynamical_ feature of Anderson-
localized matter known as the “quantum boomerang” effect was only recently
theoretically predicted [13, 14, 15]. The central prediction of the quantum
boomerang is that, for an appropriate average over disorder realizations, a
wavepacket launched with any nonzero momentum will return to and localize at
its initial position [13]. This surprising phenomenon can be understood as a
consequence of both Anderson localization and time-reversal symmetry: the
long-time density distribution for a wavepacket with an initial momentum
$k_{0}$ is (i) time-independent because of localization and (ii) does not
depend on the sign of $k_{0}$ because of time-reversal symmetry. Together,
these features imply that at long times the wavepacket must settle at its
initial position. The quantum boomerang effect is predicted to occur in a
broad class of disordered or pseudo-disordered tight-binding models [15]
including the quantum kicked rotor (QKR), a momentum-space realization of a
disordered quasi-1D wire [16, 17] which provides an ideal platform for the
study of the interplay between localization, chaos, interactions, and
transport [18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30]. In any
context, the quantum boomerang effect contravenes classical predictions and
serves as a sensitive diagnostic of the quantum nature of localization, with
the potential, for example, to map mobility edges [13, 31].

Fig. 1: Observation of the quantum boomerang effect. (a): Experimental
sequence. A BEC is loaded into one optical lattice and then kicked with a
second, phase-shifted lattice. The momentum distribution is measured after
free expansion. BEC size and timeline are not to scale. (b): Measured average
momentum versus kick number, for parameters given in the text. The return to
the origin demonstrates the quantum boomerang effect. Error bars show standard
error on ten repeated measurements. Solid line is the numerical prediction.

We report the experimental observation and characterization of the quantum
boomerang effect. Using a degenerate gas and a pair of phase-shifted optical
lattices to create a kicked rotor with tunable initial conditions (Fig. 1), we
observe the boomerang’s distinctive return dynamics. We experimentally
characterize the dependence of the dynamics on symmetry properties, Floquet
gauge, and initial state, and confirm that results agree with an analytical
theory we develop in the appendices. Finally, we measure the controlled
destruction of the boomerang effect by stochastic kicking. The atom-optics
kicked-rotor realization of the quantum boomerang offers three key advantages:
i) finite-width quasimomentum distributions naturally achieve single-shot
disorder averaging, ii) working in momentum space rather than position space
enables simple observation of boomerang dynamics with single-site resolution,
and iii) time-reversal symmetry breaking can be straightforwardly tuned by
adjusting properties of the quench [32, 33, 34].

Fig. 2: Characterization of the quantum boomerang effect. (a-c): Average
momentum vs. kick number for $V_{0}=38\,E_{\mathrm{R}}$ and $T=8\,\upmu$s, at
$\alpha=0.25$, 0.47, and 0.7 respectively. Solid curves show numerical
predictions. Dotted lines show infinite-time average momenta predicted by the
Floquet diagonal ensemble (Eq. 6). (d): Measured energy vs. kick number for
all three values of $\alpha$. Shaded areas indicate long-term energy by
averaging data points between 10 and 25 kicks. (e): Measured average momentum
after 1 kick (grey circles) and 25 kicks (blue squares) vs. $\alpha$ for the
same experimental parameters. Same-color solid lines show numerical
predictions. Dashed vertical lines indicate $\alpha$ in (a-c). Dotted purple
line shows prediction of the Floquet diagonal ensemble [13]. (f): Average
momentum after 30 kicks vs. $\phi$ for $V_{0}=70\,E_{\mathrm{R}}$,
$T=7\,\upmu$s, and $\alpha=0.1$ (diamonds), 0.45 (circles), and 1 (squares).

We begin each experiment by preparing a Bose-Einstein condensate (BEC) of
$10^{5}$ 7Li atoms and then tuning interatomic interactions to zero with a
Feshbach resonance. A typical experimental sequence from this point on is
depicted in Fig. 1a. The atoms are exposed in turn to two coaxial optical
lattices with lattice constant $d=532\,\mathrm{nm}$, wave vector
$k_{\mathrm{L}}=\pi/d$, recoil energy
$E_{\mathrm{R}}=\hbar^{2}k_{\mathrm{L}}^{2}/2M$, and tunable relative offset
$x_{0}$. $M$ is the atomic mass. To prepare an initial wavefunction which is
well-localized in position space within each lattice site, we adiabatically
load the BEC into the first optical lattice over $50\,\mathrm{ms}$, reaching a
final depth of $7\,E_{\mathrm{R}}$. At $t=0$, we suddenly turn off the first
lattice and quench into a Hamiltonian created by repeated kicking with the
second, spatially-shifted lattice:

$H_{\alpha}=\frac{p^{2}}{2M}+\frac{V_{0}}{2}\cos\left(2k_{\mathrm{L}}x\right)\sum_{n}f_{\tau}(t-nT-\alpha
T).$ (1)

As described by Eq. 1, the second, phase-shifted lattice is pulsed $n$ times
to depth $V_{0}$ with pulse shape $f_{\tau}$, pulse width
$\tau=300\,\mathrm{ns}$, and period $T$, after a variable delay time $\alpha
T$ from the first lattice switch-off. Because of the adiabatic loading into
the first lattice and the relative phase shift, the atomic ensemble starts out
centered at a nonzero position given by $x_{0}$. This is analogous to the
situation in a position-space quantum boomerang, where the initial state is
centered at nonzero momentum.

