Eur. Phys. J. C (2015) 75:89 DOI 10.1140/epjc/s10052-015-3305-0

Regular Article - Theoretical Physics

Spontaneous Lorentz violation: the case of infrared QED

A. P. Balachandran1,a, S. Kürkçüoˇglu2,b,A.R.deQueiroz3,4,c, S. Vaidya5,d 1 Physics Department, Syracuse University, Syracuse, NY 13244-1130, USA 2 Department of Physics, Middle East Technical University, 06800 Ankara, Turkey 3 Instituto de Física, Universidade de Brasília, Caixa Postal 04455, Brasília, DF 70919-970, Brazil 4 Departamento de Física Teórica, Facultad de Ciencias, Universidad de Zaragoza, 50009 Zaragoza, Spain 5 Centre for High Energy Physics, Indian Institute of Science, Bangalore 560012, India

Received: 8 January 2015 / Accepted: 4 February 2015 / Published online: 24 February 2015 © The Author(s) 2015. This article is published with open access at Springerlink.com

Abstract It is by now clear that the infrared sector of The group of automorphisms of A generated by conjuga- quantum electrodynamics (QED) has an intriguingly com- tion using unitary elements of A is the group InnA of inner plex structure. Based on earlier pioneering work on this sub- automorphisms. It is a normal subgroup of AutA . The group ject, two of us recently proposed a simple modification of AutA /InnA is the outer automorphism group OutA . QED by constructing a generalization of the U(1) charge The automorphisms generated by some global symmetries group of QED to the “Sky” group incorporating the well- may be equivalent to inner ones. If that is not the case, then known spontaneous Lorentz violation due to infrared pho- they define non-trivial elements of OutA . tons, but still compatible in particular with locality (Bal- In QFT, we choose an irreducible representation ρ or a achandran and Vaidya, Eur Phys J Plus 128:118, 2013). It superselection sector of A . It can happen that a global sym- was shown that the “Sky” group is generated by the algebra metry transformation cannot be implemented by a unitary or of angle-dependent charges and a study of its superselec- antiunitary operator in the representation space H of A .In tion sectors has revealed a manifest description of sponta- that case, the symmetry is said to be spontaneously broken. neous breaking of the Lorentz symmetry. We further elab- Since elements of A act by definition on H ,InnA cannot be orate this approach here and investigate in some detail the spontaneously broken. So that can happen only if S ∈/ InnA . properties of charged particles dressed by the infrared pho- Spontaneous breaking by a Higgs field can be understood tons. We find that Lorentz violation due to soft photons may in this framework. Thus no local operation can change the be manifestly codified in an angle-dependent fermion mass, asymptotic expression φ∞ of the Higgs field φ. Hence φ∞ is modifying therefore the fermion dispersion relations. The a label for the representation ρ. If a symmetry changes φ∞,it fact that the masses of the charged particles are not Lorentz changes the representation. Hence it is spontaneously broken. invariant affects their spin content, and time dilation formulas The mechanism of spontaneous breaking can be illustrated for decays should also get corrections. even with the 3 × 3 matrix algebra M3(C). In its irreducible representation ρ, which is three-dimensional, its unitary sub- group U is represented irreducibly by a representation we 1 Introduction 3 can name as 3. Complex conjugation is an anti-linear auto- (C) In quantum field theory (QFT), observables are local and morphism of M3 . It changes 3 to the inequivalent repre- ¯ ρ ρ¯ generate the algebra of local observables A . In contrast, time sentation 3 and hence to the inequivalent representation . evolution and global symmetries are not local. In Lagrangian This anti-linear automorphism is thus spontaneously broken ρ field theories, their generators involve integrals of fields over in the representation . all of space. They are thus not elements of A , but instead The Poincaré group is an automorphism group of the generate automorphisms of A . They are elements of the auto- observables of quantum electrodynamics (QED). It trans- A morphism group AutA of A . forms elements of nontrivially. Instead, the electric charge Q, at least classically, is the total electric flux at infinity and a e-mail: [email protected] hence commutes with all elements of A . Its different values b e-mail: [email protected] q go into the labels for the different irreducible representa- c e-mail: [email protected] tions of A . They are similar to the Casimir operators of Lie d e-mail: [email protected] algebras. Since Q is Poincaré invariant, Poincaré symmetry 123 89 Page 2 of 11 Eur. Phys. J. C (2015) 75 :89  is compatible with charge superselection. The latter does not F ∧ dα (3) cause spontaneous Poincaré violation. In QED, infrared photons accumulate at infinity and create which is close to the standard Chern–Simons term. In order a non-zero electromagnetic field fμν there as nicely shown by  to see this, we can interpret A in A ∧ F as dα. Buchholz [1,2] and Fröhlich et al. [3,4] (see also R. Haag’s Section 3 reviews the proposed mass twist introduced in book [5]). Since local operations cannot affect fμν, this field [8]. We establish that the twist is compatible with locality. also labels superselection sectors. But fμν is not Lorentz Still it