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\begin{document}
\title{Color Reflection Invariance and Monopole Condensation in QCD}
\bigskip

\author{Y. M. Cho}
\email{ymcho@yongmin.snu.ac.kr}
\affiliation{C.N.Yang Institute for Theoretical Physics,
State University of New York, Stony Brook, \\ 
New York 11794-3840}
\altaffiliation{Permanent Address: School of Physics,
College of Natural Sciences,
Seoul National University, Seoul 151-747, Korea }
\begin{abstract}
The quantum instability of the Savvidy-Nielsen-Olesen (SNO)
vacuum of the one-loop effective action of $SU(2)$ QCD 
is due to the lack of gauge invariance.
We  prove that the gauge invariance, in particular 
the color reflection invariance, 
restores the stability of the SNO vacuum, and thus 
guarantees the stability of the magnetic condensation in QCD.
\end{abstract}
\pacs{11.15.Bt, 14.80.Hv, 12.38.Aw}
\keywords{color reflection invariance, monopole condensation,
vacuum stability of QCD}
\maketitle

\section{Introduction}

The confinement problem in quantum chromodynamics (QCD) 
is probably one of the most challenging problems
in theoretical physics. It has
long been argued that the confinement phase in QCD
can be triggered by the monopole condensation \cite{nambu,cho1}.
Indeed, if one assumes monopole condensation, one can easily argue
that the ensuing dual Meissner effect could guarantee the confinement
of color \cite{cho1,cho2}. A satisfactory theoretical proof of the desired
monopole condensation in QCD, however, has remained 
elusive. 

A natural way to establish the monopole condensation in QCD
is to demonstrate that the quantum fluctuation triggers 
a phase transition through 
the Coleman-Weinberg mechanism \cite{cole},
by generating a non-trivial vacuum which can be identified
as a monopole condensation. Coleman and Weinberg have demonstrated that
the quantum effect could trigger a phase transition 
in massless scalar QED with a quartic self-interaction of
scalar electrons, by showing that the one-loop effective action generates 
a condensation of scalar electrons which defines a non-trivial new vacuum. 
To prove the monopole condensation, one need to demonstrate
such a phase transition in QCD.
There have been many attempts to do so 
with the one-loop effective action of QCD using the background
field method \cite{savv,niel,ditt}. Savvidy has first calculated 
the effective action of $SU(2)$ QCD in the presence of an {\it ad hoc}
color magnetic background, and has ``almost'' proved 
the magnetic condensation in QCD. In particular, he 
showed that the quantum effective potential obtained 
from the real part of the one-loop effective action
has the minimum at a non-vanishing magnetic background \cite{savv}. 
This is exactly what everybody was looking for. 
Unfortunately, this calculation was repeated by Nielsen and 
Olesen, who showed that the magnetic background 
generates an extra imaginary part in the effective action
which induces the pair-creation of the gluons and thus 
de-stablizes the magnetic condensation \cite{niel}.  
This instability of the ``Savvidy-Nielsen-Olesen (SNO)
vacuum'', has never been seriously
challenged, and has destroyed the hope to establish the monopole condensation
in QCD with the effective action.

A few years ago, however, there was a new attempt to calculate
the one-loop effective action of QCD 
with a gauge independent separation
of the non-Abelian monopole background from 
the quantum field \cite{cho3}. Remarkably, in this
calculation the effective action has been shown to
produce no imaginary part in the presence of
the monopole background, but a negative imaginary part
in the presence
of the pure color electric background. This means that in QCD
the non-Abelian monopole background produces a stable monopole
condensation, but the color electric background
becomes unstable by generating a pair annhilation of
the valence gluon at one-loop level.
The new result sharply contradicts with the earlier results,
in particular on the stability of the monopole condensation.
This has resurrected the old controversy on the stability of monopole
condensation. So it is
important for us to understand the origin of this discrepancy
and resolve the controversy.

Mathematically, this discrepancy follows from
the different infra-red regularizations. 
The calculation of the effective action involves the calculation 
of the functional determinant which results from
the gluon loop integral, and a ``naive'' calculation of 
the determinant shows that the determinant contains tachyonic
eigenstates which have imaginary energy. Furthermore, these
tachyonic states generate a severe infra-red divergence
in the effective action. In this case one must remove
the divergence with a proper infra-red regularization.
The SNO effective action follows from the well-known
$\zeta$-function regularization. In QCD, however,
it has been known that the $\zeta$-function regularization 
still retains the contribution of the tachyonic eigenstates
in the effective action.
Because of this the effective action has 
an imaginary part which destablizes 
the magnetic condensation \cite{niel,ditt}. On the other hand, 
one can calculate the effective action
with the infra-red regularization which respects
causality. In this case one can show that the resulting effective action
generates a phase transition with a stable monopole condensation, 
because it no longer has the imaginary part \cite{cho3}.
This is because the infra-red regularization by causality
naturally removes the tachyonic eigenstates which violate 
the causality. But since the $\zeta$-function regularization has worked so
well in quantum field theory, there seems no compelling reason
why it does not work in QCD. So one may still wonder which 
regularization is the correct one for QCD.

Fortunately, one can check this with a perturbative method \cite{sch,cho4}.
At first thought this might sound absurd, because one-loop
effective action is non-perturbative. In general one
can not derive a non-perturbative result with a perturbative method.
For the massless gauge theories, however, the imaginary part of 
the one-loop effective action becomes propotional to $g^2$, where $g$ 
is the coupling constant. This is true in 
both QCD \cite{niel,ditt} and massless QED 
\cite{cho01,cho5}. This enables us to calculate the imaginary part of
the effective action with a perturbative method. 
The perturbative calculation tells that the infra-red regularization 
by causality is indeed the correct one for QCD \cite{cho4}. 
This is not surprising because 
the perturbative calculation always respects the causality,
and thus excludes the contribution of tachyonic eigenstates 
in the calculation of the effective action.

The perturbative confirmation of the infra-red regularization by causality 
must be enough to resolve the controversy on the the stability of the
QCD vacuum. Nevertheless, since the monopole condensation is
such an important issue for the confinement in QCD, 
one can not easily dismiss the
instability of the SNO vacuum. One still need 
a deeper understanding of the origin of the instability
and a more convincing argument which can tell why the infra-red 
regularization by causality should be the correct  
regularization for QCD.
