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

Abstract

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. We however notice the existence of two fairly different discussions on the interaction energy between the solenoid and a charge particle, the one is due to Boyer and the other is due to Saldanha and others. In the present paper, based on a self-contained quantum mechanical treatment of the combined system of a solenoid, a charged particle, and the quantized electromagnetic fields, we show that both interaction energies of Boyer and of Saldanha are in fact gauge invariant at least for non-singular gauge transformations but they are destined to cancel each other. Our analysis rather shows that the origin of the AB-phase can be traced back to other part of our effective Hamiltonian. Furthermore, based on the path-integral formalism with our effective Lagrangian, we explicitly demonstrate that the AB-phase shift for a non-closed path is not a gauge-variant quantity, which means that it would not correspond to direct experimental observables.

Appendices

Appendix A: On the convergence of the integral given by Eq. (9)

As indicated in the main text, Boyer as well as Saldanha have implicitly assumed the convergence of Eq. (9) for a very long solenoid without formally demonstrating it. We point out that more rigorous argument was already given in the paper by Babiker and Lowdon [38]. They derived the following expression for the vector potential of a finite solenoid of length $2 \,L$ and a radius of R, as measured in the x-y plane. The expression can be written as

$$\begin{aligned} \varvec{A} \ = \ \frac{2 \,\Phi _{AB} \, \varvec{e}_\phi }{\pi ^2 \,R} \,\int _0^L \, \frac{(2 - k^2) \,\mathbb {K} (k) - 2 \,\mathbb {E} (k)}{k^2 \,\sqrt{(\rho + R)^2 + z^2}} \,d z, \end{aligned}$$

(A1)

where $\Phi _{AB}$ is the AB-flux corresponding to an infinitely long solenoid, $k^2 = 4 \,\pi \,R / [(\rho + R)^2 + z^2]$, and $\mathbb {K} (k)$ and $\mathbb {E} (k)$ are the complete elliptical integrals of the first and second kind, respectively. Following them, it can be shown that along the regions $\rho \gg R$ and $\rho \ll R$, one can approximate Eq. (A1) as

$$\begin{aligned} \varvec{A} = \frac{\Phi _{AB} \,\varvec{e}_\phi }{2 \,\pi \,\rho } \frac{L}{\sqrt{\rho ^2 + L^2}} \ \ (\rho \gg R), \varvec{A} = \frac{\Phi _{AB} \,\rho \,\varvec{e}_\phi }{\pi \,R^2} \frac{L}{\sqrt{R^2 + L^2} } \ \ (\rho \ll R). \end{aligned}$$

(A2)

Clearly, these potentials do not correspond with those in the AB effect since we are dealing with a finite-length solenoid, which is consistent with the fact that outside the solenoid the magnetic field is non-vanishing $\nabla \times \varvec{A} \ne 0$. Accordingly, there would be a Lorentz force outside the solenoid acting on the charge. However, from Eq. (A2) it follows that such force would be negligible compared with other forces acting on the charge (e.g., a mechanical force) as L increases. In this regard, Eq. (A2) gives a qualitative idea of what is meant by a “very long solenoid”. In fact, in the limit of an infinitely long solenoid ($L \rightarrow \infty )$ we obtain the expected result

$$\begin{aligned} \lim _{L \rightarrow \infty } \,\varvec{A} = \frac{\Phi _{AB} \,\varvec{e}_\phi }{2 \,\pi \,\rho } \ \ (\rho > R), \lim _{L \rightarrow \infty } \,\varvec{A} = \frac{\Phi _{AB} \,\rho \,\varvec{e}_\phi }{\pi \,R^2} \ \ (\rho < R). \end{aligned}$$

(A3)

On the other hand, when $L \gg \rho$ and $L \gg R$, we obtain the approximate expressions

$$\begin{aligned} \varvec{A} \vert _{L \gg \rho } \simeq \frac{\Phi _{AB} \,\varvec{e}_\phi }{2 \,\pi \rho } \ \ (\rho \gg R), \varvec{A} \vert _{L \gg R} \simeq \frac{\Phi _{AB} \,\rho \,\varvec{e}_\phi }{\pi \,R^2} \ \ (\rho \ll R). \end{aligned}$$

(A4)

For example, if $L = 10^3 \,\rho$ and $\rho = 10^3 \,R$, then outside the solenoid the vector potential takes the form

$$\begin{aligned} \varvec{A} = 0.99 \times \left( \frac{\Phi _{AB} \,\varvec{e}_\phi }{2 \,\pi \,\rho } \right) \simeq \frac{\Phi _{AB} \,\varvec{e}_\phi }{2 \,\pi \rho }, \end{aligned}$$

(A5)

This clarifies in what sense the expression of the vector potential given in Boyer’s paper is justified.

