# Lasting effects of static magnetic field on classical Brownian motion

Ashot Matevosyan1,2) and Armen E. Allahverdyan2) 1)University of Cambridge,
Cavendish Laboratory, 19 J.J. Thompson Avenue, Cambridge CB3 0HE, UK,
2)Alikhanyan National Laboratory (Yerevan Physics Institute),
Alikhanian Brothers Street 2, Yerevan 375036, Armenia

###### Abstract

The Bohr-Van Leeuwen theorem states that an external static magnetic field
does not influence the state of a classical equilibrium system: there is no
equilibrium classical magnetism, since the magnetic field does not do work. We
revisit this famous no-go result and consider a classical charged Brownian
particle interacting with an equilibrium bath. We confirm that the Bohr-Van
Leeuwen theorem holds for the long-time (equilibrium) state of the particle.
But the external static, homogeneous magnetic field does influence the long-
time state of the thermal bath, which is described via the Caldeira-Leggett
model. In particular, the magnetic field induces an average angular momentum
for the (uncharged!) bath, which separates into two sets rotating in opposite
directions. The effect relates to the bath going slightly out of equilibrium
under the influence of the Brownian particle and persists for arbitrarily long
times. In this context we studied the behavior of the two other additive
integrals of motion, energy and linear momentum. The situation with linear
momentum is different, because it is dissipated away by (and from) the bath
modes. The average energy of the bath mode retains the magnetic field as a
small correction. Thus, only the bath angular momentum really feels the
magnetic field for long times.

##  I Introduction

The Bohr-Van Leeuwen theorem governs the response of equilibrium systems to
external magnetic fields [1, 2, 3]. It states that charged classical particles
do not feel the magnetic field in equilibrium. Thus, equilibrium influences of
the magnetic field are to be restricted to the quantum domain, where the
theorem does not apply due to the non-commutativity of quantum observables
[4]. The essence of the theorem is that the equilibrium state can be
represented as a function of the energy; e.g. the Gibbs coordinate-velocity
distribution, or the microcanonical distribution. Since the magnetic field
does not do work, it does not appear in the energy that enters equilibrium
distributions [1, 4]; see section III for a reminder on this theorem and its
origin.

The Bohr-Van Leeuwen theorem prevents the existence of equilibrium classical
magnetism [1], a subject that could have potential applications in various
fields including real and complex plasma [5, 6, 7], where the influence of
magnetic fields is relevant for the fusion research, charged colloidal liquids
[6, 8], macroions [9] etc. Especially interesting are biophysical
applications[10, 11, 12, 13, 14]: since a stationary magnetic field is not
screened by a living body, it makes an interesting diagnostic tool and also a
potential abusive factor for bio-systems. Moreover, there is a massive body of
experimental results witnessing observable effects of weak, static magnetic
fields in biological systems; see e.g. [10, 11]. A natural target for the
magnetic field is metal ions (Na+, K+, Ca2+ etc) that are crucial in molecular
biology: nearly 1/3 of all proteins employ metal ions for their functioning
[13]. Ions are important in bioenergetics, communication (e.g., nerve impulse
generation), osmotic regulation, metabolism, energy storage etc [14]. However,
the translational motion of ions is always classical, hence the Bohr-Van
Leeuwen theorem prohibits their equilibrium magnetic response [12].
Biophysical responses to static magnetic fields might be looked for in the
quantum domain; see e.g. [15, 16, 17]. It is however not likely that all
biophysical influences of magnetic fields can be accounted for by quantum
models [12].

In all these fields one deals with classical charges moving in the thermal
baths. Thus, it is necessary to understand which effects are not prohibited by
the Bohr-Van Leeuwen theorem and can support long-time (i.e. lasting)
influences of a static magnetic field for classical Brownian motion of a
charged particle.

Consider a Brownian charged particle described via the Langevin equation. The
equilibrium state feels no magnetic field according to the Bohr-Van Leeuwen
theorem. This is however not the end of the story, because during its
relaxation (whatever short) the particle perturbs the equilibrium state of the
bath, which is now slightly out of equilibrium. Using the Caldeira-Leggett
(CL) model [18, 19, 20, 21, 22] that reproduces the Langevin dynamics, we show
that an external static magnetic field leads to long-time changes in the bath
state. The (uncharged) bath oscillators acquire a sizable average angular
momentum and separate (in the frequency space) into two groups that rotate in
different directions. This bath angular momentum can (but need not) be driven
by the conservation law of the total (for the bath + Brownian particle),
effective angular momentum. Thus, the magnetic field can persist in the long-
time limit, though its influence is to be looked for not in the state of the
Brownian charge, but rather in its environment.

Recall that CL is a concrete microscopic model composed of undamped harmonic
oscillators. It has a wide range of applications for baths composed of weakly
interacting particles (photons, phonons) [20, 21]. Its applicability to the
real Brownian dynamics was clarified within molecular dynamic theories, where
the oscillators refer to effective modes of a liquid bath [23, 24, 22].

Our study of angular momentum led us to a more general physical question: how
do the three additive integrals of motion (angular momentum, linear momentum
and energy) behave at the interface between the system (Brownian particle) and
bath? For example, is the bath capable of storing linear momentum in the same
way it stores angular momentum? The answer to this question is negative: the
bath angular momentum differs from the linear momentum and energy; see
sections V and VI for details. For the energy the situation is less
interesting, since the influence of the Brownian particle amounts to a small
perturbation of the bath energy. The linear momentum is more interesting,
because it is transferred to the bath can be a dominant effect for finite
times, but is eventually dissipated from its observables.

This paper is organized as follows. The next section discusses the CL model:
an ion interacting with many independent harmonic oscillators representing an
equilibrium thermal bath. Section III presents the famous Bohr-Van Leeuwen
theorem with discussion. In Section IV and V we study the angular momentum and
total energy of individual modes respectively. We compare the behaviour of
angular momentum with the linear momentum in the section VI. We summarize in
the last section and provide a perspective on future research. We relegated
detailed derivations of our results to appendixes making our conclusions self-
contained. Appendix A discusses the solution of the Langevin equation (6) and
studies pertinent correlation functions of the Brownian motion. Appendix B
discusses the system-bath (Caldeira-Leggett) model and the derivation of the
Langevin equation. in Appendix C calculated the mean angular momentum of a
single bath oscillator, while Appendix D studies its mean energy.

##  II The Model

Consider a classical particle with coordinates ${\bf R}=(X,Y,Z)$, unit charge
and unit mass that interacts with magnetic field. The particle is subject to
an external, rotation symmetric, harmonic potential with frequency
$\omega_{0}$. Particle’s Lagrangian reads

$\displaystyle\mathcal{L}_{S}=\frac{1}{2}\dot{{\bf
R}}^{2}-\frac{\omega_{0}^{2}}{2}{\bf R}^{2}+{\bf A}({\bf R})\dot{\bf R},$ (1)

where $A({\bf R})$ is a vector potential that generates a static, homogeneous
magnetic field $\bm{B}$ with the magnitude $b=|\bm{B}|$ along the $z$-axes
(the normal vector ${\bf e}_{z}$):

$\displaystyle\bm{B}={\rm rot}\,{\bf A}={\bf e}_{z}b,\quad{\bf A}({\bf
R})=\frac{1}{2}(-{bY},{bX},0).$ (2)

The particle couples with a bath made of $N$ harmonic oscillators (modes) with
coordinates ${\bf r}_{k}=(x_{k},y_{k},z_{k})$, masses $m_{k}$, frequencies
$\omega_{k}$ and coupling constants $c_{k}$. The potential energy of particle-
bath interaction is assumed to be non-negative and bi-linear over the particle
and bath coordinates (Caldeira-Leggett model). Hence the bath + interaction
Lagrangian reads [18, 21, 19, 20, 22]

$\displaystyle\mathcal{L}_{B}={\sum}_{k=1}^{N}\left[\frac{m_{k}}{2}\dot{{\bf
r}}_{k}^{2}-\frac{m_{k}\omega_{k}^{2}}{2}\left({\bf r}_{k}-\frac{c_{k}{\bf
R}}{m_{k}\omega_{k}^{2}}\right)^{2}\right],$ (3)

where the full Lagrangian is $\mathcal{L}_{S}+\mathcal{L}_{B}$. Due to (2) and
linearity of equations of motion generated by
$\mathcal{L}_{S}+\mathcal{L}_{B}$, the motion along $z$ coordinates for all
particles involved in (1, 3) decouples from the motion along $(x,y)$
coordinates.

We get from Noether’s theorem, or directly from equations of motion generated
by $\mathcal{L}_{S}+\mathcal{L}_{B}$, that the following quantity is
conserved:

$\displaystyle L$ $\displaystyle=$ $\displaystyle
X\dot{Y}-Y\dot{X}+{\sum}_{k=1}^{N}m_{k}\left(x_{k}\dot{y_{k}}-y_{k}\dot{x_{k}}\right)$
(4) $\displaystyle+$ $\displaystyle\frac{b}{2}\left(X^{2}+Y^{2}\right).$

$L$ is a sum of the particle’s angular momentum $X\dot{Y}-Y\dot{X}$ along $z$
direction, angular momenta of all bath oscillators and a contribution
$\frac{b}{2}\left(X^{2}+Y^{2}\right)$ from the charged particle related to the
magnetic field. Solving Euler-Lagrange equations of motion generated by
$\mathcal{L}_{S}+\mathcal{L}_{B}$ for ${\bf r}_{k}$ we get:

$\displaystyle{\bf r}_{k}(t)={\bf r}_{k}(0)\cos(\omega_{k}t)+\frac{\dot{{\bf
r}}_{k}(0)}{\omega_{k}}\sin(\omega_{k}t)$
$\displaystyle+\frac{c_{k}}{m_{k}\omega_{k}}\int_{0}^{t}{\rm
d}t^{\prime}\sin(\omega_{k}(t-t^{\prime})){\bf R}(t^{\prime}).$ (5)

Plugging (5) into the Euler-Lagrange equations of motion for ${\bf R}$, we get
the Langevin equation for the charged particle in magnetic field [25] (see
Appendix A):

$\displaystyle\ddot{{\bf R}}=b{\bf e}_{z}\times\dot{{\bf
R}}-\omega_{0}^{2}{\bf R}-\int_{0}^{t}{\rm d}u\zeta(t-u)\dot{{\bf
R}}(u)+\bm{\xi},$ (6)
$\displaystyle\zeta(t)={\sum}_{k=1}^{N}\frac{c_{k}^{2}}{m_{k}\omega_{k}^{2}}\cos(\omega_{k}t),$
(7) $\displaystyle\bm{\xi}(t)={\sum}_{k=1}^{N}[\,c_{k}\overline{{\bf
r}}_{k}(0)\cos(\omega_{k}t)+\frac{c_{k}{\bf
v}_{k}(0)}{\omega_{k}}\sin(\omega_{k}t)],\,\,~{}~{}$ (8)
$\displaystyle\overline{{\bf r}}_{k}(0)\equiv{\bf r}_{k}(0)-{\bf
R}(0)\frac{c_{k}}{m_{k}\omega_{k}^{2}},\qquad{\bf v}_{k}=\dot{{\bf r}}_{k}.$
(9)

