\section{Introduction}

The recent discover of the Higgs particle at LHC and Tevatron
has reconfirmed that the electroweak theory of Weinberg and
Salam provides the true unification of electromagnetic and weak
interactions \cite{LHC,tev}. Indeed the discovery of the Higgs
particle has been claimed to be the ``final" test of the standard
model. This, however, might be a premature claim. The real final
test should come from the discovery of the electroweak monopole,
because the standard model predicts it \cite{plb97,yang}. In fact
the existence of the monopole topology in the standard model tells
that the discovery of the monopole must be the topological test
of the standard model.

In this sense it is timely that the latest MoEDAL detector
(``The Magnificient Seventh") at LHC is actively searching for
such monopole \cite{pin,ijmpa14}. To detect the electroweak monopole
experimentally, however, it is important to estimate the
monopole mass in advance. {\it The purpose of this paper is
to estimate the mass of the electroweak monopole. We show
that the monopole mass is expected to be around 4 to 7 TeV.}

Ever since Dirac \cite{Dirac} has introduced the concept of
the magnetic monopole, the monopoles have remained a fascinating
subject. The Abelian monopole has been generalized to the non-Abelian
monopoles by Wu and Yang \cite{wu,prl80} who showed that the pure
$SU(2)$ gauge theory allows a point-like monopole, and by 't Hooft
and Polyakov \cite{Hooft,prasad} who have constructed a finite energy
monopole solution in Georgi-Glashow model as a topological soliton.
Moreover, the monopole in grand unification has been constructed by
Dokos and Tomaras \cite{dokos}.

In the interesting case of the electroweak theory of Weinberg and
Salam, however, it has been asserted that there exists no topological
monopole of physical interest \cite{vach}. The basis for this
``non-existence theorem'' is, of course, that with the spontaneous
symmetry breaking the quotient space $SU(2) \times U(1)_Y/U(1)_{\rm em}$
allows no non-trivial second homotopy. This has led many people to
believe that there is no monopole in Weinberg-Salam model.

This claim, however, has been shown to be false. If the electroweak
unification of Weinberg and Salam is correct, the standard model
must have a monopole which generalizes the Dirac monopole. Moreover,
it has been shown that the standard model has a new type of monopole
and dyon solutions \cite{plb97}. This was based on the observation
that the Weinberg-Salam model, with the $U(1)_Y$, could be viewed
as a gauged $CP^1$ model in which the (normalized) Higgs doublet
plays the role of the $CP^1$ field. So the Weinberg-Salam model
does have exactly the same nontrivial second homotopy as the
Georgi-Glashow model which allows topological monopoles.

Once this is understood, one could proceed to construct the desired
monopole and dyon solutions in the Weinberg-Salam model. Originally
the electroweak monopole and dyon solutions were obtained by numerical
integration. But a mathematically rigorous existence proof has been
established which endorses the numerical results, and the solutions
are now referred to as Cho-Maison monopole and dyon \cite{yang}.

It should be emphasized that the Cho-Maison monopole is completely
different from the ``electroweak monopole'' derived from the Nambu's
electroweak string. In his continued search for the string-like
objects in physics, Nambu has demonstrated the existence of a rotating
dumb bell made of the monopole anti-monopole pair connected by the
neutral string of $Z$-boson flux (actually the $SU(2)$ flux) in
Weinberg-Salam model \cite{nambu}. Taking advantage of the Nambu's
pioneering work, others claimed to have discovered another type of
electroweak monopole, simply by making the string infinitely long
and moving the anti-monopole to infinity \cite{vacha}. This
``electroweak monopole'', however, must carry a fractional magnetic
charge and can not be isolated with finite energy. Moreover, this
has no spherical symmetry which is manifest in the Cho-Maison
monopole \cite{plb97}.

The existence of the electroweak monopole makes the experimental
confirmation of the monopole an urgent issue. Till recently the
experimental effort for the monopole detection has been on the Dirac's
monopole \cite{cab}. But the electroweak unification of the Maxwell's
theory requires the modification of the Dirac monopole, and this
modification changes the Dirac monopole to the Cho-Maison monopole.
This means that the monopole which should exist in the real world is
not likely to be the Dirac monopole but the electroweak monopole.

To detect the electroweak monopole experimentally, it is important
to estimate the mass of the monopole theoretically. Unfortunately the
Cho-Maison monopole carries an infinite energy at the classical level,
so that the monopole mass is not determined. This is because it can
be viewed as a hybrid between the Dirac monopole and the 'tHooft-Polyakov
monopole, so that it has a $U(1)_{\rm em}$ point singularity at the center
even though the $SU(2)$ part is completely regular.

{\em A priori} there is nothing wrong with this. Classically
the electron has an infinite electric energy but a finite mass.
But for the experimental search for the monopole we need a
solid idea about the monopole mass. In this paper we show
how to predict the mass of the electroweak monopole. Based
on the dimensional argument we first show that the monopole
mass should be of the order of $1/\alpha$ times the W-boson
mass, or around 10 TeV. To back up this we adopt the scaling
argument to predict the mass to be around 4 TeV. Finally,
we show how the quantum correction could regularize
the point singularity of the Cho-Maison dyon, and construct
finite energy electroweak dyon solutions introducing the effective
action of the standard model. Our result suggests that the
electroweak monopole with the mass around 4 to 7 TeV could
exist, which implies that there is a very good chance that the
MoEDAL at the present LHC can detect the electroweak
monopole.

The paper is organized as follows. In Section II we review the
Cho-Maison dyon for later purpose. In Section III we provide
two arguments, the dimensional and scaling arguments, which
indicate that the mass of the electroweak monopole could be
around 4 to 10 TeV. In Section IV we discuss the Abelian
decomposition and gauge independent Abelianization of
Weinberg-Salam model and Georgi-Glashow model to help
us how to regularize the Cho-Maison monopole. In Section
V we discuss two different methods to regularize the Cho-Maison
dyon with the quantum correction which modifies the coupling
constants at short distance, and construct finite energy dyon
solutions which support the scaling argument. In Section VI
we discuss another way to make the Cho-Maison dyon regular,
by enlarging the gauge group $SU(2)\times U(1)_Y$ to
$SU(2)\times SU(2)_Y$. Finally in Section VII we discuss
the physical implications of our results.

\section{Cho-Maison Dyon in Weinberg-Salam Model: A Review}

Before we construct a finite energy dyon solution in the electroweak
theory we must understand how one can obtain the infinite energy
Cho-Maison dyon solution first. Let us start with the Lagrangian
which describes (the bosonic sector of) the Weinberg-Salam theory
\begin{gather}
{\cal L} = -|{\cal D}_{\mu} \phi|^2
 -\displaystyle\frac{\lambda}{2}\big(\phi^\dagger \phi
 -\displaystyle\frac{\mu^2}{\lambda} \big)^2-\displaystyle\frac{1}{4} \vec F_{\mu\nu}^2
 -\displaystyle\frac{1}{4} G_{\mu\nu}^2, \nonumber\\
{\cal D}_{\mu} \phi=\big(\partial_{\mu}
-i\displaystyle\frac{g}{2} \vec \tau \cdot \vec A_{\mu}
- i\displaystyle\frac{g'}{2}B_{\mu}\big) \phi \nonumber\\
= \big(D_\mu - i\displaystyle\frac{g'}{2}B_{\mu}\big) \phi,
\label{wslag1}
\end{gather}
where $\phi$ is the Higgs doublet, $\vec F_{\mu\nu}$ and $G_{\mu\nu}$
are the gauge field strengths of $SU(2)$ and $U(1)_Y$ with the
potentials $\vec A_{\mu}$ and $B_{\mu}$, and $g$ and $g'$
are the corresponding coupling constants. Notice that $D_{\mu}$
describes the covariant derivative  of the $SU(2)$ subgroup only.
With
\begin{eqnarray}
\phi = \dfrac{1}{\sqrt{2}}\rho~\xi,~~~(\xi^{\dagger} \xi = 1),
\end{eqnarray}
where $\rho$ and $\xi$ are the Higgs field and unit doublet,
we have
\begin{gather}
{\cal L}=-\displaystyle\frac{1}{2}{(\partial_{\mu}\rho)}^2
- \displaystyle\frac{\rho^2}{2} {|{\cal D}_{\mu} \xi |}^2
-\displaystyle\frac{\lambda}{8}\big(\rho^2-\displaystyle\frac{2\mu^2}{\lambda}\big)^2 \nonumber\\
-\displaystyle\frac{1}{4}{\vec F}_{\mu\nu}^2 -\displaystyle\frac{1}{4} G_{\mu\nu}^2.
\end{gather}
Notice that the $U(1)_Y$ coupling of $\xi$ makes
the theory a gauge theory of $CP^1$ field \cite{plb97}.

