# Is the Aharonov-Bohm phase shift for a non-closed path
a measurable quantity ?

Masashi Wakamatsu<EMAIL_ADDRESS>KEK Theory Center, Institute of
Particle and Nuclear Studies, High Energy Accelerator Research Organization
(KEK), 1-1, Oho, Tsukuba, Ibaraki 305-0801, Japan

###### Abstract

It is widely accepted that the Aharonov-Bohm (AB) effect is understood as
local action of the vector potential generated by the solenoid on the complex
phase of a charged particle’s wave functions. Nevertheless, because the vector
potential is a gauge-dependent quantity, there still exist not a few
researchers who claim that the AB-effect is caused by nonlocal action of the
magnetic field confined inside the solenoid. A serious drawback of such non-
locality explanation is that it is hard to offer definite mechanism of such
nonlocal interaction between separated charge and magnetic field. This is not
surprising, since the locality is the foundation of modern physics including
electromagnetism and quantum mechanics. On the other hand, there recently
appear some interesting attempts to explain the AB-effect through the
interaction between the charged particle and the solenoid current mediated by
the exchange of a virtual photon. A vital assumption of this approach is that
AB-phase shift is proportional to the change of the interaction energy between
the charged particle and solenoid along the path of the moving charge.
Accordingly, they insist that the AB-phase change along a path does not depend
on the gauge choice so that the AB-phase shift for a non-closed path is in
principle measurable. In the present paper, we critically examine this
astonishing conclusion with the utmost care and attention. At the same time,
we try to give refreshed insight into the role of the vector potential in the
interpretation of the AB-effect.

††preprint: KEK-TH-2527 and J-PARK-TH-0288

##  I Introduction

Since it was first predicted theoretically ES1949 ,AB1959 , the Aharonov-Bohm
effect (AB-effect) has raised a lot of debates which concern theoretical
foundation of modern physics. (For review, see Peshkin1981 -WKZZ2018 , for
example.) A central question is whether the magnetic vector potential is a
physical entity or it is just a convenient mathematical tool for calculating
force fields. Most popular interpretation of the AB-effect is that the vector
potential locally affects the phase of an electron wave function, thereby
causing a change of phase that can be measured through interference
experiments Tonomura1986 ,Osakabe1986 . Main objection to this understanding
is based on the fact that the vector potential is a gauge-dependent quantity
so that it has inherent arbitrariness. By this reason, not a few researchers
still hold on to nonlocal magnetic field interpretation Healey2022  -Heras2022
. The ground of this field interpretation is that the AB-phase shift for a
closed path can be expressed as a closed contour integration of the vector
potential, so that with use of the Stokes theorem, it is eventually expressed
with the magnetic flux penetrating the solenoid. Based on this fact, some
authors insist that the AB-phase shift is generated by the action of the
magnetic field on the electron, which are spatially separated from each other.
The great deficiency of such nonlocality interpretation is that it contradicts
the widely-accepted locality principle of modern physics. In fact, no one has
succeeded to give persuasive concrete mechanism which supports such
nonlocality interpretation between the spatially separated charge and the
magnetic field inside the solenoid. On the other hand, despite the
arbitrariness of the vector potential, there exists some evidence that
supports its physical reality as long as the theoretical interpretation of the
AB-phase shift corresponding to a closed path WKZZ2018 ,Adachi1992 -Li2012 .
It is based on the idea that the vector potential contains in itself a gauge-
invariant piece which can never be eliminated by any regular gauge
transformation. In fact, it has been argued by several authors that the AB-
phase shift is saturated by this gauge-invariant piece of the vector potential
and the remaining gauge arbitrariness gives no contribution to the AB-phase
shift in such standard settings WKZZ2018 ,Adachi1992 -Li2012 .

A new question arises, however, if one considers the AB phase shift for a non-
closed path. This is because the contribution of the gauge-variant component
of the vector potential would not generally vanish for the AB-phase shift for
a non-closed path. In other words, the AB-phase shift corresponding to a non-
closed path is thought to be a gauge-variant quantity. This indicates that it
would not correspond to an observable quantity as long as one believes the
celebrated gauge principle. It seems to us that this contradicts the recent
claim by several researchers that the AB phase shift for a non-closed path can
in principle be observed Vaidman2012  -LHK2022 . We recall that a vital
assumption in their argument is that the AB phase shift is proportional to the
change of interaction energy between a charged particle and the current of the
solenoid mediated along the path of a moving change. According to them, the AB
phase shift along such non-closed path can in principle be observed, since the
change of energy along such a path is a gauge-invariant quantity. This is a
striking conclusion, because, if it were true, it would mean that the AB-
effect is not necessarily a topological phenomena as is widely advertised
Kobe1991 ,CLBNGK2019 . In view of the impact of the conflict above, we think
it very important to elucidate what is the origin of this discrepancy. To get
a clear answer to this question is the main purpose of the present paper.

The paper is organized as follows. First, in sect.II, we argue that the vector
potential is not a mere mathematical tool but it necessarily contains a
physical entity which can never be eliminated by a regular gauge
transformation. It is shown by rigorisation of the idea of the transverse-
longitudinal decomposition of the vector potential based on the Helmholtz
theorem. Next, in sect.III, we briefly review some recent papers which claim
the observability of the AB-phase shift for a non-closed path. In sect IV, to
inspect the validity of such claims, we try to give a self-contained quantum
mechanical formulation of the combined system of a solenoid and a charged
particle coupled to the quantized electromagnetic fields.. The analysis will
be done based on two forms of physically equivalent Hamiltonians constructed
there. Next, this quantum mechanical analysis is compared with the analyses by
Boyer Boyer1971 , Saldanha Saldanha2021  and others Kang2017 , Marletto-
Vedral2020 , who claim that the AB-phase shit for a non-closed path can in
principle be observed. In sect.V, using the path integral formalism based on
one form of our Lagrangians, we try to show that the AB-phase for a non-closed
path is likely to be a gauge-dependent quantity at variance with the recent
claims mentioned above. Sect.VI summarizes what we can conclude from the
present study.

##  II Refreshing the vector potential interpretation of the Aharonov-Bohm
effect

Because of its gauge-variant nature, the vector potential is widely regarded
as just a convenient mathematical tool without any physical entity. However,
in spite of its arbitrariness, to think of it as totally unphysical quantity
would not be correct Feynman1963  -Semon-Taylor1996 . Here, to make the
discussion as concrete as possible, let us consider the vector potential
generated by an infinitely long solenoid corresponding to an idealized setting
of the AB-effect. Naturally, the arbitrariness of the vector potential remains
also in this case. Nevertheless, there exists long-lasting belief that the
vector potential contains in itself a sort of physical component which can
never be gauged away. To explain this idea, let us start with the familiar
transverse-longitudinal decomposition of the vector potential given as

$\bm{A}(\bm{x})\ =\ \bm{A}_{\perp}(\bm{x})\ +\ \bm{A}_{\parallel}(\bm{x}),$
(1)

where the transverse component $\bm{A}_{\perp}$ and the longitudinal component
$\bm{A}_{\parallel}$ are respectively demanded to satisfy the following
conditions :

$\nabla\cdot\bm{A}_{\perp}(\bm{x})\ =\ 0,\ \ \ \mbox{and}\ \ \
\nabla\times\bm{A}_{\parallel}(\bm{x})\ =\ 0.$ (2)

An interesting idea is that the gauge degrees of freedom (or the gauge
arbitrariness) is solely carried by the longitudinal component
$\bm{A}_{\parallel}$ alone, while the transverse component $\bm{A}_{\perp}$ is
intact under arbitrary gauge transformation. Some authors even claim that,
since the transverse component is independent of the choice of gauge, it is
qualified as a physical entity Li2012 . Furthermore, it was demonstrated by
several researchers that only the transverse part of the vector potential
contributes to the AB-phase for a closed path and the contribution of
longitudinal part vanishes exactly WKZZ2018 ,Adachi1992 \- Li2012 .

Main criticism to such assertions comes from the doubt whether the transverse-
longitudinal decomposition is truly unique or not. We recall that the existing
uniqueness proof of the transverse-longitudinal decomposition is mostly based
on the Helmholtz theorem Helmholtz1858 . (To avoid misunderstanding, we point
out the fact that the forming condition of the Helmholtz theorem is not always
satisfied. This happens, for example, in the famous Landau problem that treats
the motion of an electron in a magnetic field uniformly spreading over the
whole 2-dimensional plane. In such special circumstances, the forming
condition of the Helmholtz theorem is apparently broken, so that the
transverse-longitudinal decomposition cannot be uniquely made WKZ2018 ,WH2022
.) For the sake of completeness, here we recapitulate the body of the
Helmholtz theorem in the most rigorous form following Griffith’s textbook
Griffiths1999 . (From now on, to avoid unnecessary notational complexity, we
use natural units, i.e. the Heaviside-Lorenz units with $c=1$.)

###### Theorem 1

Suppose that a vector field $\bm{F}(\bm{x})$ is specified by its divergence
and curl as

$\displaystyle\nabla\cdot\bm{F}(\bm{x})$ $\displaystyle=$ $\displaystyle
D(\bm{x}),$ (3) $\displaystyle\nabla\times\bm{F}(\bm{x})$ $\displaystyle=$
$\displaystyle\bm{C}(\bm{x}).$ (4)

then the Helmholtz theorem claims that, if $D(\bm{x})$ and $\bm{C}(\bm{x})$ go
to zero sufficiently rapidly at infinity, the vector field $\bm{F}(\bm{x})$
can uniquely be expressed as

$\bm{F}(\bm{x})\ =\ -\,\nabla U(\bm{x})\ +\ \nabla\times\bm{W}(\bm{x}),$ (5)

where

$\displaystyle U(\bm{x})$ $\displaystyle=$
$\displaystyle\frac{1}{4\,\pi}\,\int\frac{D(\bm{x}^{\prime})}{|\bm{x}-\bm{x}^{\prime}|}\,d^{3}x^{\prime},$
(6) $\displaystyle\bm{W}(\bm{x})$ $\displaystyle=$
$\displaystyle\frac{1}{4\,\pi}\,\int\frac{\bm{C}(\bm{x}^{\prime})}{|\bm{x}-\bm{x}^{\prime}|}\,d^{3}x^{\prime}.$
(7)

The only subtlety of the Helmholtz theorem is the convergence question of the
above integrals, or equivalently the behavior of the vector field
$\bm{F}(\bm{x})$ at the spatial infinity. Once this convergence condition is
guaranteed, one can straightforwardly apply the Helmholtz theorem to the
transverse-longitudinal decomposition of the vector potential to obtain

$\displaystyle\bm{A}_{\perp}(\bm{x})$ $\displaystyle=$
$\displaystyle\nabla\times\frac{1}{4\,\pi}\,\int\,\frac{\nabla^{\prime}\times\bm{A}(\bm{x}^{\prime})}{|\bm{x}-\bm{x}^{\prime}|}\,d^{3}x^{\prime},$
(8) $\displaystyle\bm{A}_{\parallel}(\bm{x})$ $\displaystyle=$
$\displaystyle-\,\nabla\,\frac{1}{4\,\pi}\,\int\,\frac{\nabla^{\prime}\cdot\bm{A}(\bm{x}^{\prime})}{|\bm{x}-\bm{x}^{\prime}|}\,d^{3}x^{\prime},$
(9)

where we use the notation
$\nabla^{\prime}\equiv\partial\,/\,\partial\bm{x}^{\prime}$. Our next concern
is the transformation properties of $\bm{A}_{\perp}$ and $\bm{A}_{\parallel}$
under an arbitrary gauge transformation

$\bm{A}(\bm{x})\ \rightarrow\ \bm{A}^{\prime}(\bm{x})\ =\ \bm{A}(\bm{x})\ +\
\nabla\chi(\bm{x}).$ (10)

Since $\nabla^{\prime}\times\bm{A}(\bm{x}^{\prime})$ in (8) gives the magnetic
field $\bm{B}(\bm{x}^{\prime})$, the transverse component $\bm{A}_{\perp}$ is
apparently gauge invariant. The main question is therefore the gauge
transformation property of the longitudinal component. The delicacy of this
question can be understood, for example, if we remember the following proof by
Adachi et al. Adachi1992 . Under the gauge transformation, they first observe
that

$\displaystyle\bm{A}_{\parallel}(\bm{x})\ \rightarrow\
\bm{A}^{\prime}_{\parallel}(\bm{x})$ $\displaystyle=$
$\displaystyle-\,\nabla\,\frac{1}{4\,\pi}\,\int\,\frac{\nabla^{\prime}\cdot(\bm{A}(\bm{x}^{\prime})+\nabla^{\prime}\chi(\bm{x}^{\prime}))}{|\bm{x}-\bm{x}^{\prime}|}\,d^{3}x^{\prime}$
(11) $\displaystyle=$ $\displaystyle\bm{A}_{\parallel}(\bm{x})\ -\
\nabla\,\frac{1}{4\,\pi}\,\int\,\frac{\Delta^{\prime}\chi(\bm{x}^{\prime})}{|\bm{x}-\bm{x}^{\prime}|}\,d^{3}x^{\prime}.$

Carrying out the integration by parts and neglecting the surface term, they
obtain

$\displaystyle\frac{1}{4\,\pi}\,\int\,\frac{\Delta^{\prime}\chi(\bm{x}^{\prime})}{|\bm{x}-\bm{x}^{\prime}|}\,d^{3}x^{\prime}$
$\displaystyle=$
$\displaystyle\frac{1}{4\,\pi}\,\int\,\left(\Delta\,\frac{1}{|\bm{x}-\bm{x}^{\prime}|}\right)\,\chi(\bm{x}^{\prime})\,d^{3}x^{\prime}$
(12) $\displaystyle=$
$\displaystyle-\,\int\,\delta^{3}(\bm{x}-\bm{x}^{\prime})\,\chi(\bm{x}^{\prime})\,d^{3}x^{\prime}\
=\ -\,\chi(\bm{x}),$

thereby arriving at the transformation law
$\bm{A}_{\parallel}(\bm{x})\,\rightarrow\,\bm{A}^{\prime}_{\parallel}(\bm{x})\,=\,\bm{A}_{\parallel}(\bm{x})\,+\,\nabla\chi(\bm{x})$.
Strictly speaking, there is a minute incompleteness in the above proof. The
reason is that the neglect of the surface term is not generally justified
because the gauge function $\chi(\bm{x})$ need not always go to zero at the
spatial infinity.