The cumulative force from the bath is decomposed into friction with a kernel
$\zeta$ and noise $\bm{\xi}(t)$ that emerges due to the initially random state
of the bath. Let the initial state of the particle and bath is given by the
density

$\displaystyle{\cal P}({\bf r}_{k},{\bf v}_{k},{\bf R},{\bf V})\propto
e^{-H_{B}(\overline{{\bf r}}_{k},{{\bf v}}_{k})/T}{\cal P}({\bf R},{\bf
V}),~{}~{}{\bf V}=\dot{\bf R},~{}~{}~{}~{}$ (10)

where $T$ is the temperature ($k_{B}=1$), and where the interaction-dressed
bath energy $H_{B}$ reads from (3):

$\displaystyle H_{B}$ $\displaystyle=$ $\displaystyle-{\cal
L}_{B}+{\sum}_{k=1}^{N}{m_{k}}\dot{{\bf r}}_{k}^{2}$ (11) $\displaystyle=$
$\displaystyle\sum_{k=1}^{N}\left[\frac{m_{k}}{2}\dot{{\bf
r}}_{k}^{2}+\frac{m_{k}\omega_{k}^{2}}{2}\left({\bf r}_{k}-\frac{c_{k}{\bf
R}}{m_{k}\omega_{k}^{2}}\right)^{2}\right].$ (12)

Eq. (10) shows that bath’s initial density is Gibbsian. Particle’s initial
density ${\cal P}({\bf R},{\bf V})$ is arbitrary. The initial state (10) is
not independent over the particle and bath, since $e^{-H_{B}(\overline{{\bf
r}}_{k},{{\bf v}}_{k})/T}$ contains the shifted coordinate $\overline{{\bf
r}}_{k}$; cf. (9). Eq. (10) refers to the time-scale separation, where the
bath is prepared in equilibrium under fixed coordinate and momentum of the
particle. This is realistic for Brownian motion.

Eqs. (8, 10) imply the fluctuation-dissipation relation

$\displaystyle\left\langle\xi_{\alpha}(t)\xi_{\beta}(t^{\prime})\right\rangle=T\delta_{\alpha\beta}\zeta(t-t^{\prime})\qquad\alpha,\beta=x,y,z,$
(13)

where $\left\langle...\right\rangle$ is the average over initial state (10).

The thermodynamic limit $N\to\infty$ for the bath is taken together with the
dense frequency limit $\delta\omega\to 0$ ($n=1,...,N$) under condition
$N\delta\omega\gg 1$ (see Appendix B for details). Simultaneously, we take a
weak coupling $c_{n}\to 0$ to each oscillator [18, 21, 19, 20, 22]. The
specific choice

$\displaystyle\omega_{n}=\delta\omega\,n,~{}c_{n}=\sqrt{{2\gamma\omega_{n}^{2}m_{n}\delta\omega}/{\pi}},$
(14) $\displaystyle\zeta(t)=2\gamma\,\delta(t),$ (15)

reproduces in the Langevin equation (6, 7, 8) the Ohmic friction with
magnitude $\gamma$ and the white noise [20, 21, 22, 18, 19].

##  III The Bohr-Van Leeuwen theorem

Fluctuation-dissipation relation (13) ensures from (6) (together with the
thermodynamic limit for the bath) that the particle’s state relaxes to the
Gibbsian density [18, 21, 19, 20, 22]:

$\displaystyle{\cal P}({\bf R},{\bf V};t\to\infty)\propto
e^{-\frac{1}{2}({{\bf V}}^{2}+{\omega_{0}^{2}}{\bf R}^{2})/T},$ (16)

that does not (and cannot) contain the magnetic field $b$. Obviously, the
average angular momentum $\left\langle XV_{y}-YV_{x}\right\rangle$ calculated
via (16) is zero. Eq. (16) implies the Bohr-Van Leeuwen theorem for the
Brownian motion [1, 4, 2, 3]. The absence of the magnetic field $\bm{B}$ from
the equilibrium density holds for any confining potential [1, 3]. This theorem
is an equilibrium result and it does not hold for non-equilibrium steady
states [26, 27, 28, 29, 30, 31]. In particular, the theorem may be broken by a
weak white noise, which is sufficient for generating a sizable diamagnetic
angular momentum for the Brownian particle [31].

Note as well that the magnetic field shows up in correlation functions of the
Brownian particle and in its relaxation times to the equilibrium state [31];
see also Appendix A. This however is not the long-time response we seek here,
since correlation functions decay within the relaxation time.

##  IV Angular momentum of bath modes

Let us assume that the initial state of the Brownian particle in (10) holds

$\displaystyle{\cal P}({\bf R},{\bf V})={\cal P}_{1}(X){\cal P}_{1}(Y){\cal
P}_{1}(Z){\cal P}_{2}(V_{x}){\cal P}_{2}(V_{y}){\cal P}_{2}(V_{z}),$
$\displaystyle{\cal P}_{1}(-a)={\cal P}_{1}(a),\qquad{\cal P}_{2}(-a)={\cal
P}_{2}(a).$ (17)

Eqs. (17) allow a large class of initial non-equilibrium states for the
particle. Note that the initial angular momentum of the particle nullifies:
$\left\langle XV_{y}-YV_{x}\right\rangle=0$. The virtue of (17, 10) is that
all averages hold specific symmetry features related to the invariance of (17)
with respect to $\pi/2$ rotations in the $(x,y)$ plane; see Appendix B.3.

For initial second moments in (17) we denote

$\displaystyle\left\langle X^{2}\right\rangle=\left\langle
Y^{2}\right\rangle=\sigma_{X}\;{T}/{\omega_{0}^{2}},~{}~{}\left\langle
V_{x}^{2}\right\rangle=\left\langle
V_{y}^{2}\right\rangle=\sigma_{V}\;T,~{}~{}~{}$ (18)

where $\sigma_{X}=\sigma_{V}=1$ in (18) refers to equilibrium second moments
of the initial state; cf. (16).

The long-time average angular momentum $L(\omega)$ of a bath oscillator with
frequency $\omega=\omega_{k}$ is calculated from (5, 6, 17) using the Laplace
transform (see Appendix C):

$\displaystyle L(\omega)\equiv m_{k}\left\langle
x_{k}v_{yk}-y_{k}v_{xk}\right\rangle_{t\to\infty}=\delta\omega\,\frac{4\gamma
bT}{\pi}\times$ (19)
$\displaystyle\frac{(\omega^{2}-\omega_{0}^{2})\left((1-\sigma_{V})\omega^{2}+(1-\sigma_{X})\omega_{0}^{2}\right)}{\left((\omega^{2}-\omega_{0}^{2}-b\omega)^{2}+\gamma^{2}\omega^{2}\right)\left((\omega^{2}-\omega_{0}^{2}+b\omega)^{2}+\gamma^{2}\omega^{2}\right)}.$

Eq. (19) shows that even though bath oscillators are not charged, they acquire
a non-zero angular momentum: $L(\omega)\not=0$ for $t\to\infty$, i.e. for
times much larger than the relaxation time of the particle. This means that
the long-time state of the bath feels the magnetic field $b$. Eqs. (10, 17)
show that the initial mean angular momentum nullifies both for the bath
oscillator and the Brownian particle. Hence the fact of $L(\omega)\not=0$ for
$t\to\infty$ means that the final state of the bath is out of equilibrium.

Now $L(\omega_{k})\propto c_{k}^{2}\propto\delta\omega$ since $L(\omega_{k})$
is driven by the coupling of the corresponding mode with the Brownian
particle; cf. (5). In view of (19), the observable collective mode momentum is
well-defined and amounts to integral between two finite frequencies:

$\displaystyle\sum_{k=k_{1}}^{k_{2}}L(\omega_{k})=\sum_{\omega=\omega_{1}}^{\omega_{2}}L(\omega)=\int_{\omega_{1}}^{\omega_{2}}\frac{{\rm
d}\omega}{\delta\omega}\,L(\omega).$ (20)

Eq. (19) [and (27) below] can describe a finite-bath situation [32, 33, 34,
35, 36]. Here the bath is large (for Langevin’s equation (6) to apply for
certain times), but finite.

Figure 1:  $L(\omega)/\delta\omega$ versus $\omega$ for $T=1$, $\omega_{0}=1$,
$\gamma=0.1$, $\sigma_{X}=1$, $\sigma_{V}=0.1$; cf. (19, 18). Curves refer to
different magnetic fields (2) (from bottom to top): $b=0.3$ (black), $b=0.1$
(blue), $b=0.03$ (red).

We emphasize that $L(\omega)\not=0$ due to initially non-equilibrium state of
the particle, since $L(\omega)=0$ whenever $\sigma_{X}=\sigma_{V}=1$ in (17);
cf. (19). Next, we confirm that (19) is consistent with the average angular
momentum conservation (4) between $t=0$ and $t=\infty$:

$\displaystyle\int_{0}^{\infty}{\rm
d}\omega\,L(\omega)=-\frac{Tb(1-\sigma_{X})}{\omega_{0}^{2}}=b[\langle
X^{2}\rangle_{0}-\langle X^{2}\rangle_{\infty}],~{}~{}~{}~{}$ (21)

where $\langle X^{2}\rangle_{0}$ and $\langle X^{2}\rangle_{\infty}$ are the
initial and final values; cf. (18). Even for $\sigma_{X}=1$ we can have
$L(\omega)\not=0$ due to $\sigma_{V}\not=1$ in (19). Hence, $L(\omega)\not=0$
need not be driven by the conservation law; see Fig. 1. Now $L(\omega)$
changes its sign at $\omega=\omega_{0}$ and goes to zero as $\sim\omega^{-4}$
for $\omega\gg{\rm max}[\omega_{0},\gamma,b]$; i.e. two sets of oscillators
rotate in different directions, as Fig. 1 shows.