To make concrete the connection to the kicked rotor we introduce dimensionless
variables $P\equiv 2k_{\mathrm{L}}Tp/M$, $Q\equiv 2k_{\mathrm{L}}x$, and
$\mathcal{T}\equiv t/T$ as well as dimensionless parameters $\mathchar
22\relax\mkern-9.0muk\equiv 8E_{\mathrm{R}}T/\hbar$, an effective Planck’s
constant, and $K\equiv\mathchar 22\relax\mkern-9.0mukV_{0}\tau/2\hbar$,
encoding kicking strength. Substituting these into Eq. 1, rescaling the
Hamiltonian
$\mathcal{H}_{\alpha}\equiv(4k_{\mathrm{L}}^{2}T^{2}/M)H_{\alpha}$, and taking
the $\delta$-function limit of $f_{\tau}/\tau$, we recover the canonical QKR
Hamiltonian $\mathcal{H}_{\alpha}=\frac{P^{2}}{2}+{K}\cos
Q\sum_{n}\delta(\mathcal{T}-n-\alpha)$. This gives rise to a single-particle
1-cycle Floquet operator $U_{\alpha}=e^{-i(1-\alpha)P^{2}/2\mathchar
22\relax\mkern-9.0muk}e^{-iK\cos Q/\mathchar 22\relax\mkern-9.0muk}e^{-i\alpha
P^{2}/2\mathchar 22\relax\mkern-9.0muk}$.

Figure 1b illustrates the main result of this work: we experimentally observe
the departure from and return to zero average momentum which is the key
signature of the quantum boomerang effect [13]. Here
$V_{0}=38\,E_{\mathrm{R}}$, $T=8\,\upmu$s, $\phi\equiv
2k_{\mathrm{L}}x_{0}=\pi/2$ and $\alpha=0.47$, which correspond to $\mathchar
22\relax\mkern-9.0muk=10$, $K=9$. While the first kick imparts momentum
asymmetry due to the lattice phase offset, the system returns to zero average
momentum at long times. The experimental result agrees well with fit-
parameter-free time-dependent Schrödinger equation (TDSE) simulations of the
finite-pulse width Hamiltonian (1) using a split-step method. In all such
numerical calculations discussed in this work we compute the dynamics of
individual Bloch states and average over a Gaussian quasimomentum distribution
of standard deviation $0.1\,k_{\mathrm{L}}$, which both models our finite-
width BEC and realizes the disorder average in standard Anderson model
calculations 111According to the adiabatic theorem, the BEC wavefunction after
the lattice load state-preparation is a superposition of Bloch states
$e^{i\beta x}u_{\beta}(x)$, where $\beta$ denotes quasimomentum, weighted by
the initial Gaussian momentum wavefunction ($\psi(p)$ before the load becomes
$\psi(\beta)$) and with a $\beta$-dependent phase. Due to the periodicity of
$u_{\beta}(x)$, it can be shown that the $\beta$-dependent phase does not
affect the time-evolution of $\langle p\rangle$ and $\langle
p^{2}/2M\rangle$..

This demonstration of the quantum boomerang effect enables an in-depth
experimental characterization of the circumstances under which it can occur.
Our measurements reveal a rich interplay among symmetries and quench dynamics
which is well described by the analytic theory we present in the appendices.
Specifically, it is the relation between the symmetry properties of the
initial state and the Floquet gauge $\alpha$ defined by the $t=0$ quench which
dictates the presence or absence of the quantum boomerang effect in the
quantum kicked rotor. This interplay offers a contrast to static disordered
systems, in which initial state symmetries are also important but which lack a
Floquet gauge degree of freedom. We note for clarity that the QKR Hamiltonian
itself is of course time-reversal and parity symmetric independent of Floquet
gauge and spatial phase choices, so long as one references appropriate
symmetry axes. For the remainder of the paper, we will solely refer to time-
reversal and parity symmetry with respect to inversions about $t=0$ and $x=0$
unless otherwise stated. $\mathcal{H}_{\alpha}$ is then clearly parity
symmetric, and only when $\alpha=1/2$ does the evolution also become time-
reversal invariant. As we show in Appendix A, the initial Bloch state ensemble
we experimentally prepare satisfies the exact properties necessary to observe
a boomerang effect under these conditions: it is (i) parity asymmetric to
allow for nontrivial early-time momentum dynamics and (ii) time-reversal
symmetric ($\psi^{*}(x)=\psi(x)$) to ensure a complete return to 0 average
momentum at late-times. This interplay of symmetries arising from the QKR
quench protocol enables the quantum boomerang effect to be switched on or off
by adjusting the Floquet gauge.

