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# Gauge dependence of the Aharonov-Bohm phase in a quantum electrodynamics
framework

A. Hayashi<EMAIL_ADDRESS>Department of Applied Physics (Emeritus),
University of Fukui, Fukui 910-8507, Japan

###### Abstract

The Aharonov-Bohm (AB) phase is usually associated with a line integral of the
electromagnetic vector potential generated by an external current source, such
as a solenoid. According to this interpretation, the AB phase of a nonclosed
path cannot be observed, as the integral depends on the gauge choice of the
vector potential. Recent attempts to explain the AB effect through the
interaction between a charged particle and an external current, mediated by
the exchange of quantum photons, have assumed that the AB phase shift is
proportional to the change in interaction energy between the charged particle
and the external current source. As a result, these attempts argue that the AB
phase change along a path does not depend on the gauge choice, and that the AB
phase shift for a nonclosed path is in principle measurable. In this paper, we
critically examine this claim and demonstrate that the phase obtained through
this approach is actually gauge-dependent and not an observable for a
nonclosed path. We also provide a brief critical discussion of the proposed
experiment for observing the AB phase shift of a nonclosed path.

###### pacs:

PACS:03.67.Hk

##  I Introduction

The Aharonov-Bohm (AB) effect ehrenberg1949 ; aharonov1959  is a quantum
mechanical phenomenon in which a charged particle is influenced by the
electromagnetic potential even in regions where the electromagnetic field can
be neglected. For reviews, see, e.g., peshkin1981 ; olariu1985 ; peshkin1989 ;
wakamatsu2018 . The existence of the AB effect was first experimentally
demonstrated by Tonomura et al. using electron beam holography techniques
tonomura1986 ; osakabe1986 .

In a typical scenario of the AB effect, we consider an external current
source, such as a solenoid, and assume that a charged particle is moving in a
region where the magnetic field generated by the current source is negligible,
but the corresponding vector potential is finite. In the standard
interpretation of the AB effect, the Schroedinger equation for the charged
particle includes a coupling term between the particle and the background
vector potential generated by the external current. As a result, the charged
particle acquires an extra phase even in regions where no magnetic field is
present. This phase (the AB phase) is represented by a line integral of the
background vector potential along the path of the particle. Therefore it is
widely accepted that the AB phase is a gauge-dependent quantity and is not an
observable unless the particle’s path forms a closed loop.

However, there are some interesting unconventional approaches to understand
the AB phase. For instance, Vaidman made an attempt to explain the AB phase
without relying on gauge-dependent potentials vaidman2012 . According to this
explanation, the AB phase is attributed to the local interaction between the
charged particle’s field and the potential source. For further discussions on
this approach, refer to aharonov2015 ; pearle2017 ; pearle2017-2 ; li2022 .

There have been yet other proposals of new attempts based on a quantum
electrodynamics framework santos1999 ; marletto2020 ; saldanha2021 ; kang2022
. In this approach, it is considered that a charged particle and an external
current source interact by the exchange of quantum electrodynamics photons,
with the AB phase being directly proportional to the change in interaction
energy. Since energy is commonly regarded as a gauge-invariant observable, it
is asserted that the AB phase shift is also gauge invariant and can be
measured even when the particle’s path is not closed marletto2020 ;
saldanha2021 ; kang2022 . In this paper, however, we will disprove this claim
by showing that the energy correction computed in the quantum electrodynamics
framework generally depends on the choice of gauge.

In Sec. II, we first review the approach to the AB phase in the quantum
electrodynamics framework. This approach primarily involves perturbations in
both the charge of the particle, denoted as $e$, and the strength parameter of
the external current, denoted as $g$. Next, we introduce a different scheme
using the coherent state, where the assumption of $g$ being small is no longer
required. In this scheme, while the results remain unchanged, the calculations
are simplified, providing us with a deeper understanding of how the energy
correction depends on the gauge condition. The Coulomb gauge or the Lorenz
gauge is commonly used in the quantum electrodynamics approach to the AB
phase. In Sec. III, we adopt the axial gauge condition $A_{3}=0$, and
demonstrate that the energy correction is a gauge dependent quantity, implying
that the AB phase for a nonclosed path is not an observable. Discussions
including brief comments on the proposed experiment for measuring the AB phase
for a nonclosed path are given in Sec. IV.

##  II The system of a charged particle, quantum electromagnetic field, and an
external current source

We consider the combined system of a charged particle, quantum electromagnetic
field, and a static external current source such as a solenoid. The mass, the
charge and the coordinate of the particle are denoted as $m$, $e$, and $q$,
respectively. The charged particle interacting with the electromagnetic field
is also subject to the action of a certain potential $V(\mbox{\boldmath$q$})$.
The quantum electromagnetic field also couples to the current
$J_{\mu}(\mbox{\boldmath$x$})=(0,\mbox{\boldmath$J$}(\mbox{\boldmath$x$}))$
produced by the static external current source. The equations of motion for
the charged particle and the electromagnetic fields can be written as follows:

$\displaystyle m\frac{d^{2}\mbox{\boldmath$q$}}{dt^{2}}=-\frac{\partial
V}{\partial\mbox{\boldmath$q$}}+e\left(\mbox{\boldmath$E$}(\mbox{\boldmath$q$},t)+\mbox{\boldmath$\dot{q}$}\times\mbox{\boldmath$B$}(\mbox{\boldmath$q$},t)\right),$
(1a)
$\displaystyle\nabla\cdot\mbox{\boldmath$E$}(\mbox{\boldmath$x$},t)=j^{0}(\mbox{\boldmath$x$},t),$
(1b)
$\displaystyle\nabla\times\mbox{\boldmath$B$}(\mbox{\boldmath$x$},t)=\mbox{\boldmath$j$}(\mbox{\boldmath$x$},t)+g\mbox{\boldmath$J$}(\mbox{\boldmath$x$})+\frac{\partial\mbox{\boldmath$E$}(\mbox{\boldmath$x$},t)}{\partial
t},$ (1c)

where $j^{\mu}(x)$ is the current of the charged particle given by

$\displaystyle
j^{0}(\mbox{\boldmath$x$},t)=e\delta(\mbox{\boldmath$x$}-\mbox{\boldmath$q$}),$
(2)
$\displaystyle\mbox{\boldmath$j$}(\mbox{\boldmath$x$},t)=e\dot{\mbox{\boldmath$q$}}\delta(\mbox{\boldmath$x$}-\mbox{\boldmath$q$}),$
(3)

and we introduced the strength parameter $g$ for the external current
$\mbox{\boldmath$J$}(\mbox{\boldmath$x$})$ for later convenience. Note that
the static external current $\mbox{\boldmath$J$}(\mbox{\boldmath$x$})$ is
conserved: $\nabla\cdot\mbox{\boldmath$J$}=0$. Furthermore, we take into
account that the external current is localized in space. As a result, we can
safely assume that all the fields rapidly approach zero at infinity.
Throughout this paper we employ the Heaviside-Lorentz (rationalized Gaussian)
system of units with $\hbar=c=1$.