##  V Energy of bath modes

Let us now see that the magnetic field also shows up in interaction-induced
corrections to the energy of the bath mode. Eqs. (10, 11) imply that the
average energy of the bath mode with the frequency $\omega=\omega_{k}$ can be
defined as

$\displaystyle E(\omega_{k})=$
$\displaystyle\frac{m_{k}}{2}\left\langle\dot{{\bf
r}}_{k}^{2}\right\rangle+\frac{m_{k}\omega_{k}^{2}}{2}\left\langle{{\bf
r}}_{k}^{2}\right\rangle$ $\displaystyle+\left\langle{\bf
R}^{2}\right\rangle\frac{c_{k}^{2}}{2m_{k}\omega_{k}^{2}}-c_{k}\left\langle{\bf
R}\cdot{\bf r}_{k}\right\rangle,$ (22)

where the average is taken over the initial state (10). The full conserved
energy of the particle+bath amounts to the sum of (22) over all modes,
$\sum_{k}E(\omega_{k})$, plus the mean energy $\langle\frac{1}{2}\dot{{\bf
R}}^{2}+\frac{\omega_{0}^{2}}{2}{\bf R}^{2}\rangle$ of the particle; cf. (1,
11). Eq. (22) in the limit $t\to\infty$ is worked out in Appendix D:

$\displaystyle E(\omega)$ $\displaystyle=$ $\displaystyle
3T-\delta\omega\left(2\varepsilon(\omega)+\varepsilon(\omega)|_{b=0}\right),$
(23) $\displaystyle\varepsilon(\omega)$ $\displaystyle=$
$\displaystyle\frac{T\gamma}{2\pi}\times\frac{\omega^{2}(b^{2}+\gamma^{2})+\left(\omega^{2}-\omega_{0}^{2}\right)^{2}}{(\omega^{2}-\omega_{0}^{2}-b\omega)^{2}+\gamma^{2}\omega^{2}}$
(24)
$\displaystyle\qquad\times\frac{(1-\sigma_{V})\omega^{2}+(1-\sigma_{X})\omega_{0}^{2}}{(\omega^{2}-\omega_{0}^{2}+b\omega)^{2}+\gamma^{2}\omega^{2}}.$

where $\sigma_{X}$ and $\sigma_{V}$ are defined in (18). Now $3T$ in
$E(\omega)$ is the thermal energy of the free mode; cf. the equipartition
theorem from (10). The factor $\propto\delta\omega$ in (24) is due to
interaction with the Brownian particle: $2\varepsilon(\omega)$ comes from
$(x,y)$-components of (22) that feel the magnetic field $b$ [cf. (6)], while
$\varepsilon(\omega)|_{b=0}$ comes from the $z$-component that does not feel
it. Similar to (19), the interaction-induced factor in (24) nullifies for
$\sigma_{X}=\sigma_{V}=1$; see (10, 18). Thus, for long times the mode mean
energy (24) (in contrast to particle’s mean energy) depends on the magnetic
field, though this dependence is weaker than for (19). The interaction-driven
factor in (23, 24) that contains the magnetic field is a small correction to
the leading term $2T$. This is different from (19), where the average angular
momentum without interaction is zero, and hence the magnetic field induced
effect is the main one.

##  VI Linear momentum

Given the above finding for the angular momentum, it is reasonable to ask how
the linear momentum is transferred from the Brownian particle to the bath.
Answering this question has intrinsic value and contrasts with the angular
momentum’s behavior.

Note from (3, 1) that for a free Brownian particle (i.e. $\omega_{0}={\bf
B}=0$) there is a conservation of linear momentum:

$\displaystyle\frac{{\rm d}}{{\rm
d}t}\left[{\sum}_{k=1}^{N}\,\,\frac{c_{k}}{\omega_{k}^{2}}\,\dot{\bf
r}_{k}(t)+{\bf V}(t)\right]=0.$ (25)

Eq. (25) contains the full momentum ${\bf V}(t)$ of the Brownian particle
(recall that its mass is taken $1$), which is added to
$\frac{c_{k}}{\omega_{k}^{2}}\,\dot{\bf r}_{k}(t)$ for each mode. The latter
is a part of the mode momentum $m_{k}\dot{\bf r}_{k}$. We emphasize that the
full mode momentum $m_{k}\dot{\bf r}_{k}$ cannot appear in the conservation
law, since the model lacks the full translation invariance. Indeed, (3) shows
that each mode feels a harmonic potential. A relations similar to (25) is
formally mentioned in Ref. [37].

Instead of (17, 18) consider initial conditions $\langle{\bf R}(0)\rangle=0$,
$\langle{{\bf V}}(0)\rangle\not=0$ that hold together with (10). As follows
from (6, 15), and is generally well-known, the velocity density of the
particle thermalizes for long times [cf. (16)]:

$\displaystyle{\cal P}({\bf V};t\to\infty)\propto e^{-\frac{1}{2T}{{\bf
V}}^{2}}.$ (26)

But ${\bf R}$ is subject to an unbound Brownian motion and hence it is not
thermalized.

For $\gamma t\gg 1$ we find from (5, 6, 15, 10) for the average linear
momentum of a bath mode with frequency $\omega=\omega_{k}$ (see Appendix B.4):

$\displaystyle\frac{c_{k}}{\omega_{k}^{2}}\langle\dot{\bf
r}_{k}\rangle=\delta\omega\times\frac{2\gamma\langle{\bf
V}(0)\rangle}{\pi}\,\,\frac{\frac{\gamma}{\omega}\sin[\omega t]-\cos[\omega
t]}{\gamma^{2}+\omega^{2}}.$ (27)

Eq. (27) shows that the single mode (partial) linear momentum
$\frac{c_{k}}{\omega_{k}^{2}}\langle\dot{\bf r}_{k}\rangle$ is a time-
dependent oscillating function. In contrast, relaxation occurs once we
consider a collective quantity and take the bath’s thermodynamic limit, i.e.
using the integral instead of the sum:

$\displaystyle\sum_{\omega_{1}}^{\omega_{2}}\frac{c_{k}}{\omega_{k}^{2}}\langle\dot{\bf
r}_{k}\rangle=\frac{2\gamma\langle{\bf
V}(0)\rangle}{\pi}\int_{\omega_{1}}^{\omega_{2}}{\rm
d}\omega\,\frac{\frac{\gamma}{\omega}\sin[\omega t]-\cos[\omega
t]}{\gamma^{2}+\omega^{2}},~{}~{}~{}~{}$ (28)

where we used (27), and where naturally $\omega_{2}>\omega_{1}$.

The behavior of the integral in (28) for $t\to\infty$ essentially depends on
whether $\omega_{1}>0$ or $\omega_{1}=0$, i.e. whether the zero frequency is
included in the collective quantity or not. For $\omega_{1}>0$, (28) implies

$\displaystyle\sum_{\omega_{1}>0}^{\omega_{2}}\frac{c_{k}}{\omega_{k}^{2}}\langle\dot{\bf
r}_{k}\rangle=O\left(\frac{1}{t}\right)\quad{\rm for}\quad t\to\infty,$ (29)

as becomes clear after changing the variable $\omega t\to\omega$ in (28).
Hence the collective linear momentum for non-zero frequencies nullifies for
large times. Note that these times are larger than $1/\gamma$.

However, the total bath linear momentum is non-zero, as follows from (25).
Indeed, applying (25) at times $t=0$ and $t\gg 1/\gamma$ together with
conditions $\langle\dot{\bf r}_{k}\rangle=0$, $\langle{{\bf
V}}(0)\rangle\not=0$ and $\langle{{\bf V}}(t)\rangle\approxeq 0$ we find for
the total final linear momentum:

$\displaystyle\sum_{k=1}^{N}\frac{c_{k}}{\omega_{k}^{2}}\,\dot{\bf r}_{k}(t)$
$\displaystyle=$ $\displaystyle\frac{2\gamma\langle{\bf
V}(0)\rangle}{\pi}\int_{0}^{\infty}{\rm
d}\omega\,\frac{\frac{\gamma}{\omega}\sin[\omega t]-\cos[\omega
t]}{\gamma^{2}+\omega^{2}}$ (30) $\displaystyle=$ $\displaystyle\langle{\bf
V}(0)\rangle.$

Note that if $\omega_{1}$ is extended to zero in (28), we get back from the
$t\to\infty$-limit of (28) the same total momentum $\langle{\bf V}(0)\rangle$.
This means that the total momentum $\langle{\bf V}(0)\rangle$ is (for
$t\to\infty$) transferred to the zero-frequency mode $\omega=0$, which
corresponds to the homogeneous shift. Thus, the linear momentum dissipates
away from both the particle velocity and collective bath observables
$\sum_{\omega_{1}>0}^{\omega_{2}}\frac{c_{k}}{\omega_{k}^{2}}\langle\dot{\bf
r}_{k}\rangle$. However, the velocity $\langle\dot{\bf r}_{\omega\to
0}\rangle$ of the zero-frequency mode is not observable, because it is
infinitisemaly small due to a ${\rm lim}_{\omega\to
0}\frac{c_{\omega}}{\omega^{2}}=\infty$; see (14). Hence the total linear
momentum does not result into any visible motion for long times.

The comparison between the linear momentum and the angular momentum can be
carried out in several different set-ups. We choose the most reasonable set-up
for showing that the linear momentum is different from the angular momentum.
Within this set-up the behavior of the angular momentum for the Brownian
particle subject to the external potential with frequency $\omega_{0}\not=0$
and magnetic field ${\bf B}\not=0$ is compared with the linear momentum under
$\omega_{0}=0$ and ${\bf B}=0$; see (25–30). In the former case the overall
angular momentum is conserved [cf. (4)], while the final state of the Brownian
particle thermalizes; see (16). In the latter case, the partial linear
momentum is conserved, and there is a partial long-time thermalization; see
(25) and (26) below. Other possible set-up for comparing the linear and
angular momentum are discussed in Appendix B.4; see Table 1 there.

There are two major differences between the angular momentum and linear
momentum. First, the single mode angular momentum (19) converges to a well-
defined, time-independent value, in contrast to the single mode linear
momentum (27) that is an oscillating function. Consequently, the angular
momentum is sustained in the long-time limit for finite bath observables in
the thermodynamic limit, while the linear momentum dissipates away; cf. (29)
with (20). Second, (27, 28) trivialize for zero initial momentum $\langle{\bf
V}(0)\rangle=0$, which again contrasts (19) that does not need an initial
angular momentum for the particle.

##  VII Summary

We restored the influence of the static (homogeneous) magnetic field on the
long-time limit of a classical Brownian motion. For long times the influence
of the magnetic field is found not in the particle (as correctly claimed by
Bohr-Van Leeuwen theorem), but in the orbital momentum of (uncharged) bath
modes. The effect is not enforced by the conservation law of the angular
momentum, i.e. there are situations where its absence is fully consistent with
the conservation. This behavior is specific for the angular momentum, as
compared to the two other additive integrals of motion: the linear momentum is
dissipated away by (and from) bath modes, while the bath energy can feel the
magnetic field, but only as a small correction.

At this point, the major open problem is whether these effects are specific
for the Caldeira-Leggett model of the bath—which we emphasize does apply for
describing liquids [24, 22]—or it will generalize to more realistic bath
models. This question is currently under investigation, but we can mention two
preliminary hypotheses. The non-spontaneous aspect of the bath angular
momentum is based on the conservation law (4). Hence it is likely to persist
for more general bath models. The spontaneous aspect could be a specific point
of linear bath models, including the Caldeira-Leggett model.

