Effects of the Fermionic Vacuum Polarization in QED

Eur. Phys. J. C (2018) 78:12 https://doi.org/10.1140/epjc/s10052-017-5498-x

Regular Article - Theoretical Physics

Effects of the fermionic vacuum polarization in QED

M. F. X. P. Medeiros1,a, F. E. Barone2,b, F. A. Barone1,c 1 IFQ-Universidade Federal de Itajubá, Av. BPS 1303, Caixa Postal 50, Itajubá, MG CEP 37500-903, Brazil 2 Niterói, RJ, Brazil

Received: 3 December 2017 / Accepted: 22 December 2017 / Published online: 9 January 2018 © The Author(s) 2018. This article is an open access publication

Abstract Some effects of vacuum polarization in QED due radiative corrections to the Aharonov–Bohm scattering [10], to the presence of field sources are investigated. We focus the interaction between two solenoids [11], vacuum currents on effects with no counter-part in Maxwell electrodynamics. produced around a Dirac string [12,13], Bremsstrahlung and The the Uehling interaction energy between two stationary pair production in the Aharonov–Bohm potential [14] and so point-like charges is calculated exactly in terms of Meijer-G on. functions. Effects induced on a hydrogen atom by the vacuum The vacuum polarization can be modified by the presence polarization in the vicinity of a Dirac string are considered. of an external field. The most common situations studied in We also calculate the interaction between two parallel Dirac this context are the effects produced by the presence of an strings and corrections to the energy levels of a quantum par- external magnetic field in the Coulomb interaction and in the ticle constrained to move on a ring circumventing a solenoid. hydrogen-like atoms [15–20]. Modifications in the nuclear Coulomb field induced by a strong laser field were also considered in the literature [21]. 1 Introduction The vacuum polarization was also studied in coordinate space in reference [22], where the Green’s function was cal- Since the establishment of the QED, the effects regarding culated in higher orders beyond the Uehling term. the vacuum polarization had been drawing attention mainly In this paper we study some effects produced by the vac- in situations with no counterpart in classical electrodynam- uum polarization of the fermionic field. We focus on sit- ics. In this context, we can mention the Uehling potential uations with no counterpart in classical electrodynamics, [1], which is the electromagnetic potential associated with a studying setups where sources for the electromagnetic field single point-like stationary charge corrected in lowest order can interact via the vacuum polarization. In Sect. 2 we start in the fine structure constant. The calculation of this poten- by studying the standard interaction between two point-like tial is usually found in the literature perturbatively in the steady charges corrected by the vacuum polarization, in low- momentum space for a point-like source [2–5] and was also est order in the fine structure constant. We calculate exactly generalized for a charge distribution with finite radius [6]. the Uehling interaction between them. Our result has two Recently, the Uehling potential was calculated exactly in main advantages: it is an exact result (in lowest order in the terms of Bessel Integral functions [7]. In the context of non fine structure constant) valid for any distance between the relativistic quantum mechanics, the Uehling potential can charges and the result is given by a simple expression writ- lead, for instance, to effects on hydrogen-like atoms [8]. ten in terms of a K-Bessel functions and Meijer-G functions, Atomic effects of QED not accounted by Uehling potential what makes it easier to be plotted, once those functions are were also studied in the literature [9]. well known in the literature. Another interesting scenario created by the vacuum polar- In Sect. 3 we find out the field produced outside a Dirac ization, with no counterpart in classical electrodynamics, is string due to the vacuum polarization, in lowest order in the related to the effects which emerge around solenoids and fine structure constant. We show that we have a magnetic field Dirac strings. In this context we can mention, for instance, the outside the string anti-parallel to the internal magnetic flux and we calculate the corrections in order α (the fine structure constant) in the energy levels of a quantum particle con- a e-mail: [email protected] strained to move on a ring (2-D quantum rigid rotor). In Sect. b e-mail: [email protected] 4 we show that a hydrogen atom, in its ground state, interact c e-mail: [email protected] 123 12 Page 2 of 9 Eur. Phys. J. C (2018) 78 :12 with the string via a kind of Zeeman effect. This interaction Substituting the Fourier integrals for the field configura- falls down very quickly when the distance between the atom tion and for the external source, and the string increases. This force is attractive when the total 4 μ d p μ angular momentum of the electron is parallel to the internal A (x) = A˜ (p)e−ipx, M (2π)4 M magnetic flux of the string, and repulsive in the opposite case. 4 μ d p μ − In Sect. 5 we show that it emerges an interaction between J (y) = J˜ (p )e ip y, (7) two Dirac strings due to the vacuum polarization of the (2π)4 fermionic field. We calculate this interacting force exactly, and the integral (4)inEq.(6) and using the fact that α up to order , for any distance between the strings when d4 yei(p−p )y = (2π)4δ4(p − p), we can show that they are parallel to each other. Section 6 is devoted to some comments and final remarks. ˜μ ( ) = ˜ ( ) ˜μ( ). AM p DM p J p (8)