We start by adopting the Coulomb gauge, where the vector potential $A$
satisfies the transverse condition $\nabla\cdot\mbox{\boldmath$A$}=0$, and for
clarity the vector potential with this condition is denoted by
$\mbox{\boldmath$A$}_{\perp}$. The Hamiltonian of this system is then given by

$\displaystyle H_{C}=$
$\displaystyle\frac{1}{2m}(\mbox{\boldmath$p$}-e\mbox{\boldmath$A$}_{\perp}(\mbox{\boldmath$q$}))^{2}+V(\mbox{\boldmath$q$})$
$\displaystyle+\int\\!\\!d^{3}x\frac{\mbox{\boldmath$E$}_{\perp}^{2}+\mbox{\boldmath$B$}^{2}}{2}-g\int\\!\\!d^{3}x\mbox{\boldmath$J$}(\mbox{\boldmath$x$})\cdot\mbox{\boldmath$A$}_{\perp}(\mbox{\boldmath$x$}),$
(4)

where $\mbox{\boldmath$E$}_{\perp}$ is the transverse component of the
electric field $E$, and the longitudinal component
$\mbox{\boldmath$E$}_{\parallel}$ and the time component $A_{0}$ are given by

$\displaystyle\mbox{\boldmath$E$}_{\parallel}=-\nabla A_{0},\ \
A_{0}(\mbox{\boldmath$x$},t)=\frac{1}{4\pi}\frac{e}{|\mbox{\boldmath$x$}-\mbox{\boldmath$q$}|}.$
(5)

We have the following equal-time commutation relations:

$\displaystyle[q^{i},p^{j}]=i\delta^{ij},$ (6)
$\displaystyle[A_{i}^{\perp}(\mbox{\boldmath$x$},t),E_{j}^{\perp}(\mbox{\boldmath$x$}^{\prime},t)]=-i\delta_{ij}^{tr}(\mbox{\boldmath$x$}-\mbox{\boldmath$x$}^{\prime}),$
(7)

for $i,j=1,2,3$, and all the other commutation relations vanish. Here
$\delta_{ij}^{tr}(\mbox{\boldmath$x$}-\mbox{\boldmath$x$}^{\prime})$ is the
transverse delta function defined by

$\displaystyle\delta_{ij}^{tr}(\mbox{\boldmath$x$}-\mbox{\boldmath$x$}^{\prime})=\int\\!\\!\frac{d^{3}k}{(2\pi)^{3}}e^{i\mbox{\boldmath${\scriptstyle
k}$}\cdot(\mbox{\boldmath${\scriptstyle x}$}-\mbox{\boldmath${\scriptstyle
x}$}^{\prime})}\left(\delta_{ij}-\frac{k_{i}k_{j}}{\mbox{\boldmath$k$}^{2}}\right).$
(8)

The fields $\mbox{\boldmath$A$}_{\perp}$ and $\mbox{\boldmath$E$}_{\perp}$ are
expanded in terms of the photon creation and annihilation operators as

$\displaystyle\mbox{\boldmath$A$}_{\perp}(\mbox{\boldmath$x$})=$
$\displaystyle\int\\!\\!\frac{d^{3}k}{\sqrt{(2\pi)^{3}2\omega}}\sum_{\lambda=1}^{2}\mbox{\boldmath$e$}(k,\lambda)$
$\displaystyle\times\left(a(k,\lambda)e^{i\mbox{\boldmath${\scriptstyle
k}$}\cdot\mbox{\boldmath${\scriptstyle
x}$}}+a^{\dagger}(k,\lambda)e^{-i\mbox{\boldmath${\scriptstyle
k}$}\cdot\mbox{\boldmath${\scriptstyle x}$}}\right),$ (9)
$\displaystyle\mbox{\boldmath$E$}_{\perp}(\mbox{\boldmath$x$})=$
$\displaystyle
i\int\\!\\!\frac{d^{3}k}{\sqrt{(2\pi)^{3}}}\sqrt{\frac{\omega}{2}}\sum_{\lambda=1}^{2}\mbox{\boldmath$e$}(k,\lambda)$
$\displaystyle\times\left(a(k,\lambda)e^{i\mbox{\boldmath${\scriptstyle
k}$}\cdot\mbox{\boldmath${\scriptstyle
x}$}}-a^{\dagger}(k,\lambda)e^{-i\mbox{\boldmath${\scriptstyle
k}$}\cdot\mbox{\boldmath${\scriptstyle x}$}}\right),$ (10)

where $\omega=|\mbox{\boldmath$k$}|$ and $\mbox{\boldmath$e$}(k,\lambda)$ is
the transverse polarization vector satisfying
$\mbox{\boldmath$e$}(k,\lambda)\cdot\mbox{\boldmath$k$}=0$ for $\lambda=1,2$.

###  II.1 The effective Hamiltonian for the charged particle: The second order
in $g$ and $e$

The system described by the Hamiltonian of Eq. (4) consists of the charged
particle and the electromagnetic fields. To obtain an effective Hamiltonian
for the charged particle, we need to eliminate the degrees of freedom of the
electromagnetic fields. In this subsection we mainly follow the derivation by
Saldanha saldanha2021 . See also santos1999 ; marletto2020 ; kang2022 .

We write the Hamiltonian $H_{C}$ as

$\displaystyle H_{C}=$
$\displaystyle\frac{\mbox{\boldmath$p$}^{2}}{2m}+V(\mbox{\boldmath$q$})+H_{C}^{\prime},$
(11) $\displaystyle H_{C}^{\prime}=$ $\displaystyle
H_{EM}+H_{g}+H_{e}+O(e^{2}),$ (12)

where

$\displaystyle
H_{EM}=\int\\!\\!d^{3}x\frac{\mbox{\boldmath$E$}_{\perp}^{2}+\mbox{\boldmath$B$}^{2}}{2}=\sum_{\lambda=1}^{2}\int\\!\\!d^{3}k\omega
a^{\dagger}(k,\lambda)a(k,\lambda),$ $\displaystyle
H_{g}=-g\int\\!\\!d^{3}x\mbox{\boldmath$J$}(\mbox{\boldmath$x$})\cdot\mbox{\boldmath$A$}_{\perp}(\mbox{\boldmath$x$}),\
\
H_{e}=-e\frac{\mbox{\boldmath$p$}}{m}\cdot\mbox{\boldmath$A$}_{\perp}(\mbox{\boldmath$q$}).$

In this subsection we assume that the parameters $e$ and $g$ are small, and we
calculate the energy of $H_{C}^{\prime}$ up to the second order in $e$ and
$g$. The resultant energy depends on the dynamical variables of the charged
particle, $q$ and $p$, and it contributes part of the effective Hamiltonian of
the charged particle.

