# Mechanics of a ferromagnetic domain wall

Se Kwon Kim1 and Oleg Tchernyshyov2 1 Department of Physics, Korea Advanced
Institute of Science and Technology, Daejeon 34141, Republic of Korea  2
William H Miller III Department of Physics and Astronomy and Institute for
Quantum Matter, Johns Hopkins University, Baltimore, MD 21218, USA
<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

This paper gives a pedagogical introduction to the mechanics of ferromagnetic
solitons. We start with the dynamics of a single spin and develop all the
tools required for the description of the dynamics of solitons in a
ferromagnet.

††: J. Phys.: Condens. Matter

###### Contents

  1. 1 Introduction
  2. 2 Single magnetic dipole
    1. 2.1 Magnetic moment and angular momentum
    2. 2.2 Spherical geometry
    3. 2.3 Precessional dynamics
    4. 2.4 Example: Larmor precession
    5. 2.5 Conservative and gyroscopic forces
    6. 2.6 Lagrangian
    7. 2.7 Example: precession near a potential minimum
    8. 2.8 Dissipative force
    9. 2.9 Example: Larmor precession with dissipation
    10. 2.10 Transformation to canonical variables
    11. 2.11 Transformation to general variables
    12. 2.12 Rayleigh’s dissipation function
    13. 2.13 Example: strong easy-plane anisotropy
    14. 2.14 Recap: equations of motion
  3. 3 Ferromagnet in 1 dimension
    1. 3.1 Heisenberg spin chain
    2. 3.2 Continuum theory of a ferromagnet
    3. 3.3 Model with local interactions
    4. 3.4 Spin waves in a ground state
    5. 3.5 Static domain wall
    6. 3.6 Domain wall in a magnetic field
    7. 3.7 Weakly perturbed domain wall
    8. 3.8 Normal modes of a domain wall
    9. 3.9 Walker’s model of domain wall dynamics
    10. 3.10 Angular momentum of a domain wall
    11. 3.11 Linear momentum of a domain wall
    12. 3.12 Counting states of a domain wall
  4. 4 Discussion
  5. A Schrödinger equation with a Pöschl–Teller potential

##  1 Introduction

This article aims to provide a self-contained introduction to theory of
ferromagnetic solitons. The subject has been actively researched for more than
50 years and there are a number of excellent reviews and books [1, 2]. The
last decades have seen new developments in the field such as the use of
collective coordinates to describe the dynamics of solitons [3].

A ferromagnet contains a large number of atomic magnetic dipoles that tend to
line up with one another at sufficiently low temperatures. The interaction
most commonly responsible for the parallel alignment is Heisenberg’s exchange,
which represents a quantum-mechanical effect related to the fermionic
statistics of electrons. One of the most frequently used models of a
ferromagnet is the Heisenberg model with quantum spins of length $S=\hbar/2$
on a lattice with nearest-neighbor exchange interactions. The ground state has
all spins pointing in the same direction. Elementary excitations are
magnons—quasiparticles carrying spin $S_{z}=-\hbar$ along the direction of
magnetization. A magnon mode in a coherent state with a large amplitude
represents a classical wave of magnetization (a spin wave).

This widespread approach, starting with a quantum model of spins on a lattice,
is inconvenient for our purposes. Although it is suitable for obtaining weakly
excited states such as magnons, the treatment of topological solitons at the
quantum level and on a lattice is too complicated. Instead, we shall deal with
continuum models of a classical ferromagnet. This approach allows one to
obtain both linear and nonlinear excitations at the classical level and then
quantize them.

In Section 2, we discuss the dynamics of a single spin and introduce various
mathematical tools. The same tools, suitably generalized, are used for the
description of soliton dynamics in a one-dimensional ferromagnet in Section 3.

The paper is meant as a pedagogical introduction to the subject, not as a
literature review. We have therefore kept bibliographical references to an
absolute minimum.

##  2 Single magnetic dipole

###  2.1 Magnetic moment and angular momentum

The building block of a macroscopic magnet is an atom with angular momentum
$\mathbf{J}$ and magnetic dipole moment $\boldsymbol{\mu}$ related by the
gyromagnetic ratio $\gamma$,

$\boldsymbol{\mu}=\gamma\mathbf{J}.$ (1)

If the angular momentum comes solely from the electron spins, the gyromagnetic
ratio is $\gamma=e/m_{e}c$, where $e<0$ is the electron charge and $m_{e}$ its
mass. More generally, $\gamma=ge/2m_{e}c$, where the Lande $g$-factor is
determined by the lengths of spin, orbital, and total angular momenta $S$,
$L$, and $J$ [4].

We shall treat both the angular momentum $\mathbf{J}$ and the magnetic moment
$\boldsymbol{\mu}$ as classical physical variables of fixed lengths $J$ and
$\mu$. Often both are expressed in terms of a unit vector $\mathbf{m}$
parallel to $\mathbf{J}$:

$\mathbf{J}=J\mathbf{m},\quad\boldsymbol{\mu}=\gamma
J\mathbf{m},\quad|\mathbf{m}|=1.$ (2)

Because of the linear proportionality between the three quantities, we will
casually refer to $\mathbf{m}$ as the magnetic moment or spin.

###  2.2 Spherical geometry

Figure 1: The local frame defined by the unit vectors (4).

To resolve the constraint of the unit length $|\mathbf{m}|=1$, it is
convenient to introduce two spherical angles: the polar angle $\theta$ (co-
latitude) and the azimuthal angle $\phi$ (longitude):

$\mathbf{m}=(\sin{\theta}\cos{\phi},\sin{\theta}\sin{\phi},\cos{\theta}).$ (3)

They define a local frame with three mutually orthogonal unit vectors

$\displaystyle\mathbf{e}_{m}=\mathbf{m}=(\sin{\theta}\cos{\phi},\sin{\theta}\sin{\phi},\cos{\theta}),$
$\displaystyle\mathbf{e}_{\theta}=\frac{\partial\mathbf{m}}{\partial\theta}=(\cos{\theta}\cos{\phi},\cos{\theta}\sin{\phi},-\sin{\theta}),$
(4)
$\displaystyle\mathbf{e}_{\phi}=\frac{1}{\sin{\theta}}\frac{\partial\mathbf{m}}{\partial\phi}=(-\sin{\phi},\cos{\phi},0),$

pointing up, south, and east, respectively (Fig. 1).

Small deviations of magnetization from a given direction $\mathbf{m}$ can be
written in the local frame as follows:

$\delta\mathbf{m}=\mathbf{e}_{\theta}\,\delta
m_{\theta}+\mathbf{e}_{\phi}\,\delta m_{\phi},\qquad\delta
m_{\theta}=\delta\theta,\quad\delta m_{\phi}=\sin{\theta}\,\delta\phi.$ (5)

###  2.3 Precessional dynamics

The dynamics of an atomic magnetic dipole is very different from that of a
Newtonian particle. For a particle, an external force generates an
acceleration proportional to the force and inversely proportional to its mass.
In contrast, a magnetic dipole does not have inertia. Like a fast-spinning
gyroscope, it precesses with a velocity proportional to an external torque.

It is helpful to picture a magnetic dipole whose south pole is pivoted and the
north pole is allowed to move on a sphere of radius $r$ so that its position
is $\mathbf{r}=r\mathbf{m}$. The force from the potential energy may then be
calculated as
$\mathbf{F}=-\frac{dU}{d\mathbf{r}}=-\frac{1}{r}\frac{dU}{d\mathbf{m}}$ and
its torque as
$\boldsymbol{\tau}=\mathbf{r}\times\mathbf{F}=-\mathbf{m}\times\frac{dU}{d\mathbf{m}}$.
The torque determines the rate of change of angular momentum
$d\mathbf{J}/dt=Jd\mathbf{m}/dt$, hence the equation of motion

$J\dot{\mathbf{m}}=-\mathbf{m}\times\frac{dU}{d\mathbf{m}},$ (6)

where $\dot{\mathbf{m}}$ stands for $d\mathbf{m}/dt$. Note that the
geometrical size of the dipole $r$ canceled out.

It is worth keeping in mind that the derivative $dU/d\mathbf{m}$ is only taken
in the directions locally tangential to the sphere $|\mathbf{m}|=1$. In the
formal sense, this derivative is defined as the coefficient in the Taylor
expansion

$U(\mathbf{m}+d\mathbf{m})=U(\mathbf{m})+\frac{dU}{d\mathbf{m}}\cdot
d\mathbf{m}+\ldots,$ (7)

where $d\mathbf{m}$ is an infinitesimal displacement tangential to the sphere
and thus transverse to $\mathbf{m}$. Even if $dU/d\mathbf{m}$ had a component
parallel to $\mathbf{m}$, it would not contribute to the first order in the
expansion (7).

###  2.4 Example: Larmor precession

Consider the motion of a magnetic dipole in an external magnetic field
$\mathbf{H}$. Its potential energy is
$U=-\boldsymbol{\mu}\cdot\mathbf{H}=-\mu\mathbf{m}\cdot\mathbf{H}$. Upon
substituting $U$ into the equation of motion (6) we obtain

$J\dot{\mathbf{m}}=-\mu\mathbf{H}\times\mathbf{m}.$ (8)

The magnetic moment rotates uniformly about the direction of the field at the
Larmor precession frequency
$\boldsymbol{\omega}_{L}=-\mu\mathbf{H}/J=-\gamma\mathbf{H}$. The angle
between the magnetic moment and the field remains constant, implying
conservation of energy $U=-\boldsymbol{\mu}\cdot\mathbf{H}$.

To see this explicitly, it is convenient to rewrite the equation of motion in
terms of spherical angles $\theta$ and $\phi$ (3):

$-J\sin{\theta}\,\dot{\phi}-\frac{\partial U}{\partial\theta}=0,\quad
J\dot{\theta}-\frac{1}{\sin{\theta}}\frac{\partial U}{\partial\phi}=0.$ (9)

For the field along the $z$ axis, the potential energy $U=-\mu H\cos{\theta}$
is independent of the azimuthal angle $\phi$. Thus the polar angle $\theta$
remains unchanged during precession.

###  2.5 Conservative and gyroscopic forces

Conservation of energy is a general property of the precession equation (6).
After some simple algebra, it can be transformed into the form

$-J\dot{\mathbf{m}}\times\mathbf{m}-\frac{dU}{d\mathbf{m}}=0.$ (10)

Returning to the picture of a dipole pivoted at the south pole, we may
interpret this equation of motion as Newton’s second law for a massless
particle confined to move on a unit sphere (hence no inertial term
proportional to $\ddot{\mathbf{m}}$). The second term obviously represents a
conservative force.

What does the first term represent? It is similar to the Lorentz force acting
on a moving electric charge in a magnetic field as it is proportional to the
velocity $\dot{\mathbf{m}}$ and is orthogonal to it. We thus may interpret the
first term in Eq. (10) as a Lorentz force acting on a particle with unit
electric charge moving on the unit sphere $|\mathbf{m}|=1$ in a radial
“magnetic field” $\mathbf{b}(\mathbf{m})=-J\mathbf{m}$, as if there were a
magnetic monopole of strength $J$ at the sphere’s center. (We use the scare
quotes to distinguish the fictitious “magnetic field” $\mathbf{b}$ on the unit
sphere $|\mathbf{m}|=1$ from the physical magnetic field $\mathbf{H}$.)