With the aid of Eqs. (8) and (5) we can write the Fourier 2 Uehling interaction transform of the external source as a function of the Fourier transform of the gauge ﬁeld obtained from the Maxwell elec- In this section we calculate exactly the Uehling interaction trodynamics, between two point-like steady charges.

The gauge sector of the classical electrodynamics can be 2 μ p μ described by the Lagrangian J˜ (p) =− A˜ (p). (9) 4π M 1 μν μ 1 μ 2 L =− Fμν F − Jμ A − (∂μ A ) (1) μ( ) = μ( ) 16π 8π For a steady external source, J x J x , one can show that the energy stored in the electromagnetic field is where the last term is a gauge fixing one, Aμ is the electro- given by μν μ ν ν μ magnetic field, F = ∂ A − ∂ A is the field strength μ and J is the external source. 3 4 1 μ EM = d xd y Jμ(x)DM (x − y)J (y). (10) From the Lagrangian (1) one obtain the dynamical equa- 2 tion It is well known in the literature that the net QED effects ∂μ∂ ν = π ν μ AM 4 J (2) of the fermionic vacuum bubbles can be taken into account by a correction in the gauge field propagator [2,5], as follows for which the corresponding propagator DM (x − y) satisfies the differential equation 4π D˜ (p) = D˜ (p) 1 + (p) =− 1 + (p) . (11) M p2 μ 4 ∂μ∂ DM (x − y) = 4πδ (x − y). (3) where The sub-index M in (3) means that we have the quantities 1 calculated for the Maxwell theory. α v2 1 − 1 v2 ( ) =− v 3 , The solution for (3) is given by the Fourier integral in the p d 2 (12) μ π v2 + 4m − 1 four momentum p , 0 p2 4 α d p ˜ −ip(x−y) with standing for the fine structure constant and m,the DM (x − y) = DM (p)e , (4) (2π)4 mass of the electron. The corrected propagator for the electromagnetic field is where given by the Fourier integral 4π ˜ ( ) =− 4 DM p (5) d p − ( − ) p2 D(x − y) = D˜ (p)e ip x y . (13) (2π)4 is the Fourier transform of the propagator DM (x − y) The solution for the field equation (2) is given by the inte- In the presence of an external source, the corrected field gral configuration is μ ( ) = 4 ( − ) μ( ). μ( ) = 4 ( − ) μ( ). AM x d xDM x y J y (6) A x d xD x y J y (14)