Appendix B: On gauge-choice independence of Boyer’s interaction energy

Here, we address to the question whether the interaction energy of Boyer has truly a gauge-invariant meaning beyond the Coulomb gauge. To this end, we ask ourselves the following question:

• What is the most general form of vector potential generated by the steady current of an extremely long solenoid ?

To answer this question, we start with one of the (time-dependent) Maxwell equation,

$$\begin{aligned} \nabla \times \varvec{B} (\varvec{x}, t) \ - \ \frac{\partial \varvec{E} (\varvec{x}, t)}{\partial t} \ = \ \varvec{j} (\varvec{x}, t). \end{aligned}$$

(B6)

If one introduces the electromagnetic potentials $\phi$ and $\varvec{A}$ by the familiar relations $\varvec{B} = \nabla \times \varvec{A}$ and $\varvec{E} = - \,\nabla \phi - \frac{\partial \varvec{A}}{\partial t}$, the above Maxwell equation becomes

$$\begin{aligned} \nabla \left( \nabla \cdot \varvec{A} (\varvec{x}, t) \ + \, \frac{\partial \phi (\varvec{x}, t)}{\partial t} \right) \ - \ \Delta \varvec{A} (\varvec{x}, t) \ - \ \frac{\partial ^2 \varvec{A} (\varvec{x}, t)}{\partial t^2} \ = \ \varvec{j} (\varvec{x}, t), \end{aligned}$$

(B7)

Since the magnetic field inside the solenoid is thought to be generated by a steady (time-independent) surface current $\varvec{j} (\varvec{x})$ of the solenoid, we can think that $\phi$ and $\varvec{A}$ are both time-independent without loss of generality. Under such circumstances, the above equation reduces to

$$\begin{aligned} \Delta \varvec{A} (\varvec{x}) \ - \,\nabla \,( \nabla \cdot \varvec{A} (\varvec{x})) \ = \ - \,\varvec{j} (\varvec{x}). \end{aligned}$$

(B8)

independently of the gauge condition for $\varvec{A} (\varvec{x})$. The solution of the above equation can most easily be found if we impose the Coulomb gauge condition $\nabla \cdot \varvec{A}^{(C)} (\varvec{x}) = 0$ for the vector potential. In fact, under this condition, the above equation reduces to the familiar Poisson equation [37]:

$$\begin{aligned} \Delta \varvec{A}^{(C)} (\varvec{x}) \ = \ - \,\varvec{j} (\varvec{x}), \end{aligned}$$

(B9)

Here, $\varvec{A}^{(C)} (\varvec{x})$ stands for the vector potential satisfying the above Poisson equation together with the Coulomb gauge condition. As described in standard textbooks, the general solution to this Poisson equation is given as

$$\begin{aligned} \varvec{A}^{(C)} (\varvec{x}) \ = \ \varvec{A}^{(S)} (\varvec{x}) \ + \ \nabla \chi ^{(H)} (\varvec{x}), \end{aligned}$$

(B10)

where $\varvec{A}^{(S)} (\varvec{x})$ is given by⁴

$$\begin{aligned} \varvec{A}^{(S)} (\varvec{x}) \ = \ \frac{1}{4 \,\pi } \,\int \,d^3 x^\prime \,\, \frac{\varvec{j} (\varvec{x}^\prime )}{\vert \varvec{x} - \varvec{x}^\prime \vert }. \end{aligned}$$

