% \hfill{\normalsize\vbox{% \hbox{} }}\\ {A theoretical estimate of the Higgs boson mass}
Using the invariance of the electroweak vacuum in the presence of a background Z field we estimate the mass of the Higgs boson in the standard model at $m_H$=125.9 GeV.
1208.3753.tex



\documentclass[prd,aps,showpacs,preprintnumbers,amssymb]{revtex4}
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\usepackage{color}
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%\newcounter{ehsctr}
\def\e3p{$\eta \rightarrow 3 \pi$}

\begin{document}
\title{%
\hfill{\normalsize\vbox{%
\hbox{}
 }}\\
{A theoretical estimate of the Higgs boson mass}}
\author{Renata Jora $^{\it \bf a}$~\footnote[1]{Email:
 rjora@theory.nipne.ro}}
 \affiliation{$^ {\bf \it a}$ National Institute of Physics and Nuclear Engineering, PO Box MG-6, Bucharest-Magurele, Romania.}
\date{\today}

\begin{abstract}
Using  the invariance of the electroweak vacuum in the presence of a background Z field we estimate the mass of the Higgs boson in the standard model at $m_H$=125.9 GeV.
\end{abstract}
\pacs{11.30. Qc, 11.15 Ex, 12.15. Lk}
\maketitle







\section{Introduction}
%In the standard model of elementary particles  \cite{Weinberg}-\cite{Glashow} with the regular Higgs boson spontaneous symmetry breaking at zero or finite temperature can be treated using %Landau theory of phase transitions.  This employs as an order parameter the vacuum expectation values of the Higgs boson associated to the one or two loop scalar effective potential as %analogue of the Gibbs energy. From the point of view of the gauge bosons the symmetry breaking is characterized by the induction of masses for some or all of the gauge bosons. One cannot %use vacuum  expectation values for the gauge fields as we know from QCD that vacuum condensates for the gauge boson do not necessarily signal gauge symmetry breaking. However there should %be things that one can learn about spontaneous symmetry breaking from the gauge sectors of the theories at hand. In this work we will show that this is indeed the case.
In an historical event the Atlas and CMS experiments at CERN recently announced the discovery of a Higgs like particle in the mass region 125 GeV \cite{CMS} or 126 GeV \cite{Atlas}.
 It is well known that the standard model \cite{Weinberg} does not predict the Higgs couplings or the mass so it is hard to have direct comparison between a theoretical estimate and the experiment.
  The usual treatment of theories with spontaneous symmetry breaking relies on the expansion of the lagrangian around the classical constant field which correspond to the minimum
   of the potential. Gauge bosons acquire masses through the Goldstone mechanism for which also the role of scalars is crucial. Despite the huge amount of information that one
    can get about the standard model from electroweak precision data the Higgs boson mass in the standard model remains up to today undetermined in a simple theoretical set-up.
    It is the lore of the present work to fill this gap.
We will illustrate our approach for the abelian Higgs model in section II. In section III we will give a detailed estimate of the mass of the Higgs boson in the standard model.
Section IV is dedicated to conclusions.


\section{The abelian Higgs model}

We start with the simplest model for spontaneous symmetry breaking \cite{Higgs}: a U(1) gauge theory coupled with a complex scalar field. The corresponding Lagrangian is:
\begin{eqnarray}
{\cal L}=-\frac{1}{4}(F_{\mu\nu})^2+|D_{\mu}\Phi|^2-V(\Phi),
\label{lagr45}
\end{eqnarray}

where,

\begin{eqnarray}
V(\Phi)=-\mu^2(\Phi^{\dagger}\Phi)+\lambda(\Phi^{\dagger}\Phi)^2.
\label{oloop8787}
\end{eqnarray}
We write the complex scalar field as,
\begin{eqnarray}
\Phi=\frac{1}{\sqrt{2}}(\Phi_0+\Phi_1+i\Phi_2).
\label{noty6768}
\end{eqnarray}

Here $\Phi_0$ is the vacuum expectation value of the scalar field $\Phi_0=\frac{\mu^2}{\lambda}$.
It is useful to write down all the terms that appear in the lagrangian with spontaneous symmetry breaking:

\begin{eqnarray}
&&|D_{\mu}\Phi|^2\rightarrow
\frac{1}{2}(\partial_{\mu}\Phi_1)^2+\frac{1}{2}(\partial_{\mu}\Phi_2)^2+
\frac{1}{2}g^2A_{\mu}A^{\mu}(\Phi_1^2+\Phi_2^2)
+
\nonumber\\
&&+gA^{\mu}\Phi_2\partial_{\mu}\Phi_1-gA^{\mu}\Phi_1\partial_{\mu}\Phi_2
+\frac{1}{2}g^2A_{\mu}A^{\mu}(\Phi_0^2+2\Phi_0\Phi_1)-
g A^{\mu}\Phi_0\partial_{\mu}\Phi_2
\label{usef3435}
\end{eqnarray}

In the Higgs potential the partition is as follows:

\begin{eqnarray}
&&-\frac{\mu^2}{2}\Phi_1^2-\frac{\mu^2}{2}\Phi_2^2+
\frac{\lambda}{4}\Phi_1^4+\frac{\lambda}{4}\Phi_2^4+\frac{\lambda}{2}\Phi_1^2\Phi_2^2+
\nonumber\\
&&-\frac{1}{2}\mu^2\Phi_0^2-\mu^2\Phi_1\Phi_0
+\frac{\lambda}{4}(\Phi_0^4+4\Phi_0^3\Phi_1+4\Phi_0\Phi_1^3+6\Phi_0^2\Phi_1^2)
+\frac{\lambda}{2}\Phi_2^2(\Phi_0^2+2\Phi_0\Phi_1)
\label{somres43534}
\end{eqnarray}


From these one can compute the effective action which identifies with the effective potential.



Assume now that we have spontaneous symmetry breaking and we decompose the gauge field into a background gauge field $B_{\mu}$ and a quantum gauge field $A_{\mu}$.
The background gauge field can be considered as slowly varying or approximately constant.



All terms previously listed will remain in the Lagrangian with the remark that the gauge field in Eq(\ref{usef3435}) should be identified with the quantum gauge field.
 Then the extra terms corresponding to the background gauge field are:
\begin{eqnarray}
&&\frac{1}{2}g^2(B_{\mu}B^{\mu}+2B_{\mu}A^{\mu})(\Phi_1^2+\Phi_2^2)
+g^2 B_{\mu}\Phi_2\partial^{\mu}\Phi_1 -B^{\nu}\Phi_1\partial_{\nu}\Phi_2+
\nonumber\\
&&\frac{1}{2}g^2B_{\mu}B^{\mu}(\Phi_0^2+2\Phi_1\Phi_0)
+g^2B_{\mu}A^{\mu}(\Phi_0^2+2\Phi_1\Phi_0)-g B_{\mu}\Phi_0\partial^{\mu}\Phi_2.
\label{res55546}
\end{eqnarray}



From Eq (\ref{res55546}) we can select the contribution,
\begin{eqnarray}
\frac{1}{2}g^2B_{\mu}B^{\mu}\Phi_0^2+ g^2B_{\mu}A^{\mu}\Phi_0^2
\label{tr5646}
\end{eqnarray}
which will lead at tree level to,
\vspace{0.5cm}
\begin{center}
\begin{picture}(200,50)(0,0)
\Photon(10,30)(70,30){3}{4}
\Vertex(40,30){2}
\Text (80,30)[]{+}
\Photon(90,30)(180,30){3}{12}
\Vertex(120,30){2}
\Vertex(150,30){2}
\Text(190,30)[]{=}
\Text(90,0)[]{$ig^2\Phi_0^2g^{\mu\nu}+ (i)g^2\Phi_0^2\frac{-i}{-g^2\Phi_0^2}(i)g^2\Phi_0^2g^{\mu\nu}=0$}
\end{picture}
\vspace{0.5cm}
\\{\sl Fig.1. The cancelation of terms.}
\end{center}



and thus to the cancelation of these terms.


In rest we need to determine only those terms which contain the background gauge field and the Higgs field:
\begin{eqnarray}
2g^2B_{\mu}A^{\mu}\Phi_1\Phi_0+g^2B_{\mu}B^{\mu}\Phi_1^2
\label{ter445367}
\end{eqnarray}

The second term in Eq(\ref{ter445367}) can be regarded as an extra mass term for the Higgs boson.
Besides  this at one loop the only term that contributes is the first term in Eq(\ref{ter445367}).
\vspace{0.5cm}
\begin{center}
\begin{picture}(200,100)(0,0)
\Photon(10,50)(110,50){3}{5}
\Vertex(40,50){1}
\Vertex(80,50){1}
\Text(20,60)[]{$B_{\mu}$}
\Text(100,60)[]{$B_{\mu}$}
\CArc(60,50)(20,0,180)
\end{picture}
\\{\sl Fig.2. One diagram that contributes to the effective action.}
\end{center}

The main point is the determine the one loop effective action for a theory with spontaneous symmetry breaking in the presence of a background gauge field.
 For the simple case at hand we expect that this action will contain the result of the diagram in Fig.2 plus the usual one loop effective
 scalar potential where the masses of the scalars will have an extra  contribution proportional to
$B_{\mu}B^{\mu}g^2$.