is affected by the infrared cloud at S2 . Thus inertia is invariant. Therefore Lorentz symmetry is spontaneously bro- ∞ affected by the “vacuum” as in Mach’s principle. In a simi- ken in QED. lar manner, the Higgs field at infinity induces vector meson The group U(1) of QED is based on electric charge which −→ masses. Section 4 briefly indicates the changes in our con- classically is a measure of the net flux of electric field E siderations for QED . In particular, in this case the ω in the = 2 3 (Ei f0i ) on the sphere S∞ at infinity. In previous work, vertex operator is a closed one-form, ω = dα on R2,1. Sec- we extended U(1) to the “Sky” group Gsky which is sensitive 2 tion 5 summarizes the part of Fröhlich et al. [4] helpful in the to all the partial waves of E on S∞. It is superselected because construction of ω. In Sect. 6, we examine the modified dis- G / ( ) f0i is. But elements of sky U 1 are not Lorentz invariant persion relations of charged particles which violate Lorentz and cause spontaneous Lorentz violation. 2 invariance. It seems possible to measure this violation. There In quantum theory, the infrared cloud on S∞ is incorpo- are also similar violations in scattering amplitudes which are rated in the state vectors, the charged states being dressed by a also in principal measurable. coherent state of photons [3,4,6,7]. We have constructed the For related recent work with emphasis on BMS group; see ω dressing operator in [8]usingaclosedform . Previous work [10–13]. [3,4] determined the coherent state of the photon and hence Extensive work on Lorentz violation using effective ω . Thus we can write the dressed charged particle states. A Lagrangians has been done by Kostelecky and collabora- simple twist of the electron (or charged particle) mass is sen- tors [14–16]. A study of the relation of their approach and sitive to the coherent state. The twisted mass is not Lorentz ours is yet to be done. Up-to-date experimental constraints invariant. It affects the dispersion relation, the spin content on Lorentz violation may be obtained from the data tables of the particle and time dilatations in decays and life times. in [17]. An easily accessible review on both terrestrial and These can be measured. Naturally it also leads to Lorentz astrophysical constraints on Lorentz violation may be found non-invariant scattering amplitudes in charged sectors which in [18]. too can be measured. In Sect. 2, we recall the vertex operator dressing the charged vector states with a cloud of infrared photons. It 2 The vertex operator for QED4 coherent states is gauge, but not in general Lorentz invariant. A new result we describe is its incorporation in the Lagrangian. We can 2.1 Preliminaries set it equal to 1 by adding the term proportional to F ∧ ω, The gauge transformations generated by the Gauss law will be denoted as the group G∞, the subscript 0 denoting that it : ,ω: , 0 F electromagnetic field closed two-form (1) is connected to identity. Its elements g approach the identity −→ to the action. It is remarkably close to the θ-term when its spatial argument x approaches infinity,  θ ∞ −→ −→ Tr F ∧ F, F : curvature two-form, (2) G =g : R3 → U(1) : g( x ) → e as | x |→∞. 8π 2 0 (4) in QCD, which induces P and T violation. Just like the θ- term, (1) is a surface term and does not affect the equations Its Lie algebra is spanned by of motion.  3 −→ By QED, we always mean QED in (3 + 1)-dimensional G( ) = d x (−Ei ∂i + J0)( x ), (5) spacetime, or QED4. But we can also consider QED3 in ( + ) 2 1 -dimensional spacetime where analogous consider- where Ei is the electric field, J0 is the charge density, and ations regarding infrared photons apply [9]. In that case, the ω (−→) → |−→|→∞, . . ∈ C∞, dressing operator involves a closed one-form , so that on x 0as x i e 0 (6) R2,1, ω = dα, with α a scalar function. The term in the action which absorbs the vertex operator from state vectors where the subscript 0 indicates compact support and super- is proportional to script ∞ indicates infinite differentiability, both of which are 123 Eur. Phys. J. C (2015) 75 :89 Page 3 of 11 89 standard notations. The Gauss law is imposed in the theory where by requiring that G( ) vanishes for any choice of the test  ∈ C∞ G∞ →{1} χ ∞(ˆ) = χ ∞ (ˆ). function 0 so that 0 on quantum states. n lmYlm n (14) Another manner to state this requirement is the follow- lm ing. If C is the space of connections and G∞ acts on C by 0 Q(χ ∞Y ) gauge transformations, the principal bundle for gauge theo- All the lm lm commute among themselves. Hence each ries without matter is C /G∞, that is, the quotient of C by of them generates a one-dimensional abelian group. 0 G G∞ Suppose we are given a standard representation of sky, the action of 0 . The generators Q (χ) of the charge group also come from ∞ 0 Q(χ)|· = χ Q |·. (15) the Gauss law, 00 0  Then the vertex operator (χ) = 3 (− ∂ χ + χ )(−→), Q0 d x Ei i J0 x (7)  ∧ω V (ω) = ei A , (16) but now χ is required to approach a constant at infinity, ω −→ −→ where is a closed two-form, χ( x ) → χ∞ as | x |→∞. (8) dω = 0, (17) The charge group they generate will be denoted by G.The G∞ G G∞ →{1} |· |· group 0 is normal in . Since 0 on quantum maps to the infrared-dressed states ω as we show below: G/G∞ ≈ ( ) states, the effective group is 0 U 1 . The normalized  i A∧ω charge is just |·ω = e |·0, (18)