(B11)

while $\chi ^{(H)} (\varvec{x})$ is any scalar function satisfying the Laplace equation $\Delta \chi ^{(H)} (\varvec{x}) = 0$. The last condition $\Delta \chi ^{(H)} = 0$ follows, because $\varvec{A}^{(C)}$ must satisfy the Coulomb gauge condition, which dictates that

$$\begin{aligned} 0 \ = \ \nabla \cdot \varvec{A}^{(C)} \ = \ \nabla \cdot \varvec{A}^{(S)} \ + \ \Delta \chi ^{(H)} \ = \ \Delta \chi ^{(H)}. \end{aligned}$$

(B12)

Here, use has been made of the relation $\nabla \cdot \varvec{A}^{(S)} = 0$, which can be easily verified by using the conservation law $\nabla \cdot \varvec{j} (\varvec{x}) = 0$ for the steady current. Unfortunately, the above form of the vector potential holds only in the Coulomb gauge. In fact, $\varvec{A}^{(C)}$ does not satisfy more general field equation (B8) for the vector potential. A question is whether one can find more general solution that satisfies Eq.(B8), which holds in arbitrary gauge. Actually, this is not so difficult. In fact, one can easily confirm that the following form of the vector potential satisfies the equation (B8),

$$\begin{aligned} \varvec{A} (\varvec{x}) \ = \ \varvec{A}^{(S)} (\varvec{x}) \ + \ \nabla \chi (\varvec{x}), \end{aligned}$$

(B13)

where $\chi (\varvec{x})$ is an arbitrary scalar function not restricted to a Harmonic function. In fact, it holds that

$$\begin{aligned} \Delta \varvec{A} (\varvec{x}) \, - \, \nabla (\nabla \cdot \varvec{A} (\varvec{x}))&= {} \Delta \,(\varvec{A}^{(S)} (\varvec{x}) + \nabla \chi (\varvec{x})) \, - \, \nabla \,(\nabla \cdot \varvec{A}^{(S)} (\varvec{x}) + \Delta \chi (\varvec{x})) \nonumber \\&= {} \Delta \,\varvec{A}^{(S)} (\varvec{x}) \ = \ \Delta \,\frac{1}{4 \,\pi } \, \int \,d^3 x^\prime \,\,\frac{\varvec{j} (\varvec{x}^\prime )}{\vert \varvec{x} - \varvec{x}^\prime \vert } \nonumber \\ &= {} \frac{1}{4 \,\pi } \,\int \,d^3 x^\prime \,\,( - \,4 \,\pi ) \,\, \delta (\varvec{x} - \varvec{x}^\prime ) \,\varvec{j} (\varvec{x}^\prime ) \ = \ - \,\varvec{j} (\varvec{x}). \end{aligned}$$

(B14)

We thus find that the vector potential generated by the steady current of a very long solenoid can always be expressed in the form (B13) independently of the gauge choice. The part $\varvec{A}^{(S)} (\varvec{x})$ looks clearly gauge invariant since it is expressed as a definite convolution integral of the source current. On the other hand, the part $\nabla \chi (\varvec{x})$, which carries the gauge degrees of freedom of $\varvec{A} (\varvec{x})$, is totally arbitrary. Here, a natural guess is that $\varvec{A}^{(S)} (\varvec{x})$ and $\nabla \chi (\varvec{x})$ can, respectively, be identified with the transverse part $\varvec{A}_\perp (\varvec{x})$ and the longitudinal part $\varvec{A}_\parallel (\varvec{x})$ of the vector potential.⁵

Unfortunately. there still remains a delicate problem in this identification. In fact, since the function $\chi$ is an arbitrary scalar function, one can in principle consider the following singular or multi-valued gauge function,

$$\begin{aligned} \chi (\varvec{x}) \ = \ - \,\Phi \,\phi \ = \ - \,\Phi \,\arctan \left( \frac{y}{x} \right) , \end{aligned}$$

(B15)

with $\varvec{x} = (\varvec{\rho }, z) = (\rho , \phi , z)$. As shown below, the rotation of $\nabla \chi$ does not vanish, i.e. $\nabla \times (\nabla \chi ) \ne 0$. Rather, it can be shown that $\nabla \chi$ satisfies the transverse condition, $\nabla \cdot (\nabla \chi ) = 0$. This means that the transverse-longitudinal decomposition, or more precisely, the identification of the transverse component of the vector potential is not unique, once the multi-valued gauge function as above is allowed.