Fig. 2 demonstrates this dependence on Floquet gauge with a detailed
characterization of the average momentum evolution for various $\alpha$.
Measurements at $\alpha$ values of 0.25, 0.47, and 0.7 (Fig. 2a-c) exhibit
fundamental differences in the dynamical behavior. As discussed above, only
near $\alpha=1/2$ do we observe the quantum boomerang effect: a sharp initial
increase in $\langle p\rangle$ followed by a decline back to the initial value
of 0. The slight deviation of 0.47 from 1/2 is consistent with corrections to
the condition due to finite pulse width: $\delta\alpha=\tau/2T=0.019$. In
contrast, evolution under the other two Floquet gauges converges to nonzero
momentum values at long times, indicating the absence of the full quantum
boomerang effect. Each data set agrees well with fit-parameter-free numerical
predictions, and all late-time momenta match the analytical predictions of the
Floquet diagonal ensemble (Eq. 6).

Fig. 2d shows that the observed energy evolution is qualitatively consistent
for all three values of $\alpha$, with each ensemble exhibiting dynamical
localization as expected for a non-interacting quantum kicked rotor [36, 18,
24]. We note that the energy for $\alpha=0.47$ appears to saturate at a
slightly higher value than the other gauges, a result consistent with the
predictions of Ref. [15].

A key feature of quantum boomerang dynamics which can be probed with a kicked
rotor is the contrast between short- and long-time evolution. Fig. 2e maps out
the Floquet gauge dependence of momentum in these two regimes. Since the
energy saturates after only five kicks, we take 25 kicks as a reasonable proxy
for the infinite-time limit predicted by relaxation to the Floquet diagonal
ensemble (see Eq. 6). The initial momentum asymmetry imparted to the rotor
after one kick arises purely from the Talbot effect [37, 38, 39], a periodic
re-imaging of the wavefunction in time under free-evolution as a result of the
spatially periodic lattice exciting only quantized momentum modes. This Talbot
oscillation for a Bloch state is illustrated in Fig. 3d, where for our
experiment, $T_{\mathrm{Talbot}}=\pi\hbar/2E_{\mathrm{R}}\approx
9.95\,\upmu$s.

In contrast, the long-time behavior is governed by localization physics and
symmetries surrounding the quench. One feature of both experimental and
theoretical results in Fig. 2e is that the long-time momentum should be an odd
function about $\alpha=1/2$, a result we prove in appendix C. This result is
connected to the resemblance between forward time-evolution in the gauge
$\alpha$ (propagation with $U_{\alpha}$) and backward time-evolution in the
gauge $1-\alpha$ (propagation with $U_{1-\alpha}^{\dagger}$). The anti-
symmetry about $\alpha=1/2$ is dictated by the time-reversal invariance of the
initial Bloch state ensemble. We note that because this property arises due to
similarities between forward evolution in one gauge and backward evolution in
the other, it only manifests as an (anti-)symmetry in the forward evolution of
both gauges in the long-time limit (see Eq. 7); at short times, no symmetry is
generically expected, as evidenced by the 1-kick data.

The odd parity of the long-time momentum about $\alpha=0.5$ is further
reflected in Fig. 2f, which shows the long-time momentum as a function of
initial wavepacket position for three different values of $\alpha$. Here a
shift in the relative lattice phase is exactly analogous to variation in the
initial launch velocity of a position space boomerang. We observe that
regardless of initial position, $\alpha$ values near 0 and 1 result in
opposite-sign, non-zero late-time momenta. The only exception occurs when the
initial state recovers parity symmetry at $\phi=0$ or $\phi=\pi$ and the
average momentum dynamics become trivial. Near $\alpha=1/2$, the entire curve
collapses toward 0. This indicates that, as mentioned above, it is time-
reversal symmetry which plays the fundamental role in determining the _return_
of the boomerang. The initial position which controls the degree of parity
symmetry breaking is instead more closely related to the momentum-space mean
free path and _amplitude_ of the early-time boomerang dynamics.

Fig. 3: Initial state dependence of the quantum boomerang effect. (a): Average
momentum versus kick number for an initial state consisting of a BEC kicked
once by a $60\,E_{\mathrm{R}}$ lattice. Pink circles correspond to
$\alpha=0.86$ and purple diamonds correspond to $\alpha=0.57$ for
$T=7\,\upmu\mathrm{s}$. Solid lines are predictions of TDSE numerics. (b):
Average momentum after 30 kicks versus $\alpha$, for an initial condition set
by adiabatic loading of a BEC into a $7\,E_{\mathrm{R}}$ lattice (grey
circles) and an initial condition set by kicking a BEC once with a
$60\,E_{\mathrm{R}}$ lattice (blue squares). Dashed vertical lines correspond
to $\alpha$ values in (a). (c-d): Calculated density evolution over two Talbot
times of a BEC kicked by a $60\,E_{\mathrm{R}}$ lattice (left) and a a BEC
adiabatically loaded into a $7\,E_{\mathrm{R}}$ lattice which is then quenched
off (right). Dashed vertical lines represent time-reversal symmetry axes.

