A $Sim(2)$ invariant dimensional regularization

TL;DR

This paper presents a new dimensional regularization method invariant under $Sim(2)$ symmetry and applies it to compute one-loop quantum corrections in the VSRSM, advancing the understanding of quantum effects in this framework.

Contribution

It introduces a $Sim(2)$ invariant dimensional regularization technique and applies it to calculate quantum corrections in the VSRSM.

Findings

Successfully computed one-loop corrections to photon, electron self energies, and vertex.

Demonstrated the regularization method's consistency with $Sim(2)$ symmetry.

Provided a new tool for quantum calculations in Very Special Relativity models.

Abstract

We introduce a $Sim(2)$ invariant dimensional regularization of loop integrals. Then we compute the one loop quantum corrections to the photon self energy, electron self energy and vertex in the Electrodynamics sector of the Very Special Relativity Standard Model(VSRSM).

Tables

Table 1. Table 1: In the right column we list the form factor. In the left column we have the NR limit of the matrix element accompanying the form factor in ( 14 ).All form factors are evaluated at q μ = 0 subscript 𝑞 𝜇 0 q_{\mu}=0 . Here A 0 subscript 𝐴 0 A_{0} is the electric potential and A i subscript 𝐴 𝑖 A_{i} is the vector potential. φ s ′ subscript 𝜑 superscript 𝑠 ′ \varphi_{s^{\prime}} is a two dimensional constant vector that corresponds to the NR limit of the Dirac spinors.

NR limit	Form factor
$2M_e{\phi }_s^{\uparrow }{\phi }_s{A}_0$	$F_1$(0)
$\frac{3m^2}{4M^2}i{\epsilon }_{ijk}{\phi }_s^{\uparrow }{\sigma }^i{\phi }_{s^{\prime }}{\hat{n}}_j{q}_k{A}_0$	$F_1$(0)
$i{\epsilon }_{ijk}{q}_j{\phi }_s^{\uparrow }{\sigma }^k{\phi }_{s^{\prime }}{A}_i$	$F_1$(0)
$\frac{m^2}{2M^2}(-i{\hat{n}}_i{\phi }_s^{\uparrow }{\sigma }^a{\phi }_{s^{\prime }}{\epsilon }_{abc}{\hat{n}}_b{q}_c+i{\epsilon }_{ijk}{\phi }_s^{\uparrow }{\sigma }^j{\phi }_{s^{\prime }}{\hat{n}}_k\hat{n}.\overrightarrow{q})A_i$	$F_1$(0)
$-2i{n}_{0}M{\epsilon }_{ijk}{\phi }_s^{\uparrow }{\sigma }^k{\phi }_{s^{\prime }}{q}_j{A}_i$	$G_2\left(0\right)$
$-i{\epsilon }_{ijk}{n}_k\frac{m^2}{M}{\phi }_s^{\uparrow }\hat{n}.\overrightarrow{\sigma }{\phi }_{s^{\prime }}{q}_j{A}_i$	$G_2\left(0\right)$
$i\left(2M{\epsilon }_{ijk}{n}_k{\phi }_s^{\uparrow }{\sigma }^j{\phi }_{s^{\prime }}+2M_{e}i{n}_i{\phi }_s^{\uparrow }{\phi }_{s^{\prime }}\right)A_0{q}_i$	$G_2\left(0\right)$
$2M_e{n}_0{\phi }_s^{\uparrow }{\phi }_{s^{\prime }}{Q}_{\mu }{A}^{\mu }$	$G_3\left(0\right)$
$(-4M_e{\epsilon }_{ijk}{n}_k{\phi }_s^{\uparrow }\overrightarrow{n}.\overrightarrow{\sigma }{\phi }_{s^{\prime }}+4M_e{n}_0^2{\epsilon }_{ijk}{\phi }_s^{\uparrow }{\sigma }^k{\phi }_{s^{\prime }})q_j{A}_i$	$F_3\left(0\right)$
$4M_e{n}_0{\epsilon }_{ijk}{n}_j{\phi }_s^{\uparrow }{\sigma }^k{\phi }_{s^{\prime }}{A}_0{q}_i$	$F_3\left(0\right)$
$i{\epsilon }_{ijk}{\phi }_s^{\uparrow }{\sigma }^k{\phi }_{s^{\prime }}{A}_i{q}_j$	$F_2\left(0\right)$
$-i\frac{m^2}{2M^2}{\epsilon }_{ijk}{\hat{n}}_j{\phi }_s^{\uparrow }{\sigma }^k{\phi }_s{A}_0{q}_i$	$F_2\left(0\right)$