In this context, we plan to study bath models that are especially relevant for
cellular ions. Here the bath is described as a fluctuating hydrodynamic system
[38]. It is known that such a bath also reproduces the Langevin equation for
the Brownian particle immersed into it [39]. Hence we anticipate that the
effect of transferring the influence of the magnetic field to the bath will
show up also here, and will be reflected in the angular momentum of the fluid.
The major difference of this class of models compared to the Caldeira-Leggett
model is that bath modes are subject to viscosity, i.e. there is an additional
source of irreversibility.

###### Acknowledgements.

This work was supported by SCS of Armenia grants No. 21AG-1C038, No.
22AA-1C028 and No. 20TTAT-QTa003. We thank David Petrosyan for important
remarks.

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##  Appendix A Brownian stochastic motion and correlation functions

We start by presenting the solution for the following Langevin equation with
Ohmic friction and white noise [cf. (6, 15)]:

$\displaystyle\ddot{X}(t)$
$\displaystyle=~{}~{}b\dot{Y}(t)-\omega_{0}^{2}X(t)-\gamma\dot{X}(t)+\xi_{x}(t)$
(A.1) $\displaystyle\ddot{Y}(t)$
$\displaystyle=-b\dot{X}(t)-\omega_{0}^{2}Y(t)-\gamma\dot{Y}(t)+\xi_{y}(t)$
(A.2)
$\displaystyle\left\langle\xi_{\alpha}(t)\xi_{\beta}(t^{\prime})\right\rangle$
$\displaystyle=2\gamma\,T\delta_{\alpha\beta}\delta(t-t^{\prime})\qquad\alpha,\beta=x,y.$
(A.3)

Note that $X$ and $Y$ couple only via the magnetic field $b$; cf. (2). The
third equation in (6) is decoupled from (A.1, A.2).

We solve (A.1–A.3) via the Laplace transform:

$\begin{pmatrix}s^{2}+\omega_{0}^{2}+\gamma s&-bs\\\
bs&s^{2}+\omega_{0}^{2}+\gamma s\\\ \end{pmatrix}\begin{pmatrix}\hat{X}\\\
\hat{Y}\end{pmatrix}=\begin{pmatrix}\hat{\xi}_{x}+X(0)(s+\gamma)+V_{x}(0)-bY(0)\\\
\hat{\xi}_{y}+Y(0)(s+\gamma)+V_{y}(0)+bX(0)\end{pmatrix}.$ (A.4)

Inverting the matrix on the LHS amounts to transposing it and dividing it on
its determinant. Hence, we get

$\displaystyle\hat{X}(s)=$ $\displaystyle
X(0)\left(\hat{K}_{0}+b\hat{H}_{1}\right)+Y(0)\left(\hat{K}_{1}-b\hat{H}_{0}\right)$
(A.5)
$\displaystyle+V_{x}(0)\hat{H}_{0}+V_{y}(0)\hat{H}_{1}+\hat{H}_{0}\hat{\xi}_{x}+\hat{H}_{1}\hat{\xi}_{y}$
(A.6) $\displaystyle\hat{Y}(s)=$ $\displaystyle
X(0)\left(-\hat{K}_{1}+H_{0}b\right)+Y(0)\left(\hat{K}_{0}+b\hat{H}_{1}\right)$
(A.7) $\displaystyle-
V_{x}(0)\hat{H}_{1}+V_{y}(0)\hat{H}_{0}-\hat{H}_{1}\hat{\xi}_{x}+\hat{H}_{0}\hat{\xi}_{y}$
(A.8)

where

$\displaystyle{\begin{array}[]{rlrl}\hat{H}_{0}&=\frac{s^{2}+\gamma\,s+\omega_{0}^{2}}{\Delta}&\qquad\hat{H}_{1}&=\frac{b\,s}{\Delta}\\\\[10.0pt]
\hat{K}_{0}&=\hat{H}_{0}(s+\gamma)&\qquad\hat{K}_{1}&=\hat{H}_{1}(s+\gamma)\end{array}}$
(A.11)

and the $\Delta$ is the determinant of the matrix in (A.4):

$\displaystyle\Delta$
$\displaystyle=(b\,s)^{2}+\left(s^{2}+\gamma\,s+\omega_{0}^{2}\right)^{2}=\prod_{i=1}^{4}(s-p_{i})$
(A.12)

where $p_{i}$ are roots of $\Delta=0$:

$\displaystyle p_{1,2}=\frac{1}{2}\left(-\gamma+ib\pm\sqrt{(\gamma-
ib)^{2}-4\omega_{0}^{2}}\right)\qquad p_{3,4}=p_{1,2}^{*}$ (A.13)

We also define residues of $\tfrac{1}{\Delta}$ which has 4 simple poles
$\\{p_{i}\\}$ as

$r_{i}=\frac{1}{\prod_{j\neq i}(p_{i}-p_{j})}\qquad i,j=1,2,3,4$ (A.14)

thus, the inverse Laplace transform of the above kernels are (recall Cauchy’s
residue theorem)

$\displaystyle H_{0}(t)=\sum_{i=1}^{4}r_{i}(p_{i}^{2}+\gamma
p_{i}+\omega_{0}^{2})e^{p_{i}t}\qquad
H_{1}(t)=\sum_{i=1}^{4}r_{i}b\,p_{i}\,e^{p_{i}t}$ (A.15)

and similarly, we get analytical expressions for $K_{0}(t),K_{1}(t)$. Using
these functions together with the convolution theorem for the Laplace
transform, we get

$\displaystyle X(t)=$ $\displaystyle
X(0)\left(K_{0}(t)+bH_{1}(t)\right)+Y(0)\left(K_{1}(t)-bH_{0}(t)\right)+V_{x}(0)H_{0}(t)+V_{y}(0)H_{1}(t)$
(A.16) $\displaystyle+\int_{0}^{t}{\rm
d}t^{\prime}H_{0}(t-t^{\prime})\xi_{x}(t^{\prime})+\int_{0}^{t}{\rm
d}t^{\prime}H_{1}(t-t^{\prime})\xi_{y}(t^{\prime})$ $\displaystyle Y(t)=$
$\displaystyle
X(0)\left(-K_{1}(t)+bH_{0}(t)\right)+Y(0)\left(K_{0}(t)+bH_{1}(t)\right)-V_{x}(0)H_{1}(t)+V_{y}(0)H_{0}(t)$
(A.17) $\displaystyle-\int_{0}^{t}{\rm
d}t^{\prime}H_{1}(t-t^{\prime})\xi_{x}(t^{\prime})+\int_{0}^{t}{\rm
d}t^{\prime}H_{0}(t-t^{\prime})\xi_{y}(t^{\prime})$

We assume that distribution of random variables $X(0)$ and $Y(0)$ (initial
values) are unbiased, symmetric and independent from each other:

$\displaystyle{\cal P}(X(0),Y(0))={\cal P}(X(0)){\cal P}(Y(0)),\qquad{\cal
P}(-x)={\cal P}(x),$ (A.18)

which in particular implies:

$\displaystyle\left\langle X(0)\right\rangle=\left\langle
Y(0)\right\rangle=0,$ (A.19) $\displaystyle\left\langle
X(0)^{2}\right\rangle=\left\langle Y(0)^{2}\right\rangle\equiv\left\langle
X_{0}^{2}\right\rangle,$ (A.20) $\displaystyle\left\langle
V_{x}(0)^{2}\right\rangle=\left\langle
V_{y}(0)^{2}\right\rangle\equiv\left\langle V_{0}^{2}\right\rangle,$ (A.21)
$\displaystyle\left\langle X(0)Y(0)\right\rangle=\left\langle
V_{x}(0)V_{y}(0)\right\rangle=0.$ (A.22)

Also note that, $X(0)$, $Y(0)$, $\xi_{x}(t\geq 0)$ and $\xi_{y}(t\geq 0)$ are
independent random variables due to the assumed system-bath initial state.
Then the correlation function $C_{XY}(t,t^{\prime})=\left\langle
X(t)Y(t^{\prime})\right\rangle$ will be

$\displaystyle C_{XY}(t,s)=$ $\displaystyle\left\langle
V_{0}^{2}\right\rangle\left(H_{1}(t)H_{0}(s)-H_{0}(t)H_{1}(s)\right)$ (A.23)
$\displaystyle+\left\langle
X_{0}^{2}\right\rangle\Big{(}\left(K_{0}(t)+bH_{1}(t)\right)\left(-K_{1}(s)+bH_{0}(s)\right)$
(A.24)
$\displaystyle\qquad\qquad+\left(K_{1}(t)-bH_{0}(t)\right)\left(K_{0}(s)+bH_{1}(s)\right)\Big{)}$
(A.25) $\displaystyle+\int_{0}^{t}\int_{0}^{s}{\rm d}t^{\prime}{\rm
d}s^{\prime}\left\langle\xi\left(t^{\prime}\right)\xi(s^{\prime})\right\rangle\left(-H_{0}(t-t^{\prime})H_{1}(s-s^{\prime})+H_{0}(s-s^{\prime})H_{1}(t-t^{\prime})\right)$
(A.26)

where the last line can be simplified as noise is white:

$\displaystyle C_{XY}(t,s)=$ $\displaystyle\left\langle
V_{0}^{2}\right\rangle\left(H_{1}(t)H_{0}(s)-H_{0}(t)H_{1}(s)\right)$ (A.27)
$\displaystyle+\left\langle
X_{0}^{2}\right\rangle\Big{(}\left(K_{0}(t)+bH_{1}(t)\right)\left(-K_{1}(s)+bH_{0}(s)\right)$
(A.28)
$\displaystyle\qquad\qquad+\left(K_{1}(t)-bH_{0}(t)\right)\left(K_{0}(s)+bH_{1}(s)\right)\Big{)}$
(A.29) $\displaystyle+2\gamma T\int_{0}^{\text{min}(t,s)}{\rm
d}y\left(-H_{0}(t-y)H_{1}(s-y)+H_{0}(s-y)H_{1}(t-y)\right).$ (A.30)

Note that $C_{XY}$ is antisymmetric:

$\displaystyle C_{XY}(t,s)=-C_{XY}(s,t).$ (A.31)

Similarly, the correlation function $C_{XX}(t,s)=\left\langle
X(t)X(s)\right\rangle$ will be