The unperturbed state is the photon vacuum $\ket{0}$ that is the ground state
of $H_{EM}$. It is evident that the first order energy correction vanishes, as
the expectation value of $A$ in the vacuum $\ket{0}$ is 0. The second order
energy in parameters $e$ and $g$ is given by

$\displaystyle\bra{0}(H_{e}+H_{g})\frac{Q}{E_{0}-H_{EM}}(H_{e}+H_{g})\ket{0},$
(13)

where $Q=\mbox{\boldmath$1$}-\ket{0}\bra{0}$ and $E_{0}$ is the unperturbed
energy of the vacuum $\ket{0}$. Note that the terms of $O(e^{2})$ and
$O(g^{2})$ are constants; they are independent of the dynamical variables of
the charged particle. The relevant energy is, therefore, of $O(ge)$ and given
by

$\displaystyle\Delta\epsilon=$
$\displaystyle\bra{0}H_{g}\frac{Q}{E_{0}-H_{EM}}H_{e}\ket{0}+{\rm c.c.}$ (14a)
$\displaystyle=$
$\displaystyle\int\\!\\!d^{3}k\sum_{\lambda=1}^{2}\braket{0}{H_{g}}{k,\lambda}\frac{-1}{\omega}\braket{k,\lambda}{H_{e}}{0}+{\rm
c.c.},$ (14b)

with c.c. representing the complex conjugate of the terms preceding it. The
intermediate state $\ket{k,\lambda}$ is the one-photon state of momentum $k$
and polarization $\lambda=1,2$. Using the Fourier expansion of
$\mbox{\boldmath$A$}_{\perp}$ of Eq. (9), we have

$\displaystyle H_{g}=$
$\displaystyle-g\sum_{\lambda=1}^{2}\int\\!\\!d^{3}k\frac{1}{\sqrt{2\omega}}\mbox{\boldmath$e$}(k,\lambda)$
$\displaystyle\cdot\left(a(k,\lambda)\mbox{\boldmath$J$}_{\mbox{\boldmath${\scriptstyle
k}$}}^{*}+a^{\dagger}(k,\lambda)\mbox{\boldmath$J$}_{\mbox{\boldmath${\scriptstyle
k}$}})\right),$ (15)

where $\mbox{\boldmath$J$}_{\mbox{\boldmath${\scriptstyle k}$}}$ is the
Fourier transform of the external current source
$\mbox{\boldmath$J$}(\mbox{\boldmath$x$})$:

$\displaystyle\mbox{\boldmath$J$}_{\mbox{\boldmath${\scriptstyle
k}$}}=\frac{1}{\sqrt{(2\pi)^{3}}}\int\\!\\!d^{3}x\mbox{\boldmath$J$}(\mbox{\boldmath$x$})e^{-i\mbox{\boldmath${\scriptstyle
k}$}\cdot\mbox{\boldmath${\scriptstyle x}$}}.$ (16)

We can now calculate $\Delta\epsilon$ of Eq. (14). We find

$\displaystyle\Delta\epsilon=-eg\frac{\mbox{\boldmath$p$}}{m}\cdot\mbox{\boldmath$A$}^{\perp}_{ext}(\mbox{\boldmath$q$}),$
(17)

where $\mbox{\boldmath$A$}^{\perp}_{ext}(\mbox{\boldmath$x$})$ is defined to
be

$\displaystyle\mbox{\boldmath$A$}_{ext}^{\perp}(\mbox{\boldmath$x$})=$
$\displaystyle\frac{1}{\sqrt{(2\pi)^{3}}}\int\\!\\!d^{3}k\frac{\mbox{\boldmath$J$}_{\mbox{\boldmath${\scriptstyle
k}$}}}{\mbox{\boldmath$k$}^{2}}e^{i\mbox{\boldmath${\scriptstyle
k}$}\cdot\mbox{\boldmath${\scriptstyle x}$}}$ $\displaystyle=$
$\displaystyle\frac{1}{4\pi}\int\\!\\!d^{3}x^{\prime}\frac{\mbox{\boldmath$J$}(\mbox{\boldmath$x$}^{\prime})}{|\mbox{\boldmath$x$}-\mbox{\boldmath$x$}^{\prime}|},$
(18)

which is just the vector potential generated by the current source
$\mbox{\boldmath$J$}(\mbox{\boldmath$x$})$ according to the Biot-Savart law.
This $\mbox{\boldmath$A$}_{ext}^{\perp}(\mbox{\boldmath$x$})$ satisfies the
transverse condition
$\nabla\cdot\mbox{\boldmath$A$}_{ext}^{\perp}(\mbox{\boldmath$x$})=0$, which
follows from the current conservation law
$\nabla\cdot\mbox{\boldmath$J$}(\mbox{\boldmath$x$})=0$.

Thus the effective Hamiltonian for the charged particle is given by

$\displaystyle
h_{C}=\frac{\mbox{\boldmath$p$}^{2}}{2m}+V(\mbox{\boldmath$q$})-eg\frac{\mbox{\boldmath$p$}}{m}\cdot\mbox{\boldmath$A$}^{\perp}_{ext}(\mbox{\boldmath$q$})+O(e^{2}).$
(19)

Note that the $h_{C}$ takes the gauge covariant form

$\displaystyle
h_{C}=\frac{(\mbox{\boldmath$p$}-eg\mbox{\boldmath$A$}^{\perp}_{ext}(\mbox{\boldmath$q$}))^{2}}{2m}+V(\mbox{\boldmath$q$}),$
(20)

if the appropriate higher order term of $O(g^{2}e^{2})$ is added. This can be
interpreted as representing the Hamiltonian of a charged particle moving under
the influence of the external potential
$\mbox{\boldmath$A$}^{\perp}_{ext}(\mbox{\boldmath$q$})$.

In this approach, the coupling term in the effective Hamiltonian given by Eq.
(17) is the second-order energy correction. Suppose that the charged particle
moves along a path $L$: from $t=t_{1}$ to $t=t_{2}$. The AB phase shift
acquired by the particle is then given by a time integral of $\Delta\epsilon$:

$\displaystyle\Phi_{AB}=-eg\int_{t_{1}}^{t_{2}}\\!\\!dt\frac{\mbox{\boldmath$p$}}{m}\cdot\mbox{\boldmath$A$}_{ext}^{\perp}(\mbox{\boldmath$q$})=-eg\int_{L}\\!\\!d\mbox{\boldmath$q$}\cdot\mbox{\boldmath$A$}_{ext}^{\perp}(\mbox{\boldmath$q$}),$

while energy is generally considered to be a gauge-invariant observable. It is
this observation that underlies the claim that the AB phase shift is a gauge
invariant and measurable observable even when the path $L$ is not closed
marletto2020 ; saldanha2021 ; kang2022 .

In the subsequent sections, we will refute this assertion by showing that the
energy correction $\Delta\epsilon$ is generally gauge dependent. In the Lorenz
gauge, the form of $\Delta\epsilon$ given in Eq. (14a) remains unaltered,
although the intermediate states should include scalar and longitudinal photon
states alongside the transverse ones. However, it is not difficult to see that
the energy correction $\Delta\epsilon$ coincides with that in the Coulomb
gauge . In Sec. III, we will choose the axial gauge to demonstrate the energy
correction $\Delta\epsilon$ is a gauge dependent quantity. In the next
subsection, however, we present the new scheme using the coherent state where
the parameter $g$ is not assumed to be small. We will see that this scheme
gives the same result as the one presented here but provides more insight into
how the energy correction depends on the gauge condition.