Assume that Z[J,K] is the generating functional for a spontaneously broken gauge theories where J, K are generic sources corresponding to the scalars and gauge fields respectively.
Now we introduce a background gauge field $B_{\mu}$. The new generating functional reads \cite{Abbott}:
\begin{eqnarray}
&&Z'[J,K,B]=\int \delta Q \delta A \exp[i[S[Q,A+B]+JQ+KA]]
=
\nonumber\\
&&\int \delta Q \delta A\exp[iS[Q,A]+iJQ+iK(A-B)]=
Z[J,K]\exp[-iKB]
\label{someus54646}
\end{eqnarray}

The vacuum expectation value of the scalar field before introducing the background gauge field is given by the expression:
\begin{eqnarray}
Q_J=\frac{-i\ln Z[J,K]}{\delta J}
\label{r4546}
\end{eqnarray}
 whereas in the presence of the background gauge field is
 \begin{eqnarray}
 Q_J'=\frac{-i\ln Z[J,K]}{\delta J}-\frac{\delta K B}{\delta J}=\frac{-i\ln Z[J,K]}{\delta J}.
 \label{res435242}
 \end{eqnarray}

 Thus  Eq(\ref{res435242}) states that the vacuum expectation of the scalar field is preserved also in the presence of the background gauge field (Note that this is true even if
  the gauge source depends on the minimum of the scalar field).
  By setting the sources to zero one obtains
the minimum of the effective action (or potential):
\begin{eqnarray}
\frac{\delta \Gamma[B,\Phi_{cl}]}{\delta \Phi_{cl}}=
\frac{\delta \Gamma[\Phi_{cl}]}{\delta \Phi_{cl}}
\label{ewm777}
\end{eqnarray}

Here we denote by $\Gamma[B,\Phi_{cl}]$ the effective action in the presence of the background gauge field whereas $\Gamma[\Phi_{cl}]$ is the effective action in the absence of it.
As the background gauge field is arbitrary one should expand $\Gamma[B,\Phi_{cl}]$ in terms of B. In what follows we will consider only terms of order $B^2$ and neglect the higher order
contributions.
Applied to our abelian Higgs model this means that the sum of the term in Fig.2 plus the term in the scalar potential proportional to $B_{\mu}B^{\mu}$ will give zero contribution
to the vacuum  expectation of the scalar. From this one can derive various constraints and even estimate the Higgs boson mass.
We will illustrate in detail how this works for the standard model in the next section.


\section{ Higgs boson mass in the standard model}

We will work with a Higgs potential of the form:
\begin{eqnarray}
V=-m^2\Phi^{\dagger}\Phi+\lambda(\Phi^{\dagger}\Phi)^2
\label{Higgs54646}
\end{eqnarray}

which displays spontaneous symmetry breaking as the scalar develops the vacuum expectation value:
\begin{eqnarray}
\langle 0 |\Phi| 0\rangle=
\left[
\begin{array}{cc}
0\\
\frac{\Phi_0}{\sqrt{2}}
\end{array}
\right]
\label{veve435}
\end{eqnarray}.

The next step is to introduce a background gauge field for the Z boson, for the photon or the $W^{\pm}$ bosons.  The case for the W bosons is more intricate and we will not discuss it here. Assume that we introduce only a background gauge field for the photon. This does not couple to the Higgs so it cannot help us with determining the mass of the Higgs boson. However it has almost symmetric interactions (with different couplings) with the W bosons as the Z boson as it is evident from the interaction term in the Lagrangian:
\begin{eqnarray}
-D^{\dagger\mu}W^{-\nu}D_{\mu}W_{\nu}^{+}+D^{\dagger\mu}W^{-\nu}D_{\nu}W_{\mu}^{+}
\label{lagr65757}
\end{eqnarray}
where,
\begin{eqnarray}
D_{\mu}=\partial_{\mu}-ie(A_{\mu}+\cot{\theta_W}Z_{\mu}).
\label{terr54646}
\end{eqnarray}
The contribution of the background photon field to the effective action will thus contain only the interactions depicted above and the interaction with the fermions.
%We can expand the effective action in terms of the background  gauge field as in Eq(\ref{ewm777}). At zero external momenta the fermion contribution is zero.
This means that the W bosons will also give per total zero contribution to the vacuum expectation value of the scalars
(through the derivative of the corresponding term in the effective action). By the symmetry of the W-photon and W-Z interactions we conclude that
if one introduces a background gauge field only for the Z boson those terms that contain Z interaction to the W should not be taken into consideration.

Then the effective action will be simply the sum of the regular effective potential \cite{Sher},\cite{Jones},
\begin{eqnarray}
V(\Phi)&&=-\frac{1}{2}m^2\Phi^2+\frac{\lambda}{4}\Phi^4+\frac{1}{64\pi^2}(-m^2-g^2B^2+3\lambda\Phi^2)^2[\ln(\frac{-m^2-g^2B^2+3\lambda\Phi^22}{\mu_0^2})-\frac{3}{2}]+
\nonumber\\
&&+\frac{3}{64\pi^2}(-m^2-g^2B^2+\lambda\Phi^2)^2[\ln(\frac{-m^2-g^2B^2+\lambda\Phi^2}{\mu_0^2})-\frac{3}{2}]+
\frac{3}{64\pi^2}(\frac{1}{4}(g^2+g'^2)\Phi^2)^2[\ln(\frac{\frac{1}{4}(g^2+g'^2)\Phi^2}{\mu_0^2})-\frac{5}{6}]+
\nonumber\\
&&+\frac{3}{32\pi^2}(\frac{1}{4}g^4\Phi^2)[\ln(\frac{\frac{1}{4}g^2\Phi^2}{\mu_0^2})-\frac{5}{6}]-
\frac{3}{16\pi^2}(\frac{1}{2}g_t^2\Phi^2)^2[\ln\frac{\frac{1}{2}g_t^2\Phi^2}{\mu_0^2}-\frac{3}{2}]
\label{oneloop9898}
\end{eqnarray}

and the result of the one loop diagram (see Fig.2) corresponding to the vertex $2B_{\mu}Z^{\mu}\Phi_0\Phi_1$ where $\Phi_1$ is the Higgs boson,
\begin{eqnarray}
&&\Gamma_1[B^2]=\frac{6B^2}{16\pi^2}g_z^2\Phi^2(1+\frac{1}{-m^2-g_z^2\Phi^2-g_z^2B^2}[g_z^2\Phi^2\ln(\frac{gz^2\Phi^2}{\mu_0^2})-
\nonumber\\
&&-(-m^2+3\lambda\Phi^2-g_z^2B^2)\ln(\frac{-m^2+3\lambda\Phi^2-g_z^2B^2}{\mu_0^2})].
\label{loop87878}
\end{eqnarray}

Here $g_z^2=\frac{g^2+g'^2}{4}$.
There is also the one loop fermion contribution (we consider only the top quark):
\begin{eqnarray}
\Gamma_2[B^2]=B^2[6\frac{g^2g_t^2}{8\cos^2\theta_W}[(1-(1-\frac{8}{3}\sin^2\theta_W)^2)+2\ln[\frac{g_t^2\Phi^2}{2\mu_0^2}]]
\label{quarkt}
\end{eqnarray}

The full one loop effective potential is thus:
\begin{eqnarray}
\Gamma[B,\Phi]=\Gamma_1[B^2]+\Gamma_2[B^2]-V_{eff}[\Phi,B]+ {\rm higher\,order\,terms\,in\,B}.
\label{efe54646}
\end{eqnarray}

We denote $\Gamma[B^2]=\Gamma_1[B^2]+\Gamma_2[B^2]$.
The invariance of the vacuum in the presence of background gauge field translates into the constraint:
\begin{eqnarray}
\frac{d\Gamma[B,\Phi]}{d\Phi d B^2}|_{B^2=0}=0.
\label{43535}
\end{eqnarray}
where we used the expansion of the effective action in powers of $B^2$
( The sum over the $B^{2n}$ terms can be obtained formally by integrating the $B^2$ term so the condition (\ref{43535}) suffices).

 In what follows we will apply the constraint in Eq(\ref{43535}).
Furthermore we will simply take $m_H^2=2\lambda \Phi_0^2$ which is true apart for some small corrections; $\mu_0^2=m_H^2$ and the electroweak values($g_z^2=\frac{M_Z^2}{\Phi_0^2}$,
 for the coupling constants with the notation: $g_z^2=\frac{M_Z^2}{\Phi_0^2}$, $g_w^2=\frac{M_W^2}{\Phi_0^2}$, $g_y^2=\frac{m_t^2}{\Phi_0^2}$). Then
 Eq(\ref{43535}) reduces to:

 \begin{eqnarray}
 &&-6g_z^2/(g_z^2-2\lambda)^2[-\lambda(g_z^2-2\lambda)+g_z^2(g_z^2-\lambda)\ln(\frac{g_z^2}{2\lambda})+
 \nonumber\\
&&\frac{12g_y^2g_w^2}{\cos^2\theta_W}\ln(\frac{g_y^2}{2\lambda})+\frac{6g_y^2g_w^2}{\cos^2\theta_W}[1-(1-\frac{8}{3}\sin^2\theta_W)^2)]=0
\label{res546}
\end{eqnarray}

This leads to the following result for the coupling and the mass of the Higgs boson: $\lambda=0.13$; $m_H=125.9$. In this estimate we also used a small correction coming form the bottom quark.



\section{Discussion}
 Our approach is based on the introduction of a background gauge field in a Higgs theory after the spontaneous symmetry breaking.
 The background gauge is considered slowly varying such that any derivative term can be neglected.
 We show that in particular the vacuum expectation value of the scalar field remains unchanged by this transformation.
 We then apply the constraint deduced from this invariance to the standard model. In this case the most advantageous choice is to introduce a background gauge field for the Z boson.
 We obtained a mass for the standard model Higgs boson of $m_H=125.9$ in excellent agreement with the experimental results.  Our result is
 susceptible to corrections coming from the running of the coupling constants and fields. We expect these corrections to be small.