|·0 ≡|·. (19) Q0 := Q0(χ)|χ∞=1. (9) −→ But first note that V (ω) is “gauge invariant”, that is, that it For QED with spatial slice R3, we can choose χ( x ) = 1 ∞ −→ commutes with G , for all x conveniently since G( ) → 0 in quantum theory. 0      That gives the standard expression  G( ), A ∧ ω = d ∧ ω =− dω = 0, (20) 3 Q0 = d xJ0 (10) | = since S∞2 0. (ω)|· for the charge. Hence V is a Gauss law compatible vector (assum- |· The Sky group G has generators ing that is the case with ). sky (χ) (ω) ω∞  But Q does not commute with V .Let be the ω 3 −→ asymptotic expression for , Q(χ) = d x(−Ei ∂i χ + χ J0)( x ), (11) ∞ −→ ω (nˆ) = lim ω(rnˆ), (21) −→ r→∞ ∞ −→ x r→∞ χ( x ) ≡ χ(rnˆ) −→ χ (nˆ), r =|x |, nˆ = −→ (12) | x | while we let χ ∞ be the asymptotic χ as before, χ ∞ 2 where the function on S∞ need not be constant. As a ∞ (χ) χ (nˆ) = lim χ(rnˆ). (22) result, Q need not even be rotationally invariant modulo r→∞ a G( ). Hence Gsky breaks Lorentz invariance. The Sky group acts trivially on the charge zero sector Then with which has no infrared cloud. But in charged sectors, state (χ) = iQ(χ), vectors are twisted by a coherent state vector of the infrared U e (23) cloud, and hence, Gsky can act by a non-trivial representation breaking Lorentz invariance. we obtain We note that G has one-dimensional irreducible rep- sky (χ) (ω) = (χ ∞,ω∞) (ω) (χ), resentations. That is because it is abelian. In fact, since the U V c V U (24) action of Q(χ) on quantum state vectors depends only on ∞ with the central element χ (nˆ), we can write,     (χ) = (χ ∞ ), ∞ ∞ ∞ ∞ Q Q lmYlm (13) c(χ ,ω ) = exp −i ω χ . (25) 2 lm S∞ 123 89 Page 4 of 11 Eur. Phys. J. C (2015) 75 :89

These c’s generate the center of the algebra of U’s and V ’s. That is because the coefficient of ∂0 Ai in the added term Lω This algebra resembles the Weyl algebra. gives the contribution ∗ ωi to the conjugate momentum of Ai . If We observe also that Lω involves ω0i , which are new. The covariant-looking Lω, if it is not to affect the equa- iχ∞ Q U(χ)|· = e 00 0 |·, (26) tions of motion, has to be a total divergence. This means that

|· ( ) where denotes any vector transforming just by U 1 of μνλρ∂νωλρ = 0, (35) charge of Sky, then or ω as a two-form in 4-dimensions must be closed: V (ω)|· = |·ω (27) dω = 0. (36) can transform non-trivially under Sky, Hence in R4, ∞ ∞ U(χ)|·ω = c(χ ,ω )|·ω. (28) ω = dα,ˆ αˆ : a one-form in 4-dimensions. (37) ∞ If ω is not zero, then the sector Hω of the Hilbert space spanned by {|·ω} breaks Lorentz symmetry. The term ijk0 Fijωk0 in Lω may also be obtained by a Lorentz transformation of F0i and ω jk from 0ijkF0i ω jk.The 2.2 Incorporation of vertex operator in Lagrangian presence of ω0i in Lω means that it can in general dress the electron with an infrared cloud containing both electric Let E j denote the electric field. It is conjugate to the potential and magnetic fields. A j : It is natural to identify the asymptotic part of ωμν with Buchholz’s fμν [1]. −→ −→ 3 −→ −→ [Ai ( x ), E j ( y )]=iδijδ ( x − y ) (29) As observed in the introduction, Lω resembles the QCD θ-term. ω = ω k ∧ l at equal times. Now, let us write kldx dx , Note also that ωμν depends on the charged particle. Hence it transforms under CPT in a manner required to maintain −1 ω·|E |· ω = ·|V (ω) E V (ω)|·  j 0  j  0  CPT invariance of (34).

= 0 · E j + (−iQ0) ∗ ωi Ai , E j · 0 3 The twisted mass for QED4 = 0·|E j +∗ω j |· 0, (30)