The demonstrated control of symmetry properties by varying $\alpha$ opens up
the possibility of realizing quantum boomerang dynamics starting from more
exotic initial states. Fig. 3 explores the quench dynamics of an initial state
with an essentially homogeneous distribution in position space, achieved by
replacing the adiabatic lattice load with a single kick. The initial
wavefunction, approximately $\psi(x)=e^{-iV\sin(2k_{\mathrm{L}}x)}$ with $V$
the effective kick impulse, is neither time-reversal symmetric nor well-
localized in position. Intuitively one would not expect such a state to
demonstrate quantum boomerang dynamics, as our understanding of the position-
space quantum boomerang effect is based on a well-defined initial velocity.
Despite these expectations, the data in Fig. 3a reveal a clear quantum
boomerang effect, though at a Floquet gauge $\alpha=0.86$, very different from
the value of 0.5 expected for a time-reversal symmetric state. This can be
understood by observing that free evolution maps this initial state into a
time-reversal but not parity invariant wavefunction after a time
$T_{\mathrm{Talbot}}/4$, as illustrated numerically in Fig. 3c, and proven
analytically in Appendix B. Thus, one expects to observe the boomerang effect
at $\alpha=1/2+T_{\mathrm{Talbot}}/4T\simeq 0.855$ for the kicked initial
state, in good agreement with numerical and experimental results in Fig. 3.
This finding demonstrates that even for quenched states completely unlike
traditional boomerang starting points, the Floquet gauge freedom can restore
the boomerang effect in certain cases where the Talbot effect defines a new
time-inversion symmetric point.

We note that while the kicked state is not initially time-reversal symmetric,
it is invariant under simultaneous parity-time inversion
($\psi^{*}(-x)=\psi(x)$). As we show in Appendix C, this different symmetry
condition manifests in the late-time momentum exhibiting even rather than odd
symmetry about $\alpha=1/2$. These different Floquet gauge properties for the
two initial state symmetries we consider are demonstrated and contrasted in
Fig. 3b, which clearly distinguishes the different parity about $\alpha=0.5$.
The data are not perfectly symmetric on the appropriate intervals even when
taking into account finite-pulse width; we attribute this mainly to finite-
time effects modifying the infinite-time gauge symmetries. For the kicked
state data shown, the symmetry is most apparent when extending to cases where
$\alpha$ is slightly greater than 1 222The experimental delay-time definition
and Floquet gauge definition of $\alpha$ is only consistent on
$\left[0,1\right]$ (for finite pulse case, strictly on the interval
$\left[0,1-\tau/T\right]$). For $\alpha>1-\tau/T$ data and simulations, we
define $\alpha$ solely by the delay-time definition. Mathematically, the
$\delta$-kicking case ($\tau\to 0$) can then be cleanly and consistently
described by simply re-defining the quenched initial state as
$e^{-i\left\lfloor\alpha\right\rfloor P^{2}/2\mathchar
22\relax\mkern-9.0muk}\ket{\psi_{0}}$, with $\ket{\psi_{0}}$ the original
quenched state for $\absolutevalue{\alpha}\leq 1$ cases and
$\left\lfloor\alpha\right\rfloor$ the integer part of $\alpha$, and taking the
Floquet gauge to be $\alpha\bmod 1$. The finite-pulse case can be described
similarly, except for modifications in the small intervals $n-\tau/T<\alpha<n$
where $n$ is an integer (assuming a square pulse $f_{\tau}(t)=1$ when
$0<t<\tau$)..

Fig. 4: Destruction of the quantum boomerang effect by stochastic kicking.
(a): Average momentum measured after $n$ kicks for temporal disorder strength
$\Delta=0$ (blue circles) and $\Delta=0.6$ (red squares). For all data in this
figure, $T=8\,\upmu$s, $\alpha=0.51$, $V_{0}=38\,E_{\mathrm{R}}$, and the
initial condition was set by an adiabatic lattice load to $8\,E_{\mathrm{R}}$.
Circles (squares) represent the mean of measurements with 5 separate kick
disorder realizations for $\Delta=0$ ($\Delta=0.6$). Dashed lines show
individual realizations. Shaded areas show numerically calculated $\langle
p\rangle\pm\sigma_{\langle p\rangle}$ with 9000 disorder realizations and 500
initial quasimomenta. (b): Measured mean energy for $\Delta=0$ (blue circles)
and $\Delta=0.6$ (red squares) over 10 points (blue) and 5 disorder
realizations (red), illustrating the destruction of dynamical localization by
random kicking. (c): Average momentum $\langle p\rangle$ after 25 kicks versus
disorder strength $\Delta$. Dots are averages over 10 disorder realizations,
except for $\Delta=0,0.6$ (5 realizations each). Solid line is numerically
calculated $\langle p\rangle$ over 9000 disorder realizations, with shaded
area showing a one-standard-deviation range.