The “Lorentz force” in Eq. (10) does no work and thus cannot be described in
terms of potential energy. It belongs to a class of _gyroscopic_ forces first
discussed by Kelvin [5]. Precession of a spin is indeed reminiscent of the
motion of a fast-spinning gyroscope.

Eq. (10) can be used to establish conservation of potential energy during
motion:

$\frac{dU}{dt}=\dot{\mathbf{m}}\cdot\frac{dU}{d\mathbf{m}}=\dot{\mathbf{m}}\cdot(-J\dot{\mathbf{m}}\times\mathbf{m})=0.$
(11)

###  2.6 Lagrangian

The equation of motion (10) can be obtained by minimization of action $S=\int
L\,dt$. The Lagrangian

$L=\mathbf{a}(\mathbf{m})\cdot\dot{\mathbf{m}}-U(\mathbf{m})$ (12)

has no kinetic energy. The first term represents the gyroscopic (“Lorentz”)
force and contains a gauge potential $\mathbf{a}(\mathbf{m})$ whose curl gives
the “magnetic field” of the monopole,

$\nabla_{\mathbf{m}}\times\mathbf{a}(\mathbf{m})=\mathbf{b}(\mathbf{m})=-J\mathbf{m}.$
(13)

The first term in the Lagrangian (12) is peculiar for a number of reasons.

  1. 1.

Its action

$S_{g}=\int\mathbf{a}(\mathbf{m})\cdot\dot{\mathbf{m}}\,dt=\int\mathbf{a}(\mathbf{m})\cdot
d\mathbf{m}$ (14)

is independent of how fast the vector $\mathbf{m}$ moves on the unit sphere
and depends just on the geometry of its path. Hence the name _geometric
action_ for $S_{g}$ and geometric (or Berry) phase for $S_{g}/\hbar$ [6].

  2. 2.

This term is not uniquely defined. A gauge transformation

$\mathbf{a}(\mathbf{m})\mapsto\mathbf{a}(\mathbf{m})+\nabla_{\mathbf{m}}\,\chi(\mathbf{m}),$
(15)

where $\chi(\mathbf{m})$ is an arbitrary scalar function on the unit sphere,
changes the vector potential $\mathbf{a}(\mathbf{m})$ but leaves the “magnetic
field” $\mathbf{b}(\mathbf{m})$ the same.

  3. 3.

Strictly speaking, the magnetic field of a monopole $\mathbf{b}(\mathbf{m})$
cannot be represented by a gauge potential because this field configuration is
not divergence-free: it has a source with magnetic charge $-J$ at the origin
$\mathbf{m}=0$. Although the origin is not accessible, we can still observe
that there is a net “magnetic flux” $-4\pi J$ by integrating the “field” over
the unit sphere.

  4. 4.

As a result of this problem, any choice of the gauge potential
$\mathbf{a}(\mathbf{m})$ contains a singularity, e.g.,

$\mathbf{a}(\mathbf{m})=J\,\frac{\mathbf{m}_{s}\times\mathbf{m}}{1-\mathbf{m}_{s}\cdot\mathbf{m}},$
(16)

where $\mathbf{m}_{s}$ is an arbitrary unit vector. The singularity of the
gauge potential at $\mathbf{m}=\mathbf{m}_{s}$ is associated with a localized
“magnetic flux” of strength $+4\pi J$, which compensates the uniform flux
$-4\pi J$ spread over the sphere. If we were allowed to explore the entire
3-dimensional space of $m$, we would discover that there is a Dirac string
carrying “magnetic flux” $+4\pi J$ from the origin to infinity in the
direction $\mathbf{m}_{s}$. With the Dirac string attached, the “magnetic
field” becomes solenoidal, which makes it possible to describe it in terms of
a vector potential.

  5. 5.

Vector potentials (16) describe the “magnetic field” of a monopole
$\mathbf{b}(\mathbf{m})=-J\mathbf{m}$ _almost_ everywhere on the unit sphere
$\mathbf{m}$, with the exception of point $\mathbf{m}_{s}$, where the
“magnetic flux” of the Dirac string pierces the sphere. We might argue that,
as long as the trajectory $\mathbf{m}(t)$ stays away from the Dirac string, we
can use the gauge potential. This will work for a single magnetic dipole.
However, we will run into this problem head-on when we deal with the dynamics
of an extended ferromagnet. In that case, most dipoles will avoid the
singularity but there will be some that will hit it.

To complete this section, we write down the gauge term in spherical angles for
the azimuthally symmetric gauge choices with the Dirac string attached to the
north and south poles:

$\mathbf{a}(\mathbf{m})\cdot\dot{\mathbf{m}}=J(\cos{\theta}\pm 1)\dot{\phi}.$
(17)

These are the two most commonly used gauge choices [7].

###  2.7 Example: precession near a potential minimum

Choose the axes so that the energy has a minimum at the north pole,
$\mathbf{m}_{0}=\mathbf{e}_{z}$, and the potential energy has a diagonal form
for small transverse deviations $\delta\mathbf{m}=\mathbf{e}_{x}\delta
m_{x}+\mathbf{e}_{y}\delta m_{y}$:

$U(\mathbf{m})=U(\mathbf{m}_{0})+\frac{K_{x}\delta
m_{x}^{2}}{2}+\frac{K_{y}\delta m_{y}^{2}}{2}+\ldots$ (18)

Here $K_{x}$ and $K_{y}$ are positive constants and the omitted terms are of
higher orders in the transverse components of magnetization. Choose the gauge
potential (16) with the singularity at the south pole,
$\mathbf{m}_{s}=-\mathbf{e}_{z}$, so that $\mathbf{a}\approx(J\delta
m_{y}/2,-J\delta m_{x}/2,0)+\ldots$ The resulting Lagrangian

$L\approx\left[J(\delta\dot{m}_{x}\delta
m_{y}-\delta\dot{m}_{y}m_{x})-K_{x}\delta m_{x}^{2}-K_{y}\delta
m_{y}^{2}\right]/2+\ldots$ (19)

yields the equations of motion for $\delta m_{x}$ and $\delta m_{y}$:

$-J\delta\dot{m}_{y}-K_{x}\delta m_{x}=0,\quad J\delta\dot{m}_{x}-K_{y}\delta
m_{y}=0.$ (20)

The spin follows an elliptical trajectory $K_{x}\delta m_{x}^{2}+K_{y}\delta
m_{y}^{2}=\mathrm{const}$ precessing at the frequency
$\omega=-\sqrt{K_{x}K_{y}}/J$.

###  2.8 Dissipative force

In the real world, a magnetic dipole will not precess about the magnetic field
forever and will eventually line up with it, settling into a minimum of
potential energy. Energy dissipation can be caused by various processes, e.g.,
radiation of electromagnetic waves or friction. We shall add a simple
phenomenological term to the equations of motion that represents—in the
analogy with a particle on a sphere—a dissipative force $-\alpha
J\dot{\mathbf{m}}$ proportional to the velocity $\dot{\mathbf{m}}$ and
directed against it. Eq. (10) now reads

$-J\dot{\mathbf{m}}\times\mathbf{m}-\frac{dU}{d\mathbf{m}}-\alpha
J\dot{\mathbf{m}}=0.$ (21)

It expresses the balance of gyroscopic, conservative, and dissipative forces
acting on the dipole. The dimensionless constant $\alpha>0$ introduced by
Gilbert [8] characterizes the strength of energy dissipation. It is usually
small, $\alpha\ll 1$, with typical values ranging from $10^{-4}$ in insulating
magnets to $10^{-2}$ in metallic ones.

In spherical angles,

$-J\sin{\theta}\,\dot{\phi}-\frac{\partial U}{\partial\theta}-\alpha
J\dot{\theta}=0,\quad J\dot{\theta}-\frac{1}{\sin{\theta}}\frac{\partial
U}{\partial\phi}-\alpha J\sin{\theta}\,\dot{\phi}=0.$ (22)

In the presence of the dissipative force, potential energy is dissipated at
the rate

$\frac{dU}{dt}=\dot{\mathbf{m}}\cdot\frac{dU}{d\mathbf{m}}=\dot{\mathbf{m}}\cdot(-J\dot{\mathbf{m}}\times\mathbf{m}-\alpha
J\dot{\mathbf{m}})=-\alpha J|\dot{\mathbf{m}}|^{2}\leq 0.$ (23)

###  2.9 Example: Larmor precession with dissipation

We return to the problem of a dipole in a magnetic field, this time with
dissipation. With energy $U=-\mu H\cos{\theta}$, Eq. (22) reads

$-J\sin{\theta}\,\dot{\phi}-\mu H\sin{\theta}-\alpha J\dot{\theta}=0,\quad
J\dot{\theta}-\alpha J\sin{\theta}\,\dot{\phi}=0.$ (24)

Solving this pair of equations yields a slightly reduced Larmor precession
frequency $\omega_{L}$. A new element is a gradual reduction of the polar
angle:

$\displaystyle\phi(t)=\phi(0)-\frac{\gamma
Ht}{1+\alpha^{2}}\equiv\phi(0)+\omega_{L}t,$
$\displaystyle\cos{\theta(t)}=\tanh{\left(\alpha|\omega_{L}|(t-t_{0})\right)},$
(25)

where $t_{0}$ is the time when $\cos{\theta}=0$. We see that the azimuthal
motion is precessional and fast, with the frequency of order $\alpha^{0}$,
whereas the polar motion is relaxational and slow, with the relaxation rate of
order $\alpha^{1}$. For weak damping $\alpha\ll 1$, the dipole will make of
order $\alpha^{-1}$ full rotations before approaching the energy minimum at
the north pole.

###  2.10 Transformation to canonical variables

Although our physical quantities—magnetization and angular momentum—are
represented by a three-dimensional vector $\mathbf{m}$, its three components
are not independent because of the fixed length $|\mathbf{m}|=1$. It may
sometimes be convenient to work with two independent variables, which can be
introduced in a number of ways.

One possible route is to use a pair of canonical variables $(q,p)$, one of
which is a coordinate and the other its conjugate momentum. To obtain them, we
may use the Lagrangian in one of the standard gauges (17),

$L=J(\cos{\theta}\pm 1)\dot{\phi}-U(\theta,\phi).$ (26)

We choose the azimuthal angle $\phi$ as the coordinate and obtain the
corresponding canonical momentum in the usual way:

$q=\phi,\quad p=\frac{\partial L}{\partial\dot{\phi}}=J(\cos{\theta}\pm
1)=J_{z}\pm J.$ (27)

Not unexpectedly, the momentum conjugate to $\phi$ is (up to an additive
constant) the angular momentum $J_{z}=J\cos{\theta}$. It is somewhat alarming
that the polar angle $\theta$ does not have a conjugate momentum (no
$\dot{\theta}$ term in the Lagrangian). We will address this problem in a
later section. For now, we shall dismiss this concern by noting that $\theta$
is present in the canonical momentum so that both angles are represented in
the canonical pair $(q,p)$.