From (\ref{wslag1}) one has the following equations of motion
\begin{gather}
\partial^2\rho= |{\cal D}_\mu \xi |^2 \rho
+\displaystyle\frac{\lambda}{2}\big (\rho^2
- \displaystyle\frac {2\mu^2}{\lambda}\big) \rho, \nonumber\\
{\cal D}^2 \xi=-2 \dfrac{\partial_\mu \rho}{\rho}
{\cal D}_\mu \xi +\big[\xi^\dagger {\cal D}^2\xi
+2\dfrac{\partial_\mu \rho}{\rho}
(\xi^\dagger {\cal D}_\mu \xi)\big] \xi, \nonumber\\
D_{\mu} \vec F_{\mu\nu}=i \displaystyle\frac{g}{2}\rho^2 \big[\xi^{\dagger}
\vec \tau( {\cal D}_{\nu} \xi )
-({\cal D}_{\nu} \xi)^{\dagger} \vec \tau \xi \big], \nonumber\\
\partial_{\mu} G_{\mu\nu}
=i\displaystyle\frac{g'}{2}\rho^2 \big[\xi^{\dagger} ({\cal D}_{\nu} \xi)
- ({\cal D}_{\nu} \xi)^{\dagger} \xi \big].
\end{gather}
Now we choose the following ansatz in the spherical coordinates
$(t,r,\theta,\varphi)$
\begin{gather}
\rho =\rho(r),
~~~\xi=i\left(\begin{array}{cc} \sin (\theta/2)~e^{-i\varphi}\\
- \cos(\theta/2) \end{array} \right),   \nonumber\\
\vec A_{\mu}= \displaystyle\frac{1}{g} A(r)\partial_{\mu}t~\hat r
+\displaystyle\frac{1}{g}(f(r)-1)~\hat r \times \partial_{\mu} \hat r, \nonumber\\
B_{\mu} =\displaystyle\frac{1}{g'} B(r) \partial_{\mu}t
-\displaystyle\frac{1}{g'}(1-\cos\theta) \partial_{\mu} \varphi.
\label{ans1}
\end{gather}
Notice that $\xi^{\dagger} \vec \tau ~\xi = -\hat r$. Moreover,
${\vec A}_\mu$ describes the Wu-Yang monopole when $A(r)=f(r)=0$.
So the ansatz is spherically symmetric. Of course, $\xi$ and
$B_{\mu}$ have an apparent string singularity along the negative
$z$-axis, but this singularity is a pure gauge artifact which
can easily be removed making the $U(1)_Y$ bundle
non-trivial. So the above ansatz describes a most general
spherically symmetric ansatz of an electroweak dyon.

Here we emphasize the importance of the non-trivial nature of
$U(1)_Y$ gauge symmetry to make the ansatz spherically symmetric.
Without the extra $U(1)_Y$ the Higgs doublet does not allow a
spherically symmetric ansatz. This is because the spherical
symmetry for the gauge field involves the embedding of the radial
isotropy group $SO(2)$ into the gauge group that requires the
Higgs field to be invariant under the $U(1)$ subgroup of $SU(2)$.
This is possible with a Higgs triplet, but not with a Higgs
doublet \cite{Forg}. In fact, in the absence of the $U(1)_Y$
degrees of freedom, the above ansatz describes the $SU(2)$
sphaleron which is not spherically symmetric \cite{manton}.

To see this, one might try to remove the string in $\xi$ with
the $U(1)$ subgroup of $SU(2)$. But this $U(1)$ will necessarily
change $\hat r$ and thus violate the spherical symmetry. This
means that there is no $SU(2)$ gauge transformation which can
remove the string in $\xi$ and at the same time keeps the spherical
symmetry intact. The situation changes with the inclusion of the
$U(1)_Y$ in the standard model, which naturally makes $\xi$
a $CP^1$ field \cite{plb97}. This allows the spherical symmetry
for the Higgs doublet.

To understand the physical content of the ansatz we perform
the following gauge transformation on (\ref{ans1})
\begin{gather}
\xi \rightarrow U \xi = \left(\begin{array}{cc} 0 \\ 1
\end{array} \right),  \nonumber\\
U=i\left( \begin{array}{cc}
\cos (\theta/2)& \sin(\theta/2)e^{-i\varphi} \\
-\sin(\theta/2) e^{i\varphi} & \cos(\theta/2)
\end{array}  \right),
\label{gauge}
\end{gather}
and find that in this unitary  gauge we have
\begin{gather}
\hat r \rightarrow \left( \begin{array}{c} 0\\ 0\\
1 \end{array} \right),  \nonumber\\
\vec A_\mu \rightarrow \displaystyle\frac{1}{g} \left( \begin{array}{c}
-f(r)(\sin\varphi\partial_\mu\theta
+\sin\theta\cos\varphi \partial_\mu\varphi) \\
f(r)(\cos\varphi\partial_\mu \theta
-\sin\theta\sin\varphi\partial_\mu\varphi) \\
A(r)\partial_\mu t -(1-\cos\theta)\partial_\mu\varphi
\end{array} \right).
\label{unitary}
\end{gather}
So introducing the electromagnetic and neutral $Z$-boson potentials
$A_\mu^{\rm (em)}$ and $Z_\mu$ with the Weinberg angle $\theta_{\rm w}$
\begin{gather}
\left( \begin{array}{cc} A_\mu^{\rm (em)} \\ Z_{\mu}
\end{array} \right)
= \left(\begin{array}{cc}
\cos\theta_{\rm w} & \sin\theta_{\rm w}\\
-\sin\theta_{\rm w} & \cos\theta_{\rm w}
\end{array} \right)
\left( \begin{array}{cc} B_{\mu} \\ A^3_{\mu}
\end{array} \right) \nonumber\\
= \displaystyle\frac{1}{\sqrt{g^2 + g'^2}} \left(\begin{array}{cc}
g & g' \\ -g' & g \end{array} \right)
\left( \begin{array}{cc} B_{\mu} \\ A^3_{\mu}
\end{array} \right),
\label{wein}
\end{gather}
we can express the ansatz (\ref{ans1}) in terms of the physical
fields
\begin{gather}
W_{\mu} = \displaystyle\frac{1}{\sqrt{2}}(A_\mu^1 + i A_\mu^2)
=\dfrac{i}{g}\displaystyle\frac{f(r)}{\sqrt2}e^{i\varphi}
(\partial_\mu \theta +i \sin\theta \partial_\mu \varphi), \nonumber\\
A_{\mu}^{\rm (em)} = e\big( \displaystyle\frac{1}{g^2}A(r)
+ \displaystyle\frac{1}{g'^2} B(r) \big) \partial_{\mu}t  \nonumber\\
-\displaystyle\frac{1}{e}(1-\cos\theta) \partial_{\mu} \varphi,  \nonumber \\
Z_{\mu} = \displaystyle\frac{e}{gg'}\big(A(r)-B(r)\big) \partial_{\mu}t,
\label{ans2}
\end{gather}
where $W_\mu$ is the $W$-boson and $e$ is the electric charge
\begin{eqnarray*}
e=\displaystyle\frac{gg'}{\sqrt{g^2+g'^2}}=g\sin\theta_{\rm w}=g'\cos\theta_{\rm w}.
\end{eqnarray*}
This clearly shows that the ansatz is for the electroweak dyon.

The spherically symmetric ansatz reduces the equations of
motion to
\begin{gather}
\ddot{\rho}+\displaystyle\frac{2}{r} \dot{\rho}-\displaystyle\frac{f^2}{2r^2}\rho
=-\displaystyle\frac{1}{4}(A-B)^2\rho +\displaystyle\frac {\lambda}{2}\big(\rho^2
-\displaystyle\frac{2\mu^2}{\lambda}\big)\rho , \nonumber\\
\ddot{f}-\displaystyle\frac{f^2-1}{r^2}f=\big(\displaystyle\frac{g^2}{4}\rho^2
- A^2\big)f, \nonumber\\
\ddot{A}+\displaystyle\frac{2}{r}\dot{A}-\displaystyle\frac{2f^2}{r^2}A
=\displaystyle\frac{g^2}{4}\rho^2(A-B), \nonumber \\
\ddot{B} +\displaystyle\frac{2}{r} \dot{B}
=-\displaystyle\frac{g'^2}{4} \rho^2 (A-B).
\label{cmeq}
\end{gather}
Obviously this has a trivial solution
\begin{eqnarray}
\rho=\rho_0=\sqrt{2\mu^2/\lambda},~~~f=0,
~~~A=B=0,
\end{eqnarray}
which describes the point monopole in Weinberg-Salam model
\begin{eqnarray}
A_\mu^{\rm (em)}=-\displaystyle\frac{1}{e}(1-\cos \theta) \partial_\mu \varphi.
\end{eqnarray}
This monopole has two remarkable features. First, this is the
electroweak generalization of the Dirac's monopole, but not
the Dirac's monopole. It has the electric charge $4\pi/e$, not
$2\pi/e$ \cite{plb97}. Second, this monople naturally admits a
non-trivial dressing of weak bosons. Indeed, with the non-trivial
dressing, the monopole becomes the Cho-Maison dyon.

To see this let us choose the following boundary condition
\begin{eqnarray}
&\rho(0)=0,~~f(0)=1,~~A(0)=0,~~B(0)=b_0, \nonumber\\
&\rho(\infty)=\rho_0,~f(\infty)=0,~A(\infty)=B(\infty)=A_0.
\label{bc0}
\end{eqnarray}
Then we can show that the equation (\ref{cmeq}) admits
a family of solutions labeled by the real parameter $A_0$
lying in the range \cite{plb97,yang}
\begin{eqnarray}
0 \leq A_0 < {\rm min} ~\Big(e\rho_0,~\displaystyle\frac{g}{2}\rho_0\Big).
\label{boundA}
\end{eqnarray}
In this case all four functions
$f(r),~\rho(r),~A(r)$, and $B(r)$ must be positive for $r>0$, and
$A(r)/g^2+B(r)/g'^2$ and $B(r)$ become increasing functions of
$r$. So we have $0 \leq b_0 \leq A_0$. Furthermore, we have
$B(r)\ge A(r)\ge 0$ for all range, and $B(r)$ must approach to
$A(r)$ with an exponential damping. Notice that, with the
experimental fact $\sin^2\theta_{\rm w}=0.2312$, (\ref{boundA})
can be written as $0 \leq A_0 < e\rho_0$.

With the boundary condition (\ref{bc0}) we can integrate
(\ref{cmeq}). For example, with $A=B=0$, we have the
Cho-Maison monopole. In general, with $A_0\ne0$, we find
the Cho-Maison dyon \cite{plb97}.

Near the origin the dyon solution has the following behavior,
\begin{eqnarray}
&\rho \simeq \alpha_1 r^{\delta_-},
~~~~~f \simeq 1+ \beta_1  r^2,  \nonumber \\
&A \simeq a_1 r,~~~~~B \simeq b_0 + b_1 r^{2\delta_+},
\label{origin}
\end{eqnarray}
where $\delta_{\pm} =(\sqrt{3} \pm 1)/2$.
Asymptotically it has the following behavior,
\begin{eqnarray}
&\rho \simeq \rho_0 +\rho_1\dfrac{\exp(-\sqrt{2}\mu r)}{r},
~~~f \simeq  f_1 \exp(-\omega  r),  \nonumber\\
&A \simeq A_0 +\dfrac{A_1}{r},
~~~B \simeq A +B_1 \dfrac{\exp(-\nu r)}{r},
\label{infty}
\end{eqnarray}
where $\omega=\sqrt{(g\rho_0)^2/4 -A_0^2}$,
and $\nu=\sqrt{(g^2 +g'^2)}\rho_0/2$.
The physical meaning of the asymptotic behavior
must be clear. Obviously $\rho$, $f$, and $A-B$ represent
the Higgs boson, $W$-boson, and $Z$-boson whose masses
are given by $M_H=\sqrt{2}\mu=\sqrt{\lambda}\rho_0$,
$M_W=g\rho_0/2$, and $M_Z=\sqrt{g^2+g'^2}\rho_0/2$.

So (\ref{infty}) tells that $M_H$, $\sqrt{1-(A_0/M_W)^2}~M_W$,
and $M_Z$ determine the exponential damping of the Higgs boson,
$W$-boson, and $Z$-boson to their vacuum expectation values
asymptotically. Notice that it is $\sqrt{1-(A_0/M_W)^2}~M_W$,
but not $M_W$, which determines the exponential damping of the
$W$-boson. This tells that the electric potential of the dyon
slows down the exponential damping of the $W$-boson, which is
reasonable.

The dyon has the following electromagnetic charges
\begin{eqnarray}
&q_e=-\dfrac{8\pi}{e}\sin^2\theta_{\rm w} \displaystyle{\int}_0^\infty f^2 A dr
=\dfrac{4\pi}{e} A_1, \nonumber\\
&q_m = \dfrac{4\pi}{e}.
\label{eq:Charge}
\end{eqnarray}
Also, the asymptotic condition (\ref{infty}) assures
that the dyon does not carry any neutral charge,
\begin{eqnarray}
&Z_e =-\dfrac{4\pi e}{gg'}\big[ r^2 (\dot{A}-\dot{B})\big]
\Big|_{r=\infty} =0,\nonumber\\
&Z_m = 0.
\label{neutral}
\end{eqnarray}
Furthermore, notice that the dyon equation (\ref{cmeq})
is invariant under the reflection
\begin{eqnarray}
A \rightarrow -A,~~~~~~B\rightarrow -B.
\label{ref}
\end{eqnarray}
This means that, for a given magnetic charge,
there are always two dyon solutions
which carry opposite electric charges $\pm q_e$.
Clearly the signature of
the electric charge of the dyon is determined by
the signature of the boundary value $A_0$.

We can also have the anti-monopole or in general anti-dyon
solution, the charge conjugate state of the dyon, which
has the magnetic charge $q_m=-4\pi/e$ with the following
ansatz
\begin{gather}
\rho' =\rho(r),
~~~\xi'=-i\left(\begin{array}{cc} \sin (\theta/2)~e^{+i\varphi}\\
- \cos(\theta/2) \end{array} \right),   \nonumber\\
\vec A'_{\mu}= -\displaystyle\frac{1}{g} A(r)\partial_{\mu}t~\hat r'
+\displaystyle\frac{1}{g}(f(r)-1)~\hat r' \times \partial_{\mu} \hat r', \nonumber\\
B'_{\mu} =-\displaystyle\frac{1}{g'} B(r) \partial_{\mu}t
+\displaystyle\frac{1}{g'}(1-\cos\theta) \partial_{\mu} \varphi,  \nonumber\\
\hat r'=-\xi'^{\dagger} \vec \tau ~\xi'
=(\sin \theta \cos \phi,-\sin \theta \sin \phi,\cos \theta).
\label{antid1}
\end{gather}
Notice that the ansatz is basically the complex conjugation
of the dyon ansatz.