\section*{Acknowledgments} \vskip -.5cm
I am happy to thank J. Schechter and Marc Sher for useful comments on the manuscript.
This work has been supported by PN 09370102/2009.



\begin{thebibliography}{15}
\bibitem{CMS} The CMS Collaboration, arXiv:1207.7235, 2012.
\bibitem{Atlas} The ATLAS Collaboration, arXiv:1207.7214, 2012.
\bibitem{Weinberg} S. L. Glashow, Nucl. Phys. {\bf B} {\bf 22} 579, 1961.
                   S. Weinberg, Phys. Rev. Lett. {\bf 19} 1264, 1967.
                    A. Salam, p. 367 of Elementary Particle Theory, ed. N. Svartholm (almquist and Wiksells, Stockholm), 1969.
\bibitem{Higgs}    P. W. Higgs, Phys. Rev. Lett {\bf 13} 2364, 1963.
                   P. W. Higgs, Phys. Rev. {\bf 145} 1156, 1966.
                   F. Englert and R. Brout, Phys. Rev. Lett {\bf 13} 321, 1964.
                   G. S. Guralnik, C. R.Hagen and T. W. Kibble, Phys. Rev. Lett {\bf 13} 585, 1965.
                   T. W. Kibble, Phys. Rev. {\bf 155} 1554, 1967.
\bibitem{Abbott} L. F. Abbott, Ref.TH.3113-CERN, 1981.
\bibitem{Sher} M. Sher, Phys. Rep. {\bf 179} 273, 1989.
               M. Sher, Phys. Lett. {\bf B} {\bf 317} 159, 1993. Addendum ibid. {\bf 331} 448, 1994.
 \bibitem{Jones} C. Ford, D. R. T. Jones, P.W. Stephenson and M. B. Einhorn, Nucl. Phys. {\bf B}{\bf 395} 17, 1993.

%\bibitem{Weinberg} S. Weinberg, Phys. Rev. Lett. {\bf 19} 1264, 1967.
%\bibitem{Salam} A. Salam, p. 367 of Elementary Particle Theory, ed. N. Svartholm (almquist and Wiksells, Stockholm), 1969.
%\bibitem{Glashow} S. L. Glashow, J. Iliopoulos and L. Maiani, Phys. Rev. D {\bf 2} 1285, 1970.
%\bibitem{Peskin} M. E. Peskin and D.V. Schroeder, " Quantum Field Theory", Perseus Books Publishing, L. L. C., 1995.
%\bibitem{PDG} K. Nakamura et al., Journal of Physics G, {\bf 37} 075021, 2010 (PDG).
%\bibitem{Gross} D. J. Gross and F. Wilczek, Phys. Rev. Lett {\bf 30} 1343, 1973.
%\bibitem{Politzer} H. D. Politzer, Phys. Rev. Lett {\bf 30} 1346, 1973.
%\bibitem{Gross1} D. J. Gross and F. Wilczek, Phys. Rev. D {\bf 8} 3633, 1973.
%\bibitem{Jones} D. R. T. Jones, Phys. Rev. D {\bf 25} 581, 1982.
%\bibitem{Hill} M. S. Fischler and C. T. Hill, Nucl. Phys. B {\bf 193} 53, 1981.
%\bibitem{Machacek} M. E. Machacek and M. T. Vaughn, Nucl. Phys. B {\bf 222} 83, 1983.
%\bibitem{Jack} I. Jack and H. Osborn, Nucl. Phys. B {\bf 249} 472, 1985.
%\bibitem{Mihaila} L. N. Mihaila, J. Salomon and M. Steinhauser, arXiv:1201.5868, 2012.
\end{thebibliography}

\end{document}





\end{document}




\label{den4546]
\end{eqnarray}


\begin{eqnarray}
\Phi(x)=\Phi_0+\frac{1}{\sqrt{2}}(\Phi_1(x)+i\Phi_2(x)).
\label{eq324}
\end{eqnarray}

The kinetic term for the scalar field reads:
\begin{eqnarray}
&&|D_{\mu}\Phi|^2=\frac{1}{2}(\partial_{\mu}\Phi_1)^2+\frac{1}{2}(\partial_{\mu}\Phi_2)^2+
\nonumber\\
&&\sqrt{2}g\Phi_0A_{\mu}\partial^{\mu}\Phi_2+g^2\Phi_0^2A_{\mu}A^{\mu}+...
\label{exp87868}
\end{eqnarray}

This translates into a mass term for the gauge boson according to the picture:
\vspace{0.5cm}
\begin{center}
\begin{picture}(280,40)(0,0)
\Photon(20,20)(40,20){1}{3}
\GCirc(50,20){10}{0.5}
\Photon(60,20)(80,20){1}{3}
\Text(100,20)[]{=}
\Photon(120,20)(160,20){1}{6}
\Vertex(140,20){1}
\Text(180,20)[]{+}
\Photon(200,20)(220,20){1}{3}
\Line(220,20)(240,20)
\Photon(240,20)(260,20){1}{3}
\Vertex(220,20){1}
\Vertex(240,20){1}
\end{picture}
\vspace{0.5cm}
\\{\sl Fig.1. The mechanism through which the Goldstone boson makes the gauge boson transverse.}
\end{center}

But there is another way in which one can view this process.  We rewrite the last terms of the Eq (\ref{exp87868})as:
\begin{eqnarray}
&&\sqrt{2}g\Phi_0A_{\mu}\partial^{\mu}\Phi_2+g^2\Phi_0^2A_{\mu}A^{\mu}=
(g\Phi_0A_{\mu}+\frac{1}{\sqrt{2}})^2-\frac{1}{2}(\partial_{\mu}\Phi_2)^2.
\label{form6578}
\end{eqnarray}

This cancels the kinetic term for the Goldstone boson $\Phi_2$ and redefines a new field $A_{\mu}'$:
\begin{eqnarray}
A_{\mu}'=A_{\mu}+\frac{1}{g\Phi_0\sqrt{2}}\partial_{\mu}\Phi_2.
\label{rdef4345}
\end{eqnarray}

We can interpret this as a discontinuity of the wave function of the gauge boson at the spontaneous symmetry breaking scale.This result is valid in any order of perturbation theory (See Fig.2).

\vspace{0.5cm}
\begin{center}
\begin{picture}(120,120)(0,0)
\Photon(20,40)(60,60){1}{6}
\Line(20,80)(60,60)
\Photon(60,60)(100,60){1}{6}
\Line(60,20)(60,100)
\Text(30,30)[]{$A_{\mu}$}
\Text(90,50)[]{$A_{\mu}'$}
\Text(30,90)[]{$\Phi_2$}
\Text(60,10)[]{SSB}
\Vertex(60,60){1}
\end{picture}
\vspace{0.5cm}
\\{\sl Fig. 2 The wave function for the gauge boson before and after spontaneous symmetry breaking. The dot represents the vev of the Higgs bsoson.}
\end{center}

We illustrated this for the simplest case but the argument is quite general and applies very well to the standard's model $W^{\pm}_{\mu}$ and $Z_{\mu}$ bosons.

One then can deduce that in the case when a linear combination of the fields goes unaltered through the spontaneous symmetry breaking the
corresponding wave function should be continuous (part form a phase factor) in each order of perturbation theory at the transition scale. Then one can infer (assuming the background gauge field method) that the
the variation with the scale of the corresponding coupling constant(or combination of coupling constants) is the same before and after symmetry breaking. This means that beta functions are the same at the transition scale.

In what follows we will apply this to the standard model's electromagnetic gauge boson but the result can be used for studying spontaneous symmetry breaking and phase transitions for any type of gauge theories.



\section{The Weinberg angle}


One of the fundamental parameters in the standard model of elementary particles is the Weinberg angle. This can be calculated with the aid of various experimental quantities like
the masses of the $W^{\pm}$ and Z bosons, the Fermi constant etc. (\cite{PDG}).  In some of the beyond the standard model theories the Weinberg angle can be predicted at some scale but no theoretical determination of it exists in the framework of the minimal standard model. In this work we fill this gap by showing that the Weinberg angle can be deduced exactly from the gauge structure and particle content of the standard model.

The Weinberg angle appears in the orthogonal matrix which diagonalizes the mass matrix after spontaneous symmetry breaking as:
\begin{eqnarray}
&&\left[
\begin{array}{c}
Z_{\mu}\\
A_{\mu}
\end{array}
\right]
\left[
\begin{array}{cc}
\cos(\theta_W)&-\sin(\theta_W)\\
\sin(\theta_W)&\cos(\theta_W)
\end{array}
\right]
\left[
\begin{array}{c}
A_{\mu}'^3\\
B_{\mu}'
\end{array}
\right].
\label{dsesds}
\end{eqnarray}

Here $A_{\mu}'^3$ belongs to the gauge group $SU(2)$ whereas $B_{\mu}'$ is the $U(1)_Y$ gauge boson.

We will consider the inverse orthogonal transformation and implement it in the standard model before and after electroweak symmetry breaking,

\begin{eqnarray}
&&\left[
\begin{array}{c}
A_{\mu}'^3\\
B_{\mu}'
\end{array}
\right]
\left[
\begin{array}{cc}
\cos(\theta_W)&\sin(\theta_W)\\
-\sin(\theta_W)&\cos(\theta_W)
\end{array}
\right]
\left[
\begin{array}{c}
Z_{\mu}\\
A_{\mu}
\end{array}
\right].
\label{dsesds}
\end{eqnarray}

We observe that even before the symmetry breaking the combination $A_{\mu}$ appears in all interaction terms with the coupling constant $\frac{gg'}{\sqrt{g^2+g'^2}}$. Assuming we are working in the background gauge field method this coupling constant will completely characterize the wave function. We then require the continuity of the probability associated with this wave function at the spontaneous symmetry breaking scale in any order of perturbation theory.