After all, the above subtlety in the application of the Helmholtz theorem
arises because one cannot always demand that the full gauge potential
$\bm{A}(\bm{x})$ approaches zero at the spatial infinity. This indicates that
a smarter formulation of the transverse-longitudinal decomposition of the
vector potential $\bm{A}(\bm{x})$ with
$\bm{B}(\bm{x})=\nabla\times\bm{A}(\bm{x})$ should be the following. We start
with decomposing the net $\bm{A}(\bm{x})$ as

$\bm{A}(\bm{x})\ =\ \bm{A}_{\perp}(\bm{x})\ +\
\left(\bm{A}(\bm{x})\,-\,\bm{A}_{\perp}(\bm{x})\right)\ \equiv\
\bm{A}_{\perp}(\bm{x})\ +\ \bm{A}_{\parallel}(\bm{x}).$ (13)

In addition to the familiar transverse and longitudinal conditions
$\nabla\cdot\bm{A}_{\perp}(\bm{x})=0,\,\nabla\times\bm{A}_{\parallel}(\bm{x})=0$,
we impose an additional (we say, physical ) requirement on
$\bm{A}_{\perp}(\bm{x})$ as follows :

$\bm{A}_{\perp}(\bm{x})\ \rightarrow\ 0\ \ \ \mbox{as}\ \ \ |\bm{r}|\
\rightarrow\ \infty.$ (14)

We emphasize that we demand this property only for the transverse (or
physical) component not for the full vector potential. Then, from the identity

$0\ =\ \nabla\times\bm{A}_{\parallel}(\bm{x})\ =\
\nabla\times\left(\bm{A}(\bm{x})\ -\ \bm{A}_{\perp}(\bm{x})\right)$ (15)

we find that the following equality holds

$\nabla\times\bm{A}(\bm{x})\ =\ \nabla\times\bm{A}_{\perp}(\bm{x})\ =\
\bm{B}(\bm{x}).$ (16)

This means that $\bm{A}_{\perp}(\bm{x})$ satisfies both of the following two
relations

$\nabla\cdot\bm{A}_{\perp}(\bm{x})\ =\ 0,\ \ \
\nabla\times\bm{A}_{\perp}(\bm{x})\ =\ \bm{B}(\bm{x}),$ (17)

together with the condition $\bm{A}_{\perp}(\bm{x})\,\rightarrow\,0$ as
$|\bm{r}|\ \rightarrow\ \infty$. Now we can safely apply the Helmholtz theorem
to the vector field $\bm{A}_{\perp}(\bm{x})$ (not to $\bm{A}(\bm{x})$) to get

$\bm{A}_{\perp}(\bm{x})\ =\
\nabla\times\frac{1}{4\,\pi}\,\int\,\frac{\bm{B}(\bm{x}^{\prime})}{|\bm{x}-\bm{x}^{\prime}|}\,d^{3}x^{\prime}.$
(18)

This obviously shows that $\bm{A}_{\perp}(\bm{x})$ is unique as long as the
above integral converges. (It can be explicitly verified that this convergence
condition is satisfied in our infinitely-long solenoid problem.) The above
equation also convinces that $\bm{A}_{\perp}(\bm{x})$ is manifestly gauge
invariant. Now that this is confirmed, the gauge transformation property of
the longitudinal component can easily be established in the following manner,

$\displaystyle\bm{A}_{\parallel}(\bm{x})\ \rightarrow\
\bm{A}^{\prime}_{\parallel}(\bm{x})$ $\displaystyle=$
$\displaystyle\bm{A}^{\prime}(\bm{x})\ -\ \bm{A}^{\prime}_{\perp}(\bm{x})$
(19) $\displaystyle=$ $\displaystyle\bm{A}(\bm{x})\ +\ \nabla\chi(\bm{x})\ -\
\bm{A}_{\perp}(\bm{x})\ =\ \bm{A}_{\parallel}(\bm{x})\ +\ \nabla\chi(\bm{x}).$

We emphasize that the above line of proof for the gauge transformation
property of the longitudinal component as well as the uniqueness of the
transverse-longitudinal decomposition cleverly avoids the delicate convergence
problem in the proof of Adachi et al. Adachi1992 . The key factor is the
condition (14) which we impose only on the transverse (or physical) component
of the vector potential not on the full vector potential.

The transverse-longitudinal decomposition of the vector potential for the
infinitely-long solenoid can be viewed also from a somewhat different
perspective. As is widely known, if we introduce the electromagnetic
potentials $\phi$ and $\bm{A}$ by the relations $\bm{B}=\nabla\times\bm{A}$
and $\bm{E}=-\,\nabla\phi-\frac{\partial\bm{A}}{\partial t}$, one of the
Maxwell equations becomes

$\nabla\,\left(\nabla\cdot\bm{A}\ +\,\frac{\partial\phi}{\partial t}\right)\
-\ \Delta\bm{A}\ -\ \frac{\partial^{2}\bm{A}}{\partial t^{2}}\ =\ \bm{j},$
(20)

where $\bm{j}$ is the electric source current. Since the magnetic field inside
the infinitely-long solenoid is thought to be generated by steady current of
the solenoid, we can think that $\phi$ and $\bm{A}$ are both time-independent.
Under such circumstances, the above equation reduces to a simpler one

$\Delta\bm{A}(\bm{x})\ =\ -\,\bm{j}(\bm{x}),$ (21)

in either of the Coulomb gauge or the Lorenz gauge. A special solution of this
Poisson equation is given by

$\bm{A}^{(S)}(\bm{x})\ =\
\frac{1}{4\,\pi}\,\int\,\frac{\bm{j}(\bm{x}^{\prime})}{|\bm{x}-\bm{x}^{\prime}|}\,d^{3}x^{\prime}.$
(22)

Although this special solution reproduces the magnetic field as
$\nabla\times\bm{A}^{(S)}(\bm{x})=\bm{B}(\bm{x})$, it is not a general
solution of (21). The general solution of (21) is given by

$\bm{A}(\bm{x})\ =\ \bm{A}^{(S)}(\bm{x})\ +\ \bm{A}^{(H)}(\bm{x}),$ (23)

where $\bm{A}^{(H)}(\bm{x})$ is a solution of the homogeneous Laplace equation
$\Delta\bm{A}^{(H)}(\bm{x})=0$.

Now we show that $\bm{A}^{(S)}$ and $\bm{A}^{(H)}$ respectively correspond to
the transverse and longitudinal parts of $\bm{A}$ in the preceding argument.
In fact, we first see that $\bm{A}^{(S)}$ satisfies the transverse condition :

$\displaystyle\nabla\cdot\bm{A}^{(S)}(\bm{x})$ $\displaystyle=$
$\displaystyle\frac{1}{4\,\pi}\,\int\,\left(\nabla\,\frac{1}{|\bm{x}-\bm{x}^{\prime}|}\right)\cdot\bm{j}(\bm{x}^{\prime})\,\,d^{3}x^{\prime}$
(24) $\displaystyle=$
$\displaystyle-\,\frac{1}{4\,\pi}\,\int\,\left(\nabla^{\prime}\,\frac{1}{|\bm{x}-\bm{x}^{\prime}|}\right)\cdot\bm{j}(\bm{x}^{\prime})\,\,d^{3}x^{\prime}\,=\,\frac{1}{4\,\pi}\,\int\,\frac{\nabla^{\prime}\cdot\bm{j}(\bm{x}^{\prime})}{|\bm{x}-\bm{x}^{\prime}|}\,\,d^{3}x^{\prime}\,=\,0.\
\ \ \ \ $

Here, use has been made of the conservation law of the steady current, i.e.
$\nabla\cdot\bm{j}(\bm{x})=0$. We also point out that the neglect of the
surface term in a process of partial integral is justified, because the
current is localized on the surface of the solenoid. Next, from
$\nabla\times\bm{A}^{(S)}(\bm{x})=\bm{B}(\bm{x})=\nabla\times\bm{A}(\bm{x})$,
it follows that $\nabla\times\bm{A}^{(H)}(\bm{x})=0$, which means that
$\bm{A}^{(H)}(\bm{x})$ satisfies the longitudinal condition.

The above general argument can be more explicit by considering the problem of
the infinitely-long solenoid with the inside radius $R$. Since this setting is
essentially reduced to the 2-dimensional problem, it is convenient to work in
the cylindrical coordinates $\bm{x}=(\rho,\phi,z)=(\bm{\rho},z)$. The uniform
steady current is supposed to flow on the surface of the solenoid, so that it
is expressed as

$\bm{j}(\bm{\rho})\ =\ B_{0}\,\delta(\rho-R)\,\bm{e}_{\phi},$ (25)

with no dependence on the coordinate $z$. Here, $B_{0}$ denotes the strength
of uniform magnetic field inside the solenoid, while $\bm{e}_{\phi}$ is the
unit vector along the azimuthal angle direction. The total magnetic flux
through the solenoid is given by $\Phi=\pi\,R^{2}\,B_{0}$. By carrying out the
integral in (22), the special solution for the vector potential is found to
take the familiar form given as (for derivation, see Shadowitz1975 , for
example.)

$\bm{A}^{(S)}(\bm{\rho})\ =\ \left\\{\frac{\Phi}{2\,\pi
R^{2}}\,\,\rho\,\,\theta(R-\rho)\ +\
\frac{\Phi}{2\,\pi}\,\,\frac{1}{\rho}\,\,\theta(\rho-R)\,\right\\}\,\bm{e}_{\phi}.$
(26)

The general solution for the net vector potential is obtained by adding the
solution of the homogeneous equation $\Delta\bm{A}^{(H)}(\bm{\rho})=0$. As
pointed out before, since $\bm{A}^{(H)}$ satisfies the longitudinal condition
$\nabla\times\bm{A}^{(H)}=0$, it can be expressed as a gradient of some scalar
function $\chi(\bm{\rho})$ as $\bm{A}^{(H)}(\bm{\rho})=\nabla\chi(\bm{\rho})$.
If we further require that the net vector potential satisfies the Coulomb
gauge or transverse condition $\nabla\cdot\bm{A}=0$ (or the Lorenz gauge
condition in the static problem), $\chi$ must be a harmonic function
satisfying the Laplace equation,

$\Delta\chi(\bm{\rho})\ =\ 0.$ (27)

General solutions of this 2-dimensional Laplace equation can be given in the
form

$\chi(\rho,\phi)\ =\ \frac{a_{0}}{2}\ +\
\sum_{m=1}^{\infty}\,\left(a_{m}\,\cos m\phi\ +\ b_{m}\,\sin
m\phi\right)\,\rho^{m}.$ (28)

If one demands that $\chi$ vanishes at the spatial infinity, all the
coefficients $a_{m}\,(m=0,1,\cdots,\infty)$, and
$b_{m}\,(m=1,2,\cdots,\infty)$ must vanish. In fact, it is a well-known fact
that any harmonic function which vanishes at the spatial infinity must be
identically zero. In general, however, the function $\chi$ need not vanish at
the spatial infinity. For example, if we take the following special choice

$b_{2}\ =\ \frac{\Phi}{4\,\pi R},\ \ \ \mbox{other coefficients}\ =\ 0.$ (29)

the gauge function $\chi$ becomes

$\chi\ =\ \frac{\Phi}{4\,\pi\,R^{2}}\,\,\rho^{2}\,\,\sin 2\phi\ =\
\frac{1}{2}\,B_{0}\,x\,y,$ (30)

which in turn gives

$\bm{A}^{(H)}\ =\ \frac{1}{2}\,B_{0}\,y\,\bm{e}_{x}\ +\
\frac{1}{2}\,B_{0}\,x\,\bm{e}_{y}.$ (31)

This clearly does not vanish at the spatial infinity. In the argument above,
we limit the gauge function $\chi$ to a class of regular function satisfying
the Laplace equation in the whole 2-dimensional plane. In principle, we can
consider more general gauge function like BL1978

$\chi(\bm{\rho})\ =\ -\,\frac{\Phi}{2\,\pi}\,\arctan\frac{y}{x}.$ (32)

This is a multi-valued function that has a singularity at the coordinate
origin. With this choice, it is possible to completely eliminate the vector
potential outside the solenoid BL1978

$\bm{A}^{(S)}(\bm{\rho})\ \rightarrow\ \bm{A}^{\prime}(\bm{\rho})\ =\
\bm{A}^{(S)}(\bm{\rho})\ +\ \nabla\chi(\bm{\rho})\ =\ 0\ \ \ \ \ (\rho>R),$
(33)

as can be easily verified. However, it is a well-known fact that such a
singular gauge transformation alters the magnetic field inside the solenoid as
well Adachi1992

$\displaystyle\bm{B}(\bm{\rho})$ $\displaystyle=$
$\displaystyle\nabla\times\bm{A}^{\prime}(\bm{\rho})\ =\
\nabla\times\bm{A}^{(S)}(\bm{\rho})\ +\ \nabla\times\nabla\chi(\bm{\rho})$
(34) $\displaystyle=$ $\displaystyle\left(B_{0}\ -\
\Phi\,\delta^{2}(\bm{\rho})\right)\,\bm{e}_{z}\ \ \ \ \ (\rho<R).$

For that reason, such a nonsingular gauge transformation is not physically
acceptable. This conversely reconfirms the fact that there is no regular gauge
transformation, with which one can eliminate the physical component
$\bm{A}^{(S)}(\bm{\rho})$ without leaving any trace. To sum up, although the
physical component of the vector potential $\bm{A}$ is likely to be uniquely
given by the part $\bm{A}^{(S)}$ or $\bm{A}_{\perp}$, we must say that
$\bm{A}$ can still contain the part $\bm{A}^{(H)}$ or $\bm{A}_{\parallel}$,
even though the latter simply carries the residual gauge degrees of freedom,
which has little physical meaning.