Equations

$$A_{\mu }=\int dp\frac{f(p^2)p_{\mu }}{(n\cdot p)}$$

$$\frac{1}{(n\cdot p)}=\underset{\epsilon \to 0}{\lim }\frac{(p\cdot \bar{n})}{(n\cdot p)(p\cdot \bar{n})+i\epsilon }$$

$$n_{\mu }\to \lambda n_{\mu },{\bar{n}}_{\mu }\to {\lambda }^{-1}{\bar{n}}_{\mu },\lambda \ne 0,\lambda \epsilon R$$

$$0=(n\cdot n)\to {\lambda }^2(n\cdot n)=0$$

$$0=(\bar{n}\cdot \bar{n})\to {\lambda }^{-2}(\bar{n}\cdot \bar{n})=0$$

$$1=(n\cdot \bar{n})\to (n\cdot \bar{n})=1$$

$$\frac{1}{(n\cdot p)}\to \frac{1}{(n\cdot p)}{\lambda }^{-1}$$

$$A_{\mu }=a{\bar{n}}_{\mu }$$

$$\int dp\frac{f(p^2)p_{\mu }}{(n\cdot p)}={\bar{n}}_{\mu }\int dpf(p^2)$$

$$A=\int dp\frac{F(p^2,p\cdot q)}{(n\cdot p)}=(\bar{n}\cdot q)f(q^2,(n\cdot q)(\bar{n}\cdot q))$$

$$\frac{\partial A}{\partial q^{\mu }}{n}_{\mu }=\int dp{F}_{,u}=$$

$$f(x,y)+2y\frac{\partial }{\partial x}f(x,y)+y\frac{\partial }{\partial y}f(x,y)$$

$$\int dp\frac{1}{[p^2+2p\cdot q-m^2{]}^a}\frac{1}{(n\cdot p)}=(\bar{n}\cdot q)f(x,y)$$

$$g(x)=-2a\int dp\frac{1}{[p^2-x-m^2{]}^{a+1}}$$

$$f(x,y)=-\frac{1}{y}\left\{\int dp[p^2-x-m^2{]}^{-a}-\int dp[p^2-x+2y-m^2{]}^{-a}\right\}$$

$$\int dp\frac{1}{[p^2+2p\cdot q-m^2{]}^a}\frac{1}{((n\cdot p){)}^b}=(\bar{n}\cdot q{)}^b(-2{)}^b\frac{\Gamma (a+b)}{\Gamma (a)\Gamma (b)}{\int }_0^{1}dt{t}^{b-1}\int dp[p^2-q^2+2n.q\bar{n}.qt-m^2{]}^{-a-b}$$

$$\int dp\frac{1}{[p^2+2p\cdot q-m^2{]}^a}\frac{1}{((n\cdot p){)}^b}=$$

$$(-1{)}^{a+b}i(\pi {)}^{\omega }(-2{)}^b\frac{\Gamma (a+b-\omega )}{\Gamma (a)\Gamma (b)}(\bar{n}\cdot q{)}^b{\int }_0^{1}dt{t}^{b-1}\frac{1}{(m^2+q^2-2(n\cdot q)(\bar{n}\cdot q)t{)}^{a+b-\omega }},\omega =d/2$$

