# Controlling the real-time dynamics of a spin coupled to the helical edge
states of the Kane-Mele model

Robin Quade  I. Institute of Theoretical Physics, Department of Physics,
University of Hamburg, Notkestraße 9-11, 22607 Hamburg, Germany  Michael
Potthoff  I. Institute of Theoretical Physics, Department of Physics,
University of Hamburg, Notkestraße 9-11, 22607 Hamburg, Germany  The Hamburg
Centre for Ultrafast Imaging, Luruper Chaussee 149, 22761 Hamburg, Germany

###### Abstract

The time-dependent state of a classical spin locally exchange coupled to an
edge site of a Kane-Mele model in the topologically non-trivial phase is
studied numerically by solving the full set of coupled microscopic equations
of motion for the spin and the electron system. Dynamics in the long-time
limit is accessible thanks to dissipative boundary conditions, applied to all
but the zigzag edge of interest. We study means to control the state of the
spin via transport of a spin-polarization cloud through the helical edge
states. The cloud is formed at a distant edge site using a local magnetic
field to inject an electron spin density and released by suddenly switching
off the injection field. This basic process, consisting of spin injection,
propagation of the spin-polarization cloud, and scattering of the cloud from
the classical spin, can be used to steer the spin state in a controlled way.
We find that the effect of a single basic process can be reverted to a high
degree with a subsequent process. Furthermore, we show that by concatenating
several basic injection-propagation-scattering processes, the spin state can
be switched completely and that a full reversal can be achieved.

##  I Introduction

Topological quantum matter Hasan and Kane (2010); Qi and Zhang (2011); Ando
(2013); Teo and Kane (2010); Chiu  _et al._ (2016); Schnyder  _et al._ (2008)
has proven to be a fascinating concept not only because of its mathematical
beauty but also because it paves the way for novel applications. In
particular, as a consequence of the bulk-boundary correspondence, quantum
matter in a topologically non-trivial state is characterized by the existence
of topologically protected states, which are localized at the boundaries of a
sample. These edge states are accessible by local or surface-sensitive
experimental probes and feature unique physical properties. Their robustness
against certain perturbations is highly attractive for the development of
devices with new functionalities.

Symmetries are essential for the classification of topological quantum matter
Chiu  _et al._ (2016); Teo and Kane (2010); Schnyder  _et al._ (2008).
Accordingly, the topological edge modes are distinguished by their protection
against local and symmetry-preserving perturbations. This robustness property
has been exploited for various suggested applications.

A prime example is quantum computation Kane (1998); Lahtinen and Pachos
(2017), employing, e.g., fractional quantum-Hall states Nayak  _et al._ (2008)
or Majorana zero modes Sarma and Nayak (2015), where topological protection
and corresponding robustness is a decisive issue. In the field of spintronics
Wolf  _et al._ (2001), examples are given by one-dimensional spin transport in
inverted-gap semiconductor-based devices Akhmerov  _et al._ (2009), or by
spin-dependent reflection with control of the spin rotation in trilayer
junctions consisting of quantum-spin Hall (QSH) and metallic materials
Yokoyama  _et al._ (2009). The QSH effect can be utilized to create nearly
fully spin-polarized charge currents, controlled via magnetic defects Van Dyke
and Morr (2016), and all-electrical routes have been suggested to manipulate
the spin of a magnetic adatom at the edge of a QSH insulator Narayan  _et al._
(2013); Hurley  _et al._ (2013).

Our present study addresses the helical edge states of a time-reversal (TR)
symmetric two-dimensional topological insulator and concentrates on the Kane-
Mele (KM) model as a prototype Kane and Mele (2005a, b). Topologically non-
trivial properties of the KM model originate from its TR symmetric spin-orbit
coupling term. In particular, this induces the QSH effect. The model has
originally been proposed for graphene Kane and Mele (2005a) but turned out to
be more relevant for quantum-well systems Bernevig  _et al._ (2006); König
_et al._ (2007). It can also be understood to describe a class of graphene-
like two-dimensional monolayer honeycomb materials that feature significant
spin-orbit interaction, such as silicene and related systems Liu  _et al._
(2011); Ezawa (2015). It has recently been shown that an interacting KM model
emerges as an effective low-energy theory in stacked 1T-TaSe2 bilayers Pizarro
_et al._ (2020). Quite generally, the KM model represents a paradigmatic model
for two-dimensional class-AII topological insulators with phases characterized
by a $\mathbb{Z}_{2}$ index and protected by time-reversal symmetry (TRS).

It is quite natural to probe TRS protected topological states of
$\mathbb{Z}_{2}$ insulators by means of TRS breaking local perturbations Chang
_et al._ (2014). This idea has been pursued in various studies of TR symmetric
topological systems, such as Bi2Te3, Bi2Se3, Sb2Te3, by doping with magnetic
transition-metal atoms Yu  _et al._ (2010); Chen  _et al._ (2010) or by
depositing magnetic adatoms at the surface Eelbo  _et al._ (2014); Vondráček
_et al._ (2016). Locally breaking TRS may lead to rather exotic phenomena such
as an image magnetic monopole Qi  _et al._ (2009).

Especially interesting is the interaction between two magnetic adatoms that is
mediated by the helical edge states. In the limit of a weak exchange coupling
$J$ between adatoms and substrate, standard concepts of RKKY theory Kit  can
be adapted to tight-binding or continuum models for helical QSH boundary
states. In the vicinity of a classical magnetic impurity, the local (spin)
density of states is suppressed at low energies Liu  _et al._ (2009). The RKKY
interaction between two impurities becomes ferromagnetic for a chemical
potential such that the Fermi wavelength is much larger than the impurity-
impurity distance. In general the coupling is noncollinear and in-plane with a
power-law spatial decay Gao  _et al._ (2009). There is an additional
Kurilovich  _et al._ (2017) Bloembergen-Rowland-type Bloembergen and Rowland
(1955) bulk contribution decaying exponentially with distance. A weakly broken
TRS gaps out the Dirac cone and induces a strongly anisotropic RKKY coupling
with Dzyaloshinsky–Moriya and with both, in-plane and out-of-plane Ising
terms, decaying exponentially with the distance between the impurities for a
chemical potential within the gap Hosseini  _et al._ (2020). Effects of strong
electron interaction and the Doniach competition of indirect exchange with
Kondo screening in a helical Luttinger liquids limit the applicability of RKKY
theory Yevtushenko and Yudson (2018).

The real-time dynamics of magnetic impurities at surfaces of topological
insulators has been studied to a lesser extent. Recent studies have employed
time-dependent density-functional theory Bouaziz  _et al._ (2019) and, for
periodically driven impurities, Floquet theory Pradhan and Fransson (2019).
The effects of a single TRS-breaking magnetic impurity on the transport
properties of a KM zigzag ribbon mediated by the helical edge states have been
studied within scattering theory Tatsumi  _et al._ (2020). The long-time
dynamics of a single classical spin exchange coupled to the edge of a Su-
Schrieffer-Heger model has been investigated recently Elbracht and Potthoff
(2021). Apart from the latter, the full microscopic real-time dynamics beyond
the linear-response approach Bajpai and Nikolic (2019); Onoda and Nagaosa
(2006); Bhattacharjee  _et al._ (2012) has not yet been addressed so far.

In the present work we numerically study the real-time dynamics of a classical
“read-out” spin that is exchange coupled to the local spin of the electron
system at a site of a zigzag edge of the KM model on a honeycomb lattice. The
dynamical state of the read-out spin is affected by another impurity at a
distant site on the same edge, which is used to inject a local spin
excitation. The transport of the injected spin density through the helical
edge state and its impact on the classical spin is microscopically traced as
function of time. Note that local breaking of TRS due to the read-out impurity
spin enables backscattering Maciejko  _et al._ (2009); Tanaka  _et al._ (2011)
of the transported spin density and thus allows driving of the read-out spin
dynamics. Our goal is to fully control the state of the classical spin by
local excitations of the system over long distances making use of the
topological protection of the edge state. We demonstrate that this can be
achieved with a predefined precision by iterating the spin-injection and
transport process.

The studied setup is partially motivated by the further progress of
experimental techniques, e.g., in detecting states of magnetic adatoms
Wiesendanger (2009) and measuring indirect magnetic (RKKY) interactions on a
nanoscale Zhou  _et al._ (2010). Scanning-tunnelling microscopy with two or
several tips Xu  _et al._ (2006); Kolmer  _et al._ (2017); Hus  _et al._
(2017); Voigtländer  _et al._ (2018), with a separation down to the
nanometers, ideally with magnetic tips and spin resolution, would be perfectly
suited to initiate, probe and control spin-momentum-locked transport. Our
study aims at an improved understanding of manipulation of local magnetic
states through topological surface states in the time domain. As time-
dependent STM techniques van Houselt and Zandvliet (2010); Tian  _et al._
(2018) address the $\mu$s rather than the ps time regime, however, we focus on
the initial and the final-state spin configurations.

A numerically exact solution of the coupled set of equations of motion for the
classical read-out spin and for the dynamics of the entire electronic system
can only be achieved for a lattice of finite size. Here, we demonstrate that a
ribbon-shaped geometry with only 4 unit cells in the direction perpendicular
to the zigzag edge and with about 100 sites along the zigzag edge is fully
sufficient to achieve propagation times of the order of $10^{3}$ inverse
nearest-neighbor hoppings (and more) and thus allows for a complete monitoring
of the injection-transport-driving process. This, however, requires special
boundary conditions Elbracht and Potthoff (2020) for the armchair edges and
for the opposite zigzag edge. Namely, the boundaries must fully absorb the
residual propagating excitations to avoid reflections. As has been shown
recently for a spin coupled to a one-dimensional topological chain Elbracht
and Potthoff (2021), this can be achieved with properly modified Lindblad-type
edge potentials. This scheme is adapted here and applied to the considered
ribbon geometry. It allows us to easily achieve the required propagation times
without any unwanted interference effects and without any modification of the
physical real-time dynamics in the core of the system.

The rest of the paper is organized as follows: The next sections II and III
introduce the model and our approach to compute time-dependent observables. In
section IV we specify the geometry in detail and discuss a static spin
injection and the subsequent propagation of the spin-polarization cloud.
Scattering of the polarization cloud and its impact on the read-out spin are
addressed in Secs. V and VI. A basic process that consists of a dynamic spin-
injection and subsequent pumping of the read-out spin can be reverted or
iterated, as described in Sec. VII. Section VIII summarizes our findings and
gives a brief outlook.