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Substituting the second Eqs. (7) and (13)in(14), we can The right and side of Eq. (20) is the same as the one found show that in the calculations of the Uehling potential [2]. We shall cal- 4 culate it exactly in this section. μ d p ˜ ˜μ −ipx A (x) = D(p)J (p)e The first term on the right hand side of Eq. (20)is (2π)4 the coulombian interaction between the charges. This well 4 d p μ − = D˜ (p) 1 + (p) J˜ (p)e ipx. (15) known result [3,23–27] is obtained from EM (the interac- ( π)4 M 2 tion energy given by the Maxwell electrodynamics) and the where, in the second line, we used Eq. (11). propagator defined in (5) and (4). The integral in the second With the aid of Eq. (8), we can rewrite Eq. (15)intheform term can√ be calculated by changing the integration variables q = p 1 − v2, as follows 4 μ d p μ − ( ) = ˜ ( ) + ( ) ipx · A x AM p 1 p e 3 ip a (2π)4 d p e = 1 / 4 (2π)3 p2(1 − v2) + 4m2 (1 − v2)3 2 μ d p μ − = A (x) + A˜ (p)(p)e ipx. (16) √ M ( π)4 M 3 iq·a/ 1−v2 2 × d q e . ( π)3 2 + 2 In the second line of Eq. (16), the second term on the right 2 q 4m hand side can be interpreted as a correction due to the vacuum (21) μ A (x) polarization for the classical field configuration M . 3 The energy stored in the electromagnetic field due to the Now we perform the integral in d q with the results of ref- μ( ) erence [23]. The result is presence of a static external source J x , taking into account 3 ip·a the corrections imposed by the fermionic vacuum bubbles, d p e = 1 1 1 is given by ( π)3 2( − v2)+ 2 π − v2 2 p 1 4m 4 1 a ma × − √2 3 4 1 μ exp (22) E = d xd y Jμ(x)D(x − y)J (y) (17) 1 − v2 2 Substituting the result (22)inEq.(√20), performing the Substituting Eqs. (13)in(17), using expression (11), change of integration variable u = 1/ 1 − v2 and making 0 0 integrating out in dy and in dp and using the fact that some simple manipulations, we have 0 ip0 y0 0 dy e = 2πδ(p ),wehave ∞ q q q q α 1 3 E = 1 2 + 1 2 du√ 1 3 3 d p μ ˜ 0 π 2 E = EM + d xd y Jμ(x)J (y)DM (p a a u − 1 2 (2π)3 1 = , )( 0 = , ) ip(x−y). 2 1 1 − 0 p p 0 p e (18) × − − e 2mau 3 3u2 3u4 Notice that Eq. (18) is the energy obtained from the ∞ q1q2 α 1 Maxwell electrodynamics, EM , with a correction term added. = 1 + du√ Now, let us consider the external source produced by two a 3π u2 − 1 1 stationary charges, q1 and q2, placed at positions a1 and a2, 1 1 − respectively, × 2 − − e 2mau . (23) u2 u4 μ( ) = ημ0δ3( − ) + ημ0δ3( − ). J x q1 x a1 q2 x a2 (19) All integrals in Eq. (23) can be performed exactly. The first one is given by Substituting Eqs. (19), (12) and (5)in(18), discarding the terms of self energies (the ones corresponding to the inter- ∞ 1 −2mau actions of a given charge with itself) and performing some du√ e = K0(2ma), (24) u2 − 1 simple manipulations, we can write 1

where K (2ma) is the K-Bessel function of second kind 1 0 3 ip·a q1q2 1 d p e [28], and the other two ones are E = + 4q q α dvv2 1 − v2 , a 1 2 3 (2π)3 p2(1 − v2) + 4m2 0 ∞ 1 1 1 (20) du√ e−2mau = (ma)3 u2 − 1 u2 2 1 where we deﬁned the vector a = a1 − a2 and its corresponding modulus a =|a|. × MeijerG [[], [0]], [[−1/2, −1, −3/2], []],(ma)2 , 123 12 Page 4 of 9 Eur. Phys. J. C (2018) 78 :12

∞ 1 1 1 du√ e−2mau = (ma)5 u2 − 1 u4 2 1 × MeijerG [[], [0]], [[−1/2, −2, −5/2], []],(ma)2 , (25) with MeijerG standing for the G-Meijer functions [29]. Substituting the result (24)inEq.(23) we have ﬁnally the Uehling interaction

q q α ( ) E = 1 2 1 + 2K (2ma) Fig. 1 Graphic for the function g ma , which gives the induced vac- a 3π 0 uum charge around a point-like electric charge 1 − (ma)3MeijerG([[], [0]], [[−1/2, −1, internal magnetic flux . Its corresponding source is given 2 by [30–34] −3/2], []],(ma)2) μ ( ) J(D) x⊥ −1( )5 ([[], [ ]], [[− / , − , ma MeijerG 0 1 2 2 4 2 d p μ ν − = i (2π)2δ(p0)δ(p3)ε0 p e ipx. (29) (2π)4 ν3 −5/2], []],(ma)2) . (26) If >0 we have the internal magnetic field pointing in the ˆ < Notice that expression (26) is exact, (up to order α) valid z direction. For 0, the internal magnetic field points ⊥ for any distance a. in the opposite direction. The sub-index means we are It is usual to interpret the result (26) as a coulombian taking just the components of a given vector perpendicular to the string. For instance, p⊥ = (px , py, 0) is the momentum interaction between a test charge q2 and an effective one given by perpendicular to the string. From Eq. (29) we can identify the Fourier transform of the Dirac string source qef f = q1 1 + g(ma) . (27) μ μ ν J˜ (p) = i (2π)2δ(p0)δ(p3)ε0 p . (30) where we defined the function (D) ν3