Still, it seems to us that such a multi-valued gauge transformation is a little unnatural as well as artificial from a physical viewpoint. The reason will be explained below. We already know that, assuming an appropriate limiting procedure, the vector potential defined by

$$\begin{aligned} \varvec{A}^{(S)} (\varvec{x}) \ = \ \int \,\frac{\varvec{J}_{ext} (\varvec{x}^\prime )}{\vert \varvec{x} - \varvec{x}^\prime \vert } \,d^3 x^\prime \end{aligned}$$

with the surface current of the solenoid given by

$$\begin{aligned} \varvec{J}_{ext} (\varvec{x}) \ = \ B \,\delta (\rho - R) \,\varvec{e}_\phi \end{aligned}$$

takes the following axially-symmetric form

$$\begin{aligned} \varvec{A}^{(S)} (\varvec{x}) \ = \ \frac{\Phi }{2 \,\pi } \, \left[ \,\frac{\rho \,\theta (R - \rho )}{R^2} \ - \ \frac{\theta (\rho - R)}{\rho } \,\right] \,\varvec{e}_\phi , \end{aligned}$$

where $\varvec{x} = (\varvec{\rho }, z) = (\rho , \phi , z)$. Suppose that we define a new vector potential $\varvec{A}^\prime (\varvec{x})$ through the multi-valued gauge transformation

$$\begin{aligned} \varvec{A}^\prime (\varvec{x}) \ = \ \varvec{A}^{(S)} (\varvec{x}) \ + \ \nabla \chi (\varvec{x}), \end{aligned}$$

with the singular gauge function $\chi$ given by Eq. (B15). Since $\partial ^2 \chi / \partial x \,\partial y \ne \partial ^2 \chi / \partial y \,\partial x$ for the singular gauge function $\chi$, $\nabla \times (\nabla \chi )$ does not vanish. In fact, we find that

$$\begin{aligned} \nabla \times (\nabla \chi ) \ = \ - \,\Phi \,\delta (x) \,\delta (y) \,\varvec{e}_z \end{aligned}$$

Introducing the notation

$$\begin{aligned} \varvec{A}^\prime (\varvec{x}) \ = \ \varvec{A}^{(S)} (\varvec{x}) \ + \ \nabla \chi (\varvec{x}) \ \equiv \ \varvec{A}^{(S)} (\varvec{x}) \ + \ \varvec{A}_{string} (\varvec{x}), \end{aligned}$$

we therefore find that

$$\begin{aligned} \varvec{B}^\prime (\varvec{x}) &= {} \nabla \times \varvec{A}^\prime (\varvec{x}) \\ &= {} \nabla \times \varvec{A}^{(S)} (\varvec{x}) \ + \ \nabla \times \varvec{A}_{string} (\varvec{x}) \\ &= {} B \,\theta (R - \rho ) \,\varvec{e}_z \ - \ \Phi \,\delta (x) \,\delta (y) \,\varvec{e}_z \\ &\equiv {} \varvec{B}^{(S)} (\varvec{x}) \ + \ \varvec{B}_{string} (\varvec{x}). \end{aligned}$$

Note that the total magnetic flux penetrating the solenoid vanishes, since

$$\begin{aligned} \Phi ^\prime&\equiv {} \int \,\varvec{B}^{(S)} (\varvec{x}) \cdot d \varvec{S} \ + \ \int \,\varvec{B}_{string} (\varvec{x}) \cdot d \varvec{S} \\ &= {} \pi \,R^2 \ - \ \Phi \ = \ 0. \end{aligned}$$

Also widely-known is the fact that the vector potential outside the solenoid can be completely eliminated by the above multi-valued gauge transformation, i.e.

$$\begin{aligned} \varvec{A}^\prime (\varvec{x}) \ = \ 0 \ \ \ \text{ for } \ \ \ \rho > R \end{aligned}$$

(B16)

Based on this last observation, Bocchieri and Loisinger once erroneously claimed that the Aharonov–Bohm effect does not exist [41]. As shown by several researchers, the AB-effect remains to exist even after such a multi-valued gauge transformation, if one properly takes account of the change of the $2 \,\pi$ periodic boundary condition for the electron wave function [42,43,44].