First let us define,
\begin{eqnarray}
&&s^2_W=\sin^2\theta_W=\frac{g'^2}{g^2+g'^2}
\nonumber\\
&&e^2=\frac{g^2g'^2}{g^2+g'^2} \,\,\, {\rm at\, the\, SSB\, scale}
\nonumber\\
&&\beta(g'^2)=\frac{g'^4}{16\pi^2}[a_1+a_2\frac{g'^2}{16\pi^2}+x\frac{g^2}{16\pi^2}]
\nonumber\\
&&\beta(g^2)=\frac{g^4}{16\pi^2}[b_1+b_2\frac{g^2}{16\pi^2}+y\frac{g'^2}{16\pi^2}]
\nonumber\\
&&\beta(e^2)=\frac{e^4}{16\pi^2}[z_1+z_2\frac{e^2}{16\pi^2}]
\label{nn65656}
\end{eqnarray}

We require,
\begin{eqnarray}
\beta(\frac{g'^2g^2}{g^2+g'^2})=\pm \beta(e^2) \,\,\,{\rm at \, EWSB\, scale}
\label{r4353}
\end{eqnarray}

to notice immediately that the sign must be negative. We obtain at the electroweak scale:
\begin{eqnarray}
\frac{g'^2g^2}{g^2+g'^2}[\frac{g'^2g^2}{g^2+g'^2}(a_1+b_1)+((a_2+y)g'^2+(b_2+x)g^2)\frac{1}{16\pi^2}]&=&-e^2[e^2z_1+e^2z_2\frac{1}{16\pi^2}]
\nonumber\\
a_1+b_1&=&-z_1
\nonumber\\
(a_2+y)g'^4+(b_2+x)g^4&=&-(z+a_2+b_2+x+y)g'^2g^2
\label{cond54546}
\end{eqnarray}


As a check of our approach the second equation in (\ref{cond54546}) is satisfied for the full spectrum of particles in the SM with or without scalars.
The third equation in (\ref{cond54546}) gives the ratio $\frac{g'^2}{g^2}=\tan^2\theta_W$ and in consequence the Weinberg angle:
\begin{eqnarray}
&&sin^2\theta_W=0.226329 \,\,\,{\rm in\, the\, presence \,of\,scalars}
\nonumber\\
&&sin^2\theta_W=0.175909 \,\,\,{\rm in\, the\, absence \,of\,scalars}
\label{res54646}
\end{eqnarray}

The first result is obtained with the exact content of scalars of the standard model and agrees very well with other results in the literature (see Table 10.1 in \cite{PDG})
The second is the Weinberg angle in the case when one eliminates the scalars from the theory. Even from this simple result it is evident that the scalars (with the qunatum numbers of the standard model) are necessary not only for achieving the electroweak symmetry breaking but also for an agreement with the experiment of the Weinberg angle.

Note that throughout these calculations the  quarks and leptons are assumed completely massless. One can take the masses into account as corrections to the value obtained in
Eq. (\ref{res54646}).


\section{Discussion}

The historical perspective on spontaneous symmetry breaking and other phase transitions of gauge theories is based on the study of the scalar sector with its associated effective potential.
One then translates the knowledge accumulated into the gauge and fermion sectors. But there is a lot that one can learn about gauge theories also from the pure gauge point of view since phase transitions characterize mainly the gauge structure.
Using pure gauge arguments we were able to find the mixing angle for the neutral gauge bosons at the spontaneous symmetry breaking scale in the standard model. We did that without relying on any experimental results but in perfect accordance to them. Our approach can be extended for studying phase transitions and other phenomena in many possible gauge systems.


\section*{Acknowledgments} \vskip -.5cm
I am happy to thank J. Schechter for support and encouragement and for useful comments on the manuscript.
This work has been supported by PN 09370102/2009.




\appendix

\section{}


Beta functions for the gauge coupling constants in the standard model have been computed up to three loop \cite{Gross}-\cite{Mihaila}.
We define,
\begin{eqnarray}
\beta(g^2)=\frac{d g^2}{d \ln(\mu^2)}
\label{def4344}
\end{eqnarray}
and list below the first two orders relevant for our calculations:
\begin{eqnarray}
&&\beta((g'^2)=\frac{(g')^4}{16\pi^2}[\frac{20}{9}n_g+\frac{1}{6}+(\frac{95}{27}n_g+\frac{1}{2})\frac{(g')^2}{16\pi^2}+(n_g+\frac{3}{2})\frac{g^2}{16\pi^2}]
\nonumber\\
&&\beta(g^2)=-\frac{g^4}{16\pi^2}[\frac{22}{3}-\frac{4}{3}n_g-\frac{1}{6}+(\frac{136}{3}-\frac{49}{3}n_g-\frac{13}{6})\frac{g^2}{16\pi^2}+(-\frac{1}{5}n_g-
\frac{3}{10})\frac{(g')^2}{16\pi^2}].
\label{beta324324}
\end{eqnarray}
Here $n_g$ represents the number of fermion generations.

The beta function for QED with all the fermion generations and also a charged scalar is:
\begin{eqnarray}
\beta(e^2)=\frac{e^4}{16\pi^2}[\frac{32}{9}n_g+\frac{1}{3}+(\frac{44}{27}n_g+1)\frac{e^2}{16\pi^2}].
\label{qqq22}
\end{eqnarray}






\begin{thebibliography}{15}
\bibitem{Weinberg} S. Weinberg, Phys. Rev. Lett. {\bf 19} 1264, 1967.
\bibitem{Salam} A. Salam, p. 367 of Elementary Particle Theory, ed. N. Svartholm (almquist and Wiksells, Stockholm), 1969.
\bibitem{Glashow} S. L. Glashow, J. Iliopoulos and L. Maiani, Phys. Rev. D {\bf 2} 1285, 1970.
\bibitem{Peskin} M. E. Peskin and D.V. Schroeder, " Quantum Field Theory", Perseus Books Publishing, L. L. C., 1995.
\bibitem{PDG} K. Nakamura et al., Journal of Physics G, {\bf 37} 075021, 2010 (PDG).
\bibitem{Gross} D. J. Gross and F. Wilczek, Phys. Rev. Lett {\bf 30} 1343, 1973.
\bibitem{Politzer} H. D. Politzer, Phys. Rev. Lett {\bf 30} 1346, 1973.
\bibitem{Gross1} D. J. Gross and F. Wilczek, Phys. Rev. D {\bf 8} 3633, 1973.
\bibitem{Jones} D. R. T. Jones, Phys. Rev. D {\bf 25} 581, 1982.
\bibitem{Hill} M. S. Fischler and C. T. Hill, Nucl. Phys. B {\bf 193} 53, 1981.
\bibitem{Machacek} M. E. Machacek and M. T. Vaughn, Nucl. Phys. B {\bf 222} 83, 1983.
\bibitem{Jack} I. Jack and H. Osborn, Nucl. Phys. B {\bf 249} 472, 1985.
\bibitem{Mihaila} L. N. Mihaila, J. Salomon and M. Steinhauser, arXiv:1201.5868, 2012.
\end{thebibliography}
\end{document}




The instanton approach for $SU(N)$ gauge theories with or without fermions has been initiated by 't Hooft \cite{Hooft} and further developed in \cite{Gross} and
\cite{Vainshtein}. In this method the separation of quantum degrees of freedom into zero modes (spin dependent) and positive modes (spin independent) is crucial. Moreover the so called zero modes have an " antiscreening " effect which is ultimately responsible for asymptotic freedom. The presence of fermions has an opposite effect. In \cite{Jora1} we suggest that in essence the magnetic properties of the QCD vacuum play a decisive role in the chiral symmetry breaking. Furthermore we show in \cite{Jora2} that in the process of gluino
decoupling from supersymmetric QCD separation into zero and positive modes is very important.





A very useful method for computing beta functions for the gauge coupling constant is the background gauge field method \cite{Abbott} which is based on the decomposition of the gauge field into a
background gauge field and a fluctuating field, the quantum gauge field. Even from the dawn of this method the background gauge field was regarded as an alternate source. However  the regular sources $J(x)$ and $\eta(x)$, $\eta'(x)$ (corresponding to the quantum gauge fields and ghost respectively) are introduced and one uses the conventional functional formalism to derive beta function or other loop corrections. The reason is simple; the background gauge field does not couple linearly to the other fields (as linear terms are canceled) and it is not obvious how one can compute simply Green functions with the background gauge field as a source.


In the present work we determine the two loop contribution to the beta function for Yang Mills theories using the background gauge field as the only source present in
the functional formalism. Of course the beta function is known up the fourth order \cite{beta} in the MS scheme so our main interest lies in the method that we introduce and the possibility for that to be developed for higher orders. We rely on the well-known result of the one loop effective potential (derived either in the perturbative or in the instanton approach) and on the decomposition of the one loop operators into spin dependent and spin independent operators corresponding to each field. Our derivation is entirely based on an analytic functional approach (see the Appendix) that does not involve the computation of any two loop Feynman diagram.




%The instanton approach for SU(N) gauge theories with $N_f$ flavors has been initiated by 't Hooft and further developed by Gross and Shifman. In this approach one can compute transition %amplitudes and determine the one loop beta function and the various degrees of freedom contributing to it. Here the separation into zero (spin dependent) and positive (spin independent) %modes is crucial. Even so one still needs to compute one loop Feynman diagrams as in the perturbative method. In the perturbation theory beta function for QCD like theories is known up
%to the fourth order and involves the calculation of a very large number of diagrams. The first two orders however are renormalization scheme independent and one should be able to
%obtain them even in the instanton case. The functional formalism for treating various orders is well established but does not exempt someone for computing Feynman diagrams.

%In the present work we will address the derivation of the two loop effective potential and beta function in afunctional formalism suitable for both instanton and perturbative approaches.
%This relies heavily on the one loop and on the separation into zero and positive modes but has the advantage that it does not involve the calculation of one single two loop diagram.
%We hope that our method could be extended to higher orders when one would have to face the choice of renormalization scheme. We will leave this for further work. For the order which
%we are going to discuss here the choice of renormalization scheme is as deprived of content as it is in the case of the instanton approach. There is only one result that one can get and %this is the correct result.

\section{The method}

The Yang Mills Lagrangian in the background gauge field method (where the gauge field is separated into $B_{\mu}^a+ A_{\mu}^a$ and $B_{\mu}^a$ is the background gauge field) has the expression:
\begin{eqnarray}
{\cal L}&=&-\frac{1}{4g^2}[F^a_{\mu\nu}+D_{\mu}A^a_{\nu}-D_{\nu}A_{\mu}^a+f^{abc}A^b_{\mu}A^c_{\nu}]^2-
\nonumber\\
&-&\frac{1}{2g^2}(D^{\mu}A_{\mu}^a)^2+{\bar c}^a[(-D^2)^{ac}-D_{\mu}f^{abc}A_{\mu}^b]c^c
\label{mainl}
\end{eqnarray}

This lagrangian contains quantum gauge fields $A^a_{\mu}$ and ghosts $c^a$, ${\bar c}^a$ and can be separated into a quadratic contribution and a higher order one.
The procedure for extracting the quadratic terms in this lagrangian is standard and after integration leads to the one loop effective potential \cite{Peskin}.