We now realize that our two-dimensional space is a _phase space_. The number
of states in a phase space is proportional to phase-space volume:

$d\Gamma=\frac{dq\,dp}{2\pi\hbar}=\frac{J}{2\pi\hbar}\sin{\theta}\,d\theta\,d\phi.$
(28)

Integrating over the sphere, we obtain the total number of states

$\Gamma=\frac{J}{2\pi\hbar}\int_{0}^{\pi}\sin{\theta}\,d\theta\int_{0}^{2\pi}d\phi=\frac{2J}{\hbar}=2j.$
(29)

where $j=J/\hbar$ is the length of angular momentum in the units of $\hbar$.
The exact answer, which we have learned in a quantum mechanics course, is
$\Gamma=2j+1$. Our approximate answer, obtained within the scope of classical
mechanics, is correct in the limit of a large length of angular momentum,
$j\ll 1$.

With canonical variables identified, we have the Poisson bracket. In
particular,

$\\{\theta,\phi\\}=-\\{\phi,\theta\\}=\frac{\partial\theta}{\partial
q}\frac{\partial\phi}{\partial p}-\frac{\partial\theta}{\partial
p}\frac{\partial\phi}{\partial q}=\frac{1}{J\sin{\theta}}.$ (30)

The Hamiltonian is obtained in a standard way,

$H=p\dot{q}-L=U(\theta,\phi).$ (31)

It contains potential energy only. The $\dot{\phi}$ term in the Lagrangian
(26) is _not_ kinetic energy. Hamilton’s equations of motion,

$\dot{q}=\\{q,H\\}=\partial H/\partial p,\quad\dot{p}=\\{p,H\\}=-\partial
H/\partial q,$ (32)

reproduce the equations of motion for the spherical angles (9).

###  2.11 Transformation to general variables

We may not always want to use pairs of canonically conjugate variables and
instead perform a more general transformation from $\mathbf{m}$ to a pair of
coordinates $(q^{1},q^{2})$, e.g., $q^{1}=\theta$ and $q^{2}=\phi$. Equations
of motion for the new variables can be obtained by starting from the equation
(21) expressing force balance.

To obtain the equations of motion for the new variables $\\{q^{i}\\}$, where
$i=1,2$, we first note that the time evolution of $\mathbf{m}(q^{1},q^{2})$
comes solely through the time evolution of the coordinates:
$\dot{\mathbf{m}}=\dot{q}^{j}\,\partial\mathbf{m}/\partial q^{j}$, where
summation is implied over the doubly repeated index $j$. We substitute this
expression into Eq. (21) and multiply the result by
$\partial\mathbf{m}/\partial q^{i}$ to obtain an equation of motion for the
new coordinates:

$G_{ij}\dot{q}^{j}-\partial U/\partial q^{i}-D_{ij}\dot{q}^{j}=0.$ (33)

It expresses the balance of the gyroscopic, conservative, and dissipative
forces for each coordinate $q^{i}$. The coefficients $G_{ij}$ and $D_{ij}$ of
the antisymmetric gyroscopic and symmetric dissipative tensors are defined as
follows:

$G_{ij}=-J\mathbf{m}\cdot\left(\frac{\partial\mathbf{m}}{\partial
q^{i}}\times\frac{\partial\mathbf{m}}{\partial q^{j}}\right),\quad
D_{ij}=\alpha J\,\frac{\partial\mathbf{m}}{\partial
q^{i}}\cdot\frac{\partial\mathbf{m}}{\partial q^{j}}.$ (34)

The gyroscopic tensor $G_{ij}$ is the inverse of the of Poisson tensor
$\Pi^{ij}=\\{q^{i},q^{j}\\}$, i.e., $G_{ij}\Pi^{jk}=\delta_{i}^{k}$.

The number of states in an infinitesimal rectangle with sides
$(dq^{1},dq^{2})$ is

$d\Gamma=\frac{1}{2\pi\hbar}|G_{12}|\,dq^{1}\,dq^{2}=\sqrt{\det
G}\,\prod_{i=1}^{2}\frac{dq^{i}}{\sqrt{2\pi\hbar}}.$ (35)

For the choice of coordinates $q^{1}=\theta$ and $q^{2}=\phi$, the nonzero
components of the tensors are

$G_{\theta\phi}=-G_{\phi\theta}=-J\sin{\theta},\quad D_{\theta\theta}=\alpha
J,\quad D_{\phi\phi}=\alpha J\sin^{2}{\theta}.$ (36)

Upon substituting these tensor components into Eq. (33), we obtain the
equations of motion for the spherical angles (22).

###  2.12 Rayleigh’s dissipation function

Dissipative forces cannot be described as a term in a Lagrangian. It is an
emergent force, arising from interaction of a macroscopic object with numerous
microscopic degrees of freedom. In thermodynamics, it plays a major role in
the relaxation of the physical system toward thermal equilibrium. In that
context, the dissipative force can be obtained from the Rayleigh dissipation
function $R$ quadratic in velocities [9],

$R=\frac{1}{2}\alpha
J|\dot{\mathbf{m}}|^{2}=\frac{1}{2}D_{ij}\dot{q}^{i}\dot{q}^{j}.$ (37)

The modified Euler-Lagrange equations are

$\frac{\partial L}{\partial q^{i}}-\frac{d}{dt}\frac{\partial
L}{\partial\dot{q}^{i}}-\frac{\partial R}{\partial\dot{q}^{i}}=0.$ (38)

The last term is the dissipative force.

###  2.13 Example: strong easy-plane anisotropy

Let us discuss the dynamics of a spin whose potential energy

$U(\theta,\phi)=-\frac{K_{\theta}}{2}\sin^{2}{\theta}+V(\theta,\phi)$ (39)

is dominated by the easy-plane anisotropy term with $K_{\theta}>0$. Whatever
the starting state, after a period of initial relaxation the spin will find
itself near the equatorial circle $\theta=\pi/2$. Its long-term dynamics will
be influenced by the azimuthal potential landscape defined by $V(\pi/2,\phi)$.

To get a first look at the long-term dynamics, consider the motion in the
vicinity of the potential energy minimum at $\theta=\pi/2$ and some
$\phi=\phi_{0}$, where the potential energy is approximately quadratic in the
deviations $\delta\theta$ and $\delta\phi$ from the minimum:

$U(\pi/2+\delta\theta,\phi_{0}+\delta\phi)\approx\frac{K_{\theta}\delta\theta^{2}}{2}+\frac{K_{\phi}\delta\phi^{2}}{2},$
(40)

where $k_{\phi}\ll K_{\theta}$. Equations of motion (22), linearized in the
deviations from equilibrium, read

$-J\,\delta\dot{\phi}-K_{\theta}\,\delta\theta-\alpha
J\,\delta\dot{\theta}=0,\quad
J\,\delta\dot{\theta}-\,K_{\phi}\delta\phi-\,\alpha J\delta\dot{\phi}=0.$ (41)

In the absence of dissipation, $\alpha=0$, we obtain elliptical precession
near the equatorial plane with the frequency
$\omega=\sqrt{K_{\theta}K_{\phi}}/J$ and the amplitude ratio
$\delta\theta_{0}/\delta\phi_{0}=\sqrt{K_{\phi}/K_{\theta}}\ll 1$. Turning on
dissipation at first makes the precession underdamped, with the relaxation
rate $\Gamma\approx\alpha K_{\theta}/2J$ in the limit $\alpha\ll
2\sqrt{K_{\phi}/K_{\theta}}$. When the damping constant exceeds the critical
value $\alpha_{c}=2\sqrt{K_{\theta}K_{\phi}}/(K_{\theta}-K_{\phi})\ll 1$, the
spin dynamics becomes overdamped, with two distinct relaxation rates,
$\Gamma_{1}\approx\alpha K_{\theta}/J$ and $\Gamma_{2}\approx K_{\phi}/\alpha
J\ll\Gamma_{1}$ in the limit $2\sqrt{K_{\phi}/K_{\theta}}\ll\alpha\ll 1$.

Our simple analysis of the final approach to equilibrium demonstrates that,
even when dissipation is nominally weak, $\alpha\ll 1$, spin dynamics does not
necessarily retain its precessional character and can be purely relaxational.
An apparently small damping constant $\alpha\ll 1$ may represent relatively
strong dissipation when it is compared to another small parameter, the ratio
of azimuthal and polar anisotropies $\sqrt{K_{\phi}/K_{\theta}}$. Put another
way, increasing the strength of the easy-plane anisotropy $K_{\theta}$ at a
fixed $\alpha\ll 1$ will eventually bring us into a regime where the spin
motion separates into two independent components, fast relaxation, during
which the spin approaches the equatorial plane, and slow relaxation within the
equatorial plane. (The fast mode include azimuthal as well as polar motion.)

Next we shall derive the long-term dynamics of spin motion in a more general
setting, when the polar angle is already near equilibrium,
$\theta=\pi/2+\delta\theta$ with a small $\delta\phi$, but $\phi$ may not be.
The Lagrangian can then be approximated as follows:

$L\approx-J\,\delta\theta\,\dot{\phi}-K_{\theta}\delta\theta^{2}/2-V(\pi/2,\phi).$
(42)

In the Rayleigh dissipation function,

$R\approx\alpha J(\delta\dot{\theta}^{2}+\dot{\phi}^{2})/2\approx\alpha
J\dot{\phi}^{2}/2,$ (43)

we have neglected the contribution of polar motion to damping because it is
restricted by the strong easy-plane anisotropy, so
$\delta\dot{\theta}\ll\dot{\phi}$. This is an important general point, so let
us emphasize it: _dissipation is dominated by soft modes; dissipation from
hard modes can often be neglected_.

From the Lagrangian (42) and Rayleigh function (43) we obtain the equations of
motion:

$-J\dot{\phi}-K_{\theta}\delta\theta=0,\quad J\delta\dot{\theta}-\partial
V/\partial\phi-\alpha J\dot{\phi}=0.$ (44)

The first equation lets us express the polar angle (a hard mode of no interest
to us) in terms of the azimuthal angle (the soft mode we are interested in).
Then the second equation will be for the azimuthal angle alone:

$I\ddot{\phi}-\alpha J\dot{\phi}-\partial V/\partial\phi=0,$ (45)

where we introduced a “moment of inertia” $I=J^{2}/K_{\theta}$.

By integrating out the hard mode $\delta\theta$, we have generated inertia
(mass) for the soft mode $\phi$. This is a sufficiently frequent occurrence in
ferromagnetic dynamics to merit a proper name, the Döring mass [10]. The
Lagrangian of the azimuthal angle alone (with the polar angle integrated out)
acquires kinetic energy:

$L(\phi)=I\dot{\phi}^{2}/2-V(\pi/2,\phi).$ (46)

This elegant result hinges on neglect of dissipation for the hard mode
$\delta\theta$. Had we not neglected it, the equation relating $\theta$ to
$\phi$ would be more involved—non-local in time—and this simple picture would
not have emerged. The non-locality in time is negligible on time scales longer
than $\tau_{1}=1/\Gamma_{1}$.