$$\int dp\frac{1}{[p^2+2p.q-m^2{]}^a}\frac{1}{(n.p{)}^b}=$$

$$(-1{)}^{a+b}i(\pi {)}^{\omega }(-2{)}^b\frac{\Gamma (a+b-\omega )}{\Gamma (a)\Gamma (b)}{\left(\frac{q^2}{2n\cdot q}\right)}^b{\int }_0^{{}^1}dt{t}^{b-1}\frac{1}{(m^2+q^2(1-t){)}^{a+b-\omega }},$$

$$\mathcal{L}=\bar{\psi }\left(i\left(/D+\frac{1}{2}/n{m}^2(n\cdot D{)}^{-1}\right)-M\right)\psi -\frac{1}{4}{F}_{\mu \nu }{F}^{\mu \nu }-\frac{({\partial }_{\mu }{A}_{\mu }{)}^2}{4}$$

$$D_{\mu }={\partial }_{\mu }-ie{A}_{\mu },$$

$$i{\Pi }_{\mu \nu }=A({\eta }_{\mu \nu }{q}^2-q_{\mu }{q}_{\nu })+B\left(-q^2\frac{n_{\mu }{n}_{\nu }}{(n.q{)}^2}+\frac{n_{\mu }{q}_{\nu }+n_{\nu }{q}_{\mu }}{n.q}-{\eta }_{\mu \nu }\right)$$

$$A=(-ie{)}^2\frac{i}{(4\pi {)}^{\omega }}{\int }_0^{1}dx\Gamma (2-\omega )\frac{8x(1-x)}{(M_e^2-(1-x)x{q}^2{)}^{2-\omega }}$$

$$\displaystyle B=-m^{2}i{\color[rgb]{1,0,0}}\frac{e^{2}}{4\pi^{2}}\int_{0}^{1}\frac{dx}{{(1-x)}}\log\left[1-\frac{q^{2}(1-x)^{2}}{M_{e}^{2}-q^{2}(1-x)x}\right]$$

$$-i\Sigma (q)=C\frac{/n}{n.q}+D/q+E$$

$$C=(-ie{)}^2{m}^2[\frac{i}{16{\pi }^2}{\int }_0^{1}dx(1-x{)}^{-1}\ln \left(1+\frac{q^2(1-x)}{(M_e^2-q^2-i\epsilon )}\right)+$$

$$2i(4\pi {)}^{-\omega }{\int }_0^{1}dx\frac{\Gamma (2-\omega )}{[{\mu }^{2}x-x(1-x)q^2+(M^2+m^2)(1-x)-i\epsilon {]}^{2-\omega }}],$$

$$D=-2(-ie{)}^2(\omega -1)i(4\pi {)}^{-\omega }{\int }_0^{1}dx\frac{\Gamma (2-\omega )x}{[{\mu }^{2}x-x(1-x)q^2+(M^2+m^2)(1-x)-i\epsilon {]}^{2-\omega }},$$

$$E=(-ie{)}^{2}2\omega Mi(4\pi {)}^{-\omega }{\int }_0^{1}dx\frac{\Gamma (2-\omega )}{[{\mu }^{2}x-x(1-x)q^2+(M^2+m^2)(1-x){]}^{2-\omega }}$$

$$-i{q}_{\mu }{\Gamma }^{\mu }(p+q,p)=S^{-1}(p+q)-S^{-1}(p)$$

$$\text{mathbbm}P\Sigma (q)=-(-ie{)}^2\frac{1}{16{\pi }^2}\left\{2m^2\frac{/n}{n.q}-/q+4M\right\}\frac{1}{2-\omega }$$

$$\text{mathbbm}P{\Gamma }^{\mu }(p+q,p)=-(-ie{)}^2\frac{1}{16{\pi }^2}\frac{1}{2-\omega }\left({\gamma }_{\mu }+2m^2/n\frac{n_{\mu }}{n.pn.(p+q)}\right)$$