##  II Kane-Mele $s$-$d$ model

Using standard notations, the Hamiltonian of the Kane-Mele model Kane and Mele
(2005a, b) is given by:

$\displaystyle H_{\rm KM}=t_{\rm
hop}\sum_{\braket{i,j},\alpha}c_{i\alpha}^{\dagger}c_{j\alpha}+it_{\text{so}}\sum_{\braket{\braket{i,j}},\alpha,\beta}\nu_{ij}c_{i\alpha}^{\dagger}\sigma_{\alpha\beta}^{(z)}c_{j\beta}$
$\displaystyle+V\sum_{i,\alpha}\epsilon_{i}c_{i\alpha}^{\dagger}c_{i\alpha}\>.$
(1)

Here, $i$ and $j$ label the sites of a honeycomb lattice,
$\alpha,\beta=\uparrow,\downarrow$ is the spin projection, $\braket{.\,,.}$
and $\braket{\braket{.\,,.}}$ indicate summation over nearest or next-nearest
neighbors, respectively, and $\sigma^{(z)}$ is the $z$-Pauli matrix. The spin-
diagonal nearest-neighbor hopping amplitude $t_{\rm hop}\equiv 1$ sets the
energy unit and with $\hbar\equiv 1$ also the time unit. The anisotropic spin-
orbit coupling (SOC), responsible for the helicity of the topological boundary
states, is modeled as a spin-dependent next-nearest neighbor hopping with a
sign factor $\nu_{ij}=\pm 1$, which is positive (negative) for anticlockwise
(clockwise) hopping $j\to i$ within a hexagon of the lattice. The strength of
the SOC is controlled by the amplitude $it_{\rm so}$ which is purely imaginary
($t_{\rm so}>0$). In combination with the sign factor this ensures hermiticity
of the SOC term. Finally, $V>0$ is the strength of a sublattice-parity-
breaking ionic potential, which includes a sign $\epsilon_{i}=+1(-1)$ if site
$i$ belongs to sublattice $A$ ($B$). The $V$ term is used to tune the ground
state of the system between topologically distinct phases. For given $t_{\rm
so}$, we choose

$V=3\sqrt{3}t_{\rm so}\pm 0.5\Delta$ (2)

to produce a band structure with a desired (small) bulk band gap $\Delta>0$.
For positive (negative) sign this yields the topologically trivial (non-
trivial) phase. This choice of $V$ has the convenient feature that the bulk
gap is the same in both phases.

Each term of the KM model is invariant under TR, i.e., its Hamiltonian
commutes with the antiunitary TR operator
${\Theta}=e^{-i\pi{S}_{y}}{\mathcal{K}}$ where ${S}_{y}$ is the $y$-component
of the total spin and ${\mathcal{K}}$ is the complex conjugation with respect
to the spinful orbital basis, such that
${\Theta}c_{i\alpha}^{(\dagger)}{\Theta}=i\sum_{\beta}\sigma_{\alpha\beta}^{(y)}c_{i\beta}^{(\dagger)}$.
In the single-electron subspace, ${\Theta}$ squares to $-{\mathds{1}}$. This
entails Kramers degeneracy of the single-particle eigenstates of $H_{\rm KM}$.

The two topologically distinct bulk phases of the KM model are distinguished
by a topological $\mathbb{Z}_{2}$ invariant $\nu$, with $\nu=0$ and $\nu=1$
referring to the topologically trivial and nontrivial phases, respectively. It
can be defined as Kane and Mele (2005a, b)

$(-1)^{\nu}=\prod_{\Gamma_{i}\in{\boldsymbol{\Gamma}}}\frac{\sqrt{\det\omega(\Gamma_{i})}}{\text{Pf
}\omega(\Gamma_{i})}$ (3)

in terms of the set of TR symmetric invariant momenta ${\boldsymbol{\Gamma}}$
and the $2\times 2$ TRS “scattering” matrix Kaufmann  _et al._ (2016)

$\omega_{rs}(\Gamma_{i})=\braket{u_{r}(\Gamma_{i})}{\Theta}{u_{s}(\Gamma_{i})}\>.$
(4)

where the indices $r,s=1,2$ label the two Bloch states
$\ket{u_{1}(\mathbf{k})}$, $\ket{u_{2}(\mathbf{k})}$ of the occupied Kramers
pair. The bulk-boundary correspondence of the KM model is expressed as
$\nu=N_{\rm K}\;\text{mod }2$, where $N_{\rm K}$ is the number of Kramers
boundary pairs, i.e., the number is even (odd) in the topologically trivial
(non-trivial) phase. The topological edge states of the model are helical,
meaning that the two spin species are interlocked with opposite momenta and
hence opposite propagation directions. Note that helical edge states are
“immune” to TRS-preserving perturbations but scatter from TRS-breaking
perturbations Maciejko  _et al._ (2009); Tanaka  _et al._ (2011).

We consider the KM model in a nanoribbon geometry with topological zigzag
edges (see Sec. IV below). At a site $R$ of a zigzag edge, a classical (“read-
out”) spin ${\boldsymbol{S}}_{\rm R}$ of fixed length is exchange coupled to
the local spin ${\boldsymbol{s}}_{\rm R}=\frac{1}{2}\sum_{\alpha\beta}c_{{\rm
R}\alpha}^{\dagger}{\boldsymbol{\sigma}}_{\alpha\beta}c_{{\rm R}\beta}$ of the
electron system. Here, ${\boldsymbol{\sigma}}$ is the vector of Pauli
matrices. This perturbation, $J{\boldsymbol{S}}_{\rm R}{\boldsymbol{s}}_{\rm
R}$, locally breaks TRS opposed to, e.g., a Kondo coupling to a quantum-spin
$1/2$. The classical spin is a proper way to describe a “magnetic” adatom with
a well-formed spin moment that is stable on a time scale exceeding all other
relevant time scales of the system.

The classical spin ${\boldsymbol{S}}_{\rm R}$ is susceptible to a spin-density
excitation propagating along the edge through gapless helical edge states, the
presence of which is enforced in case of a topologically non-trivial phase of
the bulk electron system. Local breaking of TRS allows the spin excitation to
scatter from the impurity spin and thereby to exchange a spin torque that
drives the dynamics of ${\boldsymbol{S}}_{\rm R}$. Due to the topological
protection and due to the helicity of the edge states the electron-spin
density is transported robustly and unidirectionally.

Locally initializing a spin excitation in the electron system that is confined
to the zigzag edge likewise requires another TRS-breaking local perturbation.
This could be achieved by means of a second magnetic adatom with a spin moment
that is externally driven, e.g., by means of a spin-polarized STM tip. For
simplicity, we will here model the spin-injection process by a local magnetic
field ${\boldsymbol{B}}_{\rm I}$ which couples to the local electron spin at a
site $I$ and which is suddenly switched on and off to induce real-time
dynamics.

The Hamiltonian of the total system, consisting of the Kane-Mele electron
model and the two local perturbations, the exchange-coupled read-out spin
${\boldsymbol{S}}_{\rm R}$ at edge site $R$ and the local magnetic field
$B_{\rm I}$ at edge site $I$, then reads:

$H=H_{\rm KM}+J{\boldsymbol{S}}_{\rm R}{\boldsymbol{s}}_{\rm
R}-{\boldsymbol{B}}_{\rm I}{\boldsymbol{s}}_{\rm I}\>.$ (5)

$J$ is the strength of the local exchange interaction. $H$ represents a two-
impurity $s$-$d$-type model Zener (1951); V+  with a possibly non-trivial bulk
topological ground state.

##  III Real-Time Dynamics

Our ambition is to fully determine the microscopic real-time dynamics of the
system, Eq. (5). To this end we introduce the one-particle reduced density
matrix ${\boldsymbol{\rho}}$ with elements
$\rho_{i\alpha,j\beta}=\langle\Psi(t)|c_{j\beta}^{\dagger}c_{i\alpha}|\Psi(t))\rangle$,
where $|\Psi(t)\rangle$ is the $N$-particle quantum state of the electron
system at time $t$ which implies $\mbox{tr}{\boldsymbol{\rho}}=N$. We consider
a half-filled system, where $N=L$ is given by the number of lattice sites $L$.
For the quantum-classical hybrid system Elze (2012); Yang and Hirschfelder
(1980), there is a closed system of equations of motion Sayad and Potthoff
(2015), consisting of a Landau-Lifschitz-type equation,

$\frac{d{\boldsymbol{S}}_{\rm R}}{dt}=J\braket{{\boldsymbol{s}}_{\rm
R}}\times{\boldsymbol{S}}_{\rm R}\>,$ (6)

with $\langle{\boldsymbol{s}}_{\rm
R}\rangle=\frac{1}{2}\sum_{\alpha\beta}\rho_{R\alpha,R\beta}{\boldsymbol{\sigma}}_{\beta\alpha}$
for the read-out spin and a von Neumann-type equation,

$i\frac{d{\boldsymbol{\rho}}}{dt}=[{\boldsymbol{T}}_{\rm
eff},{\boldsymbol{\rho}}]\>,$ (7)

for the density matrix. Here, ${\boldsymbol{T}}_{\rm eff}$ is the effective
hopping matrix with elements given by

$T_{\rm
eff,i\alpha,j\beta}=T_{i\alpha,j\beta}+\delta_{iR}\delta_{jR}J\frac{1}{2}{\boldsymbol{\sigma}}_{\alpha\beta}{\boldsymbol{S}}_{\rm
R}-\delta_{iI}\delta_{jI}\frac{1}{2}{\boldsymbol{\sigma}}_{\alpha\beta}{\boldsymbol{B}}_{\rm
I}\>,$ (8)

and ${\boldsymbol{T}}$ is the bare hopping matrix of the unperturbed Kane-Mele
model, Eq. (1):

$\displaystyle T_{i\alpha,j\beta}=t_{\rm
hop}\delta_{\braket{i,j}}\delta_{\alpha\beta}+it_{\rm
so}\delta_{\braket{\braket{i,j}}}\nu_{ij}\sigma_{\alpha\beta}^{(z)}+V\epsilon_{i}\delta_{ij}\delta_{\alpha\beta}\>.$
(9)

Using standard methods for systems of ordinary differential equations, this
can be solved conveniently for finite systems with up to $L\approx 10^{3}$
sites Sayad and Potthoff (2015); Stahl and Potthoff (2017).

Initiating the real-time dynamics by a local time-dependent perturbation,
causes excitations propagating as a wave packet – predominantly along the
edge. The propagation speed is roughly given by the Fermi velocity $v_{\rm
F}=d\varepsilon_{\sigma}(k=k_{\rm F})/dk$ of the two edge states, where
$\varepsilon_{\rm F}=0$ is the Fermi energy. Fig. 1 gives an example. It
displays the bulk bandstructure of the unperturbed model, Eq. (1), projected
onto a zigzag edge. Parameters (see caption) are chosen such that the
topologically non-trivial phase is realized. We can read off $v_{\rm
F}\approx\pm 0.285$ in good agreement with $v_{\rm F}\approx\pm 0.286$ as
obtained from an analytical expression given in Ref. Cano-Cortés  _et al._
(2013)

Figure 1:  Bandstructure of the Kane-Mele model, Eq. (1), projected onto a
zigzag edge. Gray area: projected bulk bands as obtained from a calculation
for a ribbon geometry with 20 unit cells in the normal direction. Red (green)
line: dispersions $\varepsilon_{\uparrow}(k)$ ($\varepsilon_{\downarrow}(k)$)
of the two helical edge states outside the bulk continuum (edge states
localized at the opposite edge are not displayed). $t_{\rm hop}=1$ sets the
energy scale. Calculation for $t_{\rm so}=0.05$, band gap $\Delta=0.3$ and $V$
as obtained from Eq. (2) for the negative sign.