###  II.2 Coherent state scheme: The first order in $e$

In this subsection, we present the method that uses photon coherent states
without assuming that the parameter $g$ is small. The results we will obtain
are the same as those in the preceding subsection. However, the calculations
are simplified in this method. It also provides better insight into the gauge
dependence of the energy correction in question. In this method we treat
$H_{C}^{\prime}$ of Eq. (12) in the fist-order perturbation in $e$. The
unperturbed Hamiltonian is then $H_{EM}+H_{g}$, which is expressed as

$\displaystyle H_{EM}+H_{g}$ $\displaystyle=$
$\displaystyle\sum_{\lambda=1}^{2}\int\\!\\!d^{3}k\omega
a^{\dagger}(k,\lambda)a(k,\lambda)$
$\displaystyle-g\sum_{\lambda=1}^{2}\int\\!\\!\frac{d^{3}k}{\sqrt{2\omega}}\mbox{\boldmath$e$}(k,\lambda)\cdot\left(a(k,\lambda)\mbox{\boldmath$J$}_{\mbox{\boldmath${\scriptstyle
k}$}}^{*}+a^{\dagger}(k,\lambda)\mbox{\boldmath$J$}_{\mbox{\boldmath${\scriptstyle
k}$}}\right).$

This Hamiltonian can be “diagonalized” in terms of the photon annihilation and
creation operators, $\tilde{a}(k,\lambda)$ and
$\tilde{a}^{\dagger}(k,\lambda)$, which are defined through

$\displaystyle a(k,\lambda)=\tilde{a}(k,\lambda)+\alpha(k,\lambda),$ (21)

with
$\alpha(k,\lambda)=g/\sqrt{2\omega^{3}}\mbox{\boldmath$e$}(k,\lambda)\cdot\mbox{\boldmath$J$}_{\mbox{\boldmath${\scriptstyle
k}$}}$. Up to a constant the result is given by

$\displaystyle H_{EM}+H_{g}=\sum_{\lambda=1}^{2}\int
d^{3}k\,\omega\,\tilde{a}^{\dagger}(k,\lambda)\tilde{a}(k,\lambda).$ (22)

The ground state of $H_{EM}+H_{g}$, denoted by $\ket{\tilde{0}}$, is then the
state that is annihilated by $\tilde{a}(k,\lambda)$; that is
$\tilde{a}(k,\lambda)\ket{\tilde{0}}=0$. The state $\ket{\tilde{0}}$ is
therefore an eigenstate of the annihilation operator $a(k,\lambda)$, the
coherent state of photons gerry2005 .

$\displaystyle a(k,\lambda)\ket{\tilde{0}}=\alpha(k,\lambda)\ket{\tilde{0}}.$
(23)

Note that the expectation value of $\mbox{\boldmath$A$}_{\perp}$ with respect
to the coherent state $\ket{\tilde{0}}$ is not 0, but given by $g$ times
$\mbox{\boldmath$A$}_{ext}^{\perp}(\mbox{\boldmath$x$})$ of Eq. (18).

$\displaystyle\braket{\tilde{0}}{\mbox{\boldmath$A$}_{\perp}(\mbox{\boldmath$x$})}{\tilde{0}}=$
$\displaystyle\int\\!\\!\frac{d^{3}k}{\sqrt{(2\pi)^{3}2\omega}}\sum_{\lambda=1}^{2}\mbox{\boldmath$e$}(k,\lambda)$
$\displaystyle\times\left(\alpha(k,\lambda)e^{i\mbox{\boldmath${\scriptstyle
k}$}\cdot\mbox{\boldmath${\scriptstyle
x}$}}+\alpha^{\dagger}(k,\lambda)e^{-i\mbox{\boldmath${\scriptstyle
k}$}\cdot\mbox{\boldmath${\scriptstyle x}$}}\right)$ $\displaystyle=$
$\displaystyle
g\frac{1}{4\pi}\int\\!\\!d^{3}x^{\prime}\frac{\mbox{\boldmath$J$}(\mbox{\boldmath$x$}^{\prime})}{|\mbox{\boldmath$x$}-\mbox{\boldmath$x$}^{\prime}|}=g\mbox{\boldmath$A$}_{ext}^{\perp}(\mbox{\boldmath$x$}).$
(24)

The unperturbed energy $\tilde{E}_{0}$, which is a constant, can be taken to
be 0. The first order energy correction in $e$ is simply given by the
expectation value of $H_{e}$ in the unperturbed state $\ket{\tilde{0}}$:

$\displaystyle\Delta E=$
$\displaystyle\braket{\tilde{0}}{H_{e}}{\tilde{0}}=-e\frac{\mbox{\boldmath$p$}}{m}\cdot\braket{\tilde{0}}{\mbox{\boldmath$A$}_{\perp}(\mbox{\boldmath$q$})}{\tilde{0}}$
$\displaystyle=$ $\displaystyle-
eg\frac{\mbox{\boldmath$p$}}{m}\mbox{\boldmath$A$}_{ext}^{\perp}(\mbox{\boldmath$q$}).$
(25)

This is exactly equal to the energy correction $\Delta\epsilon$ of Eq. (17)
that is the result of the second order perturbation in $e$ and $g$. Thus we
have obtained the same effective particle Hamiltonian $h_{C}$ as that given in
Eq. (19).

A few remarks are now in order. First, the results obtained in the preceding
subsection II.1 hold true regardless of the magnitude of $g$. This is not a
coincidence, but is due to the following reasons: Equation (24) shows that the
expectation value
$\braket{\tilde{0}}{\mbox{\boldmath$A$}_{\perp}(\mbox{\boldmath$x$})}{\tilde{0}}$
is linear in $g$ though the coherent state $\ket{\tilde{0}}$ itself contains
higher-order terms in $g$. Suppose that we calculate the expectation value of
$\mbox{\boldmath$A$}_{\perp}(\mbox{\boldmath$x$})$ in the state $\ket{0_{1}}$
that is the approximate ground state of $H_{EM}+H_{g}$ in the first-order
perturbation in $g$.

$\displaystyle\ket{0_{1}}=\ket{0}+\frac{Q}{E_{0}-H_{EM}}H_{g}\ket{0}.$

We then have

$\displaystyle\braket{\tilde{0}}{\mbox{\boldmath$A$}_{\perp}(\mbox{\boldmath$x$})}{\tilde{0}}=$
$\displaystyle\braket{0_{1}}{\mbox{\boldmath$A$}_{\perp}(\mbox{\boldmath$x$})}{0_{1}}+\delta$
$\displaystyle=$
$\displaystyle\bra{0}H_{g}\frac{Q}{E_{0}-H_{EM}}\mbox{\boldmath$A$}_{\perp}(\mbox{\boldmath$x$})\ket{0}+{\rm
c.c.}+\delta,$

where the error $\delta$ is $O(g^{2})$. Notice that the left-hand side is
order $O(g)$. This implies that $\delta$ should be exactly 0 as the terms
other than $\delta$ on the right-hand side are order $O(g)$ . Thus we find

$\displaystyle\braket{\tilde{0}}{H_{e}}{\tilde{0}}=\bra{0}H_{g}\frac{Q}{E_{0}-H_{EM}}H_{e}\ket{0}+{\rm
c.c.},$ (26)

which is just the second-order energy correction of Eq. (14) obtained in the
preceding subsection.