Returning to the equation of motion for the azimuthal angle (45), we see that
it describes a massive rotator subject to an external potential and
dissipative friction. On long time scales, we may neglect inertia and obtain a
simpler equation of dissipative dynamics,

$-\alpha J\dot{\phi}-\partial V/\partial\phi=0.$ (47)

The azimuthal velocity $\dot{\phi}$ is proportional to the potential torque
$-\partial V/\partial\phi$. Doing so is permissible on time scales longer than
the characteristic acceleration time $\tau_{1}=I/\alpha J=J/\alpha
K_{\theta}=1/\Gamma_{1}$.

We should check that our long-term dynamics reproduces the previously obtained
approach to equilibrium, when $\phi=\phi_{0}+\delta\phi$. For small
$\delta\phi$, the equatorial torque $-\partial
V/\partial\phi\approx-k\,\delta\phi$ and we recover exponential relaxation of
$\delta\phi$ at the rate $\Gamma_{2}=K_{\phi}/\alpha J$, in agreement with our
preliminary analysis.

On a final note, we observe that the long-term dynamics of the soft mode
$\phi$ given by Eq. (47) can be obtained directly from the Lagrangian (42) and
Rayleigh dissipation function (43) by merely ignoring the hard mode
$\delta\theta$. This is a general principle that we will apply in many
contexts in what follows: _the response of a system to weak external
perturbations is dominated by its soft modes_.

###  2.14 Recap: equations of motion

We end the discussion of a single magnetic dipole by gathering in one place
different forms of its equations of motion, all of which are equivalent to one
another.

  * •

This form of the equation expresses the balance of the gyroscopic,
conservative, and dissipative forces acting on a magnetic dipole:

$-J\dot{\mathbf{m}}\times\mathbf{m}-\frac{dU}{d\mathbf{m}}-\alpha
J\dot{\mathbf{m}}=0.$ (48)

  * •

This equation expresses the rate of change of angular momentum in terms of
conservative and dissipative torques acting on the dipole:

$J\dot{\mathbf{m}}=-\mathbf{m}\times\frac{dU}{d\mathbf{m}}-\alpha
J\mathbf{m}\times\dot{\mathbf{m}}.$ (49)

  * •

The previous equation is often written in terms of an “effective field”
$\mathbf{H}_{\mathrm{eff}}=-dU/d\boldsymbol{\mu}$ as follows:

$\dot{\mathbf{m}}=-\gamma\mathbf{H}_{\mathrm{eff}}\times\mathbf{m}-\alpha\mathbf{m}\times\dot{\mathbf{m}},\quad\mathbf{H}_{\mathrm{eff}}=-\frac{1}{\mu}\frac{dU}{d\mathbf{m}}.$
(50)

This form of the equation of motion is known as the Landau-Lifshitz-Gilbert
(LLG) equation.

  * •

The LLG equation (50) contains two terms with a time derivative
$\dot{\mathbf{m}}$, which is inconvenient when the equation is integrated
numerically. Through simple algebra, it can be converted into the form used
originally by Landau and Lifshitz,

$(1+\alpha^{2})\dot{\mathbf{m}}=-\gamma\mathbf{H}_{\mathrm{eff}}\times\mathbf{m}-\alpha|\gamma|\mathbf{H}_{\mathrm{eff}}.$
(51)

This form of the equation contains a single term with the time derivative
$\dot{\mathbf{m}}$ and is used in numerical micromagnetic solvers such as
OOMMF and MuMax. One cautionary remark is on order. The effective field
$\mathbf{H}_{\mathrm{eff}}$ defined in Eq. (50) is understood as a derivative
$dU(\mathbf{m})/d\mathbf{m}$ with respect to _transverse_ variations of
$\mathbf{m}$, see Eq. (7). A sloppy evaluation may produce a longitudinal
component of $\mathbf{H}_{\mathrm{eff}}$ parallel to $\mathbf{m}$ and create a
problem for the Landau-Lifshitz equation (51), changing the length of the unit
vector $\mathbf{m}$. Just to be on the safe side, one may replace
$\mathbf{H}_{\mathrm{eff}}$ in the second term of Eq. (51) with its transverse
part $\mathbf{m}\times(\mathbf{H}_{\mathrm{eff}}\times\mathbf{m})$. The same
applies to Eq. (48).

  * •

Written in terms of spherical angles, the equations of motion are

$\displaystyle-J\sin{\theta}\,\dot{\phi}-\frac{\partial
U}{\partial\theta}-\alpha J\dot{\theta}=0,$ $\displaystyle
J\dot{\theta}-\frac{1}{\sin{\theta}}\frac{\partial U}{\partial\phi}-\alpha
J\sin{\theta}\,\dot{\phi}=0.$ (52)

  * •

For arbitrary coordinates $q^{1}$ and $q^{2}$ parametrizing the unit vector
$\mathbf{m}$, the equations of motion for these coordinates again expresses
the balance of the gyroscopic, conservative, and dissipative forces associated
with each coordinate:

$G_{ij}\dot{q}^{j}-\partial U/\partial q^{i}-D_{ij}\dot{q}^{j}=0.$ (53)

The antisymmetric gyroscopic and symmetric dissipative tensors are

$G_{ij}=-J\mathbf{m}\cdot\left(\frac{\partial\mathbf{m}}{\partial
q^{i}}\times\frac{\partial\mathbf{m}}{\partial q^{j}}\right),\quad
D_{ij}=\alpha J\,\frac{\partial\mathbf{m}}{\partial
q^{i}}\cdot\frac{\partial\mathbf{m}}{\partial q^{j}}.$ (54)

##  3 Ferromagnet in 1 dimension

This section discusses the dynamics of a one-dimensional ferromagnet, its spin
wave, and domain wall within the continuum description.

###  3.1 Heisenberg spin chain

Theory of a single magnetic dipole can be readily extended to that of many
dipoles. A convenient way to accomplish that is to start with the Lagrangian
of a single spin (12) and to generalize it to many spins
$\\{\mathbf{m}_{n}\\}$:

$L=\sum_{n}\mathbf{a}(\mathbf{m}_{n})\cdot\dot{\mathbf{m}}_{n}-U(\\{\mathbf{m}_{n}\\}),$
(55)

where once again the vector potential $\mathbf{a}(\mathbf{m})$ describes the
field of a magnetic monopole,
$\nabla_{\mathbf{m}}\times\mathbf{a}(\mathbf{m})=-J\mathbf{m}$, and
$U(\\{\mathbf{m}_{n}\\})$ is the potential energy of the system. The Rayleigh
function is generalized in the same way,

$R=\frac{1}{2}\sum_{n}\alpha J|\dot{\mathbf{m}}_{n}|^{2}.$ (56)

Then the Euler-Lagrange equations of motion for every spin $\mathbf{n}$,

$\frac{\partial L}{\partial\mathbf{m}_{n}}-\frac{d}{dt}\frac{\partial
L}{\partial\dot{\mathbf{m}}_{n}}-\frac{\partial
R}{\partial\dot{\mathbf{m}}_{n}}=0,$ (57)

yields an obvious generalization of single-spin dynamics (49),

$J\dot{\mathbf{m}}_{n}=-\mathbf{m}_{n}\times\frac{\partial
U}{\partial\mathbf{m}_{n}}-\alpha J\mathbf{m}_{n}\times\dot{\mathbf{m}}_{n},$
(58)

or its variants, Eqs. (48) and (50). The effective field acting on spin
$\mathbf{m}_{n}$ is $\mathbf{H}^{\mathrm{eff}}_{n}=-\mu^{-1}\,\partial
U/\partial\mathbf{m}_{n}$.

Here we examine a Heisenberg chain with magnetic dipoles arranged periodically
with lattice constant $a$. Adjacent dipoles $\mathbf{m}_{n}$ and
$\mathbf{m}_{n+1}$ interact with energy
$-A\,\mathbf{m}_{n}\cdot\mathbf{m}_{n+1}$, where $A$ is the exchange constant:

$U=-A\sum_{n}\mathbf{m}_{n}\cdot\mathbf{m}_{n+1}.$ (59)

The equation of motion for dipole $\mathbf{m}_{n}$ is

$J\dot{\mathbf{m}}_{n}=A\mathbf{m}_{n}\times(\mathbf{m}_{n-1}+\mathbf{m}_{n+1})-\alpha
J\mathbf{m}_{n}\times\dot{\mathbf{m}}_{n}.$ (60)

An equilibrium state, $\dot{\mathbf{m}}_{n}=0$, is reached when all dipoles
line up along the same direction, $\mathbf{m}_{n}=\mathbf{m}$. Such a state
clearly minimizes the potential energy (59) for $A>0$. The common direction of
magnetization $\mathbf{m}$ is arbitrary.

It is instructive to investigate the dynamics of magnetization close to
equilibrium, $\mathbf{m}_{n}=\mathbf{m}+\delta\mathbf{m}_{n}$, where
infinitesimal deviations from equilibrium $\delta\mathbf{m}_{n}$ are
transverse to $\mathbf{m}$ in order to preserve the unit length of
$\mathbf{m}_{n}$.

It is convenient to introduce a reference frame with three unit vectors
$\mathbf{e}_{x}$, $\mathbf{e}_{y}$, and $\mathbf{e}_{z}$ forming a right
triple, $\mathbf{e}_{i}\times\mathbf{e}_{j}=\epsilon_{ijk}\mathbf{e}_{k}$. We
align one of them with the equilibrium direction, $\mathbf{e}_{z}=\mathbf{m}$.
Then $\delta\mathbf{m}_{n}=m_{n,x}\,\mathbf{e}_{x}+\delta
m_{n,y}\,\mathbf{e}_{y}$. After expanding Eq. (60) to the first order in
$\delta\mathbf{m}_{n}$, we obtain the dynamical equation for the deviations:

$\displaystyle J\dot{m}_{n,x}$ $\displaystyle=$
$\displaystyle-A(m_{n-1,y}-2m_{n,y}+m_{n+1,y})+\alpha J\dot{m}_{n,y},$
$\displaystyle-J\dot{m}_{n,y}$ $\displaystyle=$
$\displaystyle-A(m_{n-1,x}-2m_{n,x}+m_{n+1,x})+\alpha J\dot{m}_{n,x}.$ (61)

The two transverse components can be conveniently combined into a complex
number $\psi_{n}=m_{n,x}+im_{n,y}$. These complex variables have the following
equation of motion:

$i(J+i\alpha J)\dot{\psi}_{n}=-A(\psi_{n-1}-2\psi_{n}+\psi_{n+1}).$ (62)

Solutions of this equation have the form of waves $\delta
m_{n}(t)=Ce^{ikna-i\omega t}$, where $a$ is the lattice period and $k$ is the
wavenumber; the frequency is

$\omega=\frac{2A(1-\cos{ka})}{J+i\alpha J}.$ (63)

Two things are worth noting, First, the frequency of a spin wave $\omega$
vanishes when the wavenumber $k\to 0$. This is a consequence of spontaneous
breaking of a continuous symmetry, in this case of global spin rotations, by
the ground state with spontaneous magnetization $\mathbf{m}$. Second, in the
presence of dissipation ($\alpha>0$), the frequency has an imaginary part,

$\mathop{\mathrm{Im}{\omega}}=-\frac{2\alpha A(1-\cos{ka})}{J(1+\alpha^{2})}.$
(64)

Because $\mathop{\mathrm{Im}{\omega}}<0$, a spin wave
$\psi_{n}(t)=Ce^{ikna-i\omega t}$ decays exponentially in time.