Addressing the full microscopic real-time dynamics requires a system of finite
(but large) size. In case of open boundary conditions, an initially excited
wave packet propagating along the edge with velocity $\pm v_{\rm F}$ at a
distance $d$ from one of the corners will be reflected after a time $t_{\rm
refl.}\approx d/|v_{\rm F}|$, back-propagate, and finally lead to
interferences at the position of the impurity spin. Such unwanted finite-size
effects can be suppressed almost completely with the help of absorbing
boundary conditions that have been introduced and tested in Ref. Elbracht and
Potthoff (2020). This allows us to study real-time dynamics on long time
scales even for moderate system sizes, see Sec. VII. Following Ref. Elbracht
and Potthoff (2020) we therefore replace Eq. (7) by

$\frac{d{\boldsymbol{\rho}}(t)}{dt}=-i[{\boldsymbol{T}}_{\rm
eff}(t),{\boldsymbol{\rho}}(t)]-\\{{\boldsymbol{\gamma}},{\boldsymbol{\rho}}(t)-{\boldsymbol{\rho}}(0)\\}\;,$
(10)

where $\\{\cdot,\cdot\\}$ denotes the anticommutator. This equation contains
an additional boundary term involving a diagonal and non-negative matrix
${\boldsymbol{\gamma}}$ with elements

$\gamma_{i\alpha,j\beta}=\delta_{ij}\delta_{\alpha\beta}\gamma_{i\alpha}\>.$
(11)

$\gamma_{i\alpha}$ is nonzero for sites $i$ in a thin “Lindblad shell” at the
system boundaries, except, of course, for the “physical” zigzag edge of
interest. In our case, see Fig. 2, the sites of the Lindblad shell are those
of the three (“unphysical”) edges of the ribbon coupling to the Lindblad bath
shown in red color.

In fact, Eq. (11) is derived by starting from the general Lindblad equation
Lindblad (1976); Pearle (2012) but allowing only sites in the Lindblad shell
close to the boundaries to couple to the environment. In a next step the
Lindblad theory is specialized to a system of noninteracting electrons. This
results in an equation of motion involving the one-particle reduced density
matrix only, rather than the many-body statistical operator. Finally, and most
importantly, an additional term $\propto\rho(0)$ is incorporated which is
missing in standard Lindblad-based computations and which is necessary to
avoid artificial excitations initially generated due to the coupling to the
bath. With a standard Lindblad bath extended by this extra term, reflections
of physical excitations from the boundaries as well as initial-state artifacts
can be suppressed very efficiently. This has been demonstrated recently
Elbracht and Potthoff (2020, 2021) to work well for one-dimensional tight-
binding systems with classical-spin impurities. The approach describes time
evolution respecting total-probability conservation. Energy and angular
momentum (spin), on the other hand, are only conserved locally but dissipated
at the boundaries.

Here, we adapt the method to a two-dimensional nanoribbon geometry to
construct absorbing boundaries on the zigzag edge opposite to the edge of
interest as well as on the armchair boundaries. For the geometrical setup and
the propagation time scales discussed in the sections below, it has turned out
that a uniform shell of unit thickness and a spin-independent
${\boldsymbol{\gamma}}$-matrix is fully sufficient. This means that only a
single scalar parameter $\gamma$ must be fixed. This parameter controls the
rate of dissipation at the absorbing boundaries and is set $\gamma=0.2$
throughout this study, following Refs. Elbracht and Potthoff (2020, 2021). By
comparing, for sufficiently short time scales, with the open-boundary dynamics
[Eq. (7)], we have carefully checked that the presence of absorbing Lindblad
sites with $\gamma=0.2$ does not affect the physical dynamics in the core
region of interest, i.e., on the zigzag edge to which the impurity terms in
Eq. (5) are coupled.

##  IV Ribbon geometry, spin injection and propagation

We consider the Kane-Mele model in a ribbon geometry as shown in Fig. 2. The
numerical effort for computing the full real-time dynamics of the electronic
system asymptotically scales as $L^{2}$ for a large total number of lattice
sites $L$. Computations for systems with $L\approx 500$ are convenient. For
the calculations we choose a ribbon geometry with $L=572$ sites. The short
side of the ribbon along the armchair ($y$-) direction extends over four unit
cells consisting of two sites each, while along the $x$-direction the ribbon
extends over 71.5 unit cells, as displayed in Fig. 2.

Figure 2:  Sketch of the Kane-Mele model in a ribbon geometry. Green (blue)
dots: sites of sublattice A (B). Red dots: edge sites with coupling to a
Lindblad bath (thick dashed red lines). Yellow (orange) symbol $\odot$
($\otimes$): out-of-plane magnetic field ${\boldsymbol{B}}_{\rm I}$ in
positive (negative) $z$-direction coupled (dotted black line) to an “injection
site” $I$. Here, $I$ is chosen as the central site of the “physical” zigzag
edge (without Lindblad boundary). A finite ${\boldsymbol{B}}_{\rm I}=+|B_{\rm
I}|{\boldsymbol{e}}_{z}$ (${\boldsymbol{B}}_{\rm I}=-|B_{\rm
I}|{\boldsymbol{e}}_{z}$) induces a finite spin-up (spin-down) $z$-polarized
spin density. In the spin-up case, the excitation propagates to the left
(yellow line with arrow at the bottom), once the injection field is switched
off. Vice versa, a spin-down polarized density propagates to the right (orange
line).

In the figure, the lower zigzag edge is the physical edge. Sites at the
remaining three edges are coupled to the Lindblad bath, as discussed in Sec.
III above and as shown in red color. Considering local time-dependent
perturbations at the physical edge, this setup almost perfectly simulates the
real-time dynamics of a half-infinite sample. Note that the helical edge
states are exponentially localized on the zigzag edge. We have checked that
with four unit cells in $y$-direction, the overlap of edge states with edge
states on the opposite zigzag edge, which are perturbed by the Lindblad
boundary condition, is in fact negligible. Furthermore, on the physical edge,
the edge states are predominantly localized on the A-sublattice sites, such
that it is advantageous to couple the two B-sublattice edge sites to the
Lindblad bath. This motivates the choice of an extra half unit cell in
$x$-direction. At the other edges both, B- and A-sublattice sites, are coupled
to the Lindblad bath.

Evidently, a larger system generally comes with reduced finite-size artifacts.
We have verified that the results for the real-time dynamics presented in the
following are robust against both, shape-preserving and shape-altering
rescaling of the geometry, mainly by comparing with results obtained for
smaller $L$.

We start the discussion of the results with a static spin injection process.
To this end, we consider the Kane-Mele ribbon with an additional local
magnetic field ${\boldsymbol{B}}_{\rm I}$ pointing out-of-plane, i.e., in
$+z$\- or ($-z$)-direction, see yellow (orange) $\odot$ ($\otimes$) symbols in
Fig. 2, and coupling locally to an “injection site” $I$ of the physical zigzag
edge. We first choose $I$ as the central site on this edge.

Figure 3:  Snapshots of the spatial distribution of spin-polarization
$\braket{{s}_{i}^{z}(t)}$ at selected instants of time (colors). At $t=0$ the
spin-injection field ${\boldsymbol{B}}_{\rm I}$ at the central site $I=36$ of
the zigzag edge ($A$-sublattice edge sites are enumerated from $i=1$ to
$i=71$) has suddenly been been switched off and the excitation is released.
Upper part with $\braket{{s}_{i}^{z}(t)}>0$: spin-up injection. Lower part
with $\braket{{s}_{i}^{z}(t)}<0$: spin-down injection. The time unit is set by
the inverse hopping parameter $t_{\rm hop}^{-1}=1$ ($\hbar\equiv 1$). Further
parameters: $t_{\rm so}=0.05$, $\Delta=0.3$ (as in Fig. 1), $V$ as obtained
from Eq. (2) with negative sign, initial field strength $B_{\rm I}=1$,
Lindblad parameter $\gamma=0.2$.

The pure Kane-Mele nanoribbon without the injection field, Eq. (1), is time-
reversal symmetric. Without the SOC term, the Hamiltonian is additionally
invariant under global $SU(2)$ rotations generated by the total electron spin.
The SOC introduces an anisotropy and reduces this symmetry to a $U(1)$
rotational symmetry of the Kane-Mele Hamiltonian around the $z$-axis. The
ground state, however, is an unpolarized Fermi sea of the form
$|\mbox{g.s.}\rangle=\prod_{k\leq k_{\rm
F}}c^{\dagger}_{k\uparrow}c^{\dagger}_{k\downarrow}|\mbox{vac.}\rangle$, since
the total-spin $z$-component is conserved. Thus, the ground state is a non-
degenerate SU(2)-symmetric spin singlet. Any additional local magnetic field
${\boldsymbol{B}}_{\rm I}$ reduces the SU(2) symmetry of the ground state to a
U(1) symmetry, but the ground-state energy is independent of the direction of
the field. Contrary, the U(1) symmetry of the Hamiltonian is broken by any
field with a nonzero $z$-component. Here, we choose ${\boldsymbol{B}}_{\rm
I}=\pm B_{\rm I}{\boldsymbol{e}}_{z}$. In this case, we have a $U(1)$ spin-
rotation invariance around the $z$-axis for both, the Hamiltonian and its
ground state. A field in $z$-direction induces a spin-up (spin-down)
polarization of the local magnetic moments in the vicinity of site $I$, which
we expect to show unidirectional (spin-momentum-locked) propagation.

Calculations are performed for $t_{\rm hop}=1$, $t_{\rm so}=0.05$,
$\Delta=0.3$, and with $V$ fixed by Eq. (2) for the topologically non-trivial
case, as in Fig. 1. The field strength is set to $B_{\rm I}=1$ and the
Lindblad parameter to $\gamma=0.2$. The resulting ground-state local moment at
site $I$ amounts to $\braket{{s}_{I}^{z}(t=0)}\approx 0.33<0.5$, i.e., the
local moment is not fully polarized. Fig. 3, for time $t=0$, shows the entire
ground-state polarization cloud induced by ${\boldsymbol{B}}_{\rm I}=+B_{\rm
I}{\boldsymbol{e}}_{z}$ (see upper half, purple data). Data for
${\boldsymbol{B}}_{\rm I}=-B_{\rm I}{\boldsymbol{e}}_{z}$ (shown in the lower
half) differ by the sign of $\braket{{s}_{i}^{z}(t=0)}$ only.

Figure 4:  The same as Fig. 3 but for the topologically trivial case with
$V=3\sqrt{3}t_{\rm so}+0.5\Delta$, see Eq. (2).

To release the spin excitation, we suddenly switch off the field at time
$t=0$, i.e., ${\boldsymbol{B}}_{\rm I}=0$ for $t>0$. Right after the
injection, at $t=0$, the polarization cloud is spread over about 5 sites in
both cases. In the course of time, it continuously broadens and spreads over
about 10 sites at $t=100$ inverse hoppings after the injection. The results
for spin-up and spin-down injection are completely identical apart from the
important fact that the spin-up excitation mainly propagates to the left,
while the spin-down excitations moves to the right along the zigzag chain.
This is exactly the effect of spin-momentum locking in the topologically non-
trivial state. We also note that at early times (of the order of about 10
inverse hoppings) the total weight of the cloud is approximately halved. This
is attributed to the fact that about a half of the injected spin density is
carried by bulk states and thus immediately transported away from the edge and
dissipated into the bulk.

This interpretation is corroborated by a calculation for a different strength
of the ionic potential, where in Eq. (2) the positive sign is used, i.e.,
$V=3\sqrt{3}t_{\rm so}+0.5\Delta$. This yields the same size of the bulk gap
$\Delta$ but the topologically trivial state. Fig. 4 displays the according
results. In this case one finds a much less developed polarization cloud with
about a factor of two smaller local moment $|\braket{{s}_{I}^{z}(t)}|=0.16$
right after the injection. The propagation of both, a spin-up and a spin-down
excitation, is completely identical. The polarization cloud at the edge sites
does not propagate significantly along the edge but essentially ceases to
exist after about 10 inverse hoppings. In fact, almost the whole weight is
immediately dissipated into the bulk of the system.