To understand the meaning of the anti-dyon ansatz notice that
in the unitary gauge
\begin{gather}
\xi' \rightarrow U' \xi' = \left(\begin{array}{cc} 0 \\ 1
\end{array} \right),
\nonumber\\
U'=-i\left( \begin{array}{cc}
\cos (\theta/2)& \sin(\theta/2)e^{i\varphi} \\
-\sin(\theta/2) e^{-i\varphi} & \cos(\theta/2)
\end{array} \right),
\end{gather}
we have
\begin{gather}
{\vec A}'_\mu \rightarrow \displaystyle\frac{1}{g} \left( \begin{array}{c}
f(r)(\sin\varphi\partial_\mu\theta
+\sin\theta\cos\varphi \partial_\mu\varphi) \\
f(r)(\cos\varphi\partial_\mu \theta
-\sin\theta\sin\varphi\partial_\mu\varphi) \\
-A(r)\partial_\mu t +(1-\cos\theta)\partial_\mu\varphi
\end{array} \right).
\label{unitary}
\end{gather}
So in terms of the physical fields the ansatz (\ref{antid1})
is expressed by
\begin{eqnarray}
&W'_{\mu}=\dfrac{i}{g}\dfrac{f(r)}{\sqrt2}e^{-i\varphi}
(\partial_\mu \theta -i \sin\theta \partial_\mu \varphi)
=-W_{\mu}^*, \nonumber\\
&A_{\mu}^{ \rm (em)} = -e\big( \dfrac{1}{g^2}A(r)
+ \dfrac{1}{g'^2} B(r) \big) \partial_{\mu}t  \nonumber\\
&+\dfrac{1}{e}(1-\cos\theta) \partial_{\mu} \varphi,  \nonumber \\
&Z'_{\mu} = -\dfrac{e}{gg'}\big(A(r)-B(r)\big) \partial_{\mu}t
=-Z_\mu.
\label{antid2}
\end{eqnarray}
This clearly shows that the the electric and magnetic charges
of the ansatz (\ref{antid1}) are the opposite of the dyon ansatz,
which confirms that the ansatz indeed describes the anti-dyon.

With the ansatz (\ref{antid1}) we have exactly the same
equation (\ref{cmeq}) for the anti-dyon. This assures
that the standard model has the anti-dyon as well as
the dyon.

The above discussion tells that the W and Z boson part of
the anti-dyon solution is basically the complex conjugation
of the dyon solution. This, of course, is natural from
the physical point of view. On the other hand there is
one minor point to be clarified here. Since the topological
charge of the monopole is given by the second homotopy
defined by $\hat r=-\xi^\dagger \vec \tau \xi$, one might
expect that $\hat r'$ defined by the anti-dyon ansats
$\xi'=\xi^*$ must be $-\hat r$. But this is not so, and
we have to explain why.

To understand this notice tha we can change $\hat r'$
to $-\hat r$ by a SU(2) gauge transformation, by the
$\pi$-rotation along the y-axis. With this gauge
transformation the ansatz (\ref{antid1}) changes to
\begin{gather}
\xi' \rightarrow i\left(\begin{array}{cc} \cos(\theta/2) \\
\sin (\theta/2)~e^{+i\varphi} \end{array} \right),
~~~\hat r' \rightarrow -\hat r,  \nonumber\\
\vec A_{\mu} \rightarrow  -\displaystyle\frac{1}{g} A(r)\partial_{\mu}t~\hat r
+\displaystyle\frac{1}{g}(f(r)-1)~\hat r \times \partial_{\mu} \hat r.
\label{antid3}
\end{gather}
This tells that the monopole topology defined by $\hat r'$ is
the same as that of $\hat r$.

Since the Cho-Maison solution is obtained numerically one
might like to have a mathematically rigorous existence proof
of the Cho-Maison dyon. The existence proof is non-trivial,
because the equation of motion (\ref{cmeq}) is not the
Euler-Lagrange equation of the positive definite energy
(\ref{cme}), but that of the indefinite action
\begin{gather}
{\cal L}=-4\pi \int\limits_0^\infty dr
\bigg\{\displaystyle\frac{1}{2}(r\dot\rho)^2
+\displaystyle\frac{\lambda r^2}{8}\big(\rho^2-\rho_0^2\big)^2 \nonumber\\
+\displaystyle\frac{1}{4} f^2\rho^2+ \frac1{g^2} \big(\dot f^2
-\displaystyle\frac{1}{2}(r\dot A)^2- f^2 A^2 \big)-\displaystyle\frac{1}{2g'^2}(r\dot B)^2  \nonumber\\
-\displaystyle\frac{r^2}{8} (B-A)^2 \rho^2 +\displaystyle\frac{1}{2 r^2}\big(\displaystyle\frac{1}{g'^2}
+\frac1{g^2}(f^2-1)^2\big) \bigg\}.
\label{cmlag}
\end{gather}
Fortunately the existence proof has been established by
Yang \cite{yang}.

Before we leave this section it is worth to re-address the
important question again: Does the standard model predict
the monopole? Notice that the Dirac monopole in electrodynamics
is optional: It can exist only when the $U(1)_{\rm em}$ is
non-trivial, but there is no reason why this has to be so.
If so, why can't the electroweak monopole be optional?

As we have pointed out, the non-trivial $U(1)_Y$ is crucial
for the existence of the monopole in the standard model.
So the question here is why the $U(1)_Y$ must be non-trivial.
To see why, notice that in the standard model $U(1)_{\rm em}$
comes from two $U(1)$, the $U(1)$ subgroup of $SU(2)$ and
$U(1)_Y$, and it is well known that the $U(1)$ subgroup of $SU(2)$
is non-trivial. Now, to obtain the electroweak monopole we have
to make the linear combination of two monopoles, that of the
$U(1)$ subgroup of $SU(2)$ and $U(1)_Y$. This must be clear
from (\ref{wein}).

In this case the mathematical consistency requires the two
potentials $A_\mu^3$ and $B_\mu$ (and two $U(1)$) to have
the same structure, in particular the same topology. But we
already know that $A_\mu^3$ is non-trivial. So $B_\mu$,
and the corresponding $U(1)_Y$, has to be non-trivial. In
other words, requiring $U(1)_Y$ to be trivial is inconsistent
(i.e., in contradiction with the self-consistency) in the
standard model. This tells that, unlike the Maxwell's theory,
the $U(1)_{\rm em}$ in the standard model must be non-trivial.
This assures that the standard model must have the monopole.

But ultimately this question has to be answered by the experiment.
So the discovery of the monopole must be the topological test
of the standard model, which has never been done before. This
is why MoEDAL is so important.

\section{Mass of the Electroweak Monopole}

To detect the electroweak monopole experimentally, we have to
have a firm idea on the mass of the monopole. Unfortunately,
at the classical level we can not estimate the mass of the
Cho-Maison monopole, because it has a point singularity at the
center which makes the total energy infinite.

Indeed the ansatz (\ref{ans1}) gives the following energy
\begin{gather}
E=E_0 +E_1,  \nonumber \\
E_0=4\pi\int_0^\infty \displaystyle\frac{dr}{2 r^2}
\bigg\{\displaystyle\frac{1}{g'^2}+ \frac1{g^2}(f^2-1)^2\bigg\}, \nonumber\\
E_1=4\pi \int_0^\infty dr \bigg\{\frac12 (r\dot\rho)^2
+\frac1{g^2} \big(\dot f^2 +\displaystyle\frac{1}{2}(r\dot A)^2  \nonumber\\
+ f^2 A^2 \big)+\displaystyle\frac{1}{2g'^2}(r\dot B)^2
+\displaystyle\frac{\lambda r^2}{8}\big(\rho^2-\rho_0^2 \big)^2 \nonumber\\
+\frac14 f^2\rho^2
+\displaystyle\frac{r^2}{8} (B-A)^2 \rho^2 \bigg\}.
\label{cme}
\end{gather}
The boundary condition (\ref{bc0}) guarantees that
$E_1$ is finite. As for $E_0$ we can minimize it with
the boundary condition $f(0)=1$, but even with this
$E_0$ becomes infinite. Of course the origin of this
infinite energy is obvious, which is precisely due to
the magnetic singularity of $B_\mu$ at the origin.
This means that one can not predict the mass of dyon.
Physically it remains arbitrary.

To estimate of the monopole mass theoretically, we have to
regularize the point singularity of the Cho-Maison dyon. One
might try to do that introducing the gravitational interaction,
in which case the mass is fixed by the asymptotic behavior of
the gravitational potential. But the magnetic charge of the
monopole is not likely to change the character of the singularity,
so that asymptotically the leading order of the gravitational
potential becomes of the Reissner-Nordstrom type \cite{bais}.
This implies the gravitational interaction may not help us to
estimate the monopole mass.

To make the the energy of the Cho-Maison monopole finite,
notice that the origin of the infinite energy is the first
term $1/g'^2$ in $E_0$ in (\ref{cme}). A simple way to make
this term finite is to introduce a UV-cutoff which removes
this divergence. This type of cutoff could naturally come
from the quantum correction of the coupling constants. In
fact, since the quantum correction changes $g'$ to the
running coupling $\bar g'$, $E_0$ can become finite
if $\bar g'$ diverges at short distance.

We will discuss how such quantum correction could take place
later, but before doing that we present two arguments, the
dimsnsional argument and the scaling argument, which
could give us a rough estimate of the monopole mass.

\subsection{Dimensional argument}

To have the order estimate of the monopole mass it is important
to realize that, roughly speaking, the monopole mass comes from
the Higgs mechanism which generates the mass to the W-boson.
This can easily be seen in the 'tHooft-Polyakov monopole in
Georgi-Glashow model
\begin{eqnarray}
&{\cal L}_{GG} =-\dfrac{1}{4} \vec{F}_{\mu\nu}^2
-\dfrac{1}{2}(D_\mu \vec{\Phi} )^2-\dfrac{\lambda}{4}
\big(\vec \Phi^2-\dfrac{\mu^2}{\lambda}\big)^2,
\label{ggl}
\end{eqnarray}
where $\vec \Phi$ is the Higgs triplet. Here the monopole ansatz
is given by
\begin{eqnarray}
&\vec \Phi=\rho~\hat r,~~~{\vec A}_\mu= \vec C_\mu+ {\vec W}_\mu,  \nonumber\\
&\vec C_\mu= -\dfrac{1}{g} \hat r \times \partial_\mu \hat r,
~~~{\vec W}_\mu=-f \vec C_\mu,
\label{wu}
\end{eqnarray}
where $\vec C_\mu$ represents the Wu-Yang monopole
potential \cite{wu,prd80}. Notice that the W-boson
part of the monopole is given by the Wu-Yang potential,
except for the overall amplitude $f$.

With this we clearly have
\begin{eqnarray}
|D_\mu \Phi|^2=(\partial_\mu \rho)^2
+ g^2 \rho^2 f^2 (\vec C_\mu)^2.
\end{eqnarray}
So, when the Higgs field has a non-vanishing vacuum expectation
value, $\vec C_\mu$ acquires a mass (with $f\simeq 1$). This,
of course, is the Higgs mechanism which generates the W-boson
mass. The only difference is that here the W-boson is expressed
by the Wu-Yang potential and the Higgs coupling becomes magnetic
($\vec C_\mu$ contains the extra factor $1/g$).

Similar mechanism works for the Weinberg-Salam model. Here again
${\vec A}_\mu$ (with $A=B=0$) of the asnatz (\ref{ans1}) is identical to
(\ref{wu}), and we have
\begin{gather}
D_\mu \xi=i\big(g f~\vec C_\mu+(1-\cos \theta) \partial_\mu \phi~{\hat r} \big)
\cdot \displaystyle\frac{\vec \tau}{2}~\xi,  \nonumber\\
|{\cal D}_\mu \xi|^2=|D_\mu \xi|^2- |\xi^\dagger D_\mu \xi|^2  \nonumber\\
-(\xi^\dagger D_\mu\xi-i \displaystyle\frac{g'}{2} B_\mu)^2
=\frac14 g^2 f^2 (\vec C_\mu)^2, \nonumber\\
|{\cal D}_\mu \phi|^2=\frac12 (\partial_\mu \rho)^2
+\frac12 \rho^2 |{\cal D}_\mu \xi|^2 \nonumber\\
=\frac12 (\partial_\mu \rho)^2+\frac18 g^2 \rho^2 f^2 (\vec C_\mu)^2.
\end{gather}
This (with $f\simeq 1$) tells that the electroweak monopole
acquires mass through the Higgs mechanism which generates mass
to the W-boson.