α Substituting the source (29) in expression (6), with the g(ma) = 2K0(2ma) 3π Maxwell propagator (4), we have the four-potential −1( )3 ([[], [ ]], [[− / , − , ma MeijerG 0 1 2 1 μ y x 2 A (x) = 0, − , , 0 , (31) M(D) π 2 + 2 2 + 2 −3/2], []],(ma)2) 2 x y x y 1 − (ma)5MeijerG([[], [0]], [[−1/2, −2, as expected for a Dirac string. The potential (31) produces a 2 vanishing electromagnetic field outside the z axis.1 −5/2], []],(ma)2) (28) The vacuum polarization effects in the vicinity of a Dirac string can be obtained by substituting, in the solution for field In the Fig. 1 we can see a plot for the function (28). It gives configuration (15), the definitions (5) and (12) and the source the charge distribution induced in the fermionic vacuum by for the Dirac string (30). The result is the presence of a point-like stationary electric charge. For μ ( ) = μ ( ) + μ ( ) ma → 0 the function g(ma) diverges as ln(ma) [2,5]. When A(D) x AM(D) x A(D) x (32) ma →∞the function g(ma) goes to zero as e−2ma/(ma)3/2 μ ( ) [2,5]. where AM(D) x is given by (31) and we defined 1 4 π α v2 1 − 1 v2 μ ( ) = d p 4 v 3 A(D) x d 2 3 Field of a Dirac string (2π)4 p2 π v2 + 4m − 1 0 p2 × ( π)2δ( 0)δ( 3)ε0μ ν −ipx. In this section we discuss some effects of the vacuum polar- i 2 p p ν3 p e (33) ization in the vicinity of a Dirac string. We choose a coordinate system where the Dirac string lies along the z-axis with 1 Along the z axis, the magnetic field in infinity. 123 Eur. Phys. J. C (2018) 78 :12 Page 5 of 9 12

Integrating out expression (33)inp0 and p3 and noticing that just its 1 and 2 components are non-vanishing, we can write 1 2 μ d p⊥ A (x⊥) = 4α dv (D) (2π)2 0 2 1 2 1 v 1 − v · − 3 (0, ip2, −ip1, 0)eip⊥ x⊥ 2 v2 − 4m2 − p⊥ 2 1 p⊥ 1 ∂ ∂ v2 1 − 1 v2 = 4α 0, , − , 0 dv 3 ∂ ∂ − v2 y x 1 Fig. 2 Graphic for the function f (mρ), which gives the induced mag- 0 · netic field outside a solenoid d2p⊥ eip⊥ x⊥ × . (34) (2π)2 p2 + 4m2 ⊥ 1−v2 The function (39) is always positive, as one can see in The integral above is calculated in reference [23] the Fig. 2, so the magnetic field (38) points in the opposite direction in comparison with the magnetic flux of the string.

· For small and long distances from the string, the magnetic d2p⊥ eip⊥ x⊥ 1 2m|x⊥| = K0 √ , (35) ﬁeld (38) reads (2π)2 2 + 4m2 2π − v2 p⊥ −v2 1 1 4α 1 B =∼ − zˆ, mρ<<1 3π ρ2 where K0(x) stands for the Bessel function of second kind. e−2mρ So, acting with the derivatives, performing the change of B =∼ −α , mρ>>1. (40) integration variable ρ2