where ∗ω j = jklωkl. Note that in (30), the ket |·  and bra The infrared cloud leading to Lorentz violation brightens the 2 ·| may represent different vectors. Thus the Hamiltonian of sphere S∞ at spatial infinity. We need a model for observing Fμν, its effects. In setting up this model, we can be guided by the spontaneous symmetry breakdown due to the Higgs field.  1 −→ −→ −→ In general, H = d3x( E 2 + B 2), with B : magnetic field, 2  χ( ˆ) r−→→∞ χ ∞(ˆ) = χ ∞ (ˆ). (31) rn n lm Ylm n (38) fulfills Let χˆ denote those test functions for which the l = 0termis absent, −1 ω·|H|· ω = 0·|V (ω) HV(ω)|· 0 =: 0·|H|· 0, (32)   r→∞ ∞ ∞ −→ −→ χ(ˆ rnˆ) −→χ ˆ (nˆ) = χˆ Y (nˆ). (39) 1 3 2 2 lm lm H = d x( E + B ), Ei = Ei +∗ωi . 2 l =0 (33) If the charged particle state is not dressed by the infra- photons, that is, on states |·ω=0, then The field conjugate to Ai is thus shifted from Ei to Ei for the zero-twist state vectors. We can accomplish this shift by ∞ Q(χ)|· = χ Q |· . (40) adding to the QED Lagrangian density L atermLω = 0 00 0 0 1 μνλρ Fμνωλρ , that is, writing 2 Hence 1 L = L + μνλρ Fμνωλρ := L + Lω. (34) Q(χ)ˆ |· = . 2 0 0 (41) 123 Eur. Phys. J. C (2015) 75 :89 Page 5 of 11 89

It follows that twisting the mass term mψψ¯ of spin-1/2 par- results without proof. Other important work relevant to us on ticles to this coherent state is by Kibble [19–22], Eriksson [6], and Gervais and Zwanziger [7]. m (χ)ˆ − (χ)ˆ m cos Q(χ)ˆ ψψ¯ = (eiQ + e iQ )ψψ¯ := m(χ)˜ ψψ,¯ The results of Fröhlich et al. can be described as follows. 2 In the Coulomb gauge, the free electromagnetic field has the m(0) ≡ m (42) expression ω =  does not change the physics in the sector with label 0 3 −→ −→ −→ −→ −→ −→ ¯ −→ d k i k · x † −i k · x where the mass term is unaffected. Furthermore m(χ)ˆ ψψ Ai ( x ) = (ai ( k )e + a ( k )e ), (48) 2k i is local, since mψψ¯ is local and the twist eiQ(χ)ˆ commutes 0 with local observables by gauge invariance. This property is where independent of ω labeling state vectors. −→ −→ =| |, = , ( ) = . The Weyl relation is still valid. If k0 k A0 0 ki ai k 0 (49) The commutation relations are (χ)ˆ = iQ(χ)ˆ , U e (43) −→ −→ −→ −→ † ˆ ˆ 3 ˆ ki [ai ( k ), a ( k )]=(δij−ki k j )2k0 δ ( k − k ), ki = −→ . j | | then k (50) (χ)ˆ (ω) = (χˆ ∞,ω∞) (ω) (χ).ˆ −→ U V c V U (44) Let | p , e|0 denote the tensor product of a single non- interacting charged particle vector of charge e and momen- ω∞ = −→ Therefore with da, tum p and the photon vacuum, where we have suppressed  spin labels. Then the dressed electron state vector surrounded χ ∞ = . da 00 Y00 0 (45) by its infrared cloud is S∞2 −→ V (ω )| p , e|0, (51) Now, p  −→ m iQ(χ)ˆ −iQ(χ)ˆ ω ω · (e + e ) · where, as indicated, depends on p . 2  ω   Let us write ∞ ∞  = · m cos ω χˆ · . (46) −→ −→ 0 3 i k · x S2 ∗ω (x) = d k ∗ ω (k) e . (52) ∞ 0 p i p i Therefore the effect of the ω-twist is to replace m with Then the Fourier transform ∗˜ωp(k)i is [4]   ∞ ∞ m cos ω χˆ . (47) e −→ −→ ˆ ˆ 2 ∗ωp(k)i = ( p i − p · k ki ). (53) S∞ p · k

∞ ∞ In (47), ω is known (see below), while χˆ is associ- Since ated with the charged particle. It is a new form factor for −1 the charged particle. It is neither rotationally nor Lorentz V (ωp) ai (k) V (ωp) = ai (k) + i(∗ωp(k)i ), (54) invariant. It also depends on the sum of the momenta of all the charged particles in, say, the in- or out-state vector. We we can set V (ω) ={1} and instead replace Ai by σp(Ai ) ∞ interpret χˆ as a new form factor characterizing the charged where, as in [4], particle. Let an experiment measure the mass of the charged particle in the direction xˆ. Then the resultant value for the σp(ai )(k) = ai (k) + i(∗ωp(k)i ). (55) mass is given by (47). The map

4 Calculation of ω σp : ai → σp(ai ) (56)

The vertex operator creates a coherent state of infrared pho- and its adjoint define an automorphism of the algebra of cre- tons which depends on the charged particle state |· on which ation and annihilation operators. 0 −→ it operates. This coherent state has been calculated in a form For N charged particles of momenta p a and charges ea convenient for us by Fröhlich et al. [4]. We will use their (a : 1,...,N), the dressed state vector is 123 89 Page 6 of 11 Eur. Phys. J. C (2015) 75 :89       −→ particle sector by replacing V ( ωp ) by V (ωP = ω ). V ω−→ | p , e |0, (57) a p p a a a As regards Lorentz invariance, we lose no information since a i    ∗ω where the Fourier transform of pa is ω − ω V pa P (66) a ∗ω ( ) = ea (−→ − −→ · ˆ ˆ). pa k p a p a k k (58) pa · k does not induce Lorentz invariance violation. In scattering theory, however, with widely separated parti- Lorentz invariance can be spoiled by the presence of the cles in the in- or out-state vector, the above replacement may second term in (55). For the case of a single charged particle, not be appropriate as it is nonlocal. Instead, it seems best to for Lorentz invariance, if U(1) is the unitary operator for use V ( ωp ) which dresses each charged particle with its a a Lorentz transformation on the quantum Hilbert space, we own infrared cloud. require that If is a rotation, ∗Q(k) vanishes. This suggests that −→ −→ the rotational symmetry is preserved. However, we note that U( )V (ωp)| p , e|0=V (ω p)| p , e|0. (59) composition of two boosts can produce a rotation, suggest- ing trouble with the latter too. This point requires further However, this is not the case. Instead, we have [3,4] clarification.