Once this is understood, we can use the dimensional argument
to predict the monopole energy. Since the monopole mass term
in the Lagrangian contributes to the monopole energy in the
classical solution we may expect
\begin{eqnarray}
E \simeq C \times \dfrac{4\pi}{e^2} M_W,
~~~C\simeq 1.
\label{mmass}
\end{eqnarray}
This implies that the monopole mass should be about $1/\alpha$
times bigger than the electroweak scale, around 10 TeV. But this
is the order estimate. Now we have to know how to estimate $C$.

\subsection{Scaling argument}

We can use the Derrick's scaling argument to estimate the
constant $C$ in (\ref{mmass}), assuming the existence of
a finite energy monopole solution. If a finite energy monopole
does exist, the action principle tells that it should be stable
under the rescaling of its field configuration. So consider such
a monopole configuration and let
\begin{gather}
K_A = \displaystyle{\int} d^3x ~\dfrac{1}{4} \vec{F}_{ij}^2,
\quad K_B=\displaystyle{\int} d^3x ~\dfrac{1}{4} B_{ij}^2  \nonumber\\
K_\phi=\displaystyle{\int} d^3 x ~|{\cal D}_i \phi|^2, \nonumber\\
V_\phi=\displaystyle{\int} d^3x ~\dfrac{\lambda}{2}\big( |\phi|^2
-\dfrac{\mu^2}{\lambda} \big)^2.
\end{gather}
With the ansatz (\ref{ans1}) we have (with $A=B=0$)
\begin{gather}
K_A= \displaystyle\frac{4\pi}{g^2} \int_0^\infty
\Big\{\dot{f}^2 + \displaystyle\frac{(f^2-1)^2}{2r^2} \Big\} dr, \nonumber\\
K_B=\displaystyle\frac{2\pi}{g'^2}\int\limits_0^\infty \displaystyle\frac{1}{r^2}dr,
~~~K_\phi= 2\pi \int_0^\infty (r\dot{\rho})^2 dr,  \nonumber \\
V_\phi=\displaystyle\frac{\pi}{2} \int_0^\infty \lambda r^2
\big(\rho^2 -\rho_0^2 \big)^2 dr.
\end{gather}
Notice that $K_B$ makes the monopole energy infinite.

Now, consider the spatial scale transformation
\begin{eqnarray}
\vec x \longrightarrow \lambda \vec x.
\label{scale}
\end{eqnarray}
Under this we have
\begin{eqnarray}
&{\vec A}_k(\vec x) \rightarrow \lambda {\vec A}_k(\lambda \vec x),
~~~B_k(\vec x)\rightarrow \lambda B_k(\lambda \vec x), \nonumber\\
&\phi (\vec x) \rightarrow \phi (\lambda \vec x),
\end{eqnarray}
so that
\begin{eqnarray}
&K_A \longrightarrow \lambda K_A,
~~~K_B \longrightarrow \lambda K_B,  \nonumber\\
&K_\phi \longrightarrow \lambda^{-1} K_\phi,
~~~V_\phi \longrightarrow \lambda^{-3} V_\phi.
\end{eqnarray}
With this we have the following requirement for the stable
monopole configuration
\begin{eqnarray}
K_A+K_B=K_\phi+3V_\phi.
\label{derrick}
\end{eqnarray}
From this we can estimate the finite value of $K_B$.

Now, for the Cho-Maison monopole we have (with
$M_W \simeq 80.4~{\rm GeV}$, $M_H \simeq 125~{\rm GeV}$,
and $\sin^2\theta_{\rm w}=0.2312$)
\begin{eqnarray}
&K_A \simeq 0.1904 \times\dfrac{4\pi}{e^2}{M_W},
~~~K_\phi \simeq 0.1577 \times\dfrac{4\pi}{e^2}{M_W},  \nonumber \\
&V_\phi \simeq 0.0111 \times\dfrac{4\pi}{e^2}{M_W}.
\end{eqnarray}
This, with (\ref{derrick}), tells that
\begin{eqnarray}
K_B \simeq 0.0006 \times \dfrac{4\pi}{e^2} M_W.
\end{eqnarray}
From this we estimate the energy of the monopole to be
\begin{eqnarray}
E \simeq 0.3598 \times \dfrac{4\pi}{e^2} M_W \simeq 3.96~{\rm TeV}.
\end{eqnarray}
This strongly endorses the dimensional argument. In particular, this
tells that the electroweak monopole of mass around a few TeV could
be possible.

The important question now is to show how the quantum correction
could actually make the energy of the Cho-Maison monopole finite.
To do that we have to understand the structure of the electroweak
theory, in particular the Abelian decomposition of the electroweak
theory. So we review the gauge independent Abelian decomposition
of the standard model first.

\section{Abelian Decomposition of the Electroweak Theory}

Consider the Yang-Mills theory
\begin{eqnarray}
{\cal L}_{YM} =-\dfrac{1}{4} \vec F_{\mu\nu}^2.
\end{eqnarray}
A best way to make the Abelian decomposition is to introduce
a unit $SU(2)$ triplet $\hat n$ which selects the Abelian
direction at each space-time point, and impose the isometry
on the gauge potential which determines the restricted potential
$\hat A_\mu$ \cite{prd80,prl81}
\begin{eqnarray}
&D_\mu {\hat n}=0,  \nonumber\\
&\vec A_\mu\rightarrow \hat A_\mu
=A_\mu {\hat n} -\dfrac{1}{g} {\hat n}\times\partial_\mu{\hat n}
=A_\mu {\hat n}+\vec C_\mu, \nonumber\\
&A_\mu={\hat n} \cdot {\vec A}_\mu,
~~~\vec C_\mu=-\dfrac{1}{g} {\hat n}\times\partial_\mu{\hat n} .
\label{chocon}
\end{eqnarray}
Notice that the restricted potential is precisely the connection
which leaves ${\hat n}$ invariant under parallel transport. The restricted
potential is called Cho connection or Cho-Duan-Ge (CDG)
connection \cite{fadd,shab,zucc}.

With this we obtain the gauge independent Abelian decomposition
of the $SU(2)$ gauge potential adding the valence potential
$\vec W_\mu$ which was excluded by the isometry \cite{prd80,prl81}
\begin{gather}
\vec{A}_\mu = \hat A_\mu + {\vec W}_\mu,
~~~(\hat{n}\cdot\vec{W}_\mu=0).
\label{chodecom}
\end{gather}
The Abelian decomposition has recently been referred
to as Cho (also Cho-Duan-Ge or Cho-Faddeev-Niemi)
decomposition \cite{fadd,shab,zucc}.

Under the infinitesimal gauge transformation
\begin{eqnarray}
\delta {\hat n} = - \vec \alpha \times {\hat n},
~~~~\delta {\vec A}_\mu = \displaystyle\frac{1}{g}  D_\mu \vec \alpha,
\end{eqnarray}
we have
\begin{gather}
\delta A_\mu = \displaystyle\frac{1}{g} {\hat n} \cdot \partial_\mu {\vec \alpha},
\quad \delta \hat A_\mu = \displaystyle\frac{1}{g} {\hat D}_\mu {\vec \alpha}, \nonumber\\
\delta {\vec W}_\mu = -{\vec \alpha} \times {\vec W}_\mu.
\label{gt1}
\end{gather}
This tells that $\hat A_\mu$ by itself describes an $SU(2)$
connection which enjoys the full $SU(2)$ gauge degrees of freedom.
Furthermore the valence potential $\vec W_\mu$ forms a gauge
covariant vector field under the gauge transformation. But
what is really remarkable is that the decomposition is gauge
independent. Once $\hat n$ is chosen, the decomposition follows
automatically, regardless of the choice of gauge.

Notice that $\hat{A}_\mu$ has a dual structure,
\begin{gather}
\hat{F}_{\mu\nu}
= \partial_\mu \hat A_\nu-\partial_\nu \hat A_\mu
+ g \hat A_\mu \times \hat A_\nu
= (F_{\mu\nu}+ H_{\mu\nu})\hat{n}, \nonumber \\
F_{\mu\nu} =\partial_\mu A_\nu-\partial_\nu A_\mu, \nonumber\\
H_{\mu\nu} = -\displaystyle\frac{1}{g} \hat{n}\cdot(\partial_\mu
\hat{n}\times\partial_\nu\hat{n}).
\end{gather}
Moreover, $H_{\mu \nu}$ always admits the potential because
it satisfies the Bianchi identity. In fact, replacing ${\hat n}$
with a $CP^1$ field $\xi$ (with ${\hat n}=-\xi^{\dagger} \vec \tau ~\xi$)
we have
\begin{eqnarray}
&H_{\mu\nu} = \partial_\mu \tilde C_\nu-\partial_\nu \tilde C_\mu
= \dfrac{2i}{g} (\partial_\mu \xi^{\dagger}
\partial_\nu \xi - \partial_\nu \xi^{\dagger} \partial_\mu \xi), \nonumber\\
&\tilde C_\mu = \dfrac{2i}{g} \xi^{\dagger} \partial_\mu \xi
=\dfrac{i}{g} \big(\xi^{\dagger} \partial_\mu \xi
- \partial_\mu \xi^{\dagger} \xi \big).
\end{eqnarray}
Of course $\tilde C_\mu$ is determined uniquely up to the $U(1)$
gauge freedom which leaves ${\hat n}$ invariant. To understand the
meaning of $\tilde C_\mu$, notice that with $\hat n=\hat r$ we
have
\begin{eqnarray}
\tilde C_\mu = \displaystyle\frac{1}{g} (1- \rm cos~\theta) \partial_\mu \varphi.
\end{eqnarray}
This is nothing but the Abelian monopole potential, and the
corresponding non-Abelian monopole potential is given by
the Wu-Yang monopole potential $\vec C_\mu$ \cite{wu,prl80}.
This justifies us to call $A_\mu$ and $\tilde C_\mu$ the electric
and magnetic potential.

The above analysis tells that $\hat{A}_\mu$ retains all essential
topological characteristics of the original non-Abelian potential.
First, $\hat{n}$ defines $\pi_2(S^2)$ which describes the non-Abelian
monopoles. Second, it characterizes the Hopf invariant
$\pi_3(S^2)\simeq\pi_3(S^3)$ which describes the topologically
distinct vacua \cite{bpst,plb79}. Moreover, it provides the gauge
independent separation of the monopole field from the generic
non-Abelian gauge potential.

With the decomposition (\ref{chodecom}), we have
\begin{eqnarray}
\vec{F}_{\mu\nu}
=\hat F_{\mu \nu} + {\hat D} _\mu {\vec W}_\nu - {\hat D}_\nu
{\vec W}_\mu + g{\vec W}_\mu \times {\vec W}_\nu,
\end{eqnarray}
so that the Yang-Mills Lagrangian is expressed as
\begin{gather}
{\cal L}_{YM} =-\displaystyle\frac{1}{4} {\hat F}_{\mu\nu}^2
-\displaystyle\frac{1}{4}({\hat D}_\mu{\vec W}_\nu-{\hat D}_\nu{\vec W}_\mu)^2 \nonumber\\
-\displaystyle\frac{g}{2}{\hat F}_{\mu\nu} \cdot ({\vec W}_\mu \times {\vec W}_\nu)
-\displaystyle\frac{g^2}{4} ({\vec W}_\mu \times {\vec W}_\nu)^2.
\end{gather}
This shows that the Yang-Mills theory can be viewed as
a restricted gauge theory made of the restricted potential,
which has the valence gluons as its source \cite{prd80,prl81}.

An important advantage of the decomposition (\ref{chodecom}) is
that it can actually Abelianize (or more precisely ``dualize'')
the non-Abelian gauge theory gauge independently \cite{prd80,prl81}.
To see this let$({\hat n}_1,~{\hat n}_2,~{\hat n})$ be a right-handed orthonormal
basis of $SU(2)$ space and let
\begin{gather}
\vec{W}_\mu =W^1_\mu ~\hat{n}_1 + W^2_\mu ~\hat{n}_2,
\nonumber\\
(W^1_\mu = \hat {n}_1 \cdot \vec W_\mu,~~~W^2_\mu =
\hat {n}_2 \cdot \vec W_\mu).
\nonumber
\end{gather}
With this we have
\begin{eqnarray}
&\hat{D}_\mu \vec{W}_\nu =\Big[\partial_\mu W^1_\nu
-g(A_\mu+ \tilde C_\mu)W^2_\nu \Big]\hat n_1  \nonumber\\
&+\Big[\partial_\mu W^2_\nu
+ g (A_\mu+ \tilde C_\mu)W^1_\nu \Big]\hat{n}_2,
\end{eqnarray}
so that with
\begin{eqnarray}
{\cal A}_\mu = A_\mu+ \tilde C_\mu,
\quad
W_\mu = \displaystyle\frac{1}{\sqrt{2}} ( W^1_\mu + i W^2_\mu ), \nonumber
\end{eqnarray}
we can express the Lagrangian explicitly in terms of the dual
potential ${\cal A}_\mu$ and the complex vector field $W_\mu$,
\begin{eqnarray}
\label{eq:Abelian}
{\cal L}_{YM} = -\displaystyle\frac{1}{4} {\cal F}_{\mu\nu}^2
-\displaystyle\frac{1}{2}|\hat{D}_\mu{W}_\nu-\hat{D}_\nu{W}_\mu|^2 \nonumber\\
+ ig {\cal F}_{\mu\nu} W_\mu^*W_\nu
+ \displaystyle\frac{g^2}{4}(W_\mu^* W_\nu - W_\nu^* W_\mu)^2,
\end{eqnarray}
where ${\cal F}_{\mu\nu} = F_{\mu\nu} + H_{\mu\nu}$ and
$\hat{D}_\mu=\partial_\mu + ig {\cal A}_\mu$. This shows
that we can indeed Abelianize the non-Abelian theory
with our decomposition.

Notice that in the Abelian formalism the Abelian potential
${\cal A}_\mu$ has the extra magnetic potential $\tilde C_\mu$.
In other words, it is given by the sum of the electric and magnetic
potentials $A_\mu+\tilde C_\mu$. Clearly $\tilde C_\mu$
represents the topological degrees of the non-Abelian
symmetry which does not show up in the naive Abelianization that
one obtains by fixing the gauge \cite{prd80,prl81}.

Furthermore, this Abelianization is gauge independent,
because here we have never fixed the gauge to obtain
this Abelian formalism. So one might ask how the non-Abelian
gauge symmetry is realized in this Abelian formalism. To
discuss this let
\begin{gather}
\vec \alpha = \alpha_1~{\hat n}_1 + \alpha_2~{\hat n}_2 + \theta~\hat n,
\quad
\alpha = \displaystyle\frac{1}{\sqrt 2} (\alpha_1 + i ~\alpha_2),
\nonumber\\
\vec C_\mu = - \displaystyle\frac {1}{g} {\hat n} \times \partial_\mu {\hat n}
= - C^1_\mu {\hat n}_1 - C^2_\mu {\hat n}_2,
\nonumber\\
C_\mu = \displaystyle\frac{1}{\sqrt 2} (C^1_\mu + i ~ C^2_\mu).
\end{gather}
Certainly the Lagrangian (\ref{eq:Abelian}) is invariant under
the active (classical) gauge transformation (\ref{gt1}) described
by
\begin{gather}
\delta A_\mu = \displaystyle\frac{1}{g} \partial_\mu \theta
- i (C_\mu^* \alpha - C_\mu \alpha^*),
\nonumber\\
\delta \tilde C_\mu = - \delta A_\mu,
\quad
\delta W_\mu = 0.
\label{eq:active}
\end{gather}
But it has another gauge invariance, the invariance under the
following passive (quantum) gauge transformation
\begin{gather}
\delta A_\mu = \displaystyle\frac{1}{g} \partial_\mu \theta
-i (W_\mu^* \alpha - W_\mu \alpha^*),
\nonumber\\
\delta \tilde C_\mu = 0,
\quad
\delta W_\mu = \displaystyle\frac{1}{g} {\hat D}_\mu \alpha - i \theta W_\mu.
\label{eq:passive}
\end{gather}
Clearly this passive gauge transformation assures the desired
non-Abelian gauge symmetry for the Abelian formalism.
This tells that the Abelian theory not only retains
the original gauge symmetry, but actually has an enlarged (both
active and passive) gauge symmetries.

The reason for this extra (quantum) gauge symmetry is that
the Abelian decomposition automatically put the theory in
the background field formalism which doubles the gauge
symmetry \cite{dewitt}. This is because in this decomposition
we can view the restricted and valence potentials as the
classical and quantum potentials, so that we have freedom to
assign the gauge symmetry either to the classical field
or to the quantum field. This is why we have the extra gauge
symmetry.

The Abelian decomposition has played a crucial role in QCD
to demonstrate the Abelian dominance and the monopole condensation
in color confinement \cite{prd00,prd13,kondo}. This is because
it separates not only the Abelian potential but also the monopole
potential gauge independently.

Now, consider the Georgi-Glashow model (\ref{ggl}). With
\begin{eqnarray}
\vec{\Phi} = \rho~\hat{n},
~~~{\vec A}_\mu=\hat A_\mu +\vec W_\mu,
\end{eqnarray}
we have the Abelian decomposition,
\begin{eqnarray}
&{\cal L}_{GG} =-\dfrac{1}{2} (\partial_\mu \rho)^2
-\dfrac{g^2}{2} {\rho}^2 ({\vec W}_\mu)^2-\dfrac{\lambda}{4}
\big(\rho^2 -\dfrac{\mu^2}{\lambda}\big)^2 \nonumber\\
&-\dfrac{1}{4} {\hat F}_{\mu\nu}^2
-\dfrac{1}{4}({\hat D}_\mu{\vec W}_\nu-{\hat D}_\nu{\vec W}_\mu)^2 \nonumber\\
&-\dfrac{g}{2}{\hat F}_{\mu\nu} \cdot ({\vec W}_\mu \times {\vec W}_\nu)
-\dfrac{g^2}{4} ({\vec W}_\mu \times {\vec W}_\nu)^2.
\label{gglag1}
\end{eqnarray}
With this we can Abelianize it gauge independently,
\begin{gather}
{\cal L}_{GG}= -\displaystyle\frac{1}{2} (\partial_\mu \rho)^2
- g^2 {\rho}^2 |W_\mu |^2-\displaystyle\frac{\lambda}{4}\big(\rho^2
-\displaystyle\frac{\mu^2}{\lambda}\big)^2 \nonumber\\
- \displaystyle\frac{1}{4} {\cal F}_{\mu\nu}^2
-\displaystyle\frac{1}{2} |{\hat D}_\mu W_\nu-{\hat D}_\nu W_\mu|^2
+ ig {\cal F}_{\mu\nu} W_\mu^*W_\nu \nonumber\\
+ \displaystyle\frac{g^2}{4}(W_\mu^* W_\nu - W_\nu^* W_\mu)^2.
\label{gglag2}
\end{gather}
This clearly shows that the theory can be viewed as a
(non-trivial) Abelian gauge theory which has a charged
vector field as a source.

