# Dipoles in blackbody radiation: Momentum fluctuations, decoherence, and drag
force

Kanupriya Sinha  School of Electrical, Energy and Computer Engineering,
Arizona State University, Tempe, Arizona 85281 USA<EMAIL_ADDRESS>Peter
W. Milonni  Theoretical Division (T-4), Los Alamos National Laboratory, Los
Alamos, New Mexico 87545 USA  Department of Physics and Astronomy, University
of Rochester, Rochester, NY 14627 USA<EMAIL_ADDRESS>
###### Abstract

A general expression is derived for the momentum diffusion constant of a small
polarizable particle in blackbody radiation, and is shown to be closely
related to the long-wavelength collisional decoherence rate for such a
particle in a thermal environment. We show how this diffusion constant appears
in the steady-state photon emission rate of two dipoles induced by blackbody
radiation. We consider in addition the Einstein–Hopf drag force on a small
polarizable particle moving in a blackbody field, and derive its fully
relativistic form from the Lorentz transformation of forces.

##  I Introduction

One of the ways in which the quantized nature of the electromagnetic field is
manifested is in the recoil of particles in scattering, absorption, and
emission processes [1]. In spontaneous emission, for example, a classical
treatment of the field would not account for the recoil of an atom because the
field from the atom would have zero linear momentum, a consequence of the
inversion symmetry of the field about the atom. Einstein [2] showed that atoms
in an isotropic field with spectral energy density $\rho(\omega)$ undergo a
mean-square momentum gain resulting from the recoil momentum of magnitude
$\hbar\omega/c$ in each emission and absorption event. Assuming the atoms have
a mean kinetic energy $\left\langle\frac{1}{2}m{\bf
v}^{2}\right\rangle=\frac{3}{2}k_{B}T$ in equilibrium, this mean-square
momentum increase is compensated by a drag force in just such a way as to
yield the Planck distribution for $\rho(\omega)$. In earlier work, Einstein
and Hopf [3, 4] obtained classical expressions for the variance of the
fluctuating momentum and the drag force on a small polarizable particle in
blackbody radiation and used these expressions to derive the Rayleigh–Jeans
spectrum. In this paper we derive a simple quantum-mechanical generalization
of their expression for the momentum variance and diffusion constant. A
similar generalization of the Einstein–Hopf formula for the drag force was
obtained earlier [5]. We show that, not surprisingly, these two
generalizations of Einstein’s expressions can be used to extend Einstein’s
analysis for atoms [2] to any small polarizable particle in thermal
equilibrium with radiation.

The expression we obtain in the Heisenberg picure for the momentum diffusion
constant is an ensemble average over random momentum kicks. We show that this
momentum diffusion constant appears also in the rate of photon emission from
two localized dipoles in blackbody radiation, and has the same form as the
center-of-mass decoherence rate of a particle in a thermal photon environment.
Such blackbody-induced decoherence was previously analyzed in a scattering-
theory framework [6], wherein which-path information carried by the thermal
photons scattering off the particle causes the particle’s center of mass to
decohere and localize in the position basis. If the characteristic thermal
wavelength $\lambda_{\mathrm{th}}$ of the electromagnetic environment is
larger than the coherence length $\Delta x$, each scattered thermal photon
only gains partial information about the particle’s position (‘long-wavelength
limit’); for shorter wavelengths a single scatterer has sufficient information
to resolve a coherent superposition (‘short-wavelength limit’). It will be
shown that the momentum diffusion constant is related specifically to the
long-wavelength limit of the blackbody-induced center-of-mass decoherence
rate.

The rest of the paper is organized as follows. In Section II we calculate the
momentum variance and diffusion constant based first on a model in which the
particle experiences random, instantaneous, and statistically independent
momentum kicks. This is followed by a more rigorous calculation which is
simplified by treating the blackbody field at first as a classical stochastic
field as in the original Einstein–Hopf theory [3, 4]. Following Einstein and
Hopf we treat the electric field and its spatial derivative as independent
stochastic processes; a proof of this independence is given in Appendix A. In
Section III we calculate the photon emission rate for two dipoles in blackbody
radiation. We show how the reduction with the dipole separation of an
interference term in this rate involves the diffusion constant. In Section IV
we derive the relativistic form of the drag force on a moving particle in
blackbody radiation directly from the Lorentz transformation of forces. We
discuss in Section V some aspects of our calculations as they relate to
previous work, and conclude with remarks relating to Einstein’s derivation of
the Planck spectrum and the applicability of our results to atoms in blackbody
radiation. Appendix B gives a brief derivation of the momentum diffusion
constant due to scattering of air molecules and the corresponding center-of-
mass decoherence rate.

##  II  Momentum Fluctuations

###  II.1  Momentum Kicks

We begin by assuming that the momentum diffusion of a particle results from
statistically independent momentum kicks of magnitude $\hbar\omega/c$ due to
photons of frequency $\omega$ (see Fig. 1). Together with a frictional force
$-\beta p(t)$, this implies, for the cumulative change in the linear momentum
$p(t)$,

$\dot{p}(t)+\beta p(t)=\sum_{t_{j}}\frac{\hbar\omega}{c}\delta(t-t_{j}),$ (1)

for motion along, say, the $x$ direction, where $t_{j}$ is the (random) time
at which the $j$th instantaneous momentum kick occurs. Thus in the steady-
state limit

$p(t)=\sum_{t_{j}}\frac{\hbar\omega}{c}G(t-t_{j}),$ (2)

where $G(t)\equiv e^{-\beta t}$ is the response to a unit impulse.

The field energy density in the interval $[\omega,\omega+d\omega]$ is given by
$\rho(\omega)d\omega$. The contribution of photons with energy between
$[\omega,\omega+d\omega]$ to the steady-state, mean-square momentum of the
particle is given, according to Campbell’s theorem [7, 8], by

$d\langle\Delta
p^{2}\rangle=\left(\frac{\hbar\omega}{c}\right)^{2}\times{\rm{(average\
photon\ rate}}\ u(\omega)d\omega)\times\int_{0}^{\infty}G^{2}(t)dt.$ (3)

For $u(\omega)d\omega$ we use the photon rate per unit area per unit time,
$c\rho(\omega)d\omega/\hbar\omega$, times the Rayleigh cross section

$\sigma(\omega)=\frac{8\pi}{3}\left(\frac{\omega}{c}\right)^{4}\left|\alpha(\omega)\right|^{2}=\frac{4\pi\omega}{c}\alpha_{I}(\omega),$
(4)

where $\alpha(\omega)$ is the particle’s polarizability [9],
$\alpha_{I}(\omega)$ is its imaginary part, and we have used the optical
theorem for Rayleigh scattering [10]:

$\alpha_{I}(\omega)=\frac{2}{3}\left(\frac{\omega}{c}\right)^{3}|\alpha(\omega)|^{2}.$
(5)

We integrate Eq. (3) over all field frequencies, using the relation [10]

$\displaystyle\rho(\omega)=\frac{\hbar\omega^{3}}{\pi^{2}c^{3}}n(\omega)$ (6)

between $\rho(\omega)$ and the average number $n(\omega)$ of photons at
frequency $\omega$, to obtain

Figure 1: Schematic representation of a polarizable particle interacting with
a thermal electromagnetic field.

$\langle\Delta p^{2}\rangle=\frac{4\hbar^{2}}{\pi
c^{5}}\int_{0}^{\infty}d\omega\omega^{5}\alpha_{I}(\omega)n(\omega)\times
1/(2\beta).$ (7)

$2\beta$ is the rate at which $\langle\Delta p^{2}\rangle$ decreases due to
the drag force, and in steady state the rate $\langle\Delta
p^{2}\rangle/\Delta t$ at which $\langle\Delta p^{2}\rangle$ increases must
equal $2\beta\langle\Delta p^{2}\rangle$. It follows that

$\frac{\langle\Delta p^{2}\rangle}{\Delta t}=\frac{4\hbar^{2}}{\pi
c^{5}}\int_{0}^{\infty}d\omega\omega^{5}\alpha_{I}(\omega)n(\omega).$ (8)

We have assumed momentum kicks $\hbar\omega/c$, but for any particular kick
the recoil momentum depends on the photon scattering angle $\theta$. Let
$\hat{k}$ and $\hat{k}^{\prime}$ be the unit vectors for the directions of
incoming and scattered photons, respectively. The recoil momentum of the
particle in the $\hat{x}$ direction for this scattering event is

$\delta
p=\frac{\hbar\omega}{c}\left(\hat{k}-\hat{k}^{\prime}\right)\cdot\hat{x}.$ (9)

In terms of the scattering angle $\theta$,
$|\hat{k}-\hat{k}^{\prime}|^{2}=4\sin^{2}(\theta/2)$ and

$\overline{\delta
p^{2}}=\frac{1}{3}\left(\frac{\hbar\omega}{c}\right)^{2}\left(4\sin^{2}\left(\theta/2\right)\right)$
(10)

when we average over the directions of $\hat{k}-\hat{k}^{\prime}$. The average
of this expression over all scattering angles is then
$(2/3)(\hbar\omega/c)^{2}$. We therefore account for all possible scattering
angles by multiplying the expression (8) by 2/3:

$\frac{\langle\Delta p^{2}\rangle}{\Delta t}=\frac{8\hbar^{2}}{3\pi
c^{5}}\int_{0}^{\infty}d\omega\omega^{5}\alpha_{I}(\omega)n(\omega).$ (11)

This derivation of the momentum diffusion constant ${\langle\Delta
p^{2}\rangle}/{\Delta t}$ based on Campbell’s theorem is simpler and more
intuitive than the more rigorous derivation given below. However, it does not
account for the Bose–Einstein statistics of blackbody photons, the effect of
which is to replace $n(\omega)$ in Eq. (11) by $n(\omega)[n(\omega)+1]$. The
application of Campbell’s theorem requires that successive impulses are
statistically independent, whereas the “photon bunching” effect in blackbody
radiation implies that successive momentum kicks are not in fact independent.
The effect of the Bose–Einstein factor $n(\omega)+1$ is considered for two
examples in the following section. For these examples the effect of the
Bose–Einstein factor is negligible and Campbell’s theorem provides an accurate
estimate of the momentum diffusion constant.

It might be worth noting that while Campbell’s theorem does not account for
Bose-Einstein statistics, it can be used, as shown in Appendix B, to describe
momentum diffusion from other environmental scatterers such as air molecules.