After all these preparations, let us now reexamine the standard vector
potential interpretation of the Aharonov-Bohm effect. As is well-known, the
AB-phase difference for the two paths surrounding the solenoid is proportional
to the line-integral of the vector potential along a closed path $C$ as

$\phi_{AB}\ \propto\ \oint_{C}\,\bm{A}(\bm{x})\cdot d\bm{x}.$ (35)

Using the Stokes theorem, this closed line-integral is expressed by the net
magnetic flux penetrating the solenoid as

$\oint_{C}\,\bm{A}(\bm{x})\cdot d\bm{x}\ =\ \pi\,R^{2}\,B_{0}\ =\ \Phi.$ (36)

For some researchers, it seems that this very fact gives the ground of the
nonlocal interpretation of the AB-effect. Note however that the closed line-
integral of the vector potential can be divided into two pieces as

$\oint_{C}\,\bm{A}(\bm{x})\cdot d\bm{x}\ =\
\oint_{C}\,\bm{A}_{\perp}(\bm{x})\cdot d\bm{x}\ +\
\oint_{C}\,\bm{A}_{\parallel}(\bm{x})\cdot d\bm{x}.$ (37)

An important observation is that the transverse (or physical) part of $\bm{A}$
saturates this integral and the longitudinal (or unphysical) component never
contributes to it Adachi1992 ,WKZZ2018 . This is because the longitudinal part
is expressed as a gradient of some scalar function as
$\bm{A}_{\parallel}(\bm{x})=\nabla\chi(\bm{x})$, so that it follows that

$\oint_{C}\,\bm{A}_{\parallel}(\bm{x})\cdot d\bm{x}\ =\
\oint_{C}\,\nabla\chi(\bm{x})\cdot d\bm{x}\ =\
\int_{S}\,\left(\nabla\times\nabla\chi(\bm{x})\right)\cdot\bm{n}\,dS\ =\ 0,$
(38)

owing to the identity of vector calculus $\nabla\times\nabla\chi(\bm{x})=0$.
This is a relatively well-known fact, but if it is combined with our
discussion on the uniqueness as well as the meaning of the transverse-
longitudinal decomposition of the vector potential, we believe it gives a
strong support to the standard interpretation that the AB-effect
(corresponding to a closed path) is explained by the local action of the
vector potential on the phase of a charged particle’s wave function, even
though there remains an arbitrariness in the net vector potential.

However, this is not the end of the story. If one considers the AB phase shift
corresponding to a non-closed path, a new question arises. This is because the
contribution of the longitudinal component to the non-closed line integral
does not vanish in general. Consequently the AB-phase shift corresponding to a
non-closed path is thought to be a gauge-variant quantity. This appears to
contradict the recent claims by several researchers Kang2017 -Saldanha2021 .
They claim that the AB-phase shift for a non-closed path is a gauge-invariant
quantity, so that it can in principle be observed. Remember that the vital
assumption in their argument is that the AB phase shift is proportional to the
change of interaction energy between a charged particle and the current of the
solenoid along the path of a moving change. We think that the validity of this
ad hoc assumption needs more careful investigation. Such studies will be
carried out in the following sections.

##  III Brief review of recent locality interpretations of the Aharonov-Bohm
phase shift

First, we briefly introduce Boyer’s argument, which is based on the framework
of classical electromagnetism Boyer1971 . The central object in his study is
the interaction energy between the magnetic field $\bm{B}^{s}$ generated by
the current of a infinitely long solenoid and the magnetic field
$\bm{B}^{\prime}$ generated by a moving charge with a constant velocity
$\bm{v}$. Explicitly, it is given as

$\Delta\varepsilon\ =\
\frac{1}{4\,\pi}\,\int\,d^{3}x\,\,\bm{B}^{s}(\bm{x})\cdot\bm{B}^{\prime}(\bm{x},t).$
(39)

Here, $\bm{B}^{s}(\bm{x})$ generated by the surface current $\bm{j}(\bm{x})$
of the solenoid satisfies the equation

$\nabla\times\bm{B}^{s}(\bm{x})\ =\ \bm{j}(\bm{x}).$ (40)

On the other hand, in the nonrelativistic limit, the electric and magnetic
fields generated by a moving charge with a constant velocity $\bm{v}$ is known
to be given as Heaviside1888 -RPS2006

$\displaystyle\bm{E}^{\prime}(\bm{x},t)$ $\displaystyle=$
$\displaystyle\frac{e\,(\bm{x}-\bm{x}^{\prime})}{|\bm{x}-\bm{x}^{\prime}|}\ =\
-\,\nabla\,\frac{e}{|\bm{x}-\bm{x}^{\prime}|},$ (41)
$\displaystyle\bm{B}^{\prime}(\bm{x},t)$ $\displaystyle=$
$\displaystyle\bm{v}\times\bm{E}^{\prime}(\bm{x},t),$ (42)

with $d\bm{x}^{\prime}(t)/dt=\bm{v}$. (We recall that Boyer started with the
relativistic expression for the electric and magnetic field, but he eventually
took non-relativistic limit Boyer1971 .) The interaction energy is therefore
written as

$\Delta\varepsilon\ =\
\frac{1}{4\,\pi}\,\int\,d^{3}x\,\,\bm{B}^{s}\cdot(\bm{v}\times\bm{E}^{\prime})\
=\
\bm{v}\cdot\frac{1}{4\,\pi}\,\int\,d^{3}x\,\,\bm{E}^{\prime}\times\bm{B}^{s}.$
(43)

Here, the relevant integral can be transformed as

$\displaystyle\frac{1}{4\,\pi}\,\int\,d^{3}x\,\,\bm{E}^{\prime}(\bm{x})\times\bm{B}^{s}(\bm{x})\
=\
-\,\frac{1}{4\,\pi}\,\int\,d^{3}x\,\,\left(\nabla\frac{e}{|\bm{x}-\bm{x}^{\prime}|}\right)\times\bm{B}^{s}(\bm{x})$
(44) $\displaystyle=$
$\displaystyle-\,\frac{e}{4\,\pi}\,\int_{S}\,dS\,\,\left(\frac{1}{|\bm{x}-\bm{x}^{\prime}|}\,\,\bm{n}\times\bm{B}^{s}(\bm{x})\right)\
+\
\frac{e}{4\,\pi}\,\int\,d^{3}x\,\,\frac{1}{|\bm{x}-\bm{x}^{\prime}|}\,\nabla\times\bm{B}^{s}(\bm{x}).\hskip
14.22636pt$

The surface term above can safely be dropped, since the magnetic field by the
solenoid current is confined inside the solenoid. Then, using
$\nabla\times\bm{B}^{s}(\bm{x})=\bm{j}(\bm{x})$, he obtains

$\frac{1}{4\,\pi}\,\int\,d^{3}x\,\,\bm{E}^{\prime}(\bm{x})\cdot\bm{B}^{s}(\bm{x})\
=\
\frac{e}{4\,\pi}\,\int\,d^{3}x\,\,\frac{\bm{j}(\bm{x})}{|\bm{x}-\bm{x}^{\prime}|}.$
(45)

In this way, Boyer arrives at a remarkable relation

$\Delta\varepsilon\ =\ e\,\bm{v}\cdot\bm{A}^{(S)}(\bm{x}^{\prime}),$ (46)

with

$\bm{A}^{(S)}(\bm{x}^{\prime})\ =\
\frac{1}{4\,\pi}\,\int\,d^{3}x\,\,\frac{\bm{j}(\bm{x})}{|\bm{x}-\bm{x}^{\prime}|}.$
(47)

Boyer called this $\bm{A}^{(S)}$ the gauge potential in the Coulomb gauge.
This naming is not very accurate, however, because any vector potential
$\bm{A}$ obtained from $\bm{A}^{(S)}$ by means of regular gauge transformation
satisfies the Coulomb gauge condition (or the transverse condition)
$\nabla\cdot\bm{A}=0$. (In other words, it should be clearly recognized that
there generally remain residual gauge degrees of freedom in the Coulomb gauge
potential.) Rather, if one remembers our discussion before, the above
$\bm{A}^{(S)}$ should be identified with the axially-symmetric gauge potential
or the transverse component of $\bm{A}$, which is uniquely determined by the
solenoid current. Apart from this minor caution, what is important here is the
following observation by Boyer. Since the starting point of the derivation
above is the interaction energy $\Delta\varepsilon$ expressed with the
magnetic fields alone, he claims that the final expression given by the r.h.s.
of (46) must also be independent of the gauge choice. In fact, this appears to
be consistent with our preceding argument, which shows that the transverse
part of $\bm{A}$ (or the axially-symmetric potential $\bm{A}^{(S)}(\bm{x})$)
around the infinitely-long solenoid is a gauge-invariant quantity. Boyer made
step further by assuming that the AB-phase shift $\phi_{AB}$ is proportional
to the change of the above interaction energy along the path of the moving
charge, even when the path is not necessarily closed. According to him, since
this energy change is a gauge-invariant quantity, the AB-phase shift for such
a non-closed path is independent of the gauge choice. Based on this
observation, he claims that the corresponding AB-phase can in principle be
observed.

Partially motivated by the above-mentioned observation by Boyer, several
researchers investigated the interaction energy between the solenoid and a
moving charge within the framework of quantum electrodynamics Marletto-
Vedral2020  -LHK2022 . Most comprehensive for our later discussion is the
analysis by Saldanha Saldanha2021 . (We point out that his quantum mechanical
analysis was done on the basis of the preceding work by Santos and Gonzalo
Santos-Gonzalo1999 .) Saldanha evaluated the interaction energy between the
solenoid and the charged particle mediated by the exchange of a virtual photon
within the framework of the quantum electrodynamics in the Lorenz gauge. He
starts with the local interaction between the quantized electromagnetic field
$\bm{A}$ and the solenoid current $\bm{j}$ given as

$V_{1}\ =\ -\,\int\,d^{3}x\,\,\bm{j}(\bm{x})\cdot\bm{A}(\bm{x}),$ (48)

where

$\bm{A}(\bm{x})\ =\
\int\,d^{3}k\,\,\sqrt{\frac{1}{(2\,\pi)^{3}\,2\,\omega}}\,\,\,\sum_{\lambda=0}^{3}\,\,\bm{\epsilon}_{\bm{k},\lambda}\,\,a(\bm{k},\lambda)\,\,e^{\,i\,\bm{k}\cdot\bm{x}}\
+\ h.c.,$ (49)

with $\omega=|\bm{k}|$. Here, $\bm{\epsilon}_{\bm{k},\lambda}$ and
$a(\bm{k},\lambda)$ respectively stand for the polarization vector and
annihilation operator of a photon with momentum $\bm{k}$ and polarization
$\lambda$. The polarization indices $\lambda=1,2$ correspond to the transverse
photon, $\lambda=3$ to the longitudinal photon, and $\lambda=0$ to the scalar
photon, respectively. On the other hand, the local interaction between the
electromagnetic field and a moving charge is given by

$V_{2}\ =\ -\,e\,\bm{v}\cdot\bm{A}(\bm{x}^{\prime}),$ (50)

where $\bm{v}$ is the velocity of the moving charge. Next, based on the
framework of 2nd order perturbation theory, the interaction energy is
evaluated as

$\Delta\varepsilon\ =\ \sum_{n\neq 0}\,\frac{|\langle n\,|\,V_{1}\ +\
V_{2}\,|\,vac\rangle|^{2}}{E_{0}^{(0)}\ -\ E^{(0)}_{n}},$ (51)

where $|vac\rangle\equiv|n=0\rangle$ represents the vacuum of the photon,
while $|n\rangle$ with $n\neq 0$ stand for the exited states of the free
electromagnetic Hamiltonian. Retaining only the one-photon intermediate states
for $|n\rangle$ and dropping the self-energy terms, $\Delta\varepsilon$
reduces to

$\Delta\varepsilon\ =\
\mbox{Re}\,\left[\,\int\,d^{3}k\,\sum_{\lambda}\,\frac{\langle
vac\,|\,V_{2}\,|\,\bm{k},\lambda\rangle\,\langle\bm{k},\lambda\,|\,V_{1}\,|vac\rangle}{-\,\omega}\,\right]\
+\ \left(V_{1}\leftrightarrow V_{2}\right),$ (52)

where $|\bm{k},\lambda\rangle$ represents a one-photon state with the momentum
$\bm{k}$ and the polarization $\lambda$. Using

$\displaystyle\langle\bm{k},\lambda\,|\,V_{1}\,|vac\rangle$ $\displaystyle=$
$\displaystyle-\,\int\,d^{3}k\,\sqrt{\frac{1}{(2\,\pi)^{3}\,2\,\omega}}\,\,\,\bm{j}(\bm{x})\cdot\bm{\epsilon}_{\bm{k},\lambda}\,\,e^{\,-\,i\,\bm{k}\cdot\bm{x}},$
(53) $\displaystyle\langle vac\,|\,V_{2}\,|\,\bm{k},\lambda\rangle$
$\displaystyle=$
$\displaystyle-\,e\,\,\sqrt{\frac{1}{(2\,\pi)^{3}\,2\,\omega}}\,\,\,\bm{v}\cdot\bm{\epsilon}_{\bm{k},\lambda}\,\,e^{\,i\,\bm{k}\cdot\bm{x}^{\prime}}.$
(54)

and carrying out the polarization sum for the virtual photon states with

$\sum_{\lambda=0,1,2,3}\,\epsilon^{\mu}_{\bm{k},\lambda}\,\epsilon^{\nu}_{\bm{k},\lambda}\
=\ -\,g^{\mu\nu},$ (55)

Saldanha obtains

$\Delta\varepsilon\ =\
-\,e\,\,\mbox{Re}\,\left\\{\,\int\,d^{3}x\,\left[\int\,d^{3}k\,\frac{e^{\,i\,\bm{k}\cdot(\bm{x}^{\prime}-\bm{x})}}{(2\,\pi)^{3}\,k^{2}}\,\right]\,\bm{v}\cdot\bm{j}(\bm{x})\right\\}.$
(56)

Then, with the use of the identity

$\int\,\frac{d^{3}k}{(2\,\pi)^{3}}\,\,\frac{e^{\,i\,\bm{k}\cdot(\bm{x}^{\prime}-\bm{x})}}{k^{2}}\
=\ \frac{1}{4\,\pi}\,\,\frac{1}{|\bm{x}^{\prime}-\bm{x}|},$ (57)

he eventually arrives at

$\Delta\varepsilon\ =\ -\,e\,\bm{v}\cdot\bm{{\cal A}}(\bm{x}^{\prime}),$ (58)

with

$\bm{{\cal A}}(\bm{x}^{\prime})\ =\
\frac{1}{4\,\pi}\,\int\,d^{3}x\,\,\frac{\bm{j}(\bm{x}^{\prime})}{|\bm{x}^{\prime}-\bm{x}|}.$
(59)

With the standard setting of the infinitely-long solenoid, the vector
potential resulting from the above integral is nothing but the axially-
symmetric potential, i.e. $\bm{{\cal
A}}(\bm{x}^{\prime})=\bm{A}^{(S)}(\bm{x}^{\prime})$, which was also the case
in Boyer’s analysis Boyer1971 . We point out that, although the above result
by Saldanha was obtained in the Lorenz gauge Saldanha2021 , it can be easily
shown that exactly the same answer is obtained also in the Coulomb gauge. The
key ingredient here is the conservation law $\nabla\cdot\bm{j}(\bm{x})=0$ for
the time-independent (or steady) current of the solenoid.