Although mathematically allowable, however, we point out that very peculiar nature of the above-mentioned multi-valued gauge transformation. To explain it, let us recall the basic definition of the familiar transverse-longitudinal decomposition of the vector field $\varvec{F}$ is given as

$$\begin{aligned} \varvec{F} \ = \ \varvec{F}_\perp \ + \ \varvec{F}_\parallel . \end{aligned}$$

where the transverse component $\varvec{F}_\perp$ and the longitudinal component $\varvec{F}_\parallel$ are, respectively, demanded to satisfy the following transverse condition and the longitudinal condition:

$$\begin{aligned} \nabla \cdot \varvec{F}_\perp \ = \ 0, \ \ \ \nabla \times \varvec{F}_\parallel \ = \ 0. \end{aligned}$$

Note that, since $\nabla \times \varvec{A}_{string} \ne 0$, the part $\varvec{A}_{string} = \nabla \chi$ with $\chi = - \,\Phi \,\arctan (y / x)$ is not categorized into a longitudinal component any more, but it is a transverse component. In fact, one can verify that it satisfies the transverse condition,

$$\begin{aligned} \nabla \cdot \varvec{A}_{string} (\varvec{x}) \ = \ 0. \end{aligned}$$

This means that the transverse part of the vector potential is neither gauge invariant nor unique, if multi-valued gauge transformation is allowed. However, we continue to argue unphysical nature of such multi-valued gauge transformation. Suppose that we calculate rotation of the magnetic field $\varvec{B}^{(S)} (\varvec{x})$ and $\varvec{B}_{string} (\varvec{x})$, the sum of which gives the new magnetic field $\varvec{B}^\prime (\varvec{x}) = \varvec{B}^{(S)} (\varvec{x}) + \varvec{B}_{string} (\varvec{x})$ after singular gauge transformation. We find that

$$\begin{aligned} \nabla \times \varvec{B}^{(S)} (\varvec{x})= & {} B \,\delta (\rho - R) \,\varvec{e}_\phi \ = \ \varvec{J}_{ext} (\varvec{x}), \\ \nabla \times \varvec{B}_{string} (\varvec{x})= & {} \Phi \,\left( \, \delta (x) \,\delta ^\prime (y) \,\varvec{e}_x \ - \ \delta ^\prime (x) \,\delta (y) \,\varvec{e}_y \,\right) \ \equiv \ \varvec{J}_{string} (\varvec{x}). \end{aligned}$$

This means that the new magnetic field $\varvec{B}^\prime (\varvec{x})$ satisfies the following equation

$$\begin{aligned} \nabla \times \varvec{B}^\prime (\varvec{x}) \ = \ \varvec{J}_{ext} (\varvec{x}) \ + \ \varvec{J}_{string} (\varvec{x}), \end{aligned}$$

which is different from the original Maxwell equation

$$\begin{aligned} \nabla \times \varvec{B}^{(S)} (\varvec{x}) \ = \ \varvec{J}_{ext} (\varvec{x}). \end{aligned}$$

In this way, we now realize that the multi-valued gauge transformation changes the magnetic field distribution inside the solenoid, and formally this change of the magnetic field distribution is thought to be generated by a peculiar effective source current $\varvec{J}_{string} (\varvec{x})$ induced by the above singular gauge transformation. A logical consequence of this consideration is that it might also affect Boyer’s interaction energy between the solenoid and the charged particle.

To confirm it, let us first remember the expression of Boyer’s interaction energy written as

$$\begin{aligned} \Delta \varepsilon (Boyer) \ = \ \frac{1}{4 \,\pi } \,\int \,d^3 x \, \varvec{B}^s \cdot (\varvec{v} \times \varvec{E}^\prime ) \ = \ \varvec{v} \cdot \frac{1}{4 \,\pi } \,\int \,d^3 x \,\varvec{E}^\prime \times \varvec{B}^s. \end{aligned}$$

with

$$\begin{aligned} \varvec{E}^\prime (\varvec{x}, t)= & {} - \,\nabla \,\frac{e}{\vert \varvec{x} - \varvec{x}^\prime \vert }, \varvec{B}^s (\varvec{x}) \ = \ B \,\theta (R - \rho ) \,\varvec{e}_z. \end{aligned}$$