The Abelianized Lagrangian looks very much like the
Georgi-Glashow Lagrangian written in the unitary gauge.
But we emphasize that this is the gauge independent
Abelianization which has the full (quantum) $SU(2)$
gauge symmetry.

Obviously we can apply the same Abelian decomposition
to the Weinberg-Salam theory
\begin{eqnarray}
&{\cal L}=-\dfrac{1}{2}{(\partial_{\mu}\rho)}^2
-\dfrac{\rho^2}{2} {|{\cal \hat D}_{\mu} \xi |}^2
-\dfrac{\lambda}{8}(\rho^2-\rho_0^2)^2 \nonumber\\
&-\dfrac{1}{4} {\hat F}_{\mu\nu}^2
-\dfrac{1}{4} G_{\mu\nu}^2
-\dfrac{1}{4}({\hat D}_\mu{\vec W}_\nu-{\hat D}_\nu{\vec W}_\mu)^2
-\dfrac{g^2}{8}\rho^2 ({\vec W}_\mu)^2  \nonumber\\
&-\dfrac{g}{2}{\hat F}_{\mu\nu} \cdot ({\vec W}_\mu \times {\vec W}_\nu)
-\dfrac{g^2}{4} ({\vec W}_\mu \times {\vec W}_\nu)^2,  \nonumber\\
&{\cal \hat D}_\mu=\partial_\mu
-i\dfrac{g}{2} \vec{\tau}\cdot\hat {A}_\mu-i\dfrac{g'}{2}B_\mu.
\label{wslag2}
\end{eqnarray}
Moreover, with
\begin{gather}
\left( \begin{array}{cc} A_\mu^{\rm (em)} \\  Z_{\mu}
\end{array} \right)
= \displaystyle\frac{1}{\sqrt{g^2 + g'^2}} \left(\begin{array}{cc} g & g' \\
-g' & g
\end{array} \right)
\left( \begin{array}{cc}
B_{\mu} \\ {\cal A}_{\mu}
\end{array} \right),
\label{mixing}
\end{gather}
we can Abelianize it gauge independently
\begin{gather}
{\cal L}= -\displaystyle\frac{1}{2}(\partial_\mu \rho)^2
-\displaystyle\frac{\lambda}{8}\big(\rho^2-\rho_0^2 \big)^2 \nonumber\\
-\displaystyle\frac{1}{4} {F_{\mu\nu}^{\rm (em)}}^2
-\displaystyle\frac{1}{4} Z_{\mu\nu}^2-\displaystyle\frac{g^2}{4}\rho^2 |W_\mu|^2
-\displaystyle\frac{g^2+g'^2}{8} \rho^2 Z_\mu^2 \nonumber\\
-\displaystyle\frac{1}{2}|(D_\mu^{\rm (em)} W_\nu - D_\nu^{\rm (em)} W_\mu)
+ ie \displaystyle\frac{g}{g'} (Z_\mu W_\nu - Z_\nu W_\mu)|^2  \nonumber\\
+ie F_{\mu\nu}^{\rm (em)} W_\mu^* W_\nu
+ie \displaystyle\frac{g}{g'}  Z_{\mu\nu} W_\mu^* W_\nu \nonumber\\
+ \displaystyle\frac{g^2}{4}(W_\mu^* W_\nu - W_\nu^* W_\mu)^2,
\label{wslag3}
\end{gather}
where $D_\mu^{\rm (em)}=\partial_\mu+ieA_\mu^{\rm (em)}$.
Again we emphasize that this is not the Weinberg-Salam Lagrangian
in the unitary gauge. This is the gauge independent Abelianization
which has the extra quantum (passive) non-Abelian gauge degrees of
freedom. This can easily be understood comparing (\ref{mixing})
with (\ref{wein}). Certainly (\ref{mixing}) is gauge independent, while
(\ref{wein}) applies to the unitary gauge.

This provides us important piece of information. In the absence
of the electromagnetic interaction (i.e., with $A_\mu^{\rm (em)}
= W_\mu = 0$) the Weinberg-Salam model describes a spontaneously
broken $U(1)_Z$ gauge theory,
\begin{gather}
{\cal L} = -\displaystyle\frac{1}{2}(\partial_\mu \rho)^2
-\displaystyle\frac{\lambda}{8}\big(\rho^2-\rho_0^2\big)^2  \nonumber\\
-\displaystyle\frac{1}{4} Z_{\mu\nu}^2-\displaystyle\frac{g^2+g'^2}{8} \rho^2 Z_\mu^2,
\end{gather}
which is nothing but the Ginsburg-Landau theory of superconductivity.
Furthermore, here $M_H$ and $M_Z$ corresponds to the coherence length
(of the Higgs field) and the penetration length (of the magnetic
field made of $Z$-field). So, when $M_H > M_Z$ (or $M_H < M_Z$),
the theory describes a type II (or type I) superconductivity,
which is well known to admit the Abrikosov-Nielsen-Olesen
vortex solution. This confirms the existence of Nambu's string
in Weinberg-Salam model. What Nambu showed was that he could make
the string finite by attaching the fractionally charged monopole
anti-monopole pair to this string \cite{nambu}.

\section{Comparison with Julia-Zee Dyon}

The Cho-Maison dyon looks very much like the well-known
Julia-Zee dyon in the Georgi-Glashow model. Both can be
viewed as the Wu-Yang monopole dressed by the weak boson(s).
However, there is a crucial difference. The the Julia-Zee
dyon is completely regular and has a finite energy, while
the Cho-Maison dyon has a point singularity at the center
which makes the energy infinite.