−1 U( )V (ωp)U( ) = V (ω p + e), (60) 5 Consequences of twisted electron mass where  The consistency of twisting the charged spinor mass to cap- −→ −→ 3  i k · x ture Lorentz breaking raises issues about its effect on locality ∗e(x)i = d k ∗ e(k)i e , (61) and perturbation theory. We tentatively suggest the following e ∗ (k) =− (( −1) − ( −1) kˆ kˆ ). answers. e i ( ) 0i 0 j j i (62) k 0 As we observed, Q(χ)ˆ commutes with local observables. Hence locality seems unaffected. ˆ The unit vector k denotes the direction in which we observe In perturbation theory, we encounter the charged particle the “sky” of the source. Thus Lorentz symmetry is broken. propagator It is remarkable that the Lorentz invariance breaking sec- ond term in (62) has no dependence on the charged particle T (ψ(x)ψ(y)) (67) momentum. As a consequence, in the N charged particle sec- ˆ tor, this term depends only on the total charge Q and k. Thus, in the internal lines. This is a vacuum expectation value and the twisted mass becomes the untwisted one on vacuum. So it     appears that no internal lines are affected. Hence the Lorentz   −1 breaking term affects only external lines and thereby scatter- U( )V ω U( ) = V ω( ) +  , pa p a Q ing amplitudes. i a This conclusion is supported by the Lagrangian (34). The (63) Lorentz breaking term is a surface term at infinity. The twisted mass term where  m cos(Q(χ))ˆ ψψ¯ (68) Q = ea (64) a is a local operator. Its correlators restricted to local space-  time regions should not be affected by the twist. and ∗Q(k) is given by (62) with Q for e. We can try to observe the effect of mass twist in the dis- Writing persion relation and in scattering where the electron is the  dressed one. Then the mass term m of the electron with P = p , (65) a a momentum p in the scattering amplitude is changed to a    as the total charged particle momentum, we can capture the −→ 2 ∞ ∞ m( p , χ)ˆ ≡ m cos lim r dxˆ ωˆ (x)χˆ (xˆ) , (69) r→∞ p features of Lorentz invariance breaking in the N charged S∞2 123 Eur. Phys. J. C (2015) 75 :89 Page 7 of 11 89 where 5.1 Spontaneous breakdown of symmetries: internal ∞ 2 ∞ and spacetime lim ωˆ (x) = lim r dxˆ ∗ ω (x)i xˆi . (70) r→∞ p r→∞ p Equation (69) is different for incident and outgoing momenta, In the standard model, the Higgs field breaks SU(2) × U(1) if they differ. to a U(1) subgroup. We can understand this result by noting Here, we are twisting each charged particle mass by its that, in quantum field theory, the group generators of the own infrared cloud. As discussed earlier, we can also twist the broken transformations diverge because of the asymptotic entire N-particle state vector depending on the total momen- Higgs field. −→ −→ 2 − 2 We shall now demonstrate that the Lorentz boost gener- tum P and the total charge Q. Then P0 P is the variable s of scattering theory. The results below can easily be adapted ators diverge in the charged sectors with infrared dressing. to this case. Hence the mechanism as regards breaking Lorentz boosts is The dispersion relation reads similar to spontaneous breaking of internal symmetries. This calculation also shows that angular momenta and −→2 + 2( ˆ , χ)ˆ = 2. p m p p0 (71) four-momenta do not show such a divergence. This result is Now, as noted previously, we have consistent with those of [4].  Below we give a give a quick review of spontaneous break- −→ ·−→ down of U(1) by a Higgs field using the collective coordinate ∗ω (x) = d3k ∗ ω (k) ei k x , p i p i approximation. We then examine the cases of boosts and the e −→ remaining Poincaré generators. ∗ω (k) = (p − p · kˆ kˆ ) (72) p i p · k i i We begin with some general remarks regarding sponta- = e ( − −→ · ˆ ˆ ). neous symmetry breakdown. It is followed up with the col- −→ ˆ pi p k ki (73) k0(p0 − p · k) lective coordinate calculation. Standard spontaneous symmetry breakdown is caused by We can take the large r limit of the expression (72)fol- the Higgs field at spatial infinity. One may imagine then that it lowing Gervais and Zwanziger [7]. Thus, we write cannot be observed by local physics. But that is not the case.  It makes itself felt in two ways, under two circumstances: ∗ω ( ) =  ω2 ω(∗ω (ˆ) iωr(kˆ·ˆx+i )), > , p x i d kˆ d p k i e 0 – In a non-gauge theory, it creates Goldstone bosons. They are described by quantizing the local fluctuations around (74) the constant Higgs field configuration. They affect the spectrum of the Hamiltonian by creating massless par- whereweusek0 = ω and ticles and eliminating the spectral gap of the Hamiltonian ∗ω (ˆ) = e ( − −→ · ˆ ˆ ), between vacuum and a particle state. p k i −→ ˆ pi p k ki (75) ω(p0 − p · k) – In a gauge theory, the appropriate gauge fields consume the Higgs fluctuations and become massive. Thus the opposite and >0 gives the high frequency cut-off. Hence happens regarding the Hamiltonian spectrum: a gapless   spectrum with massless vector fields gets gapped. Mach ∗ω ( ) =  ω ω e suggested that inertia is affected by the ambient back- p x i lim d kˆ d −→ →0+ (p − p · kˆ) 0  ground [23]. That is what happens here to the inertial mass of the vector fields. It is important to observe that ×( − −→ · ˆ ˆ ) iωr(kˆ·ˆx+i ) , pi p k ki e (76) the acquired mass of vector fields depend on their quan- tum numbers. For instance, the W ± and Z masses are or, defining ω = ωr, different.  – In theories with spontaneous symmetry breaking, the pas- ∗ω ( ) =−1  e sage from the massless to the massive phase of the gauge p x i lim d kˆ −→ r 2 →0+ (p − p · kˆ) field involves the gauge transformation to the U-gauge.  0  2 It is affected by the gauge group element obtained from −→ ˆ ˆ 1 × (p − p · k k ) . (77) the polar decomposition of the Higgs field. It preserves i i ˆ ·ˆ+ k x i locality so that the massive vector theory is a local theory. −→ ˆ We regard this also as an important fact. This gives (69). The pole in (77)atp0 = p · k can be treated as a principal value, while the pole due to the (kˆ · xˆ + i )−2 term gives a well-defined integral by the → 0+ Now we turn our attention to the collective coordinate prescription. calculation. 123 89 Page 8 of 11 Eur. Phys. J. C (2015) 75 :89