Second, as can be seen in Eq. (25), the coupling term in the effective
Hamiltonian, which is identified with the energy correction $\Delta
E=\Delta\epsilon$, is expressed in terms of the expectation value of the
vector potential in the coherent state. This expectation value should reflect
the gauge condition imposed on the vector potential. This strongly suggests
that the coupling term depends on the gauge that we choose. We will examine
specifically the case of the axial gauge in Sec. III.

Third, this concerns the change in magnetic field energy due to the motion of
the charged particle. Using classical electromagnetism, Boyer calculated this
quantity and obtained the following result boyer1971 :

$\displaystyle\Delta{\cal E}_{\rm
Boyer}=eg\frac{\mbox{\boldmath$p$}}{m}\mbox{\boldmath$A$}_{ext}^{\perp}(\mbox{\boldmath$q$}),$
(27)

which has the same form as the coupling term between the charged particle and
the external potential
$\mbox{\boldmath$A$}_{ext}^{\perp}(\mbox{\boldmath$q$})$. However, no
attention seems to have been paid to the fact that the signs are different.
Where does the Boyer’s energy (27) appear in our treatment, in which the
energy correction of $O(e)$ is given by $\braket{\tilde{0}}{H_{e}}{\tilde{0}}$
only? The answer is that it is hidden as part of the expectation value of the
unperturbed Hamiltonian. Let $\ket{\tilde{\phi}_{1}}$ be the first-order
perturbed eigenstate of $H_{EM}+H_{g}+H_{e}$.

$\displaystyle\ket{\tilde{\phi}_{1}}=\ket{\tilde{0}}+\frac{\tilde{Q}}{\tilde{E}_{0}-(H_{EM}+H_{g})}H_{e}\ket{\tilde{0}},$
(28)

with $\tilde{Q}=\mbox{\boldmath$1$}-\ket{\tilde{0}}\bra{\tilde{0}}$. Now
consider the expectation value of the unperturbed Hamiltonian in the state
$\ket{\tilde{\phi}_{1}}$. The result should be given by

$\displaystyle\braket{\tilde{\phi}_{1}}{H_{EM}+H_{g}}{\tilde{\phi}_{1}}=\tilde{E}_{0}+O(e^{2}),$
(29)

with no terms of order $O(e)$. This, however, does not necessarily imply that
neither $\braket{\tilde{\phi}_{1}}{H_{EM}}{\tilde{\phi}_{1}}$ nor
$\braket{\tilde{\phi}_{1}}{H_{g}}{\tilde{\phi}_{1}}$ contains a contribution
of order $O(e)$. It turns out that
$\braket{\tilde{\phi}_{1}}{H_{EM}}{\tilde{\phi}_{1}}$ is given by $\Delta{\cal
E}_{\rm Boyer}$ up to a constant, but is canceled by the contribution from
$\braket{\tilde{\phi}_{1}}{H_{g}}{\tilde{\phi}_{1}}$. See the Appendix A for
details.

##  III The AB-phase of a nonclosed path is not observable

In this section we show that the coupling term in the effective Hamiltonian
depends on the gauge. To do so we treat the same system as that discussed in
the preceding section by imposing the axial gauge condition, $A_{3}=0$. Note
that this condition completely fixes the vector potential as we assume all the
fields should approach zero at infinity. For canonical quantization of
electromagnetic fields in the axial gauge, we refer the reader to Ref.
haller1994 . The Hamiltonian is given by

$\displaystyle H_{X}=$
$\displaystyle\frac{1}{2m}(\mbox{\boldmath$p$}-e\mbox{\boldmath$A$}(\mbox{\boldmath$q$}))^{2}+V(\mbox{\boldmath$q$})+\int\\!\\!d^{3}x\frac{\mbox{\boldmath$E$}^{2}+\mbox{\boldmath$B$}^{2}}{2}$
$\displaystyle-g\int\\!\\!d^{3}x\mbox{\boldmath$J$}(\mbox{\boldmath$x$})\cdot\mbox{\boldmath$A$}(\mbox{\boldmath$x$}).$
(30)

Here, for $\mbox{\boldmath$q$},\mbox{\boldmath$p$}$, and the $x$\- and $y$\-
components of $A$ and $E$, the following typical canonical commutation
relations are assumed:

$\displaystyle[q^{k},q^{l}]=[p^{k},p,^{l}]=0,\ [q^{k},p^{l}]=i\delta_{kl},$
$\displaystyle[A_{i}(\mbox{\boldmath$x$}),A_{j}(\mbox{\boldmath$x$}^{\prime})]=[E_{i}(\mbox{\boldmath$x$}),E_{j}(\mbox{\boldmath$x$}^{\prime})]=0,$
$\displaystyle[A_{i}(\mbox{\boldmath$x$}),E_{j}(\mbox{\boldmath$x$}^{\prime})]=-i\delta_{ij}\delta(\mbox{\boldmath$x$}-\mbox{\boldmath$x$}^{\prime}),$
$\displaystyle[q^{k},A_{i}(\mbox{\boldmath$x$})]=[q^{k},E_{i}(\mbox{\boldmath$x$})]=[p^{k},A_{i}(\mbox{\boldmath$x$})]=[p^{k},E_{i}(\mbox{\boldmath$x$})]=0,$

for $k,l=1,2,3$ and $i,j=1,2$. The other commutation relations are determined
by the constraint conditions in the axial gauge. First, we have the gauge
fixing condition: $A_{3}(\mbox{\boldmath$x$})=0$. Second, the $z$ component of
$E$ is constrained to be

$\displaystyle
E_{3}(\mbox{\boldmath$x$})=-\frac{1}{\partial_{3}}\left(\sum_{i=1}^{2}\partial_{i}E_{i}(\mbox{\boldmath$x$})+j_{0}(\mbox{\boldmath$x$})\right),$
(31)

so that Gauss’s law $\nabla\cdot\mbox{\boldmath$E$}=j_{0}$ is fulfilled.
Third, the component $A_{0}$ is given by
$A_{0}(\mbox{\boldmath$x$})=\partial_{3}^{-1}E_{3}(\mbox{\boldmath$x$})$ ,
which respects the relations $E_{3}=\partial_{3}A_{0}-\partial_{0}A_{3}$ with
$A_{3}=0$. Under these commutation relations and constraints , one can verify
that the Hamiltonian $H_{X}$ correctly reproduces the equations of motion of
Eq. (1).