As already pointed out, under the multi-valued gauge transformation, the magnetic field distribution changes as

$$\begin{aligned} \varvec{B}^s (\varvec{x}) \ \rightarrow \ \varvec{B}^s (\varvec{x}) \ + \ \varvec{B}_{string} (\varvec{x}). \end{aligned}$$

This means that Boyer’s energy also changes as

$$\begin{aligned} \Delta \varepsilon (Boyer) \ \rightarrow \ \Delta \varepsilon ^\prime (Boyer) \ = \ \Delta \varepsilon (Boyer) \ + \ \Delta \varepsilon (string), \end{aligned}$$

where

$$\begin{aligned} \Delta \varepsilon (string) \ = \ \varvec{v} \cdot \frac{1}{4 \,\pi } \,\int \,d^3 x \,\varvec{E}^\prime \times \varvec{B}_{string}. \end{aligned}$$

After some algebra, we can show that

$$\begin{aligned} \Delta \varepsilon (string) \ = \ - \,e \,\varvec{v} \cdot \left( \,\frac{\Phi }{2 \,\pi } \, \frac{1}{\rho ^\prime } \right) \,\varvec{e}_{\phi ^\prime }. \end{aligned}$$

Comparing this with the expression of $\Delta \varepsilon (Boyer)$ given as

$$\begin{aligned} \Delta \varepsilon (Boyer) \ = \ e \,\varvec{v} \cdot \frac{\Phi }{2 \,\pi } \, \left[ \frac{\rho ^\prime \,\theta (R - \rho ^\prime )}{R^2} \ + \ \frac{\theta (\rho ^\prime - R)}{\rho ^\prime } \right] \,\varvec{e}_{\phi ^\prime }, \end{aligned}$$

we find that

$$\begin{aligned} \Delta \varepsilon ^\prime (Boyer) \ = \ \Delta \varepsilon (Boyer) \ + \ \Delta \varepsilon (string) \ = \ 0. \end{aligned}$$

outside the solenoid, i.e. for $\rho ^\prime > R$, which is the region where the charged particle is making a rectilinear motion. That is, Boyer’s interaction energy becomes zero after the multi-valued gauge transformation. As is discussed in the main text, the expression of Saldanha’s energy is just the same as Boyer’s energy except for the sign difference, so that it is clear that the Saldanha’s energy also becomes zero after the multi-valued gauge transformation. This means that the cancellation between the energies of Boyer and of Saldanha remains intact even after the multi-valued gauge transformation.

In this way, we realize that that the interaction energies of Boyer (and also of Saldanha) is not gauge-invariant, if a multi-valued gauge transformation is allowed. However, remember that the multi-valued gauge transformation induces peculiar effective current and it changes the form of the Maxwell equation for the magnetic field. Then, even though such a multi-valued gauge transformation is mathematically allowed, we feel it very unnatural from a physical viewpoint. On the other hand, if we confine to regular gauge transformation, we can say that the transverse part of the gauge potential is gauge-invariant as well as unique, and the energies of Boyer and Saldanha are certainly expressed with this transverse component, even though these two energies are destined to cancel each other out.

Appendix C: On the identity for $H_{EM} + H_j$

With the use of Eqs. (90) and (97), we get

$$\begin{aligned} H_{EM} + H_j= & {} \tilde{H}_{EM} + \int d^3 x \,\, \tilde{\varvec{B}} (\varvec{x}) \cdot \varvec{B}_{ext} (\varvec{x}) \nonumber \\&- \int d^3 x \,\,\tilde{\varvec{A}} (\varvec{x}) \cdot \varvec{j}_{ext} (\varvec{x}) + \text{ constant }. \end{aligned}$$

(C17)

Note that the 2nd and 3rd terms on the r.h.s. of the above equation cancel out owing to the nontrivial identity (74) in the text. Thus, we have

$$\begin{aligned} H_{EM} + H_j &= {} \tilde{H}_{EM} + \text{ constant } \nonumber \\ &= {} \sum _{\lambda = 1}^2 \,\int d^3 k \,\,\omega \, \tilde{a}^\dagger (\varvec{k}, \lambda ) \,\tilde{a} (\varvec{k}, \lambda ) + \text{ constant }. \end{aligned}$$

(C18)

This means that the ground state of $H_{EM} + H_j$ is the vacuum $\vert \tilde{0} \rangle$ of the quanta $\tilde{a} (\varvec{k}, \lambda )$, so that it holds that

$$\begin{aligned} \left( H_{EM} + H_j \right) \,\vert \tilde{0} \rangle \ = \ \tilde{E}_0 \,\vert \tilde{0} \rangle . \end{aligned}$$

(C19)

Here, $\tilde{E}_0$ represents the ground-state energy of $H_{EM} + H_j$, which may in principle contain self-energies of the solenoid and the moving charge.