After some simplification the quadratic part of the  Lagrangian can be written as:
\begin{eqnarray}
{\cal L}_2=-\frac{1}{2g^2}[A^a_{\mu}(-(D^2)^{ac}g^{\mu\nu}-2f^{abc}F^{b \mu\nu})A^c_{\nu}]+{\bar c}^a(-(D^2)^{ac})c^c
\label{qadrmodes}
\end{eqnarray}

The trilinear and quadrilinear contributions are summarized below:
\begin{eqnarray}
{\cal L}_{3,4}&=&-\frac{1}{2g^2}(D_{\mu}A^a_{\nu}-D_{\nu}A^a_{\mu})f^{abc}A^{b\mu}A^{c\nu}-
\nonumber\\
&-&\frac{1}{4g^2}f^{abc}f^{ade}A_{\mu}^bA_{\nu}^cA^{d\mu}A^{e\nu}+{\bar c}^a(-D_{\mu}f^{abc}A_{\mu}^b)c^c
\label{therest54}
\end{eqnarray}

Then the effective action reduces to:
\begin{eqnarray}
e^{i \Gamma[B]}&=&\int {\cal D}A {\cal D}c\exp[i\int d^4x({\cal  L}+{\cal L}_{ct}]=
\nonumber\\
&=&\int {\cal D}{\cal c}\exp[i\int d^4 x[-\frac{1}{4g^2}(F^a_{\mu\nu})^2+{\cal L}_{ct}+{\cal L}_2+{\cal L}_{3,4}]]=
\nonumber\\
&=&\int {\cal D}A {\cal D}c \exp[i \int d^4x[-\frac{1}{4g^2}(F^a_{\mu\nu})^2+{\cal L}_2+{\cal L}_{ct}]]
\nonumber\\
&\times&[1+i\int d^4x {\cal L}_{3,4}-\frac{1}{2} \int \int d^4 x d^4 y{\cal L}_{3,4}(x){\cal L}_{3,4}(y)+....]
\label{exp6578}
\end{eqnarray}

where ${\cal L}_{ct}$ is the counterterm Lagrangian.

The quadratic term alone leads to the one loop effective potential whereas using standard functional procedures one can find higher order Feynman diagrams and solve for the higher order
contributions to it. Since the first two orders of the beta function are renormalization scheme independent one may wonder if it not possible to practically deduce two loop contributions using only Eq(\ref{exp6578}) and an entirely functional approach without considering and calculating any two loop Feynman diagram. In what follows we will show that this is indeed the case
by considering the background gauge field as a source in the functional approach.
For that first we need to express in a suitable form the quadratic operators:
\begin{eqnarray}
-\frac{1}{2g^2}[-(D^2)^{ac}-2f^{abc} F^{b\mu\nu}]
=-\frac{1}{2g^2}[-\partial^2+\Delta^{1}+\Delta^{2}+\Delta^J]=-\frac{1}{2g^2}\Delta,
\label{dec65748}
\end{eqnarray}

where $\Delta^1+\Delta^2$ is a spin independent operator and $\Delta^J$ is a spin dependent operator,
\begin{eqnarray}
\Delta^1&=&i[\partial^{\mu}B^b_{\mu}f^{abc}+B_{\mu}^bf^{abc}\partial^{\mu}]
\nonumber\\
\Delta^2&=&B^{a\mu}t^aB^b_{\mu}t^b
\nonumber\\
\Delta^j&=&-2f^{abc}F^{b\mu\nu}.
\label{res324}
\end{eqnarray}

No spin dependent operator acts on ghosts.


The one loop effective potential for a Yang Mills theory is obtained by computing,
\begin{eqnarray}
\exp[i\Gamma[B]]=\exp[i\int d^4 x [-\frac{1}{4g^2}(F^a_{\mu\nu})^2][\det(\Delta_{G,1})]^{-1/2}\det(\Delta_{G,0}),
\label{det55}
\end{eqnarray}

where $\Delta_{G,1}$ refers to the gauge fields  and $\Delta_{G,0}$ to the ghost fields.
This leads to:
\begin{eqnarray}
\Gamma[B]=-\frac{1}{4}(\frac{1}{g^2}\int d^4 x (F^a_{\mu\nu})^2+\frac{1}{2}\ln[\det(\Delta_{G,1})]-\ln[\det(\Delta_{G,0})]).
\label{loop8989}
\end{eqnarray}


Each operator $\Delta$ has the decomposition from Eq (\ref{res324})with the following calculated contribution to the one loop effective potential:
\begin{eqnarray}
&&-\frac{1}{4g^2}\int d^4x (F^a_{\mu\nu})^2\longrightarrow -\frac{1}{4}[-4N\ln{\frac{M^2}{k^2}}]\int d^4x (F^a_{\mu\nu})^2\,\,\,{\rm zero\, modes\, contribution\,for\,quantum\,gauge\,fields}
\nonumber\\
&&-\frac{1}{4g^2}\int d^4x (F^a_{\mu\nu})^2\longrightarrow -\frac{1}{4}[\frac{2}{3}N\ln{\frac{M^2}{k^2}}]\int d^4x (F^a_{\mu\nu})^2\,\,\,{\rm positive\, modes\, contribution\,for\,quantum\,gauge\,fields}
\nonumber\\
&&-\frac{1}{4g^2}\int d^4x (F^a_{\mu\nu})^2\longrightarrow -\frac{1}{4}[-\frac{1}{3}N\ln{\frac{M^2}{k^2}}]\int d^4x (F^a_{\mu\nu})^2\,\,\,{\rm positive\, modes\, contribution\,for\,ghost\,fields}.
\label{contr657}
\end{eqnarray}

However it is more convenient for us to represent these results as,
\begin{eqnarray}
&&\ln[\det(\Delta^1+\Delta^2)_{G,1}]=\frac{1}{3}NX\int d^4x (F^a_{\mu\nu})^2
\nonumber\\
&&\ln[\det(\Delta^J)_{G,1}]=-2NX\int d^4x (F^a_{\mu\nu})^2
\nonumber\\
&&\ln[\det(\Delta^1+\Delta^2)_{G,0}]=\frac{1}{12}NX\int d^4x (F^a_{\mu\nu})^2,
\label{somelist767}
\end{eqnarray}

to keep track of the proper regularization procedures. The exact role of X will be revealed in section VI.

 Note that in the case of dimensional regularization each term in Eq(\ref{somelist767}) will be multiplied by a different X factor such that the infinite parts are the same whereas the finite parts are different.

In the next section we will implement a procedure for computing the two loop beta function by considering the background gauge field as a source. Knowing the one loop effective potential
before and after the path integration we use simple results as those in the Appendix to derive the relevant contributions.  The key point in the whole approach is the separation of the one loop operators in spin dependent and spin independent ones both in the integrand and in the final results. In order to justify that let us consider the general form of the integrals with
$D_{ij}$ and $B_{ij}$ the spin independent and spin dependent operators respectively:
\begin{eqnarray}
&&\int \prod_i d\xi_i\exp[-\xi_i(B_{ij}+D_{ij})\xi_j]=
\nonumber\\
&&\int \prod_k d x_k\exp[-x_i(O^t(B+D)O)_{ii}x_i]= \sqrt{\pi}[\det[O^t(B+D)O]]^{-1/2}
\label{someres7867}
\end{eqnarray}

Here O is the orthogonal operator which realizes the diagonalization. Taking the logarithm of the expression we observe that at first order at least there is no interference between the eigenmodes of D and B such that $O^t(B+D)O=B_d+D_d+...$ where $B_d$ and $D_d$ are the diagonalized operators. This means that at least in the first order the operators B and D are diagonalized by the same orthogonal matrix. We will use this feature in our computation.






\section{The quadrilinear term}

We start with the simplest contribution to the effective potential, the quadrilinear term:
\begin{eqnarray}
\int {\cal D} A \frac{-i}{4g^2}f^{abc}f^{ade}A_{\mu}^bA_{\nu}^cA^{d\mu}A^{e\nu}\exp[i\int d^4 x[-\frac{1}{4g^2}(F^a_{\mu\nu})^2+{\cal L}_2]]
\label{qop657}
\end{eqnarray}


The correct structure can be obtained from
\begin{eqnarray}
&&\int {\cal D} A \frac{ig^2}{4}\int d^4 x d^4y \delta(x-y)\frac{\delta^2}{\delta F^b_{\mu\nu}(x) \delta F^{b\mu\nu}(y)}\exp[\int d^4 x \frac{i}{g^2}f^{abc}F^{b\mu\nu}A^a_{\mu}A^c_{\nu}]=
\nonumber\\
&&=\int {\cal D} A \frac{-i}{4g^2}f^{abc}f^{ade}A_{\mu}^bA_{\nu}^cA^{d\mu}A^{e\nu}\exp[\int d^4 x \frac{i}{g^2}f^{abc}F^{b\mu\nu}A^a_{\mu}A^c_{\nu}]
\label{res45367}
\end{eqnarray}

so it is clear that this operator comes only from the spin dependent part in the one loop effective potential.
 Then quite clearly using the results in the appendix one finds:
\begin{eqnarray}
&&\int {\cal D} A \frac{-i}{4g^2}f^{abc}f^{ade}A_{\mu}^bA_{\nu}^cA^{d\mu}A^{e\nu}\exp[i\int d^4 x[-\frac{1}{4g^2}(F^a_{\mu\nu})^2+{\cal L}_2]]=
\nonumber\\
&&ig^2[-\frac{1}{8}\int d^4 x d^4 y \delta(x-y)\frac{\delta^2}{\delta F^b_{\mu\nu}(x)F^{b\mu\nu}(y)}\exp[-2NX\int d^4 z F^2(z)]\exp[\frac{5}{3}N X \int d^4 z F^2(z)]+
\nonumber\\
&+&\int d^4 x d^4 y \delta(x-y)\frac{3}{16}\frac{\delta}{\delta F^b_{\mu\nu}(x)}\exp[-2NX \int d^4 z F^2(z)]\times\exp[\frac{10}{3}N X\int d^4 z F^2(z)]\times
\nonumber\\
&&\times \frac{\delta}{\delta F^b_{\mu\nu}(x)}\exp[-2NX\int d^4 z F^2(z)]\times
{\rm one\,loop\, contribution}
\nonumber\\
&=&ig^2 4X^2N^2\int d^4 z F^2(z)\times{\rm one\,loop\, contribution}\times \exp[-\frac{1}{3}NX\int d^4 z F^2(z)][1+...]
\label{firsteerm55}
\end{eqnarray}

For each stage of the calculations we select only the term proportional to the background gauge field invariant $F^a_{\mu\nu}F^{a\mu\nu}$ and drop conveniently terms proportional to higher order invariants.