The mathematics can be simplified for waves whose wavelength is long compared
to the lattice spacing, $ka\ll 1$. For such waves, magnetizations of adjacent
dipoles are almost the same, $\mathbf{m}_{n+1}\approx\mathbf{m}_{n}$. In this
limit, we may treat discrete variables $\mathbf{m}_{n}$ as a slowly varying
and continuous function $\mathbf{m}(z)$, where $z=na$ is the coordinate along
the chain. Then finite differences can be transformed into spatial derivatives
via Taylor expansion:

$\displaystyle\mathbf{m}_{n+1}-\mathbf{m}_{n}=\mathbf{m}(z+a)-\mathbf{m}(z)\approx\frac{\partial\mathbf{m}(z)}{\partial
z}a,$
$\displaystyle\mathbf{m}_{n+1}-2\mathbf{m}_{n}+\mathbf{m}_{n-1}\approx\frac{\partial^{2}\mathbf{m}(z)}{\partial
z^{2}}a^{2},$ (65)

and so on.

In the continuum approximation, the equation of motion for transverse
fluctuations $\psi(z,t)=m_{x}(z,t)+im_{y}(z,t)$ (62) reads

$iJ\frac{\partial\psi}{\partial t}=-Aa^{2}\frac{\partial^{2}\psi}{\partial
z^{2}}.$ (66)

For simplicity, we switched off dissipation, $\alpha\to 0$. The resulting wave
equation resembles the Schrödinger equation for a free particle with a
nonrelativistic energy-momentum relation $E=p^{2}/2m$ and the mass
$m=J\hbar/(2a^{2}A)$. These particles were first introduced by Felix Bloch
[11] and are now known as magnons, the quanta of spin waves.

###  3.2 Continuum theory of a ferromagnet

To develop a continuum description of a ferromagnetic medium, on a formal
level, we promote the unit vector $\mathbf{m}$ encoding the magnetic moment of
a single dipole to a unit-vector field $\mathbf{m}(\mathbf{r})$ varying
smoothly in space $\mathbf{r}$. The mathematical apparatus developed in the
previous section is generalized accordingly.

For example, the equation of motion (48) undergoes two very minor changes:

$-\mathcal{J}\dot{\mathbf{m}}\times\mathbf{m}-\frac{\delta
U}{\delta\mathbf{m}}-\alpha\mathcal{J}\dot{\mathbf{m}}=0.$ (67)

First, the “length” of angular momentum $J$ has been replaced with the
corresponding intensive quantity, the angular momentum $\mathcal{J}$ per unit
volume. Second, the potential energy function $U(\mathbf{m})$ has become the
potential energy functional $U[\mathbf{m}(\mathbf{r})]$ and thus the ordinary
derivative $dU/d\mathbf{m}$ (7) has turned into the functional derivative
$\delta U/\delta\mathbf{m}(\mathbf{r})$.

The Landau-Lifshitz equations (50) and (51) retain their forms, albeit with an
effective field

$\mathbf{H}_{\mathrm{eff}}(\mathbf{r})=-\frac{1}{\mathcal{M}}\frac{\delta
U}{\delta\mathbf{m}(\mathbf{r})}$ (68)

containing the functional derivative.

Equations of motion for the spherical-angle fields $\theta(\mathbf{r})$ and
$\phi(\mathbf{r})$ are

$\displaystyle-\mathcal{J}\sin{\theta}\,\dot{\phi}-\frac{\delta
U}{\delta\theta}-\alpha\mathcal{J}\dot{\theta}=0,$
$\displaystyle\mathcal{J}\dot{\theta}-\frac{1}{\sin{\theta}}\frac{\delta
U}{\delta\phi}-\alpha\mathcal{J}\sin{\theta}\,\dot{\phi}=0.$ (69)

A field $\mathbf{m}(\mathbf{r})$ has infinitely many degrees of freedom, so if
we attempt to describe it in terms of some coordinates $q^{i}$, the number of
these coordinates will be infinite. Nonetheless, the equations of motion for
these coordinates remain unchanged,

$G_{ij}\dot{q}^{j}-\partial U/\partial q^{i}-D_{ij}\dot{q}^{j}=0,$ (70)

and only the gyroscopic and dissipative tensors are promoted to functionals of
the magnetization field:

$G_{ij}=-\mathcal{J}\int
dV\,\mathbf{m}\cdot\left(\frac{\partial\mathbf{m}}{\partial
q^{i}}\times\frac{\partial\mathbf{m}}{\partial q^{j}}\right),\quad
D_{ij}=\alpha\mathcal{J}\int dV\,\frac{\partial\mathbf{m}}{\partial
q^{i}}\cdot\frac{\partial\mathbf{m}}{\partial q^{j}}.$ (71)

Our model is a ferromagnet in one spatial dimension with coordinate $z$. It
can describe a ferromagnetic wire whose transverse dimensions are smaller than
the characteristic length $\lambda_{0}$ defined in Eq. (74). Exchange
interaction tends to align spins, so a ground state should have a uniform
magnetization field $\mathbf{m}(z)=\mathrm{const}$. However, some of these
uniform states will have lower energies than others. In particular, long-range
dipolar interactions prefer that the dipoles line up along the direction of
the wire, leaving just two ground states,

$\mathbf{m}(z)=\pm\mathbf{e}_{z}=(0,0,\pm 1).$ (72)

###  3.3 Model with local interactions

We will use a simple phenomenenological form of the potential energy
functional minimized by these two ground states:

$U[\mathbf{m}(z)]=\int
dz\left(A|\mathbf{m}^{\prime}|^{2}+K|\mathbf{m}\times\mathbf{e}_{z}|^{2}\right)/2.$
(73)

Here $\mathbf{m}^{\prime}\equiv d\mathbf{m}/dz$. The first term models the
exchange interaction and penalizes spatially non-uniform states. It can be
derived as a continuum approximation to the Heisenberg chain. The second term
penalizes states with magnetization deviating from the direction of the wire
$\mathbf{e}_{z}$.

Both the exchange constant $A$ and anisotropy strength $K$ are positive.
Together with the density of angular momentum $\mathcal{J}$, they define the
characteristic scales of length $\lambda_{0}$, time $\tau_{0}$, and energy
$\epsilon_{0}$:

$\lambda_{0}=\sqrt{A/K},\quad\tau_{0}=\mathcal{J}/K,\quad\epsilon_{0}=\sqrt{AK}.$
(74)

We stress that this model is not very realistic as it relies on a local
anisotropy term to select the two ground states (72). A more realistic model
with dipolar interactions would require a non-local energy functional as spins
would interact over large distances. The advantage of our toy model is in its
relative simplicity: with not too much effort, we can study its linear and
nonlinear excitations.

###  3.4 Spin waves in a ground state

To obtain spin waves near one of the uniform ground states,
$\mathbf{m}_{0}=\mathbf{e}_{z}$, we consider a weakly excited state with
$\mathbf{m}(z)=\mathbf{m}_{0}+\delta\mathbf{m}(z)$, where
$\delta\mathbf{m}=\mathbf{e}_{x}m_{x}+\mathbf{e}_{y}m_{y}$ represents small
transverse deviations of magnetization. Expand the Lagrangian density to the
second order in $\delta\mathbf{m}$ as in Sec. 2.7:

$L=\frac{1}{2}\int
dz\,\left(\dot{m}_{x}m_{y}-\dot{m}_{y}m_{x}-{m_{x}^{\prime}}^{2}-{m_{y}^{\prime}}^{2}-m_{x}^{2}-m_{y}^{2}\right)+\ldots$
(75)

Here we have switched to natural units (74), in which $\mathcal{J}=A=K=1$. In
the absence of dissipation, $\alpha=0$, the equations of motion for the
transverse magnetization are

$-\dot{m}_{y}+m_{x}^{\prime\prime}-m_{x}=0,\quad\dot{m}_{x}+m_{y}^{\prime\prime}-m_{y}=0.$
(76)

It is convenient to combine the two transverse components into a complex field
$\psi=m_{x}+im_{y}$, whose equation of motion,

$i\dot{\psi}=-\psi^{\prime\prime}+\psi,$ (77)

resembles the Schrödinger equation $i\dot{\psi}=\mathcal{H}\psi$ with the
Hamiltonian

$\mathcal{H}=-\frac{d^{2}}{dz^{2}}+1,$ (78)

whose energy spectrum is $\epsilon_{k}=1+k^{2}$. A traveling spin wave
$\psi(t,z)=\psi(0,0)e^{-i\omega t+ikz}$ has the frequency

$\omega=\epsilon_{k}=\frac{K+Ak^{2}}{\mathcal{J}},$ (79)

where we have restored the units. The frequency spectrum has a gap with a
minimal excitation frequency $\lim_{k\to 0}\omega=K/\mathcal{J}=1/\tau_{0}$.

###  3.5 Static domain wall

The uniform ground states $\mathbf{m}(z)=\pm\mathbf{e}_{z}$ are global minima
of the potential energy (73). There are also local minima, in which
magnetization interpolates between $-\mathbf{e}_{z}$ on one end of the wire
and $+\mathbf{e}_{z}$ on the other. In such a state, magnetization
$\mathbf{m}(z)$ forms two (nearly) uniform domains, in which the energy
density is (nearly) zero. The region in between has non-uniform magnetization
deviating from the easy axis has a positive energy density and is known as a
domain wall.

Figure 2: A domain wall with the $Z_{2}$ topological charge $\sigma=-1$. The
$z$-axis is parallel to the chain.

To obtain a formal solution for a domain wall, it is convenient to write the
potential energy (73) in terms of spherical angles,

$U[\theta(z),\phi(z)]=\int
dz\left[A({\theta^{\prime}}^{2}+\sin^{2}{\theta}\,{\phi^{\prime}}^{2})+K\sin^{2}{\theta}\right]/2.$
(80)

Minimization of the energy $\delta U=0$ yields field equations

$-A\theta^{\prime\prime}+\frac{1}{2}(A{\phi^{\prime}}^{2}+K)\sin{2\theta}=0,\quad
A(\sin^{2}{\theta}\,\phi^{\prime})^{\prime}=0.$ (81)

Solutions with a uniform azimuthal angle, $\phi^{\prime}=0$ automatically
satisfy the second of these equations. The first one can be integrated once to
obtain $-A{\theta^{\prime}}^{2}+K\sin^{2}{\theta}=C$. The constant of
integration can be obtained by noting that at $z\to\pm\infty$ we recover
uniform states with $\theta^{\prime}=0$ and $\sin{\theta}=0$. Therefore $C=0$.
Integrating this equation yields domain-wall solutions,

$\cos{\theta(z)}=\sigma\tanh{\frac{z-Z}{\lambda_{0}}},\quad\phi(z)=\Phi.$ (82)

A domain wall with the topological charge $\sigma=-1$ is shown in Fig. 2.