The final set of results we describe (Fig. 4) probes the question: what
happens to quantum boomerang dynamics as localization is destroyed? In the QKR
context, dynamical localization can be tunably disrupted by introducing
temporal stochasticity into the kicking sequence [41, 24]. To implement this,
we add a deviation $\delta_{n}$ to the delay time after the $n^{\mathrm{th}}$
kick, where $\delta_{n}$ is randomly selected from a uniform distribution in
the range $\left[-W/2,W/2\right]$. The parameter $\Delta=W/T$ quantifies the
degree of randomness, and can range from 0 to 1. Fig. 4a shows the time
evolution of average momentum for $\Delta=0$ (no randomness) and $\Delta=0.6$,
with the first kick fixed so that $\alpha=0.51$, and with the same
experimental parameters as in Fig. 2b. In contrast to the characteristic
boomerang-like evolution when $\Delta=0$, each of the five experimental
realizations of randomness for $\Delta=0.6$ exhibits a different behavior. The
mean momentum over these disorder realizations demonstrates the destruction of
the quantum boomerang effect, with a long-time average momentum significantly
different from zero, and agrees reasonably well with the result of disorder-
averaged simulations. Fig. 4b shows the evolution of the average energy for
the same experiments: the destruction of dynamical localization by random
kicking and the approach to classical linear-time energy growth are clearly
visible in both experiment and theory, and are associated with the destruction
of the quantum boomerang effect in Fig. 4a. To directly probe the dependence
of the quantum boomerang on the disorder strength, in Fig. 4c we plot the
disorder-averaged mean momentum after 25 kicks as a function of $\Delta$.
Measured long-time $\langle p\rangle$ increases monotonically up to
$\Delta\approx 0.5$, and agrees reasonably well with disorder-averaged
numerical predictions. An interesting topic for further study is the
relationship between the degree of destruction of the boomerang effect
(measured, for example, by long-time average momentum) and the diffusive
exponent [42, 43, 44] or, more speculatively, the one-parameter universal
scaling parameter of quantum transport [45, 22].

In this work we have presented the first experimental evidence for the quantum
boomerang effect in an Anderson-localized system. The measurements confirm
theoretical predictions [13, 15] and demonstrate that boomerang dynamics can
occur from a variety of initial conditions including even a homogeneous
density distribution. These experimental results, along with the analytical
theory we present in the appendices, elucidate the crucial role of parity and
time-reversal symmetry in determining the presence or absence of the quantum
boomerang effect. We have probed the dependence of boomerang dynamics upon
Anderson localization itself by measuring the effect of tunably random kicking
sequences: results confirm that classical diffusion induced by stochasticity
destroys both dynamical localization and the quantum boomerang effect. These
experiments validate a new, powerful probe of the uniquely quantum-mechanical
nature of localization applicable to a general class of disordered systems,
and suggest a variety of intriguing topics for future exploration. These
include boomerang phenomena in higher-dimensional systems, different symmetry
classes [46], and more exotic initial states, and the presence or absence of
many-body boomerang effects in interacting systems [14] or even many-body
localized states. Applications of our demonstration that adjusting the Floquet
gauge can switch on or off the boomerang effect are an intriguing possibility.
Beyond the context of the quantum kicked rotor, the investigation of
boomerang-like phenomena in condensed matter systems and implications of these
results for ultrafast electron dynamics in disordered solids represent
unexplored frontiers.

## Acknowledgments

We thank Toshihiko Shimasaki, Esat Kondakci, and Yifei Bai for helpful
conversations. D.W. acknowledges support from the Air Force Office of
Scientific Research (FA9550-20-1-0240), the Army Research Office
(W911NF-20-1-0294), the National Science Foundation (CAREER 1555313), and the
Eddleman Center for Quantum Innovation, and from the NSF QLCI program through
grant number OMA-2016245. T.M. acknowledges CNPq (grant no. 311079/2015-6).
T.M. and F.S. are supported by the Serrapilheira Institute (grant number
Serra-1812-27802). R.S. and E.N.-M. acknowledge support from the UCSB NSF
Quantum Foundry through the Q-AMASEi program (Grant No. DMR-1906325). A.C.
acknowledges support from the NSF Graduate Research Fellowship Program (grant
DGE2040434).

## Author Contributions

R.S., J.T., H.M., E.N.-M. and E.S. performed the measurements. R.S., J.T., and
H.M. analyzed the data. R.S., H.M., J.T., E.S., F.S., and A.C. performed
numerical modeling. A.C., T.M., and F.S. performed the theoretical symmetry
analysis with assistance from P.V. R.S., J.T., H.M., A.C., and D.W. drafted
the manuscript with input from all authors. D.W. supervised the experiments.
All authors contributed insights in discussions on the project.