Very interestingly, one realizes that Boyer’s analysis based on the classical
electromagnetism and Saldanha’s analysis based on the quantum electrodynamics
give almost the same form of answer for the interaction energy between the
solenoid current and a moving charge. However, one should notice that there is
a delicate difference between their results. That is, the overall sign of the
interaction energy is just opposite, which also means that the predictions for
the AB-phase shift would be different in sign. To our knowledge, no one has
paid attention to this subtle but remarkable difference. In the following
sections, we show that there is an unmissable deep physical meaning in this
sign difference.

##  IV Consistent quantum mechanical treatment of the combined system of a
solenoid and a charged particle

In pursuit of the root of the mysterious observation made at the end of the
last section, we develop here a self-contained formulation of the combined
system of a moving charge and a solenoid. The mass, the charge and the
coordinate of a moving particle are respectively denoted as $m$, $e$, and
$\bm{q}$. This charged particle not only interacts with the electromagnetic
field, but also subjects to the action of a certain potential $V(\bm{q})$,
which is supposed to work for preventing the charged particle from entering
the interior of the solenoid. The electromagnetic field also couples to a
given external current $j^{\mu}_{ext}(x)=(0,\bm{j}_{ext}(\bm{x}))$, where
$\bm{j}_{ext}(\bm{x})$ stands for the uniform steady current flowing on the
surface of the infinitely-long solenoid. The equations of motion for the
charged particle and the electromagnetic fields can straightforwardly be
written down as follows :

$\displaystyle\,m\,\frac{d^{2}\bm{q}}{dt^{2}}\ =\ -\,\frac{\partial
V(\bm{q})}{\partial\bm{q}}\ +\ e\,\left(\bm{E}(\bm{q},t)\ +\
\dot{\bm{q}}\times\bm{B}(\bm{q},t)\right),$ (60a)
$\displaystyle\,\nabla\cdot\bm{E}(\bm{x},t)\ =\ \tilde{j}^{0}(\bm{x},t),$
(60b) $\displaystyle\,\nabla\times\bm{B}(\bm{x},t)\ =\
\tilde{\bm{j}}(\bm{x},t)\ +\ \bm{j}_{ext}(\bm{x})\ +\
\frac{\partial\bm{E}(\bm{x},t)}{\partial t},$ (60c)

with $\dot{\bm{q}}\equiv d\bm{q}/\partial t$. Here,
$\tilde{j}^{\mu}(x)=(\tilde{j}^{0}(x),\tilde{\bm{j}}(x))$ represents the
current density generated by the moving charge with a constant velocity
$\dot{\bm{q}}=\bm{v}$ given as

$\displaystyle\tilde{j}^{0}(\bm{x},t)$ $\displaystyle=$ $\displaystyle
e\,\delta(\bm{x}-\bm{q}),$ (61) $\displaystyle\tilde{\bm{j}}(\bm{x},t)$
$\displaystyle=$ $\displaystyle e\,\dot{\bm{q}}\,\delta(\bm{x}-\bm{q}).$ (62)

It is a textbook exercise to show that the above equations of motion can be
derived from the following Lagrangian $L_{A}$ :

$L_{A}\ =\ \frac{1}{2}\,m\,\dot{\bm{q}}^{2}\ -\ V(\bm{q})\ -\
\frac{1}{4}\,\int\,d^{3}x\,F_{\mu\nu}\,F^{\mu\nu}\ -\
\int\,d^{3}x\,\left(\tilde{j}_{\mu}(x)+j^{ext}_{\mu}(x)\right)\,A^{\mu}(x),$
(63)

where $F_{\mu\nu}(x)=\partial_{\mu}\,A_{\nu}\ -\ \partial_{\nu}\,A_{\mu}$ is
the familiar field-strength tensor of the electromagnetic field. To obtain the
Hamiltonian corresponding to the above Lagrangian $L_{A}$ by means of the
standard canonical procedure, we need to fix gauge. For simplicity, we choose
the Coulomb gauge specified by $\nabla\cdot\bm{A}=0$. This leads to the
following Hamiltonian $H_{A}$ :

$\displaystyle H_{A}$ $\displaystyle=$
$\displaystyle\frac{1}{2\,m}\,\left(\,\bm{p}-e\,\bm{A}(\bm{q},t)\right)^{2}\
+\ V(\bm{q})\ +\ \int\,d^{3}x\,\,\frac{1}{2}\,\left(\bm{E}_{\perp}^{2}\ +\
\bm{B}^{2}\right)\ -\ \int\,d^{3}x\,\,\bm{j}_{ext}\cdot\bm{A},\ \ \ \ \ $ (64)

where $\bm{E}_{\perp}$ denotes the transverse component of the electric field
$\bm{E}$, while the longitudinal component of $\bm{E}$ is given by

$\bm{E}_{\parallel}(\bm{x},t)\ =\ -\,\nabla A_{0}(\bm{x},t),\ \ \
A_{0}(\bm{x},t)\ =\ \frac{1}{4\,\pi}\,\frac{e}{|\bm{x}-\bm{q}|}.$ (65)

The equal-time commutation relations for the relevant dynamical variables are
given as

$\displaystyle\left[\,p_{i},q_{j}\right]\ =\ -\,i\,\delta_{ij},$ (66)
$\displaystyle\left[A_{i}(\bm{x},t),E^{\perp}_{j}(\bm{x}^{\prime},t)\right]\
=\ -\,i\,\delta^{tr}_{ij}(\bm{x}-\bm{x}^{\prime}),$ (67)

while other commutation relations are all zero. Here,
$\delta^{tr}_{ij}(\bm{x}-\bm{x}^{\prime})$ is the familiar transverse delta
function defined by

$\delta^{tr}_{ij}(\bm{x}-\bm{x}^{\prime})\ =\
\int\,\frac{d^{3}k}{(2\,\pi)^{3}}\,\,e^{\,i\,\bm{k}\cdot(\bm{x}-\bm{x}^{\prime})}\,\left[\delta_{ij}\
-\ \frac{k_{i}\,k_{j}}{|\bm{k}|^{2}}\right].$ (68)

We have confirmed that that the Heisenberg equations corresponding to the
above Hamiltonian $H_{A}$ properly reproduces the equations of motion (60).
Hereafter, the treatment based on $H_{A}$ will be called the scheme (A).

For our following discussion, it is useful to divide the total electromagnetic
field into two parts as

$A^{\mu}(x)\ =\ A^{\mu}_{ext}(x)\ +\ \tilde{A}^{\mu}(x).$ (69)

Here, $A^{\mu}_{ext}(x)$ is the part generated by the external current of the
solenoid, while $\tilde{A}^{\mu}(x)$ is the part generated by the moving
charged particle. As already mentioned, in view of the time-independence of
the solenoid current, we can set $A_{ext}^{\mu}(x)=(0,\bm{A}_{ext}(\bm{x}))$.
Denoting the magnetic field generated by the external current as
$\bm{B}_{ext}$, it obeys the following equations :

$\nabla\times\bm{B}_{ext}(\bm{x})\ =\ \bm{j}_{ext}(\bm{x}),\ \ \
\bm{B}_{ext}(\bm{x})\ =\ \nabla\times\bm{A}_{ext}(\bm{x}).$ (70)

From these relations, one finds that $\bm{A}_{ext}(\bm{x})$ satisfies the
equation :

$\nabla^{2}\,\bm{A}_{ext}\ -\ \nabla\,(\nabla\cdot\bm{A}_{ext})\ =\
-\,\bm{j}_{ext}.$ (71)

It is convenient to work in the Coulomb gauge $\nabla\cdot\bm{A}_{ext}=0$ also
for the external vector potential. The equation for $\bm{A}_{ext}$ then
reduces to the Poisson equation $\Delta\bm{A}_{ext}=-\,\bm{j}_{ext}$. As
already discussed, the general solution to this Poisson equation is given as

$\bm{A}_{ext}(\bm{x})\ =\ \bm{A}_{ext}^{(S)}(\bm{x})\ +\ \nabla\chi(\bm{x})\
\equiv\ \bm{A}^{\perp}_{ext}(\bm{x})\ +\ \bm{A}^{\parallel}_{ext}(\bm{x}),$
(72)

with $\Delta\chi(\bm{x})=0$. Here, $\bm{A}^{(S)}_{ext}(\bm{x})$ denotes the
axially-symmetric gauge potential generated by the current of the infinitely-
long solenoid, while the residual gauge degrees of freedom in the Coulomb
gauge is carried by the longitudinal part $\bm{A}^{\parallel}_{ext}(\bm{x})$
or $\nabla\chi(\bm{x})$.

For aid in understanding the discussion below, we point out that, in general,
the vector potential $\tilde{\bm{A}}(\bm{x},t)$ generated by the moving charge
is also made up of the transverse and longitudinal parts as

$\tilde{\bm{A}}(\bm{x},t)\ =\ \tilde{\bm{A}}^{\perp}(\bm{x},t)\ +\
\tilde{\bm{A}}^{\parallel}(\bm{x},t).$ (73)

Thus, the total vector potential is generally made up of the following pieces
:

$\displaystyle\bm{A}(\bm{x},t)$ $\displaystyle=$
$\displaystyle\bm{A}_{ext}(\bm{x})\ +\ \tilde{\bm{A}}(\bm{x},t)$ (74)
$\displaystyle=$ $\displaystyle\left(\bm{A}^{\perp}_{ext}(\bm{x})\ +\
\bm{A}^{\parallel}_{ext}(\bm{x})\right)\ +\
\left(\tilde{\bm{A}}^{\perp}(\bm{x},t)\ +\
\tilde{\bm{A}}^{\parallel}(\bm{x},t)\right).$

Note that the the above decomposition can also be grouped in the following
form :

$\displaystyle\bm{A}(\bm{x},t)$ $\displaystyle=$
$\displaystyle\bm{A}^{\perp}(\bm{x},t)\ +\ \bm{A}^{\parallel}(\bm{x},t)$ (75)
$\displaystyle=$ $\displaystyle\left(\bm{A}^{\perp}_{ext}(\bm{x})\ +\
\tilde{\bm{A}}^{\perp}(\bm{x},t)\right)\ +\
\left(\bm{A}^{\parallel}_{ext}(\bm{x})\ +\
\tilde{\bm{A}}^{\parallel}(\bm{x},t)\right).$

In the following, to avoid unnecessary complexity, we may set
$\tilde{\bm{A}}^{\parallel}(\bm{x},t)=0$, which corresponds to the complete
Coulomb gauge fixing for the part $\tilde{\bm{A}}(\bm{x},t)$. Alternatively,
we can think as if this longitudinal part
$\tilde{\bm{A}}^{\parallel}(\bm{x},t)$ is absorbed into the part
$\bm{\bm{A}}_{ext}^{\parallel}(\bm{x})$. In any case, the corresponding
electromagnetic fields are represented as

$\displaystyle\tilde{\bm{E}}_{\perp}(\bm{x},t)$ $\displaystyle=$
$\displaystyle\bm{E}_{\perp}(\bm{x},t),$ (76)
$\displaystyle\tilde{\bm{B}}(\bm{x},t)$ $\displaystyle=$
$\displaystyle\nabla\times\tilde{\bm{A}}(\bm{x},t),$ (77)
$\displaystyle\bm{B}_{ext}(\bm{x})$ $\displaystyle=$
$\displaystyle\nabla\times\bm{A}_{ext}(\bm{x}).$ (78)

After the above separation, the equations of motion (60) can be transformed
into the form

$\displaystyle\,m\,\frac{d^{2}\bm{q}}{dt^{2}}\ =\ -\,\frac{\partial
V}{\partial\bm{q}}\ +\ e\,\left[\tilde{\bm{E}}_{\perp}(\bm{q},t)\ +\
\dot{\bm{q}}\times(\tilde{\bm{B}}(\bm{q},t)\ +\ \bm{B}_{ext}(\bm{q}))\right],$
(79a) $\displaystyle\,\nabla\cdot\tilde{\bm{E}}_{\perp}(\bm{x},t)\ =\
\tilde{j}^{0}(\bm{x},t),$ (79b)
$\displaystyle\,\nabla\times\tilde{\bm{B}}(\bm{x},t)\ =\
\tilde{\bm{j}}(\bm{x},t)\ +\
\frac{\partial\tilde{\bm{E}}_{\perp}(\bm{x},t)}{\partial t}.$ (79c)

One can verify that the above equations of motion (79) can be derived from the
following Lagrangian $L_{B}$ :

$L_{B}\ =\ \frac{1}{2}\,m\,\dot{\bm{q}}^{2}\ -\ V(\bm{q})\ -\
\frac{1}{4}\,\int\,d^{3}x\,\tilde{F}_{\mu\nu}\,\tilde{F}^{\mu\nu}\ -\
\int\,d^{3}x\,\,\tilde{j}_{\mu}(x)\,\left(\tilde{A}^{\mu}(x)\ +\
A^{\mu}_{ext}(x)\right),$ (80)

with
$\tilde{F}_{\mu\nu}=\partial_{\mu}\,\tilde{A}_{\nu}-\partial_{\nu}\,\tilde{A}_{\mu}$.
The Hamiltonian corresponding to this Lagrangian $L_{B}$ can be constructed by
imposing the Coulomb gauge condition $\nabla\cdot\tilde{\bm{A}}=0$ for
$\tilde{\bm{A}}$. It is given as

$H_{B}\ =\
\frac{1}{2\,m}\,\left(\bm{p}-e\,\left(\tilde{\bm{A}}(\bm{q},t)+\bm{A}_{ext}(\bm{q})\right)\right)^{2}\
+\ V(\bm{q})\ +\
\int\,d^{3}x\,\,\frac{1}{2}\,\left(\tilde{\bm{E}}^{2}_{\perp}+\tilde{\bm{B}}^{2}\right).$
(81)

The equal-time commutation relations for the relevant dynamical variables are
given by

$\displaystyle\left[\,p_{i},q_{j}\right]\ =\ -\,i\,\delta_{ij},$ (82)
$\displaystyle\left[\tilde{A}^{\perp}_{i}(\bm{x},t),\tilde{E}^{\perp}_{j}(\bm{x}^{\prime},t)\right]\
=\ -\,i\,\delta^{tr}_{ij}(\bm{x}-\bm{x}^{\prime}),$ (83)

while other commutation relations are all zero. One can also verify that the
Heisenberg equations based on the Hamiltonian $H_{B}$ reproduce the equations
of motion (79). Let us call this treatment based on $H_{B}$ the scheme (B). We
point out that this scheme corresponds to the treatment described in the
textbook of Cohen-Tannoudji et al. Cohen-Tannoudji1989 . (See Complement
$\mbox{C}_{II}$ titled “Electrodynamics in the presence of an external
field”.)