Solutions (82) have three parameters, one discrete and two continuous.

  * •

Topological charge $\sigma=\pm 1$ determines the ground states on the two
sides of the domain wall: $\mathbf{m}=-\sigma\mathbf{e}_{z}$ at $z=-\infty$
and $\mathbf{m}=\sigma\mathbf{e}_{z}$ at $z=+\infty$. The word “topological”
indicates that this characteristic cannot be changed by small continuous
deformations of the soliton.

  * •

Position $Z$ determines the location of the domain wall. The change from
$\mathbf{m}=-\sigma\mathbf{e}_{z}$ to $\sigma\mathbf{e}_{z}$ happens around
$z=Z$, in a region of width $\lambda_{0}$ (74).

  * •

Angle $\Phi$ defines an azimuthal plane in which the spins interpolate between
$\mathbf{m}=-\sigma\mathbf{e}_{z}$ and $\sigma\mathbf{e}_{z}$.

Note that the domain-wall solution (82) contains two parameters, $Z$ and
$\Phi$, defining the position of the domain wall and its azimuthal plane.
These _collective coordinates_ of a domain wall quantify its rigid
translations and rotations, transformations that leave the energy (73)
invariant and thus known as _zero modes_.

###  3.6 Domain wall in a magnetic field

A magnetic field applied along the easy axis, $\mathbf{H}=H\mathbf{e}_{z}$,
exerts a force on a domain wall, as can be seen from the following
consideration. The field couples to magnetization through the Zeeman term with
energy density $-\mathcal{M}\mathbf{H}\cdot\mathbf{m}$, where the
magnetization density is $\mathcal{M}=\gamma\mathcal{J}$. By moving the domain
wall, we can reduce its Zeeman energy by increasing the length of the domain
where $\mathbf{m}$ points along $\mathbf{H}$. Displacing the domain wall (82)
by $\Delta Z$ changes the Zeeman energy by $2\sigma\mathcal{M}H\Delta Z$,
which indicates that the applied field exerts force

$F_{Z}=-2\sigma\mathcal{M}H$ (83)

on the wall.

It is natural to expect that the domain wall will move in the direction of the
applied force. However, we shall see shortly, that the primary effect of the
force will be precession of the domain wall, rather than translational motion.
In fact, in the absence of dissipation the domain wall will not move at all,
$\dot{Z}=0$, because conservation of energy precludes that. The presence of
dissipation will allow translational motion with a velocity proportional to
the damping constant $\alpha$.

To determine the dynamics of the domain wall in a magnetic field, we use the
equations of motion for the fields $\theta(z)$ and $\phi(z)$ (69) with the
energy functional

$U=\int
dz\left\\{\frac{A}{2}({\theta^{\prime}}^{2}+\sin^{2}{\theta}\,{\phi^{\prime}}^{2})+\frac{K}{2}\sin^{2}{\theta}-\mathcal{M}H\cos{\theta}\right\\}.$
(84)

The equations of motion are

$\displaystyle-\mathcal{J}\sin{\theta}\,\dot{\phi}+A\theta^{\prime\prime}-\frac{1}{2}(A{\phi^{\prime}}^{2}+K)\sin{2\theta}-\mathcal{M}H\sin{\theta}-\alpha\mathcal{J}\dot{\theta}=0,$
$\displaystyle\mathcal{J}\dot{\theta}-A(\sin^{2}{\theta}\,\phi^{\prime})^{\prime}-\alpha\mathcal{J}\sin{\theta}\,\dot{\phi}=0.$
(85)

Although this looks like a big mess, it turns out that our previous solution
can still be made to work, with one minor modification. We simply assume that
the collective coordinates $Z$ and $\Phi$ in Eq. (82) become dependent on
time. Eqs. (85) are then satisfied if the linear and angular velocities obey
the relations

$-\mathcal{J}\dot{\Phi}-\mathcal{M}H-\sigma\alpha\mathcal{J}\dot{Z}/\lambda_{0}=0,\quad\sigma\mathcal{J}\dot{Z}-\alpha\mathcal{J}\lambda_{0}\dot{\Phi}=0.$
(86)

We thus obtain the velocities

$\dot{\Phi}=-\frac{\gamma
H}{1+\alpha^{2}},\quad\dot{Z}=-\sigma\frac{\alpha\lambda_{0}\gamma
H}{1+\alpha^{2}}.$ (87)

Two things immediately stand out. First, as we have already mentioned, the
primary effect of the applied force (83) is precession: the angular velocity
$\dot{\Phi}$ is of order $\alpha^{0}$, whereas the linear velocity $\dot{Z}$
is of order $\alpha^{1}$. Second, the precession frequency of a domain wall is
the same as that of a single spin (25). The complexity brought by interactions
of spins has no effect on the precession frequency. Simple results call for
simple explanations, but our purely mathematical approach to the problem—find
a solution to coupled partial differential equations—obscures the physical
picture. Furthermore, closed-form solutions are available only in a very
restricted set of problems. It worked for a magnetic field along the easy axis
but it would fail for a field orthogonal to it. We need to find another
perspective.

###  3.7 Weakly perturbed domain wall

Fortunately, there is a general approach that will provide a path to solution
for all kinds of perturbations and will be provide more physical intuition. Of
course, there is a price that we will pay for the universality of the
approach: it is only guaranteed to work for weak perturbations.

The method is based on rigidity of a domain wall (and solitons in general). It
takes a finite amount of energy to alter the width of a domain wall from its
equilibrium value $\lambda_{0}$. On the other hand, no energy is required to
translate a domain wall or to rotate its azimuthal plane. One can make an
analogy with a tennis ball that is hard to squeeze but easy to move. Thus we
may expect that a weak enough perturbation will produce substantial response
only in the zero modes $Z$ and $\Phi$ of a domain wall but not in other modes
(such as its width $\lambda_{0}$).

On a formal level, although a domain wall can be deformed in an infinite
number of ways and is therefore described by infinitely many coordinates
$q^{i}$, we only need to take into account two of them—position $Z$ and
azimuthal angle $\Phi$—when we consider the motion of a domain wall under a
weak external perturbation.

Thus the recipe is to use the equations of motion (70) with $q^{i}=Z$ and
$\Phi$ only. To that end, we need the gyrotropic and dissipative tensors
$G_{ij}$ and $D_{ij}$ (71). The gyroscopic tensor $G_{ij}$ is antisymmetric,
so with 2 coordinates we just need to compute one of its components, say,
$G_{Z\Phi}$. The dissipative tensor $D_{ij}$ is symmetric and, for this choice
of coordinates, diagonal; so it only has 2 nonzero components.

We compute in detail only the gyroscopic coefficient

$G_{Z\Phi}=-\mathcal{J}\int_{-\infty}^{\infty}dz\,\mathbf{m}\cdot\left(\frac{\partial\mathbf{m}}{\partial
Z}\times\frac{\partial\mathbf{m}}{\partial\Phi}\right).$ (88)

As $\mathbf{m}$ is a function of $z-Z$, we may replace
$\partial\mathbf{m}/\partial Z$ with $-\partial\mathbf{m}/\partial z$. On the
other hand, $\phi=\Phi$ everywhere, so
$\partial\mathbf{m}/\partial\Phi=\partial\mathbf{m}/\partial\phi$. Next,
because only $\theta$ varies in space and $\phi$ does not, application of the
chain rule yields $\partial\mathbf{m}/\partial
z=(\partial\mathbf{m}/\partial\theta)(\partial\theta/\partial z)$. We then
make use of a local spin frame with three mutually orthogonal vectors
(pointing locally up, south, and east)

$\mathbf{e}_{m}=\mathbf{m},\quad\mathbf{e}_{\theta}=\frac{\partial\mathbf{m}}{\partial\theta},\quad\mathbf{e}_{\phi}=\frac{1}{\sin{\theta}}\frac{\partial\mathbf{m}}{\partial\phi},\quad\mathbf{e}_{m}=\mathbf{e}_{\theta}\times\mathbf{e}_{\phi},$
(89)

to obtain

$G_{Z\Phi}=-\mathcal{J}\int_{-\infty}^{\infty}dz\,\frac{\partial\cos{\theta}}{\partial
z}=-2\sigma\mathcal{J}=-G_{\Phi Z}.$ (90)

It is remarkable that the gyroscopic coefficient $G_{Z\Phi}$ depends just on
the density of angular momentum $\mathcal{J}$ and the topological charge
$\sigma$. It is completely insensitive to the precise shape $\theta(z-Z)$ of
the domain wall or even to the energetic parameters such as exchange and
anisotropy!

The dissipative coefficients are computed along similar lines:

$D_{ZZ}=2\alpha\mathcal{J}/\lambda_{0},\quad
D_{\Phi\Phi}=2\alpha\mathcal{J}\lambda_{0},\quad D_{\Phi Z}=D_{Z\Phi}=0$ (91)

We substitute the gyroscopic and dissipative coefficients into the equations
of motion to obtain

$-2\sigma\mathcal{J}\dot{\Phi}+F_{Z}-2\alpha\mathcal{J}\dot{Z}/\lambda_{0}=0,\quad
2\sigma\mathcal{J}\dot{Z}+F_{\Phi}-2\alpha\mathcal{J}\lambda_{0}\dot{\Phi}=0,$
(92)

where $F_{Z}=-\partial U/\partial Z$ is conservative force and
$F_{\Phi}=-\partial U/\partial\Phi$ conservative torque.

Returning to the previously solved problem of a domain wall in an applied
field parallel to the easy axis, we have a force $F_{Z}=-2\sigma\mathcal{M}H$
and torque $F_{\Phi}=0$. With these, the new approximate equations of motion
(92) reproduce the exact ones (86).

The new approach also provides an insight into the vexing question: why is the
precession frequency of a domain wall the same as that of a single spin in the
same magnetic field? In the absence of friction, $\alpha=0$, the precession
frequency can be obtained from the first equation (92):

$\dot{\Phi}=-\frac{F_{Z}}{G_{Z\Phi}}=-\frac{2\sigma\mathcal{M}H}{2\sigma\mathcal{J}}=-\gamma
H.$ (93)

Both the Zeeman force on the domain wall $F_{Z}=-2\sigma\mathcal{M}H$ and the
gyroscopic coefficient $G_{Z\Phi}=-2\sigma\mathcal{J}$ are insensitive to the
detailed structure of the domain wall. Their ration determines the precession
frequency.

Equations of motion (92) also paint a simple physical picture of the mechanics
of a domain wall in a ferromagnetic wire. With two degrees of freedom,
position $Z$ and angle $\Phi$, it resembles a bead on a string, which can move
along the string and rotate. However, the dynamics of the two objects are very
different. For a bead, a force creates linear acceleration and a torque
creates angular acceleration. For a domain wall, a force generates angular
velocity, whereas a torque produces linear velocity. _When pushed, a domain
wall rotates, and when twisted, it moves!_

We can also understand now how the addition of dissipation enables a domain
wall to move under an external force. When it rotates (its primary response to
a force), dissipation creates a dissipative torque directed against the
direction of rotation. This torque causes linear motion. The proportionality
of the dissipative torque to the damping constant $\alpha$ explains why the
velocity of the soliton is also of order $\alpha$.