So, to regularize the Cho-Maison dyon it is important
to understand the difference between the two dyons.
To do that notice that, in the absence of the
$Z$-boson, (\ref{wslag3}) reduces to
\begin{eqnarray}
&{\cal L}= -\dfrac{1}{2}(\partial_\mu \rho)^2
-\dfrac{\lambda}{8}\big(\rho^2-\rho_0^2\big)^2
-\dfrac{g^2}{4}\rho^2 |W_\mu|^2 \nonumber\\
&-\dfrac{1}{4} {F_{\mu\nu}^{\rm (em)}}^2
-\dfrac{1}{2}|D_\mu^{\rm (em)} W_\nu-D_\nu^{\rm (em)} W_\mu|^2 \nonumber\\
&+ie F_{\mu\nu}^{\rm (em)} W_\mu^* W_\nu
+ \dfrac{g^2}{4}(W_\mu^* W_\nu - W_\nu^* W_\mu)^2.
\label{wslag4}
\end{eqnarray}
This should be compared with (\ref{gglag2}), which shows that
the two theories have exactly the same type of interaction in
the absence of the $Z$-boson, if we identify $\cal F_{\mu\nu}$ in
(\ref{gglag2}) with $F_{\mu\nu}^{\rm (em)}$ in (\ref{wslag4}). The
only difference is the coupling strengths of the $W$-boson
quartic self-interaction and Higgs interaction of $W$-boson
(responsible for the Higgs mechanism). This difference, of
course, originates from the fact that the Weinberg-Salam model
has two gauge coupling constants, while the Georgi-Glashow model
has only one.

This tells that, in spite of the fact that the Cho-Maison dyon
has infinite energy, it is not much different from the Julia-Zee
dyon. To amplify this point notice that the spherically symmetric
ansatz of the Julia-Zee dyon
\begin{gather}
\vec \Phi=\rho(r)~\hat r,
~~~\hat A_\mu=\displaystyle\frac{1}{g}A(r)\partial_\mu t~\hat{r}
- \displaystyle\frac{1}{g}\hat{r}\times\partial_\mu \hat{r} \nonumber\\
{\vec W}_\mu= \displaystyle\frac{1}{g}f(r) \hat{r}\times\partial_\mu \hat{r} ,
\label{ggdans}
\end{gather}
can be written in the Abelian formalism as
\begin{gather}
\rho = \rho(r),
~~~W_{\mu} = \displaystyle\frac{i}{g}\displaystyle\frac{f(r)}{\sqrt2}e^{i\varphi}
(\partial_\mu \theta +i \sin\theta \partial_\mu \varphi), \nonumber\\
{\cal A}_{\mu} = \displaystyle\frac{1}{g}A(r) \partial_{\mu}t
-\displaystyle\frac{1}{g}(1-\cos\theta) \partial_{\mu} \varphi.
\end{gather}
In the absence of the $Z$-boson this is identical to the
ansatz (\ref{ans2}).

With the ansatz we have the following equation for the dyon
\begin{gather}
\ddot{\rho}+\displaystyle\frac{2}{r}\dot{\rho} - 2\displaystyle\frac{f^2}{r^2}\rho
=\lambda \big(\rho^2 - \displaystyle\frac{\mu^2}{\lambda} \big)\rho,
\nonumber \\
\ddot{f}- \displaystyle\frac{f^2-1}{r^2}f=(g^2\rho^2-A^2)f,
\nonumber\\
\ddot{A}+\displaystyle\frac{2}{r}\dot{A} -2\displaystyle\frac{f^2}{r^2} A=0.
\label{ggdeq}
\end{gather}
This should be compared to the equation of motion (\ref{cmeq})
for the Cho-Maison dyon. They are not much different.

With the boundary condition
\begin{gather}
\rho(0)=0, \quad  f(0)=1, \quad A(0)=0, \nonumber\\
\rho(\infty)=\bar \rho_0=\sqrt{\mu^2/\lambda},~~f(\infty)=0,
~~A(\infty)=A_0,
\label{ggbc}
\end{gather}
one can integrate (\ref{ggdeq}) and obtain the Julia-Zee
dyon which has a finite energy. Notice that the boundary condition
$A(0)=0$ and $f(0)=1$ is crucial to make the solutions regular
at the origin. This confirms that the Julia-Zee dyon is nothing
but the Abelian monopole regularized by $\rho$ and $W_\mu$, where
the charged vector field adds an extra electric charge to the
monopole. Again it must be clear from (\ref{ggdeq}) that,
for a given magnetic charge, there are always two dyons with
opposite electric charges.

Moreover, for the monopole (and anti-monopole) solution
with $A=0$, the equation reduces to the following
Bogomol'nyi-Prasad-Sommerfield equation in the limit
$\lambda=0$
\begin{gather}
\dot{\rho}\pm \displaystyle\frac{1}{gr^2}(f^2-1)=0,
~~~\dot{f} \pm g \rho f=0.
\label{pseq}
\end{gather}
This has the analytic solution
\begin{gather}
\rho= \bar \rho_0\coth(g\bar \rho_0 r)-\dfrac{1}{er},
~~~f= \dfrac{g\bar \rho_0 r}{\sinh(g \bar \rho_0 r)},
\end{gather}
which describes the Prasad-Sommerfield monopole \cite{prasad}.

Of course, the Cho-Maison dyon has a non-trivial dressing of
the $Z$-boson which is absent in the Julia-Zee dyon. But notice
that the $Z$-boson plays no role in the Cho-Maison monopole.
This confirms that the Cho-Maison monopole and the `tHooft-Polyakov
monopole are not so different, so that the Cho-Maison monopole
could be modified to have finite energy.

For the anti-dyon we can have the following ansatz
\begin{gather}
\vec \Phi=\rho(r)~{\hat r}',
~~~\hat A_\mu'=-\displaystyle\frac{1}{g}A(r)\partial_\mu t~{\hat r}'
- \displaystyle\frac{1}{g}{\hat r}' \times\partial_\mu {\hat r}' \nonumber\\
{\vec W}_\mu'= \displaystyle\frac{1}{g}f(r)~{\hat r}' \times\partial_\mu {\hat r}', \nonumber\\
{\hat r}'=(\sin \theta \cos \phi,-\sin \theta \sin \phi,\cos \theta),
\label{antiggd}
\end{gather}
or equivalently
\begin{gather}
\rho' = \rho(r),
~~~W_{\mu} = \displaystyle\frac{i}{g}\displaystyle\frac{f(r)}{\sqrt2}e^{-i\varphi}
(\partial_\mu \theta -i \sin\theta \partial_\mu \varphi), \nonumber\\
{\cal A}'_{\mu} = -\displaystyle\frac{1}{g}A(r) \partial_{\mu}t
+\displaystyle\frac{1}{g}(1-\cos\theta) \partial_{\mu} \varphi.
\end{gather}
This ansatz looks different from the popular ansatz described
by $\vec \Phi=-\rho(r)~{\hat r}$, but we can easily show that they
are gauge equivalent. With this we have exactly the same equation
(\ref{ggdeq}) for the anti-dyon, which assures that the theory
has both dyon and anti-dyon.

\section{Ultraviolet Regularization of Cho-Maison Dyon}

Since the Cho-Maison dyon is the only dyon in the standard
model, it is impossible to regularize it within the model.
However, the Weinberg-Salam model is the ``bare" theory
which should change to the ``effective" theory after the
quantum correction, and the ``real" electroweak dyon must
be the solution of such theory. So we may hope that the
quantum correction could regularize the Cho-Maison dyon.

The importance of the quantum correction in classical solutions
is best understood in QCD. The ``bare" QCD Lagrangian has no
confinement, so that the classical solutions of the bare QCD
can never describe the quarkonium or hadronic bound states.
Only the effective theory can do.

To see how the quantum modification could make the energy of
the Cho-Maison monopole finite, notice that after the quantum
correction the coupling constants change to the scale dependent
running couplings. So, if this quantum correction makes $1/g'^2$
in $E_0$ in (\ref{cme}) vanishing in the short distance limit, the
Cho-Maison monopole could have finite energy.

To do that consider the following effective Lagrangian which
has the non-canonical kinetic term for the $U(1)_Y$ gauge field
\begin{gather}
{\cal L}_{eff} = -|{\cal D} _\mu \phi|^2
-\displaystyle\frac{\lambda}{2} \Big(\phi^2 -\displaystyle\frac{\mu^2}{\lambda}\Big)^2
-\displaystyle\frac{1}{4} \vec F_{\mu\nu}^2  \nonumber\\
-\displaystyle\frac{1}{4} \epsilon(|\phi|^2 ) G_{\mu\nu}^2,
\label{effl}
\end{gather}
where $\epsilon(|\phi|^2)$ is a positive dimensionless function
of the Higgs doublet which approaches to one asymptotically.
Clearly $\epsilon$ modifies the permittivity of the $U(1)_Y$
gauge field, but the effective action still retains the
$SU(2)\times U(1)_Y$ gauge symmetry. Moreover, when
$\epsilon \rightarrow 1$ asymptotically, the effective
action reproduces the standard model.

This type of effective theory which has the field dependent
permittivity naturally appears in the non-linear electrodynamics
and higher-dimensional unified theory, and has been studied
intensively in cosmology to explain the late-time accelerated
expansion \cite{prd87,prl92,babi}.

\begin{figure}
\includegraphics[height=4cm, width=8cm]{fcdyon}
\caption{\label{fcdyon} The finite energy electroweak dyon
solution obtained from the effective Lagrangian (\ref{effl}).
The solid line represents the finite energy dyon
and dotted line represents the Cho-Maison dyon, where
$Z=A-B$ and we have chosen $f(0)=1$ and $A(\infty)=M_W/2$.}
\label{fig2}
\end{figure}

From (\ref{effl}) we have the equations for $\rho$ and $B_\mu$
\begin{gather}
\partial^2 \rho =|{\cal D}_\mu \xi|^2 \rho
+\displaystyle\frac{\lambda}{2}(\rho^2 - \rho_0^2) \rho
+\displaystyle\frac{1}{2} \epsilon' \rho G_{\mu\nu}^2,  \nonumber \\
\partial_\mu G_{\mu\nu}= i \displaystyle\frac{g'}{2 \epsilon} \rho^2
[\xi^\dagger {\cal D}_\nu \xi - ({\cal D}_\nu\xi)^\dagger \xi]
-\displaystyle\frac{\partial_\mu \epsilon}{\epsilon} G_{\mu\nu},
\label{meq}
\end{gather}
where $\epsilon' = d\epsilon/d{\rho^2}$. This changes the dyon
equation (\ref{cmeq}) to
\begin{gather}
\ddot{\rho} + \displaystyle\frac{2}{r}\dot{\rho}-\displaystyle\frac{f^2}{2r^2}\rho
=-\displaystyle\frac{1}{4} (A-B)^2 \rho
+\displaystyle\frac{\lambda}{2} (\rho^2- \rho_0^2) \rho \nonumber\\
+ \displaystyle\frac{\epsilon'}{g'^2}\Big(\displaystyle\frac{1}{r^4}-\dot{B}^2 \Big) \rho,  \nonumber\\
\ddot{f}-\displaystyle\frac{f^2-1}{r^2}f=\big(\displaystyle\frac{g^2}{4}\rho^2
- A^2\big)f, \nonumber\\
\ddot{A}+\displaystyle\frac{2}{r}\dot{A}-\displaystyle\frac{2f^2}{r^2}A
=\displaystyle\frac{g^2}{4}\rho^2(A-B), \nonumber \\
\ddot{B} + 2\big(\displaystyle\frac{1}{r}+
\displaystyle\frac{\epsilon'}{\epsilon} \rho \dot{\rho} \big) \dot{B}
=-\displaystyle\frac{g'^2}{4 \epsilon} \rho^2 (A-B).
\end{gather}
This tells that effectively $\epsilon$ changes the
$U(1)_Y$ gauge coupling $g'$ to the ``running" coupling
$\bar g'=g' /\sqrt{\epsilon}$. This is because with
the rescaling of $B_\mu$ to $B_\mu/g'$, $g'$ changes
to $g' /\sqrt{\epsilon}$. So, by making $\bar g'$
infinite (requiring $\epsilon$ vanishing) at the
origin, we can regularize the Cho-Maison monopole.