###  II.2  Calculation of the Momentum Diffusion Constant from the Force on an
Induced Dipole

We start from the classical expression for the force on an electric dipole
moment ${\bf d}$ in an electromagnetic field:

${\bf F}=({\bf d}\cdot\nabla){\bf E}+\frac{1}{c}\frac{\partial{\bf
d}}{\partial t}\times{\bf B}.$ (12)

For motion along the $x$ axis and a dipole moment ${\bf d}=d_{0}\hat{z}$,

$F_{x}=d_{0}\frac{\partial E_{x}}{\partial z}-\frac{1}{c}\frac{\partial
d_{0}}{\partial t}B_{y},$ (13)

and the impulse experienced by the dipole in a time interval from $t=0$ to
$t=\Delta t$ is

$\displaystyle\Delta p$ $\displaystyle=$ $\displaystyle\int_{0}^{\Delta
t}dtd_{0}\frac{\partial E_{x}}{\partial z}-\frac{1}{c}\int_{0}^{\Delta
t}dt\frac{\partial d_{0}}{\partial t}B_{y}=\int_{0}^{\Delta
t}dtd_{0}\frac{\partial E_{x}}{\partial z}-\frac{1}{c}\int_{0}^{\Delta
t}dt\Big{[}\frac{\partial}{\partial t}(B_{y}d_{0})-d_{0}\frac{\partial
B_{y}}{\partial t}\Big{]}$ (14) $\displaystyle=$
$\displaystyle\int_{0}^{\Delta t}dtd_{0}\frac{\partial E_{z}}{\partial
x}-\frac{1}{c}(B_{y}d_{0})\Big{|}_{0}^{\Delta t},$

where we have used the Maxwell equation $(1/c)\partial B_{y}/\partial
t=\partial E_{z}/\partial x-\partial E_{x}/\partial z$. The last term vanishes
in a steady state, in which case

$\Delta p=\int_{0}^{\Delta t}dtd_{0}\frac{\partial E_{z}}{\partial x}.$ (15)

For the blackbody field at the point dipole we write

$E_{z}(t)=i\sum_{{\bf
k}\lambda}\Big{(}\frac{2\pi\hbar\omega}{V}\Big{)}^{1/2}\big{[}\tilde{a}_{{\bf
k}\lambda}e^{-i\omega t}-\tilde{a}^{*}_{{}_{{\bf k}\lambda}}e^{i\omega
t}\big{]}e_{{\bf k}\lambda z}$ (16)

in the conventional notation [11], except that in the classical approach we
have taken thus far we regard $\tilde{a}_{{\bf k}\lambda}$ and
$\tilde{a}^{*}_{{\bf k}\lambda}$ as classical stochastic variables rather than
photon annihilation and creation operators. Similarly,

$\frac{\partial E_{z}}{\partial x}=-\sum_{{\bf
k}\lambda}\Big{(}\frac{2\pi\hbar\omega}{V}\Big{)}^{1/2}k_{x}\big{[}\tilde{a}_{{\bf
k}\lambda}e^{-i\omega t}+\tilde{a}^{*}_{{}_{{\bf k}\lambda}}e^{i\omega
t}\big{]}e_{{\bf k}\lambda z}.$ (17)

Introducing the complex polarizability $\alpha(\omega)$ for the induced dipole
along the direction $E_{z}$ of the electric field, we have

$d_{0}=i\sum_{{\bf
k}\lambda}\left(\frac{2\pi\hbar\omega}{V}\right)^{1/2}\left[\alpha(\omega)\tilde{a}_{{\bf
k}\lambda}e^{-i\omega t}-\alpha^{*}(\omega)\tilde{a}^{*}_{{}_{{\bf
k}\lambda}}e^{i\omega t}\right]e_{{\bf k}\lambda z},$ (18)

and therefore we obtain, after doing the integral over time in Eq. (15),

$\displaystyle\Delta p$ $\displaystyle=$ $\displaystyle-2i\sum_{{\bf
k}_{1}{\lambda_{1}}}\sum_{{\bf
k}_{2}{\lambda_{2}}}\left(\frac{2\pi\hbar\omega_{1}}{V}\right)^{1/2}\left(\frac{2\pi\hbar\omega_{2}}{V}\right)^{1/2}{k_{1}}_{x}\frac{\sin\left(\frac{1}{2}(\omega_{1}-\omega_{2})\Delta
t\right)}{\omega_{1}-\omega_{2}}$
$\displaystyle\mbox{}\times\left[\alpha(\omega_{2})\tilde{a}_{{\bf
k}_{2}{\lambda_{2}}}\tilde{a}^{*}_{{\bf
k}_{1}{\lambda_{1}}}e^{i(\omega_{1}-\omega_{2})\Delta
t/2}-\alpha^{*}(\omega_{2})\tilde{a}^{*}_{{\bf
k}_{2}{\lambda_{2}}}\tilde{a}_{{\bf
k}_{1}{\lambda_{1}}}e^{-i(\omega_{1}-\omega_{2})\Delta t/2}\right]{e_{{\bf
k}_{1}{\lambda_{1}}z}}{e_{{\bf k}_{2}{\lambda_{2}}z}},$

where we neglect non-resonant terms varying as $1/(\omega_{1}+\omega_{2})$.
The terms denoted by subscripts 1 and 2 in Eq. (LABEL:eq107) derive from
$\partial E_{z}/\partial x$ and $E_{z}$, respectively. Einstein and Hopf [3]
showed that these terms are statistically independent. We give a simplified
proof of this independence, which significantly simplifies the calculation of
$\langle\Delta p^{2}\rangle$, in Appendix A. From this independence it follows
from Eq. (LABEL:eq107) that

$\displaystyle\langle\Delta p^{2}\rangle$ $\displaystyle=$ $\displaystyle
4\sum_{{\bf k}_{1}{\lambda_{1}}}\sum_{{\bf
k}_{2}{\lambda_{2}}}\frac{2\pi\hbar\omega_{1}}{V}\frac{2\pi\hbar\omega_{2}}{V}k^{2}_{1x}\frac{\sin^{2}\left(\frac{1}{2}(\omega_{1}-\omega_{2})\Delta
t\right)}{(\omega_{1}-\omega_{2})^{2}}\left|\alpha(\omega_{2})\right|^{2}\times$
$\displaystyle\mbox{}\Big{[}\langle\tilde{a}_{{\bf
k}_{2}{\lambda_{2}}}\tilde{a}^{*}_{{\bf
k}_{2}{\lambda_{2}}}\rangle\langle\tilde{a}^{*}_{{\bf
k}_{1}{\lambda_{1}}}\tilde{a}_{{\bf
k}_{1}{\lambda_{1}}}\rangle+\langle\tilde{a}^{*}_{{\bf
k}_{2}{\lambda_{2}}}\tilde{a}_{{\bf
k}_{2}{\lambda_{2}}}\rangle\langle\tilde{a}_{{\bf
k}_{1}{\lambda_{1}}}\tilde{a}^{*}_{{\bf
k}_{1}{\lambda_{1}}}\rangle\Big{]}{e^{2}_{{\bf
k}_{1}{\lambda_{1}}z}}{e^{2}_{{\bf k}_{2}{\lambda_{2}}z}}$
$\displaystyle\mbox{}=8\sum_{{\bf k}_{1}{\lambda_{1}}}\sum_{{\bf
k}_{2}{\lambda_{2}}}\frac{2\pi\hbar\omega_{1}}{V}\frac{2\pi\hbar\omega_{2}}{V}k^{2}_{1x}\frac{\sin^{2}\left(\frac{1}{2}(\omega_{1}-\omega_{2})\Delta
t\right)}{(\omega_{1}-\omega_{2})^{2}}A(\omega_{1})A(\omega_{2})\left|\alpha(\omega_{2})\right|^{2}{e^{2}_{{\bf
k}_{1}{\lambda_{1}}z}}{e^{2}_{{\bf k}_{2}{\lambda_{2}}z}},$

where we have assumed the ensemble averages

$\displaystyle\langle\tilde{a}_{{\bf k}_{1}{\lambda_{1}}}\tilde{a}^{*}_{{\bf
k}_{1}^{\prime}\lambda_{1}^{\prime}}\rangle$ $\displaystyle=$ $\displaystyle
A(\omega_{1})\delta^{3}_{{\bf k}_{1},{\bf
k}_{1}^{\prime}}\delta_{\lambda_{1},\lambda_{1}^{\prime}},\ \ \
\omega_{1}=|{\bf k}_{1}|c,$ $\displaystyle\langle\tilde{a}_{{\bf
k}_{1}{\lambda_{1}}}\tilde{a}_{{\bf
k}_{1}^{\prime}\lambda_{1}^{\prime}}\rangle$ $\displaystyle=$ $\displaystyle
0,$ (21)

for the classical stochastic variables describing blackbody radiation, and
likewise for the averages of the terms denoted by subscript 2. Here
$A(\omega)$ is a positive number defined by the first line of Eq. (21).

We have assumed as in the model of Einstein and Hopf that the induced dipole
is along the $z$ direction, and have therefore dealt with only the $z$
component of the electric field. To account for all three components of the
electric field we replace the polarization sums by

$\sum_{\lambda_{1}}\big{[}e^{2}_{{\bf k}_{1}\lambda_{1}x}+e^{2}_{{\bf
k}_{1}\lambda_{1}y}+e^{2}_{{\bf
k}_{1}\lambda_{1}z}\big{]}\sum_{\lambda_{2}}\big{[}e^{2}_{{\bf
k}_{2}\lambda_{2}x}+e^{2}_{{\bf k}_{2}\lambda_{2}y}+e^{2}_{{\bf
k}_{2}\lambda_{2}z}\big{]}=\sum_{\lambda_{1}}{\bf e}^{2}_{{\bf
k}_{1}\lambda_{1}}\sum_{\lambda_{2}}{\bf e}^{2}_{{\bf k}_{2}\lambda_{2}}=4,$
(22)

since ${\bf e}_{{\bf k}\lambda}$ is a unit vector, and then average over the
contributions from the three electric field components. Taking
$V\rightarrow\infty$ in the usual fashion, we then have

$\displaystyle\langle\Delta p^{2}\rangle$ $\displaystyle=\frac{1}{3}\times$
$\displaystyle\frac{32(2\pi\hbar)^{2}}{(8\pi^{3})^{2}c^{8}}\frac{4\pi}{3}4\pi\int_{0}^{\infty}d\omega_{1}\omega_{1}^{5}\int_{0}^{\infty}d\omega_{2}\omega_{2}^{3}\frac{\sin^{2}\left(\frac{1}{2}(\omega_{1}-\omega_{2})\Delta
t\right)}{(\omega_{1}-\omega_{2})^{2}}$ (23) $\displaystyle\mbox{}\times
A(\omega_{1})A(\omega_{2})\left|\alpha(\omega_{2})\right|^{2}.$