{\it The purpose of this paper is to show that it is 
the incorrect calculation of the functional determinant, 
not the $\zeta$-function regularization, which causes 
the instability of the SNO vacuum. More importantly,
we show that a proper implementation of 
the gauge invariance can naturally restore
the stability for the SNO vacuum, because the gauge invariance 
excludes the tachyons from the physical eigenstates}.
This means that tachyons should not have been 
there in the first place. They were there to create 
a mirage, not physical states. In the absence of
the tachyons, of course, there is no infra-red divergence
and no need to regularize the divergence.
So one can not blame the $\zeta$-function regularization
for the instability of the SNO vacuum.
It was the incorrect calculation of the functional determinant
which was responsible.

From the physical point of view, the difference 
between the two results originates from 
the different classical backgrounds they start from. 
In the old approach Savvidy starts from an {\it ad hoc} magnetic 
background which is not gauge invariant \cite{niel}.
Because of this the functional determinant of the gluon loop
contains tachyonic eigenstates, which in turn 
develops an imaginary part in the effective action
which destabilizes the SNO vacuum. To cure this defect
Nielsen and Olesen have proposed the so-called
``Copenhagen vacuum'', which is made of gauge invariant 
combination of blockwise randomly oriented
color magnetic fields in space.
But this Copenhagen vacuum, although conceptually appealing,
does not allow a mathematical description and 
clearly is not so useful in practical purposes.
On the other hand in the new approach the authors start from a gauge invariant 
non-Abelian monopole background from the beginning \cite{cho3}.
This precludes the tachyonic eigenstates 
to enter in the calculation of the effective action.
Nevertheless, so far how this gauge invariance affects 
the calculation of the functional determinant and can actually
bring the stability to the QCD vacuum has not been explained 
in physical terms.  
In particular, how (if at all) one can cure
the defect of SNO vacuum and make it
gauge invariant has not been made clear. 
{\it In this paper we show that a natural way 
to make the SNO vacuum gauge invariant is to impose the color reflection 
invariance to the vacuum. Indeed we 
show how the color reflection invariance
removes the contribution of tachyonic modes from the effective action,
and makes the magnetic condensation stable in QCD}.

The paper is organized as follows. In Section II
we review the background field method to calculate 
the quantum effective action of QCD. In Section III
we compare the old SNO background and the new monopole background,
and review the calculation of the functional determinant of 
the gluon loop. In Section IV we compare 
the $\zeta$-function regularization and 
the recently proposed infra-red regularization by causality 
in QCD effective action, to clarify 
the difference between the two approaches. In Section V
we review the color reflection invarince in QCD, and show how 
the reflection invarince affects the calculation of 
the effective active action and cures the instability of 
the SNO vacuum. In Section VI we discuss
the physical meaning of our analysis, in particular 
the color reflection invarince, in connection with 
the confinement in QCD.


\section{Background Field Method}

In this section we review the background field method \cite{dewitt,pesk} 
to obtain the one-loop effective action of QCD, and
derive the integral expression of the QCD effective action.
For simplicity we will concentrate on $SU(2)$ QCD in this paper.
To obtain the one-loop effective action one must first divide
the gauge potential $\vec A_\mu$ into two parts, the slow-varying
classical background $\vec B_\mu$ and the fluctuating quantum
part $\vec Q_\mu$, 
\bea
\vec A_\mu = \vec B_\mu + \vec Q_\mu,
\label{d}
\eea
and integrate out the quantum part
with a functional integration. To do this remember that 
(\ref{d}) allows two types of gauge transformations, the background
gauge transformation described by
\bea
&\delta \vec B_\mu = \dfrac{1}{g} \bar D_\mu \vec \alpha,
~~~~~\delta \vec Q_\mu = - \vec \alpha \times \vec Q_\mu, \nn\\
&\bar D_\mu = \partial_\mu + g \vec B_\mu \times,
\label{bgt}
\eea
and the physical gauge transformation described by
\bea 
\delta \vec B_\mu = 0,
~~~~~\delta \vec Q_\mu = \dfrac{1}{g} D_\mu \vec \alpha,
\label{pgt}
\eea
where $\vec \alpha$ is an infinitesimal gauge parameter.
Notice that both (\ref{bgt}) and (\ref{pgt}) satisfy
\bea
\delta \vec A_\mu = \dfrac{1}{g} D_\mu \vec \alpha.
\label{gt}
\eea
To integrate out the quantum field one may impose 
the following gauge condition to the quantum fields,
\bea
&\vec F = \bar D_\mu \vec Q_\mu = 0, \nn\\
&{\cal L}_{gf}=- \dfrac{1}{2\xi} (\bar D_\mu \vec Q_\mu)^2.
\label{gc}
\eea
The corresponding Faddeev-Popov determinant is given by
\bea
M^{ab}_{FP} = \dfrac {\delta F^a}{\delta \alpha^b} = (\bar D_\mu D_\mu)^{ab}.
\label{fpd}
\eea
With this gauge fixing
the effective action takes the following form,
\begin{widetext}
\bea
&\exp ~\Big[iS_{eff}(\vec B_\mu) \Big] = \dfrac{}{} \int {\cal D}
\vec Q_\mu {\cal D} \vec{c} ~{\cal D}\vec{c}^{~*}
\exp \Big\{~i \dfrac{}{} \int \Big[-\dfrac {1}{4} \vec G_{\mu\nu}^2
-\dfrac{1}{4} (\bar D_\mu \vec Q_\nu 
- \bar D_\nu \vec Q_\mu)^2 \nn\\
&-\dfrac{g}{2} \vec G_{\mu\nu} \cdot (\vec Q_\mu \times \vec Q_\nu)
-\dfrac{g^2}{4}(\vec Q_\mu \times \vec Q_\nu)^2
+\vec{c}^{~*}\bar {D}_\mu D_\mu\vec{c}
-\frac{1}{2\xi} (\bar D_\mu \vec Q_\mu)^2 \Big] d^4x \Big\},
\label{ea}
\eea
\end{widetext}
where $\vec c$ and ${\vec c}^{~*}$ are the ghost fields.
Notice that the effective action
(\ref{ea}) is explicitly invariant under the background
gauge transformation (\ref{bgt}) which involves
only $\vec B_\mu$, if we add the following
gauge transformation of the ghost fields to (\ref{bgt}),
\bea
\delta \vec c = - \alpha \times \vec c,
~~~~~\delta \vec c^{~*} = - \alpha \times \vec c^{~*}.