$$q_{\mu }\text{mathbbm}P{\Gamma }^{\mu }(p+q,p)=\text{mathbbm}P\Sigma (p)-\text{mathbbm}P\Sigma (p+q)=-(-ie{)}^2\frac{1}{16{\pi }^2}\frac{1}{2-\omega }\left\{/q-2m^2/n\left(\frac{1}{n.p+n.q}-\frac{1}{n.p}\right)\right\}$$

$$\bar{u}(p+q)\left\{G_2\left[-i{\sigma }_{\mu \nu }{q}_{\nu }/n\right]+G_3/n{Q}_{\mu }+F_3/n{\sigma }_{\mu \nu }{q}_{\nu }/n+{\tilde{\gamma }}_{\mu }{F}_1+F_{2}i\frac{{\sigma }_{\mu \nu }}{2M}{q}_{\nu }\right\}u(p)$$

Full text

The Weinberg-Salam model(SM) is a very successfull description of Nature, that is being verified at the LHC with a great precission. Moreover, until now, neither new particles nor new interactions have been discovered at the LHC[1]. This cannot be the whole story, though. The SM assumes that the neutrino is a massless particle, whereas we know that the neutrino is massive in order to describe the observed neutrino oscillations[2]

If we assume that Lorentz’s is an exact symmetry of Nature, we have to introduce new particles and interactions in order to give masses to the observed neutrinos through, for instance, the seesaw mechanism[3].

A new possibility to have a massive neutrino arises in Very Special Relativity(VSR)[4].  Instead of the 6 parameter Lorentz group, a 4 parameters subgroup($Sim(2)$) is assumed to be the symmetry of Nature. $Sim(2)$ transformations change a fixed null four vector $n_{\mu}$ at most by a scale factor, so ratios of scalar quantities containing the same number of $n_{\mu}$ in the numerator as in the denominator are $Sim(2)$ invariant, although  they are not Lorentz invariant. In this way it is possible to write a VSR mass term for left handed neutrinos[5].

Recently, we have proposed the SM with VSR[6] (VSRSM).It contains the same particles and interactions as  the SM, but neutrinos can have a VSR mass without lepton number violation. Since the electron and the electron neutrino form a $SU(2)_{L}$ doublet, the VSR neutrino mass term will modify the QED of the electron.

A main obstacle in exploring the loop corrections in the VSRSM is the non-existence of a gauge invariant regulator that preserve the $Sim(2)$ symmetry of the model.

In this letter, we define an appropriate regulator, based on the calculation of integrals using the Mandelstam-Leibbrandt(ML)[7] [8]prescription introduced in [9].We want to emphasize that our method directly lead to the ML prescription, the only one compatible with canonical quantum field theory[10].The regulator preserve gauge invariance, a property inherited from the ML prescription, as well as the $Sim(2)$ symmetry.

Then we proceed to compute  one loop corrections. We find the divergent and  finite part of the vacuum polarization and electron self energy. Moreover we compute the leading correction to the standard QED result for the anomalous magnetic model of the electron.

We want to emphasize that meanwhile no new particles or interactions are discovered at the LHC or elsewhere, we have to consider the VSRSM as a very strong candidate to describe weak and electromagnetic interactions. It contains all the predictions of the SM plus neutrino masses and neutrino oscillations. It is renormalizable(as we show explicitly in this letter) and unitarity of the $S$ metric is preserved. If future experiments validates the predictions of the model, it would be the first evidence of Lorentz Symmetry violation.

1 Mandelstam-Leibbrandt(ML) prescription from a hidden symmetry

In this section we review the results of [9].

Let us compute the following simple integral:

[TABLE]

where $f$ is an arbitrary function.$dp$ is the integration measure in $d$ dimensional space and $n_{\mu}$ is a fixed null vector($(n\cdot n)=0$). This integral is infrared divergent when $(n\cdot p)=0$.