The factors of $4\pi/3$ and $4\pi$ result from the angular integration over
${\bf k}_{1}$ and ${\bf k}_{2}$, respectively in Eq. (LABEL:eq108). The term
involving $\omega_{1}-\omega_{2}$ in Eq.(23) is sharply peaked at
$\omega_{1}=\omega_{2}$ compared with the remaining factors in the integrand.
We therefore evaluate these other factors at $\omega_{2}=\omega_{1}$ and use

$\int_{0}^{\infty}d\omega_{2}\frac{\sin^{2}\left(\frac{1}{2}(\omega_{2}-\omega_{1})\Delta
t\right)}{(\omega_{2}-\omega_{1})^{2}}\cong\frac{1}{2}\int_{-\infty}^{\infty}dx\frac{\sin^{2}{x\Delta
t}}{x^{2}}=\pi\Delta t/2$ (24)

in the remaining part of the $\omega_{2}$ integral to obtain

$\frac{\langle\Delta p^{2}\rangle}{\Delta t}=\frac{16\hbar^{2}}{9\pi
c^{8}}\int_{0}^{\infty}d\omega_{1}\omega_{1}^{8}A^{2}(\omega_{1})|\alpha(\omega_{1})|^{2}.$
(25)

Using again the optical theorem Eq. (5), we have finally

$\frac{\langle\Delta p^{2}\rangle}{\Delta t}=\frac{8\hbar^{2}}{3\pi
c^{5}}\int_{0}^{\infty}d\omega\omega^{5}A^{2}(\omega)\alpha_{I}(\omega)$ (26)

for the momentum diffusion constant.

The expression Eq. (26) was obtained from a classical stochastic treatment of
the field. In the quantum treatment of the field $\tilde{a}_{{\bf
k}_{1}{\lambda_{1}}}$ and $\tilde{a}^{*}_{{\bf k}_{1}{\lambda_{1}}}$ in Eq.
(LABEL:eq108) are replaced by photon annihilation and creation operators
${a}_{{\bf k}_{1}{\lambda_{1}}}$ and ${a}^{{\dagger}}_{{\bf
k}_{1}{\lambda_{1}}}$, respectively. Then the quantized-field expression for
the momentum diffusion constant can be obtained by replacing classical
ensemble averages by quantum expectation values:

$\displaystyle\langle\tilde{a}_{{\bf k}_{1}{\lambda_{1}}}\tilde{a}^{*}_{{\bf
k}_{1}{\lambda_{1}}}\rangle$ $\displaystyle\rightarrow$
$\displaystyle\langle{a}_{{\bf k}_{1}{\lambda_{1}}}{a}^{{\dagger}}_{{\bf
k}_{1}{\lambda_{1}}}\rangle=n(\omega_{1})+1,$
$\displaystyle\langle\tilde{a}^{*}_{{\bf k}_{2}{\lambda_{2}}}\tilde{a}_{{\bf
k}_{2}{\lambda_{2}}}\rangle$ $\displaystyle\rightarrow$
$\displaystyle\langle{a}^{{\dagger}}_{{\bf k}_{2}{\lambda_{2}}}{a}_{{\bf
k}_{2}{\lambda_{2}}}\rangle=n(\omega_{2}),$
$\displaystyle\langle\tilde{a}^{*}_{{\bf k}_{1}{\lambda_{1}}}\tilde{a}_{{\bf
k}_{1}{\lambda_{1}}}\rangle$ $\displaystyle\rightarrow$
$\displaystyle\langle{a}^{{\dagger}}_{{\bf k}_{1}{\lambda_{1}}}{a}_{{\bf
k}_{1}{\lambda_{1}}}\rangle=n(\omega_{1}),$
$\displaystyle\langle\tilde{a}_{{\bf k}_{2}{\lambda_{2}}}\tilde{a}^{*}_{{\bf
k}_{2}{\lambda_{2}}}\rangle$ $\displaystyle\rightarrow$
$\displaystyle\langle{a}_{{\bf k}_{2}{\lambda_{2}}}{a}^{{\dagger}}_{{\bf
k}_{2}{\lambda_{2}}}\rangle=n(\omega_{2})+1.$ (27)

Then, following essentially the same arguments that led from Eq. (LABEL:eq108)
to Eq. (26), we obtain the quantized-field expression for the momentum
diffusion constant:

$\frac{\langle\Delta p^{2}\rangle}{\Delta t}=\frac{8\hbar^{2}}{3\pi
c^{5}}\int_{0}^{\infty}d\omega\omega^{5}\alpha_{I}(\omega)[n^{2}(\omega)+n(\omega)],$
(28)

which differs from Eq. (11) by the Bose–Einstein factor $n(\omega)+1$. This is
our general expression for the momentum diffusion constant of a small particle
in blackbody radiation. We note that, with a bit more algebra, it follows
directly of course from a fully quantized-field calculation; this involves the
symmetrization of $\Delta p$ in Eq. (LABEL:eq107) to make it manifestly
Hermitian, and taking $\tilde{a}_{{\bf k}_{1}{\lambda_{1}}}$ and
$\tilde{a}^{*}_{{\bf k}_{2}{\lambda_{2}}}$ to be the operators ${a}_{{\bf
k}_{1}{\lambda_{1}}}$ and ${a}^{{\dagger}}_{{\bf k}_{2}{\lambda_{2}}}$,
respectively. The physical significance of the $n^{2}(\omega)$ and $n(\omega)$
contributions to the momentum diffusion constant is discussed further in
Section V.

Defining a wave number $K=p/\hbar$ associated with a de Broglie wavelength
$\lambda_{{\rm dB}}=2\pi/K$, and using again the optical theorem Eq. (5), we
can express Eq. (28) as

$\displaystyle\frac{\langle\Delta K^{2}\rangle}{\Delta t}$ $\displaystyle=$
$\displaystyle\frac{16}{9\pi
c^{8}}\int_{0}^{\infty}d\omega\omega^{8}|\alpha(\omega)|^{2}[n^{2}(\omega+n(\omega)]$
(29) $\displaystyle=$ $\displaystyle\frac{8}{3\pi
c^{5}}\int_{0}^{\infty}d\omega\omega^{5}\alpha_{I}(\omega)[n^{2}(\omega)+n(\omega)],$

As discussed below, $(\langle\Delta K^{2}\rangle/\Delta t)\Delta x^{2}$ is
closely related to $\Lambda\Delta x^{2}$, the measure of the rate at which
interference, or coherent superposition, is lost between two localized wave
packets separated by a distance $\Delta x$. $\Lambda$ here is the “scattering
constant” appearing in the theory of collisional decoherence due to thermal
photon scattering.

####  II.2.1 Electron in blackbody radiation

Consider as an example an electron in blackbody radiation. From the classical
Abraham-Lorentz equation of motion

$m_{e}\ddot{{\bf r}}-m_{e}\tau_{e}\stackrel{{\scriptstyle...}}{{{\bf r}}}={\bf
E},$ (30)

we deduce a “polarizability”

$\alpha(\omega)=-\frac{e^{2}/m_{e}}{\omega^{2}+i\tau_{e}\omega^{3}}.$ (31)

Since $\tau_{e}=2e^{2}/3m_{e}c^{3}\cong 6.3\times 10^{-24}$ s, we approximate
$\alpha_{I}(\omega)$ by

$\alpha_{I}(\omega)=\frac{e^{2}\tau_{e}}{m_{e}\omega}=\frac{c\sigma_{T}}{4\pi\omega},$
(32)

where $\sigma_{T}=(8\pi/3)(e^{2}/m_{e}c^{2})^{2}$ is the Thomson scattering
cross section. Then, from Eq. (28) with
$n(\omega)=(e^{\hbar\omega/k_{B}T}-1)^{-1}$,

$\frac{\langle\Delta p_{e}^{2}\rangle}{\Delta t}=\frac{16\hbar^{2}}{9\pi
c^{4}}\left(\frac{e^{2}}{m_{e}c^{2}}\right)^{2}\int_{0}^{\infty}d\omega\frac{\omega^{4}e^{\hbar\omega/k_{B}T}}{(e^{\hbar\omega/k_{B}T}-1)^{2}}=\frac{64\pi^{3}}{135}\left(\frac{e^{2}}{m_{e}c^{2}}\right)^{2}\frac{(k_{B}T)^{5}}{\hbar^{3}c^{4}},$
(33)

exactly as obtained by Oxenius [13] in a different approach based directly on
Thomson scattering. $\langle\Delta K_{e}^{2}\rangle/\Delta t$, aside from a
numerical prefactor due to a different averaging procedure, has the same form
as the decoherence rate for free electrons obtained by Joos and Zeh [6].

The role of the Bose–Einstein factor in this example turns out to be very
small. If we replace $n^{2}(\omega)+n(\omega)$ by $n(\omega)$ in Eq. (29) we
obtain

$\displaystyle\frac{\left\langle\Delta p_{e}^{2}\right\rangle}{\Delta
t}\approx\frac{128}{3\pi}\zeta\left(5\right)\left(\frac{e^{2}}{m_{e}c^{2}}\right)^{2}\frac{\left(k_{B}T\right)^{5}}{\hbar^{3}c^{4}},$
(34)

with $\zeta(5)\approx 1.0369$, which is about 96% of the complete result with
both terms retained. $n^{2}(\omega)$ is much larger than $n(\omega)$ at low
frequencies, but this dominance is suppressed by the $\omega^{5}$ factor in
the integrand of Eq. (29); this is due in part to the fact that the field mode
density decreases rapidly with decreasing frequency. The suppression of the
$n^{2}(\omega)$ contribution is even more pronounced in the following example.