\eea
This guarantees the gauge invariance of the resulting 
effective action.
 
The gluon loop and the ghost loop integrals
give the following functional deteminants (at one-loop level)
\bea
&{\rm Det}^{-\frac{1}{2}} K_{\mu \nu}^{ab} =
{\rm Det}^{-\frac{1}{2}} \Big(-g_{\mu \nu}
\bD^2_{ab}- 2g \epsilon_{abc} G_{\mu \nu}^c \Big),\nn \\
& {\rm Det} M_{ab} = {\rm Det} \Big(-\bD^2_{ab} \Big),
\label{fd}
\eea
from which one has
\bea
\Delta S = \dfrac{i}{2} \ln {\rm Det} K - i \ln {\rm Det} M.
\label{ea0}
\eea
So the correct calculation of the determinants becomes crucial
for us to obtain the effective action.

\section{SNO Background versus Monopole Background}

Savvidy, and Nielsen and Olesen have chosen a covariantly constant 
color magnetic field as the classical background 
in their calculation of the effective action \cite{savv,niel,ditt}  
\bea
&\vec B_\mu = \bar B_\mu \n_0 = \dfrac{1}{2} \bar H_{\mu\nu} x_\nu \n_0,
~~~~~\vec G_{\mu\nu} = \bar H_{\mu\nu} \n_0, \nn\\
&\bar D_\mu \vec G_{\mu\nu} = 0, 
\label{sb}
\eea
where $\bar H_{\mu\nu}$ is a constant magnetic field and $\n_0$ is 
a constant unit isovector ($\n_0^2=1$).
With the background, one can calculate the functional
determinant (\ref{fd}). 
The calculation of the determinant amounts to the calculation of
the eigenvalues of the determinant. Nielsen and Olesen have
correctly pointed out that 
this reduces to the calculation of the energy eigenvalues of
a massless charged vector field $X_\mu$ in a constant external magnetic field
$\bar H_{\mu\nu}$ \cite{niel},
\bea
&E^2 X_\mu = [-\bar D^2 g_{\mu\nu} + \bar D_\mu \bar D_\nu
+ 2ig \bar H_{\mu\nu}] X_\nu =0, \nn\\
& \bar D_\mu = \partial_\mu + ig \bar B_\mu.
\label{eveh}
\eea
This has the well-known eigenvalues
\bea
&E^2 = 2gH (n + \dfrac{1}{2}) + k^2 \mp 2gH, \nn\\
& H = \sqrt{\bar H_{\mu\nu}^2 /2},
\label{ev}
\eea
where $k$ is the momentum of the eigen-function in
the direction of the background magnetic field.
Notice that the $\mp$ signature correspond to the spin $S_3=\mp 1$
of the charged vector field (in the direction of the magnetic field).
So, when $n=0$, 
the eigen-function with $S_3=-1$ has an imaginary energy when
$k^2<gH$, and thus becomes a tachyon which violates
the causality. This point will become very important
when we calculate the functional determinant later. 

From this they obtained
\bea
&\Delta S = i\ln {\rm Det} \Big[(-\bD^2+2gH)(-\bD^2-2gH) \Big],
\label{fdhx}
\eea
and the integral expression of the effective action
\bea
&\Delta{\cal L} = \dfrac{}{} \lim_{\epsilon \rightarrow 0}
\dfrac{1}{16 \pi^2}\int_{0}^{\infty}
\dfrac{dt}{t^{2-\epsilon}} \dfrac{gH \mu^2}{\sinh (gHt/\mu^2)} \nn\\
&\times \Big[\exp (-2gHt/\mu^2 )
+ \exp (+2gHt/\mu^2) \Big],
\label{eahx}
\eea
where $\mu^2$ is a dimensional parameter. Notice that the 
effective action has a severe infra-red divergence, 
which originates from the unstable tachyonic eigenvalues
of the functional determinant (\ref{fdhx}).

On the ther hand in the recent calculation the authors
have chosen the monopole background given by \cite{cho1,cho2}
\bea
&\vec B_\mu = \vec C_\mu,
~~~~~\vec G_{\mu\nu} = \vec H_{\mu\nu}, \nn\\
&\vec C_\mu= -\dfrac{1}{g}\hat n \times \partial_\mu\hat n, \nn\\
&\vec H_{\mu\nu}=\partial_\mu \vec C_\nu-\partial_\nu \vec C_\mu+ g
\vec C_\mu \times \vec C_\nu \nn\\
&= -\dfrac{1}{g}
\partial_\mu\hat{n}\times\partial_\nu\hat{n},
\label{ccon}
\eea
where $\n$ is a unit isovector ($\n^2 =1$) which selects 
the color charge direction everywhere in space-time.
The advantage of (\ref{ccon}) over (\ref{sb}) is that in (\ref{sb})
one can not tell the origin of the magnetic background,
whereas in (\ref{ccon}) one can tell for sure that 
it comes exclusively from the non-Abelian monopole.
Furthermore, the Savvidy background is {\it ad hoc},
but here $\vec C_\mu$ provides a gauge independent separation
of the monopole field from the generic non-Abelian gauge
field.

To see this consider the gauge-independent decomposition 
of the gauge potential into the binding gluon $\hat A_\mu$
and the valence gluon $\X_\mu$ \cite{cho1,cho2}, which has recently
been referred to as the Cho decomposition 
(or Cho-Faddeev-Niemi-Shabanov decomposition) \cite{fadd,shab},
\bea
& \vec{A}_\mu =A_\mu \n - \oneg \n\times\pro_\mu\n+\X_\mu\nonumber
         = \hat A_\mu + \X_\mu, \nn\\
& (A_\mu = \n\cdot \vec A_\mu,~~~ \hat{n}\cdot\vec{X}_\mu=0),
\label{cdec}
\eea
where $A_\mu$ is the ``electric'' potential.
Notice that $\hat A_\mu$ is precisely
the connection which leaves $\n$ invariant under parallel transport,
\bea
\D_\mu \n = \pro_\mu \n + g {\hat A}_\mu \times \n = 0.
\eea
Under the infinitesimal gauge transformation
\bea
\delta \n = - \vec \alpha \times \n  \,,\,\,\,\,
\delta \A_\mu = \oneg  D_\mu \vec \alpha,
\eea
one has
\bea
&&\delta A_\mu = \oneg \n \cdot \pro_\mu \valpha,\,\,\,\
\delta \hat A_\mu = \oneg \D_\mu \valpha  ,  \nn \\
&&\hspace{1.2cm}\delta \X_\mu = - \valpha \times \X_\mu  .