The ML is:

[TABLE]

where $\bar{n}_{\mu}$ is a new null vector with the property $(n\cdot\bar{n})=1$.

To compute $A_{\mu}$ we must know what $f$ is , provide an specific form of  $n_{\mu}$ and $\bar{n}_{\mu}$, and evaluate the residues of all poles of  $\frac{f(p^{2})}{(n\cdot p)}$ in the $p_{0}$ complex plane, a difficult task for an arbitrary $f$.

Instead we want to point out the following symmetry:

[TABLE]

It preserves the definitions of $n_{\mu}$ and $\bar{n}_{\mu}$:

[TABLE]

We see from (1) that:

[TABLE]

Now we compute $A_{\mu}$, based on its symmetries. It is a Lorentz vector which scales under (2) as $\lambda^{-1}$. The only Lorentz vectors we have available in this case are $n_{\mu}$ and $\bar{n}_{\mu}$. But (2) forbids $n_{\mu}$. That is:

[TABLE]

Multiply by  $n_{\mu}$ to find $(A\cdot n)=a$. Thus $a=\int dpf(p^{2})$. Finally:

[TABLE]

By the same arguments, we can compute the generic integral:

[TABLE]

$q_{\mu}$ is an external momentum, a Lorentz vector. $F$ is an arbitrary function. The last relation follows from (2), for a certain $f$ we will find in the following.

Taking the partial derivative respect to $q_{\mu}$ in both sides of (3), we obtain that

[TABLE]

We defined $u=p\cdot q$,$x=q^{2}$,$y=(n\cdot q)(\bar{n}\cdot q)$. $()_{,u}$ means derivative respects to $u$.

Assuming that the solution and its partial derivatives are finite in the neighborhood of $y=0$, it follows from the equation that $f(x,0)=g(x)$. That is the partial differential equation has a unique regular solution.

Now we apply this result to compute integrals that appear in gauge theory loops:

[TABLE]

In this case

[TABLE]

The unique regular solution of (4) is:

[TABLE]

We can check  that $f(x,0)=-2a\int dp[p^{2}-x-m^{2}]^{-a-1}=g(x)$.

In the same way we can compute the whole family of loop integrals:

[TABLE]

Using dimensional regularization, we obtain:

[TABLE]

2 $Sim(2)$ invariant regulator

The prescription to regularize the infrared divergences that we have reviewed in chapter 1, always produces finite results depending on two fixed null vectors $\bar{n}_{\mu},n_{\mu}$. Moreover it preserves gauge invariance because it respects the shift symmetry of the loop integral $\int dpf(p_{\mu})=\int dpf(p_{\mu}+q_{\mu})$ for arbitrary $q_{\mu}$. However ML does not respect $Sim(2)$ symmetry of VSRSM. Below we show how to remedy this.

We start from the ML result for the integral(5).

We trade $\bar{n}_{\mu}$ by $q_{\mu}$. i.e. $\bar{n}_{\mu}=an_{\mu}+bq_{\mu}$111This is the more general form for $\bar{n}_{\mu}$ compatible with reality, right scaling under (2) and $\bar{n}_{\mu}$ dimensionless. For instance in $d=3$ we must have $\bar{n}_{\mu}=a\frac{q^{2}n_{\mu}}{(n.q)^{2}}+b\frac{q_{\mu}}{n.q}+c\varepsilon_{\mu\nu\lambda}\frac{n_{\nu}}{(n.q)^{2}}q_{\lambda}\sqrt{q^{2}}$ with $a,b,c$ pure numbers. This fails to be real for $q^{2}<0$.. From the conditions: $\bar{n}.\bar{n}=0$,$\bar{n}.n=1$ we get $\bar{n}_{\mu}=-\frac{q^{2}}{2(n.q)^{2}}n_{\mu}+\frac{q_{\mu}}{n.q}$. Moreover, we see that $\bar{n}_{\mu}$ satisfies the scaling (2) and is real for any value of $q^{2}$ in Minkowsky space. So, all the conditions to apply the procedure reviewed in section 1 are satisfied. Therefore,,