####  II.2.2 Dielectric sphere in blackbody radiation

For a small dielectric sphere of radius $a$,

$\alpha(\omega)=\left(\frac{\epsilon-1}{\epsilon+2}\right)a^{3},$ (35)

and, if $n(\omega)=(e^{\hbar\omega/k_{B}T}-1)^{-1}$ and the variation of the
permittivity $\epsilon$ with $\omega$ is negligible,

$\displaystyle\frac{\langle\Delta K_{s}^{2}\rangle}{\Delta t}=$
$\displaystyle\frac{16}{9\pi
c^{8}}a^{6}\left|\frac{\epsilon-1}{\epsilon+2}\right|^{2}\int_{0}^{\infty}d\omega\omega^{8}\frac{e^{\hbar\omega/k_{B}T}}{(e^{\hbar\omega/k_{B}T}-1)^{2}}=\frac{1024\pi^{7}}{135}a^{6}c\left|\frac{\epsilon-1}{\epsilon+2}\right|^{2}\left(\frac{k_{B}T}{\hbar
c}\right)^{9}.$ (36)

If as above we retain only the term $n(\omega)$ in Eq. (29), we obtain,
independently of the temperature, 99.8% of the full result, i.e.,

$\displaystyle\frac{\langle\Delta K_{s}^{2}\rangle}{\Delta t}$
$\displaystyle\cong$ $\displaystyle\frac{16}{9\pi
c^{8}}a^{6}\left|\frac{\epsilon-1}{\epsilon+2}\right|^{2}\int_{0}^{\infty}d\omega\omega^{8}\frac{1}{e^{\hbar\omega/k_{B}T}-1}$
(37) $\displaystyle=$ $\displaystyle
8!\frac{16}{9\pi}\zeta(9)a^{6}c\left|\frac{\epsilon-1}{\epsilon+2}\right|^{2}\left(\frac{k_{B}T}{\hbar
c}\right)^{9}\equiv 2\Lambda,$

$\zeta(9)\cong 1.002$. For this example $\Lambda$ as defined by Eq. (37) is
the scattering constant derived, for example, in the book by Schlosshauer [6,
14].

The momentum diffusion rate we calculated is a mean-square ensemble average
over momentum fluctuations and makes no reference to entanglement with the
environment that is generally understood to be the hallmark of decoherence
[14]. $\Lambda$ as defined in our Eq. (37) is nevertheless consistent with
rigorous calculations of the decoherence rate [16, 17, 18, 14]. In a similar
vein Adler [18] has compared a diffusion rate with results of “decoherence-
based” calculations of collisional decoherence to provide more generally an
“independent cross-check” of those calculations. Such a correspondence between
the center-of-mass decoherence rate and the momentum diffusion has also been
deduced from the variance of momentum as governed by the position-
localization-decoherence master equation [12].

##  III  Field Interference and Decoherence

Consider two point dipoles ${\bf d}_{1}$ and ${\bf d}_{2}$ at fixed positions
${\bf x}_{1}$ and ${\bf x}_{2}$. The interaction Hamiltonian may be taken for
our purposes to be

$H_{I}=-{\bf d}_{1}\cdot{\bf E}({\bf x}_{1},t)-{\bf d}_{2}\cdot{\bf E}({\bf
x}_{2},t).$ (38)

The electric field operator at a point ${\bf x}$ may be expanded in plane-wave
mode functions as in Eq. (16), but here $a_{{\bf k}\lambda}(t)$ and
$a^{{\dagger}}_{{\bf k}\lambda}(t)$ are photon annihilation and creation
operators:

${\bf E}({\bf r},t)=i\sum_{{\bf
k}\lambda}\left(\frac{2\pi\hbar\omega}{V}\right)^{1/2}\big{[}a_{{\bf
k}\lambda}(t)e^{i{\bf k}\cdot{\bf x}}-a^{{\dagger}}_{{\bf
k}\lambda}(t)e^{-i{\bf k}\cdot{\bf x}}\big{]}{\bf e}_{{\bf k}\lambda}.$ (39)

From the Heisenberg equation of motion

$i\hbar\dot{a}_{{\bf k}\lambda}=[a_{{\bf k}\lambda},H_{F}]+[a_{{\bf
k}\lambda},H_{I}]$ (40)

with $H_{F}=\sum_{{\bf k}\lambda}a^{{\dagger}}_{{\bf k}\lambda}a_{{\bf
k}\lambda}$ and the commutation relations for the photon operators, it follows
that

$a_{{\bf k}\lambda i}(t)=a_{{\bf k}\lambda}(0)e^{-i\omega
t}-\left(\frac{2\pi\omega}{\hbar V}\right)^{1/2}\sum_{i}e^{-i{\bf k}\cdot{\bf
x}_{i}}\int_{0}^{t}dt^{\prime}\big{[}{\bf d}_{i}(t^{\prime})\cdot{\bf e}_{{\bf
k}\lambda}\big{]}e^{i\omega(t^{\prime}-t)}.$ (41)

Figure 2: Schematic representation of two dipoles located at positions
$\mathbf{x}_{1}$ and $\mathbf{x}_{2}$, interacting with a thermal field. We
consider the rate $R$ of the average number of photons emitted from the dipole
moments $\mathbf{d}_{1}$ and $\mathbf{d}_{2}$ induced by the blackbody field.

We define a rate of change of the average number of photons radiated by the
dipole moments induced at ${\bf x}_{1}$ and ${\bf x}_{2}$:

$R=\frac{d}{dt}\sum_{{\bf k}\lambda}\big{\langle}\big{[}a^{{\dagger}}_{{\bf
k}\lambda}(t)-a^{{\dagger}}_{{\bf k}\lambda}(0)e^{i\omega
t}\big{]}\big{[}a_{{\bf k}\lambda}(t)-a_{{\bf k}\lambda}(0)e^{-i\omega
t}\big{]}\big{\rangle}.$ (42)

From Eq. (41),

$R=\frac{2\pi}{\hbar V}\times 2{\rm Re}\sum_{i,j}\sum_{{\bf k}}\omega e^{i{\bf
k}\cdot({\bf x}_{i}-{\bf
x}_{j})}\int_{0}^{t}dt^{\prime}e^{i\omega(t^{\prime}-t)}\sum_{\lambda}\left\langle\left[{\bf
d}_{i}^{{\dagger}}(t)\cdot{\bf e}_{{\bf k}\lambda}\right]\left[{\bf
d}_{j}(t^{\prime})\cdot{\bf e}_{{\bf k}\lambda}\right]\right\rangle.$ (43)

Consider first

$X_{ij}\equiv\int_{0}^{t}dt^{\prime}e^{i\omega(t^{\prime}-t)}\sum_{\lambda}\Big{\langle}[{\bf
d}_{i}^{{\dagger}}(t)\cdot{\bf e}_{{\bf k}\lambda}][{\bf
d}_{j}(t^{\prime})\cdot{\bf e}_{{\bf k}\lambda}]\Big{\rangle},$ (44)

where ${\bf d}_{i}(t)$ is the dipole moment induced by the blackbody field at
${\bf x}_{i}$:

$\displaystyle{\bf d}_{i}(t)\cdot{\bf e}_{{\bf k}\lambda}$ $\displaystyle=$
$\displaystyle i\sum_{{\bf
k}^{\prime}\lambda^{\prime}}\left(\frac{2\pi\hbar\omega^{\prime}}{V}\right)^{1/2}\Big{[}\alpha(\omega{{}^{\prime}})a_{{\bf
k}^{\prime}\lambda^{\prime}}(0)e^{-i\omega{{}^{\prime}}t}e^{i{\bf
k}^{\prime}\cdot{\bf
x}_{i}}-\alpha^{*}\left(\omega{{}^{\prime}}\right)a^{{\dagger}}_{{\bf
k}^{\prime}\lambda^{\prime}}(0)e^{i\omega{{}^{\prime}}t}e^{-i{\bf
k}^{\prime}\cdot{\bf x}_{i}}\Big{]}$ (45) $\displaystyle\mbox{}\times{\bf
e}_{{\bf k}^{\prime}\lambda^{\prime}}\cdot{\bf e}_{{\bf k}\lambda}.$

For $X_{ij}$ we obtain

$\displaystyle X_{ij}$ $\displaystyle=$ $\displaystyle\sum_{{\bf
k}^{\prime}\lambda^{\prime}}\Big{(}\frac{2\pi\hbar\omega{{}^{\prime}}}{V}\Big{)}\Big{\\{}|\alpha(\omega{{}^{\prime}})|^{2}\Big{[}\langle
a_{{\bf k}^{\prime}\lambda^{\prime}}^{{\dagger}}(0)a_{{\bf
k}^{\prime}\lambda^{\prime}}(0)\rangle e^{-i{\bf k}^{\prime}\cdot({\bf
x}_{i}-{\bf
x}_{j})}\int_{0}^{t}dt^{\prime}e^{i(\omega-\omega{{}^{\prime}})(t^{\prime}-t)}$
(46) $\displaystyle\mbox{}+\langle a_{{\bf
k}^{\prime}\lambda^{\prime}}(0)a_{{\bf
k}^{\prime}\lambda^{\prime}}^{{\dagger}}(0)\rangle e^{i{\bf
k}^{\prime}\cdot({\bf x}_{i}-{\bf
x}_{j})}\int_{0}^{t}dt^{\prime}e^{i(\omega+\omega{{}^{\prime}})(t^{\prime}-t)}\Big{]}|\hat{e}_{{\bf
k}\lambda}\cdot\hat{e}_{{\bf k}^{\prime}\lambda^{\prime}}|^{2}\Big{\\}}.$

We have used the expectation values

$\big{\langle}a^{{\dagger}}_{{\bf k}^{\prime}\lambda^{\prime}}(0)a_{{\bf
k}^{\prime\prime}\lambda^{\prime\prime}}(0)\big{\rangle}=n_{{\bf
k}^{\prime}\lambda^{\prime}}\delta^{3}_{{\bf k}^{\prime}{\bf
k}^{\prime\prime}}\delta_{\lambda^{\prime}\lambda^{\prime\prime}}=n_{k^{\prime}}\delta^{3}_{{\bf
k}^{\prime}{\bf
k}^{\prime\prime}}\delta_{\lambda^{\prime}\lambda^{\prime\prime}},\ \ \
\big{\langle}a_{{\bf k}^{\prime}\lambda^{\prime}}(0)a_{{\bf
k}^{\prime\prime}\lambda^{\prime\prime}}(0)\big{\rangle}=0$ (47)

for the blackbody field. From the form of the time integrals in Eq. (46) it is
seen that only the normally ordered expectation value will contribute to the
final result, and therefore we write