$\displaystyle C_{XX}(t,s)=$ $\displaystyle\left\langle
V_{0}^{2}\right\rangle\left(H_{0}(t)H_{0}(s)+H_{1}(t)H_{1}(s)\right)$ (A.32)
$\displaystyle+\left\langle
X_{0}^{2}\right\rangle\Big{(}\left(K_{0}(t)+bH_{1}(t)\right)\left(K_{0}(s)+bH_{1}(s)\right)$
(A.33)
$\displaystyle\qquad\qquad+\left(K_{1}(t)-bH_{0}(t)\right)\left(K_{1}(s)-bH_{0}(s)\right)\Big{)}$
(A.34) $\displaystyle+2\gamma T\int_{0}^{\text{min}(t,s)}{\rm
d}y\left(H_{0}(t-y)H_{0}(s-y)+H_{1}(t-y)H_{1}(s-y)\right)$ (A.35)

Therefore, the position-velocity autocorrelation functions can be derived by:

$\displaystyle C_{XU_{x}}(t,s)\equiv\left\langle X(t)U_{x}(s)\right\rangle$
$\displaystyle=\frac{{\rm d}}{{\rm d}s}\left\langle
X(t)X(s)\right\rangle=\frac{{\rm d}}{{\rm d}s}C_{XX}(t,s)$ (A.36)
$\displaystyle C_{XU_{y}}(t,s)\equiv\left\langle X(t)U_{y}(s)\right\rangle$
$\displaystyle=\frac{{\rm d}}{{\rm d}s}\left\langle
X(t)Y(s)\right\rangle=\frac{{\rm d}}{{\rm d}s}C_{XY}(t,s)$ (A.37)

##  Appendix B Equations of motion and Langevin equations

Recall the full Lagrangian $\mathcal{L}_{S}+\mathcal{L}_{B}$ of our model [cf.
(1, 3)]:

$\displaystyle\mathcal{L}=$ $\displaystyle~{}\frac{1}{2}\dot{{\bf
R}}^{2}-\frac{\omega_{0}^{2}}{2}{\bf R}^{2}+{\bf A}{\bf R}$
$\displaystyle+\sum_{k=1}^{N}\left[\frac{m_{k}}{2}\dot{{\bf
r}}_{k}^{2}-\frac{m_{k}\omega_{k}^{2}}{2}\left({\bf r}_{k}-\frac{c_{k}{\bf
R}}{m_{k}\omega_{k}^{2}}\right)^{2}\right]$ (A.38)

with ${\bf R}=(X,Y)$, ${\bf r}_{k}=(x_{k},y_{k})$ and the vector potential for
the stationary and homogeneous magnetic field reads ${\bf
A}=(A_{x},A_{y})=(-\frac{bY}{2},\frac{bX}{2})$; cf. (1–3). For simplicity we
assumed that the motion is two-dimensional [cf. (A.1, A.2)], i.e. we did not
account explicitly for the third coordinate of the bath, since for the
harmonic external potential of the Brownian particle (central oscillator), the
third coordinates decouple from the other two coordinates.

From Lagrange equations we get equations of motion for the Brownian particle:

$\displaystyle\ddot{X}$
$\displaystyle=b\dot{Y}-\omega_{0}^{2}X+\sum_{k}c_{k}\left(x_{k}-X\frac{c_{k}}{m_{k}\omega_{k}^{2}}\right),$
(A.39) $\displaystyle\ddot{Y}$
$\displaystyle=-b\dot{X}-\omega_{0}^{2}Y+\sum_{k}c_{k}\left(y_{k}-Y\frac{c_{k}}{m_{k}\omega_{k}^{2}}\right),$
(A.40)

and for bath oscillators:

$\displaystyle m_{k}\ddot{x}_{k}+m_{k}\omega_{k}^{2}x_{k}$
$\displaystyle=c_{k}X(t)$ (A.41) $\displaystyle
m_{k}\ddot{y}_{k}+m_{k}\omega_{k}^{2}y_{k}$ $\displaystyle=c_{k}Y(t)$ (A.42)

We can find exact solutions for $x_{k}(t),y_{k}(t)$ in (A.41, A.42) assuming
that $X(t),Y(t)$ are given. Employ the Laplace transform in (A.41):

$\displaystyle\hat{x}_{k}(s)=v_{xk}(0)\frac{1}{s^{2}+\omega_{k}^{2}}+x_{k}(0)\frac{s}{s^{2}+\omega_{k}^{2}}+\frac{c_{k}}{m_{k}\left(s^{2}+\omega_{k}^{2}\right)}\hat{X}(s).$
(A.43)

Then the inverse Laplace transform produces:

$\displaystyle x_{k}(t)$
$\displaystyle=x_{k}(0)\cos(\omega_{k}t)+\frac{v_{xk}(0)}{\omega_{k}}\sin(\omega_{k}t)+\frac{c_{k}}{m_{k}\omega_{k}}\int_{0}^{t}{\rm
d}t^{\prime}\sin(\omega_{k}(t-t^{\prime}))X(t^{\prime}).$ (A.44)

A similar formula holds for $y(t)$. Now integrate by parts the final integral
in (A.44):

$\displaystyle x_{k}(t)-X(t)\frac{c_{k}}{m_{k}\omega_{k}^{2}}=$
$\displaystyle\left(x_{k}(0)-X(0)\frac{c_{k}}{m_{k}\omega_{k}^{2}}\right)\cos(\omega_{k}t)+\frac{v_{xk}(0)}{\omega_{k}}\sin(\omega_{k}t)$
(A.45) $\displaystyle-\frac{c_{k}}{m_{k}\omega_{k}^{2}}\int_{0}^{t}{\rm
d}t^{\prime}\cos(\omega_{k}(t-t^{\prime}))V_{x}(t^{\prime}),$ (A.46)

and insert these results back to (A.39, A.40)

$\displaystyle\ddot{X}$
$\displaystyle=b\dot{Y}-\omega_{0}^{2}X-r_{x}(t)+\xi_{x}(t),$ (A.47)
$\displaystyle\ddot{Y}$
$\displaystyle=-b\dot{X}-\omega_{0}^{2}Y-r_{y}(t)+\xi_{y}(t).$ (A.48)

Eqs. (A.47, A.48) are Langevin equations for the Brownian particle, where we
defined for friction $r_{s}$ and noise $\xi_{s}$:

$\displaystyle r_{s}(t)$ $\displaystyle=\int_{0}^{t}{\rm
d}t^{\prime}\zeta(t-t^{\prime})V_{s}(t^{\prime}),\qquad s=x,y$ (A.49)
$\displaystyle\zeta(t)$
$\displaystyle=\sum_{k}\frac{c_{k}^{2}}{m_{k}\omega_{k}^{2}}\cos(\omega_{k}t)$
(A.50) $\displaystyle\xi_{x}(t)$
$\displaystyle=\sum_{k}c_{k}\overline{x_{k}}\cos(\omega_{k}t)+\frac{c_{k}}{\omega_{k}}\overline{v_{xk}}\sin(\omega_{k}t),$
(A.51) $\displaystyle\xi_{y}(t)$
$\displaystyle=\sum_{k}c_{k}\overline{y_{k}}\cos(\omega_{k}t)+\frac{c_{k}}{\omega_{k}}\overline{v_{yk}}\sin(\omega_{k}t),$
(A.52)

where we define

$\displaystyle\left(x_{k}(0)-X(0)\frac{c_{k}}{m_{k}\omega_{k}^{2}}\right)=\overline{x_{k}},\qquad$
$\displaystyle v_{xk}(0)=\overline{v_{xk}},\qquad$ (A.53)
$\displaystyle\left(y_{k}(0)-Y(0)\frac{c_{k}}{m_{k}\omega_{k}^{2}}\right)=\overline{y_{k}},\qquad$
$\displaystyle v_{yk}(0)=\overline{v_{yk}}.$ (A.54)

###  B.1 Equilibrium bath

If we take Gibbs distribution for initial conditions for oscillators, i.e.

$\displaystyle(x_{k}(0),y_{k}(0),v_{xk}(0),v_{yk}(0))\sim\mathcal{N}e^{-H/T}$
(A.55)

where $\mathcal{N}$ normalisation constant and

$\displaystyle H=\sum_{k}$
$\displaystyle\left[\frac{1}{2}m_{k}\omega_{k}^{2}\;\overline{y_{k}}^{2}+\frac{1}{2}m_{k}\omega_{k}^{2}\;\overline{y_{k}}^{2}+\frac{1}{2}m_{k}\;\overline{v_{xk}}^{2}+\frac{1}{2}m_{k}\;\overline{v_{yk}}^{2}\right]$
(A.56)

we get

$\displaystyle\left\langle\xi_{\alpha}(t)\xi_{\beta}(t^{\prime})\right\rangle=T\delta_{\alpha\beta}\zeta(t-t^{\prime})\qquad\alpha,\beta=x,y$
(A.57)

We see that $\zeta(t)$ appears both in friction memory (A.49) and noise
correlation (A.57), so this implies fluctuation–dissipation relation holds.

###  B.2 Continuous spectrum of oscillators

Now we consider some specific bath, which has uniformly spaced frequencies
with Drude-Ullersma’s spectrum:

$\displaystyle\omega_{n}=\delta\omega\;n\qquad n=1,2,3,\dots$ (A.58)
$\displaystyle c_{n}=\sqrt{\frac{2\gamma
m_{n}\omega_{n}^{2}\delta\omega}{\pi}\frac{\theta^{2}}{\omega_{n}^{2}+\theta^{2}}}$
(A.59)

we are going to take continuum limit where the spacing of frequencies
$\delta\omega\rightarrow 0$. The memory kernel (which is also the noise
correlation function) will be:

$\displaystyle\zeta(t)=$
$\displaystyle\sum_{n=1}^{\infty}\frac{2\gamma\,\delta\omega}{\pi}\cos(\omega_{n}t)\frac{\theta^{2}}{\omega_{n}^{2}+\theta^{2}}$
(A.60) $\displaystyle\rightarrow$
$\displaystyle\int_{0}^{\infty}\frac{2\gamma}{\pi}\cos(\omega
t)\frac{\theta^{2}}{\omega^{2}+\theta^{2}}{\rm
d}\omega\qquad\text{as}~{}~{}\delta\omega\rightarrow 0$ (A.61)
$\displaystyle=$ $\displaystyle\gamma\;\theta e^{-\theta t}$ (A.62)

This memory function represents memory time $1/\theta$. In the limit
$\theta\rightarrow\infty$ we are going to have memoryless/Ohmic friction and
white noise:

$\displaystyle\zeta(t)=$ $\displaystyle 2\gamma\,\delta(t)$ (A.63)
$\displaystyle c(\omega)=$ $\displaystyle\sqrt{\frac{2\gamma\omega^{2}}{\pi}}$
(A.64) $\displaystyle m\equiv$ $\displaystyle 1$ (A.65)

here we took $m=1$ for all the oscillators; cf. (15).