Let φ be a complex Higgs field with φ → φ = 0as The finiteness of the remaining field-dependent terms −→ 0 | x |→∞. Let us calculate the charge operator Q by the depends on the states they act on. For instance, they give collective coordinate method. If the classical configuration finite answers on Fock state states. is, say On the other hand, for the transformed Hamiltonian  −→ −→ −→ φ( ) = φ ≡ , 1 3 −→ 2 2 x 0 constant (78) H = d x(( E (x) +∗ω p(x)) + B ), (86) 2 we put the field-independent term gives a finite integral. The cross- −→ iθ(t) term in the above integral, that is, φ( x , t) = e φ0 (79)  −→ in the Lagrangian L. Then 3 −→ d x E (x) ·∗ω p(x), (87)   = 3 (|φ˙|2 −|φ |2) = 3 | θφ˙ |2 L d x d x i 0 is also well defined for a free electric field and acting on Fock  space states. Indeed, recall that for a free field φ, the inte- ˙2 3 2 3 −→ −→ = θ d x|φ0| =∞. (80) gral d xφ( x )α( x ) acting on the vacuum gives a square- integrable state if α is square-integrable. This operator is  similarly checked to be well defined on any Fock space state. Thus the “moment of inertia” d3x|φ |2 diverges, showing −→ 0 In our case ∗ ω plays the role of α is square-integrable. We that the symmetry φ → eiθ φ is spontaneously broken. This conclude that H is a well-defined operator. Actually, it is argument also implies that the charge  unitarily equivalent to H. ∗ ∗ The fact that the energy stored in the infrared cloud is finite Q = d3x(π φ + πφ ) (81) is well known and explained in QFT textbooks as Itzykson −→ and Zuber [24, Section 4.1.2] and Peskin and Schroeder [25, is divergent if φ → φ as | x |→∞even if φ depends on −→ 0 Section 6.1]. x . Next consider the angular momenta A conceptually identical argument applies to the Lorentz  boost generators, but since the details are a little different, = 3 (− δ ∂ − ) . we give them below. Ji d xEj i jk i i ijk Ak (88) Consider the infrared-dressed state (18)  3 ( )∗ω ( ) It is transformed as (ω )|· = i d xAi x p x i |· =|· . V p 0 e 0 ωp (82)   −1 −→ Ji → Ji = V (ωp)Ji V (ωp) (89) ∗ω ( x ) = O 1 r →∞ From (77), we see that p i r2 as . The boosts Ki are given by by V (ω ) with no field-dependent term. The same is true for  p −→ −→ the momenta 3 1 2 2 Ki = d xxi H(x), H(x) = ( E (x) + B (x)). (83)  2 3 Pi = d xEj (−i∂i )A j . (90) |· ( ) = In the state ωp , the electric field Ei is shifted to Ei x E (x) +∗ω (x) , while the magnetic field B is unchanged. i p i i So we can show as in the case of (87) that these operators Thus the boost operators become are well defined in the Fock space. Finally we infer that only boosts are spontaneously bro- K = V −1(ω )K V (ω ) i  p i p ken. This is compatible with [4]. −→ −→ 1 3 −→ 2 2 = d xxi (( E (x) +∗ω p(x)) + B (x)) (84) 2 5.2 On Lorentz-violating mass: dispersion relation and particle spin are changed when the vertex operator is absorbed in the redefined boosts