$X_{ij}=\frac{2\pi\hbar}{V}\sum_{{\bf
k}^{\prime}}\omega{{}^{\prime}}|\alpha(\omega{{}^{\prime}})|^{2}n_{k^{\prime}}e^{-i{\bf
k}^{\prime}\cdot({\bf x}_{i}-{\bf
x}_{j})}\sum_{\lambda}\sum_{\lambda^{\prime}}|{\bf e}_{{\bf
k}^{\prime}\lambda^{\prime}}\cdot{\bf e}_{{\bf
k}\lambda}|^{2}\int_{0}^{t}dt^{\prime}e^{i(\omega-\omega{{}^{\prime}})(t^{\prime}-t)},$
(48)

or, since

$\sum_{\lambda}\sum_{\lambda^{\prime}}|{\bf e}_{{\bf
k}^{\prime}\lambda^{\prime}}\cdot{\bf e}_{{\bf k}\lambda}|^{2}=1+({\bf
k}\cdot{\bf k}^{\prime})^{2}/(kk^{\prime})^{2},$ (49)
$X_{ij}=\frac{2\pi\hbar}{V}\frac{V}{8\pi^{3}}\int
d^{3}k^{\prime}\omega{{}^{\prime}}|\alpha(\omega{{}^{\prime}})|^{2}n_{k^{\prime}}e^{-i{\bf
k}^{\prime}\cdot({\bf x}_{i}-{\bf x}_{j})}\left[1+({\bf k}\cdot{\bf
k}^{\prime})^{2}/(kk^{\prime})^{2}\right]\int_{0}^{t}dt^{\prime}e^{i(\omega-\omega{{}^{\prime}})(t^{\prime}-t)}$
(50)

in the mode continuum limit. Therefore,

$\displaystyle R$ $\displaystyle=$ $\displaystyle\frac{4\pi}{\hbar
V}\frac{2\pi\hbar}{V}\frac{V}{8\pi^{3}}\frac{V}{8\pi^{3}}\sum_{i,j}\int
d^{3}k\omega\int
d^{3}k^{\prime}\omega{{}^{\prime}}|\alpha(\omega{{}^{\prime}})|^{2}n_{k}e^{i({\bf
k}-{\bf k}^{\prime})\cdot({\bf x}_{i}-{\bf x}_{j})}\left[1+({\bf k}\cdot{\bf
k}^{\prime})^{2}/(kk^{\prime})^{2}\right]$ (51)
$\displaystyle\mbox{}\times\int_{0}^{t}dt^{\prime}\cos\left[(\omega-\omega{{}^{\prime}})(t^{\prime}-t)\right].$

For $t\rightarrow\infty$,

$\int_{0}^{t}dt^{\prime}\cos\left[(\omega-\omega{{}^{\prime}})(t^{\prime}-t)\right]\rightarrow\pi\delta(\omega-\omega{{}^{\prime}})=\frac{\pi}{c}\delta(k-k^{\prime}),$
(52)

and

$\displaystyle R$ $\displaystyle=$
$\displaystyle\sum_{i,j}\frac{c}{8\pi^{3}}\int d^{3}kk\int
d^{3}k^{\prime}k^{\prime}|\alpha(\omega{{}^{\prime}})|^{2}n_{k}e^{i({\bf
k}-{\bf k}^{\prime})\cdot({\bf x}_{i}-{\bf x}_{j})}\big{[}1+({\bf k}\cdot{\bf
k}^{\prime})^{2}/(kk^{\prime})^{2}\Big{]}\delta(k-k^{\prime})$ (53)
$\displaystyle=$ $\displaystyle\sum_{i,j}\frac{c}{8\pi^{3}}\int
dkk^{6}|\alpha(\omega)|^{2}n_{k}\int d\Omega_{\hat{{\bf k}}}\int
d\Omega_{\hat{{\bf k}}^{\prime}}e^{ik(\hat{{\bf k}}-\hat{{\bf
k}}^{\prime})\cdot({\bf x}_{i}-{\bf x}_{j})}\big{(}1+\cos^{2}\theta\big{)}$
$\displaystyle\equiv$ $\displaystyle\sum_{i,j}R_{ij},$

where $\hat{{\bf k}}$ is a unit vector in the direction of ${\bf k}$, $\theta$
is the angle between $\hat{{\bf k}}$ and $\hat{{\bf k}}^{\prime}$, and
$d\Omega_{\hat{{\bf k}}}$ is a differential element of solid angle about
$\hat{{\bf k}}$ ($\int d^{3}k=\int dkk^{2}\int d\Omega_{\hat{{\bf k}}}$).

We rewrite

$R_{12}=\frac{c}{8\pi^{3}}\int dkk^{6}|\alpha(\omega)|^{2}n_{k}\int
d\Omega_{\hat{{\bf k}}}\int d\Omega_{\hat{{\bf k}}^{\prime}}e^{ik(\hat{{\bf
k}}-\hat{{\bf k}}^{\prime})\cdot({\bf x}_{i}-{\bf
x}_{j})}\big{(}1+\cos^{2}\theta\big{)}$ (54)

using $n_{k}=(e^{\hbar kc/k_{B}T}-1)^{-1}$ and the differential cross section
for Rayleigh scattering:

$|f({\bf k},{\bf
k}^{\prime})|^{2}=\frac{d\sigma}{d\Omega}=k^{4}|\alpha(\omega)|^{2}\frac{1}{2}(1+\cos^{2}\theta).$
(55)

Thus

$R_{12}=\frac{c}{4\pi^{3}}\int dk\frac{k^{2}}{e^{\hbar kc/k_{B}T}-1}\int
d\Omega_{\hat{{\bf k}}}\int d\Omega_{\hat{{\bf k}}^{\prime}}e^{ik(\hat{{\bf
k}}-\hat{{\bf k}}^{\prime})\cdot({\bf x}_{i}-{\bf x}_{j})}\big{|}f(k\hat{{\bf
k}},k\hat{{\bf k}^{\prime}})\big{|}^{2}.$ (56)

We express this in terms of $q=\hbar k$, $v(q)=c$, and

$dq\rho(q)=dk\rho(k)=\frac{k^{2}dk/\pi^{2}}{e^{\hbar kc/k_{B}T}-1},$ (57)

and, to compare with Reference [14], for instance, we also change notation,
writing ${\bf x}_{1}-{\bf x}_{2}={\bf x}-{\bf x}^{\prime}$,
$d\Omega_{\hat{{\bf k}}}d\Omega_{\hat{{\bf k}}^{\prime}}=d\hat{\bf n}d\hat{\bf
n}^{\prime}$, and replacing $\big{|}f(k\hat{{\bf k}},k\hat{{\bf
k}}^{\prime})\big{|}^{2}$ by $\big{|}f(q\hat{\bf n},q\hat{\bf
n}^{\prime})\big{|}^{2}$. Then

$R_{12}=\int dq\rho(q)v(q)\int\frac{d\hat{\bf n}d\hat{\bf
n}^{\prime}}{4\pi}e^{iq(\hat{\bf n}-\hat{\bf n}^{\prime})\cdot({\bf x}-{\bf
x}^{\prime})/\hbar}\big{|}f(q\hat{\bf n},q\hat{\bf n}^{\prime})\big{|}^{2},$
(58)

and similarly

$R_{11}=\int dq\rho(q)v(q)\int\frac{d\hat{\bf n}d\hat{\bf
n}^{\prime}}{4\pi}\big{|}f(q\hat{\bf n},q\hat{\bf n}^{\prime})\big{|}^{2}.$
(59)

The photon scattering rate $R_{12}$, as opposed to $R_{11}$ involves
interference between the fields of the two dipoles. The difference
$R_{11}-R_{12}$ is therefore a measure of the rate at which this interference
decreases with increasing separation of the dipoles. In other words, it
characterizes the loss of spatial coherence between the two dipole fields. In
fact

$F({\bf x}-{\bf x}^{\prime})\equiv R_{11}-R_{12}=\int
dq\rho(q)v(q)\int\frac{d\hat{\bf n}d\hat{\bf
n}^{\prime}}{4\pi}\Big{(}1-e^{iq(\hat{\bf n}-\hat{\bf n}^{\prime})\cdot({\bf
x}-{\bf x}^{\prime})/\hbar}\Big{)}\big{|}f(q\hat{\bf n},q\hat{\bf
n}^{\prime})\big{|}^{2}$ (60)

happens to have exactly the form of the more general collisional decoherence
rate [16, 17, 18, 14]. In the long-wavelength limit the decay rate of the
reduced density matrix in the case of a dielectric sphere reduces to
$-\Lambda({\bf x}-{\bf x}^{\prime})^{2}$, where $\Lambda$ is defined in Eq.
(37). The “decoherence factor” $F\left({\bf x}-{\bf x}^{\prime}\right)$, as it
appears in the theory of collisional decoherence [16, 17, 18, 14], represents
the overlap between the states of environmental scatterers that scatter off
the positions ${\bf x}$ and ${\bf x}^{\prime}$ of the particle’s center of
mass, denoting their distinguishability. Though, it is to be noted that the
scattering model of decoherence considers two points ${\bf x}$ and ${\bf
x}^{\prime}$ located within the same particle whose center of mass is
spatially delocalized; in the above calculation ${\bf x}$ and ${\bf
x}^{\prime}$ correspond to the location of two distinct dipoles, that are not
necessarily constrained to constitute the same particle.

##  IV  Drag Force

As noted in the Introduction, Einstein and Hopf [3, 4] and later Einstein [2]
showed that there is a drag force $F=-\xi mv$ on a polarizable particle moving
with velocity $v$ ($v\ll c$) with respect to a frame in which there is an
isotropic (blackbody) field of spectral energy density $\rho(\omega)$. A more
general expression for $\xi$ was derived much later [5], and more recently the
drag force has been derived for relativistic particles [19, 20, 21, 22]. We
now derive the relativistic form of the drag force based in part on “energy
kicks” $\hbar\omega$, in the same heuristic spirit as in the treatment of
momentum kicks in Section II.1. The particle is assumed to have a uniform
velocity $v$ along the $x$ axis in the laboratory frame S, and we are
interested in the force $F_{x}$ on the particle in this frame. The Lorentz
transformations relating the spacetime coordinates and the momenta
$(p_{x},p_{x}^{\prime}$) and energies ($\mathcal{E},\mathcal{E}^{\prime}$) in
the laboratory frame and the particle’s rest frame S′ are

$\displaystyle x^{\prime}$ $\displaystyle=$ $\displaystyle\gamma(x-vt),$ (61)
$\displaystyle t^{\prime}$ $\displaystyle=$
$\displaystyle\gamma\left(t-\frac{vx}{c^{2}}\right),$ (62) $\displaystyle
p_{x}^{\prime}$ $\displaystyle=$
$\displaystyle\gamma\left(p_{x}-\frac{v}{c^{2}}\mathcal{E}\right),$ (63)
$\displaystyle\mathcal{E}^{\prime}$ $\displaystyle=$
$\displaystyle\gamma\left(\mathcal{E}-vp_{x}\right),$ (64)

where $dx/dt=v$ and as usual $\gamma=(1-v^{2}/c^{2})^{-1/2}$. These
transformations relate the force $F_{x}$ in S to the force $F_{x}^{\prime}$ in
S′:

$F_{x}=\frac{dp_{x}}{dt}=\frac{d}{dt^{\prime}}\left[\gamma\left(p_{x}^{\prime}+\frac{v}{c^{2}}\mathcal{E}^{\prime}\right)\right]\frac{dt^{\prime}}{dt}=\frac{dp_{x}^{\prime}}{dt^{\prime}}+\frac{v}{c^{2}}\frac{d\mathcal{E}^{\prime}}{dt^{\prime}}=F_{x}^{\prime}+\frac{v}{c^{2}}\frac{d\mathcal{E}^{\prime}}{dt^{\prime}}.$
(65)

The force on the particle has two contributions, $F_{\rm ind}$ resulting from
the dipole moment induced by the fluctuating blackbody field, and $F_{\rm dd}$
resulting from the dipole’s own radiation field and fluctuations [19, 20, 21,
22]. $F_{\rm ind}$ is the force originally considered by Einstein and Hopf [2,
3, 4], while $F_{\rm dd}$ is needed to obtain the fully relativistic
expression for the force that has been of interest in recent work [19, 20, 21,
22].