\eea
This tells that $\hat A_\mu$ by itself describes an $SU(2)$
connection which enjoys the full $SU(2)$ gauge degrees of
freedom. Furthermore $\vec X_\mu$ forms a
gauge covariant vector field under the gauge transformation.
{\it But what is really remarkable is that the decomposition is
gauge independent. Once the gauge covariant topological field
$\hat n$ is chosen, the decomposition follows automatically,
regardless of the choice of gauge} \cite{cho1,cho2}.

Remember that $\hat{A}_\mu$ retains all the essential
topological characteristics of the original non-Abelian potential.
First, $\hat{n}$ defines $\pi_2(S^2)$
which describes the non-Abelian monopoles.
Indeed, it is well-known that $\vec C_\mu$ with $\hn=\hat r$
describes precisely the Wu-Yang monopole \cite{wu,cho80}. 
Secondly, it characterizes
the Hopf invariant $\pi_3(S^2)\simeq\pi_3(S^3)$ which describes
the topologically distinct vacua. So the topologically distinct vacua 
can be described exclusively by $\hat{n}$ \cite{bpst,cho79}.
Furthermore $\hat{A}_\mu$ has a dual
structure,
\begin{eqnarray}
& \hat{F}_{\mu\nu} = (F_{\mu\nu}+ H_{\mu\nu})\hat{n}\mbox{,}
\nonumber \\
& F_{\mu\nu} = \partial_\mu A_{\nu}-\partial_{\nu}A_\mu \mbox{,}
\nonumber \\
& H_{\mu\nu} = -\dfrac{1}{g} \hat{n}\cdot(\partial_\mu
\hat{n}\times\partial_\nu\hat{n})
= \partial_\mu \tilde C_\nu-\partial_\nu \tilde C_\mu,
\end{eqnarray}
where $\tilde C_\mu$ is the ``magnetic'' potential of the monopoles
(Notice that one can always introduce the magnetic
potential since $H_{\mu \nu}$ forms a closed two-form
locally sectionwise) \cite{cho1,cho2}.
Now, evidently the monopole background (\ref{ccon}) is written as
\bea
\vec H_{\mu\nu} = H_{\mu\nu}\hat n.
\eea
This demonstrates that indeed (\ref{ccon}) does describe 
the gauge independent separation of the monopole
field $\vec H_{\mu\nu}$ from the generic non-Abelian 
gauge field $\vec F_{\mu\nu}$. The monopole potential $\vec C_\mu$
has been referred to as the Cho connection 
by Faddeev and Niemi \cite{fadd,shab}.

An important feature of the decomposition (\ref{cdec})
is that it must be invariant under the color reflection \cite{cho1,cho2}
\bea
\n \rightarrow -\n,
\label{cri}
\eea
because $\hn$ is gauge equivalent to $-\hn$.
In fact $\hat A_\mu$ is explicitly invariant under the color reflection.
To understand what this means to the valence gluon $\vec X_\mu$, let
$(\hn_1, \hn_2, \hn)$ be a right-handed orthonormal basis in 
$SU(2)$ space, and let
\bea
\vec X_\mu = X_\mu^1 \hn_1 +  X_\mu^2 \hn_2.
\eea
In the Abelian formalism of QCD \cite{cho3,cho4}, the valence gluon
can be expressed as a charged vector field
\bea
X_\mu = \dfrac{X_\mu^1+iX_\mu^2}{\sqrt{2}}.
\eea
Then, under the color reflection, $X_\mu$ should transform
to the charge conjugate state
\bea
X_\mu^* = \dfrac{X_\mu^1-iX_\mu^2}{\sqrt{2}}.
\eea
This amounts to changing $\vec X_\mu$ to its charge conjugate state
$\vec X_\mu^{(c)}$, 
\bea
\vec X_\mu^{(c)}= X_\mu^1 \hn_1 - X_\mu^2 \hn_2,
\eea
which is equivalent to changing $\hn_2$ to $-\hn_2$.
Indeed this is exactly what we need to induce the color 
reflection (\ref{cri}), because $(\hn_1, -\hn_2, -\hn)$ now 
forms a right-handed orthonormal basis. This means that 
the color reflection transforms  $\vec X_\mu$
to its charge conjugate state $\vec X_\mu^{(c)}$. 
More importantly, $\vec X_\mu$ and $\vec X_\mu^{(c)}$ 
must be indistinguishable,
because they are gauge equivalent to each other.
This point will become very important in the following.

With the monopole background (\ref{ccon}) one can calculate 
the functional determinant (\ref{fd}). But for the generality
we will calculate the the functional determinant with an arbitrary
background $\hat A_\mu$. The calculation of the ghost loop
determinant (the Faddeev-Popov determinant) $M_{ab}$ is rather 
straightforward, but that of the gluon loop $K_{\mu\nu}^{ab}$
is tricky. In this case we have
\begin{widetext}
\bea
&\ln {\rm Det} K = {\rm Tr} \ln \Big(-g_{\mu\nu} \hD^2_{ab}\Big)
+ {\rm Tr} \ln \Big[g_{\mu\nu} \delta_{ab} 
+ 2g G_{\mu\nu} \Big(\dfrac{N}{\hD^2}\Big)_{ab}\Big] \nn\\
&= 4 ~{\rm Tr} \ln \Big(-\hD^2_{ab}\Big)
+ {\rm Tr} \dfrac{}{}\sum_{n=1}^{\infty} 
\dfrac{(-1)^{n+1}}{n} \Big(2g\dfrac{N}{\hD^2}G_{\mu\nu}\Big)^n,
\label{gdet1}
\eea
where 
\bea
&G_{\mu\nu} = \partial_\mu B_\nu-\partial_\nu B_\mu 
~~~(B_\mu =A_\mu + \tilde C_\mu),
~~~~~N_{ab} = \epsilon_{abc} n_c.