[TABLE]

Notice that now (6) respects the $Sim(2)$ invariance of the original integral. The same procedure can be applied to other integrals found in [9]. Notice that first we keep $\bar{n}$ fixed, derive (5) with respect to $q_{\mu}$ as many times as necessary and then replace $\bar{n}_{\mu}=-\frac{q^{2}}{2(n.q)^{2}}n_{\mu}+\frac{q_{\mu}}{n.q}$. The rationale for this prescription derives from the observation that we could compute the integral with whatever power of $p_{\mu}$ in the numerator using Cauchy theorem of residues in $p_{0}$ complex plane. In this way it doesn’t matter whether $\bar{n}_{\mu}$ depends on $q_{\mu}$ or not.

Once we have obtained (6), we notice that it provides a unique analytic continuation of the integral from $b<0$ to $b>0$. Since for $b<0$ we do not need an infrared regulator, we can compute the integral using standard dimensional regularization. By integration by parts in the integral over $t$,we can check that (6) gives the right answer for $b<0$.

3 The model

The leptonic sector of VSRSM consists of three $SU(2)$ doublets L_{a}=\left(\begin{array}[]{c}\nu^{0}_{aL}\\ e^{0}_{aL}\end{array}\right), where $\nu^{0}_{aL}=\frac{1}{2}(1-\gamma_{5})\nu^{0}_{a}$ and $e^{0}_{aL}=\frac{1}{2}(1-\gamma_{5})e^{0}_{a}$, and three $SU(2)$ singlet $R_{a}=e^{0}_{aR}=\frac{1}{2}(1+\gamma_{5})e^{0}_{n}$. We assume that there is no right-handed neutrino. The index $a$ represent the different families and the index [math] say that the fermionic fields are the physical fields before breaking the symmetry of the vacuum.

In this letter we restrict ourselves to the electron family.

$m$ is the VSR mass of both electron and neutrino.

After spontaneous symmetry breaking(SSB), the electron adquires a mass term $M=\frac{G_{e}v}{\sqrt{2}}$, where $G_{e}$ is the electron Yukawa coupling and $v$ is the VEV of the Higgs. Please see equation (52) of [6]. The neutrino mass is not affected by SSB:$M_{\nu_{e}}=m$.

Restricting the VSRSM after SSB to the interactions between photon and electron alone, we get the VSR QED action.$\psi$ is the electron field. $A_{\mu}$ is the photon field. We use the Feynman gauge.

[TABLE]

We see that the electron mass is $M_{e}=\sqrt{M^{2}+m^{2}}$, where $m$ is the electron neutrino mass.

3.1 Feynman rules

To draw the Feynman graphs we used [11]

In the following sections, we have used extensively the program FORM [12].

4 Photon Self Energy in VSRSM

In this section we present the computation of the photon self-energy. In VSRSM it is given by two graphs:

Applying the $Sim(2)$ invariant regulator to the addition of the graphs of Figure (2), and after a long calculation, we get:

[TABLE]

with

[TABLE]

Here $-e$ is the electron electric charge, $m$ the electron neutrino mass and $M_{e}$ is the electron mass.$q_{\mu}$ is the virtual photon momentum.

We first notice that $q^{\mu}\Pi_{\mu\nu}=0$ as required by $U(1)$ gauge invariance of the photon field. It is obtained by a straightforward application of the regularized integrals of . Moreover $B(q^{2}=0)=0$,therefore the photon remains massless. Also the photon wave function divergence is the same as in QED.

5 Electron Self Energy in VSRSM

Here we calculate the electron self-energy. Again we have two graphs contributing to the 2-proper vertex. See Figure(3).