###  IV.1 Calculation of the force $F_{\rm ind}$

In Section II.2 we calculated the force when the electric field is along a
single direction $z$ and then exploited the isotropy of the field to include
the contributions from all three components of the field. The situation is
more complicated for a particle in motion because in its reference frame (S′)
the fields are different in different directions and are related to those in S
by

$\displaystyle E_{x}^{\prime}({\bf x}^{\prime},t^{\prime})$ $\displaystyle=$
$\displaystyle E_{x}({\bf x},t),$ $\displaystyle E_{y}^{\prime}({\bf
x}^{\prime},t^{\prime})$ $\displaystyle=$ $\displaystyle\gamma\left[E_{y}({\bf
x},t)-\frac{v}{c}B_{z}({\bf x},t)\right],$ $\displaystyle E_{z}^{\prime}({\bf
x}^{\prime},t^{\prime})$ $\displaystyle=$ $\displaystyle\gamma\left[E_{z}({\bf
x},t)+\frac{v}{c}B_{y}({\bf x},t)\right],$ (66)

along with the corresponding transformations for the magnetic fields. The
force $d_{0}\partial E_{z}/\partial x$ given by (15) applies when the dipole
moment $d_{0}$ ($d_{0}=d_{z}$) and the electric field both have only $z$
components. More generally the force along the $x$ direction is

$F_{x}=d_{x}\frac{\partial E_{x}}{\partial x}+d_{y}\frac{\partial
E_{y}}{\partial x}+d_{z}\frac{\partial E_{z}}{\partial x},$ (67)

as is easily shown. For the quantum calculation it will be convenient to
consider first a single field mode polarized along the $x$ direction:

$\displaystyle E^{\prime}({\bf x}^{\prime},t^{\prime})$ $\displaystyle=$
$\displaystyle E_{x}({\bf
x},t)=i\left(\frac{2\pi\hbar\omega}{V}\right)^{1/2}\left[a_{{\bf
k}\lambda}e^{-i(\omega t-{\bf k}\cdot{\bf x})}-{a}^{{\dagger}}_{{\bf
k}\lambda}e^{i(\omega t-{\bf k}\cdot{\bf x})}\right]e_{{\bf k}\lambda x}$ (68)
$\displaystyle=$ $\displaystyle
i\left(\frac{2\pi\hbar\omega}{V}\right)^{1/2}\left[a_{{\bf
k}\lambda}e^{-i(\omega_{{\bf k}^{\prime}}t^{\prime}-{\bf k}^{\prime}\cdot{\bf
x}^{\prime})}-{a}^{{\dagger}}_{{\bf k}\lambda}e^{i(\omega_{{\bf
k}^{\prime}}t^{\prime}-{\bf k}^{\prime}\cdot{\bf x}^{\prime})}\right]e_{{\bf
k}\lambda x}.$

We have used the fact that $e^{i(\omega t-{\bf k}\cdot{\bf
x})}=e^{i(\omega_{{\bf k}}^{\prime}t^{\prime}-{\bf k}^{\prime}\cdot{\bf
x}^{\prime})}$ since we will require $\partial E_{x}^{\prime}/\partial
x^{\prime}$. The primed and unprimed frequencies and wave vectors are related
by

$\displaystyle\omega{{}^{\prime}}$ $\displaystyle=$
$\displaystyle\gamma(\omega-vk_{x}),$ $\displaystyle k_{x}^{\prime}$
$\displaystyle=$
$\displaystyle\gamma\left(k_{x}-\frac{v}{c^{2}}\omega\right),\ \ \
k_{y}^{\prime}=k_{y},\ \ \ k_{z}^{\prime}=k_{z}.$ (69)

The induced dipole moment $d_{x}^{\prime}({\bf x}^{\prime},t^{\prime})$ is
given by

$d_{x}^{\prime}({\bf
x}^{\prime},t^{\prime})=i\left(\frac{2\pi\hbar\omega}{V}\right)^{1/2}\left[\alpha(\omega{{}^{\prime}})a_{{\bf
k}\lambda}e^{-i(\omega{{}^{\prime}}t^{\prime}-{\bf k}^{\prime}\cdot{\bf
x}^{\prime})}-\alpha^{*}(\omega{{}^{\prime}}){a}^{{\dagger}}_{{\bf
k}\lambda}e^{i(\omega{{}^{\prime}}t^{\prime}-{\bf k}^{\prime}\cdot{\bf
x}^{\prime})}\right]e_{{\bf k}\lambda x},$ (70)

since $\alpha(-\omega_{{\bf k}^{\prime}})=\alpha^{*}(\omega_{{\bf
k}^{\prime}})$.

Consider first the force

$[F^{\prime}_{\rm ind}]_{xx}=\frac{1}{2}\left\langle
d^{\prime}_{x}\frac{\partial E^{\prime}_{x}}{\partial
x^{\prime}}+\frac{\partial E^{\prime}_{x}}{\partial
x^{\prime}}d^{\prime}_{x}\right\rangle={\rm Re}\left\langle
d^{\prime}_{x}\frac{\partial E^{\prime}_{x}}{\partial
x^{\prime}}\right\rangle,$ (71)

corresponding quantum mechanically to the first term in (67). The expectation
value is over the thermal equilibrium state of the field. It follows from (68)
and 70) that

$[F^{\prime}_{\rm
ind}]_{xx}=\frac{4\pi\hbar\omega}{V}k_{x}^{\prime}\alpha_{I}(\omega_{{\bf
k}}^{\prime})n(\omega)e^{2}_{{\bf k}\lambda x},$ (72)

where we have defined $n(\omega)=\langle{a}^{{\dagger}}_{{\bf
k}\lambda}a_{{\bf k}\lambda}\rangle$ for the (unpolarized) blackbody field and
used the fact that $\langle a_{{\bf k}_{1}\lambda_{1}}{a}_{{\bf
k}_{2}\lambda_{2}}\rangle=0$ for this field. Next we add up the contributions
to $F_{1x}^{\prime}$ from all field modes:

$\displaystyle[F^{\prime}_{\rm ind}]_{xx}$ $\displaystyle\rightarrow$
$\displaystyle\frac{V}{8\pi^{3}}\int
d^{3}k\left(\frac{4\pi\hbar\omega}{V}\right)k_{x}^{\prime}\alpha_{I}(\omega_{{\bf
k}^{\prime}})n(\omega)\sum_{\lambda=1}^{2}e^{2}_{{\bf k}\lambda x}$ (73)
$\displaystyle=$ $\displaystyle\frac{\hbar}{2\pi^{2}}\int d^{3}k\omega
k_{x}^{\prime}\left(1-\frac{k_{x}^{2}}{k^{2}}\right)\alpha_{I}(\omega_{{\bf
k}^{\prime}})n(\omega).$

To compare with previous work we wish to write this force as an integral over
the primed variables. To this end we note from the Jacobian that follows from
Eq. (69) that

$d^{3}k=\gamma\left(1+\frac{v}{c}\frac{k_{x}^{\prime}}{k^{\prime}}\right)d^{3}k^{\prime},$
(74)

and that, again from Eq. (69),

$\displaystyle 1-\frac{k_{x}^{2}}{k^{2}}$ $\displaystyle=$
$\displaystyle\frac{1}{\gamma^{2}}\frac{1-{k_{x}^{\prime}}^{2}/{k^{\prime}}^{2}}{\big{(}1+\frac{v}{c}\frac{k_{x}^{\prime}}{k^{\prime}}\big{)}^{2}},$
$\displaystyle\omega$ $\displaystyle=$
$\displaystyle\gamma(\omega{{}^{\prime}}+vk_{x}^{\prime}).$ (75)

It then follows that

$\displaystyle[F^{\prime}_{\rm ind}]_{xx}$ $\displaystyle=$
$\displaystyle\frac{\hbar c}{2\pi^{2}}\int
d^{3}k^{\prime}k^{\prime}k_{x}^{\prime}\left(1-\frac{{k_{x}^{\prime}}^{2}}{{k^{\prime}}^{2}}\right)\alpha_{I}(\omega{{}^{\prime}})n(\gamma(\omega{{}^{\prime}}+vk_{x}^{\prime}))$
(76) $\displaystyle=$ $\displaystyle\frac{\hbar c}{2\pi^{2}}\int
d^{3}kkk_{x}\left(1-\frac{k^{2}_{x}}{k^{2}}\right)\alpha_{I}(\omega)n(\gamma(\omega+vk_{x})),$

where in the last step we simply replace primed dummy integration variables by
unprimed ones.