\section{The trilinear pure gauge terms}



This contribution corresponds to the term:

\begin{eqnarray}
&&-\frac{1}{2g^4}[\int d^4x (A^{a\nu}D_{\rho}A^c_{\nu}A^{m\rho}f^{acm})(x)\int d^4y (A^{d\mu}D_{\sigma}A^e_{\mu}A^{n\sigma}f^{den})(y)]\exp[i\int d^4 x[-\frac{1}{4g^2}(F^a_{\mu\nu})^2+{\cal L}_2]]
\label{optwo9786}
\end{eqnarray}

 It is simpler in this case to work with the gauge tensor $F^a_{\mu\nu}t^a$ where $t^a$ is the generator in the adjoint representation such that,
 \begin{eqnarray}
 Tr(F^a_{\mu\nu}t^a F^{a\mu\nu}t^a)=NF^a_{\mu\nu}F^{a\mu\nu}
 \label{s333}
 \end{eqnarray}

 and $B^{\mu}=b^{\mu c}t^c$.

 First we notice that the part of the term in Eq (\ref{optwo9786}) that contains covariant derivatives can be easily obtained from the spin independent quadratic operator
 in accordance to:

 \begin{eqnarray}
 \frac{\delta[\int d^4 x \exp[-\frac{i}{2g^2}A^a_{\mu}(\Delta^1+\Delta^2)^{ac}A^c_{\nu}]]}{\delta (B^{\rho})_{ac}}=
 \frac{1}{g^2}A^a_{\mu}D_{\rho}A^c_{\nu}g^{\mu\nu}\exp[-\frac{i}{2g^2}\int d^4 x A^a_{\mu}(\Delta^1+\Delta^2)^{ac}A^c_{\nu}]
 \label{somef4567}
 \end{eqnarray}

 Then,
 \begin{eqnarray}
 &&\frac{\delta^2}{\delta B^{\rho}_{ac} \delta B^{\sigma}_{de}}\exp[-\frac{i}{2g^2}\int d^4 x A^a_{\mu}(\Delta^1+\Delta^2)^{ac}A^c_{\nu}]=
 \nonumber\\
 &&[\frac{-i}{g^2}A^a_{\nu}A^c_{\nu}\delta^{\rho\sigma}\delta_{ad}\delta_{ce}+
 \frac{1}{g^4}A^a_{\nu}D_{\rho}A^c_{\nu}A_{\mu}^dD_{\sigma}A^e_{\mu}]\times
 \nonumber\\
 &&\times\exp[-\frac{i}{2g^2}\int d^4 x A^a_{\mu}(\Delta^1+\Delta^2)^{ac}A^c_{\nu}].
 \label{double657}
 \end{eqnarray}


We need two more component gauge fields which can be simply obtained from the spin dependent operator.
 The desired result is finally obtained from:
 \begin{eqnarray}
&& \frac{\delta^2}{\delta B^{\rho}_{ac}(x)\delta B^{\sigma}_{de}(y)}\exp[-\frac{i}{2g^2}\int d^4 xA^a_{\mu}(\Delta^1+\Delta^2)A^c_{\nu}]
 \frac{\delta}{\delta F^{mn}_{\rho\sigma}(u)}\exp[\frac{1}{g^2}\int d^4 x A^a_{\mu}F^{\mu\nu}_{ac}A^c_{\nu}]=
 \nonumber\\
 &&=[\frac{i}{g^4}A^{m\rho}A^{n\sigma}A^{a\nu}A^c_{\nu}\delta^{\rho\sigma}\delta^{\mu\nu}\delta_{ad}\delta_{ce}\delta(x-y)+
 \frac{1}{g^6}(A^{a\nu}D_{\rho}A^c_{\nu})(x)A^{m\rho}(u)(A^{d\mu}D_{\sigma}A^e_{\mu})(y)A^{n\sigma}(u)]\times
 \nonumber\\
 &&\times\exp[-\frac{i}{2g^2}\int d^4 xA^a_{\mu}(\Delta^1+\Delta^2)A^c_{\nu}+\frac{1}{g^2}\int d^4 x A^a_{\mu}F^{\mu\nu}_{ac}A^c_{\nu}].
 \label{almostf3245}
 \end{eqnarray}


 In order to get the desired term we need to multiply by $f^{acm}f^{den}$ which will lead to the cancelation by symmetry of the unwanted  first term in the last line
 of Eq (\ref{almostf3245}).


 Eq(\ref{almostf3245}) has the correct structure except for the space time dependence. We will use a small artifice in order to correct that. First we write:
 \begin{eqnarray}
 \frac{1}{g^2}\int d^4 x A^{m \rho}(x)F^{mn}_{\rho\sigma}(x)A^{n\sigma}(x)=
 \frac{1}{g^2} \int d^4 u d^4 v A^{m\rho}(u)F^{mn}_{\rho\sigma}(u)A^{n\sigma}(v)\delta(u-v).
 \label{art33425}
 \end{eqnarray}

 Then,

 \begin{eqnarray}
 \frac{\delta^2}{\delta(\delta(w_1-w_2)\delta F^{mn}_{\rho\sigma}(w_1)}\frac{1}{g^2}
 \int d^4 u d^4 v A^{m\rho}(u)F^{mn}_{\rho\sigma}(u)A^{n\sigma}(v)\delta(u-v)=\frac{1}{g^2}A^{m\rho}(w_1)A^{n\sigma}(w_2).
 \label{der454}
 \end{eqnarray}

 The desired contribution can be computed from:

 \begin{eqnarray}
 &&-\frac{1}{2g^4}[\int d^4x (A^{a\nu}D_{\rho}A^c_{\nu}A^{m\rho}f^{acm})(x)\int d^4y A^{d\mu}D_{\sigma}A^e_{\mu}A^{n\sigma}f^{den})(y)]\exp[i\int d^4 x[-\frac{1}{4g^2}(F^a_{\mu\nu})^2+{\cal L}_2]]=
 \nonumber\\
 &&-\frac{1}{2}g^2\int d^4x d^4 y d^4u d^4v \delta(x-u)\delta(y-v)f^{acm}f^{den}\times
 \nonumber\\
 &&\times\frac{\delta^2}{\delta (B^{\rho})_{ac}(x) \delta (B^{\sigma})_{de}(y)}\exp[-i\frac{1}{2g^2}\int d^4x A^a_{\nu}(\Delta^1+\Delta^2)A^c_{\nu}]\times
 \nonumber\\
 &&\times\frac{\delta^2}{\delta F^{mn}_{\rho\sigma}(u) \delta(\delta(u-v))}\exp[\frac{1}{g^2}\int d^4 x A^{m\rho}F^{mn}_{\rho\sigma}A^{n\sigma}]\times {\rm one\,loop\,ghost\, term}.
 \label{somres543}
 \end{eqnarray}




 Using Eq(\ref{somres543}) and the results from Appendix A we obtain:
 \begin{eqnarray}
 &&-\frac{1}{2g^4}[\int d^4x (A^{a\nu}D_{\rho}A^c_{\nu}A^{m\rho}f^{acm})(x)\int d^4y (A^{d\mu}D_{\sigma}A^e_{\mu}A^{n\sigma}f^{den})(y)]\exp[i\int d^4 x[-\frac{1}{4g^2}(F^a_{\mu\nu})^2+{\cal L}_2]]=
 \nonumber\\
&&=-\frac{g^2}{2}\int d^4x d^4 y d^4u d^4v \delta(x-u)\delta(y-v)f^{acm}f^{den}\times
\nonumber\\
&&-\frac{1}{2}g^2\frac{\delta^2}{\delta (B^{\rho})_{ac}(x)\delta (B^{\sigma})_{de}(y)}\exp[\frac{N}{3}X\int d^4x F^2]\times
\nonumber\\
&&\times\frac{\delta^2}{\delta F^{mn}_{\rho\sigma}(u) \delta(\delta(u-v))}\times\exp[-2NX\int d^4x F^2]\times\exp[\frac{5}{3}NX\int d^4x F^2]\times{\rm ghost\,contribution}
\nonumber\\
&&=-4ig^2N^2X^2\int d^4x F^2(x)\times{\rm one \,loop\,contribution}.
\label{fin6578}
\end{eqnarray}








\section{Terms that include ghosts}

 There is one quadratic term which contains ghosts and two higher order contributions.
  We will need to determine two terms, respectively:
  \begin{eqnarray}
&&-\frac{1}{2}( D_{\mu}{\bar c}^a f^{abc} A^b_{\mu}c^c)^2\exp[i\int d^4 x[-\frac{1}{4g^2}(F^a_{\mu\nu})^2+{\cal L}_2]]
\nonumber\\
 &&-\frac{1}{2g^2}D_{\mu}{\bar c}^a f^{abc} A^b_{\mu}c^c(D_{\rho}A_{\sigma}^df^{def}A^{e\rho}A^{f\sigma})\exp[i\int d^4 x[-\frac{1}{4g^2}(F^a_{\mu\nu})^2+{\cal L}_2]].
 \label{ter4356}
 \end{eqnarray}

 We start by analyzing the first term in Eq(\ref{ter4356}).