Since the two forms of Hamiltonians $H_{A}$ and $H_{B}$ are naturally expected
to describe the same dynamical system, which consists of the moving charge,
the solenoid and the electromagnetic field, they must be related to each
other. In fact, we can show that the two Hamiltonians $H_{A}$ and $H_{B}$
basically coincide aside from physically irrelevant constant terms. To see it,
we first notice that, with new field variables, $H_{A}$ can be expressed as

$\displaystyle H_{A}$ $\displaystyle=$
$\displaystyle\frac{1}{2\,m}\,\left[\bm{p}-e\,\left(\tilde{\bm{A}}(\bm{q},t)+\bm{A}_{ext}(\bm{q})\right)\right]^{2}\
+\ V(\bm{q})$ (84) $\displaystyle+$
$\displaystyle\int\,d^{3}x\,\,\frac{1}{2}\,\left[\tilde{\bm{E}}^{2}_{\perp}\
+\ \left(\tilde{\bm{B}}+\bm{B}_{ext}\right)^{2}\right]\ -\
\int\,d^{3}x\,\,\bm{j}_{ext}\cdot\left(\tilde{\bm{A}}+\bm{A}_{ext}\right).$

Here, we note that

$\displaystyle\frac{1}{2}\,\int\,d^{3}x\,\left(\tilde{\bm{B}}\ +\
\bm{B}_{ext}\right)^{2}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\,\int\,d^{3}x\,\left(\tilde{\bm{B}}^{2}\ +\
2\,\tilde{\bm{B}}\cdot\bm{B}_{ext}\ +\ \bm{B}^{2}_{ext}\right)$ (85)
$\displaystyle=$
$\displaystyle\frac{1}{2}\,\int\,d^{3}x\,\,\tilde{\bm{B}}^{2}\ +\
\int\,d^{3}x\,\,\tilde{\bm{B}}\cdot\bm{B}_{ext}\ +\ \mbox{constant},$

that is, the last term above represents the physically uninteresting self-
energy of the solenoid. On the other hand, the 2nd term of the above equation
can be rewritten as

$\displaystyle\int\,d^{3}x\,\tilde{\bm{B}}\cdot\bm{B}_{ext}$ $\displaystyle=$
$\displaystyle\int\,d^{3}x\,\nabla\times\tilde{\bm{A}}\cdot\bm{B}_{ext}\ =\
\int\,d^{3}x\,\tilde{\bm{A}}\cdot\nabla\times\bm{B}_{ext},$ (86)

with the use of integration by parts. The surface term of this integration by
parts can safely be neglected, since the magnetic field $\bm{B}_{ext}$ is
confined in a limited area, i.e. inside the solenoid. Then, using
$\nabla\times\bm{B}_{ext}=\bm{j}_{ext}$, we find that the following nontrivial
identity holds :

$\int\,d^{3}x\,\tilde{\bm{B}}\cdot\bm{B}_{ext}\ =\
\int\,d^{3}x\,\tilde{\bm{A}}\cdot\bm{j}_{ext}.$ (87)

A little surprisingly, this piece of the magnetic interaction energy precisely
cancels the part $-\,\int\,d^{3}x\,\bm{j}_{ext}\cdot\tilde{\bm{A}}$ in the
last term of (84). Besides, the part
$-\,\int\,d^{3}x\,\bm{j}_{ext}\cdot\bm{A}_{ext}$ in the last term of (84)
reduces to a physically uninteresting self-interaction term. As a consequence,
we have succeeded to show that

$H_{A}\ =\ H_{B}\ +\ \mbox{constants},$ (88)

which confirms that the two Hamiltonian $H_{A}$ in (64) and $H_{B}$ in (81)
coincide except for physically unimportant constant terms.

We are now ready to carry out a quantum mechanical analysis of the combined
system of a solenoid and a moving charge based on the two forms of Hamiltonian
$H_{A}$ and $H_{B}$. The general strategy is that, in view of the smallness of
the charge $e$, it is possible to neglect the second and higher order terms in
$e$.

###  IV.1 Analysis based on the Hamiltonian $H_{A}$

First, we split the Hamiltonian $H_{A}$ into the following pieces :

$H_{A}\ =\ \frac{1}{2\,m}\,\bm{p}^{2}\ +\ V(\bm{q})\ +\ H^{\prime}_{A},$ (89)

where

$\displaystyle H^{\prime}_{A}$ $\displaystyle=$
$\displaystyle\int\,d^{3}x\,\,\frac{1}{2}\,\left(\bm{E}_{\perp}^{2}+\bm{B}^{2}\right)\
-\ \int\,d^{3}x\,\,\bm{j}_{ext}\cdot\bm{A}\ -\
\frac{e}{m}\,\bm{p}\cdot\bm{A}(\bm{q})\ +\ O(e^{2})$ (90) $\displaystyle=$
$\displaystyle\hskip 22.76219ptH_{EM}\ +\ H_{j}\ +\ H_{e}\ +\ O(e^{2}),$

where we have defined as

$\displaystyle H_{EM}$ $\displaystyle\equiv$
$\displaystyle\int\,d^{3}x\,\,\frac{1}{2}\,\left(\bm{E}^{2}_{\perp}\ +\
\bm{B}^{2}\right),$ (91) $\displaystyle H_{j}$ $\displaystyle\equiv$
$\displaystyle-\,\int\,d^{3}x\,\,\bm{j}_{ext}\cdot\bm{A},$ (92) $\displaystyle
H_{e}$ $\displaystyle\equiv$
$\displaystyle-\,\frac{e}{m}\,\,\bm{p}\cdot\bm{A}(\bm{q}).$ (93)

It can be shown that the lowest-energy eigenstate (or ground state) of
$H_{EM}+H_{j}$ is the vacuum of the quanta of the electromagnetic field
$\tilde{\bm{A}}$. To see it, we begin with the expansion of the transverse
component of $\bm{A}$,

$\displaystyle\bm{A}_{\perp}(\bm{x},t)$ $\displaystyle=$
$\displaystyle\int\frac{d^{3}k}{\sqrt{(2\,\pi)^{3}}}\,\frac{1}{\sqrt{2\,\omega}}\,\,\sum_{\lambda=1}^{2}\,\,\bm{\epsilon}_{\bm{k},\lambda}\,\left(a(\bm{k},\lambda)\,e^{\,i\,(\bm{k}\cdot\bm{x}-\omega\,t)}\
+\
a^{\dagger}(\bm{k},\lambda)\,e^{\,-\,i\,(\bm{k}\cdot\bm{x}-\omega\,t)}\right),\hskip
14.22636pt$ (94) $\displaystyle\bm{E}_{\perp}(\bm{x},t)$ $\displaystyle=$
$\displaystyle-\,\frac{\partial}{\partial t}\,\bm{A}_{\perp}(\bm{x},t)$ (95)
$\displaystyle=$ $\displaystyle
i\,\int\frac{d^{3}k}{\sqrt{(2\,\pi)^{3}}}\,\sqrt{\frac{\omega}{2}}\,\,\,\sum_{\lambda=1}^{2}\,\,\bm{\epsilon}_{\bm{k},\lambda}\,\left(a(\bm{k},\lambda)\,e^{\,i\,(\bm{k}\cdot\bm{x}-\omega\,t)}\
-\
a^{\dagger}(\bm{k},\lambda)\,e^{\,-\,i\,(\bm{k}\cdot\bm{x}-\omega\,t)}\right),$

with $\omega=|\bm{k}|$. With the standard algebra, $H_{EM}$ reduces to

$H_{EM}\ =\
\sum_{\lambda=1}^{2}\,\int\,d^{3}k\,\,\omega\,\,a^{\dagger}(\bm{k},\lambda)\,a(\bm{k},\lambda)\
+\ \mbox{constant}.$ (96)

Taking care of the fact that $\bm{A}$ in $H_{j}$ generally consists of the
transverse and longitudinal parts as
$\bm{A}=\bm{A}_{\perp}+\bm{A}_{\parallel}$, $H_{j}$ can be divided into two
pieces as

$H_{j}\ =\
-\,\int\,d^{3}x\,\,\bm{j}_{ext}(\bm{x})\cdot\bm{A}_{\perp}(\bm{x},t)\ -\
\int\,d^{3}x\,\,\bm{j}_{ext}(\bm{x})\cdot\bm{A}_{\parallel}(\bm{x},t).$ (97)

Here, with the use of the current conservation
$\nabla\cdot\bm{j}_{ext}(\bm{x})=0$, one can show that the 2nd term of the
above equation vanishes as

$\displaystyle\int\,d^{3}x\,\,\bm{j}_{ext}(\bm{x})\cdot\bm{A}_{\parallel}(\bm{x},t)$
$\displaystyle=$
$\displaystyle\int\,d^{3}x\,\,\bm{j}_{ext}(\bm{x})\cdot\nabla\chi(\bm{x})$
(98) $\displaystyle=$
$\displaystyle-\,\int\,d^{3}x\,\,(\nabla\cdot\bm{j}_{ext}(\bm{x}))\,\chi(\bm{x})\
=\ 0.$

Then, using (94), we can show that

$\displaystyle H_{j}$ $\displaystyle=$
$\displaystyle-\,\int\,d^{3}x\,\,\bm{j}_{ext}(\bm{x})\cdot\bm{A}_{\perp}(\bm{x},t)$
(99) $\displaystyle=$ $\displaystyle-\
\sum_{\lambda=1}^{2}\,\int\,d^{3}k\,\,\frac{1}{\sqrt{2\,\omega}}\,\,\bm{\epsilon}_{\bm{k},\lambda}\cdot\left(\,\bm{j}^{*}_{ext}(\bm{k})\,a(\bm{k},\lambda)\
+\ \bm{j}_{ext}(\bm{k})\,a^{\dagger}(\bm{k},\lambda)\right),$

with the definition

$\bm{j}_{ext}(\bm{k})\ =\
\frac{1}{\sqrt{(2\,\pi)^{3}}}\,\int\,d^{3}x\,\,\bm{j}_{ext}(\bm{x})\,e^{\,-\,i\,\bm{k}\cdot\bm{x}}.$
(100)

In this way, up to a constant, $H_{EM}+H_{j}$ can be expressed as

$H_{EM}\ +\ H_{j}\ =\
\sum_{\lambda=1}^{2}\,\int\,d^{3}k\,\,\omega\,\,\tilde{a}^{\dagger}(\bm{k},\lambda)\,\tilde{a}(\bm{k},\lambda)\
+\ \mbox{constant},$ (101)

where we have defined the operator $\tilde{a}(\bm{k},\lambda)$ by

$\tilde{a}(\bm{k},\lambda)\ =\ a(\bm{k},\lambda)\ -\
\frac{1}{\sqrt{2\,\omega^{3}}}\,\,\bm{\epsilon}_{\bm{k},\lambda}\cdot\bm{j}_{ext}(\bm{k}).$
(102)

Here, $\tilde{a}(\bm{k},\lambda)$ corresponds to annihilation operator for the
electromagnetic field $\tilde{\bm{A}}_{\perp}$, with which
$\tilde{\bm{A}}_{\perp}$ can be expanded as

$\displaystyle\tilde{\bm{A}}_{\perp}(\bm{x},t)\ =\
\int\frac{d^{3}k}{\sqrt{(2\,\pi)^{3}}}\,\frac{1}{\sqrt{2\,\omega}}\,\,\sum_{\lambda=1}^{2}\,\,\bm{\epsilon}_{\bm{k},\lambda}\,\left(\tilde{a}(\bm{k},\lambda)\,e^{\,i\,(\bm{k}\cdot\bm{x}-\omega\,t)}\
+\
\tilde{a}^{\dagger}(\bm{k},\lambda)\,e^{\,-\,i\,(\bm{k}\cdot\bm{x}-\omega\,t)}\right).\hskip
14.22636pt$ (103)

Now, taking the difference of $\bm{A}_{\perp}(\bm{x},0)$ and
$\tilde{\bm{A}}_{\perp}(\bm{x},0)$, we find the relation

$\displaystyle\bm{A}_{\perp}(\bm{x},0)\ -\ \tilde{\bm{A}}_{\perp}(\bm{x},0)$
(104) $\displaystyle=$
$\displaystyle\int\,\frac{d^{3}k}{(2\,\pi)^{3}}\,\frac{1}{\sqrt{2\,\omega}}\,\,\sum_{\lambda=1}^{2}\,\epsilon_{\bm{k},\lambda}\,\left[(a(\bm{k},\lambda)-\tilde{a}(\bm{k},\lambda))\,e^{\,i\,\bm{k}\cdot\bm{x}}\
+\
(a^{\dagger}(\bm{k},\lambda)-\tilde{a}^{\dagger}(\bm{k},\lambda))\,e^{\,-\,i\,\bm{k}\cdot\bm{x}}\right]\hskip
17.07164pt$ $\displaystyle=$
$\displaystyle\int\,\frac{d^{3}k}{\sqrt{(2\,\pi)^{3}}}\,\,\frac{1}{2\,|\bm{k}|^{2}}\,\left(\bm{j}_{ext}(\bm{k})\,e^{\,i\,\bm{k}\cdot\bm{x}}\
+\ \bm{j}^{*}_{ext}(\bm{k})\,e^{\,-\,i\,\bm{k}\cdot\bm{x}}\right).$