###  3.8 Normal modes of a domain wall

So far we have approximated the dynamics of a domain wall by its two most
prominent modes—position $Z$ and global azimuthal angle $\Phi$. We have
justified the neglect of all other modes (such as the width of the domain wall
$\lambda$) by their hard nature. In this subsection, we shall make a
quantitative analysis of the hard modes and deduce time scales on which their
dynamics can be neglected. These results were first obtained by Yan _et al._
[12].

In natural units of length, time, and energy (74), the Lagrangian density is

$\mathcal{L}(\theta,\phi)=(\cos{\theta}-1)\dot{\phi}-({\theta^{\prime}}^{2}+\sin^{2}{\theta}\,{\phi^{\prime}}^{2}+\sin^{2}{\theta})/2.$
(94)

A static domain wall with $\sigma=+1$, $Z=0$ and $\Phi=0$ is

$\cos{\theta_{0}(z)}=\tanh{z},\quad\phi_{0}(z)=0.$ (95)

Expand the Lagrangian density in small deviations from this state
$\delta\theta(z)$ and $\delta\phi(z)$ to the second order:

$\mathcal{L}(\theta_{0}+\delta\theta,\phi_{0}+\delta\phi)=\mathcal{L}_{0}+\mathcal{L}_{1}+\mathcal{L}_{2}+\ldots,$
(96)

where $\mathcal{L}_{0}=\mathcal{L}(\theta_{0},\phi_{0})$. The first-order term
$\mathcal{L}_{1}=(\cos{\theta_{0}}-1)\delta\dot{\phi}$ is a time derivative;
it thus does not contribute to the equations of motion and can be neglected.
The second-order term is

$\mathcal{L}_{2}=-\sin{\theta_{0}}\,\delta\theta\,\delta\dot{\phi}-({\delta\theta^{\prime}}^{2}+\sin^{2}{\theta_{0}}\,{\delta\phi^{\prime}}^{2}+\cos{2\theta_{0}}\,\delta\theta^{2})/2$
(97)

Rewriting it in terms of transverse increments $\delta m_{\theta}$ and $\delta
m_{\phi}$ (5) reveals a hidden $O(2)$ symmetry:

$\mathcal{L}_{2}=-\delta m_{\theta}\,\delta\dot{m}_{\phi}-(\delta
m_{\theta}\,\mathcal{H}\,\delta m_{\theta}+\delta
m_{\phi}\,\mathcal{H}\,\delta m_{\phi})/2,$ (98)

where we have introduced the operator

$\mathcal{H}=-\frac{d^{2}}{dz^{2}}+1-2\mathrm{\,sech}^{2}{z}.$ (99)

To utilize the symmetry, we switch to complex fields

$\psi=\frac{\delta m_{\theta}+\,i\,\delta
m_{\phi}}{\sqrt{2}},\quad\psi^{*}=\frac{\delta m_{\theta}-i\,\delta
m_{\phi}}{\sqrt{2}}.$ (100)

Their Lagrangian density

$\mathcal{L}_{2}=i\psi^{*}\dot{\psi}-\psi^{*}\mathcal{H}\psi$ (101)

yields the equations of motion

$i\dot{\psi}=\mathcal{H}\psi,\quad-i\dot{\psi}^{*}=\mathcal{H}\psi^{*}.$ (102)

The spin-wave “Hamiltonian” $\mathcal{H}$ (99) is similar to its counterpart
for the ground state (78). The main difference is the presence of an
attractive Pöschl–Teller potential well
$\mathcal{U}(z)=-2\mathrm{\,sech}^{2}{z}$ [13] in Eq. (99) caused by the
presence of the domain wall. This Hamiltonian has remarkable properties. (See
A for a brief review and Ref. [14] for much more.) Its energy spectrum has a
single bound state with exactly zero energy, $\mathcal{H}\Psi_{0}=0$, and a
localized wavefunction

$\Psi_{0}(z)=2^{-1/2}\mathrm{\,sech}{z}.$ (103)

Its continuum states are reflectionless and are in a one-to-one correspondence
with the plane waves $e^{ikz}$ of the free spin-wave Hamiltonian (78):

$\mathcal{H}\psi_{k}=\epsilon_{k}\psi_{k},\quad\epsilon_{k}=1+k^{2},\quad\psi_{k}(z)=\frac{\tanh{z}-ik}{\sqrt{1+k^{2}}}e^{ikz}.$
(104)

Eigenstates with different $k$ are orthogonal to the zero mode $\Psi_{0}(z)$
and to one another,

$\int dz\,\psi_{k^{\prime}}^{*}(z)\psi_{k}(z)=2\pi\delta(k-k^{\prime}),$ (105)

The zero mode $\Psi_{0}(z)$ corresponds to the infinitesimal translations and
rotations of the domain wall, whereas the free states $\psi_{k}(z)$ describe
spin waves in the presence of the domain wall.

An arbitrary infinitesimal deformation of the domain wall can be expanded in
terms of the eigenmodes,

$\psi(z,t)=Q_{0}(t)\Psi_{0}(z)+\int\frac{dk}{2\pi}\,q_{k}(t)\psi_{k}(z),$
(106)

Complex amplitudes $Q_{0}(t)$ and $q_{k}(t)$ form a complete set of collective
coordinates for a domain wall. The amplitude of the zero mode is related to
position and azimuthal angle,

$Q_{0}=Z+i\Phi,$ (107)

whereas $q_{k}$ are complex amplitudes of the spin waves (104).

The Lagrangian of these collective coordinates is

$L_{2}=\int
dz\,\mathcal{L}_{2}=iQ_{0}^{*}\dot{Q}_{0}+\int\frac{dk}{2\pi}\,\left(iq_{k}^{*}\dot{q}_{k}-\epsilon_{k}|q_{k}|^{2}\right).$
(108)

The Rayleigh function is

$R=\alpha|\dot{Q}_{0}|^{2}+\alpha\int\frac{dk}{2\pi}\,|\dot{q}_{k}|^{2}.$
(109)

From these we obtain the equations of motion:

$i\dot{Q}_{0}-\alpha\dot{Q}_{0}=0,\quad
i\dot{q}_{k}-\epsilon_{k}q_{k}-\alpha\dot{q}_{k}=0,$ (110)

which yields the eigenfrequencies

$\Omega_{0}=0,\quad\omega_{k}=\frac{\epsilon_{k}}{1+i\alpha}=\frac{1+k^{2}}{1+i\alpha}.$
(111)

Upon restoring the units, we obtain the spectrum

$\omega_{k}=\frac{K+Ak^{2}}{\mathcal{J}(1+i\alpha\mathrm{sgn}{\mathcal{J}})}.$
(112)

The relaxation rate
$\gamma_{k}=-\mathrm{Im}{\omega_{k}}\approx\alpha|\epsilon_{k}|$ for
$\alpha\ll 1$ is slowest for modes with $k\to 0$,

$\lim_{k\to 0}{\gamma_{k}}=\alpha K/\mathcal{J}=\alpha/\tau_{0}.$ (113)

Thus the influence of spin waves can be neglected on time scales longer than
$\tau_{0}/\alpha$.

This criterion can be further relaxed. If external perturbations and the
response of the domain wall occur on time scales longer than $\tau_{0}$ then
spin waves, whose spectrum has a gap $|\omega_{k}|\geq 1/\tau_{0}$, are not
excited.

###  3.9 Walker’s model of domain wall dynamics

Schryer and Walker [15] considered a 1-dimensional model with additional
anisotropy that breaks the axial rotational symmetry and prefers two azimuthal
orientations, say $\Phi=\pm\pi/2$. The potential energy excluding the Zeeman
term is

$U_{0}=\int
dz\left[\frac{A}{2}({\theta^{\prime}}^{2}+\sin^{2}{\theta}{\phi^{\prime}}^{2})+\frac{K}{2}\sin^{2}{\theta}+\frac{\mathcal{M}^{2}}{8\pi}\sin^{2}{\theta}\cos^{2}{\phi}\right].$
(114)

The Zeeman term is given by

$U_{1}=-\int dz\,\mathcal{M}H\cos{\theta}\,.$ (115)

A static domain wall with a given azimuthal angle $\phi(z)=\Phi$ and
minimizing $U_{0}$ is

$\cos{\theta(z)}=\sigma\tanh{\frac{z-Z}{\lambda(\Phi)}},\quad\lambda^{-2}(\Phi)=\frac{1}{A}\left(K+\frac{\mathcal{M}^{2}}{4\pi}\cos^{2}{\Phi}\right).$
(116)

Here, the dynamic variables are position $Z$ and angle $\Phi$. The width of
the domain wall $\lambda(\Phi)$ is a slave of the angle $\Phi$, not an
independent variable.

The energy of the domain wall, including the Zeeman term is given by

$U(Z,\Phi)=2\sigma\mathcal{M}HZ+2\sqrt{A\left(K+\frac{\mathcal{M}^{2}}{4\pi}\cos^{2}{\Phi}\right)}.$
(117)

The gyrotropic and dissipative tensors (nonzero coefficients) are given by

$\displaystyle G_{Z\Phi}=-G_{\Phi Z}=-2\sigma\mathcal{J},$ $\displaystyle
D_{ZZ}=2\alpha\mathcal{J}/\lambda,\quad
D_{\Phi\Phi}=2\alpha\mathcal{J}\lambda\left[1+\left(\frac{1}{\lambda}\frac{\partial\lambda}{\partial\Phi}\right)^{2}\right].$
(118)

###  3.10 Angular momentum of a domain wall

Symmetry with respect to rotations about the $z$ axis implies conservation of
angular momentum $J_{z}$. In our one-dimensional model, there is no orbital
contribution to this quantity. Therefore angular momentum comes solely from
spin. This gives

$J_{z}(Z)=\int dz\,\mathcal{J}\mathbf{m}(z-Z)\cdot\mathbf{e}_{z}.$ (119)

This quantity is not well defined in an infinite wire, where the integration
region extends from $z=-\infty$ to $+\infty$ because the integrand approaches
the constant values $\pm\mathcal{J}$ away from the domain wall,
$|z-Z|\gg\lambda$. It is convenient to consider a finite wire of finite length
$\ell\gg\lambda$ so that the integration range is $-\ell/2<z<\ell/2$. If the
domain wall is located exactly in the middle of this wire, $Z=0$, then the
contributions of spins with positive and negative projections onto the $z$
axis cancel each other and we find $J_{z}=0$. Shifting the domain wall to a
position $Z\neq 0$ shifts the balance by elongating one domain and shortening
the other, creating a net angular momentum proportional to the displacement of
the domain wall, $J_{z}(Z)\propto Z$. To compute the proportionality constant,
we may take the derivative

$\displaystyle\frac{\partial J_{z}}{\partial Z}$ $\displaystyle=$
$\displaystyle\mathcal{J}\int_{-\ell/2}^{+\ell/2}dz\,\frac{\partial\mathbf{m}(z-Z)}{\partial
Z}\cdot\mathbf{e}_{z}=-\mathcal{J}\int_{-\ell/2}^{+\ell/2}dz\,\frac{\partial\mathbf{m}(z-Z)}{\partial
z}\cdot\mathbf{e}_{z}$ (120) $\displaystyle=$
$\displaystyle-\left.\mathcal{J}\mathbf{m}(z-Z)\cdot\mathbf{e}_{z}\right|_{-\ell/2}^{+\ell/2}\approx-2\sigma\mathcal{J}.$

In the last step, we assumed that the magnetization at the ends of the wire
returns to the ground-state values $\pm\sigma e_{z}$, which is justified if
the domain wall is not too close to the ends.