Both these expressions contain the ghost fields mixed with quantum gauge fields. For the sake of simplicity we write:
 \begin{eqnarray}
 {\bar c}^a(-D^{\mu}f^{abc}A^b_{\mu})c^c \equiv
 D_{\mu}{\bar c}^a f^{abc} A^b_{\mu}c^c
 \label{gh5467}
 \end{eqnarray}
which is true up to a total derivative.
Moreover the quadratic term must also be written in a similar manner as:
 \begin{eqnarray}
 {\bar c}^a(-D^2)_{ac}c^c \equiv
 (-D^2{\bar c}^ac^c)
 \label{e32456}
 \end{eqnarray}

We can switch in all these terms the order of the ghost field without problem since we are dealing with the square of the trilinear operator. Then the analogy with the previous case is obvious and with exactly the same derivation we obtain:

\begin{eqnarray}
 &&-\frac{1}{2g^4}[\int d^4x (c^{a}D_{\rho}c^cA^{m\rho}f^{acm})(x)\int d^4yc^{d}D_{\sigma} c^eA^{n\sigma}f^{den})(y)]\exp[i\int d^4 x[-\frac{1}{4g^2}(F^a_{\mu\nu})^2+{\cal L}_2]]=
 \nonumber\\
&&=-\frac{g^2}{2}\int d^4x d^4 y d^4u d^4v \delta(x-u)\delta(y-v)f^{acm}f^{den}\times
\nonumber\\
&&-\frac{1}{2}g^2\frac{\delta^2}{\delta (B^{\rho})_{ac}(x)\delta (B^{\sigma})_{de}(y)}\exp[\frac{1}{12}N X\int d^4x F^2]\times
\nonumber\\
&&\times\frac{\delta^2}{\delta F^{mn}_{\rho\sigma}(u) \delta(\delta(u-v))}\exp[-2NX\int d^4x F^2]\times
\nonumber\\
&&\times \exp[(\frac{5}{3}N-\frac{1}{12}N)X\int d^4F^2]\times{\rm one\,loop\,spin\,independent\,gauge\,contribution}
\nonumber\\
&&=ig^2\frac{1}{6}N^2X^2\int d^4x F^2(x)\times\exp[-(\frac{1}{3}NX\int d^4F^2]\times{\rm one \,loop\,contribution}.
\label{fin657888}
\end{eqnarray}


In this approach the second term in Eq (\ref{ter4356}) will give no contribution since it will appear as a product of three functional derivatives corresponding to the spin dependent, spin independent and  ghost terms in the one loop potential and this would lead to a result proportional to a gauge invariant (in the background gauge field) of order higher than two.





\section{Connecting the dots}

We add the results from Eq(\ref{firsteerm55}), Eq (\ref{fin6578}) and Eq(\ref{fin657888}) to obtain for the second order correction:
\begin{eqnarray}
-ig^2\frac{17}{6}N^2 X^2\int d^4 x(F^a_{\mu\nu})^2
\label{rez3232}
\end{eqnarray}

 Here X is just the result of the regularization at one loop. After taking into account all gauge and internal indices X amounts to a one loop scalar integral so one can write schematically for
the proper loop result:
\begin{eqnarray}
\approx \int d^4 x d^4 y F_{\mu\nu}^a U(x-y)F^{a\mu \nu} (y)
\label{reg554}
\end{eqnarray}

The two loop expression then corresponds to:
\begin{eqnarray}
\approx \int d^4 x d^4 y F_{\mu\nu}^a (U U)(x-y)F^{a\mu \nu} (y)
\label{res4343}
\end{eqnarray}

where $(UU)(x-y)$ is the result of the scalar two loop diagram with two bubbles and two external legs. But this regularized is just the square of U regularized at one loop so practically we do not need it. So finally we will take for X the expression:
\begin{eqnarray}
X=i\frac{1}{(4\pi)^2}\int_0^1\ln(\frac{x\Lambda^2}{-x(1-x)k^2})=i\frac{1}{(4\pi)^2}(1+\ln{\Lambda^2/k^2})
\label{fin545}
\end{eqnarray}
We multiply by a loop factor $\frac{1}{2}$ to obtain the second order contribution to the coupling constant:
\begin{eqnarray}
i\frac{g^2(k)}{4}\int
d^4 x (F_{\mu\nu}^a)^2 =i\frac{g^2}{4}[1-\frac{11N}{3}\frac{1}{(4\pi)^2}\ln{M^2/k^2}-\frac{34}{3}N^2\frac{g^2}{(4\pi)^4}\ln{M^2/k^2}+...]\int d^4 x (F_{\mu\nu}^a)^2
\label{y677}
\end{eqnarray}

From that the known result for the two loop beta function is obtained:
\begin{eqnarray}
\beta(g^2)=\frac{g^4}{(4\pi)^2}[-\frac{11}{3}N-\frac{34}{3}N^2\frac{g^2}{16\pi^2}].
\label{res43332}
\end{eqnarray}
Here we defined $\beta(g)=\frac{d g^2}{d\ln(\mu^2)}$.

\section{Discussion}

It is important to know the beta function for non-abelian gauge theories for several reasons. First the one loop coefficient of beta function was the main clue that these theories are endowed with asymptotic freedom. Second higher order coefficients can reveal information about the phase structure of these type of models. And it is always useful to learn more about the mathematical structures that lie at the basis of contemporary particle physics.

In the present work we do not aim to obtain higher order correction to the beta function but rather to introduce a new method that can ease their calculation. All previous computations of
order higher than two of beta function rely heavily on computer technique, rightly so because for example the fourth order involves about 50.000 Feynman diagrams. We determine the two loop
coefficient of the beta function as a check of the method. In the process we do not compute any two loop diagram and the computation in entirely analytic.

Our procedure can be easily extended to higher orders and be adjusted to work  with dimensional regularization and in MS scheme. This would amount to small changes in the general set-up.
We leave all these for future work.













\section*{Acknowledgments} \vskip -.5cm
I am happy to thank J. Schechter for support and encouragement and for useful comments on the manuscript.
This work has been supported by PN 09370102/2009.




\appendix

\section{}

First let us review some basic steps regarding integration in the functional approach. We start with the simple formula:
\begin{eqnarray}
\prod_k \int d\xi_k \exp[-\xi_i B_{ij}\xi_j]=
\prod_k\int dx_k \exp[-b_ix_i^2]=
\prod_i\sqrt{\frac{\pi}{b_i}}={\rm const}[\det{B}]^{-1/2}
\label{firstform}
\end{eqnarray}
In what follows we will drop the constant factors.

We extend this to a slightly more complicated case; assume the following:
\begin{eqnarray}
\prod_k \int   d \xi_k \exp[\xi_i B_{ij}\xi_j]\exp[\xi_i D_{ij} \xi_j]=\prod_i\sqrt{\frac{\pi}{\det(B+D)}}
\label{somfo7680}
\end{eqnarray}

Note that B and D correspond in our case to the spin dependent and spin independent operators respectively in the quadratic part of the Lagrangian.
Let us now differentiate one of the above factors with respect to a quantity $H_m$ where the index includes any type of subscript (Note that in the end this $H_m$ will be the background gauge field tensor or any component of it).
\begin{eqnarray}
&&\prod_k \int  \xi_k \exp[-\xi_i B_{ij} \xi_j]\frac{\delta}{\delta H_m} \exp[-\xi_i D_{ij} \xi_j]=
\nonumber\\
&&-\frac{1}{2}[\det(B+D)]^{-1/2}\sum_i\frac{\delta d_i}{\delta H_m}\frac{1}{d_i+b_i}
\label{res333}
\end{eqnarray}

Here $d_i$ and $b_i$ are the eigenmodes of D and B. The operators  B and D  are diagonalized together in the one loop effective potential such that,
\begin{eqnarray}
&&-\xi_i B_{ij} \xi_j\Rightarrow -bx_i^2+x_i K_{ij} x_j
\nonumber\\
&&- \xi_i D_{ij} \xi_j\Rightarrow -d_ix_i^2-x_i K_{ij}'x_j
\label{someform9989}
\end{eqnarray}

where $d_i$ and $b_i$ are proportional to the square of the background gauge field tensor $(F^a_{\mu\nu})^2$ whereas $K_{ij}$ and $K_{ij}'$ contain higher powers of the background gauge field. It is perfectly safe to ignore corrections of order $K_{ij}$ or $K_{ij}'$.