\eea
Using the relation
\bea
&G_{\mu \alpha} G_{\nu\beta} G_{\alpha \beta} = \dfrac{1}{2} G^2 G_{\mu \nu}
+\dfrac{1}{2}(G \tilde G) {\tilde G}_{\mu \nu}
~~~~~({\tilde G}_{\mu \nu}=\dfrac{1}{2}{\epsilon}_{\mu\nu\rho\sigma}
G_{\rho\sigma}),
\eea
one can simplify the functional determinant to
\bea
& \ln {\rm Det} K = 4 ~{\rm Tr} \ln \Big(-\hD^2_{ab}\Big) 
+ {\rm Tr} \ln \Big[\delta_{ab}
+4a^2\Big(\dfrac{N}{\hD^2}\Big)^2_{ab}\Big] 
+{\rm Tr} \ln \Big[\delta_{ab}
-4b^2\Big(\dfrac{N}{\hD^2}\Big)^2_{ab}\Big] \nn\\
&=\ln {\rm Det} \Big[(-\hD^2+2iaN)(-\hD^2-2iaN) \Big]_{ab}
+\ln {\rm Det} \Big[(-\hD^2+2bN)(-\hD^2-2bN)\Big]_{ab},
\label{gdet2}
\eea
where
\bea
a = \dfrac{g}{2} \sqrt {\sqrt {G^4 + (G \tilde G)^2} + G^2},
~~~~~b = \dfrac{g}{2} \sqrt {\sqrt {G^4 + (G \tilde G)^2} - G^2}.
\eea
Notice that
\bea
(-\hD^2 \pm 2iaN) \hn = 0, ~~~~~(-\hD^2 \pm 2bN) \hn = 0.
\eea
From this we can assume, without loss of generality,
that the eigenfunction of the determinants
is of the form 
\bea
\vec \phi = \phi_1 \hat n_1 + \phi_2 \hat n_2.
\eea
Furthermore, we have
\bea
&\hD_\mu \hat n_1 = gB_\mu \hat n_2,
~~~~~\hD_\mu \hat n_2 = -gB_\mu \hat n_1.
\eea
With this we can simplify the eigenvalue equation
$(-\hD^2 \pm 2iaN) \vec \phi = \lambda \vec \phi$ 
to 
\bea
\left( \begin{array}{cc}
\pro_\mu^2-g^2B_\mu^2 & -g(\pro_\mu B_\mu + 2 B_\mu \pro_\mu) \pm 2ia \\
g(\pro_\mu B_\mu + 2 B_\mu \pro_\mu) \mp 2ia & \pro_\mu^2-g^2B_\mu^2
\end{array} \right)
\left( \begin{array}{cc}
\phi_1 \\ \phi_2
\end{array} \right) 
= \lambda \left(\begin{array}{cc}
\phi_1 \\ \phi_2
\end{array} \right),
\eea
which can be diagonalized to the following Abelian form
\bea
\left( \begin{array}{cc}
-\tilde D_{+}^2 \pm 2a & 0\\ 
0 & -\tilde D_{-}^2 \mp 2a
\end{array} \right)
\left( \begin{array}{cc}
\phi_+ \\ \phi_-
\end{array} \right)
= \lambda \left(\begin{array}{cc}
\phi_+ \\ \phi_-
\end{array} \right),
~~~~~\phi_{\pm} = \dfrac{\phi_1 \pm i\phi_2}{\sqrt{2}},
\label{evea}
\eea
where $\tilde D_{\pm}^2= (\pro_\mu \pm igB_\mu)^2$.
Similarly, we can diagonalize the equation
$(-\hD^2 \pm 2bN) \vec \phi = \lambda \vec \phi$
to
\bea
\left( \begin{array}{cc}
-\tilde D_{+}^2 \mp 2ib & 0 \\
0 &-\tilde D_{-}^2 \pm 2ib
\end{array} \right)
\left( \begin{array}{cc}
\phi_+ \\ \phi_-
\end{array} \right)
= \lambda \left(\begin{array}{cc}
\phi_+ \\ \phi_-
\end{array} \right).
\label{eveb}
\eea
From this we have
\bea
&{\rm Det} (-\hD^2+2iaN)_{ab} 
= {\rm Det} (-\tilde D_{+}^2+2a)(-\tilde D_{-}^2-2a),
~~~{\rm Det} (-\hD^2-2iaN)_{ab} 
= {\rm Det} (-\tilde D_{+}^2-2a)(-\tilde D_{-}^2+2a), \nn\\
&{\rm Det} (-\hD^2+2bN)_{ab} 
= {\rm Det} (-\tilde D_{+}^2-2ib)(-\tilde D_{-}^2+2ib),
~~~{\rm Det} (-\hD^2-2bN)_{ab} 
= {\rm Det} (-\tilde D_{+}^2+2ib)(-\tilde D_{-}^2-2ib).
\label{agdet}
\eea
\end{widetext}
At this point it is important to realize that 
$(-\tilde D_{-}^2 \mp 2a)$ and
$(-\tilde D_{-}^2 \pm 2ib)$ are what one obtains from
$(-\tilde D_{+}^2 \pm 2a)$ and
$(-\tilde D_{+}^2 \mp 2ib)$ by replacing $g$ to $-g$, 
so that they are charge conjugate to the others.
Furthermore, the two eigenfunctions
$\phi_{+}$ and $\phi_{-}$ are also
charge conjugate to each other (although they are not 
complex conjugate to each other). To see this notice that 
they are related by changing $\phi_2$ to $-\phi_2$.
This, viewed in terms of $\vec \phi$,
amounts to changing $\hn_2$ to $-\hn_2$. But, 
as we have noted before, this change is equivalent
to the change of the color direction $\hn$ to $-\hn$.
This is nothing but the color reflection (\ref{cri}).
This means that $(-\tilde D_{+}^2 \pm 2a)\phi_+$ and 
$(-\tilde D_{-}^2 \mp 2a)\phi_-$, and $(-\tilde D_{+}^2 \mp 2ib)\phi_+$
and $(-\tilde D_{-}^2 \pm 2ib)\phi_-$, are
not only charge conjugate to each other, but also gauge
equivalent to each other.
This means that they must have 
identical eigenvalues, so that
\bea
&{\rm Det} (-\tilde D_{+}^2\pm2a) = {\rm Det} (-\tilde D_{-}^2\mp2a), \nn\\
&{\rm Det} (-\tilde D_{+}^2\mp2ib) = {\rm Det} (-\tilde D_{-}^2\pm2ib).
\label{ccagdet}
\eea
From this we have
\bea
&\ln {\rm Det} (-\hD^2\pm2iaN)_{ab} 
= 2 \ln {\rm Det} (-\tilde D_{+}^2\pm2a), \nn\\
&\ln {\rm Det} (-\hD^2\pm2bN)_{ab} = 2 \ln {\rm Det} (-\tilde D_{+}^2\mp2ib).