Ki . We can write (69)asfollows: This has the divergent field-independent term     −→ ∞ 3 −→ 2 m( p , χ)ˆ ≡ m e  ˆ ˆχˆ (xˆ) d xxi (∗ ω p(x)) . (85) cos d x d k −→ −→ ˆ ˆ  −→ 2 p ·ˆx − p · k k ·ˆx This diverges because xi ∼ O(r) and (∗ ω p(x)) ∼ × . (91) −→ ˆ ˆ 2 O(1/r 4) as r →∞(see 77). (p0 − p · k)(k ·ˆx + i ) 123 Eur. Phys. J. C (2015) 75 :89 Page 9 of 11 89

We would like to remark that the photon cloud surrounding the muon acquires all its orbital excitations, which depend ∞ n the charged particle has the electric field ∗ωp(x)i as alluded on χˆ .Its(2n)th power is suppressed by the coefficient α , to after (47). The function χˆ ∞ takes moments of this field: with α being the fine structure constant. This phenomenon they are determined by the experimental arrangement mea- will affect decay selection rules (and of courses scattering). suring say the mass. Further analysis of this observation is called for. The physical interpretation of χˆ ∞ is that it is a new form For sensitive experiments on the isotropy of space, see factor of the charged particle. [26–28]. Let R ∈ SO(3). It acts on the function χˆ as (Rχ)(ˆ x) = χ(ˆ R−1x). Then by using the rotational invariance of the two measures, we get 6 Non-abelian superselection and Higgs symmetry −→ −→ breaking m(R p , Rχ)ˆ = m( p , χ).ˆ (92) Thus m is a rotationally invariant function of its arguments. Non-abelian superselection rules play a role even in the famil- From each χˆ , we can form a scalar by coupling it to a U( ) lm −→ iar phenomenon where a complex Higgs field breaks a 1 polynomial of degree l in p . The integral in (91) is a linear symmetry spontaneously. We conclude this paper with this ∞ combination of these scalars. The simplest χˆ to consider observation. has only l = 1, its l = 0 component being 0. Retaining just We consider U(1) gauge symmetry broken by a complex = l 1, we henceforth set scalar field ϕ(x).Let fR be test functions supported in r ≥ R. ∞ Then χˆ (xˆ) = χi xˆi ,χi = real constants. (93)  = 3 ¯ ( )ϕ( ) Then the integral takes the form SR d x f R x x (97) (−→2)χ C p i pi (94) commutes with all observables supported in r < R.SoSR → + (−→2) for R →∞, denoted by S∞, is superselected. We note that for 0 . We evaluate C p in the appendix. It is finite  −→ → as p 0: 3 Q(ξ) = d x(∂i ξ Ei + ξ J0), (98)   2 −→ 2 −→ −→ p −|p | p0 +|p | p0 C( p 2) = 8π 2e 0 −→ ln −→ −16π 2e −→ . is superselected too and (97) and (98) do not commute. We | |3 −| | | |2 p p0 p p have (95) iQ(ξ) iQ(ξ) iξ∞ e S∞e = e S∞,ξ∞ = ξ|∞. (99) Thus finally for → 0+, We assume that S∞ = 0. −→2 (ξ) m(pˆ, χ)ˆ = m cos(C( p )χi pi ). (96) Both S∞ and Q commute with all local observables. χ Therefore, our previous arguments [8] lead to the conclusion √ i For normalized χi , that is, for χ˜i = , we can, if desired, χk χk that one of them must be spontaneously broken. plot the twisted mass say for pˆ = (0, 0, 1). We explain the above argument briefly in our context. If ˆ A generic Lorentz boost of p brings in more compli- we diagonalize S∞, then that defines a superselection sector. cated functions of p . For example, (94) is changed to −→ i However, Q(ξ) then changes it by (98). Hence it is sponta- 2 −→ I = C(( pˆ) )χi ( p )i . neously broken. If we diagonalize S∞, as we do in superconductivity, we 5.3 Mass twist smears mass and spin get a domain D1 for the Hamiltonian H which makes it self- adjoint. The operator eiQ(ξ) changes this domain: it is spon- In the Poincaré representation theory for a massive particle, taneously broken. mass is assumed to be a scalar and spin is introduced by We can also opt to diagonalize eiQ(ξ). Then U(1) is pre- attaching an irreducible representation (IRR) of SU(2) to the served, but at the expense that H is no longer defined: the vector state in the rest frame. The twisted mass, however, integral of energy density diverges classically. We can still − H depends on p or P and is not a rotational scalar. Its value define e it , finding first a domain in which S∞ is diagonal depends on the state vector it is associated with. Thus mass and then extending this unitary operator to the vectors with gets smeared, depending on p or P. Such a smearing may be eiQ(ξ) diagonal. (A unitary operator can act on all vectors of compatible with the results of Buchholz [1]. the Hilbert space.) We call such a domain D2. P H D Further, as mentioned, the twisted−→ mass is not a rotational If E is the projector for energies less than E for on 1, scalar. By (96), it depends on χi P i and all its powers for the it is bounded and hence defined in D2. A legitimate question choice made above for χ. Thus standard spin such as 1/2of is whether, if we reconstruct H from D2, then we will get the 123 89 Page 10 of 11 Eur. Phys. J. C (2015) 75 :89 same or similar low energy spectrum of H as from D1. If that Here, α and β are rotational scalars by rotational invariance. is the case, then there may be a new approach to spontaneous To find α and β, multiply the LHS of (104)byδij and sum symmetry breaking. over i, j to get  Acknowledgments We acknowledge discussions on possible experi- 1 4π 3α+β = lim dxˆ =−lim =−4π. mental realization of our results, specially muon decay as a function →0 (kˆ ·ˆx +i )2 →0 1 + 2 of direction towards sky with Catalina Curceanu (INFN, Frascati), Edoardo Milotti (Università di Trieste) and Sudhor Vempati (CHEP, (105) IISc, Bangalore). A.P.B. thanks TÜBiTAK and Seckin Kürkçüoˇglu for ˆ ˆ hospitality at the Middle East Technical University, Ankara. He also Next, multiply the LHS of (104)byki k j and sum over i, j thanks the group at the Centre for High Energy physics, Indian Institute to get of Science, Bangalore, and especially Sachin Vaidya, for hospitality.  S.K. thanks A. Behtash for useful discussions. S.K. is supported by (kˆ ·ˆx)2 TÜBA-GEBiP program of the Turkish Academy of Sciences. A.R.Q. α + β = lim dxˆ = 4π. (106) is supported by CAPES process number BEX 8713/13-8. →0 (kˆ ·ˆx + i )2