By using (100), this can be further transformed as

$\displaystyle\bm{A}_{\perp}(\bm{x},0)\ -\ \tilde{\bm{A}}_{\perp}(\bm{x},0)$
(105) $\displaystyle=$
$\displaystyle\frac{1}{(2\,\pi)^{3}}\,\int\,d^{3}x^{\prime}\,\int\,d^{3}k\,\,\frac{1}{2\,|\bm{k}|^{2}}\,\left(\bm{j}_{ext}(\bm{x}^{\prime})\,e^{\,i\,\bm{k}\cdot(\bm{x}-\bm{x}^{\prime})}\
+\
\bm{j}^{*}_{ext}(\bm{x}^{\prime})\,e^{\,-\,i\,\bm{k}\cdot(\bm{x}-\bm{x}^{\prime})}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{2}\,\int\,d^{3}x^{\prime}\,\left(\bm{j}_{ext}(\bm{x}^{\prime})\
+\
\bm{j}^{*}_{ext}(\bm{x}^{\prime})\right)\,\,\frac{1}{4\,\pi}\,\frac{1}{|\bm{x}-\bm{x}^{\prime}|}\
=\
\frac{1}{4\,\pi}\,\int\,d^{3}x^{\prime}\,\,\frac{\bm{j}_{ext}(\bm{x}^{\prime})}{|\bm{x}-\bm{x}^{\prime}|}.\hskip
22.76219pt$

This last expression is just the axially-symmetric gauge potential
$\bm{A}^{(S)}(\bm{x})$ generated by the stationary solenoid current. In this
way, we find the relation

$\bm{A}_{\perp}(\bm{x},0)\ -\ \tilde{\bm{A}}_{\perp}(\bm{x},0)\ =\
\bm{A}^{(S)}_{ext}(\bm{x}),$ (106)

which shows that the difference of $\bm{A}_{\perp}$ and
$\tilde{\bm{A}}_{\perp}$ is just the classical (non-quantized) vector
potential $\bm{A}^{(S)}_{ext}$. Now, as is obvious from (101), the ground
state of $H_{EM}+H_{j}$ is the vacuum $|\tilde{0}\rangle$ of the quanta of
$\tilde{\bm{A}}$, which satisfies

$\tilde{a}(\bm{k},\lambda)\,|\tilde{0}\rangle\ =\ 0.$ (107)

Then, using
$\langle\tilde{0}\,|\,\tilde{\bm{A}}_{\perp}(\bm{q},t)\,|\tilde{0}\rangle=0$,
we obtain

$\langle\tilde{0}\,|\,\bm{A}(\bm{q},t)\,|\,\tilde{0}\rangle\ =\
\langle\tilde{0}\,|\,\tilde{\bm{A}}_{\perp}(\bm{q},t)\,+\,\bm{A}_{ext}(\bm{q})\,|\tilde{0}\rangle\
=\ \bm{A}_{ext}(\bm{q})\ =\ \bm{A}^{(S)}_{ext}(\bm{q})\ +\
\bm{A}^{\parallel}_{ext}(\bm{q}),$ (108)

which gives

$\langle\tilde{0}\,|\,H_{e}\,|\,\tilde{0}\rangle\ =\
-\,\frac{e}{m}\,\bm{p}\cdot\langle\tilde{0}\,|\,\bm{A}(\bm{q},t)\,|\,\tilde{0}\rangle\
=\ -\,\frac{e}{m}\,\,\bm{p}\cdot\bm{A}_{ext}(\bm{q}).$ (109)

It is also clear from (101) that

$\langle\tilde{0}\,|\,H_{EM}\ +\ H_{j}\,|\,\tilde{0}\rangle\ =\ 0,$ (110)

which just reconfirms that $|\tilde{0}\rangle$ is the ground state of
$H_{EM}+H_{j}$. In this way, we finally arrive at

$\langle\tilde{0}\,|\,H_{A}\,|\tilde{0}\rangle\ =\ \frac{1}{2\,m}\,\bm{p}^{2}\
+\ V(\bm{q})\ -\ \frac{e}{m}\,\,\bm{p}\cdot\bm{A}_{ext}(\bm{q})\ +\ O(e^{2}).$
(111)

If we dare to revive a term of $O(e^{2})$ so as to make clear the gauge-
covariance of the effective Hamiltonian, we may be able to write as

$\langle\tilde{0}\,|\,H_{A}\,|\tilde{0}\rangle\ =\
\frac{1}{2\,m}\,\left(\bm{p}\,-\,e\,\bm{A}_{ext}(\bm{q})\right)^{2}\ +\
V(\bm{q})\ +\ O(e^{2})\ \equiv\ H_{particle}.$ (112)

This can be interpreted as representing an effective Hamiltonian for a charged
particle moving under the influence of the external potential
$\bm{A}_{ext}(\bm{q})$. One might notice that this form of Hamiltonian
$H_{particle}$ provides the basis of the standard treatment of the AB-effect
under the influence of the classical external potential
$\bm{A}_{ext}(\bm{q})$.

###  IV.2 Analysis based on the Hamiltonian $H_{B}$

Interestingly, the result obtained in the previous subsection can be more
straightforwardly obtained in the scheme (B). As before, let us first split
$H_{B}$ into the following pieces :

$H_{B}\ =\ \frac{1}{2\,m}\,\bm{p}^{2}\ +\ V(\bm{q})\ +\ H^{\prime}_{B},$ (113)

where

$H^{\prime}_{B}\ =\
\int\,d^{3}x\,\,\frac{1}{2}\,\left(\tilde{\bm{E}}^{2}_{\perp}\ +\
\tilde{\bm{B}}^{2}\right)\ -\
\frac{e}{m}\,\,\bm{p}\cdot\left(\tilde{\bm{A}}(\bm{q},t)\ +\
\bm{A}_{ext}(\bm{q})\right)\ +\ O(e^{2}).$ (114)

Remembering the fact that $|\tilde{0}\rangle$ is the vacuum of the quantized
electromagnetic field $\tilde{\bm{A}}$, we immediately get

$\langle\tilde{0}\,|\,H^{\prime}_{B}\,|\,\tilde{0}\rangle\ =\
-\,\frac{e}{m}\,\,\bm{p}\cdot\bm{A}_{ext}(\bm{q})\ +\ O(e^{2}).$ (115)

This gives

$\displaystyle\langle\tilde{0}\,|\,H_{B}\,|\tilde{0}\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{2\,m}\,\bm{p}^{2}\ +\ V(\bm{q})\ -\
\frac{e}{m}\,\,\bm{p}\cdot\bm{A}_{ext}(\bm{q})\ +\ O(e^{2})$ (116)
$\displaystyle=$
$\displaystyle\frac{1}{2\,m}\,\left(\bm{p}\,-\,e\,\bm{A}_{ext}(\bm{q})\right)^{2}\
+\ V(\bm{q})\ +\ O(e^{2}),$

which precisely coincides with the effective Hamiltonian $H_{particle}$
obtained in the scheme (A).

As already pointed out, this form of Hamiltonian $H_{particle}$ is expected to
provide us with the basis of the standard treatment of the AB-effect. In fact,
under the gauge transformation of the external gauge potential

$\bm{A}_{ext}(\bm{q})\ \rightarrow\ \bm{A}^{\prime}_{ext}(\bm{q})\ =\
\bm{A}_{ext}(\bm{q})\ +\ \nabla\chi(\bm{q}),$ (117)

$H_{particle}$ transform gauge-covariantly together with the wave function of
the charged particle, which is the essential mechanism of the AB-effect. (For
more detail, see the later discussion based on the path integral formalism.)
This immediately raises the following question. Where in the Hamiltonian
$H_{A}$ or $H_{B}$ are the interaction energies as discussed by Boyer and/or
Saldanha and others hidden ? We try to answer this question in the following
subsection.

###  IV.3 Relation with the analyses by Boyer, Saldanha and others

To begin with, we carefully reexamine Boyer’s analysis based on our
Hamiltonian $H_{A}$. What he calculated is the interaction energy between the
magnetic field generated by the moving charge and that generated by the
solenoid current. The question is where we can find such an interaction term
in our Hamiltonian $H_{A}=1/(2\,m)\,\bm{p}^{2}+V(\bm{q})+H^{\prime}_{A}$ with
$H^{\prime}_{A}=H_{EM}+H_{j}+H_{e}+O(e^{2})$ ? To find it out, it is useful to
treat the Hamiltonian $H^{\prime}_{A}$ in the following manner. That is,
suppose that we look for the ground state of $H_{EM}+H_{j}+H_{e}$ using the
1st order perturbation theory with respect to the charge $e$. It is given by

$|\tilde{0}_{1}\rangle\ =\ |\tilde{0}\rangle\ +\
\frac{1}{\tilde{E}_{0}\,-\,(H_{EM}\,+\,H_{j})}\,H_{e}\,|\tilde{0}\rangle,$
(118)

where $\tilde{E}_{0}$ stands for the energy of the vacuum state
$|\tilde{0}\rangle$. With the use of $|\tilde{0}_{1}\rangle$, let us calculate
the expectation value
$\langle\tilde{0}_{1}\,|\,H_{EM}+H_{j}+H_{e}\,|\tilde{0}_{1}\rangle$ up to the
1st order in $e$. To this end, we first split $H_{EM}$ as follows :

$\displaystyle H_{EM}$ $\displaystyle=$
$\displaystyle\int\,d^{3}x\,\,\frac{1}{2}\,\left\\{\tilde{\bm{E}}^{2}_{\perp}\
+\ (\tilde{\bm{B}}+\bm{B}_{ext})^{2}\right\\}$ (119) $\displaystyle=$
$\displaystyle\int\,d^{3}x\,\,\frac{1}{2}\,\left(\tilde{\bm{E}}^{2}_{\perp}\
+\ \tilde{\bm{B}}^{2}\right)\ +\
\int\,d^{3}x\,\tilde{\bm{B}}\cdot\bm{\bm{B}}_{ext}\ +\ \mbox{constant}$
$\displaystyle\equiv$ $\displaystyle\tilde{H}_{EM}\ +\
\int\,d^{3}x\,\tilde{\bm{B}}\cdot\bm{\bm{B}}_{ext}\ +\ \mbox{constant}.$

Here, we have defined $\tilde{H}_{EM}$ by

$\tilde{H}_{EM}\ \equiv\
\int\,d^{3}x\,\,\frac{1}{2}\,\left(\tilde{\bm{E}}^{2}_{\perp}\ +\
\tilde{\bm{B}}^{2}\right).$ (120)

Obviously, this $\tilde{H}_{EM}$ can be expressed as

$\tilde{H}_{EM}\ =\
\sum_{\lambda=1}^{2}\,\int\,d^{3}k\,\,\omega\,\,\tilde{a}^{\dagger}(\bm{k},\lambda)\,\tilde{a}(\bm{k},\lambda),$
(121)

so that it holds that
$\langle\tilde{0}\,|\,\tilde{H}_{EM}\,|\tilde{0}\rangle=0$. Then, using this
relation together with the fact that the difference between
$\langle\tilde{0}_{1}\,|\,\tilde{H}_{EM}\,|\tilde{0}_{1}\rangle$ and
$\langle\tilde{0}\,|\,\tilde{H}_{EM}\,|\tilde{0}\rangle=0$ is $O(e^{2})$, we
find that

$\langle\tilde{0}_{1}\,|\,H_{EM}\,|\tilde{0}_{1}\rangle\ =\
\int\,d^{3}x\,\,\langle\tilde{0}_{1}\,|\,\tilde{\bm{B}}\,|\,\tilde{0}_{1}\rangle\cdot\bm{B}_{ext}\
+\ O(e^{2}).$ (122)

The expectation value
$\langle\tilde{0}_{1}\,|\,\tilde{\bm{B}}(\bm{x})\,|\tilde{0}_{1}\rangle$ above
can be evaluated as follows :

$\displaystyle\langle\tilde{0}_{1}\,|\,\tilde{\bm{B}}(\bm{x})\,|\tilde{0}_{1}\rangle$
$\displaystyle=$
$\displaystyle\langle\tilde{0}_{1}\,|\,\nabla\times\tilde{\bm{A}}(\bm{x})\,|\tilde{0}_{1}\rangle$
(123) $\displaystyle=$
$\displaystyle-\,\frac{e}{m}\,\langle\tilde{0}\,|\,\nabla\times\tilde{\bm{A}}(\bm{x})\,\,\frac{1}{\tilde{E}_{0}\,-\,(H_{EM}\,+\,H_{j})}\,\,\bm{p}\cdot\tilde{\bm{A}}(\bm{q})\,|\tilde{0}\rangle\
+\ c.c.$

By using the form (101) for $H_{EM}+H_{J}$ and the expansion (103) for
$\tilde{\bm{A}}(\bm{x})$, it is not difficult to show that

$\langle\tilde{0}_{1}\,|\,\tilde{\bm{B}}(\bm{x})\,|\tilde{0}_{1}\rangle\ =\
-\,\frac{\bm{p}}{m}\times\nabla\,\frac{1}{4\,\pi}\,\frac{e}{|\bm{x}-\bm{q}|}\
+\ O(e^{2}).$ (124)

This is nothing but the non-relativistic expression of the magnetic field
generated by the charge moving with the velocity $\bm{v}=\bm{p}/m$, which was
used in Boyer’ analysis. The calculation hereafter proceeds just in the same
manner as Boyer did. We obtain

$\displaystyle\langle\tilde{0}_{1}\,|\,H_{EM}\,|\tilde{0}_{1}\rangle$
$\displaystyle=$
$\displaystyle-\,\frac{e}{m}\,\frac{1}{4\,\pi}\,\int\,d^{3}x\,\left(\bm{p}\times\nabla\,\frac{1}{|\bm{x}-\bm{q}|}\right)\cdot\bm{B}_{ext}(\bm{x})$
(125) $\displaystyle=$
$\displaystyle-\,e\,\bm{v}\cdot\frac{1}{4\,\pi}\,\int\,d^{3}x\,\nabla\,\frac{1}{|\bm{x}-\bm{q}|}\times\bm{B}_{ext}(\bm{x})$
$\displaystyle=$ $\displaystyle
e\,\bm{v}\cdot\frac{1}{4\,\pi}\,\int\,d^{3}x\,\,\frac{1}{|\bm{x}-\bm{q}|}\,\nabla\times\bm{B}_{ext}(\bm{x})$
$\displaystyle=$ $\displaystyle
e\,\bm{v}\cdot\frac{1}{4\,\pi}\,\int\,d^{3}x\,\,\frac{\bm{j}_{ext}(\bm{x})}{|\bm{x}-\bm{q}|}\
=\ e\,\bm{v}\cdot\bm{A}^{(S)}_{ext}(\bm{q}).$