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##  Appendix A Deriving Conditions for the QKR Boomerang Effect

Here we show that the symmetry conditions on the initial state and Floquet
gauge described in the main text indeed enable the boomerang effect in the QKR
model. Explicitly, we consider the canonical QKR Hamiltonian considered in the
main text (reproduced here and reverting to standard variable notations for
clarity)

$\displaystyle\mathcal{H}_{\alpha}(t)=\frac{p^{2}}{2}+K\cos
x\sum_{n=-\infty}^{\infty}\delta(t-n-\alpha).$ (2)

To be explicit, here we define $\alpha$ on the interval $\left[0,1\right]$,
taking the cases $\alpha=0$ and 1 to be distinct from each other and
consistent with the definition in the Floquet operator

$\displaystyle U_{\alpha}=e^{-i(1-\alpha)p^{2}/2\mathchar
22\relax\mkern-9.0muk}e^{-iK\cos x/\mathchar 22\relax\mkern-9.0muk}e^{-i\alpha
p^{2}/2\mathchar 22\relax\mkern-9.0muk}.$ (3)

That is for $\alpha=0\,(1)$ the kicks occur just after (before) $t=n$. Here we
continue to refer to symmetry operations only about fixed axes.
$\mathcal{H}_{\alpha}$ is evidently parity inversion ($x\to-x$, $p\to-p$)
symmetric at all times

$\displaystyle\left[\mathcal{P},\mathcal{H}_{\alpha}(t)\right]=0.$ (4)

The operator $\mathcal{P}$ obeys $\mathcal{P}^{\dagger}p\mathcal{P}=-p$. Only
for $\alpha=1/2$ does the system also become time-reversal ($t\to-t$,
$p\to-p$) symmetric. The complex conjugation operation $\mathcal{K}$
associated with time-reversal acts on the position space wave-function as
$\bra{x}\mathcal{K}\ket{\psi}=\psi^{*}(x)$. For the $\alpha=1/2$ case,
$\mathcal{H}_{\alpha}$ is additionally parity-time inversion ($t\to-t$,
$x\to-x$) symmetric.

We use the following intuitive definition for the QKR boomerang effect in
momentum space. (i) The boomerang should exhibit non-trivial average momentum
dynamics at short times, in particular a departure from followed by a
relaxation toward the initial value. (ii) At late times, a true boomerang
should have an average momentum equivalent to the initial momentum.
Explicitly, we will show that (i) is only compatible with parity asymmetric
states. Restricting ourselves to the case of 0 initial momentum, we then show
that time-reversal symmetry of the quenched initial state and Floquet gauge
guarantees (ii).

We consider solutions to the time-dependent Schrodinger equation $i\mathchar
22\relax\mkern-9.0muk\partial_{t}\ket{\psi(t)}=\mathcal{H}_{\alpha}(t)\ket{\psi(t)}$,
where $\ket{\psi(t=0)}$ is the initial state quenched into the Hamiltonian.
Our analysis will concern the behavior of $\langle
p(t)\rangle=\bra{\psi(t)}p\ket{\psi(t)}$. First we consider the evolution of
an arbitrary state $\ket{\psi(t=0)}$ and its parity inverted counterpart
$\ket{\phi(t=0)}\equiv\mathcal{P}\ket{\psi(t=0)}$. Let $\langle p(t)\rangle$
and $\langle\tilde{p}(t)\rangle$ correspond to the time-evolved states
$\ket{\psi(t)}$ and $\ket{\phi(t)}$ respectively. From parity symmetry (4), we
have $\ket{\phi(t)}=\mathcal{P}\ket{\psi(t)}$, and from this can conclude

$\displaystyle\langle\tilde{p}(t)\rangle=\bra{\psi(t)}\mathcal{P}^{\dagger}p\mathcal{P}\ket{\psi(t)}=-\langle
p(t)\rangle.$ (5)

This holds for any initial state and $\alpha$, as well as at all times $t$.
From this, it immediately follows that a necessary condition for observing
(i), and thus the boomerang, is that
$\ket{\psi(t=0)}\neq\mathcal{P}\ket{\psi(t=0)}$; otherwise
$\langle\tilde{p}(t)\rangle=\langle p(t)\rangle=0$, guaranteeing that (ii) is
satisfied but in a trivial way. Since all of the remaining analysis is also
concerned with $\langle p(t)\rangle$, from here on out we assume that parity
asymmetry is satisfied to derive non-trivial results.

Now we begin to investigate the $t\to\infty$ behavior concerning (ii).
Considering the Floquet spectral decomposition of the initial state
$\ket{\psi(t=0)}=\sum_{m}c_{m}\ket{\varphi_{m}^{\alpha}}$ with
$U_{\alpha}\ket{\varphi_{m}^{\alpha}}=e^{-i\omega_{m}^{\alpha}}\ket{\varphi_{m}^{\alpha}}$,
the long-time average momentum given by the Floquet diagonal ensemble is

$\displaystyle\langle
p(t\to\infty)\rangle=\sum_{m}\absolutevalue{c_{m}}^{2}\bra{\varphi_{m}^{\alpha}}p\ket{\varphi_{m}^{\alpha}}.$
(6)

One can also consider the opposite limit $t\to-\infty$ by decomposing in the
basis of $U_{\alpha}^{\dagger}$. Since the eigenstates of a unitary operator
and its Hermitian-conjugate are identical (both are eigenstates of the same
self-adjoint, time-independent Floquet Hamiltonian), the prediction of the
Floquet diagonal ensemble is the same. This leads to the conclusion that

$\displaystyle\langle p(t\to\infty)\rangle=\langle p(t\to-\infty)\rangle.$ (7)

This also holds for any initial state and $\alpha$.