The vector potential $\mbox{\boldmath$A$}(\mbox{\boldmath$x$})$ and the
electric field $\mbox{\boldmath$E$}(\mbox{\boldmath$x$})$ in the axial gauge
are related to the transverse vector potential
$\mbox{\boldmath$A$}^{\perp}(\mbox{\boldmath$x$})$ and the transverse electric
field $\mbox{\boldmath$E$}^{\perp}(\mbox{\boldmath$x$})$, respectively:

$\displaystyle A_{i}(\mbox{\boldmath$x$})=$ $\displaystyle
A_{i}^{\perp}(\mbox{\boldmath$x$})-\frac{\partial_{i}}{\partial_{3}}A_{3}^{\perp}(\mbox{\boldmath$x$})\
\ \ i=1,2,3,$ (32) $\displaystyle E_{i}(\mbox{\boldmath$x$})=$ $\displaystyle
E_{i}^{\perp}(\mbox{\boldmath$x$})-\delta_{i3}\frac{1}{\partial_{3}}j_{0}(\mbox{\boldmath$x$})\
\ \ i=1,2,3,$ (33)

with the Fourier expansions of
$\mbox{\boldmath$A$}^{\perp}(\mbox{\boldmath$x$})$ and
$\mbox{\boldmath$E$}^{\perp}(\mbox{\boldmath$x$})$ given in Eqs. (9) and (10),
respectively. Using these Fourier expansions, one can verify that the three
aforementioned constraints in the axial gauge are satisfied and all the
related commutation relations follow. The Hamiltonian of Eq. (30) can be
written as

$\displaystyle H_{X}=$
$\displaystyle\frac{1}{2m}(\mbox{\boldmath$p$}-e\mbox{\boldmath$A$}(\mbox{\boldmath$q$}))^{2}+V(\mbox{\boldmath$q$})$
$\displaystyle+\int\\!\\!d^{3}k\,\omega\sum_{\lambda=1}^{2}a^{\dagger}(k,\lambda)a(k,\lambda)$
$\displaystyle-e\frac{1}{(\partial_{3})^{2}}\left(\sum_{i=1}^{2}\partial_{i}E_{i}(\mbox{\boldmath$q$})\right)-g\int\\!\\!d^{3}x\mbox{\boldmath$J$}(\mbox{\boldmath$x$})\cdot\mbox{\boldmath$A$}(\mbox{\boldmath$x$}).$

We are now ready to calculate the effective Hamiltonian for the charged
particle in the axial gauge. As in the case of the Coulomb gauge, we write

$\displaystyle H_{X}=$
$\displaystyle\frac{\mbox{\boldmath$p$}^{2}}{2m}+V(\mbox{\boldmath$q$})+H_{X}^{\prime},$
(34) $\displaystyle H_{X}^{\prime}=$ $\displaystyle
H_{EM}+H_{g}+H_{e}^{(1)}+H_{e}^{(2)}+O(e^{2}),$ (35)

where

$\displaystyle H_{EM}=\sum_{\lambda=1}^{2}\int\\!\\!d^{3}k\omega
a^{\dagger}(k,\lambda)a(k,\lambda),$ $\displaystyle
H_{g}=-g\int\\!\\!d^{3}x\mbox{\boldmath$J$}(\mbox{\boldmath$x$})\cdot\mbox{\boldmath$A$}(\mbox{\boldmath$x$}),$
$\displaystyle
H_{e}^{(1)}=-e\frac{\mbox{\boldmath$p$}}{m}\cdot\mbox{\boldmath$A$}(\mbox{\boldmath$q$}),$
$\displaystyle
H_{e}^{(2)}=-e\frac{1}{(\partial_{3})^{2}}\left(\sum_{i=1}^{2}\partial_{i}E_{i}(\mbox{\boldmath$q$})\right).$

We evaluate $H_{X}^{\prime}$ in terms of the first-order perturbation theory
of $e$, as we did for the Coulomb gauge in Subsec. II.2. The Hamiltonian
$H_{g}$ is expanded as

$\displaystyle H_{g}=$
$\displaystyle-g\sum_{\lambda=1}^{2}\int\\!\\!d^{3}k\frac{1}{\sqrt{2\omega}}\mbox{\boldmath$e$}^{X}(k,\lambda)$
$\displaystyle\cdot\left(a(k,\lambda)\mbox{\boldmath$J$}_{\mbox{\boldmath${\scriptstyle
k}$}}^{*}+a^{\dagger}(k,\lambda)\mbox{\boldmath$J$}_{\mbox{\boldmath${\scriptstyle
k}$}})\right),$ (36)

where

$\displaystyle
e^{X}_{i}(k,\lambda)=e_{i}(k,\lambda)-\frac{k_{i}}{k_{3}}e_{3}(k,\lambda)\ \
i=1,2,3.$ (37)

is the polarization vector of the vector potential
$\mbox{\boldmath$A$}(\mbox{\boldmath$x$})$ in the axial gauge. In the above
expression of $H_{g}$, however, this polarization vector
$\mbox{\boldmath$e$}^{X}(k,\lambda)$ can be replaced by the transverse
polarization vector $\mbox{\boldmath$e$}(k,\lambda)$, as the current
conservation of the external current,
$\mbox{\boldmath$k$}\cdot\mbox{\boldmath$J$}_{\mbox{\boldmath${\scriptstyle
k}$}}=0$, implies
$\mbox{\boldmath$e$}^{X}(k,\lambda)\cdot\mbox{\boldmath$J$}_{\mbox{\boldmath${\scriptstyle
k}$}}=\mbox{\boldmath$e$}(k,\lambda)\cdot\mbox{\boldmath$J$}_{\mbox{\boldmath${\scriptstyle
k}$}}$. Thus the unperturbed Hamiltonian $H_{EM}+H_{g}$ is the same as that ,
given by Eq. (22), in the Coulomb gauge. The unperturbed ground state is
therefore the coherent state $\ket{\tilde{0}}$ defined through
$a(k,\lambda)\ket{\tilde{0}}=\alpha(k,\lambda)\ket{\tilde{0}}$ with
$\alpha(k,\lambda)=g/\sqrt{2\omega^{3}}\mbox{\boldmath$e$}(k,\lambda)\cdot\mbox{\boldmath$J$}_{\mbox{\boldmath${\scriptstyle
k}$}}$.