Here, $\bm{A}^{(S)}_{ext}(\bm{q})$ represents the axially-symmetric gauge
potential or the transverse part of $\bm{A}_{ext}(\bm{q})$. Obviously, the
expectation value of $H_{EM}$ derived above corresponds to the interaction
energy between the magnetic field $\tilde{\bm{B}}$ generated by the moving
charge and the magnetic field $\bm{B}_{ext}$ generated by the external
solenoid current in Boyer’s analysis. We recall, however, the fact that this
interaction energy is different in sign from that of Saldanha. Now, we shall
make clear the reason of this perplexing observation. To see it, we first
rewrite $H_{j}$ as

$H_{j}\ =\ -\,\int\,d^{3}x\,\,\bm{j}_{ext}\cdot\bm{A}(\bm{x})\ =\
-\,\int\,d^{3}x\,\,\bm{j}_{ext}(\bm{x})\cdot\tilde{\bm{A}}(\bm{x})\ +\
\mbox{constant}.$ (126)

The next task is to calculate the expectation value
$\langle\tilde{0}_{1}\,|\,H_{j}\,|\,\tilde{0}_{1}\rangle$ by using (118). Up
to constants, we have

$\displaystyle\langle\tilde{0}_{1}\,|\,H_{j}\,|\tilde{0}_{1}\rangle=\langle\tilde{0}\,|\,H_{j}\,\frac{1}{\tilde{E}_{0}-(H_{EM}+H_{j})}\,H_{e}\
+\
H_{e}\,\frac{1}{\tilde{E}_{0}-(H_{EM}+H_{j})}\,H_{j}\,|\,\tilde{0}\rangle.\hskip
8.53581pt$ (127)

Undoubtedly, this cross term of $H_{j}$ and $H_{e}$ corresponds to Saldanha’s
energy, i.e. the interaction energy between the solenoid and the moving charge
mediated by the exchange of a virtual photon. It can be confirmed by the
direct calculation as follows. First, we get

$\langle\tilde{0}_{1}\,|\,H_{j}\,|\tilde{0}_{1}\rangle\ =\
2\,\,\frac{e}{m}\,\langle\tilde{0}\,|\,\int\,d^{3}x\,\,\bm{j}_{ext}(\bm{x})\cdot\tilde{\bm{A}}(\bm{x})\,\frac{1}{\tilde{E}_{0}-(H_{EM}+H_{J})}\,\bm{p}\cdot\tilde{\bm{A}}(\bm{q})\,|\tilde{0}\rangle.$
(128)

Then, with the help of the manipulation

$\displaystyle\langle\tilde{0}\,|\,\tilde{A}_{i}(\bm{x})\,\frac{1}{\tilde{E}_{0}\,-\,(H_{EM}\,+\,H_{j})}\,\bm{p}\cdot\tilde{\bm{A}}(\bm{q})\,|\tilde{0}\rangle$
(129) $\displaystyle=$ $\displaystyle
p_{j}\,\int\,d^{3}k\,\frac{1}{(2\,\pi)^{3}\,2\,\omega}\,\sum_{\lambda=1,2}\,\frac{1}{-\,\omega}\,\,\,\langle\tilde{0}\,|\,\epsilon_{i}(\bm{k},\lambda)\,\left(\tilde{a}(\bm{k},\lambda)\,e^{\,i\,\bm{k}\cdot\bm{x}}\
+\ \tilde{a}^{\dagger}(\bm{k},\lambda)\,e^{\,-\,i\,\bm{k}\cdot\bm{x}}\right)$
$\displaystyle\hskip 159.33542pt\times\
\epsilon_{j}(\bm{k},\lambda)\,\left(\tilde{a}(\bm{k},\lambda)\,e^{\,i\,\bm{k}\cdot\bm{x}}\
+\
\tilde{a}^{\dagger}(\bm{k},\lambda)\,e^{\,-\,i\,\bm{k}\cdot\bm{x}}\right)\,|\tilde{0}\rangle\hskip
22.76219pt\hskip 22.76219pt$ $\displaystyle=$
$\displaystyle-\,p_{j}\,\int\,\frac{d^{3}k}{(2\,\pi)^{3}\,2\,\omega^{2}}\,\left(\delta_{ij}\
-\
\frac{k_{i}\,k_{j}}{|\bm{k}|^{2}}\right)\,e^{\,i\,\bm{k}\cdot(\bm{x}-\bm{q})},$

we obtain

$\displaystyle\langle\tilde{0}_{1}\,|\,H_{j}\,|\tilde{0}_{1}\rangle$
$\displaystyle=$
$\displaystyle-\,\frac{e}{m}\,\int\,d^{3}x\,\,j^{i}_{ext}(\bm{x})\,p^{j}\,\int\,\frac{d^{3}k}{(2\,\pi)^{3}\,\bm{k}^{2}}\,\left(\delta_{ij}\
-\
\frac{k_{i}\,k_{j}}{|\bm{k}|^{2}}\right)\,e^{\,i\,\bm{k}\cdot(\bm{x}-\bm{q})}.$
(130)

Using integration by parts and the current conservation
$\nabla\cdot\bm{j}_{ext}(\bm{x})=0$, we can show that the term
$k_{i}\,k_{j}/|\bm{k}|^{2}$ drops out. In this way, we eventually find that

$\displaystyle\langle\tilde{0}_{1}\,|\,H_{j}\,|\tilde{0}_{1}\rangle$
$\displaystyle=$
$\displaystyle-\,\frac{e}{m}\,\int\,d^{3}x\,\,\bm{p}\cdot\bm{j}_{ext}(\bm{x})\,\,\frac{1}{4\,\pi}\,\frac{1}{|\bm{x}-\bm{q}|}$
(131) $\displaystyle=$
$\displaystyle-\,\frac{e}{m}\,\,\bm{p}\cdot\frac{1}{4\,\pi}\,\int\,d^{3}x\,\frac{\bm{j}_{ext}(\bm{x})}{|\bm{x}-\bm{q}|}\
=\ -\,\frac{e}{m}\,\,\bm{p}\cdot\bm{A}^{(S)}_{ext}(\bm{q})$

This is precisely the interaction energy obtained by Saldanha. Note that the
sign of this energy is just opposite to the interaction energy of Boyer.

Finally, the expectation value of $H_{e}$ becomes

$\displaystyle\langle\tilde{0}_{1}\,|\,H_{e}\,|\,\tilde{0}_{1}\rangle$
$\displaystyle=$
$\displaystyle-\,e\,\,\frac{\bm{p}}{m}\,\cdot\,\langle\tilde{0}_{1}\,|\,\bm{A}(\bm{q})\,|\,\tilde{0}_{1}\rangle$
(132) $\displaystyle=$
$\displaystyle-\,e\,\,\frac{\bm{p}}{m}\,\cdot\,\langle\tilde{0}\,|\,\bm{A}(\bm{q})\,|\,\tilde{0}\rangle\
+\ O(e^{2})\ =\ -\,e\,\,\frac{\bm{p}}{m}\cdot\bm{A}_{ext}(\bm{q})\ +\
O(e^{2}).\hskip 11.38109pt$

We emphasize that the vector potential appearing in this last equation is the
full external potential $\bm{A}_{ext}(\bm{q})$ not its gauge-invariant part
$\bm{A}^{(S)}_{ext}(\bm{q})$.

To sum up, we find that

$\displaystyle\langle\tilde{0}_{1}\,|\,H_{EM}\,|\,\tilde{0}_{1}\rangle$
$\displaystyle=$
$\displaystyle+\,\frac{e}{m}\,\bm{p}\cdot\bm{A}^{(S)}_{ext}(\bm{q})\ =\
\Delta\epsilon\,(\mbox{Boyer}),$ (133)
$\displaystyle\langle\tilde{0}_{1}\,|\,H_{j}\,|\,\tilde{0}_{1}\rangle$
$\displaystyle=$
$\displaystyle-\,\frac{e}{m}\,\bm{p}\cdot\bm{A}^{(S)}_{ext}(\bm{q})\ =\
\Delta\epsilon\,(\mbox{Saldanha}),$ (134)
$\displaystyle\langle\tilde{0}_{1}\,|\,H_{e}\,|\,\tilde{0}_{1}\rangle$
$\displaystyle=$
$\displaystyle+\,\frac{e}{m}\,\bm{p}\cdot\bm{A}_{ext}(\bm{q})\ +\ O(e^{2}).$
(135)

This confirms that the interaction energy of Boyer and that of Saldanha
precisely cancel each other in the consistent treatment of the system of the
solenoid and the moving charge. As a consequence, we are eventually left with

$\langle\tilde{0}_{1}\,|\,H_{EM}\ +\ H_{j}\ +\ H_{e}\,|\,\tilde{0}_{1}\rangle\
=\ \langle\tilde{0}_{1}\,|\,H_{e}\,|\tilde{0}_{1}\rangle\ =\
-\,e\,\,\frac{\bm{p}}{m}\cdot\bm{A}_{ext}(\bm{q})\ +\ O(e^{2}),$ (136)

within the approximation up to the first order in $e$. In this way, we again
find that

$\langle\tilde{0}\,|\,H_{A}\,|\,\tilde{0}\rangle\ =\
\frac{1}{2\,m}\,\left(\bm{p}\ -\ e\,\bm{A}_{ext}(\bm{q})\right)^{2}\ +\
V(\bm{q})\ +\ O(e^{2})\ =\ H_{particle}.$ (137)

Remember that this effective Hamiltonian for a charged particle in the
presence of the external magnetic field is more straightforwardly obtained if
we started with the Hamiltonian $H_{B}$. (See Eq.(116).) However, somewhat
redundant analysis as above was mandatory for explicitly showing the
cancellation of the interaction energies of Boyer and of Saldanha and others.

##  V Path integral formulation based on the Lagrangian $L_{B}$

What can we learn after all from the analyses so far concerning the question
of the AB-phase shift for a non-closed path? In a formal sense, the question
can most clearly be answered by using Feynman’s path integral formalism based
on the Laglangian $L_{B}$ as will be explained below. First, by making use of
the relation

$\displaystyle\int\,d^{3}x\,\,\tilde{j}_{\mu}(x)\,A^{\mu}_{ext}(x)$
$\displaystyle=$
$\displaystyle-\,\int\,d^{3}x\,\,\tilde{\bm{j}}(\bm{x},t)\cdot\bm{A}_{ext}(\bm{x})$
(138) $\displaystyle=$
$\displaystyle-\,\int\,d^{3}x\,\,e\,\dot{\bm{q}}\,\delta(\bm{x}-\bm{q}(t))\cdot\bm{A}_{ext}(\bm{x})\
=\ -\,e\,\dot{\bm{q}}\cdot\bm{A}_{ext}(\bm{x}),\ \ \ $

it is convenient to rewrite $L_{B}$ in the form

$\displaystyle L_{B}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\,m\,\dot{\bm{q}}^{2}\ -\ V(\bm{q})\ -\
\frac{1}{4}\,\int\,d^{3}x\,\,\tilde{F}_{\mu\nu}(x)\,\tilde{F}^{\mu\nu}(x)$
(139) $\displaystyle\hskip
88.20354pt-\,\int\,d^{3}x\,\,\tilde{j}_{\mu}(x)\,\tilde{A}^{\mu}(x)\ +\
e\,\dot{\bm{q}}\cdot\bm{A}_{ext}(\bm{q}).$

Before explaining the path-integral formulation based on this form of
Lagrangian $L_{B}$, we think it instructive to reconfirm the following
features of $L_{B}$. First, there is no piece in $L_{B}$ that represents the
cross term of the magnetic field $\bm{B}^{s}$ generated by the solenoid and
the magnetic field $\tilde{\bm{B}}$ generated by the moving charge. This is
obvious from the form of the electromagnetic energy
$-\frac{1}{4}\,\int\,d^{3}x\,\tilde{F}_{\mu\nu}(x)\,\tilde{F}^{\mu\nu}(x)$,
which contains only the field strength tensor
$\tilde{F}_{\mu\nu}\equiv\partial_{\mu}\tilde{A}_{\nu}-\partial_{\nu}\tilde{A}_{\mu}$
with $\tilde{A}_{\mu}$ representing the piece of the electromagnetic potential
generated by the moving charge. Also noteworthy is the following feature.
First, the interaction term
$-\,\int\,d^{3}x\,\tilde{j}_{\mu}(x)\,\tilde{A}^{\mu}(x)$ between the solenoid
current and the electromagnetic field contains only the piece
$\tilde{A}^{\mu}$ amongst the total electromagnetic field
$A^{\mu}=A_{ext}^{\mu}\,+\,\tilde{A}^{\mu}$. On the other hand, the
interaction term $e\,\dot{\bm{q}}\cdot\bm{A}_{ext}(\bm{q})$ contains only the
piece $A_{ext}^{\mu}$ amongst the total electromagnetic field. This means
that, in $L_{B}$, there is no terms which mediate the interaction between the
solenoid and the moving charge by the exchange of a virtual photon. From the
above two observations, we can say again that the interaction energy of Boyer
and that of Saldanha are cancelling from the beginning in the scheme based on
the Laglangian $L_{B}$. Although this cancellation was already pointed out in
subsection $C$ of the previous section, we nevertheless think it important to
clearly reconfirm this fact in order not to misunderstand the significance of
the following path-integral analysis of the AB-phase shift based on the self-
contained Lagrangian $L_{B}$.