The remaining contributions to $F_{1x}^{\prime}$ corresponding to the last two
terms in Eq. (67), are found similarly. We simply state the results of the
straightforward calculations:

$[F^{\prime}_{\rm ind}]_{xy}=\frac{\hbar c}{2\pi^{2}}\int
d^{3}kkk_{x}\left(1-\frac{k^{2}_{y}}{k^{2}}\right)\alpha_{I}(\omega)n(\gamma(\omega+vk_{x})),$
(77) $[F^{\prime}_{\rm ind}]_{xz}=\frac{\hbar c}{2\pi^{2}}\int
d^{3}kkk_{x}\left(1-\frac{k^{2}_{z}}{k^{2}}\right)\alpha_{I}(\omega)n(\gamma(\omega+vk_{x})).$
(78)

Summing up all the components, one obtains:

$\displaystyle F^{\prime}_{\rm ind}$ $\displaystyle=$
$\displaystyle[F^{\prime}_{\rm ind}]_{xx}+[F^{\prime}_{\rm
ind}]_{xy}+[F^{\prime}_{\rm ind}]_{xz}$ (79) $\displaystyle=$
$\displaystyle\frac{\hbar c}{2\pi^{2}}\int
d^{3}kkk_{x}\left(1-\frac{k_{x}^{2}}{k^{2}}+1-\frac{k_{y}^{2}}{k^{2}}+1-\frac{k_{z}^{2}}{k^{2}}\right)\alpha_{I}(\omega)n(\gamma(\omega+vk_{x}))$
$\displaystyle=$ $\displaystyle\frac{\hbar c}{\pi^{2}}\int
d^{3}kkk_{x}\alpha_{I}(\omega)n(\gamma(\omega+vk_{x}))$

$F^{\prime}_{\rm ind}$ is by definition the average force in the particle
frame due to the fluctuating dipole moment induced by the blackbody field in
the particle frame. There is no contribution to this force from the dipole’s
fluctuations from radiation reaction. Furthermore there is no contribution to
$F^{\prime}_{\rm ind}$ of the type
$(v/c^{2})d\mathcal{E}^{\prime}/dt^{\prime}=(v/c^{2})({\bf F}\cdot{\bf u})$,
since the velocity ${\bf u}=0$ in the particle frame. Equation (65) therefore
implies that $F_{\rm ind}$, the force in the laboratory frame due to the
fluctuation dipole moment induced by the blackbody field, is identical to the
force $F^{\prime}_{\rm ind}$ in the particle frame:

$F_{\rm ind}=\frac{\hbar c}{\pi^{2}}\int
d^{3}kkk_{x}\alpha_{I}(\omega)n(\gamma(\omega+vk_{x})),$ (80)

where $n(\omega)=(e^{\hbar\omega/k_{b}T}-1)^{-1}$ and $T$ is the temperature
in the laboratory frame.

Before proceeding we consider the nonrelativistic limit of Eq. (79), using

$n(\gamma(\omega+vk_{x}))\cong
n(\omega)+vk_{x}\frac{dn}{d\omega}=n(\omega)-vk_{x}\frac{3\pi^{2}c^{3}}{\hbar\omega^{4}}\left[\rho(\omega)-\frac{\omega}{3}\frac{d\rho}{d\omega}\right],$
(81)

which follows from Eq. (6). Then, carrying out the integration in Eq. (79)
under this approximation, we obtain

$F_{\rm ind}=\frac{4\hbar}{3\pi
c^{5}}v\int_{0}^{\infty}d\omega\omega^{5}\alpha_{I}(\omega)\frac{dn}{d\omega}=-\frac{4\pi}{c^{2}}v\int_{0}^{\infty}d\omega\omega\alpha_{I}(\omega)\left[\rho(\omega)-\frac{\omega}{3}\frac{d\rho}{d\omega}\right]\equiv-\xi
mv$ (82)

for $v\ll c$. In this limit this is the total force on the moving particle in
blackbody radiation in the laboratory frame. It has the basic form obtained by
Einstein and Hopf [3] and is equivalent to the more general expression
obtained by Mkrtchian et al. [5]. The obtained drag force can be physically
understood as follows: the moving particle creates an asymmetric density of
modes due to a relativistic aberration effect [10], the preferential emission
into forward modes results in a friction force due to recoil in the opposite
direction.

One can consider the thermal drag for the previous examples of an electron and
a dielectric sphere. The non-relativistic thermal drag coefficient for an
electron in blackbody radiation can be obtained from Eq. (82) and Eq. (32) as

$\displaystyle\xi_{e}=\frac{32\pi^{3}\hbar}{135m_{e}}\left(\frac{e^{2}}{m_{e}c^{2}}\right)^{2}\left(\frac{k_{B}T}{\hbar
c}\right)^{4}.$ (83)

Similarly, the non-relativistic thermal drag on a dielectric sphere is found
from Eq. (82) and Eq. (35) to be

$\displaystyle\xi_{s}=\frac{512\pi^{7}\hbar}{135m_{s}}\left|\frac{\epsilon-1}{\epsilon+2}\right|^{2}\left(\frac{k_{B}T}{\hbar
c}\right)^{8},$ (84)

where $m_{s}$ represents the mass of the dielectric sphere.

We note that the friction coefficients ($\xi_{e}$ and $\xi_{s}$) and the
momentum diffusion constants (Eq. (33) and Eq. (36)) for an electron and a
dielectric sphere, respectively, obey the fluctuation-dissipation relation

$\frac{\langle\Delta p^{2}\rangle}{\Delta t}=2m\xi k_{B}T.$ (85)

We will discuss this relation further in Section V.

###  IV.2 Calculation of the force $F_{\rm dd}$

$F_{\rm dd}$ is by definition the force due to the dipole’s radiation
reaction, and is therefore attributable to radiation by the dipole. It is
given according to Eq. (65) by $(v/c^{2})d\mathcal{E}^{\prime}/dt^{\prime}$,
with $d\mathcal{E}^{\prime}/dt^{\prime}$ the rate of change due to radiation
of the particle’s energy in its reference frame.
$d\mathcal{E}^{\prime}/dt^{\prime}$ cannot of course depend on the velocity
$v$, and so it must be just minus the energy absorption rate $R_{\rm abs}$ of
the particle if it were in equilibrium with an isotropic Planck spectrum
$\rho(\omega)$ at the temperature in S′. For $R_{\rm abs}$ we write, in close
analogy to our treatment of momentum kicks in Section II.1,

$R_{\rm
abs}=-d\mathcal{E}^{\prime}/dt^{\prime}=\int_{0}^{\infty}\hbar\omega\big{[}u(\omega)d\omega\big{]}=\int_{0}^{\infty}\hbar\omega\big{[}c\rho(\omega)/\hbar\omega\big{]}\sigma(\omega)d\omega,$
(86)

and so, from Eqs. (4), (6), and (5),

$F_{\rm dd}=-\frac{4v\hbar}{\pi
c^{5}}\int_{0}^{\infty}d\omega\omega^{4}\alpha_{I}(\omega)n(\omega)=-\frac{4v\hbar}{\pi
c^{5}}\int_{0}^{\infty}d\omega\omega^{4}\alpha_{I}(\omega)\frac{1}{e^{\hbar\omega/k_{B}T^{\prime}}-1},$
(87)

where $T^{\prime}$ is the thermal equilibrium temperature in S′.

We note that a force of the form $(v/c^{2})d\mathcal{E}^{\prime}/dt^{\prime}$
is consistent with the insightful analysis by Sonnleitner et al. [24] of a
“vacuum friction force” for an excited atom moving with uniform velocity $v$
in a vacuum. For example, for a two-level atom in its excited state at $t=0$,

$F_{x}=(v/c^{2})d\mathcal{E}^{\prime}/dt^{\prime}\cong(v/c^{2})\frac{d}{dt}\left[\hbar\omega_{0}e^{-\Gamma
t}\right]=-\frac{v}{c^{2}}\hbar\omega_{0}\Gamma e^{-\Gamma t}$ (88)

to order $v/c$, where $\omega_{0}$ and $\Gamma$ are respectively the atom’s
transition frequency and spontaneous emission rate in S′ [25].

The total force $F_{x}=F_{\rm ind}+F_{\rm dd}$ in the laboratory frame follows
from Eqs. (65), (79), and (87). It takes the form given by Milton et al. [22]:

$\displaystyle F_{x}$ $\displaystyle=$
$\displaystyle-\frac{2\hbar}{\pi\gamma^{2}v^{2}}\int_{0}^{\infty}d\omega\omega^{4}\alpha_{I}(\omega)\int_{u_{-}}^{u_{+}}dy\left(y-\frac{1}{\gamma}\right)\left[\frac{1}{e^{\hbar\omega/k_{B}T^{\prime}}-1}-\frac{1}{e^{\hbar\omega
y/k_{B}T}-1}\right]$ $\displaystyle=$
$\displaystyle-\frac{\hbar}{\pi\gamma^{2}v^{2}}\int_{0}^{\infty}d\omega\omega^{4}\alpha_{I}(\omega)\int_{u_{-}}^{u_{+}}dy\left(y-\frac{1}{\gamma}\right)\big{[}\coth{(\hbar\omega/2k_{B}T^{\prime})}-\coth{(\hbar\omega
y/2k_{B}T)}\big{]},$

where $u_{+}=\sqrt{\frac{1+v/c}{1-v/c}}$, $u_{-}=\sqrt{\frac{1-v/c}{1+v/c}}$
[22].

##  V  Discussion

The average photon number $n(\omega)$ in thermal equilibrium appears
differently in the momentum diffusion constant, the decoherence rate, and the
drag force. This is because the momentum diffusion constant has a fourth-order
dependence on the electric field, whereas the decoherence rate and drag force
have a second-order dependence.

The momentum diffusion constant we have obtained for a small polarizable
particle in blackbody radiation, for example, involves the factor
$n^{2}(\omega)+n(\omega)$. The two contributions correspond respectively to
the “wave” and “particle” contributions to the Einstein fluctuation formula
for blackbody radiation [10]. But in the two examples we considered—a free
electron and a dielectric sphere—the $n^{2}(\omega)$ contribution is
negligible and the momentum diffusion constant has effectively only a
“particle” contribution from the field. It then has the same form as the
collisional decoherence rate $\Lambda$.

In the case of decoherence due to collisions with gas molecules [16, 18], in
contrast, no “wave” or Bose–Einstein factor appears because the de Broglie
wavelengths of the gas molecules are so small that quantum statistics are
irrelevant.