###  B.3 Symmetries of the model

We have seen that by the initial bath distribution (A.55) and the spectrum of
the oscillator couplings (A.59) we recover the same Langevin equation
(A.1-A.3). Note that these Langevin equations are invariant under rotation in
$X-Y$ plane. Additionally, the initial distribution of $X(0)$, $Y(0)$,
$V_{x}(0)$, $V_{y}(0)$ is invariant under $90^{\circ}$ rotations; see (A.19).
Thus, the solution of these equations must also be invariant under
$90^{\circ}$ rotations. More specifically, for any function
$f(X,V_{x},Y,V_{y})$ we have

$\left\langle f(X(t),V_{x}(t),Y(s),V_{y}(s)\right\rangle=\left\langle
f(Y(t),V_{y}(t),-X(s),-V_{x}(s))\right\rangle$ (A.66)

which corresponds to the $90^{\circ}$ rotation. Eq. (A.31) is a consequence of
it.

Similarly, the Lagrangian (A.38) and initial distribution (A.55, A.19) are
rotation symmetric, and for any function
$f\left(X,V_{x},\\{x_{k},v_{xk}\\},Y,V_{y},\\{y_{k},v_{yk}\\}\right)$ with
$\\{x_{k},v_{xk}\\}\equiv\\{x_{1},x_{2},\dots,v_{x1},v_{x2},\dots\\}$ we have

$\displaystyle\left\langle
f\left(X(t),V_{x}(t),\\{x_{k}(t),v_{xk}(t)\\},Y(s),V_{y}(s),\\{y_{k}(s),v_{yk}(s)\\}\right)\right\rangle$
(A.67) $\displaystyle\qquad=\left\langle
f\left(Y(t),V_{y}(t),\\{y_{k}(t),v_{yk}(t)\\},-X(s),-V_{x}(s),\\{-x_{k}(s),-v_{xk}(s)\\}\right)\right\rangle$
(A.68)

Again, this follows from $90^{\circ}$ rotation. Or in other words, when we
take the average over the distribution of the solution, we can replace
$X\rightarrow Y$ and $Y\rightarrow-X$ and the time derivatives accordingly.
For example

$\displaystyle\left\langle
x_{k}(t)v_{yk}(t)-y_{k}(t)v_{xk}(t)\right\rangle=2\left\langle
x_{k}(t)\,v_{yk}(t)\right\rangle$ (A.69) $\displaystyle\left\langle
X(t)v_{yk}(s)\right\rangle=-\left\langle Y(t)v_{xk}(s)\right\rangle$ (A.70)
$\displaystyle\left\langle X(t)\overline{y_{k}}\right\rangle=-\left\langle
Y(t)\overline{x_{k}}\right\rangle$ (A.71) $\displaystyle\left\langle
X^{2}\right\rangle=\left\langle Y^{2}\right\rangle,\quad\left\langle
Xx_{k}\right\rangle=\left\langle Yy_{k}\right\rangle,\quad\left\langle
v_{xk}^{2}\right\rangle=\left\langle v_{yk}^{2}\right\rangle\quad\text{etc.}$
(A.72)

###  B.4 Linear momentum of bath modes

Starting from equations of motion generated by (A.38) [see also (1, 3)] we can
derive for $\omega_{0}$ and ${\bf B}={\bf A}=0$—i.e. no external potential and
no magnetic field for the Brownian particle—the following conservation law for
the linear momentum:

$\displaystyle\frac{{\rm d}}{{\rm
d}t}\left[\sum_{k}\frac{c_{k}}{\omega_{k}^{2}}\,\dot{\bf r}_{k}+\dot{\bf
R}\right]=0,$ (A.73)

where we note that only a part of the mode linear momentum $m_{k}\dot{\bf
r}_{k}$ participates in the conservation law; $m_{k}\dot{\bf r}_{k}$ is
obviously not conserved, since each mode feels an external potential. We now
employ (A.45) with initial conditions [cf. (10)]:

$\displaystyle\langle{\bf r}_{k}(0)\rangle=\langle\dot{{\bf
r}}_{k}(0)\rangle=\langle{\bf R}(0)\rangle=0,$ (A.74)
$\displaystyle\langle\dot{{\bf R}}(0)\rangle\not=0,$ (A.75)

and also use the Langevin equation that under (A.74, A.75) and (A.63, A.65)
reads

$\displaystyle\frac{{\rm d}}{{\rm d}t}\langle{\bf V}\rangle=-\gamma\langle{\bf
V}\rangle,\qquad\langle{\bf V}(t)\rangle=e^{-\gamma t}\langle{\bf
V}(0)\rangle,\qquad{\bf V}(t)=\dot{{\bf R}}(t).$ (A.76)

Recall that for a free Brownian particle the thermalization according to the
Langevin equation is incomplete: the velocity distribution relaxes towards
Maxwell’s density (in particular, $\langle{\bf V}(t)\rangle$ goes to zero
according to (A.76)), while the coordinate does not relax, since it makes a
free Brownian motion.

We find

$\displaystyle\langle\dot{{\bf r}}_{k}(t)\rangle$ $\displaystyle=$
$\displaystyle\frac{c_{k}}{m_{k}\omega_{k}}\int_{0}^{t}{\rm
d}s\,\sin[\omega_{k}s]\langle{\bf V}(t-s)\rangle$ (A.77) $\displaystyle=$
$\displaystyle\frac{c_{k}\langle{\bf
V}(0)\rangle}{m_{k}\omega_{k}}\left[\frac{\gamma\sin[\omega_{k}t]-\omega_{k}\cos[\omega_{k}t]}{\gamma^{2}+\omega_{k}^{2}}+\frac{\omega_{k}e^{-\gamma
t}}{\gamma^{2}+\omega_{k}^{2}}\right].$ (A.78)

Hence we obtain from (A.78):

$\displaystyle\sum_{k}\frac{c_{k}}{\omega_{k}^{2}}\langle\dot{{\bf
r}}_{k}(t)\rangle=\frac{2\gamma}{\pi}\langle{\bf
V}(0)\rangle\int_{0}^{\infty}{\rm
d}\omega\,\left[\frac{\frac{\gamma}{\omega}\sin[\omega t]-\cos[\omega
t]}{\gamma^{2}+\omega^{2}}+\frac{e^{-\gamma
t}}{\gamma^{2}+\omega^{2}}\right],$ (A.79)

where employing [$t>0$ and $\gamma>0$]

$\displaystyle\int_{-\infty}^{\infty}{\rm d}\omega\,\frac{\cos[\omega
t]}{\gamma^{2}+\omega^{2}}=\frac{\pi e^{-\gamma
t}}{\gamma},\qquad\int_{-\infty}^{\infty}{\rm
d}\omega\,\frac{\frac{\gamma}{\omega}\sin[\omega
t]}{\gamma^{2}+\omega^{2}}=\frac{\pi}{\gamma}(1-e^{-\gamma t}),$ (A.80)

we confirm from (A.79) that the conservation (A.73) indeed holds.

Returning to (A.78) we see that the mean linear momentum of each mode
oscillates in time. This fact contrasts the mean angular momentum of each mode
that (under a non-zero magnetic field) relaxes in time to a well-defined,
frequency-dependent value. Another pertinent difference is that a non-zero
linear momentum for bath modes is achieved only due to $\langle{\bf
V}(0)\rangle\not=0$, i.e. the Brownian particle should have a non-zero initial
momentum for transferring it to the bath modes according to the conservation
law (A.73). Such a direct relation need not hold for the angular momentum,
where the bath modes acquire angular momentum also without the initial angular
momentum of the particle.

Note that generic finite collective observables of the bath lose the non-zero
linear momentum in the long-time limit. Indeed, we note from (A.78):

$\displaystyle\sum_{\omega_{1}}^{\omega_{2}}\frac{c_{k}}{\omega_{k}^{2}}\langle\dot{{\bf
r}}_{k}(t)\rangle=\frac{2\gamma}{\pi}\langle{\bf
V}(0)\rangle\int_{\omega_{1}}^{\omega_{2}}{\rm
d}\omega\,\left[\frac{\frac{\gamma}{\omega}\sin[\omega t]-\cos[\omega
t]}{\gamma^{2}+\omega^{2}}+\frac{\omega_{k}e^{-\gamma
t}}{\gamma^{2}+\omega^{2}}\right],$ (A.81)

where $\omega_{1}<\omega_{2}$ are finite frequencies that define the
collective observable. For $t\gg 1/\gamma$ the contribution $O(e^{-\gamma t})$
in (A.81) can be neglected. The remaining two integrals converge to zero as
$O(1/t)$ for long times $t$ and for finite and non-zero values of $\omega_{1}$
and $\omega_{2}$:

$\displaystyle\int_{\omega_{1}}^{\omega_{2}}{\rm
d}\omega\,\frac{\frac{\gamma}{\omega}\sin[\omega
t]}{\gamma^{2}+\omega^{2}}\simeq\frac{1}{\gamma}\int_{\omega_{1}t}^{\omega_{2}t}{\rm
d}\omega\,\frac{\sin[\omega]}{\omega}=\frac{1}{\gamma}\,O(\frac{1}{t}),$
(A.82) $\displaystyle\int_{\omega_{1}}^{\omega_{2}}{\rm
d}\omega\frac{\cos[\omega
t]}{\gamma^{2}+\omega^{2}}\simeq\frac{\sin[\omega_{2}t]-\sin[\omega_{1}t]}{t\gamma^{2}},$
(A.83)

where we note that the last relation in (A.82) does not apply for
$\omega_{1}=0$. At that specific (non-generic) value of $\omega_{1}$ the
integral in (A.82) converges to a finite value for long-times, since
$\int_{0}^{\infty}{\rm d}\omega\,\frac{\sin[\omega]}{\omega}=\frac{\pi}{2}$.

Altogether, we conclude that for the Brownian particle and finite observables
of the bath the linear momentum is dissipated away. This discussion can be
conveniently summarized as follows: If we define linear momentum density over
the modes at time $t$:

$\displaystyle{\bf\Pi}(\omega;\,t)=\langle{\bf
V}(0)\rangle\frac{2\gamma}{\pi}\left[\frac{\frac{\gamma}{\omega}\sin[\omega
t]-\cos[\omega t]}{\gamma^{2}+\omega^{2}}+\frac{e^{-\gamma
t}}{\gamma^{2}+\omega^{2}}\right],$ (A.84)

then for $t\rightarrow\infty$ this quantity weakly converges to

${\bf\Pi}(\omega;\,t\rightarrow\infty)=2\langle{\bf
V}(0)\rangle\delta(\omega),\qquad\int_{0}^{\infty}{\rm
d}\omega{\bf\Pi}(\omega;\,t\rightarrow\infty)=\langle{\bf V}(0)\rangle.$
(A.85)

For completeness, let us also consider a situation where (A.74) holds, but
instead of (A.75) we take $\langle{\bf V}(0)\rangle=0$. However, now there is
a constant and homogeneous electric field ${\bf E}$ that is acting on the
Brownian particle. Hence the Langevin equation reads instead of (A.76):

$\displaystyle\frac{{\rm d}}{{\rm d}t}\langle{\bf V}\rangle=-\gamma\langle{\bf
V}\rangle+{\bf E},\qquad\langle{\bf V}(t)\rangle=\frac{{\bf
E}}{\gamma}(1-e^{-\gamma t}),$ (A.86)

where we assumed a unit charge. Note that this situation is different from
that of (A.76), since now there is no thermalization even for the velocity of
the Brownian particle, because now this particle moves for long times with a
constant velocity ${\bf E}/\gamma$; cf. (A.86). Also, the conservation law
(A.73) changes to

$\displaystyle\frac{{\rm d}}{{\rm
d}t}\left[\sum_{k}\frac{c_{k}}{\omega_{k}^{2}}\,\dot{\bf r}_{k}+\dot{\bf
R}-{\bf E}\,t\right]=0,$ (A.87)

which means that the overall momentum
$\sum_{k}\frac{c_{k}}{\omega_{k}^{2}}\,\dot{\bf r}_{k}+\dot{\bf R}$ increases
monotonously in time.