\label{cci}
\eea
This tells that we can reduce 
the eigenvalue problem of a non-Abelian
functional determinants $K_{\mu\nu}^{ab}$ and $M_{ab}$
to the eigenvalue problem of the following Abelian determinants,
\bea
&\ln {\rm Det} K = 2 \ln {\rm Det} (-\tilde D^2+2a)(-\tilde D^2-2a) \nn\\
&+2 \ln {\rm Det} (-\tilde D^2-2ib)(-\tilde D^2+2ib), \nn\\
&\ln {\rm Det} M = 2 \ln {\rm Det} (-\tilde D^2),
\label{agdetx}
\eea
where
\bea
\tilde D_\mu = \pro_\mu +igB_\mu. \nn
\eea
Notice that in the Lorentz frame where the
electric field becomes parallel to the magnetic field, $a$ becomes
purely magnetic and $b$ becomes purely electric.

From this we obtain \cite{niel,ditt,cho1,cho2}
\begin{widetext}
\bea
\Delta S = i \ln {\rm Det} (-\tilde D^2+2a)(-\tilde D^2-2a)
+ i \ln {\rm Det} (-\tilde D^2-2ib)(-\tilde D^2+2ib)
- 2i \ln {\rm Det}(-\tilde D^2),
\label{fdabx}
\eea
and
\bea
&\Delta {\cal L} =  \dfrac{}{} \lim_{\epsilon\rightarrow0}
\dfrac{1}{16 \pi^2}  \int_{0}^{\infty}
\dfrac{dt}{t^{3-\epsilon}} \dfrac{ab t^2}{\sinh (at/\mu^2)
\sin (bt/\mu^2)} \Big[ \exp(-2at/\mu^2)+\exp(+2at/\mu^2) \nn\\
&+\exp(+2ibt/\mu^2)+\exp(-2ibt/\mu^2)-2 \Big].
\label{eaabx}
\eea
Notice that for the monopole background (\ref{ccon}) we have $b=0$,
so that the integral (\ref{eaabx}) is reduced to
\bea
&\Delta{\cal L} = \dfrac{}{} \lim_{\epsilon\rightarrow0}
\dfrac{1}{16 \pi^2}\int_{0}^{\infty}
\dfrac{dt}{t^{2-\epsilon}} \dfrac{a \mu^2}{\sinh (at/\mu^2)}
\Big[\exp (-2at/\mu^2 ) + \exp (+2at/\mu^2) \Big].
\label{eaax}
\eea
\end{widetext}
With $a=gH$ this reduces to (\ref{eahx}).

The integral (\ref{eaabx}) has a severe infra-red 
divergence. This is due to the fact that ${\rm Det}(-\tilde D^2-2a)$
and ${\rm Det}(-\tilde D^2+2ib)$ have unstable eigenstates.
To see this notice that, to calculate the eigenvalues,
one often has to choose a particular gauge and a particular 
Lorentz frame. Indeed a best way to calculate 
the determinants is to go to the gauge and Lorentz frame 
where the color electromagnetic field assumes
a particular direction. In this case one can easily show that 
${\rm Det}(-\tilde D^2-2a)$ has negative eigenvalues
and thus tachyonic eigenstates when $k<2a$, where $k$ is the momentum 
of the eigenstate in the direction of the color magnetic field. 
Similarly, ${\rm Det}(-\tilde D^2+2ib)$ has
acausal eigenstates which propagate backward in time when $k<2b$.
These acausal states create the instability which causes 
the infra-red divergence in (\ref{eaabx}).

Notice, however, that
there is a subtle but very important point that we have
overlooked in the calculation of the determinants (\ref{fdhx}) 
and (\ref{fdabx}). The classical backgrounds (\ref{sb}) and (\ref{ccon})
must be gauge invariant, so that in the calculation of
the functional determinant (\ref{fd}) we should make 
sure that the gauge invariance, in particular
the color reflection invariance, is properly implemented. 
Unfortunately we did not. When we do this correctly, the acausal tachyonic
states disappear from the physical eigenstates of the determinants,
and the effective action no longer has the infra-red divergence.
In Section V we will explain how this happens.

Fortunately, we still have a chance to correct this mistake
when we make the infra-red regularization. Now we show how to do this.

\section{Infra-red Regularization: A Review}

The evaluation of the integral (\ref{eaabx}) for arbitrary 
$a$ and $b$ has been notoriously difficult \cite{savv,niel,ditt}.
Even in the case of ``simpler'' QED, the integration of the
effective action has been completed only recently \cite{cho01,cho5}.
Fortunately the integral for the pure electric ($a=0$) and
pure magnetic ($b=0$) background has now 
been correctly performed \cite{cho3,cho4}.

Consider the pure magnetic background (\ref{eaax}) first.