Of course the $n^{2}(\omega)$ and $n(\omega)$ terms in the momentum variance
and the drag force are both essential in determining the Planck spectrum. In
thermal equilibrium the rate of change of kinetic energy due to momentum
diffusion, $(1/2m)\langle\Delta p^{2}\rangle/\Delta t$, plus the rate of
change $F_{x}v=-m\xi v^{2}=-2\xi\langle\frac{1}{2}mv^{2}\rangle=-\xi k_{B}T$
due to the drag force, must add to zero. ($m$ is the particle mass and $\xi$
is defined by Eq. (82).) Since for our purposes only the nonrelativistic
momentum diffusion rate was needed, we consider here the nonrelativistic drag
force. The condition for thermal equilibrium of the field, assuming thermal
equipartition of the particles’ kinetic energy, is then

$\frac{1}{2m}\frac{8\hbar^{2}}{3\pi
c^{5}}\int_{0}^{\infty}d\omega\omega^{5}\alpha_{I}(\omega)[n^{2}(\omega)+n(\omega)]+k_{B}T\frac{4\hbar}{3\pi
c^{5}}\int_{0}^{\infty}d\omega\omega^{5}\alpha_{I}(\omega)\frac{dn}{d\omega}=0,$
(90)

or

$\frac{dn}{d\omega}=-\frac{\hbar}{k_{B}T}(n^{2}+n).$ (91)

This follows independently of $\alpha_{I}(\omega)$, assuming
$\alpha_{I}(\omega)>0$ as implied by our use of the optical theorem for
Rayleigh scattering. If we assume the form of the momentum diffusion constant
obtained in Section II.1, we obtain

$\frac{dn}{d\omega}=-\frac{\hbar}{k_{B}T}n$ (92)

instead of Eq. (91), with the solution

$n(\omega)=(\rm const.)\times e^{-\hbar\omega/k_{B}T}.$ (93)

Together with Eq. (6), this implies the Wien form of the spectrum. If instead
we assume the momentum diffusion constant $\propto n^{2}$ obtained with
classical stochastic wave theory by Einstein and Hopf, we obtain

$\frac{dn}{d\omega}=-\frac{\hbar}{k_{B}T}n^{2}$ (94)

instead of Eq. (91), with the solution

$n(\omega)=k_{B}T/\hbar\omega,$ (95)

when a constant of integration is chosen to give the Rayleigh–Jean spectrum.
The solution of Eq. (91) is of course

$n(\omega)=(e^{\hbar\omega/k_{B}T}-1)^{-1},$ (96)

when a constant interpretable in terms of a chemical potential is chosen so
that the chemical potential is zero for photons, resulting in the Planck
spectrum for $\rho(\omega)$. Expressed in terms of $\rho(\omega)$, Eq. (91) is
the equation obtained by Einstein [2]).

The fluctuation-dissipation relation (Eq. (85)), can be used in the
Fokker–Planck equation for the velocity distribution function $f(v,t)$ to
obtain the Maxwell–Boltzmann distribution. That is, the Fokker–Planck equation
with the drift and diffusion terms satisfying Eq. (85),

$\frac{\partial f}{\partial t}=-\xi\frac{\partial}{\partial v}(vf)+\frac{\xi
k_{B}T}{m}\frac{\partial^{2}f}{\partial v^{2}},$ (97)

has the solution

$f(v,t)=\left(\frac{m}{2\pi k_{B}T}\right)^{1/2}e^{-mv^{2}/2k_{B}T}$ (98)

for $\xi t\gg 1$. The condition that the momentum diffusion and the drag force
together determine the blackbody spectrum, subject to the assumption
$\langle\frac{1}{2}mv^{2}\rangle=\frac{1}{2}k_{B}T$, therefore implies the
full Maxwell–Boltzmann distribution for the particle velocities [26].

As noted above, we have assumed throughout that $\alpha_{I}(\omega)>0$,
consistent with the optical theorem for Rayleigh scattering. This assumption
does not allow for stimulated emission. However, we can derive the momentum
diffusion constant and the drag force for atoms, allowing for stimulated
emission, by a similar analysis in which $\alpha_{I}(\omega)$ is introduced
not through the optical theorem but through the response of the field to the
atoms. The same expression Eq. (82) then follows for the drag force, for
example, on an atom. For a two-level atom, for example, the polarizability in
the rotating-wave approximation is

$\alpha(\omega)=\frac{{{\mu}}^{2}}{3\hbar}\frac{p_{1}-p_{2}}{\omega_{0}-\omega+i\beta},$
(99)
$\alpha_{I}(\omega)=\frac{{{\mu}}^{2}}{3\hbar}\frac{\beta(p_{1}-p_{2})}{(\omega_{0}-\omega)^{2}+\beta^{2}},$
(100)

where $\omega_{0}$ and $\mu$ are the transition frequency and transition
dipole moment, respectively, $\beta$ is proportional to the homogeneous
linewidth, and $p_{1}$ and $p_{2}$ are respectively the lower- and upper-level
occupation probabilities. Then, from Eq. (82),

$\displaystyle F_{\rm ind}$ $\displaystyle\cong$
$\displaystyle-\frac{4\pi\omega_{0}v}{c^{2}}\big{[}\rho(\omega_{0})-\frac{\omega_{0}}{3}\frac{d\rho(\omega_{0})}{d\omega}\big{]}\int_{0}^{\infty}d\omega\alpha_{I}(\omega)\cong-\frac{4\pi^{2}\mu^{2}\omega_{0}v}{3\hbar
c^{2}}(p_{1}-p_{2})\big{[}\rho(\omega_{0})-\frac{\omega_{0}}{3}\frac{d\rho(\omega_{0})}{d\omega}\big{]}$
(101) $\displaystyle=$
$\displaystyle-\frac{\hbar\omega_{0}}{c^{2}}(p_{1}-p_{2})B\big{[}\rho(\omega_{0})-\frac{\omega_{0}}{3}\frac{d\rho(\omega_{0})}{d\omega}\big{]}v,$

where $B=4\pi^{2}\mu^{2}/3\hbar^{2}$ is the Einstein $B$ coefficient. This is
equivalent to the expression for the drag force obtained by Einstein [2].

##  VI Acknowledgment

We are grateful to Bruce Shore for his many and varied contributions to
atomic, molecular, and optical science. Peter Milonni is especially thankful
for having been a friend of Bruce’s for many years.

##  Appendix A Statistical independence of $E_{z}$ and $\partial
E_{z}/\partial x$

Consider first the following elementary example. Let $X$ and $Y$ be two
classical Gaussian random processes such that $\langle X\rangle=\langle
Y\rangle=0$ and $\langle XY\rangle=0$. Then $X$ and $Y$ are statistically
independent, i.e., the joint probability distribution
$p_{XY}(X,Y)=p_{X}(X)p_{Y}(Y)$. The simple proof of this goes as follows.
Consider the characteristic function

$C_{XY}(\xi,\eta)=\Big{\langle}\exp{i(\xi X+\eta Y)}\Big{\rangle}$ (102)

whose Fourier transform is $p_{XY}(X,Y)$. Since $X$ and $Y$ are Gaussian, so
are linear combinations of $X$ and $Y$. So $Z=\xi X+\eta Y$ is Gaussian and
$\langle Z\rangle=0$. Then

$\displaystyle\Big{\langle}\exp{(iZ)}\big{\rangle}$ $\displaystyle=$
$\displaystyle 1-\frac{1}{2}\langle
Z^{2}\rangle+\frac{1}{24}3{\big{\langle}Z^{2}\big{\rangle}}^{2}-...=\exp\big{(}-\langle
Z^{2}\rangle/2\big{)}$ (103) $\displaystyle=$
$\displaystyle\exp\big{(}-\langle(\xi X+\eta
Y)^{2}/2\rangle\big{)}=\exp\big{(}-\langle\xi^{2}X^{2}+\eta^{2}Y^{2}+2\xi\eta
XY\rangle/2\big{)}$ $\displaystyle=$
$\displaystyle\exp\big{(}-\langle\xi^{2}X^{2}\rangle/2\big{)}\exp(-\langle\eta^{2}Y^{2}\rangle/2\big{)},$

since $\Big{\langle}XY\Big{\rangle}=0$ by assumption. We have used the fact
that for Gaussian statistics $\langle Z^{n}\rangle=0$ for $n$ an odd positive
integer and

$\Big{\langle}Z^{2n}\Big{\rangle}=1\cdot 3\
...\cdot(2n-1)\Big{\langle}Z^{2}\Big{\rangle}^{n}.$ (104)

to write
$\Big{\langle}Z^{4}\Big{\rangle}=3\Big{\langle}Z^{2}\Big{\rangle}^{2}$ etc. in
(103). Thus $C_{XY}(\xi,\eta)=C_{X}(\xi)C_{Y}(\eta)$ and therefore
$P_{XY}(X,Y)=p_{X}(X)p_{Y}(Y)$.

In the case of blackbody radiation, treated classically or quantum
mechanically, the moments $\big{\langle}E_{z}^{n}\big{\rangle}$ have the
Gaussian form as above. Since $E_{z}$ is Gaussian, so is its derivative
$\partial E_{z}/\partial x$ according to a well-known property of Gaussian
random processes. Furthermore $\big{\langle}E_{z}(\partial E_{z}/\partial
x)\big{\rangle}=0$. Therefore, in accordance with the example above, $E_{z}$
and $\partial E_{z}/\partial x$ for blackbody radiation can be regarded as
statistically independent in calculating expectation values, as shown by
Einstein and Hopf [3].

##  Appendix B Momentum diffusion and decoherence from scattering by air
molecules

According to Campbell’s theorem the momentum variance of dielectric particles
due to momentum kicks from air molecules is

$\displaystyle d\left(\Delta
p^{2}\right)\approx\mathrm{d}q(q^{2})\rho\left(q\right)\frac{q}{m_{\mathrm{air}}}\sigma_{\mathrm{air}},$
(105)

where $q=m_{\mathrm{air}}v$ is the momentum of an air molecule,
$\sigma_{\mathrm{air}}=2\pi a^{2}/3$ is the scattering cross section, and

$\displaystyle\rho(q)=\frac{N_{\mathrm{air}}}{V}4\pi q^{2}\left(\frac{1}{2\pi
m_{\mathrm{air}}k_{B}T}\right)^{3/2}e^{-q^{2}/(2m_{\mathrm{air}}k_{B}T)}$
(106)

corresponds to the Maxwell-Boltzmann distribution for the air molecules.
Integrating over the velocities and using similar arguments as in the case of
blackbody radiation, we obtain

$\displaystyle\frac{\left\langle\Delta p^{2}\right\rangle}{\Delta t}\approx$
$\displaystyle\frac{8\pi}{m_{\mathrm{air}}}\frac{N_{\mathrm{air}}}{V}\left(\frac{1}{2\pi
m_{\mathrm{air}}k_{B}T}\right)^{3/2}\int_{0}^{\infty}\mathrm{d}qq^{5}e^{-q^{2}/(2m_{\mathrm{air}}k_{B}T)}$
(107) $\displaystyle=$
$\displaystyle\frac{16a^{2}}{3}\frac{N_{\mathrm{air}}}{V}\sqrt{2\pi
m_{\mathrm{air}}}\left(k_{B}T\right)^{3/2},$ (108)

which is twice the center-of-mass decoherence rate for scattering of air
molecules in the long-wavelength limit [14], similar to the case of thermal
photon scattering.

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