The expectation value of $\mbox{\boldmath$A$}(\mbox{\boldmath$x$})$ in the
coherent state $\ket{\tilde{0}}$ can be calculated as

$\displaystyle\braket{\tilde{0}}{\mbox{\boldmath$A$}(\mbox{\boldmath$x$})}{\tilde{0}}=$
$\displaystyle\braket{\tilde{0}}{A^{\perp}(\mbox{\boldmath$x$})-\frac{\nabla}{\partial_{3}}A_{3}^{\perp}(\mbox{\boldmath$x$})}{\tilde{0}}$
$\displaystyle=$ $\displaystyle
g\left(\mbox{\boldmath$A$}^{\perp}_{ext}(\mbox{\boldmath$x$})-\frac{\nabla}{\partial_{3}}A_{\perp,ext}^{3}(\mbox{\boldmath$x$})\right)$
$\displaystyle\equiv$ $\displaystyle
g\mbox{\boldmath$A$}^{X}_{ext}(\mbox{\boldmath$x$}),$ (38)

where $\mbox{\boldmath$A$}^{\perp}_{ext}(\mbox{\boldmath$x$})$ is defined in
Eq. (18). Remember that
$\mbox{\boldmath$A$}^{\perp}_{ext}(\mbox{\boldmath$x$})$ is the transverse
vector potential generated by the external current. This implies that
$\mbox{\boldmath$A$}^{X}_{ext}(\mbox{\boldmath$x$})$ is also the vector
potential generated by the external current, but with the axial gauge
condition $\mbox{\boldmath$A$}^{X,3}_{ext}(\mbox{\boldmath$x$})=0$. As for the
electric field $E$, it can be readily checked that
$\braket{\tilde{0}}{E_{i}(\mbox{\boldmath$x$})}{\tilde{0}}=0$ for $i=1,2$.

The unperturbed energy is constant and taken to be 0, and the first-order
energy correction in $e$ is given by

$\displaystyle\Delta
E=\braket{\tilde{0}}{H_{e}^{(1)}+H_{e}^{(2)}}{\tilde{0}}=-eg\frac{\mbox{\boldmath$p$}}{m}\cdot\mbox{\boldmath$A$}^{X}_{ext}(\mbox{\boldmath$q$}),$
(39)

which leads to the particle effective Hamiltonian:

$\displaystyle
h_{X}=\frac{1}{2m}\mbox{\boldmath$p$}^{2}+V(\mbox{\boldmath$q$})-eg\frac{\mbox{\boldmath$p$}}{m}\cdot\mbox{\boldmath$A$}_{ext}^{X}(\mbox{\boldmath$q$})+O(e^{2}).$
(40)

In general the vector potentials
$\mbox{\boldmath$A$}_{ext}^{\perp}(\mbox{\boldmath$q$})$ and
$\mbox{\boldmath$A$}_{ext}^{X}(\mbox{\boldmath$q$})$ are different for a given
external current $\mbox{\boldmath$J$}(\mbox{\boldmath$x$})$, implying that the
AB phase acquired by the charged particle along a nonclosed path depends on
the gauge that we choose, and it is not an observable.

We obtained the effective Hamiltonian $h_{X}$ by using the first-order
perturbation theory of $e$, as we did in Subsec. II.2 for the case of Coulomb
gauge. The result should be unchanged if we perform the second-order
perturbation in $g$ and $e$ as the calculation described in Subsec. II.1.
Before concluding this subsection, we will confirm this equivalence.

In the perturbation in $g$ and $e$, the unperturbed state is $\ket{0}$ given
by the ground state of $H_{EM}$, and $H_{g}+H_{e}^{(1)}+H_{e}^{(2)}$ is the
perturbation. Since the first-order energy correction is 0, we consider the
energy correction of order $O(ge)$:

$\displaystyle\Delta\epsilon=\braket{0}{H_{g}\frac{1}{E_{0}-H_{EM}}(H_{e}^{(1)}+H_{e}^{(2)})}{0}+{\rm
c.c.}$ (41)

As in the case of the Coulomb gauge, this $\Delta\epsilon$ can be expressed as

$\displaystyle\Delta\epsilon=\braket{0_{1}}{H_{e}^{(1)}+H_{e}^{(2)}}{0_{1}}+O(g^{2}),$
(42)

with $\ket{0_{1}}$ being the approximate ground state of $H_{EM}+H_{g}$ in the
first-order perturbation in $g$. In this way, we can see that the calculation
of perturbation in $g$ and $e$ will eventually lead to the result of Eq. (39),
but let us now analyze $\Delta\epsilon$ in Eq. (41) explicitly. First, after
some involved calculation, we find

$\displaystyle\braket{0}{H_{g}\frac{1}{E_{0}-H_{EM}}H_{e}^{(1)}}{0}$
$\displaystyle=$
$\displaystyle-\frac{eg}{2m}\mbox{\boldmath$p$}\cdot\left(\mbox{\boldmath$A$}_{ext}^{\perp}(\mbox{\boldmath$q$})-\frac{\nabla}{\partial_{3}}A_{\perp,ext}^{3}(\mbox{\boldmath$q$})\right)$
$\displaystyle=$
$\displaystyle-\frac{eg}{2m}\mbox{\boldmath$p$}\cdot\mbox{\boldmath$A$}_{ext}^{X}(\mbox{\boldmath$q$}).$
(43)

Second, the following matrix element turns out to be purely imaginary:

$\displaystyle\braket{0}{H_{g}\frac{1}{E_{0}-H_{EM}}H_{e}^{(2)}}{0}.$ (44)

Combining these results, we conclude that $\Delta\epsilon$ is eventually given
by $\Delta E$ of Eq. (39), which is as expected from Eq. (42).

##  IV Discussion and conclusions

The purpose of this paper is to investigate whether the AB phase shift in the
case of a nonclosed path is independent of the gauge and therefore measurable.
In the recent approach santos1999 ; marletto2020 ; saldanha2021 ; kang2022 ,
the coupling term between the charged particle and the electromagnetic
potential is identified with the energy change $\Delta E$ resulting from the
quantum mechanical electromagnetic interaction between the charged particle
and the external current source. Since energy is generally thought to be
gauge-invariant, this leads to the claim that the AB phase shift is a gauge
invariant and measurable observable even for a nonclosed path marletto2020 ;
saldanha2021 ; kang2022 . We have disproved this claim by explicitly showing
that the energy correction $\Delta E$ is generally gauge dependent.

One may still wonder why the energy change $\Delta E$ can be gauge-dependent.
Let $H_{C}$ and $H_{X}$ be the Hamiltonians in the Coulomb gauge and the axial
gauge, respectively. As operators we know that $H_{C}\neq H_{X}$, though we
believe that their eigenvalues are the same. However, we do not try to
directly determine the eigenvalues of $H_{C}$ or $H_{X}$. What we do in this
approach is to eliminate the degrees of freedom of the electromagnetic field
and obtain an effective Hamiltonian $h$ for the charged particle:

$\displaystyle
h=\frac{1}{2m}\mbox{\boldmath$p$}^{2}+V(\mbox{\boldmath$q$})-eg\frac{\mbox{\boldmath$p$}}{m}\cdot\braket{\tilde{0}}{\mbox{\boldmath$A$}(\mbox{\boldmath$q$})}{\tilde{0}}+\cdots.$
(45)

Starting from $H_{C}$ yields $h_{C}$, and starting from $H_{X}$ yields
$h_{X}$. Generally, we have $h_{C}\neq h_{X}$, as
$\mbox{\boldmath$A$}_{ext}^{\perp}(\mbox{\boldmath$x$})\neq\mbox{\boldmath$A$}^{X}_{ext}(\mbox{\boldmath$x$})$.
To calculate the energy of the system, one must still determine the
eigenvalues of $h_{C}$ or $h_{X}$. They should be the same. In conclusion, it
may be misleading to say that this approach ”calculates the energy change of
the system.” It would be more appropriate to say that it ”derives the
effective Hamiltonian for the charged particle.” From this perspective, it
seems natural for the effective Hamiltonian to depend on the gauge choice.