1 3.1 The 2-D quantum rigid rotor ξ = √ , (36) 1 − v2 The vector potential (37) induces a modification in the Aharonov–Bohm bound states. It can be seen by taking a and using the cylindrical coordinates, with ρ =|x⊥|= x2 + y2, we can write the vector potential (34)intheform very simple example of a 2-dimensional quantum rigid rotor composed by a non-relativistic quantum particle of mass M ∞ 23α m ξ 2 − 1 constrained to move along a circular ring surrounding the A(D)(x⊥) = dξ Dirac string. Let us take the ring on the plane z = 0 centered 3π ξ at the origin with radius b. 1 1 As stated in Eq. (32), the total vector potential produced × 1 + K (2mξρ)φ,ˆ (37) 2ξ 2 1 by the string is composed by the one obtained from Maxwell theory (31) added by the correction (37). Performing the inte- with φˆ standing for the unitary vector for the azimuthal coor- gral in Eq. (37), we can write dinate. α The potential (37) is static, has vanishing zero component 4 ˆ A(D) = 1 + F(mρ) φ, (41) and does not produce any electric field. Its rotational gives 2πρ 9 a magnetic field outside the solenoid. It is simpler, first, to calculate the relevant derivatives for the rotational and, after, where we defined the function ξ perform the integration over the variable. The result is, ( ) = ( 2 + 4) 2( ) − ( 2 + 4) 2( ) F x 3x 4x K0 x 5x 4x K1 x + (6x + 4x3)K (x)K (x) (42) 4α m2 0 1 B =∇× A(D) =− f (mρ)zˆ, (38) 3π which is always positive, as one can see in the Fig. 3. It is well known in the literature [35,36] that the energy where we defined the function levels of a two dimensional quantum rigid rotor are modified when it circumvents an infinite solenoid. In this case we f (x) = K 2(x)(1 + 2x2) − 2xK (x) K (x) + xK (x) , 1 0 1 0 have a very simplified version of the so called Aharonov– (39) Bohm bound states [36]. In this section we consider the 123 12 Page 6 of 9 Eur. Phys. J. C (2018) 78 :12

h¯ 2 q 2 E =∼ n − 2 π ¯ 2Mb 2 h hq¯ α − q − e 2mb n − , mb >> 1. (46) 2Mmb3 2πh¯

4 String-atom interaction

The external magnetic field created by a Dirac string can lead to physical phenomena. Let us consider some of its effects produced on a single hydrogen atom. For this task, we take Fig. 3 Graphic for the function F(mρ) a coordinates system where the atom is placed at the origin and the Dirac string, parallel to the z axis, along the line modification introduced by the vacuum polarization in the (d, 0, z). In this setup, the Dirac string is placed at a distance energy levels of a quantum rigid rotor. For this task, we take d from the atom. We shall restrict to the situation where d is a quantum rigid rotor composed by a particle with mass M much higher in comparison with the atomic distances. The and electric charge q, restricted no move along a ring of magnetic field produced in this case can be written by shifting radius b. We adopt a coordinate system where the ring lies the coordinate x in Eq. (38), as follows on the plane z = 0, centered at the origin. We also consider 2 4α m a Dirac string placed along the z axis, with internal magnetic B =− f m (x − d)2 + y2 zˆ 3π flux . In this case, the Hamiltonian for the charged particle 4α m2 reads =∼ − f (md)zˆ, (47) 3π h¯ 2 d2 ihq¯ 4α d H =− + 1 + F(mb) where, in the last line, we used the fact that the coordinates x 2Mb2 dφ2 2π Mb2 9 dφ >> , and y are evaluated in the atomic distances and d x y. q2 2 4α 2 In a typical experiment d is a macroscopic distance in + 1 + F(mb) , (43) 8π 2b2 9 order of centimeters, what makes the values of the product md very large. So it is legitimate to approximate expression where F(mb) is defined by (42). (47)formd >> 1, as follows The energy eigenfunctions of the hamiltonian (43)are 4α m2 3π e−2md given by B =∼ − zˆ 3π 4 (md)2 −2md inφ e ψ(φ) = Ae , (44) =∼ −α zˆ. (48) d2 where n = 0, ± 1, ± 2,...is any integer and A is a normal- So, in this setup, the Dirac string produces a kind of Zee- ization constant. Up to order α, the corresponding energy man effect on the atom, once the field (48) is, approximately, levels are constant and uniform along the atom. In this regime, the field produced by the Dirac string is lower than the atomic magnetic field, so we shall use the results of the Zeeman effect h¯ 2 q 2 2hq¯ α q −ˆ E = n − − F(mb) n − . (with external magnetic field pointing along the z direc- 2Mb2 2πh¯ 9π Mb2 2πh¯ tion) for weak external field approximation to study the sys- (45) tem, where the fine structure constant dominates the energy corrections [37]. We shall take the hydrogen atom on its The first term on the right hand side of (45)isthewell ground state (n = 1, = 0, m = 0, j = s = 1/2). In known Aharonov–Bohm energy [35] and the second term is this case, the degeneracy is broken as follows a correction due to the vacuum polarization. α2 For small and large values of mb, the energy (45) reads E = E0 1 + − 2m j μB| B | (49) 4 h¯ 2 q 2 E =∼ n − 2Mb2 2πh¯ where E0 is the non-perturbed ground state energy of the hydrogen atom, μ is the Bohr magneton and m =±1/2is 4hq¯ α q B j + ln(mb) n − , mb << 1 the azimuthal quantum number for the total angular momen- 3π Mb2 2πh¯ tum (for the ground state, m j = ms). The minus signal for 123 Eur. Phys. J. C (2018) 78 :12 Page 7 of 9 12