The data for the topologically non-trivial state in Fig. 3 also prove that
after the early dissipation of the spin-density portion carried by bulk
states, the total weight of the cloud almost stays at a constant value. This
is as expected, since after the early stage almost the whole weight of the,
say, spin-up excitation is carried by the respective helical edge state only.
Furthermore, the speed of the spin-up wave packet moving to the left amounts
to $v\approx 0.29$ sites per inverse hopping, as can be read off from the peak
maxima. This almost perfectly matches with the Fermi velocity $v_{\rm
F}\approx 0.285$ of the edge state (see discussion of Fig. 1). Interestingly,
there is a low-weight peak visible for $t=20$ around site $i=43$ in the spin-
up case (upper part of the figure) found to the right of the injection site
$I=36$ and moving further to the right (see $t=40$). This is due to the
locally broken TRS which thus undermines spin-momentum locking to some extent.

##  V Spin injection and propagation in the presence of the read-out spin

Figure 5:  The same as Fig. 2 but with an additional classical “read-out spin”
${\boldsymbol{S}}_{\rm R}$ (black arrow), antiferromagnetically exchange
coupled (wavy black line) to the local spin of the electron system at site $R$
of the “physical” zigzag edge. $R$ is chosen to be the central site of the
edge. The injection field ${\boldsymbol{B}}_{\rm I}$ couples to the edge site
$I$ half way between $R$ and one of the corners. In case of
${\boldsymbol{B}}_{\rm I}=+B_{\rm I}{\boldsymbol{e}}_{z}$, the induced a spin-
up polarized density propagates to the left (yellow line) and scatters from
${\boldsymbol{S}}_{\rm R}$, if $I$ is located to the right of $R$. Vice versa
for a spin-down polarized density: scattering if $I$ is located to the left of
$R$.

In the next step, the discussion of spin injection and propagation is extended
to the full setup sketched in Fig. 5. In addition to the local injection
field, a classical read-out spin ${\boldsymbol{S}}_{\rm R}$ with $S_{\rm
R}\equiv|{\boldsymbol{S}}_{\rm R}|=1/2$ is exchange coupled
antiferromagnetically at the central site $R$ of the physical zigzag edge. We
choose a generic coupling strength $J=2$ such that $J|{\boldsymbol{S}}_{\rm
R}|=1$. The injection field is placed at a site $I$ half way between $R$ and
one of the corners.

We again consider a static spin injection and fix the injection field aligned
to the $z$-axis, ${\boldsymbol{B}}_{\rm I}=\pm B_{\rm I}{\boldsymbol{e}}_{z}$.
Furthermore, the initial direction of the classical spin is assumed to be in-
plane, say, ${\boldsymbol{S}}_{\rm R}=S_{\rm R}{\boldsymbol{e}}_{x}$. The
motivation of this choice is to maximize the torque exerted on the read-out
spin by the propagating spin excitation that is released when switching off
${\boldsymbol{B}}_{\rm I}$.

First, we concentrate on the initial state of the electron system, which is
taken to be the ground-state Fermi sea for fixed ${\boldsymbol{B}}_{\rm I}$
and ${\boldsymbol{S}}_{\rm R}$. Such an initial state is “stressed”, i.e., it
differs from the total system ground state that also minimizes the total
energy with respect to the direction of ${\boldsymbol{S}}_{\rm R}$. The total
system ground state would be realized if both, ${\boldsymbol{S}}_{\rm R}$ and
${\boldsymbol{B}}_{\rm I}$, were lying in the $x$-$y$-plane enclosing a
possibly inter-impurity distance-dependent azimuthal angle $\Delta\phi(R)$.
Following Ref. Gao  _et al._ (2009) the angle is given by
$\Delta\phi(R)=\pi-\alpha(R)$, where $\alpha(R)=2R\varepsilon_{\rm F}/v_{\rm
F}$ with the inter-impurity distance $R$, the Fermi energy $\varepsilon_{\rm
F}$ and the Fermi velocity $v_{\rm F}$. Note that in our case $\Delta\phi(R)$
is finite since $\varepsilon_{\rm F}=0$ should be obtained for $L\to\infty$
only.

Here, with the injection field aligned to the $z$-direction, a total-system
excited state is prepared at $t=0$. Hence, switching off the injection field
for $t>0$ to release the polarization cloud in the vicinity of site $I$ leads
to a propagation of the spin density (as in Fig. 3), but additionally we
expect an immediate dynamics of the read-out spin as well.

Figure 6:  The same as Fig. 3 but for a spin-down injection at site $I=18$
left to the additional classical read-out spin at site $R=36$ coupled
antiferromagnetically with $J=2$. Initially, at time $t=0$, the electron
system is in its ground state for fixed ${\boldsymbol{B}}_{\rm I}=-B_{\rm
I}{\boldsymbol{e}}_{z}$ and ${\boldsymbol{S}}_{\rm R}=S_{\rm
R}{\boldsymbol{e}}_{x}$. For $t>0$, the injection field is swichted off,
${\boldsymbol{B}}_{\rm I}=0$. Other paramters: see Fig. 3.

Let us first concentrate on the dynamics of the injected spin-polarization
cloud as shown in Fig. 6. Initially, at time $t=0$, the polarization cloud,
although formed at site $I=18$ left to the read-out spin, is almost
indistinguishable from the cloud initially formed at $I=36$ for the spin-down
injection shown in Fig. 3, i.e., finite-size effects are negligible. For $t>0$
the spin density predominantly propagates to the right reflecting again spin-
momentum locking. This is worth mentioning, as the presence of the read-out
spin implies a broken TRS of the electronic Hamiltonian. However, one still
expects Maciejko  _et al._ (2009) helical transport to prevail as TRS is
broken only locally. Also for $t=20$ and $t=40$ the polarization cloud does
not show a sizeable perturbation caused by the read-out spin at site $R=36$.
The snapshot for $t=60$, on the other hand, does show a strongly deformed
cloud. Due to the locally broken TRS, the spin excitation strongly scatters
from the read-out spin and is seen to back-propagate to the left for $t=80$
and $t=100$, predominantly as a spin-up excitation. A portion of clearly lower
weight is transmitted through the classical-spin impurity and continues as a
spin-down excitation to the right.

In the vicinity of the read-out spin, the polarization cloud does not change
much for $t\geq 80$ (compare orange and red data points close to $R=36$). This
hints to a finite $z$-component of the read-out spin, which has trapped part
of the polarization cloud close to site $R$ and whose dynamics has almost come
to an end for $t=100$. The finite $z$-component of ${\boldsymbol{S}}_{\rm R}$
must result from the torque exerted on ${\boldsymbol{S}}_{\rm R}$ by the
polarization cloud during the scattering event. It must further compensate the
countertorque exerted by the read-out spin on the polarization cloud.

##  VI Impact on the read-out spin

This interpretation is in fact corroborated by looking at the dynamics of the
classical spin. Fig. 7 displays the trajectory of the classical read-out spin
on the classical “Bloch” sphere, starting from the point
${\boldsymbol{S}}_{\rm R}=S_{\rm R}\cdot(1,0,0)$ at $t=0$. The initial state
of the electron system and of the injection field is prepared as above, namely
a static spin injection with a finite ${\boldsymbol{B}}_{\rm I}=\pm B_{\rm
I}{\boldsymbol{e}}_{z}$ and a ground-state Fermi sea.

We first discuss a spin-down injection (${\boldsymbol{B}}_{\rm I}=-B_{\rm
I}{\boldsymbol{e}}_{z}$) for two different cases: For the “proper process” A,
the injection site $I$ is located to the left of the site $R$, to which the
spin coupled (see Fig. 5), so that the right-moving polarization cloud
eventually hits the spin, while for the “improper process” B, $I$ is located
right of $R$.

However, even in case B, there is a non-trivial real-time dynamics after
switching off ${\boldsymbol{B}}_{\rm I}$ for $t>0$. There are two reasons for
this. First, the electron ground state is stressed for the chosen directions
of ${\boldsymbol{B}}_{\rm I}$ and ${\boldsymbol{S}}_{\rm R}(t=0)$. Namely, the
local spins of two neighboring edge sites enclose an angle that, depending on
the distance between $I$ and $R$, weakly deviates from the angle in the
electronic ground state at ${\boldsymbol{B}}_{\rm I}=0$. This implies a spin
relaxation in the electron system, which starts immediately and leads to a
finite torque on ${\boldsymbol{S}}_{\rm R}$. However, the result is actually
small. As can be seen in Fig. 7 for case B, the spin moves a few degrees on
the Bloch sphere and mainly within the $x$-$y$-plane only, until, at $t\approx
50$, its dynamics changes qualitatively, and a finite out-of-plane ($z$-)
component builds up. Also in this second stage of the dynamics, the spin does
not move far. After 150 inverse hoppings, the spin is still located close to
its starting point, and there is no significant further dynamics detectable
for longer propagation times. The cause for this second stage of the dynamics
is different from the first, as this results from scattering of the low-weight
portion of the polarization cloud that has started from $I$ and moved towards
$R$ (while the overwhelming part of the cloud moves away from $R$, consistent
with spin-momentum locking). This scattering event only sets in after the
necessary time required to propagate from $I$ to $R$, here after $t\approx 50$
inverse hoppings. Its effect is weak as the weight of the excitation is small.

Figure 7:  The path traced out by the tip of the classical read-out spin on
its Bloch sphere during various processes A-D, all started with a static spin
injection, switching off the injection field ${\boldsymbol{B}}_{\rm I}$ at
$t=0$ and propagating the system’s state up to $t=150$. The read-out spin
${\boldsymbol{S}}_{\rm R}$ is located at the center $R=36$ of edge. In all
cases, it points in $+x$-direction initially (black dot). A: spin-down
injection at site $I=18$ left to $R$ (“proper process”). B: spin-down
injection at site $I=54$ right to $R$ (“improper process”). C and D: spin-up
injection at $I=18$ (improper) and $I=54$ (proper), respectively. Parameters
as in Fig. 6. Colors serve to guide the eye. The four arrows indicate the
direction of the respective processes.

In case A, the proper process, on the other hand, the second stage of the
dynamics must be much more effective, since now it is almost the full weight
of the polarization cloud that propagates towards site $R$ and, again in a
time window around $t\approx 50$, scatters from the spin. In fact, as can be
seen in Fig. 7, the trajectory of the spin for case A is almost identical to
that for case B at early times, where the dynamics is dominated by stress
relaxation in the electron system. In the second stage at later times, where
the spin experiences the torque of the scattering polarization cloud, is makes
substantial progress on the Bloch sphere and develops a sizable $z$-component
$S_{\rm R,z}=S_{\rm R}\sin\vartheta$ with $\vartheta\approx-0.36\cdot\pi/2$.

Cases C and D, obtained by starting with a spin-up injection, are perfectly
symmetric to A and B, see Fig. 7. Here, for ${\boldsymbol{B}}_{\rm I}=B_{\rm
I}{\boldsymbol{e}}_{z}$ but for unchanged initial direction of
${\boldsymbol{S}}_{\rm R}$, the initial electron ground state is stressed with
opposite helicity. This results in a sign change of the torque on
${\boldsymbol{S}}_{\rm R}$. Similarly, for the proper process D, the sign of
the torque is opposite to that of case A, as now it is a spin-up cloud that
scatters from the spin.

##  VII Dynamic spin injection and iterating the process

As the spin dynamics comes to a halt once the polarization cloud has passed by
and the spin torque has diminished, the system is, at least locally at the
edge, in a state close to its ground state, and thus one may start over with a
second process thereafter. Some obvious questions arise in this context: Can
we undo the rotation of the read-out spin by a subsequent second process? Can
we achieve a “complete” switching process ${\boldsymbol{S}}_{\rm
R}\to-{\boldsymbol{S}}_{\rm R}$ by iterating the process considered so far?
Can we undo the whole iterated process?