Open Access This article is distributed under the terms of the Creative Solving for α and β, we get Commons Attribution License which permits any use, distribution, and reproduction in any medium, provided the original author(s) and the α =−4π, β = 8π, (107) source are credited. 3 Funded by SCOAP /LicenseVersionCCBY4.0. and  −→ −→ −4π p 2 + 8π( p · kˆ)2 I = e dˆ . (108) Appendix A: Calculation of the integral in (91) 1 k −→ ˆ (p0 − p · k)

We wish to evaluate This can be evaluated by elementary methods to give   −→ ·ˆ−−→ · ˆ ˆ·ˆ     ∞ p x p k k x 2 +|−→| I =e dxˆ dˆχˆ (xˆ) . (100) 2 2p0 −→ p0 p 2 k −→ ˆ ˆ 2 I = 8π e −|p | ln − 32π ep . (p0 − p ·k)(k ·ˆx +i ) 1 −→ −→ 0 | p | p0 −|p | ∞ −→2 −→2 For χˆ (x) = χi xˆi , I is C( p )χi pi . To evaluate C( p ), (109) we choose χi = pi to find The integral I is −→ −→ 2 C( p 2) p 2   (−→ · ˆ) ˆ ˆ ˆ   −→ −→ ˆ ˆ p k pi k j xi x j p ·ˆx − p · k k ·ˆx I = e ˆ  ˆ . −→ 2 d k −→ ˆ d x ˆ (110) =e dxˆ dˆ( p ·ˆx) , ( − · ) ( ·ˆ+ )2 k −→ ˆ ˆ 2 p0 p k k x i (p0 − p · k)(k ·ˆx +i )   −→ ( ·ˆ)2 The angular integral over xˆ can be done as for (104). One =  1  p x e d ˆ −→ d xˆ − πδ + π ˆ ˆ k (p − p · kˆ) (kˆ ·ˆx + i )2 finds that it is 4 ij 8 ki k j ,giving  0  −→ · ˆ −→ ·ˆˆ ·ˆ  −→ p k p x k x ( p · kˆ)2 − e dˆ dxˆ , k −→ ˆ ˆ 2 I2 = 4πe dˆ (p0 − p · k) (k ·ˆx + i ) k −→ ˆ (p0 − p · k) ≡ − .     I1 I2 (101) 2 +|−→| 2 p0 p0 p = 8π e −→ ln −→ − 2p0 . (111) | p | p0 −|p | The integral I1 is   pi p j xˆi xˆ j Finally, I1 = e dˆ dxˆ k −→ ˆ ˆ 2 (p0 − p · k) (k ·ˆx + i )  −→ I1 − I2 1 C(| p |2) = −→ = e dˆ K (102) | p |2 k ( − −→ · ˆ) 1   p0 p k 2 −|−→|2 +|−→| 2 p0 p p0 p 2 p0 where = 8π e −→ ln −→ −16π e −→ . | p |3 p −| p | | p |2  0 ˆ ˆ (112) xi x j ˆ ˆ K1 = pi p j dxˆ = pi p j (αδij + βki k j ), (ˆ ·ˆ+ )2 k x i It is easy to see that (103)  xˆ xˆ −→ 32π 2 αδ + β ˆ ˆ =  i j . (| |2) =− . ij ki k j d xˆ (104) −→lim C p (113) (kˆ ·ˆx + i )2 | p |→0 3p0

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