Appendix D: Proof of Eq. (71).

Up to $O (e^2)$, we have

$$\begin{aligned}&\langle \tilde{0}_1 \vert \,\tilde{\varvec{B}} (\varvec{x}) \vert \, \tilde{0}_1 \rangle \nonumber \\ &\quad = - \ \frac{e}{m} \,\langle \tilde{0} \,\vert \, \nabla \times \tilde{\varvec{A}} (\varvec{x}) \, \frac{1}{\tilde{E}_0 - (H_{EM} + H_j )} \, \varvec{p} \cdot \tilde{\varvec{A}} (\varvec{q}) \,\vert \tilde{0} \rangle \ + \ c.c \end{aligned}$$

(D20)

In view of the Fourier expansion of $\tilde{\varvec{A}} (\varvec{x})$, this gives

$$\begin{aligned}&\langle \tilde{0}_1 \vert \,\tilde{\varvec{B}} (\varvec{x}) \vert \, \tilde{0}_1 \rangle \nonumber \\ &\quad = \ - \,2 \,\,\frac{e}{m} \,\int \frac{d^3 k}{(2 \,\pi )^3} \,\sum _{\lambda = 1}^2 \,\, \frac{\langle \tilde{0} \,\vert \,\nabla \times \tilde{\varvec{A}} (\varvec{x}) \, \vert \varvec{k}, \lambda \rangle \,\langle \varvec{k}, \lambda \,\vert \, \varvec{p} \cdot \tilde{\varvec{A}} (\varvec{q}) \,\vert \tilde{0} \rangle }{- \,\omega } \,\, \frac{1}{2 \,\omega }. \end{aligned}$$

(D21)

Now, we proceed as

$$\begin{aligned}&\int \frac{d^3 k}{(2 \,\pi )^3} \,\frac{1}{- \,\omega ^2} \, \sum _{\lambda = 1}^2 \,\langle \tilde{0} \,\vert \, \nabla \times \tilde{\varvec{A}} (\varvec{x}) \,\vert \varvec{k}, \lambda \rangle \, \langle \varvec{k}, \lambda \,\vert \varvec{p} \cdot \tilde{\varvec{A}} (\varvec{q}) \, \tilde{0} \rangle \nonumber \\ &\quad = - \, 2 \,i \,\int \frac{d^3 k}{(2 \,\pi )^3} \,\frac{1}{\omega ^2} \, \sum _{\lambda = 1}^2 \,\varvec{k} \times \varvec{\epsilon }_{\varvec{k}, \lambda } \, e^{\,i \,\varvec{k} \cdot \varvec{x}} \,\varvec{p} \cdot \varvec{\epsilon }_{\varvec{k}, \lambda } \,e^{\,- \,i \,\varvec{k} \cdot \varvec{q}}. \end{aligned}$$

(D22)

Then, with the use of the formula for the polarization sum, we get

$$\begin{aligned} \sum _{\lambda = 1}^2 \,\varvec{k} \times \varvec{\epsilon }_{\varvec{k}, \lambda } \,\, \varvec{p} \cdot \varvec{\epsilon }_{\varvec{k}, \,\lambda } \ = \ - \,\varvec{p} \times \varvec{k}, \end{aligned}$$

(D23)

we finally obtain

$$\begin{aligned} \langle \tilde{0}_1 \,\vert \tilde{\varvec{B}} (\varvec{x}) \,\vert \tilde{0}_1 \rangle \ = \ - \,\frac{\varvec{p}}{m} \times \nabla \,\frac{1}{4 \,\pi } \, \frac{1}{\vert \varvec{x} - \varvec{q} \vert } \ + \ O (e^2), \end{aligned}$$

(D24)

which proves Eq. (95) in the text.