From the equations of motion we find that we need the
following condition near the origin to make the monopole
energy finite
\begin{gather}
\epsilon \simeq \Big(\displaystyle\frac{\rho}{\rho_0}\Big)^n,
~~~n > 4+2\sqrt 3 \simeq 7.46.
\end{gather}
With $n=8$ we have
\begin{eqnarray}
\rho(r) \simeq  r^\delta,
~~~\delta={\displaystyle\frac{\sqrt{3}-1}{2}},
\end{eqnarray}
near the origin, and have the finite energy dyon solution
shown in Fig. \ref{fcdyon}. It is really remarkable
that the regularized solutions look very much like the
Cho-Maison solutions, except that for the finite energy
dyon solution $Z(0)$ becomes zero. This confirms that
the ultraviolet regularization of the Cho-Maison
monopole can indeed be possible.

\begin{figure}
\includegraphics[width=8cm, height=4cm]{effg}
\caption{\label{effg} The running coupling $\bar g'$ of $U(1)_Y$
gauge field induced by the effective Lagrangian (\ref{effl}).}
\end{figure}

As expected with $n=8$ the running coupling $\bar g'$ becomes
divergent at the origin, which makes the energy contribution
from the $U(1)_Y$ gauge field finite. The scale dependence of
the running coupling is shown in Fig. \ref{effg}. With $A=B=0$
we can estimate the monopole energy to be
\begin{gather}
E \simeq 0.65 \times \displaystyle\frac{4\pi}{e^2} M_W \simeq 7.19 ~{\rm TeV}.
\end{gather}
This tells that the estimate of the monopole energy based on
the scaling argument is reliable. The finite energy monopole
solution is shown in Fig.~\ref{fcmono}.

\begin{figure}
\includegraphics[width=8cm, height=4cm]{fcmono}
\caption{\label{fcmono}The finite energy electroweak monopole
solution obtained from the effective Lagrangian (\ref{modl}).
The solid line (red) represents the regularized monopole and
the dotted (blue) line represents the Cho-Maison monopole.}
\end{figure}

There is another way to regularize the Cho-Maison monopole.
Suppose we have the following ultraviolet modification of
(\ref{wslag2}) from the quantum correction
\begin{gather}
\delta {\cal L}=ie \alpha F_{\mu\nu}^{\rm (em)} W_\mu^* W_\nu
+\beta \displaystyle\frac{g^2}{4}(W_\mu^*W_\nu-W_\nu^*W_\mu)^2  \nonumber\\
-\gamma \displaystyle\frac{g^2}{4} \rho^2 |W_\mu|^2.
\end{gather}
where $\alpha,~\beta,~\gamma$ are the scale dependent parameters
which vanish asymptotically (and modify the theory only at short distance).
With this we have the modified Weinberg-Salam Lagrangian
\begin{gather}
{\cal L}'= -\displaystyle\frac{1}{2}(\partial_\mu \rho)^2
-\displaystyle\frac{\lambda}{8}\big(\rho^2-\rho_0^2\big)^2
-\displaystyle\frac{1}{4} {F_{\mu\nu}^{\rm (em)}}^2
-\displaystyle\frac{1}{4} Z_{\mu\nu}^2  \nonumber\\
-\displaystyle\frac{1}{2}\big|(D_\mu^{\rm (em)} W_\nu - D_\nu^{\rm (em)} W_\mu)
+ie \displaystyle\frac{g}{g'}(Z_\mu W_\nu - Z_\nu W_\mu)\big|^2 \nonumber \\
+ie(1+\alpha) F_{\mu\nu}^{\rm (em)} W_\mu^* W_\nu
+ie \displaystyle\frac{g}{g'} Z_{\mu\nu} W_\mu^* W_\nu \nonumber\\
+(1+\beta)\displaystyle\frac{g^2}{4}(W_\mu^* W_\nu -W_\nu^* W_\mu)^2 \nonumber\\
-(1+\gamma) \displaystyle\frac{g^2}{4}\rho^2 |W_\mu|^2
-\displaystyle\frac{g^2+g'^2}{8} \rho^2 Z_\mu^2.
\label{modl}
\end{gather}
Of course, this modification is supposed to hold only in the
short distance, so that asymptotically  $\alpha,~\beta,~\gamma$
should vanish to make sure that ${\cal L}'$ reduces to the
standard model. But we will treat them as constants, partly
because it is difficult to make them scale dependent, but
mainly because asymptotically the boundary condition automatically
makes them irrelevant and assures the solution to converge to
the Cho-Maison solution.

To understand the physical meaning of (\ref{modl}) notice
that in the absence of the $Z$-boson the above Lagrangian
reduces to the Georgi-Glashow Lagrangian where the $W$-boson
has an extra ``anomalous" magnetic moment $\alpha$ when
$(1+\beta)=e^2/g^2$ and $(1+\gamma)=4e^2/g^2$, if we
identify the coupling constant $g$ in the Georgi-Glashow model
with the electromagnetic coupling constant $e$. Moreover, the
ansatz (\ref{ans1}) can be written as
\begin{eqnarray}
&{\vec A}_\mu=\hat A_\mu^{\rm (em)} +\vec W_\mu, \nonumber\\
&\hat A_\mu^{\rm (em)}=e \big[\dfrac{1}{g^2}A(r)
+\dfrac{1}{g'^2}B(r)\big] \partial_\mu t ~\hat r
-\dfrac1e \hat r\times \partial_\mu \hat r , \nonumber\\
&\vec W_\mu=\dfrac{f(r)}{g} \hat r\times \partial_\mu \hat r, \nonumber\\
&Z_{\mu} = \dfrac{e}{gg'}\big(A(r)-B(r)\big) \partial_{\mu}t.
\label{ans3}
\end{eqnarray}
This shows that, for the monopole (i.e., for $A=B=0$)
the ansatz becomes formally identical to (\ref{ggdans})
if $\vec W_\mu$ is rescaled by a factor $g/e$. This tells
that, as far as the monopole solution is concerned, in
the absence of the $Z$-boson the Weinberg-Salam model and
Georgi-Glashow model are not so different.

With (\ref{modl}) the energy of the dyon is given by
\begin{gather}
\hat E =\hat E_0 +\hat E_1, \nonumber\\
\hat E_0 =\displaystyle\frac{2\pi}{g^2}\int_0^\infty
\displaystyle\frac{dr}{r^2}\Big\{\displaystyle\frac{g^2}{g'^2}+1 -2(1+\alpha) f^2
+(1+\beta)f^4 \Big\} \nonumber\\
=\displaystyle\frac{2\pi}{g^2}\int_0^\infty
\displaystyle\frac{dr}{r^2}\Big\{\displaystyle\frac{g^2}{e^2}-\displaystyle\frac{(1+\alpha)^2}{1+\beta}
+(1+\beta)\big(f^2-\displaystyle\frac{1+\alpha}{1+\beta}\big)^2 \Big\}, \nonumber\\
\hat E_1 =\displaystyle\frac{4\pi}{g^2} \int_0^\infty dr
\bigg\{\displaystyle\frac{g^2}{2}(r\dot\rho)^2
+\displaystyle\frac{\lambda g^2r^2}{8}\big(\rho^2-\rho_0^2 \big)^2 \nonumber\\
+\dot f^2 +\displaystyle\frac{1}{2}(r\dot A)^2
+\displaystyle\frac{g^2}{2g'^2}(r\dot B)^2
+(1+\gamma) \displaystyle\frac{g^2}{4} f^2\rho^2  \nonumber\\
+\displaystyle\frac{g^2r^2}{8} (B-A)^2 \rho^2 +f^2 A^2 \bigg\}.
\label{energy_2}
\end{gather}
Notice that $\hat E_1$ remains finite with the modification,
and $\gamma$ plays no role to make the monopole energy
finite.

\begin{figure}
\includegraphics[height=4cm, width=8cm]{fecdyon}
\caption{\label{fecdyon} The finite energy electroweak dyon
solution obtained from the modified Lagrangian (\ref{modl}).
The solid line represents the finite energy dyon and dotted
line represents the Cho-Maison dyon.}
\label{fig2}
\end{figure}