We thus obtain an important result:

$J_{z}(Z)=-2\sigma\mathcal{J}Z.$ (121)

The angular momentum of a domain wall in a ferromagnetic wire is proportional
to its position $Z$ relative to the center of the wire. The proportionality
coefficient is insensitive to the detailed structure of the domain wall and
depends only on the density of angular momentum $\mathcal{J}$ (per unit
length) and the topological charge of the domain wall $\sigma$ defined in Eq.
(82).

This result has interesting consequences. Angle $\Phi$ and angular momentum
$J_{z}$ form a canonical pair with the commutator $[\Phi,J_{z}]=i\hbar$. With
the aid of Eq. (121), we find that coordinate $Z$ and azimuthal angle $\Phi$
do not commute,

$[\Phi,Z]=-\frac{i\hbar}{2\sigma\mathcal{J}}.$ (122)

It follows that the position and azimuthal orientation of a domain wall cannot
be simultaneously measured. The uncertainty product is $\Delta
Z\Delta\Phi\geq\hbar/4\mathcal{J}$. In an atomic chain with the lattice
constant $a$ and atomic spins of length $\hbar S$ (with $2S$ an integer),
$\mathcal{J}=\hbar S/a$ and the uncertainty product $\Delta Z\Delta\Phi\geq
a/4S$ is of the order of the interatomic distance.

###  3.11 Linear momentum of a domain wall

In a uniform wire, translational symmetry implies conservation of linear
momentum $P_{z}$. We cannot derive it along the lines leading to Eq. (121).
Instead, we will reverse-engineer it from the commutator of $Z$ and $\Phi$
(122).

To that end, we recall that coordinate $Z$ and its momentum $P_{z}$ have the
commutator $[P_{z},Z]=-i\hbar$. Comparison with Eq. (122) shows that a
plausible answer would be

$P_{z}=2\sigma\mathcal{J}\Phi.$ (123)

The linear momentum of a domain wall in a ferromagnetic wire is thus
proportional to its azimuthal angle.

Although this result is hard to internalize, we can check that it is
compatible with the dynamics of a domain wall derived earlier. As we have seen
previously, an external magnetic field $\mathbf{H}$ parallel to the easy axis
$\mathbf{e}_{z}$ exerts on the domain wall the force
$F_{z}=-2\sigma\mathcal{M}H$. This external force determines the rate of
change for the linear momentum, $\dot{P}_{z}=-2\sigma\mathcal{M}H$. Our
conjectured result for $P_{z}$ (123) tells us that the domain wall will
precess at the rate $\dot{\Phi}=-\gamma H$, in accordance with Eq. (93).

Eqs. (121) and (123) reveal a pleasing symmetry. The _angular_ momentum is
proportional to the _linear_ coordinate of the domain wall and vice versa,
with the same proportionality coefficient, up to a sign.

There is, however, a subtle difference between the linear and angular
coordinates: whereas the former lives on a straight line $-\infty<Z<+\infty$
and is unambiguously defined, the latter lives on a circle, so that values of
$\Phi$ differing by $2\pi$ describe the same physical state and must be
identified. This leads to an ambiguity in linear momentum (123): it is defined
modulo $4\pi\mathcal{J}$. There is no such problem with the angular momentum.

The paradox of linear momentum was resolved by Haldane [16]. It turns out that
the continuum and classical theory of the ferromagnet used throughout this
section “remembers” its discrete and quantum origins. In a periodic atomic
chain with the lattice constant $a$, linear momentum is defined modulo
$2\pi\hbar/a$ thanks to Bragg scattering. If the atomic spins have length
$\hbar S$, the spin density is $\mathcal{J}=\hbar S/a$ and the momentum
ambiguity is $4\pi\hbar S/a$. This is indeed an integer multiple of
$2\pi\hbar/a$ because $2S$ is an integer.

###  3.12 Counting states of a domain wall

At sufficiently low energies (defined below), position $Z$ and angle $\Phi$ of
a domain wall fully describe its dynamics. Since the two variables are, up to
a multiplicative constant, momenta of one another, the low-energy
configuration space $(Z,\Phi)$ can be viewed as a phase space, in which the
number of states is proportional to the volume:

$d\Gamma=\frac{dZ\,dP_{z}}{2\pi\hbar}=\frac{d\Phi\,dJ_{z}}{2\pi\hbar}=\frac{\mathcal{J}\,dZ\,d\Phi}{\pi\hbar}.$
(124)

The same result can be obtained if we extrapolate Eq. (35) from a single spin
to a domain wall by using $q^{1}=Z$, $q^{2}=\Phi$, and
$G_{12}=G_{Z\Phi}=-2\sigma\mathcal{J}$.

In our count, we included the two zero modes of a domain wall and discarded
all of the other modes, which (at low amplitudes) are spin waves with finite
frequencies $\omega\geq\omega_{0}=K/\mathcal{J}$ (112). The spin waves are
frozen out at energies below $\hbar\omega_{0}$.

##  4 Discussion

The method of collective coordinates can be applied to more complex domain
walls, e.g., composite [17] and extended [18] ones, as well as to other types
of solitons such as vortices [19] and skyrmions [20]. Even non-magnetic
degrees of freedom can be included (Tchernyshyov, this issue). An extension to
antiferromagnets [21] endows solitons with kinetic energy and inertia; they
can also arise sometimes in ferromagnets [22, 23, 20, 24]. The method has been
successful in describing the dynamics of magnetic solitons in classical spin
systems. One potentially promising direction for further development is to
generalize the collective-coordinate approach to quantum spin systems where
quantum effects are prominent in the dynamics of solitons.

## Acknowledgments

S.K.K. was supported by Brain Pool Plus Program through the National Research
Foundation of Korea funded by the Ministry of Science and ICT
(NRF-2020H1D3A2A03099291) and by the National Research Foundation of Korea
funded by the Korea Government via the SRC Center for Quantum Coherence in
Condensed Matter (NRF-2016R1A5A1008184). O.T. was supported by the U.S.
Department of Energy, Office of Science, Basic Energy Sciences under Award No.
DE-SC0019331.

##  Appendix A Schrödinger equation with a Pöschl–Teller potential

The Schrödinger equation $\mathcal{H}_{1}\psi=\epsilon\psi$ with a
Pöschl–Teller potential [13],

$\mathcal{H}_{1}=-\frac{d^{2}}{dx^{2}}+1-2\mathrm{\,sech}^{2}{x},$ (125)

can be solved with the aid of supersymmetric quantum mechanics [14]. The
Hamiltonian is factorized, $\mathcal{H}_{1}=A^{\dagger}A$, in terms of non-
Hermitian operators,

$A=\frac{d}{dx}+\tanh{x},\quad A^{\dagger}=-\frac{d}{dx}+\tanh{x},$ (126)

with the commutator $[A,A^{\dagger}]=2\mathrm{\,sech}^{2}{x}$.

The supersymmetric partner of $\mathcal{H}_{1}$ is

$\mathcal{H}_{2}=AA^{\dagger}=A^{\dagger}A+[A,A^{\dagger}]=\mathcal{H}_{1}+2\mathrm{\,sech}^{2}{x}=-\frac{d^{2}}{dx^{2}}+1,$
(127)

the Hamiltonian of a free non-relativistic particle. Its eigenstates are plane
waves, $\psi_{2}(x)=e^{ikx}$, with energy eigenvalues $\epsilon_{k}=1+k^{2}$.

From each eigenstate of $\mathcal{H}_{2}$ we can construct an eigenstate of
$\mathcal{H}_{1}$ with the same eigenvalue as follows. Suppose
$AA^{\dagger}|\psi_{2}\rangle=\epsilon|\psi_{2}\rangle$. Make a new state
$|\psi\rangle=A^{\dagger}|\psi_{2}\rangle$. It is an eigenstate of
$\mathcal{H}_{1}$:

$\displaystyle A^{\dagger}A|\psi\rangle$ $\displaystyle=$ $\displaystyle
A^{\dagger}A(A^{\dagger}|\psi_{2}\rangle)=A^{\dagger}(AA^{\dagger}|\psi_{2}\rangle)$
(128) $\displaystyle=$ $\displaystyle
A^{\dagger}(\epsilon|\psi_{2}\rangle)=\epsilon
A^{\dagger}|\psi_{2}\rangle=\epsilon|\psi\rangle.$

The norm of the new state is

$\langle\psi|\psi\rangle=\langle\psi_{2}|AA^{\dagger}\psi_{2}\rangle=\langle\psi_{2}|\epsilon\psi_{2}\rangle=\epsilon\psi_{2}|\psi_{2}\rangle.$
(129)

Therefore, a properly normalized eigenstate of $\mathcal{H}_{1}$ would be

$|\psi_{1}\rangle=\frac{A^{\dagger}}{\sqrt{\epsilon}}|\psi_{2}\rangle.$ (130)

We thus obtain eigenstates of $\mathcal{H}_{1}$ with eigenvalues
$\epsilon_{k}=1+k^{2}$:

$\psi_{k}(x)=\frac{1}{\sqrt{1+k^{2}}}\left(-\frac{d}{dx}+\tanh{x}\right)e^{ikx}=\frac{\tanh{x}-ik}{\sqrt{1+k^{2}}}e^{ikx}.$
(131)

Note that these states become plane waves far away from the potential well,

$\psi_{k}(x)\sim\frac{\pm
1-ik}{\sqrt{1+k^{2}}}e^{ikx}=e^{ikx+i\phi_{\pm}}\mbox{ as }x\to\pm\infty,$
(132)

where $\phi_{\pm}=-\pi/2\pm\arctan{(1/k)}$. It is remarkable that the
Pöschl–Teller potential is reflectionless: a plane wave with wavenumber $k$ is
fully transmitted and merely experiences a phase shift
$\phi_{+}-\phi_{-}=2\arctan{(1/k)}$.

Eigenstates of $\mathcal{H}_{1}$ have the same inner product as their
supersymmetric partners, the plane waves:

$\int dx\,\psi_{k^{\prime}}^{*}(x)\psi_{k}(x)=\int
dx\,e^{-ik^{\prime}x}e^{ikx}=2\pi\delta(k-k^{\prime}).$ (133)

In addition to free states (131), $\mathcal{H}_{1}$ has a bound state with
exactly zero energy, $A^{\dagger}A\Psi_{0}=0$. This state is annihilated by
the operator $A$, which yields a first-order differential equation

$\Psi_{0}^{\prime}(x)+\tanh{x}\Psi_{0}(x)=0.$ (134)

Its solution, properly normalized, is

$\Psi_{0}(x)=2^{-1/2}\mathrm{\,sech}{\,x}.$ (135)

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