We need to compute higher order derivatives of various types:
\begin{eqnarray}
&&\prod_k  \int d\xi_k \exp[-\xi_i B_{ij} \xi_j]\frac{\delta^2}{\delta H_m \delta H_n}\exp[-\xi_i D_{ij}\xi_j]=
\nonumber\\
&&[-\frac{1}{2}\frac{\delta^2 d_i}{\delta H_n \delta H_m}\frac{1}{b_i+d_i}+\frac{\delta d_i}{\delta H_m}\frac{\delta d_j}{\delta H_n}
(\frac{1}{4}\frac{1}{(b_i+d_i)(b_j+d_j)}+\frac{1}{2}\frac{1}{(b_i+d_i)^2}\delta_{ij})]
\nonumber\\
&&\times[\det(B+D)]^{-1/2}.
\label{rr4343}
\end{eqnarray}


Furthermore,

\begin{eqnarray}
&&\prod_k \int d \xi_k \frac{\delta}{\delta H_p}\exp[-\xi_i B_{ij} \xi_j]\frac{\delta^2}{\delta H_m \delta H_n} \exp[-\xi_i D_{ij} \xi_j]=
\nonumber\\
&&[\frac{\delta b_k}{\delta H_p}\frac{\delta^2 d_i}{\delta H_m \delta H_n}(\frac{1}{4}\frac{1}{(b_i+d_i)(b_j+d_j)}+\frac{1}{2}\frac{1}{(b_i+d_i)^2}\delta_{ik})+
\nonumber\\
&&\frac{\delta b_k}{\delta H_p}\frac{\delta d_i}{\delta H_m}\frac{\delta d_j}{\delta H_n}(-\frac{1}{8}\frac{1}{(b_i+d_i)(b_j+d_j)(b_k+d_k)}
-\frac{1}{4}\frac{1}{(b_i+d_i)^2(b_j+d_j)}\delta_{ik}
-\frac{1}{4}\frac{1}{(b_i+d_i)^2(b_j+d_j)}\delta_{jk}
\nonumber\\
&&-\frac{1}{4}\frac{1}{(b_i+d_i)^2(b_k+d_k)}\delta_{ij}+\frac{1}{2}\frac{1}{(b_i+d_i)^3}\delta_{ij}\delta_{ik})][\prod_l(b_l+d_l)]^{-1/2}.
\label{uyuy76}
\end{eqnarray}



We treat separately the ghost terms. Thus,
\begin{eqnarray}
\prod_k  \int d\theta_k d \theta_k^* \exp[\theta_i^* C_{ij} \theta_j]=
\prod_k \int d z_i dz_i^*  \exp[-c_i |z_i|^2]=\prod_i c_i=\det[C]
\label{gh456}
\end{eqnarray}

from which we can deduce,

\begin{eqnarray}
\prod_k \int d \theta_k d \theta_k^* \frac{\delta}{\delta H_m}\exp[\theta_i^* C_{ij} \theta_i]=
\prod_k \int dz_k dz^*_k \frac{\delta}{\delta H_m} \exp[-z_i^*c_i z_i]=
\frac{\delta c_i}{\delta H_m}\frac{1}{c_i}[\prod_k c_k]
\label{firste43526}
\end{eqnarray}

and further,

\begin{eqnarray}
&&\prod_k \int d \theta_k d \theta_k^* \frac{\delta^2}{\delta H_m \delta H_n}\exp[\theta_i^* C_{ij} \theta_i]=
\prod_k \int dz_k dz^*_k \frac{\delta^2}{\delta H_m \delta H_n} \exp[-z_i^*c_i z_i]=
\nonumber\\
&&\frac{\delta^2 c_i}{\delta H_m \delta H_n}\frac{1}{c_i} [\prod_k c_k]
+\frac{\delta c_i}{\delta H_m}\frac{\delta c_j}{\delta H_n}[-\frac{1}{c_i^2}\delta_{ij}+\frac{1}{c_ic_j}][\prod_k c_k].
\label{finform65789}
\end{eqnarray}





It is useful to give in what follows some results regarding the functional derivatives of the square of the gauge tensor (Here $F^a_{m\nu}t^a=F_{\mu\nu}$, where $t^a$ is the generator
in the adjoint representation).
\begin{eqnarray}
&&\frac{\delta^2}{\delta B^{\rho}_{ac}(x) \delta B^{\sigma}_{de}(y)}[\int d^4 z Tr(F_{\mu\nu})^2(z)]=
\nonumber\\
&&\int d^4 z 2(F_{\mu\nu})_{gh}(z)\frac{\delta^2 (F^{\mu\nu})_{hg}}{\delta B^{\rho}_{ac}(x) \delta B^{\sigma}_{de}(y)}+
\frac{\delta (F_{\mu\nu})_{gh}(z)}{\delta B^{\rho}_{ac}(x)}\frac{\delta (F^{mu\nu}_{hg}(z)}{\delta B^{\sigma}_{de}(y)}=
\nonumber\\
&&8i[(F_{\rho\sigma})_{ae}(x)\delta_{cd}\delta(x-y)-(F_{\rho\sigma})_{dc}\delta_{ae}\delta(x-y)].
\label{someusefrs56}
\end{eqnarray}

Furthermore from this one can deduce:
\begin{eqnarray}
&&\int d^4x d^4 y d^4u d^4v \delta(x-u)\delta(y-v)f^{acm}f^{den}\times
\nonumber\\
&&\frac{\delta^2}{\delta (B^{\rho})_{ac}(x)\delta (B^{\sigma})_{de}(y)}\exp[k1\int d^4z (F^a_{\mu\nu}F^{a\mu\nu})(z)]\times
\nonumber\\
&&\times\frac{\delta^2}{\delta F^{mn}_{\rho\sigma}(u) \delta(\delta(u-v))}\exp[k2\int d^4z (F^a_{\mu\nu}F^{a\mu\nu})(z)]=
\nonumber\\
&&=-16ik_1k_2\int d^4 x (F_{\mu\nu}^a)^2\times\exp[(k_1+k2)\int d^4z (F^a_{\mu\nu}F^{a\mu\nu})(z)]
\label{fin657899}
\end{eqnarray}




















\begin{thebibliography}{15}
\bibitem{Hooft} G 't Hooft, Phys. Rev. D {\bf 14}, 3432, 1976.
\bibitem{Gross} C. G. Callan, R. Dashen and D. J. Gross, Phys. Rev. D {\bf17}, 2717, 1978;{\bf 19}, 1826, 1979.
\bibitem{Vainshtein} A. Vainshtein, V. Zakharov, V. Novikov and M. Shifman, Sov. Phys. Usp. {\bf 25}, 195, 1982.
\bibitem{Jora1} R. Jora, Phys. Rev. D {\bf 82}, 056005, 2010; arXiv:1004.3660.
\bibitem{Jora2} R. Jora, arXiv:1101.1395, 2011.
\bibitem{Abbott} L. F. Abbott, Acta Phys. Polonica, B{\bf 13}, 33, 1982.
\bibitem{beta} W. E. Caswell, Phys. Rev. Lett. {\bf 33}, 244, 1974;  D. R. T. Jones, Nucl. Phys. B {\bf 75}, 531, 1974;
E. S. Egorian, O. V. Tarasov, Theor. Mat. Fiz. {\bf 41}, 26, 1979; O. V. Tarasov, A. A. Vladimirov, A. Yu Zharkov, Phys. Lett B {\bf 93}, 429, 1980;
S. A. Larin, J. A. M. Vermaseren, Phys. Lett. B {\bf 303}, 334, 1993;  T.van Ritbergen, J. A. M. Vermaseren and S. A. Larin, Phys. Lett B {\bf 400}, 379, 1997.
\bibitem{Peskin} M. E. Peskin and D.V. Schroeder, " Quantum Field Theory", Perseus Books Publishing, L. L. C., 1995(pg
533-543).
\end{thebibliography}
\end{document}




















1208.3753.extracted_bib
\begin{thebibliography}{15}
\bibitem{CMS} The CMS Collaboration, arXiv:1207.7235, 2012.
\bibitem{Atlas} The ATLAS Collaboration, arXiv:1207.7214, 2012.
\bibitem{Weinberg} S. L. Glashow, Nucl. Phys. {\bf B} {\bf 22} 579, 1961.
                   S. Weinberg, Phys. Rev. Lett. {\bf 19} 1264, 1967.
                    A. Salam, p. 367 of Elementary Particle Theory, ed. N. Svartholm (almquist and Wiksells, Stockholm), 1969.
\bibitem{Higgs}    P. W. Higgs, Phys. Rev. Lett {\bf 13} 2364, 1963.
                   P. W. Higgs, Phys. Rev. {\bf 145} 1156, 1966.
                   F. Englert and R. Brout, Phys. Rev. Lett {\bf 13} 321, 1964.
                   G. S. Guralnik, C. R.Hagen and T. W. Kibble, Phys. Rev. Lett {\bf 13} 585, 1965.
                   T. W. Kibble, Phys. Rev. {\bf 155} 1554, 1967.
\bibitem{Abbott} L. F. Abbott, Ref.TH.3113-CERN, 1981.
\bibitem{Sher} M. Sher, Phys. Rep. {\bf 179} 273, 1989.
               M. Sher, Phys. Lett. {\bf B} {\bf 317} 159, 1993. Addendum ibid. {\bf 331} 448, 1994.
 \bibitem{Jones} C. Ford, D. R. T. Jones, P.W. Stephenson and M. B. Einhorn, Nucl. Phys. {\bf B}{\bf 395} 17, 1993.

%\bibitem{Weinberg} S. Weinberg, Phys. Rev. Lett. {\bf 19} 1264, 1967.
%\bibitem{Salam} A. Salam, p. 367 of Elementary Particle Theory, ed. N. Svartholm (almquist and Wiksells, Stockholm), 1969.
%\bibitem{Glashow} S. L. Glashow, J. Iliopoulos and L. Maiani, Phys. Rev. D {\bf 2} 1285, 1970.
%\bibitem{Peskin} M. E. Peskin and D.V. Schroeder, " Quantum Field Theory", Perseus Books Publishing, L. L. C., 1995.
%\bibitem{PDG} K. Nakamura et al., Journal of Physics G, {\bf 37} 075021, 2010 (PDG).
%\bibitem{Gross} D. J. Gross and F. Wilczek, Phys. Rev. Lett {\bf 30} 1343, 1973.
%\bibitem{Politzer} H. D. Politzer, Phys. Rev. Lett {\bf 30} 1346, 1973.
%\bibitem{Gross1} D. J. Gross and F. Wilczek, Phys. Rev. D {\bf 8} 3633, 1973.
%\bibitem{Jones} D. R. T. Jones, Phys. Rev. D {\bf 25} 581, 1982.
%\bibitem{Hill} M. S. Fischler and C. T. Hill, Nucl. Phys. B {\bf 193} 53, 1981.
%\bibitem{Machacek} M. E. Machacek and M. T. Vaughn, Nucl. Phys. B {\bf 222} 83, 1983.
%\bibitem{Jack} I. Jack and H. Osborn, Nucl. Phys. B {\bf 249} 472, 1985.
%\bibitem{Mihaila} L. N. Mihaila, J. Salomon and M. Steinhauser, arXiv:1201.5868, 2012.
\end{thebibliography}