Now finally we consider time-reversal symmetric states
$\mathcal{K}\ket{\psi(t=0)}=\ket{\psi(t=0)}$. Under parity-time inversion, we
then have the relation

$\displaystyle\mathcal{P}\mathcal{K}\ket{\psi(t=0)}=\mathcal{P}\ket{\psi(t=0)}\equiv\ket{\phi(t=0)}.$
(8)

Now consider the full solution $\ket{\psi(t)}$ and $\langle p(t)\rangle$. For
$\alpha=1/2$, the parity-time inversion symmetry of the Hamiltonian implies
that $\mathcal{P}\mathcal{K}\ket{\psi(-t)}$ is another solution with a
corresponding momentum evolution $\langle p(-t)\rangle$. From (8), it is
straightforward to see that this solution corresponds to the evolution of the
parity-inverted initial state, i.e.
$\ket{\phi(t)}=\mathcal{P}\mathcal{K}\ket{\psi(-t)}$; note that while a
parity-time inverted state about a different time-axis can be defined for
arbitrary $\alpha$, such a state will not coincide with $\ket{\phi(t)}$ and
prevent the following conclusions. Defining $\langle\tilde{p}(t)\rangle$ in
the same way as before, this result implies
$\langle\tilde{p}(t)\rangle=\langle p(-t)\rangle$. Combining this with (5), we
thus have

$\displaystyle\langle p(-t)\rangle=-\langle p(t)\rangle.$ (9)

One can deduce this relation intuitively by noting that backward propagation
under $U_{\alpha=1/2}$ for a time-reversal symmetric initial state simply
corresponds to receiving kicks of the opposite sign. Finally, the combination
of (7) and (9) yields the final result

$\displaystyle\langle p(t\to\infty)\rangle=0.$ (10)

We conclude by discussing the applicability of this argument to the presented
experiments and the role of quasi-momentum averaging. In the primary results
presented, the initial state is an ensemble of ground-band Bloch states
$\psi_{\beta}(x-x_{0})$, where $\beta$ denotes quasi-momentum and we reference
the functional form $\psi_{\beta}(x)$ with respect to a cosine external
potential. Generically, the entire Bloch ensemble will be parity asymmetric
($\psi(-x)\neq\psi(x)$) due to the non-zero shift $x_{0}$ of the peak density
away from $x=0$. However, time-reversal symmetry demands a real spatial wave-
function ($\psi^{*}(x)=\psi(x)$), and this is only true for $\beta=0$. In
order for the dynamical evolution to relax to the diagonal ensemble
prediction, it is crucial even at the theoretical level to perform quasi-
momentum averaging. The solution is to observe that the superposition
$\psi_{\beta}(x)+\psi_{-\beta}(x)$ is real, and thus the averaging can be
performed so long as the quasi-momentum distribution is symmetric.

##  Appendix B Time-Reversal Symmetry for Kicked State under Talbot Dynamics

Here we explicitly prove that the kicked initial state preparation indeed
leads to a time-reversal symmetric state after free-evolution by
$T_{\mathrm{Talbot}}/4$. For this analysis, we will ignore the role of
quasimomentum. Using the Jacobi-Anger expansion for the initial wavefunction
$\psi(x)=e^{-iV\sin x}$ described in the main text, we can decompose the
initial kicked state in a discrete momentum basis indexed by integers $l$ as

$\displaystyle\ket{\psi(t=0)}=\sum_{l=-\infty}^{\infty}(-1)^{l}\mathcal{J}_{l}(V)\ket{l},$
(11)

where $\mathcal{J}_{l}$ is the $l^{\mathrm{th}}$ Bessel function of the first
kind. The quarter-Talbot period free evolution is described by
$e^{-ip^{2}/2\mathchar 22\relax\mkern-9.0muk}$ with $\mathchar
22\relax\mkern-9.0muk=\pi$ and $p=\mathchar 22\relax\mkern-9.0mukl$. Noting
that $(-1)^{l}=(-1)^{l^{2}}$ we have

$\displaystyle\ket{\psi(t=T_{\mathrm{Talbot}}/4T)}=\sum_{l=-\infty}^{\infty}i^{l^{2}}\mathcal{J}_{l}(V)\ket{l}.$
(12)

The condition on the momentum coefficients for a time-reversal symmetric or
real position-space wavefunction is $c^{*}_{-l}=c_{l}$, where
$c_{l}=\bra{l}\ket{\psi}$. Using
$\mathcal{J}_{-l}(V)=(-1)^{l}\mathcal{J}_{l}(V)$, we straightforwardly confirm
this condition

$\displaystyle
c^{*}_{-l}=(-i)^{(-l)^{2}}\mathcal{J}_{-l}(V)=i^{l^{2}}\mathcal{J}_{l}(V)=c_{l}.$
(13)

Taken with the proof in appendix A, this supports the claim that the kicked
state should exhibit a boomerang effect for
$\alpha=1/2+T_{\mathrm{Talbot}}/4T$. We note that upon the inclusion of non-
zero quasimomentum, this argument is no longer exact; however in considering
corrections to this argument, one should keep in mind that the initial state
(11) considered in this analysis is already only an approximate description of
the experimental reality.