Now (A.77) still holds and putting there (A.86) we find

$\displaystyle\langle\dot{{\bf r}}_{k}(t)\rangle$ $\displaystyle=$
$\displaystyle\frac{c_{k}}{m_{k}\omega_{k}}\,\frac{{\bf
E}}{\gamma}\,\frac{1-\cos[\omega_{k}t]}{\omega_{k}}$ (A.88) $\displaystyle-$
$\displaystyle\frac{{\bf
E}}{\gamma}\,\frac{c_{k}}{m_{k}\omega_{k}}\left[\frac{\gamma\sin[\omega_{k}t]-\omega_{k}\cos[\omega_{k}t]}{\gamma^{2}+\omega_{k}^{2}}+\frac{\omega_{k}e^{-\gamma
t}}{\gamma^{2}+\omega_{k}^{2}}\right].$ (A.89)

The contribution coming from (A.89) was studied by us above, hence we focus on
(A.88). Note that

$\displaystyle\sum_{k}\frac{c_{k}}{\omega_{k}^{2}}\,\langle\dot{\bf
r}_{k}\rangle=\frac{2{\bf E}}{\pi}\int_{0}^{\infty}{\rm
d}\omega\,\frac{1-\cos[\omega t]}{\omega^{2}}=t\,{\bf E}.$ (A.90)

This relation validates the conservation law (A.87). Now for
$\omega_{2}>\omega_{1}>0$ we find from (A.88):

$\displaystyle\sum_{\omega_{1}}^{\omega_{2}}\frac{c_{k}}{\omega_{k}^{2}}\,\langle\dot{\bf
r}_{k}\rangle=\frac{2{\bf E}}{\pi}\int_{\omega_{1}}^{\omega_{2}}{\rm
d}\omega\,\frac{1-\cos[\omega t]}{\omega^{2}},$ (A.91)

which converges for $t\to\infty$ to a finite positive value (for
$\omega_{2}>\omega_{1}>0$):

$\displaystyle\sum_{\omega_{1}}^{\omega_{2}}\frac{c_{k}}{\omega_{k}^{2}}\,\langle\dot{\bf
r}_{k}\rangle\to\frac{2{\bf
E}}{\pi}\left[\frac{1}{\omega_{1}}-\frac{1}{\omega_{2}}\right].$ (A.92)

After (A.83) we concluded that for the Brownian particle and finite
observables of the bath the linear momentum is dissipated away. Here the
situation is different: the linear momentum cannot be dissipated away, since
for long times $t$ the Brownian particle and finite bath observables move with
a constant linear momentum $\propto{\bf E}$; cf. (A.86, A.92). The explanation
of this is straightforward: at long times the particle moves with a constant
linear momentum and drags along the bath modes. Instead, the acceleration
induced by ${\bf E}$ is dissipated away from the particle and finite bath
observables. A detailed comparison between the linear momentum and angular
momentum is presented in Table 1.

Table 1: Comparison between the linear momentum and angular momentum is convoluted. Hence we separated 4 cases depending on the following 3 factors. (i) Initial conditions for the Brownian particle, where $\langle{\bf V}(0)\rangle$ and $\langle{\bf L}(0)\rangle$ mean the initial velocity and angular momentum, respectively. (ii) External fields electric ${\bf E}$ and magnetic fields ${\bf B}$. Both are constant and homogeneous in space and only one of them is present for each case. (iii) The external harmonic potential with magnitude $\omega_{0}$ for the particle. For each of these 4 cases we indicate whether and which conservation law is present, whether and to which extent the particle thermalizes for long times, and how collective bath observable behave. It is seen that the most reasonable comparison is to be done between the second and fifth columns, and this is the road we followed in the main text.  | $\omega_{0}=0$ | $\omega_{0}=0$ | $\omega_{0}\not=0$ | $\omega_{0}\not=0$
---|---|---|---|---
| ${\bf E}=0$ | ${\bf E}\not=0$ | ${\bf B}=0$ | ${\bf B}\not=0$
| ${\bf B}=0$ | ${\bf B}=0$ | ${\bf E}=0$ | ${\bf E}=0$
| $\langle{\bf V}(0)\rangle\not=0$ | $\langle{\bf V}(0)\rangle=0$ | $\langle{\bf L}(0)\rangle\not=0$ | $\langle{\bf L}(0)\rangle=0$
Conservation law | yes, effective | no | yes | yes, effective
| linear momentum |  | orbital momentum | orbital momentum
Thermalization of particle | yes, for velocity | no | yes | yes
| distribution only | do not forget ${\bf E}$ |  | Bohr-Van Leeuwen theorem
Collective bath observables | forget $\langle{\bf V}(0)\rangle$ | do not forget ${\bf E}$ | forget $\langle{\bf L}(0)\rangle$ | do not forget ${\bf B}$

##  Appendix C Angular momentum of a bath oscillator

Here we consider a specific oscillator from (A.44) and (A.53):

$\displaystyle x(t)=$ $\displaystyle\overline{x}\,\cos(\omega
t)+X(0)\frac{c}{m\omega^{2}}\cos(\omega
t)+\frac{\overline{v_{x}}}{\omega}\sin(\omega t)$ (A.93)
$\displaystyle+\frac{c}{m\omega}\int_{0}^{t}{\rm
d}t^{\prime}\sin(\omega(t-t^{\prime}))X(t^{\prime})$ (A.94)

where we dropped subscript $k$’s for simplicity. For the velocity we
differentiate (A.93) (written in terms of $y$)

$\displaystyle v_{y}(t)=$ $\displaystyle-\overline{y}\,\omega\sin(\omega
t)-Y(0)\frac{c}{m\omega}\sin(\omega t)+\overline{v_{y}}\,\cos(\omega t)$
(A.95) $\displaystyle+\frac{c}{m}\int_{0}^{t}{\rm
d}t^{\prime}\cos(\omega(t-t^{\prime}))Y(t^{\prime})$ (A.96)

Eventually, we want to calculate the correlation $\left\langle
xv_{y}-yv_{x}\right\rangle$ at time $t$. But as the system is rotation
symmetric, it is enough to consider $\left\langle xv_{y}\right\rangle$; see
(A.69).

$\displaystyle\left\langle xv_{y}\right\rangle_{t}=$
$\displaystyle-\frac{c}{m}\int_{0}^{t}\left\langle
X(t^{\prime})\overline{y}\right\rangle\cos(\omega t^{\prime}){\rm
d}t^{\prime}$ (A.97) $\displaystyle-\frac{c}{m\omega}\int_{0}^{t}\left\langle
X(t^{\prime})\overline{v_{y}}\right\rangle\sin(\omega t^{\prime}){\rm
d}t^{\prime}$ (A.98)
$\displaystyle-\frac{c^{2}}{m^{2}\omega^{2}}\int_{0}^{t}\left\langle
X(t)Y(0)\right\rangle\cos(\omega t^{\prime}){\rm d}t^{\prime}$ (A.99)
$\displaystyle+\frac{c^{2}}{m^{2}\omega}\int_{0}^{t}{\rm
d}t^{\prime}\int_{0}^{t}{\rm d}t^{\prime\prime}\left\langle
X(t^{\prime})Y(t^{\prime\prime})\right\rangle\sin(\omega(t-t^{\prime}))\cos(\omega(t-t^{\prime\prime}))$
(A.100)

where we used the relations (A.70, A.69) following from rotation symmetry and
the independence of the initial conditions (A.55, A.19).

We can calculate $\left\langle
X(t)\left(y(0)-\frac{Y(0)c}{m\omega^{2}}\right)\right\rangle$ and similar
terms using solutions for $X$ and $Y$ (A.16, A.17). In (A.16) only
$\xi_{y}(t)$ depends on
$\overline{y}\equiv\left(y(0)-\frac{Y(0)c}{m\omega^{2}}\right)$ and $v_{y}$;
i.e. the initial state of the bath, see (A.52). So we get

$\displaystyle\left\langle X(t)\overline{y}\right\rangle=$ $\displaystyle
T\frac{c}{m\omega^{2}}\int_{0}^{t}{\rm
d}t^{\prime}H_{1}(t-t^{\prime})\cos(\omega t^{\prime})$ (A.101)
$\displaystyle\left\langle X(t)\overline{v_{y}}\right\rangle=$ $\displaystyle
T\frac{c}{m\omega}\int_{0}^{t}{\rm d}t^{\prime}H_{1}(t-t^{\prime})\sin(\omega
t^{\prime})$ (A.102)

Now we can simplify first two lines of (A.97, A.98):

$\displaystyle-T\frac{c^{2}}{m^{2}\omega^{2}}\int_{0}^{t}{\rm
d}t^{\prime}\int_{0}^{t^{\prime}}{\rm
d}t^{\prime\prime}H_{1}(t^{\prime}-t^{\prime\prime})\left(\cos(\omega
t^{\prime})\cos(\omega t^{\prime\prime})+\sin(\omega t^{\prime})\sin(\omega
t^{\prime\prime})\right)$ (A.103)
$\displaystyle\quad=-T\frac{c^{2}}{2m^{2}\omega^{2}}\int_{0}^{t}\int_{0}^{t}{\rm
d}t^{\prime}{\rm
d}t^{\prime\prime}H_{1}\left(|t^{\prime}-t^{\prime\prime}|\right)\cos(\omega(t^{\prime}-t^{\prime\prime}))$
(A.104) $\displaystyle\quad=-T\frac{c^{2}}{m^{2}\omega^{2}}\int_{0}^{t}{\rm
d}t^{\prime}H_{1}\left(t^{\prime}\right)\,(t-t^{\prime})\cos(\omega
t^{\prime})$ (A.105)