Now, we are ready to carry out a path-integral analysis of the AB-phase shift
based on $L_{B}$. First, note that, the term
$\int\,d^{3}x\,\tilde{F}_{\mu\nu}(x)\,\tilde{F}^{\mu\nu}(x)$ in (139) is
apparently gauge-invariant. We can also show that the term
$\int\,d^{3}x\,\tilde{j}_{\mu}(x)\,\tilde{A}^{\mu}(x)$ is gauge-invariant. In
fact, under the gauge transformation
$\tilde{A}^{\mu}(x)\rightarrow\tilde{A}^{\mu}(x)+\partial^{\mu}\tilde{\chi}(x)$,
this term transforms as

$\displaystyle\int\,d^{3}x\,\tilde{j}_{\mu}(x)\,\tilde{A}^{\mu}(x)$
$\displaystyle\rightarrow$
$\displaystyle\int\,d^{3}x\,\,\tilde{j}_{\mu}(x)\,\left(\tilde{A}^{\mu}(x)\ +\
\partial^{\mu}\tilde{\chi}(x)\right)$ (140) $\displaystyle=$
$\displaystyle\int\,d^{3}x\,\,\tilde{j}_{\mu}(x)\,\tilde{A}^{\mu}(x)\ -\
\int\,d^{3}x\,\,\partial^{\mu}\tilde{j}_{\mu}(x)\,\tilde{\chi}(x)$
$\displaystyle=$
$\displaystyle\int\,d^{3}x\,\,\tilde{j}_{\mu}(x)\,\tilde{A}^{\mu}(x),$

which confirms that this term is certainly gauge-invariant. In the above
proof, we have used the conservation of the current
$\partial^{\mu}\tilde{j}_{\mu}(x)=0$, which can readily be verified from the
relations

$\displaystyle\frac{\partial}{\partial t}\,\tilde{j}^{0}(x)$ $\displaystyle=$
$\displaystyle-\,e\,\dot{\bm{q}}\,\delta^{\prime}(\bm{x}-\bm{q}(t)),$ (141)
$\displaystyle\nabla\cdot\tilde{\bm{j}}(x)$ $\displaystyle=$
$\displaystyle+\,e\,\dot{\bm{q}}\,\delta^{\prime}(\bm{x}-\bm{q}(t)).$ (142)

In this way, the Lagrangian $L_{B}$ can be expressed in the form

$L_{B}\ =\ L^{0}_{B}\ +\ L^{int}_{B},$ (143)

with

$\displaystyle L^{0}_{B}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\,m\,\dot{\bm{q}}^{2}\ -\ V(\bm{q})\ -\
\frac{1}{4}\,\int\,d^{3}x\,\,\tilde{F}_{\mu\nu}(x)\tilde{F}^{\mu\nu}(x)\ -\
\int\,d^{3}x\,\,\tilde{j}_{\mu}(x)\,\tilde{A}^{\mu}(x),$ (144)

and

$L^{int}_{B}\ =\ e\,\dot{\bm{q}}\cdot\bm{A}_{ext}(\bm{q}),$ (145)

The corresponding action is given by

$S_{B}\ \equiv\ \int_{t_{i}}^{t_{f}}\,L_{B}\,dt\ =\ S^{0}_{B}\ +\
S^{int}_{B},$ (146)

with

$\displaystyle S^{0}_{B}$ $\displaystyle=$
$\displaystyle\int_{t_{i}}^{t_{f}}\,L^{0}_{B}\,dt,$ (147) $\displaystyle
S^{int}_{B}$ $\displaystyle=$
$\displaystyle\int_{t_{i}}^{t_{f}}\,e\,\dot{\bm{q}}\cdot\bm{A}_{ext}(\bm{q})\,dt.$
(148)

Note that the part $S^{0}_{B}$ is gauge-invariant, but $S^{int}_{B}$ is not.

Now the path-integral formulation for the AB-phase based on the Lagrangian
$L_{B}$ goes just in the standard manner. We start with the expression for the
transition amplitude when the initial and final states are respectively
specified by the wave functions of the charged particle
$\psi_{i}(\bm{q})=\langle\bm{q}|\psi_{i}\rangle$ and
$\psi_{f}(\bm{q})=\langle\bm{q}|\psi_{f}\rangle$. It is given by

$\displaystyle\langle\psi_{f},t_{f}\,|\,\psi_{i},t_{i}\rangle\ =\ \int\int
d\bm{q}_{f}\,d\bm{q}_{i}\,\,\psi^{*}_{f}(\bm{q}_{f}(t_{f}))\,K\left(\bm{q}_{f},t_{f}\,;\,\bm{q}_{i},t_{i}\right)\,\psi_{i}(\bm{q}_{i}(t_{i})).$
(149)

To be more precise, the initial and final states are specified as
$|i\rangle=|\psi_{i}\rangle\,|\Phi^{(S)}_{i}\rangle\,|0\rangle$ and
$|f\rangle=|\psi_{f}\rangle\,|\Phi^{(S)}_{f}\rangle\,|0\rangle$, where
$|\psi_{i}\rangle$ represents the initial state of the charged particle,
$|\Phi^{(S)}_{i}\rangle$ the initial state of the solenoid, and $|0\rangle$
does the vacuum of the photon, and similarly for the final state Santos-
Gonzalo1999 . However, the solenoid always stays in the same stationary state,
so that the initial and final states are basically specified by those of the
charge particle. In (149),
$K(\bm{q}_{f},t_{f}\,;\,\bm{q}_{i},t_{i})\equiv\langle\bm{q}_{f},t_{f}\,|\,\bm{q}_{i},t_{i}\rangle$
stands for the so-called Feynman kernel (or Feynman propagator) Feynman-
Hibbs1965 , the path-integral representation of which is given as

$\displaystyle K(\bm{q}_{f},t_{f}\,;\,\bm{q}_{i},t_{i})$ (150) $\displaystyle\
=\ \int_{\bm{q}_{i}(t_{i})}^{\bm{q}_{f}(t_{f})}\,\,{\cal D}\bm{q}\,\int\,{\cal
D}\tilde{A}^{\mu}\,\,\delta(\nabla\cdot\tilde{\bm{A}})\,\,\exp\left[\,iS_{B}\right]$
$\displaystyle\ =\ \int_{\bm{q}_{i}(t_{i})}^{\bm{q}_{f}(t_{f})}\,\,{\cal
D}\bm{q}\,\int\,{\cal
D}\tilde{A}^{\mu}\,\,\delta(\nabla\cdot\tilde{\bm{A}})\,\,\exp\,i\left[\,S^{0}_{B}\
+\
e\,\int_{t_{i}}^{t_{f}}\,\dot{\bm{q}}\cdot\bm{A}_{ext}(\bm{q})\,dt\,\right]$
$\displaystyle\ =\ \int_{\bm{q}_{i}(t_{i})}^{\bm{q}_{f}(t_{f})}\,\,{\cal
D}\bm{q}\,\int\,{\cal
D}\tilde{A}^{\mu}\,\,\delta(\nabla\cdot\tilde{\bm{A}})\,\,\exp\,i\left[\,S^{0}_{B}\
+\ e\,\int_{\bm{q}(t_{i})}^{\bm{q}(t_{f})}\,\bm{A}_{ext}(\bm{q})\cdot
d\bm{q}\,\right].\ \ \ \ \ $

Here, we have inserted the gauge-fixing delta function corresponding to the
Coulomb gauge $\nabla\cdot\tilde{\bm{A}}=0$ in view of the fact that the
quantum theory of gauge field necessarily requires gauge-fixing. However, it
is obvious that the Coulomb gauge is not mandatory in the analysis below. Any
gauge choice works equally well.

Important here is the gauge-transformation property of the Feynman kernel
$K(\bm{q}_{f},t_{f}\,;\,\bm{q}_{i},t_{i})$. Although the part $S^{0}_{B}$ of
the action is invariant under the gauge transformation
$\bm{A}_{ext}(\bm{q})\rightarrow\bm{A}_{ext}(\bm{q})+\nabla\chi(\bm{q})$, the
part $S^{int}_{B}$ transforms as

$\displaystyle S^{int}_{B}$ $\displaystyle=$ $\displaystyle
e\,\int_{\bm{q}_{i}}^{\bm{q}_{f}}\,\bm{A}_{ext}(\bm{q})\cdot d\bm{q}$ (151)
$\displaystyle\rightarrow$ $\displaystyle
e\,\int_{\bm{q}_{i}}^{\bm{q}_{f}}\,\,\left(\bm{A}_{ext}(\bm{q})\ +\
\nabla\chi(\bm{q})\right)\cdot d\bm{q}\ =\ S^{int}_{B}\ +\
e\,\left[\chi(\bm{q}_{f})\ -\ \chi(\bm{q}_{i})\right].$

Since the transition amplitude
$\langle\psi_{f},t_{f}\,|\,\psi_{i},t_{i}\rangle$ should be gauge-invariant,
it follows that the initial and final wave functions of the charged particle
must respectively transform as

$\displaystyle\psi_{i}(\bm{q}_{i})$ $\displaystyle\rightarrow$
$\displaystyle\psi^{\prime}_{i}(\bm{q}_{i})\ =\
e^{\,i\,e\,\chi(\bm{q}_{i})}\,\psi_{i}(\bm{q}_{i}),$ (152)
$\displaystyle\psi^{*}_{f}(\bm{q}_{f})$ $\displaystyle\rightarrow$
$\displaystyle\psi^{\prime*}_{f}(\bm{q}_{f})\ =\
\psi^{*}_{f}(\bm{q}_{f})\,e^{\,-\,i\,e\,\chi(\bm{q}_{f})}.$ (153)

This means that, when the charged particle travels from $\bm{q}_{A}$ to
$\bm{q}_{B}$ along a certain path $C$, it receives a phase change given by

$\Delta\phi\ =\ e\,\int_{\bm{q}_{A}}^{\bm{q}_{B}}\,\bm{A}_{ext}(\bm{q})\cdot
d\bm{q}\ +\ e\,\left[\chi(\bm{q}_{B})-\chi(\bm{q}_{B})\right]$ (154)

which generally depends on the gauge function $\chi(\bm{q})$ representing the
residual gauge degrees of freedom within the prescribed gauge. Naturally, in
the special case where the initial and final positions of the charged particle
are the same, i.e. $\bm{q}_{B}=\bm{q}_{A}$, which corresponds to the case
where the above path $C$ is closed, the gauge-dependent part
$\chi(\bm{q}_{B})-\chi(\bm{q}_{A})$ vanishes and the phase change is just
given by

$e\,\oint_{C}\,\bm{A}_{ext}(\bm{q})\cdot d\bm{q}\ =\ e\,\Phi,$ (155)

where $\Phi$ is the total magnetic flux penetrating the closed path $C$. This
is just the familiar AB-phase shift observed in the standard interference
experiment Tonomura1986 ,Osakabe1986 .

##  VI Summary and conclusion

On the basis of the two forms of Hamiltonians $H_{A}$ and $H_{B}$, which are
constructed so as to self-consistently describe the combined system of a
solenoid and a moving charge, both of which are interacting with the quantum
electromagnetic field, we have carefully examined the recent claims by several
authors that the AB-phase shift of a charged particle along a non-closed path
is gauge-invariant and can in principle be observed. The basic assumption
taken for granted by these authors is that the AB-phase shift is proportional
to the change of interaction energy between the solenoid and the charged
particle along the path of the moving charge. In our treatment based on the
Hamiltonian $H_{A}$, however, we find that the change of energy obtained by
Boyer and that obtained by Saldanha and others are certainly gauge-invariant
but they precisely cancel each other, which means that their claim loses its
ground. This cancellation demonstrated based on the Hamiltonian $H_{A}$
becomes even more obvious if we start the analysis from the physically
equivalent Hamiltonian $H_{B}$. In fact, if one carefully looks into the form
of Hamiltonian $H_{B}$, one would take notice of the following fact. First,
there is no piece in $H_{B}$ that represents the cross term of the magnetic
field $\bm{B}^{s}$ generated by the solenoid and the magnetic field
$\tilde{\bm{B}}$ generated by the moving charge. Remember that this cross term
is the origin of the interaction energy obtained by Boyer. Second, in the
Hamiltonian $H_{B}$, interactions terms that mediates the interaction between
the solenoid and the moving charge by the exchange of a virtual photon are
simply missing. Remember that they are the origin of the interaction energy
investigated by Saldanha. Putting it another way, the interaction energy of
Boyer and that of Saldanha and others are cancelling from the outset in the
scheme based on the Hamiltonian $H_{B}$. What is the origin of the AB-phase
shift, then ? To answer this question, we have carried out a path integral
analysis based on the Lagrangian $L_{B}$ (corresponding to the Hamiltonian
$H_{B})$. It was conformed that the AB-phase shift of a charged particle for a
non-closed path is in fact a gauge-dependent quantity in contrast to the
recent claims by several researchers.

In the present paper, we have also advocated the following viewpoint on the
role of the vector potential in the AB-effect. That is, despite the gauge
arbitrariness of the vector potential, to think of it as totally unphysical
quantity would not be correct. As we have elaborately shown, it certainly
contains something we might be able to call the physical component, which is
unique and can never be eliminated by any regular gauge transformation. In
fact, it was shown that the standard AB-phase shift for a closed path is
saturated by this physical component of the gauge potential. We believe that
this demonstration gives a strong support to the widely-accepted view that the
AB-phase shift is generated by the local action of the vector potential
despite its gauge arbitrariness. Nevertheless, we cannot still say that the
physical component of the vector potential can be seen purely experimentally.
Somewhat ironically, the reason is because the contribution of the unphysical
(or longitudinal) component of the vector potential for a closed path vanishes
exactly so that any vector potential, which is made up of the transverse and
longitudinal components, give the same AB-phase shift. Conversely speaking, to
directly see the physical component of the vector potential, we need to
measure the AB-phase shift for a non-closed path. Since the gauge ambiguity of
the AB-phase for a non-closed path as shown by (154) is totally due to the
unphysical longitudinal component of the vector potential, one might be
tempted to conclude that only the first term containing the physical component
contributes to the measured AB-phase shift. As we have discussed above,
however, the AB-phase shift for a non-closed path is nevertheless a gauge-
dependent quantity, so that its observation appears to contradict the famous
gauge principle. This paradoxical observation let us realize again truly
mysterious nature of the Aharonov-Bohm effect.

###### Acknowledgements.

The author of the paper would like to thank Akihisa Hayashi for his valuable
advices and enlightening discussions. Without his help, the paper would have
never been accomplished.

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