##  Appendix C Floquet Gauge Symmetry Properties of Late-Time Momentum

Here we prove that the infinite-time momentum is either an even or odd
function of $\alpha$ about 1/2 for states with the given symmetry properties
explored in the experiment. First we consider initial states with time-
reversal symmetry such that

$\displaystyle\mathcal{K}\ket{\psi(t=0)}=\ket{\psi(t=0)}.$ (14)

We note the relation

$\displaystyle\mathcal{K}^{\dagger}U_{\alpha}\mathcal{K}=e^{i(1-\alpha)p^{2}/2\mathchar
22\relax\mkern-9.0muk}e^{iK\cos x/\mathchar 22\relax\mkern-9.0muk}e^{i\alpha
p^{2}/2\mathchar 22\relax\mkern-9.0muk}=U_{1-\alpha}^{\dagger}.$ (15)

For compactness, we will henceforth denote the state after $n$ kicks in the
gauge $\alpha$ as $\ket{\psi_{n}}_{\alpha}\equiv
U_{\alpha}^{n}\ket{\psi(t=0)}$ ($\ket{\psi_{0}}=\ket{\psi(t=0)}$), and use
similar notation for expectation values. The momentum evolution in the gauge
$1-\alpha$ is then given by

$\displaystyle\langle
p_{n}\rangle_{1-\alpha}=\bra{\psi_{0}}U_{1-\alpha}^{n\dagger}\,p\,U_{1-\alpha}^{n}\ket{\psi_{0}}.$
(16)

Using (14), (15) and $\mathcal{K}^{\dagger}\,p\,\mathcal{K}=-p$, we find

$\displaystyle\langle p_{n}\rangle_{1-\alpha}$
$\displaystyle=\bra{\psi_{0}}\mathcal{K}^{\dagger}U_{1-\alpha}^{n\dagger}\mathcal{K}\mathcal{K}^{\dagger}p\mathcal{K}\mathcal{K}^{\dagger}U_{1-\alpha}^{n}\mathcal{K}\ket{\psi_{0}}$
(17)
$\displaystyle=-\bra{\psi_{0}}U_{\alpha}^{n}\,p\,U_{\alpha}^{n\dagger}\ket{\psi_{0}}=-\langle
p_{-n}\rangle_{\alpha}.$ (18)

That is, forward quench dynamics in the gauge $1-\alpha$ possesses an opposite
momentum evolution to backward evolution under the gauge $\alpha$. Note that
this is essentially a generalization of equation (9) to arbitrary $\alpha$;
evaluating this property at $\alpha=1/2$ in conjuction with (7) serves as an
alternative proof that the late-time zero momentum feature of the boomerang
effect only depends on simultaneous time-reversal symmetry of the Floquet
gauge and quenched state. Leaving $\alpha$ arbitrary, taking the limit
$n\to\infty$ and applying (7), we derive the odd function result for time-
reversal symmetric states

$\displaystyle\langle p_{\infty}\rangle_{1-\alpha}=-\langle
p_{\infty}\rangle_{\alpha}.$ (19)

Next we consider a state with the symmetry

$\displaystyle\mathcal{P}\mathcal{K}\ket{\psi(0)}=\ket{\psi(0)}.$ (20)

Replacing $\mathcal{K}\to\mathcal{P}\mathcal{K}$ in (17), making use of the
parity symmetry of the evolution (4) and noting that
$\mathcal{K}^{\dagger}\mathcal{P}^{\dagger}p\mathcal{P}\mathcal{K}=p$, we
instead find

$\displaystyle\langle p_{n}\rangle_{1-\alpha}=\langle
p_{-n}\rangle_{1-\alpha}.$ (21)

Taking the infinite-time limit in the same way as before, we arrive at the
desired even function result

$\displaystyle\langle p_{\infty}\rangle_{1-\alpha}=\langle
p_{\infty}\rangle_{\alpha}.$ (22)

Here we suggest an intuitive understanding for the difference between the two
cases. Whereas the backward evolution in $\alpha$ for a time-reversal
symmetric state only differs from the forward evolution in $1-\alpha$ by the
sign of the kicks, which leads to opposite sign momentum impulses and
subsequently (19), the backward free-evolution in $\alpha$ of the parity-time
reversal symmetric state produces an additional parity inversion with respect
to the forward free-evolution in $1-\alpha$ (for instance, see Fig. 3c),
cancelling the kick inversion to yield (22).