Eq. (A.100) can also be simplified via (A.31):

$\displaystyle\frac{c^{2}}{m^{2}\omega}\int_{0}^{t}{\rm
d}t^{\prime}\int_{0}^{t}{\rm d}t^{\prime\prime}\left\langle
X(t^{\prime})Y(t^{\prime\prime})\right\rangle\frac{1}{2}\left(\sin\left(2\omega
t-\omega(t^{\prime}+t^{\prime\prime}))\right)+\sin(\omega(t^{\prime\prime}-t^{\prime}))\right)$
(A.106) $\displaystyle\quad=\frac{c^{2}}{m^{2}\omega}\int_{0}^{t}{\rm
d}t^{\prime}\int_{0}^{t^{\prime}}{\rm d}t^{\prime\prime}\left\langle
X(t^{\prime})Y(t^{\prime\prime})\right\rangle\sin(\omega(t^{\prime\prime}-t^{\prime}))$
(A.107)

Putting it all together we get a simplified result for $\left\langle
xv_{y}\right\rangle$:

$\displaystyle\left\langle xv_{y}\right\rangle_{t}=$
$\displaystyle-T\frac{c^{2}}{m^{2}\omega^{2}}\int_{0}^{t}{\rm
d}t^{\prime}H_{1}\left(t^{\prime}\right)\,(t-t^{\prime})\cos(\omega
t^{\prime})$ (A.108)
$\displaystyle-\frac{c^{2}}{m^{2}\omega^{2}}\int_{0}^{t}C_{XY}(t^{\prime},0)\cos(\omega
t^{\prime}){\rm d}t^{\prime}$ (A.109)
$\displaystyle-\frac{c^{2}}{m^{2}\omega}\int_{0}^{t}{\rm
d}t^{\prime}\int_{0}^{t^{\prime}}{\rm
d}t^{\prime\prime}C_{XY}(t^{\prime},t^{\prime\prime})\sin(\omega(t^{\prime}-t^{\prime\prime}))$
(A.110)

and the expressions for $C_{XY}(t^{\prime},t^{\prime\prime})$ and
$H_{1}(t^{\prime})$ are given in (A.15, A.27). Taking the limit
$t\rightarrow\infty$ we get 111 The expression that we are interested in is
expressed only by the Laplace kernels (A.15) and its integrals w.r.t. time.
These kernels are exponential functions of time and the integrals can be
easily evaluated. However, the expression contains many exponential terms, the
exact result is found using Mathematica.

$\displaystyle L=2\left\langle xv_{y}\right\rangle_{t\rightarrow\infty}=$
$\displaystyle\delta\omega\,\frac{4\gamma
bT}{\pi}\frac{(\omega-\omega_{0})(\omega+\omega_{0})\left((1-\sigma_{V})\omega^{2}+(1-\sigma_{X})\omega_{0}^{2}\right)}{b^{4}\omega^{4}+\left(\gamma^{2}\omega^{2}+\left(\omega^{2}-\omega_{0}^{2}\right)^{2}\right)^{2}+2b^{2}\left(\gamma^{2}\omega^{4}-\left(\omega^{3}-\omega\omega_{0}^{2}\right)^{2}\right)}$
(A.111) $\displaystyle=$ $\displaystyle\delta\omega\,\frac{4\gamma
bT}{\pi}\frac{(\omega^{2}-\omega_{0}^{2})\left((1-\sigma_{V})\omega^{2}+(1-\sigma_{X})\omega_{0}^{2}\right)}{\left(\left(\omega^{2}-\omega_{0}^{2}+i\gamma\omega\right)^{2}-b^{2}\omega^{2}\right)\left(\left(\omega^{2}-\omega_{0}^{2}-i\gamma\omega\right)^{2}-b^{2}\omega^{2}\right)}$
(A.112) $\displaystyle=$ $\displaystyle\delta\omega\,\frac{4\gamma
bT}{\pi}\frac{(\omega^{2}-\omega_{0}^{2})\left((1-\sigma_{V})\omega^{2}+(1-\sigma_{X})\omega_{0}^{2}\right)}{\left((\omega^{2}-\omega_{0}^{2}-b\omega)^{2}+\gamma^{2}\omega^{2}\right)\left((\omega^{2}-\omega_{0}^{2}+b\omega)^{2}+\gamma^{2}\omega^{2}\right)}$
(A.113)

where

$\displaystyle\left\langle X(0)^{2}\right\rangle=\left\langle
Y(0)^{2}\right\rangle=\sigma_{X}\;\frac{T}{\omega_{0}^{2}}\quad\text{and}\quad\left\langle
V_{x}(0)^{2}\right\rangle=\left\langle
V_{y}(0)^{2}\right\rangle=\sigma_{V}\;T$ (A.114)

##  Appendix D Energy of a bath oscillator

From the Lagrangian (A.38) we can see that conserved energy is

$\displaystyle E_{\text{tot}}=$
$\displaystyle\frac{1}{2}\left(\dot{X}^{2}+\dot{Y}^{2}\right)+\frac{1}{2}\omega_{0}^{2}\left(X^{2}+Y^{2}\right)$
(A.115) $\displaystyle-\sum_{k}c_{k}\left(Xx_{k}+Yy_{k}\right)$ (A.116)
$\displaystyle+\sum_{k}\frac{1}{2}m_{k}\left(\dot{x}_{k}^{2}+\dot{y}_{k}^{2}\right)+\frac{1}{2}m_{k}\omega_{k}^{2}\left(x_{k}^{2}+y_{k}^{2}\right)$
(A.117)
$\displaystyle+\left(X^{2}+Y^{2}\right)\sum_{k}\frac{1}{2}\frac{c_{k}^{2}}{m_{k}\omega_{k}^{2}}$
(A.118)

Our concern is the average energy of the oscillator (again we drop subscript
$k$):

$\displaystyle E=$
$\displaystyle\frac{1}{2}m\left\langle\dot{x}^{2}+\dot{y}^{2}\right\rangle+\frac{1}{2}m\omega^{2}\left\langle
x^{2}+y^{2}\right\rangle$ (A.119) $\displaystyle+\left\langle
X^{2}+Y^{2}\right\rangle\frac{1}{2}\frac{c^{2}}{m\omega^{2}}-c\left\langle
Xx+Yy\right\rangle$ (A.120)

The system is rotation invariant, see (A.72), so the above expression is
equivalent to

$\displaystyle\frac{1}{2}E=$
$\displaystyle\frac{1}{2}m\left\langle{v}_{x}^{2}\right\rangle+\frac{1}{2}m\omega^{2}\left\langle
x^{2}\right\rangle$ (A.121) $\displaystyle+\left\langle
X^{2}\right\rangle\frac{1}{2}\frac{c^{2}}{m\omega^{2}}-c\left\langle
Xx\right\rangle$ (A.122)

The first line can be found by integration of the equation of motion (A.41):

$\frac{1}{2}m\left\langle{v}_{x}^{2}\right\rangle+\frac{1}{2}m\omega^{2}\left\langle
x^{2}\right\rangle\Big{|}_{0}^{t}=c\int_{0}^{t}X(t^{\prime})v_{x}(t^{\prime}){\rm
d}t^{\prime}$ (A.123)

Using (A.95) (in terms of $v_{x}$) we write the first line

$\displaystyle(\ref{E-line1})=$ $\displaystyle
T+\sigma_{X}\frac{Tc^{2}}{2m\omega^{2}\omega_{0}^{2}}$ (A.124)
$\displaystyle+\frac{c^{2}}{m}\int_{0}^{t}{\rm
d}t^{\prime}\int_{0}^{t^{\prime}}{\rm
d}t^{\prime\prime}\;C_{XX}(t^{\prime},t^{\prime\prime})\cos(\omega(t^{\prime}-t^{\prime\prime}))$
(A.125) $\displaystyle-\frac{Tc^{2}}{m\omega}\int_{0}^{t}{\rm
d}t^{\prime}\int_{0}^{t^{\prime}}{\rm
d}t^{\prime\prime}\;H_{0}(t^{\prime}-t^{\prime\prime})\sin(\omega(t^{\prime}-t^{\prime\prime}))$
(A.126) $\displaystyle-\frac{c^{2}}{m\omega}\int_{0}^{t}{\rm
d}t^{\prime}C_{XX}(t^{\prime},0)\sin(\omega t^{\prime})$ (A.127)

where the line (A.124) comes from averages at $t=0$. The second line (A.122)
can be worked out using (A.93)

$\displaystyle(\ref{E-line2})=$
$\displaystyle\frac{Tc^{2}}{2m\omega^{2}\omega_{0}^{2}}-C_{XX}(t,0)\frac{c^{2}}{m\omega^{2}}\cos(\omega
t)$ (A.128) $\displaystyle-\frac{c^{2}}{m\omega}\int_{0}^{t}{\rm
d}t^{\prime}\;C_{XX}(t,t^{\prime})\sin(\omega(t-t^{\prime}))$ (A.129)
$\displaystyle-\frac{Tc^{2}}{m\omega^{2}}\int_{0}^{t}{\rm
d}t^{\prime}\;H_{0}(t^{\prime})\cos(\omega t^{\prime})$ (A.130)

At this stage, all expressions in (A.121, A.122) can be integrated
analytically by symbolic computation software (c.f. footnote 1 on page 1).
Then we take the limit $t\rightarrow\infty$. So we get

$\displaystyle E=$ $\displaystyle
2T-\delta\omega\,\frac{T\gamma}{\pi}\left(\frac{(1-\sigma_{V})\omega^{2}+(1-\sigma_{X})\omega_{0}^{2}}{\left(\gamma^{2}+(\omega-b)^{2}\right)\omega^{2}-2(\omega-b)\omega\omega_{0}^{2}+\omega_{0}^{4}}+\frac{(1-\sigma_{V})\omega^{2}+(1-\sigma_{X})\omega_{0}^{2}}{\left(\gamma^{2}+(\omega+b)^{2}\right)\omega^{2}-2(b+\omega)\omega\omega_{0}^{2}+\omega_{0}^{4}}\right)$
(A.131) $\displaystyle=$ $\displaystyle
2T-\delta\omega\,\frac{T\gamma}{\pi}\frac{\left(\omega^{2}(b^{2}+\gamma^{2})+\left(\omega^{2}-\omega_{0}^{2}\right)^{2}\right)\left((1-\sigma_{V})\omega^{2}+(1-\sigma_{X})\omega_{0}^{2}\right)}{\left((\omega^{2}-\omega_{0}^{2}-b\omega)^{2}+\gamma^{2}\omega^{2}\right)\left((\omega^{2}-\omega_{0}^{2}+b\omega)^{2}+\gamma^{2}\omega^{2}\right)}$
(A.132)

where $\sigma_{X}$ and $\sigma_{V}$ are defined in (A.114).