We start the discussion with a single additional “basic injection-pump” (BIP)
process. This BIP process requires (i) a spin injection, which, however, must
be treated fully dynamically, and (ii) the dynamical pump of the read-out
spin.

At, say, time $t=0$, the dynamic spin injection starts from an arbitrary
electronic state and an arbitrary state (direction) of the read-out spin
${\boldsymbol{S}}_{\rm R}$. One may assume, that initially the state of the
electron system is the ground state (for a given ${\boldsymbol{S}}_{\rm R}$),
but after the first or after several BIP processes, the system may not have
fully relaxed to its ground state before the next BIP process starts. At
$t=0$, the spin-injection field ${\boldsymbol{B}}_{\rm I}=\pm B_{\rm
I}{\boldsymbol{e}}_{z}$ is switched on. The spin-injection process lasts up to
a specified time $t=t_{\rm inj}$, when the injection field is switched off.

Figure 8:  Real-time evolution of the $z$-component $\braket{{s}_{i}^{z}(t)}$
of the local magnetic spin moment at site $I=18$ of the physical zigzag edge
during a spin-down injection process steered by an injection field
${\boldsymbol{B}}_{\rm I}$ in $-z$-direction and applied for $t_{\rm inj}=50$
inverse hoppings. Parameters for the topologically non-trivial case as in Fig.
6.

The magnitude of the final polarization decisively depends on the injection
time $t_{\rm inj}$. This is demonstrated with Fig. 8 for a spin-down injection
(${\boldsymbol{B}}_{\rm I}=-B_{\rm I}{\boldsymbol{e}}_{z}$). In this example
the dynamical spin injection starts from the electron ground state in the
presence of the classical spin pointing in $+x$-direction (see the setup shown
in Fig. 5 with $I=18$ and $R=36$). It shows the temporal evolution of the
local magnetic spin moment $\braket{{s}_{i}^{z}(t)}$ at the injection site
$i=I$. The results are basically independent of the choice of $I$, provided
that the distance to the corners and to $R$ is large enough. A few sites has
proven to be sufficient. We find that the formation of the local spin moment
at $I$ (and of the entire polarization cloud) is very fast: After a few
inverse hoppings, a considerable magnetic moment at site $I$ has formed,
pointing in field direction, with a magnitude that is already close to the
final saturation value. The field strength ($B_{\rm I}=1$) is not sufficient
to fully polarize the moment at $I$, i.e., $|\braket{{s}_{i}^{z}(t)}|<0.5$.
However, for the given strength of the field, we choose the end of the
injection step such that saturation is (almost) reached. For $B_{\rm I}=1$
this is the case for $t=t_{\rm inj}=50$, where we get
$\braket{{s}_{i}^{z}(t_{\rm pump})}\approx-0.33$. We will stick to this
injection time in the following.

The second part of the BIP process starts at $t=t_{\rm inj}$ by switching off
the injection field, thus releasing the accumulated polarization cloud. If the
location of site $I$ is properly chosen, the cloud propagates towards and
scatters from the read-out spin thereby pumping the spin direction. The
backscattered and also the transmitted part of the cloud are eventually
dissipated into the Lindblad bath at the corners of the edge and partially to
the bulk (and then to the Lindblad bath at other edges) simulating total-
energy-conserving dissipation in a macroscopically large sample. Finally, the
pump part can be terminated at $t=t_{\rm inj}+t_{\rm pump}$, when there is no
longer a significant spin dynamics, and another BIP process may follow. A
pumping time of $t_{\rm pump}=150$ is found to be sufficient, see the
discussion in Secs. V and VI. Generally, a sensible estimate for the minimal
$t_{\rm pump}$ is given by the distance between sites $I$ and $R$ divided by
the Fermi velocity $v_{\rm F}$ of the helical edge states.

The concatenated process on the left of Fig. 9 demonstrates that a spin-up BIP
process (B), to a large extent, can in fact undo a preceding spin-down process
(A). The whole process A$+$B starts at time $t=0$ (see black dot) from a
ground state in the presence of a local spin-down injection field, i.e., we
consider a static spin injection. Switching off the field, the system evolves
and the read-out spin moves towards the south pole of the Bloch sphere (green
line), this is the same as process A shown in Fig. 7. Thereafter, the system
is in a state, where there is essentially no more dynamics of
${\boldsymbol{S}}_{\rm R}$ and where locally, in the vicinity of sites $I$ and
$R$ and the edge sites in between, the electron system has relaxed to a state
that is “close” to the ground state, as can be monitored by tracing the
temporal evolution of the local spin moments
$\langle{\boldsymbol{s}}_{i}(t)\rangle$ at all sites in the nanoribbon. After
$t_{\rm pump}=150$, there are still outgoing wave packets visible that
continue to be absorbed, however, by the dissipative boundary conditions.

Figure 9:  Process cycle on the left: path traced out by the tip of the
classical read-out impurity spin during two concatenated processes A and B. A
(green): identical with D in Fig. 7, static spin-down injection, switching off
the injection field at $t=0$, propagation of the system’s state up to $t_{\rm
pump}=150$, proper process with $I=54$ and $R=36$. B: consists of B1 and B2,
proper BIP process with $I=16$ and $R=36$. B1 (red): dynamic spin-up
injection, $t_{\rm inj}=50$, starting from the final state of process A. B2
(blue): release of the injected spin-up density, propagation, scattering at
${\boldsymbol{S}}_{\rm R}$, $t_{\rm pump}=150$. Black dot: initial read-out
spin position ${\boldsymbol{S}}_{\rm R}(t=0)=S_{\rm R}\cdot(1,0,0)$. Process
cycle on the right: the same but starting, for better visibility, from
${\boldsymbol{S}}_{\rm R}=S_{\rm R}\cdot(1,1,0)/\sqrt{2}$ at $t=0$, see gray
dot, and with A replaced by A′ consisting a dynamic injection
A${}^{\prime}_{1}$ and, as before, of a pump part A${}^{\prime}_{2}$.
B${}^{\prime}_{1}$ and B${}^{\prime}_{2}$: the same as B1 and B2 but starting
from the final state of process A′.

The spin-up BIP (process B on the left) starts from this state with a
dynamical spin-up injection (first part B1, red line), which lasts for 50
inverse hoppings but does not affect the orientation of the spin very much,
and is completed with the second part B2 (blue line), which lasts for another
150 inverse hopping. During $B_{2}$, after a short time of further downward
deflection, the scattering of the polarization cloud efficiently drives the
spin back towards the north pole of the Bloch sphere. However, it does not
quite get to the original position ${\boldsymbol{S}}_{\rm R}=S_{\rm
R}\cdot(1,0,0)$ at the equator but stays a little below and has drifted
somewhat to the east.

There are two reasons for this “imperfection”: First, processes A and B are
not constructed completely inverse to each other, since in A we have employed
the “static” injection step. We have therefore repeated the cycle, see
concatenated process on the right of Fig. 9, by replacing A with a spin-down
BIP process A′, which, as described above, consists of a dynamic injection
A${}^{\prime}_{1}$ and a pump part A${}^{\prime}_{2}$ and initially starts
from the ground state of the electron system for a given spin orientation of
${\boldsymbol{S}}_{\rm R}$ in the $x$-$y$-plane. In addition, we choose a
state rotated around ${\boldsymbol{e}}_{z}$ by $\pi/4$ with
${\boldsymbol{S}}_{\rm R}=S_{\rm R}(1,1,0)/\sqrt{2}$ as a starting point, see
gray dot in Fig. 9 on the right, just to separate in the plot the process from
process A (left) on the sphere. A′ is concatenated with the inverse one, i.e.,
with the spin-up BIP process B′=B${}^{\prime}_{1}$+B${}^{\prime}_{2}$. The
duration of the subprocesses of A′ and of B′ are chosen identical (50 and 150
inverse hoppings each). As can be seen in Fig. 9 (right), the deviation from
the starting point, which remains after completion of A${}^{\prime}+$B′, is
smaller as compared to the cycle A+B on the left.

Opposed to thermodynamic processes that are steered in the state space of
equilibrium or, at zero temperature, of ground states, the processes
considered here involve non-equilibrium states. However, if one would ideally
extend the second (pump) part of the BIP process until the electron system is
fully relaxed to its ground state, at least locally in a sufficiently extended
spatial region around $I$ and $R$, the end state of process A would be
equivalent to the start state of A, and one could perfectly undo A with B. The
mentioned equivalence is just a rotation. If it was a fully relaxed, the end
state of A (of ${\boldsymbol{S}}_{\rm R}$ and of the electron system as well),
would merely be the start state of A, but rotated around an axis
${\boldsymbol{e}}$ and by some angle $\varphi$, where the unit vector
${\boldsymbol{e}}$ is perpendicular to the plane spanned by the center inside
the Bloch sphere and the start and end state on the sphere. Note that the
electronic ground state $|{\boldsymbol{S}}_{\rm R}\rangle$ for given
orientation of ${\boldsymbol{S}}_{\rm R}$ and the ground state
$|{\boldsymbol{S}}^{\prime}_{\rm R}\rangle$ of the model for a different
orientation ${\boldsymbol{S}}^{\prime}_{\rm R}$ have the same energy and that
$|{\boldsymbol{S}}^{\prime}_{\rm
R}\rangle=U({\boldsymbol{e}},\varphi)|{\boldsymbol{S}}_{\rm R}\rangle$ with
the according unitary rotation operator $U({\boldsymbol{e}},\varphi)$. This is
due to the fact that (as discussed in Sec. IV) the ground state of the pure
Kane-Mele model (without coupling to ${\boldsymbol{S}}_{\rm R}$) is an
$SU(2)$-invariant singlet, despite the SOC anisotropy. We conclude that
incomplete relaxation is exactly the second reason for the imperfect
reversibility.

Figure 10:  Trajectory of the read-out spin in a process consisting of 6 spin-
down followed by 7 spin-up BIP processes starting from a state (open black
dot) close to the north pole, $\eta=0.05$, see Eq. (12). Filled black dot:
position $S_{\rm R}\cdot(1,0,0)$ marking the longitude of the initial spin
orientation. Red lines: spin injection (50 inverse hoppings). Green lines:
proper spin-down pump (150 inverse hoppings). Blue lines: proper spin-up pump
(150 inverse hoppings). Parameters as in Figs. 3 and 6.  Figure 11:  Temporal
evolution of $S_{\rm R,z}$ corresponding to Fig. 10 with tolerance level
$\eta=0.05$, see red data points. Labels 1, 2, 4 indicate start, reversal and
termination of the switching cycle. Horizontal red lines: threshold levels
$\pm S_{\rm R}(1-\eta)$. Blue data points and lines: same starting point (1)
but now $\eta=0.001$ defines reversal and termination of the process, see
labels 3,5. Vertical lines separate the different BIP processes, each lasting
200 inverse hoppings..