We have found that the AB phase for a nonclosed path cannot be a measurable
physical quantity. On the other hand, some experiments have been proposed
marletto2020 ; saldanha2021 ; kang2022  that appear to allow the measurement
of the AB phase along a nonclosed path. This is a contradiction that
necessitates an examination of the proposed experiments. Here we take the
experiment proposed in saldanha2021  because it appears to be the simplest
conceptually. Essentially, it involves abruptly cutting off the current in a
solenoid before the charged particle completes a closed path. The assumption
here is that this would cause the charged particle to remember its phase
change at the time of the interruption, and the interference pattern at the
closed path would reflect the phase change at the intermediate point in the
path. The problem here, however, is that a time-varying magnetic field
produces an electric field. If $A_{\mu}(x)$ is time dependent, then the closed
curve drawn by the charged particle would be a closed curve in four-
dimensional space-time. In other words, when the solenoid current changes, the
charged particle also undergoes a phase change with a contribution of
$\int_{C}A_{0}(x)dt$, and it does not retain the phase difference at the time
when the solenoid current is suddenly cut. One way to demonstrate this would
be to assume that the current $J_{\mu}(x)$ in the solenoid is time dependent
and to show that the phase difference after drawing a closed curve in four-
dimensional space-time is independent of the gauge, while the phase
accumulated during the path is gauge dependent. For a more comprehensive
discussion on the four-dimensional loop integral of the time-dependent vector
potential, we refer the reader to Ref. horvat2020 .

Note added: Recently, Wakamatsu wakamatsu2023  showed that the AB phase
depends on the residual gauge freedom in the Coulomb gauge within a quantum
electrodynamics approach.

###### Acknowledgements.

The author would like to thank M. Wakamatsu for numerous useful discussions
and for critical comments on the manuscript. It is our pleasure to thank
Sebastian Horvat for bringing Ref. horvat2020  to our attention.

*

##  Appendix A The Boyer’s energy change

We first rewrite $H_{EM}$ as

$\displaystyle H_{EM}=$
$\displaystyle\int\\!\\!d^{3}x\frac{\mbox{\boldmath$E$}_{\perp}^{2}+\mbox{\boldmath$B$}^{2}}{2}$
$\displaystyle=$
$\displaystyle\int\\!\\!d^{3}x\frac{\mbox{\boldmath$E$}_{\perp}^{2}+\tilde{\mbox{\boldmath$B$}}^{2}}{2}+\int\\!\\!d^{3}x\tilde{\mbox{\boldmath$B$}}\cdot\mbox{\boldmath$B$}_{ext}+\text{const},$

where
$\mbox{\boldmath$B$}_{ext}=g\nabla\times\mbox{\boldmath$A$}_{ext}^{\perp}$ and
$\tilde{\mbox{\boldmath$B$}}=\mbox{\boldmath$B$}-\mbox{\boldmath$B$}_{ext}$.
Then we find that, up to a constant,

$\displaystyle\braket{\tilde{\phi}_{1}}{H_{EM}}{\tilde{\phi}_{1}}=\int\\!\\!d^{3}x\braket{\tilde{\phi}_{1}}{\tilde{\mbox{\boldmath$B$}}}{\tilde{\phi}_{1}}\cdot\mbox{\boldmath$B$}_{ext},$

with $\ket{\tilde{\phi}_{1}}$ defined in Eq. (28). After some calculation, we
find that

$\displaystyle\braket{\tilde{\phi}_{1}}{\tilde{\mbox{\boldmath$B$}}(\mbox{\boldmath$x$})}{\tilde{\phi}_{1}}=-\frac{\mbox{\boldmath$p$}}{m}\times\nabla\frac{1}{4\pi}\frac{e}{|\mbox{\boldmath$x$}-\mbox{\boldmath$q$}|}+O(e^{2}),$
(46)

which is the nonrelativistic expression of the magnetic field generated by the
charged particle moving at velocity $\mbox{\boldmath$p$}/m$.

Using this expression, we have

$\displaystyle\braket{\tilde{\phi}_{1}}{H_{EM}}{\tilde{\phi}_{1}}=-\frac{e}{4\pi
m}\int\\!\\!d^{3}x\left(\mbox{\boldmath$p$}\times\nabla\frac{1}{|\mbox{\boldmath$x$}-\mbox{\boldmath$q$}|}\right)\cdot\mbox{\boldmath$B$}_{ext}(\mbox{\boldmath$x$})$
$\displaystyle=\frac{e}{4\pi
m}\mbox{\boldmath$p$}\cdot\int\\!\\!d^{3}x\frac{1}{|\mbox{\boldmath$x$}-\mbox{\boldmath$q$}|}\nabla\times\mbox{\boldmath$B$}_{ext}(\mbox{\boldmath$x$})$
$\displaystyle=\frac{eg}{4\pi
m}\mbox{\boldmath$p$}\cdot\int\\!\\!d^{3}x\frac{\mbox{\boldmath$J$}(\mbox{\boldmath$x$})}{|\mbox{\boldmath$x$}-\mbox{\boldmath$q$}|}=eg\frac{\mbox{\boldmath$p$}}{m}\cdot\mbox{\boldmath$A$}_{ext}^{\perp}(\mbox{\boldmath$q$}),$

where $\mbox{\boldmath$A$}_{ext}^{\perp}$ is defined in Eq. (18). This is the
Boyer’s energy change $\Delta{\cal E}_{\rm Boyer}$ given in Eq. (27). We now
observe that $H_{g}$ can be rewritten as follows:

$\displaystyle H_{g}=$
$\displaystyle-g\int\\!\\!d^{3}x\mbox{\boldmath$J$}\cdot\mbox{\boldmath$A_{\perp}$}=-\int\\!\\!d^{3}x\nabla\times\mbox{\boldmath$B$}_{ext}\cdot\mbox{\boldmath$A$}^{\perp}$
$\displaystyle=$
$\displaystyle-\int\\!\\!d^{3}x\mbox{\boldmath$B$}_{ext}\cdot\nabla\times\mbox{\boldmath$A$}^{\perp}=-\int\\!\\!d^{3}x\mbox{\boldmath$B$}_{ext}\cdot\tilde{\mbox{\boldmath$B$}}+{\rm
const}.$

Therefore, up to a constant, we conclude

$\displaystyle\braket{\tilde{\phi}_{1}}{H_{g}}{\tilde{\phi}_{1}}=-eg\frac{\mbox{\boldmath$p$}}{m}\cdot\mbox{\boldmath$A$}_{ext}^{\perp}(\mbox{\boldmath$q$}),$
(47)

which cancels the Boyer’s energy
$\braket{\tilde{\phi}_{1}}{H_{EM}}{\tilde{\phi}_{1}}$.

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