[TABLE]

with:

[TABLE]

6 Electron-Electron-Photon Proper vertex $\Gamma^{\mu}(p+q,p)$

In this subsection we discuss the 3 points proper vertex and verify the Ward-Takahashi identity. This is an important test of the gauge invariance of the regulator. The one loop contribution to $\Gamma^{\mu}(p^{\prime}=p+q,p)$ consists of the addition of 3 graphs(Figure(4)):

As a result of the shift symmetry which is respected by the regulator, $\int dpf(p_{\mu})=\int dpf(p_{\mu}+q_{\mu})$ for arbitrary $q_{\mu}$, we can prove the Ward-Takahashi identity:

[TABLE]

Here $S(p)=\frac{i}{\not{p}-M-\Sigma(p)}$ is the full electron propagator and $\Gamma^{\mu}(p+q,p)$ is the three proper vertex.

Below we explicitly verified that the pole at $d=4$ satisfies (11)222The finite part of the Ward-Takahashi identity is true also, but the computation is too long to show it in this letter.

Pole contribution:

[TABLE]

[TABLE]

The divergent piece satisfies the Ward identity:

[TABLE]

6.1 Form factors

The on-shell proper vertex can be written as follows:

[TABLE]

where:

[TABLE]

$F_{1},F_{2}.F_{3},G_{2},G_{3}$ are forms factors(Lorentz scalar combinations of $n_{\mu},p_{\mu},q_{\mu}$). Under the $Sim(2)$ scaling $n_{\mu}\rightarrow\lambda n_{\mu}$, $F_{1},F_{2}$ are invariants,$F_{3}\rightarrow\lambda^{-2}F_{3}$,$G_{3}\rightarrow\lambda^{-1}G_{3}$,$G_{2}\rightarrow\lambda^{-1}G_{2}$.

In the Non-Relativistic(NR) limit we get Table 1, keeping terms that are at most linear in $q_{\mu}$.

To show the power of the $Sim(2)$ invariant regularization prescription presented in this letter, we will compute the one loop contribution to the (isotropic)anomalous magnetic moment of the electron. It is given by $F_{2}(0)-2n_{0}MG_{2}(0)-4F_{3}(0)M_{e}n_{0}^{2}i$(See rows 11, 5  and 9  of Table 1).

Introduce the following integrals:

[TABLE]

We get:

[TABLE]

Evaluating the integrals according to the $Sim(2)$ invariant prescription to $o(m^{2})$, we get:

[TABLE]

where $\alpha$ is the fine structure constant. Therefore to this order the QED result holds.

Notice that already at tree level, the model predicts the existence of an anisotropic electric moment of the electron, corresponding to the second line of the list and an anisotropic magnetic moment of the electron, corresponding to the fourth row of the list, both of the order of $\frac{m^{2}}{M_{e}^{2}}$. The electric dipole moment is:

[TABLE]

where $\lambda=2.4\times 10^{-12}m$ is the Compton wave length of the electron.

Using the best bound on the electric dipole moment of the electron[13], $|\vec{p}|<8.7\times 10^{-29}\leavevmode\nobreak\ e\dot{c}m.$,we get:

[TABLE]

For the muon $\lambda=1.17\times 10^{-14}m$. Using the best bound on the muon electric dipole moment[14],$|\vec{p}_{\mu}|<1.8\times 10^{-19}\leavevmode\nobreak\ e\dot{c}m.$,we get:

[TABLE]

Bounds using the experimental values of the magnetic moments are much weaker[15].

The $Sim(2)$ invariant regularization opens the way to explore the full quantum possibilities of VSR. They should be systematically studied, in Particle Physics models as well as in Quantum Gravity models.

Acknowledgements

The work of J. A. is partially supported by Fondecyt 1150390, Anillo ACT 1417. J.A. wants to thank H. Morales-Técotl, L.F. Urrutia, D. Espriu  and  R. Soldati for useful remarks.