Of course, this imperfection could be controlled by simply extending the pump
part of the BIP processes. This, however, becomes irrelevant for the question
whether a complete switching process ${\boldsymbol{S}}_{\rm R}=+S_{\rm
R}{\boldsymbol{e}}_{z}\to-S_{\rm R}{\boldsymbol{e}}_{z}$ and, vice versa, a
complete reversal ${\boldsymbol{S}}_{\rm R}=-S_{\rm
R}{\boldsymbol{e}}_{z}\to+S_{\rm R}{\boldsymbol{e}}_{z}$ can be achieved by
iterating several BIP processes. There is one small complication to overcome.
Namely, a state ${\boldsymbol{S}}_{\rm R}=\pm S_{\rm R}{\boldsymbol{e}}_{z}$,
together with the corresponding electronic ground state is non-responsive to a
spin-up oder spin-down polarized wave packet passing by, since the torque on
the read-out spin, $J{\boldsymbol{S}}_{\rm
R}\times\langle{\boldsymbol{s}}_{\rm R}\rangle$, vanishes in this case. We
therefore start from an initial state with ${\boldsymbol{S}}_{\rm R}$ close to
the north pole of the Bloch sphere but slightly tilted towards to
$x$-direction:

${\boldsymbol{S}}_{\rm R}(t=0)=S_{\rm
R}\left(\begin{array}[]{ccc}\sqrt{1-(1-\eta)^{2}}\\\ 0\\\ 1-\eta\\\
\end{array}\right)\>.$ (12)

The parameter $\eta$ controls the deviation of the $z$-component from its
maximal value in the initial state.

Fig. 10 displays the path of the classical read-out spin on the Bloch sphere
in an attempted full switching process. This process consists of 6 spin-down
BIP processes, during which the $z$-component of ${\boldsymbol{S}}_{\rm R}$ is
indeed found to decrease step by step in a controlled way. After the last
spin-down BIP process, we have $S^{z}_{\rm R}<-S_{\rm R}(1-\eta)$, i.e., the
spin has passed beyond the southern latitude that corresponds to the starting
point on according the northern latitude, i.e., the parameter $\eta$ also
serves to define a termination criterion for the switching process. This is
necessary because adding further spin-down BIP processes would bring the spin
less and less efficiently closer to the south pole. Fig. 10 demonstrates that
within a 5% tolarance ($\eta=0.05$), a full switching process is in fact
possible with 6 BIP processes, where each BIP process consists of 50 inverse
hoppings of spin injection and 150 inverse hoppings of free time evolution,
scattering and pumping the read-out spin.

When using the same tolerance level, about the same number of BIP processes
(in this case seven) are needed to subsequently reverse the switching and to
bring the spin back close to the north pole of the Bloch sphere again. It goes
without saying that we need spin-up rather than spin-down BIP processes in
this case. The near-to complete switching (green) and reversal (blue) is shown
in Fig. 10.

Fig. 11 shows the time evolution of the $z$-component of
${\boldsymbol{S}}_{\rm R}$ in detail, see red data. Note that each BIP process
lasts for $50+150=200$ inverse hoppings as indicated by the vertical lines. A
comparison with a process for a different, stricter tolerance level is
instructive: With $\eta=0.001$, see blue data in the figure, but using the
same starting point, the first part of the switching process is trivially the
same as before, but then it takes more effort to drive the spin towards the
south pole and to reach the threshold set by $\eta$. After reversal of the
switching process, it also takes more effort, i.e., more BIP processes, to
finally reach the north pole with the stricter tolerance level again. In total
8 spin-down and 13 spin-up BIP processes are needed for a complete switching
cycle.

##  VIII Conclusions

The state of a magnetic adatom with a well-formed and stable spin moment which
is exchange-coupled to the surface of a TR symmetric topological insulator can
be controlled to a large extent by making use of the helical character of the
topologically protected edge modes. We have set up a simple model of a
classical spin coupled to the zigzag edge of a Kane-Mele nanoribbon and
studied various control protocols in detail. The technical key ingredient for
the numerical study is the dissipative Lindblad boundary conditions imposed on
all but the physically relevant edge of the ribbon. This has allowed us to
study the coupled microscopic real-time dynamics of the spin and the electron
system up to time scales of thousands of inverse hoppings without disturbing
effects due to interference with propagating wave packets reflected from the
boundaries of the system.

In this study we have considered a spin-switching mechanism built from several
basic injection-pump (BIP) processes, each consisting of a dynamic spin
injection at a distant injection site $I$ of the edge and a subsequent pump
part driving the read-out spin at site $R$. The main idea is to exploit the
topological properties of the system at all stages of the process:

(i) A local spin-up or spin-down excitation at $I$ aligned to the
$z$-direction will selectively induce a spin-polarized excitation carried by
the edge states. Their presence and their helical character is ensured by the
fundamental bulk-boundary correspondence. A local magnetic field, used here to
describe the dynamical spin injection, also couples to bulk states. The bulk-
state-supported part of the polarization cloud, however, does no longer play a
role, since it is quickly dissipated into the bulk of the lattice. A sizable
edge-state-supported spin-polarization cloud remains. We found that this can
be built up quickly. After a femtosecond time scale of 50 (or more) inverse
hoppings, one finds saturation of the total spin density injected locally. One
can thus profit from a reproducible preparation step.

(ii) Switching off the coupling to the injection field releases the spin
excitation, which subsequently propagates along the edge and is unidirectional
thanks to the helicity of the edge states. Its group velocity is given by the
Fermi velocity of the according edge state. During the propagation the cloud
broadens but does not lose weight so that the injected spin density, apart
from the dissipated bulk contribution, entirely interacts with the read-out
spin. As a consequence of the topological protection of the edge modes, this
feature is also robust against various possible TR symmetric local
perturbations that could be present in a real sample. Mesoscopic distances
between sites $I$ and $R$ might therefore be conceivable.

(iii) The classical spin at $R$, on the other hand, does break TRS locally,
and hence the polarization cloud must scatter and, again due to the
topological principle of spin-momentum locking, back-propagate with
essentially opposite spin polarization. I.e., the read-out spin must exert a
spin torque on the approaching polarization cloud. Vice versa, the spin is
driven to the $\pm z$-direction due to the counter-torque of the cloud exerted
on the spin. We found that a single BIP process can change the $z$-component
of the spin to some degree, depending on its initial position. If this pump
part of the BIP process is sufficiently long, the process is revertible and
one may return to the initial spin state using exactly the same spin-injection
but with opposite spin direction. Here, we found 150 inverse hoppings to be
more or less sufficient to reach a fully relaxed final state that is necessary
for reversibility. Again, we conclude that a time scale of femto- to
picoseconds (or longer) is relevant for this part of the process.

(iv) Importantly, one may concatenate various BIP processes. As we have
demonstrated, a full switching of the spin direction between the north and the
south pole can be realized within, in principle, arbitrarily strict
tolerances. Five to ten basic processes prove sufficient for a reasonably
complete spin switching. Furthermore, reversing the polarization of the spin
in the injection part of the BIP processes, the spin switching can be inverted
as well. Assuming a nearest-neighbor hopping scale in the range of a hundred
meV, a single full switching (or back-switching) process would take place in
the picosecond regime. However, this must be seen as a lower limit. Since the
system approaches a fully relaxed state after each individual BIP process, the
next injection step does not have to follow immediately but can be delayed.
This also means that, in principle, the intermediate relaxed states of the
switching process can be controlled experimentally by techniques that do not
allow for a time resolution on the pico- or sub-pico-second time scale.

It goes without saying that the present study has a strong model character.
When addressing real $Z_{2}$ topological materials, there are a couple of
issues to be considered in addition, such as the effects of a realistic multi-
band electronic structure or electron-correlation effects. Another highly
important question is, for example, to which extent the conclusions remain
valid for time-reversal-symmetric, Kondo-type magnetic impurities, which can
be modeled, e.g., by quantum rather than classical spins (see, e.g.,
corresponding recent studies of ground-state properties AFM17 ; AFM19 ) This
question poses a formidable correlation problem to be treated in the
nonequilibrium regime and in the long-time limit. Furthermore, it would be
interesting to study the effects of imperfections regarding the structure of
the material at its surface or the precise location and type of coupling of
the magnetic impurities, even though the physics is expected to be largely
protected by the non-trivial topology of the states involved. For a realistic
description of materials, one would also have to address the effects of
magnetic anisotropies, such as the single-ion anisotropy or the
Dzyaloshinskii-Moriya interaction. We expect that those have a considerable
impact on the time scales, and further studies in this direction are well
conceivable with the present computational techniques.

###### Acknowledgements.

This work was supported by the Deutsche Forschungsgemeinschaft (DFG) through
the Cluster of Excellence “Advanced Imaging of Matter” - EXC 2056 - project ID
390715994, and through the Sonderforschungsbereich “Light-induced dynamics and
control of correlated quantum systems” - SFB 925 - project ID 170620586.

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# Exploring Mount Neverest

Michiel de Bondt

In one of the columns in the series ‘Perplexities’ in 1922, Henry Ernest
Dudeney formulated the following problem:

> Professor Walkingholme, one of the exploring party, was allotted the special
> task of making a complete circuit of the base of the mountain at a certain
> level. The circuit was exactly a hundred miles in length and he had to do it
> all alone on foot. He could walk twenty miles a day, but he could only carry
> rations for two days at a time, the rations for each day being packed in
> sealed boxes for convenience in dumping. He walked his full twenty miles
> every day and consumed one day’s ration as he walked. What is the shortest
> time in which he could complete the circuit?

This problem can be found in the book ‘536 Puzzles & Curious Problems’ from
Henry Ernest Dudeney, edited by Martin Gardner. This far, I did not find an
optimal solution to the problem. Albeit Martin Gardner is making fun on it,
the right interpretation of the problem is not clear to me at all. Let us
first formulate some ways to tackle the problem.

## Solution of the problem

One way to make the circuit is doing the same as with a straight distance of
one hundred miles. No matter how you interpret this problem, this takes
between $82$ and $87$ days (Dudeney found a solution of $86$ days, but it can
be done in $82\frac{6097}{6144}$ days). A better approach is to walk two round
trips to the $50$ miles distance point on the other side of the mountain,
since it requires between $41$ and $43$ days (Dudeney would have found a
solution of $42\frac{1}{2}$ days if he had considered this problem). But
Dudeney found the following solution:

######  Algorithm 1.

($23\frac{1}{2}$ days)

  1. 1.

Dump 5 rations at 90-mile point and return to base (5 days).

  2. 2.

Dump 1 at 85 and return to 90 (1 day).

  3. 3.

Dump 1 at 80 and return to 90 (1 day).

  4. 4.

Dump 1 at 80, return to 85, pick up 1 and dump at 80 (1 day).

  5. 5.

Dump 1 at 70 and return to 80 (1 day).

  6. 6.

Return to base (1 day). We have thus left one ration at 70 and one at 90.

  7. 7.

Dump 1 at 5 and return to base (1 day). If he must walk 20 miles he can do so
by going to 10 and returning to base.

  8. 8.

Dump 4 at 10 and return to base (4 days).

  9. 9.

Dump 1 at 10 and return to 5; pick up 1 and dump at 10 (1 day).

  10. 10.

Dump 2 at 20 and return to 10 (2 days).

  11. 11.

Dump 1 at 25 and return to 20 (1 day).

  12. 12.

Dump 1 at 30, return to 25, pick up 1 and dump at 30 (1 day).

  13. 13.

March to 70 (2 days).

  14. 14.

March to base ($1\frac{1}{2}$ days).

Now let us look at step 7. In my opinion, it only takes half a day. You just
throw away half of the content of one of the boxes. Apparently, Dudeney and I
have a disagreement on the interpretation of the problem. The problem might be
that unsealed boxes can not pass through the night, since the forest ants
creep into it. But then, you just can start walking around the mountain in the
middle of the day and do step 7 prior to the others. If Dudeney would have
allowed not to start at dawn, it is likely that he would have formulated his
solution accordingly, so we can add the following items to the analysis of
Gardner:

  1. 13.