Clearly the integral (\ref{eaax}) 
contains a severe infra-red divergence,
and to perform the integral we have to regularize it
first. There have been two competing infra-red regularizations,
the $\zeta$-function regularization and the regularization by
causality. Let us discuss the $\zeta$-function regularization 
first. From the definition of the generalized $\zeta$-function \cite{table}
\bea
&\zeta (s,\lambda) = \dfrac{}{}\sum_{n=0}^{\infty} 
\dfrac{1}{(n+\lambda)^s} \nn\\
&= \dfrac{1}{\Gamma(s)} \int_0^{\infty} \dfrac{x^{s-1} \exp(-\lambda x)}
{1-\exp(-x)} dx, 
\label{zeta}
\eea
we have
\begin{widetext}
\bea
&\Delta {\cal L} = \dfrac{}{} \lim_{\epsilon \rightarrow 0}
\dfrac{1}{16 \pi^2} \int_{0}^{\infty} \dfrac{dt}{t^{2-\epsilon}} 
\dfrac{a \mu^2}{\sinh (at/\mu^2)}
\Big[\exp (-2at/\mu^2) + \exp (+2at/\mu^2) \Big] \nn\\
&= \dfrac{}{} \lim_{\epsilon \rightarrow 0}
\dfrac{a \mu^2}{8 \pi^2} \int_{0}^{\infty} \dfrac{dt}{t^{2-\epsilon}}
\dfrac{\exp (-3at/\mu^2) + \exp (+at/\mu^2)}{1-\exp(-2at/\mu^2)} \nn\\
&= \dfrac{}{} \lim_{\epsilon \rightarrow 0}
\dfrac{a^2}{4 \pi^2} (\dfrac{2a}{\mu^2})^{-\epsilon} \Gamma(\epsilon-1)
\Big[\zeta(\epsilon-1,\dfrac{3}{2}) + \zeta(\epsilon-1,-\dfrac{1}{2})\Big]\nn\\
&= \dfrac{}{} \lim_{\epsilon \rightarrow 0}
\dfrac{a^2}{4 \pi^2} (1-\epsilon \ln \dfrac{2a}{\mu^2}) 
\big(\dfrac{1}{\epsilon} -\gamma +1 \big) 
\Big[\big(\zeta(-1,\dfrac{3}{2}) + \zeta(-1,-\dfrac{1}{2})\big)
+ \epsilon \big(\zeta'(-1,\dfrac{3}{2}) 
+ \zeta'(-1,-\dfrac{1}{2})\big)\Big] \nn\\
&= \dfrac{11 a^2}{48 \pi^2} \big(\dfrac{1}{\epsilon} - \gamma 
+1 - \ln \dfrac{2a}{\mu^2} \big) 
- \dfrac{a^2}{4 \pi^2} \big(2 \zeta'(-1,\dfrac{3}{2})
- i \dfrac{\pi}{2}\big),
\label{zetareg}
\eea
where $\zeta' = \dfrac{d\zeta}{ds} (s,\lambda)$, 
and we have used  the fact \cite{table}
\bea 
&\zeta(-1,\dfrac{3}{2}) = \zeta(-1,-\dfrac{1}{2}) = -\dfrac{11}{24},
~~~~~\zeta'(-1,-\dfrac{1}{2}) = \zeta'(-1,\dfrac{3}{2}) - i \dfrac{\pi}{2}. \nn
\eea
So, with the ultra-violet regularization by modified 
minimal subtraction we obtain the SNO effective action
\bea
&{\cal L}_{eff}=-\dfrac{a^2}{2g^2} -\dfrac{11a^2}{48\pi^2}(\ln
\dfrac{a}{\mu^2}-c) + i \dfrac {a^2} {8\pi},
~~~~~c=1-\ln 2 -\dfrac {24}{11} \zeta'(-1, \frac{3}{2})=0.94556... ,
\label{snoea}
\eea
which contains the well-known imaginary part. Observe that
the imaginary part comes from the infra-red divergence,
which originates from the tachyonic eigenstates.

On the other hand, one can evaluate the integral (\ref{eaax}) 
with infra-red regularization by causality. 
For this we go to the Minkowski time with
the Wick rotation, and find \cite{cho3,cho4}
\bea 
\label{eaam}
&\Delta {\cal L}=  \Delta{\cal L_+} + \Delta{\cal L_-}, \nn\\
& \Delta{\cal L_+} =  - \dfrac{}{} \lim_{\epsilon \rightarrow 0}
\dfrac{1}{16 \pi^2}\int_{0}^{\infty}
\dfrac{dt}{t^{2-\epsilon}} \dfrac{a \mu^2}{\sin (at/\mu^2)}
\exp (-2i a t/\mu^2 ), \nn\\
& \Delta{\cal L_-} =  - \dfrac{}{} \lim_{\epsilon \rightarrow 0}
\dfrac{1}{16 \pi^2}\int_{0}^{\infty}
\dfrac{dt}{t^{2-\epsilon}} \dfrac{a\mu^2}{\sin (at/\mu^2)}
\exp (+2i a t/\mu^2 ).
\eea
\end{widetext}
In this form the infra-red divergence has disappeared,
but now we face an ambiguity in choosing the correct contours
of the integrals in (\ref{eaam}). Fortunately this ambiguity can
be resolved by causality. To see this notice that the two integrals
$\Delta{\cal L_+}$ and $\Delta{\cal L_-}$ originate from the
two determinants $(-\tilde D^2+2a)$ and $(-\tilde D^2-2a)$,
and the standard causality argument requires us to
identify $2 a$ in the first determinant as
$2 a -i\epsilon$ but in the second determinant as
$2 a +i\epsilon$. {\it This tells that
the poles in the first integral in (\ref{eaam}) should lie above
the real axis, but the poles in the second integral should lie
below the real axis. From this we conclude
that the contour in $\Delta{\cal L_+}$ should pass below the
real axis, but the contour in $\Delta{\cal L_-}$ should pass above the
real axis}. With this causality requirement the two integrals
become complex conjugate to each other. This guarantees that
$\Delta{\cal L}$ is explicitly real, without any imaginary part.
This suggests that in the Euclidian time $\Delta{\cal L_+}$ 
and $\Delta{\cal L_-}$ must have exactly the same expression,
\bea
&\Delta{\cal L_+} = \Delta{\cal L_-} = \dfrac{}{} \lim_{\epsilon \rightarrow 0}
\dfrac{1}{16 \pi^2}\int_{0}^{\infty}
\dfrac{dt}{t^{2-\epsilon}} \dfrac{a\mu^2}{\sinh (at/\mu^2)} \nn\\
&\times \exp (-2a t/\mu^2 ).
\eea
We will see that this is exactly what one obtains
when one calculates the functional determinant (\ref{gdet1})
correctly.

With this infra-red regularization by causality we obtain
\bea
&{\cal L}_{eff} = - \dfrac{a^2}{2g^2} -\dfrac{11a^2}{48\pi^2}(\ln
\dfrac{a}{\mu^2}-c),
\label{ceaa}
\eea
for a pure monopole background. 
This is identical to the SNO effective action (\ref{snoea}),
except that it no longer contains the imaginary part.

For the pure electric background (i.e., with $a=0$)
(\ref{eaabx}) is reduced to
\bea
&\Delta {\cal L}  =  \dfrac{}{} \lim_{\epsilon \rightarrow 0}
\dfrac{1}{16 \pi^2}  \int_{0}^{\infty}
\dfrac{ d t}{t^{2-\epsilon}} \dfrac{b\mu^2}{\sin (bt/\mu^2)} \nn\\
&\times \Big[\exp(+2ibt/\mu^2) +\exp(-2ibt/\mu^2) \Big],
\label{eabx}
\eea
so that, with the infra-red regularization by causality,
we obtain \cite{cho3,cho4}
\bea \label{ceab}
{\cal L}_{eff} = \dfrac{b^2}{2g^2} +\dfrac{11b^2}{48\pi^2}
(\ln \dfrac{b}{\mu^2}-c)-i\dfrac{11b^2}{96\pi}.