strings parallel to each other with the ﬁrst one lying along the z axis and the second one lying along the line (a1, a2, z). Deﬁning the vector a = (a1, a2, 0), we can identify the distance between the strings by the modulus of a, | a |= (a1)2 + (a2)2.FromEq.(29), we can write the sources for the two strings, as follows

4 μ d k 2 0 3 0μ ν −ikx J (x⊥) = i (2π) δ(k )δ(k )ε k e (D,1) (2π)4 1 ν3 4 μ d k 2 0 3 0μ ν −ik x −ik ⊥·a J (x⊥) = i (2π) δ(k )δ(k )ε k e e (D,2) (2π)4 2 ν3 Fig. 4 Graphic for the modulus of the force (51) multiplied by (52) 3 −1 |2m j α μB m | as a function of md. Expression (51) is valid just for large values of md Substituting the sources (52)inEq.(18), discarding the self energies of each string, noticing that EM = 0 (there is no the Zeeman contribution is due to the fact that the external interaction energy between two Dirac strings in the Maxwell magnetic field (48) points along the −ˆz direction. Electrodynamics), integrating out in dy0, dp0, dk0, dk0, 3 3 3 3 As usual, in this case, the contribution of the nuclear mag- dk , dk , dx and dp , identifying the length of a Dirac netic moment (proton) is not relevant [37]. string L = dy3 and performing some simple manipula- Substituting Eq. (48)in(49)wehave tions, we have the interaction energy between the strings per unit of length α2 −2md e E = E0 1 + − 2m j α μB . (50) 2 4 d2 E d p⊥ ˜ 0 3 E = = D p = p = 0, p⊥ 1 2 ( π)2 M L 2 The energy (50) exhibits a dependence on the distance d × 0 = 3 = , ε0μ ε0 α β ip⊥·a between the Dirac string and the atom and produces a force p p 0 p⊥ α3 μβ3 p p e on the atom (taking the string as fixed) given by 2 d p⊥ ˜ 0 3 =− DM p = p = 0, p⊥ ∂ E e−2md(md + 1) 1 2 (2π)2 F =− =−4m j α μB ∂ 3 0 3 2 ip⊥·a d d ×(p = p = 0, p⊥)p⊥e . (53) e−2md =∼ − 2m α μ (51) j B d3 Now we define the differential operator where, in the last line, we discarded a term of order md. = / ∂ ∂ When m j 1 2, the projection of the total electronic ∇a = , , 0 , (54) angular momentum of the electron points in the same direc- ∂a1 ∂a2 tion of the internal magnetic flux of the string and the force 2 ip⊥·a =−∇2 ip⊥·a (51) becomes negative, what means that it exhibits an attrac- use the fact that p⊥e a e and substitute the tive nature. When the total angular electronic momentum definitions (5) and (12)inEq.(53), what leads to points in the opposite direction, with respect to the internal 1 2 1 2 magnetic flux, the force (51) is positive, what means a repul- v 1 − v 2 ip⊥·a 3 2 d p⊥ e sive behavior. E = 4α 1 2 dv ∇ . − v2 a ( π)2 2 4m2 1 2 p⊥ + In a scattering experiment with a beam of unpolarized 0 1−v2 atoms propagating in the vicinity of a solenoid, we must (55) have a bifurcation of the beam, according to the total angular ∇2 momentum of each atom in the beam. Using the result (35), acting with the operator a and per- From the Fig. 4 we can see the behavior of the magnitude forming the change of integration variables (36), the energy of the force (51)asafunctionofmd. per unit length (55) reads