To make $\hat E_0$ finite we must have
\begin{gather}
1+\alpha=\dfrac1{f(0)^2} \dfrac{g^2}{e^2},
~~~1+\beta=\dfrac1{f(0)^4} \dfrac{g^2}{e^2},
\label{fecon}
\end{gather}
so that the constants $\alpha$ and $\beta$ are fixed by $f(0)$.
With this the equation of motion is given by
\begin{gather}
\ddot \rho+\displaystyle\frac{2}{r}\dot\rho-\displaystyle\frac{(1+\gamma)f^2}{2r^2}\rho
=-\displaystyle\frac{1}{4}(A-B)^2 \rho
+\displaystyle\frac{\lambda}{2}\big(\rho^2-\rho_0^2 \big)\rho, \nonumber \\
\ddot f -\displaystyle\frac{(1+\alpha)}{r^2}\Big( \dfrac{f^2}{f^2(0)}-1 \Big) f
=\Big( (1+\gamma)\displaystyle\frac{g^2}{4}\rho^2-A^2 \Big) f, \nonumber\\
\ddot A +\displaystyle\frac{2}{r} \dot A - \displaystyle\frac{2f^2}{r^2} A
=\displaystyle\frac{g^2}{4}(A-B)\rho^2,  \nonumber \\
\ddot B+\displaystyle\frac{2}{r}\dot B =-\displaystyle\frac{g'^2}{4}(A-B) \rho^2 .
\label{eqm3}
\end{gather}
The solution has the following behavior near the origin,
\begin{eqnarray}
&\rho \simeq  \alpha_1 r^{\delta_1},
~~~~\dfrac{f}{f(0)} \simeq 1 + \beta_1 r^{\delta_2}, \nonumber \\
&A \simeq a_1 r^{\delta_3},~~~~B \simeq b_0 + b_1 r^{\delta_4},
\label{origin1}
\end{eqnarray}
where
\begin{eqnarray}
&\delta_1 = \dfrac{1}{2}(\sqrt{1+2(1+\gamma)f^2(0)} -1), \nonumber\\
&\delta_2 = \dfrac{1}{2}(1+\sqrt{8\alpha+9}),
~~~\delta_3 = \dfrac{1}{2}(\sqrt{1+8f^2(0)} -1),\nonumber\\
&\delta_4 = \sqrt{1+2f^2(0)} +1. \nonumber
\end{eqnarray}
Notice that all four deltas are positive (as far as $(1+\alpha)>0$),
so that the four functions are well behaved at the origin.

If we assume $\alpha=\gamma=0$ we have $f(0)=g/e$,
and we can integrate (\ref{eqm3}) with the boundary condition
\begin{eqnarray}
&\rho(0)=0,~~~f(0)=g/e,~~~A(0)=0,~~~B(0)=b_0, \nonumber\\
&f(\infty)=0,~\rho(\infty)=\rho_0,
~A(\infty)=B(\infty)=A_0.
\label{bc1}
\end{eqnarray}
The finite energy dyon solution is shown in Fig. \ref{fecdyon}.
It should be emphasized that the solution is an approximate
solution which is supposed to be valid only near the origin,
because the constants $\alpha,~\beta,~\gamma$ are supposed
to vanish asymptotically. But notice that asymptotically the
solution automatically approaches to the Cho-Maison solution
even without making them vanish, because we have the same
boundary condition at the infinity. Again it is remarkable that
the finite energy solution looks very similar to the Cho-Maison
solution.

Of course, we can still integrate (\ref{eqm3}) with arbitrary
$f(0)$ and have a finite energy solution.  The monopole energy for
$f(0)=1$ and $f(0)=g/e$ (with $\alpha=\gamma=0$)  are given
by
\begin{eqnarray}
&E(f(0)=1) \simeq 0.61 \times \dfrac{4\pi }{e^2} M_W
\simeq 6.73~{\rm TeV},  \nonumber\\
&E(f(0)=\dfrac{g}{e})\simeq 1.27 \times \dfrac{4\pi }{e^2} M_W
\simeq 13.95~{\rm TeV}.
\end{eqnarray}
In general the energy of dyon depends on $f(0)$, but must be
of the order of $(4\pi/e^2) M_W$. The energy dependence of
the monopole on $f(0)$ is shown in Fig. \ref{edf0}. This
strongly supports our prediction of the monopole mass based
on the scaling argument.

\begin{figure}
\includegraphics[height=4cm, width=8cm]{edf0}
\caption{\label{edf0} The energy dependence of the electroweak
monopole on $f(0)$.}
\label{fig3}
\end{figure}

As we have emphasized, in the absence of the $Z$-boson
(\ref{modl}) reduces to the Georgi-Glashow theory with
\begin{eqnarray}
\alpha = 0, \quad
1+\beta=\displaystyle\frac{e^2}{g^2}, \quad
1+\gamma=\displaystyle\frac{4e^2}{g^2}.
\label{cond4}
\end{eqnarray}
In this case (\ref{eqm3}) reduces to the following
Bogomol'nyi-Prasad-Sommerfield equation in the limit
$\lambda=0$ \cite{prasad}
\begin{gather}
\dot{\rho}\pm \displaystyle\frac{1}{er^2}\big(\displaystyle\frac{e^2}{g^2}f^2
-1 \big)=0,
~~~\dot{f}\pm e\rho f=0.
\label{self2}
\end{gather}
This has the analytic monopole solution
\begin{gather}
\rho=\rho_0\coth(e\rho_0r)-\displaystyle\frac{1}{er},
~~~f= \displaystyle\frac{g\rho_0 r}{\sinh(e\rho_0r)},
\end{gather}
whose energy is given by the Bogomol'nyi bound
\begin{eqnarray}
E=\sin \theta_{\rm w} \times \displaystyle\frac{8\pi}{e^2} M_{W}
\simeq 5.08~{\rm TeV}.
\end{eqnarray}
From this we can confidently say that the mass of the electroweak
monopole could be around 4 to 7 TeV.

This confirms that we can regularize the Cho-Maison dyon with
a simple modification of the coupling strengths of the existing
interactions which could be caused by the quantum correction.
This provides a most economic way to make the energy of the dyon
finite without introducing a new interaction in the standard model.

\section{Embedding $U(1)_Y$ to $SU(2)_Y$}

Another way to regularize the Cho-Maison dyon, of course, is
to enlarge $U(1)_Y$ and embed it to another SU(2). This type of
generalization of the standard model could naturally arise in the
left-right symmetric grand unification models, in particular in the
SO(10) grand unification, although this generalization may be too
simple to be realistic.

To construct the desired solutions we introduce a hypercharged vector
field $X_\mu$ and a Higgs field $\sigma$, and generalize the Lagrangian
(\ref{wslag2}) adding the following Lagrangian
\begin{eqnarray}
&\Delta {\cal L}=-\dfrac{1}{2}|\tilde D_\mu X_\nu-\tilde D_\nu X_\mu|^2
+ig' G_{\mu\nu}X_\mu^* X_\nu \nonumber\\
&+\dfrac{1}{4}g'^2(X_\mu^* X_\nu -X_\nu^* X_\mu)^2  \nonumber\\
&-\dfrac{1}{2}(\partial_\mu\sigma)^2 -g'^2\sigma^2 |X_\mu|^2
-\dfrac{\kappa}{4}\big( \sigma^2-\dfrac{m^2}{\kappa}\big)^2,
\label{lag4}
\end{eqnarray}
where $\tilde D_\mu = \partial_\mu +ig' B_\mu$. To understand
the meaning of it let us introduce a hypercharge $SU(2)$ gauge field
$\vec B_\mu$ and a scalar triplet ${\vec \Phi}$, and consider
the $SU(2)_Y$ Georgi-Glashow model
\begin{eqnarray}
&{\cal L}'=-\dfrac{1}{2}(D_\mu {\vec \Phi})^2
-\dfrac{\kappa}{4}\big({\vec \Phi}^2-\dfrac{m^2}{\kappa}\big)^2
-\dfrac{1}{4} \vec {G}_{\mu\nu}^2.
\end{eqnarray}
Now we can have the Abelian decomposition of this Lagrangian
with $\vec \Phi=\sigma {\hat n}$, and have (identifying $B_\mu$ and $X_\mu$
as the Abelian and valence parts)
\begin{eqnarray}
&{\cal L}'=-\dfrac14 G_{\mu\nu}^2+ \Delta {\cal L}.
\end{eqnarray}
This clearly shows that Lagrangian (\ref{lag4}) describes nothing but
the embedding of the hypercharge $U(1)$ to an $SU(2)$ Georgi-Glashow
model.

Now for a static spherically symmetric ansatz
we choose (\ref{ans1}) and let
\begin{eqnarray}
&\sigma =\sigma(r), \nonumber \\
&X_\mu =\dfrac{i}{g'}\dfrac{h(r)}{\sqrt{2}}e^{i\varphi}
(\partial_\mu \theta+i\sin\theta\partial_\mu \varphi).
\label{ansatz3}
\end{eqnarray}
With the spherically symmetric ansatz the equations of motion are
reduced to
\begin{gather}
\ddot{f}-\displaystyle\frac{f^2-1}{r^2}f
=\big(\dfrac{g^2}{4}\rho^2-A^2\big)f, \nonumber\\
\ddot{\rho}+\displaystyle\frac{2}{r} \dot{\rho} - \displaystyle\frac{f^2}{2r^2}\rho
=-\displaystyle\frac{1}{4}(A-B)^2\rho + \displaystyle\frac{\lambda}{2}\big(\rho^2
-\displaystyle\frac{2\mu^2}{\lambda}\big)\rho, \nonumber\\
\ddot{A} + \displaystyle\frac{2}{r}\dot{A} -\displaystyle\frac{2f^2}{r^2}A
= \displaystyle\frac{g^2}{4} \rho^2(A-B), \nonumber\\
\ddot{B} + \displaystyle\frac{2}{r} \dot{B}- \displaystyle\frac{2h^2}{r^2} B
=\displaystyle\frac{g'^2}{4} \rho^2 (B-A),  \nonumber\\
\ddot h -\displaystyle\frac{h^2-1}{r^2} h =(g'^2\sigma^2-B^2) h, \nonumber\\
\ddot\sigma +\displaystyle\frac{2}{r}\dot\sigma -\displaystyle\frac{2h^2}{r^2} \sigma
= \kappa\big(\sigma^2-\displaystyle\frac{m^2}{\kappa}\big)\sigma.
\label{eom4}
\end{gather}
Furthermore, the energy of the above configuration is given by
\begin{gather}
E=E_W +E_X,  \nonumber\\
E_{W}= \displaystyle\frac{4\pi}{g^2}\int_0^\infty dr
\Big\{\dot f^2 +\displaystyle\frac{(f^2-1)^2}{2r^2}
+\displaystyle\frac{1}{2}(r\dot A)^2  \nonumber\\
+f^2A^2+\displaystyle\frac{g^2}{2}(r\dot\rho)^2 + \displaystyle\frac{g^2}{4} f^2\rho^2
+\displaystyle\frac{g^2r^2}{8}(A-B)^2\rho^2  \nonumber\\
+\displaystyle\frac{\lambda g^2r^2}{8}\big(\rho^2-\displaystyle\frac{2\mu^2}{\lambda}\big)^2
\Big\}=\displaystyle\frac{4\pi}{g^2}~C_1~M_W,  \nonumber\\
E_X=\displaystyle\frac{4\pi}{g'^2}\int_0^\infty dr\Big\{\dot h^2
+\displaystyle\frac{(h^2-1)^2}{2r^2}+\displaystyle\frac{1}{2}(r\dot B)^2  \nonumber\\
+h^2B^2 +\displaystyle\frac{g'^2}{2}(r\dot\sigma)^2+g'^2 h^2\sigma^2  \nonumber\\
+\displaystyle\frac{\kappa g'^2r^2}{4}(\sigma^2-\sigma_0^2)^2 \Big\}
=\displaystyle\frac{4\pi }{g'^2}~C_2~M_X,
\end{gather}
where $\sigma_0=\sqrt{m^2/\kappa}$, $M_X=g' \sigma_0$, $C_1$ and $C_2$
are constants of the order one. The boundary conditions
for a regular field configuration can be chosen as
\begin{eqnarray}
&f(0)=h(0)=1,~~ A(0)=B(0)=\rho(0)=\sigma(0)=0, \nonumber\\
&f(\infty)=h(\infty)=0,~A(\infty)=A_0,~B(\infty)=B_0,\nonumber\\
&\rho(\infty)=\rho_0,~\sigma(\infty)=\sigma_0.
\label{bound3}
\end{eqnarray}
Notice that this guarantees the analyticity of
the solution everywhere, including the origin.

\begin{figure}
\includegraphics[height=4cm, width=8cm]{su2embed}
\caption{The $SU(2)\times SU(2)$ monopole solution with
$M_H/M_W=1.56$, $M_X=10~M_W$, and $\kappa=0$.}
\label{fig4}
\end{figure}

With the boundary condition (\ref{bound3}) one may try to
find the desired solution. From the physical point of view
one could assume $M_X \gg M_W$, where $M_X$ is an intermediate
scale which lies somewhere between the grand unification scale
and the electroweak scale. Now, let $A=B=0$ for simplicity.
Then (\ref{eom4}) decouples to describe two independent systems
so that the monopole solution has two cores, the one with the
size $O(1/M_W)$ and the other with the size $O(1/M_X)$. With
$M_X=10M_W$ we obtain the solution shown in Fig. \ref{fig4}
in the limit $\kappa=0$ and $M_H/M_W=1.56$.

In this limit we find
$C_1=1.53$ and $C_2=1$ so that the  energy of the solution
is given by
\begin{eqnarray}
&E=\dfrac{4\pi}{e^2}\Big( \cos^2\theta_{\rm w}
+0.153~\sin^2\theta_{\rm w}\Big)~M_X  \nonumber\\
&\simeq 110.17~M_X.
\end{eqnarray}
Clearly the solution describes the Cho-Maison monopole whose singularity
is regularized by a Prasad-Sommerfield monopole of the size $O(1/M_X)$.

Notice that, even though the energy of the monopole is fixed by
the intermediate scale, the size of the monopole is determined
by the electroweak scale. Furthermore from the outside the monopole
looks exactly the same as the Cho-Maison monopole. Only the inner core
is regularized by the hypercharged vector field.

\section{Conclusions}

In this paper we have discussed three ways to estimate
the mass of the electroweak monopole, the dimensional argument,
the scaling argument, and the ultraviolet regularization of
the Cho-Maison monopole. As importantly, we have shown that
the standard model has the anti-dyon as well as the dyon
solution, so that they can be produced in pairs.

It has generally been believed that the finite energy monopole
could exist only at the grand unification scale \cite{dokos}.
But our result tells that the genuine electroweak monopole
of mass around 4 to 10 TeV could exist. This strongly implies
that there is an excellent chance that MoEDAL could actually
detect such monopole in the near future, because the 14 TeV
LHC upgrade now reaches the monopole-antimonopole pair production
threshold. But of course, if the mass of the monopole exceeds
the LHC threshold 7 TeV, we may have to look for the monopole
from cosmic ray with the ``cosmic" MoEDAL.

The importance of the electroweak monopole is that it is the
electroweak generalization of the Dirac monopole, and that it is
the only realistic monopole which can be produced and detected.
A remarkable aspect of this monopole is that mathematically it
can be viewed as a hybrid between the Dirac monopole and the
'tHooft-Polyakov monopole.

However, there are two crucial differences. First, the magnetic
charge of the electroweak monopole is two times bigger than that
of the Dirac's monopole, so that it satisfes the Schwinger quantization
condition $q_m=4\pi n/e $. This is because the electroweak generalization
requires us to embed $U(1)_{\rm em}$ to the U(1) subgroup of SU(2),
which has the period of $4\pi$. So the magnetic charge of the electroweak
monopole has the unit $4\pi/e$.

Of course, the finite energy dyon solutions we discussed in
the above are not the solutions of the ``bare" standard model.
Nevertheless they tell us how the Cho-Maison dyon could be
regularized and how the regularized electroweak dyon would
look like. From the physical point of view there is no doubt
that the finite energy solutions should be interpreted as
the regularized Cho-Maison dyons whose mass (and size) is
fixed by the electroweak scale.

We emphasize that, unlike the Dirac's monopole which can exist
only when $U(1)_{\rm em}$ becomes non-trivial, the electroweak
monopole must exist in the standard model. So, if the standard
model is correct, we must have the monopole. {\it In this sense,
the experimental discovery of the electroweak monopole should
be viewed as the final topological test of the standard model.}

Clearly the electroweak monopole invites more difficult questions.
How can we justify the perturbative expansion and the renormalization
in the presence of the monopole? What are the new physical processes
which can be induced by the monopole? Most importantly, how can we
construct the quantum field theory of the monopole?

Moreover, the existence of the finite energy electroweak monopole
should have important physical implications. In particular, it could
have important implications in cosmology, because it can be produced
after inflation. The physical implications of the monopole will be
discussed in a separate paper \cite{cho}.

\textbf{Acknowledgments}

The work is supported in part by the National Research
Foundation (2012-002-134) of the Ministry of
Science and Technology and by Konkuk University.