If he does not finish his unsealed box(es) before the night, the forest ants
will.

  2. 14.

Having a long lie-in is a waste of time.

We can divide Dudeney’s solution in three parts:

Part A:

a round trip to the 70 miles point from the base, in which boxes are dumped on
positions 70 and 90 for later use (steps 1 to 6, $10$ days).

Part B:

a one-way trip to the 70 miles point from the base around the mountain (steps
7 to 13, $11\frac{1}{2}$ or $12$ days).

Part C:

walking from 70 to 100, using the boxes dumped in part A (step 14,
$1\frac{1}{2}$ days).

Since it is almost proved that Dudeneys solution can not be optimal if we
count $12$ days for part B, we allow starting on mid-day. Only for part C, it
is immediately clear that is optimal. Later, we will see that part A is
optimal as well, but part B can be improved, to $11\frac{1}{7}$ days exactly.

Before trying to find a better solution, it is always a good idea what others
did. When I searched the internet with google, I found the homepage of a
youngster called Nightvid Cole, who presents better solutions than that of
Dudeney. In the first one, he allows throwing away partially used boxes (and
this solution can easily be adapted to overcome the ants, starting on mid-
day). Albeit only part C of this solution is optimal, it requires
$22\frac{10}{11}$ days instead of $23$. The improvement is that he dumps boxes
on positions $70\frac{10}{11}$ and $90\frac{10}{11}$ rather than $70$ and
$90$. Now part B must reach further, whence it takes $12$ days now.

But probably, he reasoned the other way around: he reserved 12 days for part B
and then thought out an according solution. That was a very good idea, but
since he failed to optimize parts A and B, his solution is not optimal. So I
optimized parts A and B, which resulted in the following solution, a solution
that turned out to be optimal later on:

######  Algorithm 2.

($22\frac{9}{16}$ days)

  1. 1.

Dump one ration at $98\frac{3}{4}$ point and return to base ($\frac{1}{8}$
day).

  2. 2.

Dump one ration at $97\frac{1}{2}$, return to $98\frac{3}{4}$, pick up one,
dump at $91\frac{1}{4}$ and return to base (1 day).

  3. 3.

Dump one ration at $93\frac{3}{4}$, return to $97\frac{1}{2}$, pick up one,
dump at $93\frac{3}{4}$ also and return to base (1 day).

  4. 4.

Dump two rations at $90$ and return to base (2 days).

  5. 5.

Dump one ration at $86\frac{7}{8}$ and return to $93\frac{3}{4}$ (1 day).

  6. 6.

Dump one ration at $82\frac{1}{2}$, return to $86\frac{7}{8}$, pick up one,
dump at $86\frac{1}{4}$ and return to $90$ (1 day).

  7. 7.

Dump one ration at $80$, return to $86\frac{1}{4}$, pick up one and get to
$82\frac{1}{2}$ (1 day).

  8. 8.

Dump one ration at $71\frac{1}{4}$ and return to $80$ (1 day).

  9. 9.

Return to base (1 day). We have thus left one ration at $71\frac{1}{4}$ and
one at $91\frac{1}{4}$.

  10. 10.

Dump five rations at $10$ and return to base (5 days).

  11. 11.

Dump one ration at $12\frac{1}{2}$, return to $10$, pick up one, dump at
$12\frac{1}{2}$ also and return to $10$ (1 day).

  12. 12.

Dump one ration at $20$ and return to $10$ (1 day).

  13. 13.

Dump one ration at $20\frac{5}{8}$, return to $20$, pick up one, dump at
$20\frac{5}{8}$ also and return to $12\frac{1}{2}$ (1 day).

  14. 14.

Dump one ration at $26\frac{9}{16}$ and return to $20\frac{5}{8}$ (1 day).

  15. 15.

Dump one ration at $31\frac{1}{4}$, return to $26\frac{9}{16}$, pick up one
and get to $31\frac{1}{4}$ (1 day).

  16. 16.

March to $71\frac{1}{4}$ (2 days).

  17. 17.

March to base ($1\frac{7}{16}$ day).

If you look at the above algorithm, then one thing immediately strikes: it
would have been nicer if the circuit would have been 160 kilometers, with a
unit distance of 32 kilometers a day. Dudeney only considered solutions from
which the eating and turning points were a multiple of 5 miles. This is
however impossible for a solution of $22\frac{9}{16}$ days (for
$\frac{20}{5}\cdot 22\frac{9}{16}$ is not integral).

If professor Walkingholme must start at dawn, then the extra $\frac{7}{8}$
days must be used to increase the points where boxes are dumped in part A for
use in part C. Nightvid Cole found a solution of $23\frac{1}{3}$ days in this
context, but again, parts A and B are not optimal. The following optimal
solution not only has ugly positions, but also both part A and part B are
partially done on the first day.

######  Algorithm 3.

($23\frac{25}{116}$ days)

  1. 1.

Dump one ration at $8\frac{18}{29}$ and return to base ($\frac{25}{29}$ days).

  2. 2.

Dump two rations at $99\frac{9}{29}$ and return to base ($\frac{4}{29}$ days).

  3. 3.

Dump one ration at $96\frac{26}{29}$, return to $99\frac{9}{29}$, pick up two
in turn, dump both at $95\frac{25}{29}$ and return to base (1 day).

  4. 4.

Dump one ration at $90$ and return to base (1 day).

  5. 5.

Dump one ration at $88\frac{28}{29}$, return to $90$, pick up one, dump at
$88\frac{28}{29}$ also and return to $95\frac{25}{29}$ (1 day).

  6. 6.

Dump one ration at $82\frac{12}{29}$ and return to $88\frac{28}{29}$ (1 day).

  7. 7.

Dump one ration at $75\frac{20}{29}$ and return to $82\frac{12}{29}$ (1 day).

  8. 8.

Return to $96\frac{26}{29}$, pick up one, dump at $95\frac{20}{29}$ and return
to base (1 day). We have thus left one ration at $8\frac{18}{29}$, one at
$75\frac{20}{29}$ and another one at $95\frac{20}{29}$.

  9. 9.

Dump one ration at $9\frac{9}{29}$, return to $8\frac{18}{29}$, pick up one,
dump at $9\frac{9}{29}$ also and return to base (1 day).

  10. 10.

Dump five rations at $10$ and return to base (5 days).

  11. 11.

Dump one ration at $12\frac{19}{58}$, return to $10$, pick up one, dump at
$12\frac{19}{58}$ also and return to $9\frac{9}{29}$ (1 day).

  12. 12.

Dump one ration at $19\frac{19}{29}$ and return to $10$ (1 day).

  13. 13.

Dump one ration at $19\frac{24}{29}$, return to $19\frac{19}{29}$, pick up
one, dump at $19\frac{24}{29}$ also and return to $10$ (1 day).

  14. 14.

Dump one ration at $21\frac{19}{116}$ and return to $12\frac{19}{58}$ (1 day).

  15. 15.

Dump one ration at $23\frac{18}{29}$, return to $21\frac{19}{116}$, pick up
one, dump at $23\frac{18}{29}$ also and return to $19\frac{24}{29}$ (1 day).

  16. 16.

Dump one ration at $31\frac{21}{29}$ and return to $23\frac{18}{29}$ (1 day).

  17. 17.

Dump one ration at $35\frac{20}{29}$, return to $31\frac{21}{29}$, pick up one
and get to $35\frac{20}{29}$ (1 day).

  18. 18.

March to $75\frac{20}{29}$ (2 days).

  19. 19.

March to base ($1\frac{25}{116}$ days).

## Estimates for part A

If professor Walkingholme replaces his unsealed box by a new full box each
time he passes the base, then no rations carried in part A are used in part B
and vice versa. So we can see part A and part B as separate problems.

The optimality of part B of both solutions is almost proved in [1] and [4]. In
both articles, the problem of how far you can get in $N$ days is solved for
integers $N$. But to cross a certain distance, it is very unlikely that you
need an integral number of days. The non-integral case is an easy variation,
however. We will use the techniques of these articles here.

Since part B is almost done in the above references, we only prove the
optimality of part A in full detail. For convenience, we measure the distance
in units of 20 miles from now and indicate the positions the other way around.

But before starting, it is always a good idea to determine what must be done.
In part A, professor Walkingholme must put boxes on positions $\gamma$ and
$\gamma-1$, to be used in part C. If $\gamma>2$, then another box on
$\gamma-2$ and maybe more boxes are needed, but taking $\gamma>2$ is so bad
that remission of the additional costs of getting to the base does not affect
the estimate. If on the other hand $\gamma<1$, then there does not need to be
carried a box to $\gamma-1$, but taking $\gamma<1$ will turn out to be a bad
idea as well.

Suppose for now that $\gamma>1$ and let $m$ be the moment of the first
unsealing after dumping a box on position $\gamma$ to be used in part C, say
at position $r>0$. Assuming that boxes are not carried back and forth
unnecessarily, there is a box on position $\gamma$ and another box on a
position between $0$ and $\gamma-1$ inclusive, to be carried to $\gamma-1$ at
the end.

Let

$0<e_{l}\leq e_{l-1}\leq\cdots\leq e_{2}\leq e_{1}$

be the positions $>0$ where a box is unsealed in part A before dumping a box
at $\gamma$ (prior to $m$), and define

$e_{l+1}=e_{l+2}=\cdots=0$

Suppose that before moment $m$, professor Walkingholme unseals the last box on
position $e_{j}$. Then $(\gamma-e_{j})+(\gamma-r)\leq 1$. Together with
$e_{1}\geq e_{j}$, we get

$\gamma\leq\frac{1}{2}e_{1}+\frac{1}{2}r+\frac{1}{2}$ (1)

This estimates $r$ to be $2\gamma-e_{1}-1$ at least. In case $\gamma\leq 1$,
$r$ might be negative. Therefore we define $r=0$ in case moment $m$ takes
place on a negative position. $e_{1}$ is always nonnegative by definition,
thus (1) is also valid when $\gamma\leq 1$.

Assume that $\gamma>1$ and let $t$ be the time professor Walkingholme uses for
part A. Then professor Walkingholme walks $\frac{1}{2}t$ units in forward
direction in part A, whence by elementary logistics

$\gamma+(\gamma-1)+r+\sum_{i=1}^{k}e_{i}\leq\frac{1}{2}t$ (2)

for all $k$. Since $\gamma-1\leq 0$ when $\gamma\leq 1$, the above estimate is
also valid when $\gamma\leq 1$. This estimate can only be effective when we
show the optimality of an algorithm for part A with $\gamma\geq 1$ and $r\leq
1$ . If $r>1$, then there must be an additional box on position $r-1$ at
least, whence

$\gamma+(\gamma-1)+r+(r-1)+\sum_{i=1}^{k}e_{i}\leq\frac{1}{2}t$

for all $k$. But this estimate is satisfied for $r\leq 1$ as well, since then
you just subtract $1-r$ from the left hand side of (2). It is only not
effective for $r<1$.

If we take the average of the above estimate and (2), we get

$\gamma+(\gamma-1)+r+\frac{r-1}{2}+\sum_{i=1}^{k}e_{i}\leq\frac{1}{2}t$ (3)

for all $k$, which can only be effective for $r=1$. Notice that both (2) and
(3) can only be effective if $k\geq l$.