∞ 16 2 2 5 The interaction between two strings E = α 1 2m dξ ξ − 1 3π 1 In this section we study the interaction between two Dirac 1 × 1 + K (2m|a|ξ). (56) strings due to the vacuum polarization. We take two Dirac 2ξ 2 0 123 12 Page 8 of 9 Eur. Phys. J. C (2018) 78 :12

due to the presence of field sources for the electromagnetic field. All these effects were obtained from the vacuum polarization tensor of the QED. We obtained an exact expression for the Uehling interaction (Uehling potential) between two point-like steady charges in terms of a K-Bessel function and MeijerG functions. Our results are compatible with that ones obtained in the literature, approximately, for long and small distances between the charges. We also investigated some phenomena produced outside a Dirac string. One of these effects are the modifications induced in the energy levels of a quantum rigid rotor which circumvents a Dirac string. We have calculated these modi- 8 α 3 Fig. 5 Graphic for the force (58)dividedby 3π 1 2m , which gives fications exactly (for any radius of the quantum rigid rotor) the induced vacuum charge around a point-like electric charge up to order α. We have shown that a hydrogen atom interact with a Dirac string via a kind of Zeeman effect. The The integral in Eq. (56) can be calculated. The final result nature (attractive or repulsive) of this interaction depends on for the energy is the orientation of the total magnetic moment of the electron with respect to the internal magnetic flux of the string. We 4 E = α m2 K 2(ma) + (ma)2 restricted to the case where the atom is in its ground state and π 1 2 1 1 3 far away from the string. When the total magnetic moment − ( )2 2( ) − ( ) ( ) ( ) 2 ma K0 ma 2 ma K0 ma K1 ma (57) of the electron is parallel to the internal magnetic flux of the string, the interaction is attractive, on the contrary, it is The gradient of the energy (57) with respect to a (with a repulsive. As expected, this interaction is very small and falls minus signal) gives the force between the strings per unity down very fast when the distance between the string and the of length atom increases. F =−∇ E = 8 α 2 1 2( ) − ( )2 We have also investigated the interaction which emerges a 1 2m K1 ma 1 ma 3π a between two Dirac strings due to the vacuum polarization. We restricted to the case where the strings are parallel or + maK0(ma) maK0(ma) + K1(ma) aˆ, (58) anti-parallel to each other. We showed that the strings attract which is repulsive when 1 and 2 have the same sign and each other when their internal magnetic fluxes are anti- attractive when the magnetic fluxes have opposite signs. In parallel. This interaction is repulsive when the internal mag- Eq. (58), aˆ stands for the unit vector in the direction of a and netic fluxes are parallel to each other. We have computed the function inside brackets is positive along 0 ≤ ma ≤∞. this interacting force exactly (up to order α), showing that it Figure 5 shows the graphic for the modulus of the force falls down vary fast when the distance between the strings 8 α 3 increases. (58) divided by 3π 1 2m as a function of ma. For small and large values of ma, the behavior of the force (58)is Acknowledgements F.A. Barone thanks to CNPq (Brazilian agency) 8 1 F =∼ α aˆ, ma << 1 under the Grant 311514/2015-4 for financial support. 3π 1 2 a3 Open Access e−2ma This article is distributed under the terms of the Creative ∼ Commons Attribution 4.0 International License (http://creativecomm F = 2α 1 2m aˆ, ma >> 1. (59) a2 ons.org/licenses/by/4.0/), which permits unrestricted use, distribution, As a last comment, we highlight that in some other gauge and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative theories, even without radiative corrections, we can have Commons license, and indicate if changes were made. interactions between Dirac strings and other sources for the Funded by SCOAP3. gauge field, what is the case of Lee–Wick electrodynamics [30] and theories with explicit Lorentz symmetry breaking [31,32]. References

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