Put $e_{0}:=r$ and define

$d_{i}:=\frac{1}{2}e_{i}+\frac{1}{2}e_{i+1}+\frac{1}{2}$

for all $i\geq 0$. In order to get more information, we ask the following
question: how many units does professor Walkingholme walk within the interval
$[\beta,\infty)$, before unsealing the box on position $r$ on moment $m$,
where $\beta\geq\frac{1}{2}$? To get the right idea on this question, we
assume that part A must satisfy the Dudeney rule that boxes are only unsealed
at dawn, albeit we need a more general result (which is left to the reader,
the days are too short to write it down).

To answer the question, we slice the total walk up to moment $m$ into parts
$w_{i}$, such that $w_{i}$ ends on the middle of the day that the box at
$e_{i}$ is unsealed (at dawn) and starts on the middle of the day before.
$w_{0}$ is just the last half-day walk and also the first slice does not need
to have length one. Now we can estimate how many units professor Walkingholme
walks within the interval $[\beta,\infty)$ in $w_{i}$ for all $i>0$:

  * •

Case $\beta\leq e_{i}$: At most the length of $w_{i}$, i.e. one unit.

  * •

Case $\beta-\frac{1}{2}\leq e_{i}\leq\beta$: At most $1-2(\beta-e_{i})$ units,
since a round trip from $\beta$ to $e_{i}$ is $2(\beta-e_{i})$ units.

  * •

Case $e_{i}\leq\beta-\frac{1}{2}$: No units.

If $i=0$, then we get half of the above estimates, since $w_{0}$ is only half
a mile. Now take $\beta=e_{2k+2}+\frac{1}{2}$. Then $i\geq 2k+2$ implies the
last case and $i<2k+2$ implies one of the first two cases of the above, i.e.
$2(e_{i}+\frac{1}{2}-\max\\{e_{i},\beta\\})$ units. This makes a total of

$\displaystyle\alpha$ $\displaystyle:=$
$\displaystyle\left(e_{0}+\frac{1}{2}-\max\\{e_{0},\beta\\}\right)+2\sum_{i=1}^{2k+1}\left(e_{i}+\frac{1}{2}-\max\\{e_{i},\beta\\}\right)$
$\displaystyle=$
$\displaystyle\sum_{i=0}^{2k+1}\left(e_{i}+\frac{1}{2}-\max\\{e_{i},\beta\\}\right)+\sum_{i=1}^{2k+2}\left(e_{i}+\frac{1}{2}-\max\\{e_{i},\beta\\}\right)$
$\displaystyle=$ $\displaystyle
2\sum_{i=0}^{2k+1}d_{i}-\left(\max\\{e_{0},\beta\\}+2\sum_{i=1}^{2k+2}\max\\{e_{i},\beta\\}-\max\\{e_{2k+2},\beta\\}\right)$
$\displaystyle=$ $\displaystyle
2\sum_{i=0}^{2k+1}d_{i}-\max\\{e_{0},\beta\\}-2\sum_{i=1}^{2k+2}\max\\{e_{i},\beta\\}+\beta$

units.

Scratching all the $[\beta,\infty)$-parts of the $w_{i}$’s together gives a
walk that starts at $\beta$ and ends at $\max\\{e_{0},\beta\\}$, of which

$\frac{\alpha-(\max\\{e_{0},\beta\\}-\beta)}{2}$

units are in backward direction and

$\displaystyle\frac{\alpha+(\max\\{e_{0},\beta\\}-\beta)}{2}$ $\displaystyle=$
$\displaystyle\sum_{i=0}^{2k+1}d_{i}-\sum_{i=1}^{2k+2}\max\\{e_{i},\beta\\}$
(4) $\displaystyle\leq$
$\displaystyle\sum_{i=0}^{2k+1}d_{i}-\sum_{i=1}^{k}e_{i}-(k+2)\beta$

units are in forward direction. In order to get boxes on
$\gamma,e_{0},e_{1},\ldots$, professor Walkingholme needs to march from
$\beta$ to $\gamma$ if $\gamma>\beta$ and from $\beta$ to $e_{i}$ for all $i$
with $e_{i}>\beta$, whence at least

$\displaystyle(\gamma-\min\\{\gamma,\beta\\})+\sum_{i=0}^{\infty}(e_{i}-\min\\{e_{i},\beta\\})$
(5) $\displaystyle\geq$
$\displaystyle(\gamma-\beta)+\sum_{i=0}^{k}(e_{i}-\beta)+\sum_{i=k+1}^{\infty}(e_{i}-e_{i})$
$\displaystyle=$ $\displaystyle\gamma+r+\sum_{i=1}^{k}e_{i}-(k+2)\beta$

units in forward direction within $[\beta,\infty)$ are required. Combining
(4), (5) and $e_{0}=r$ gives

$\gamma+r+2\sum_{i=1}^{k}e_{i}\leq\sum_{i=0}^{2k+1}d_{i}$ (6)

## Optimality of algorithms 2 and 3

Using (2) for $k=2,3,4$ and (6) for $k=0,1$ gives

$t\geq 12\frac{4}{7}\gamma-9\frac{1}{7}$ (7)

(add variables $C_{k}\geq 0$ on the smaller sides of the inequalities to get
equations), which proves the optimality of part A of the second solution.
Using (3) instead of (2) gives

$t\geq 14\gamma-11$ (8)

which proves the optimality of part A of both Dudeney’s solution and the first
solution.

Next we sketch the optimality of part B. Let $e_{1}\geq e_{2}\geq
e_{3}\geq\cdots$ be the positions where boxes are unsealed in part B before
reaching $\gamma$ from the other side, and define $d_{i}$ as above. In [1],
the last inequality but one reads

$e_{1}+2\sum_{i=2}^{k}e_{i}\leq\sum_{i=1}^{2k-1}d_{i}$ (9)

and the first inequality of lemma B, together with the above definition, looks
like

$e_{1}+2\sum_{i=2}^{k}e_{i}\leq t-1$ (10)

except that the right hand side is $N-1$ instead of $t-1$. But an algorithm
with $N$ boxes takes $N$ days, thus (10) seems correct if $t$ is the time part
B takes in days.

These inequalities can be proved with the techniques of the previous section.
Together with $e_{1}+1=5-\gamma$, we get

$\displaystyle t$ $\displaystyle\geq$ $\displaystyle
13\frac{5}{7}(5-\gamma)-36\frac{6}{7}$ (11) $\displaystyle t$
$\displaystyle\geq$ $\displaystyle 16(5-\gamma)-45$ (12) $\displaystyle t$
$\displaystyle\geq$ $\displaystyle 19\frac{1}{5}(5-\gamma)-56\frac{4}{5}$ (13)

using (9) for $k=1,2,\ldots,n$, and (10) for $k=n+1,\ldots,2n$ to prove
inequality $(n+6)$.

Notice that part C takes $\gamma$ days. Both (11) and (12) prove the
optimality of part B of algorithm 2 individually, but they need to cooperate
to get the optimality of algorithm 2 as a whole. (11) gives the bound
$11\frac{1}{7}$ day on getting to $3\frac{1}{2}$ as well.

If we charge part C for two days, i.e. the number of boxes unsealed in it,
which Nightvid Cole preferred, then it is also clear that algorithm 2 is
optimal among the solutions with $\gamma>1$. But if $\gamma\leq 1$, then it
follows from (13) and (8) that more than

$\left(12\frac{4}{7}\gamma-9\frac{1}{7}\right)+\left(19\frac{1}{7}(5-\gamma)-57\right)+\gamma=29\frac{4}{7}-5\frac{4}{7}\gamma\geq
24$

days are necessary, so algorithm 2 is optimal in Coles’s way of measuring as
well. Furthermore, the straight line solution is also proved to be non-optimal
now.

Since the time used for part A and B together is optimal in the first
solution, the only way to improve it to an optimal solution in Dudeney’s way
of measuring is to decrease $\gamma$. It follows that algorithm 3 is optimal
in Dudeney’s way of measuring.

At last, the round trips to $2\frac{1}{2}$, or actually to some position
$\gamma$. The round trip from the other side is in fact a round trip to
$5-\gamma$. Let $e_{1}$ be the largest position where a box is unsealed. This
can be before or after reaching $\gamma$, but no other box need to be unsealed
in between. Notice that the box for $e_{1}$ can be transported on the road to
$\gamma$, thus $e_{1}$ does not need to be counted for dropping.

Let $r$ be the position where the first box after that on $e_{1}$ and
$e_{2}\geq e_{3}\geq\cdots$ be the positions where the boxes before that on
$e_{1}$ are unsealed. The last box before that on $e_{1}$ is unsealed on a
position $\leq e_{2}$, after which a walk of two units to $r$ which meets
$\gamma$ follows. Hence

$\gamma\leq\frac{e_{2}+r+2}{2}$ (14)

Set $d_{i}:=\frac{1}{2}e_{i}+\frac{1}{2}e_{i+1}+\frac{1}{2}$ for all $i\geq
2$. By way of the techniques for estimating part $A$, one can get the
following inequalities for the round trip to $\gamma$.

$\gamma+r+2\sum_{i=2}^{k}e_{i}\leq\frac{e_{2}+r+2}{2}+\sum_{i=2}^{2k}d_{i}$
(15)
$\gamma+r+(r-1)+2\sum_{i=2}^{k}e_{i}\leq\frac{e_{2}+r+2}{2}+\sum_{i=2}^{2k+1}d_{i}$
(16)

and

$\gamma+r+(r-1)+2\sum_{i=2}^{k}e_{i}\leq\frac{t}{2}$ (17)

Using (14), (15) for $k=2,3,4$, (16) for $k=4,5,6,7,8$, and (17) for
$k=9,10,\allowbreak\ldots,18$, we get

$t\leq 27\gamma-46\frac{7}{8}$

whence at least
$(27\gamma-46\frac{7}{8})+(27(5-\gamma)-46\frac{7}{8})=41\frac{1}{4}$ days are
necessary. This bound can be attained with my way of measuring, with two round
trips to $2\frac{1}{2}$, but in order to take into account the ants as well,
different round trips are necessary, e.g. to $2\frac{1}{2}\pm\frac{1}{72}$,
starting $\frac{1}{3}$ way during the day and ending $\frac{7}{12}$ way during
another day. The reader may verify this.

If the box on $e_{1}$ is not unsealed before reaching $\gamma$, then we have
$\gamma\geq r+1$, and by adding (14), (15) for $k=2,3,\ldots,8$, and (17) for
$k=9,10,\allowbreak\ldots,17$, to the inequality $\gamma\geq r+1$, we can
derive

$t\leq 25\frac{6}{7}\gamma-44$

If we use (16) instead of (15) for $k=8$ (also for $k=7$ and/or $k=6$ when
desired), we get

$t\leq 26\frac{1}{7}\gamma-\frac{313}{7}$

Both bounds on $t$ can be attained simultaneously, if and only if
$e_{1}=\gamma=2\frac{1}{2}$.

## References

  * [1] M. de Bondt, The camel-banana problem, Nieuw Archief voor Wiskunde, 14 (1996) 415-426.
  * [2] N. Cole, Mount Neverest Problem,
http://users.aristotle.net/${}_{\mbox{\~{}}}$nightvid/neverest.htm

  * [3] H. E. Dudeney, 536 Puzzles & Curious Problems, ed. Martin Gardner Charles Scribner’s Sons, New York, 1967.
  * [4] G. Rote, G. Zhang, Optimal logistics for expeditions: the jeep problem with complete refilling, Optimierung und Kontrolle, 71 (1996), Spezialforschungsbereich 003, Karl-Franzens-Universität Graz & Technische Universität Graz.