\eea
Observe that (\ref{ceaa}) and (\ref{ceab}) are
related by the duality. In fact we can obtain one from the other
simply by replacing $a$ with $-ib$ and $b$ with $ia$.
This duality, which states that the effective action should be
invariant under the replacement
\bea
a \rightarrow - ib,~~~~~~~b \rightarrow ia,
\label{dt}
\eea
was recently established as a
fundamental symmetry of the effective action of gauge theory,
both Abelian and non-Abelian \cite{cho3,cho01}. We emphasize that
the duality provides a very useful tool to check the self-consistency
of the effective action. The fact that (\ref{ceaa}) 
and (\ref{ceab}) are related by the duality
assures that the infra-red regularization by causality
is self-consistent.

Obviously the difference between (\ref{snoea}) and (\ref{ceaa})
originates from the different infra-red regularizations. 
To find out which one is the correct one, one must first understand
the origin of the infra-red divergence. Nielsen and Olesen
have correctly argued that the infra-red divergence follows from 
the unstable tachyonic eigenstates of the functional determinant
(\ref{fd}), and the $\zeta$-function regularization 
show how the unstable eigenstates create the instability 
of the SNO vacuum. On the other hand the authors of the recent
calculation have pointed out that the unstable eigenstates 
are not physical, because they violate the causality. 
More importantly, they showed how to remove the unphysical
states with the infra-red regularization by causality \cite{cho3}.
With the infra-red regularization by causality 
one obtains the stable monopole condensation.  

The central question now is whether one should include the 
unstable tachyonic modes in the evaluation of 
the functional determinant (\ref{fd}) or not,
and how one can actually do so. The tachyonic modes 
have been the big puzzle, the Gordian knot, in QCD. It was there,
and nobody knew how to resolve this puzzle. Nielsen and Olesen 
treated them as a sacred feature of QCD, which made it more
mysterious. But the authors of the recent
calculation pointed out that they are unphysical
and thus should be discarded, because they violate 
the causality. Now, we show that they are 
indeed an unphysical mirage which should not
have been there in the first place, because they are not gauge invariant.

\section{Color Reflection Invariance: 
Recalculation of Functional Determinant}

The effective action (\ref{ea}) is nothing but the vacuum to vacuum
amplitude in the presence of the external field $\vec B_\mu$ \cite{pesk}, 
\bea
&\exp \Big[i S_{eff} (\vec B_\mu)\Big]
= <\Omega_+|~\Omega_-> \Big|_{\vec B_\mu} \nn\\
&= \dfrac{}{} \sum_{|n_i>}<\Omega_+|~n_i><n_i~|~\Omega_-> \Big|_{\vec B_\mu},
\label{vtv}
\eea
where $|\Omega>$ is the vacuum and $|n_i>$ is 
a complete set of orthonormal states
of QCD. In the integral expression (\ref{ea}) the summation 
in terms of the complete set of states is expressed 
by the functional integration (at one-loop level). 
Now, to calculate this vacuum to vacuum amplitude 
one must use the physical vacuum and physical spectrum.  
And clearly the physical spectrum should not include 
the unphysical tachyons, unless one wants to assert that
the physical spectrum of QCD must contain the tachyonic modes. 
This means that $|n_i>$ should not include 
the unphysical acausal states. 
This justifies the exclusion of the unphysical states
in the calculation of the effective action.
The question now is how one can do so.

There are two ways to exclude
the unphysical modes. One could either do this when
one calculates the functional determinant (\ref{fd}),
or when one makes the infra-red regularization.
We have already shown how to do this with 
the infra-red regularization by causality \cite{cho1,cho2}.

Now we are ready to show that a correct evaluation of 
the functional determinant (\ref{fd}) automatically excludes 
the tachyonic modes. In fact we will show that
the gauge invariance forbids them to qualify as the physical
eigenstates of the functional determinant (\ref{fd}).
In the absence of the tachyonic modes, of course, 
we have no infra-red divergence,
and thus no need of any infra-red regularization.
This tells that it is not the $\zeta$-function
regularization, but the wrong evaluation of the functional 
determinant (\ref{fd}), which causes us the trouble.

When one evaluates the functional determinant (\ref{fd}), one must start from
a gauge invariant background. Otherwise the effective action
will not remain gauge invariant. This means that the backgrounds
(\ref{sb}) and (\ref{ccon}) must be invariant
under the background gauge transformation (\ref{bgt}).
But obviously the background (\ref{sb}) is not
gauge invariant, so that one must impose the gauge invariance
by hand in the calculation of the functional determinant (\ref{fdhx}).
Of course, one could choose a particular gauge to calculate 
the determinant, but one should make sure that the result
is gauge invariant. 

To discuss the gauge invariance it is important to understand
the color reflection invariance in QCD \cite{cho1,cho2}.
Consider the decomposition (\ref{cdec}) again. 
As we have already noted, the decomposition (\ref{cdec}) 
can be defined only up to the color reflection (\ref{cri}).
This tells that the electric potential $A_\mu$ in (\ref{cdec}),
and thus the color charge in QCD, can uniquely be defined only up to 
this reflection degrees of freedom, even after one has chosen one's
color direction. Furthermore, this means that 
any physical state must be invariant
under the the color reflection group generated by (\ref{cri}).  
To understand this in physical terms, consider a $q \bar q$ state.
There are two color neutral states
\bea
&|C,C_3> = |0,0> = \dfrac {|r\bar r>+~|b\bar b>}{\sqrt 2}, \nn\\
&|C,C_3> = |1,0> = \dfrac {|r\bar b>-~|b\bar r>}{\sqrt 2}, 
\eea
where $r$ and $b$ represent the red and blue quark.
But obviously only the first one is color singlet, and thus 
qualifies as a physical state.
Now, notice that under the color reflection (\ref{cri})
the role of red and blue colors are interchanged, 
and only the color singlet state remains invariant
under the color reflection. This shows that the 
color reflection invariance is an important
symmetry, a prerequisite for a physical state, of QCD.
In $SU(2)$ (\ref{cri}) generates a 4-element color reflection group,
and in $SU(3)$ the color reflection group forms a 24-element
symmetry group \cite{cho1,cho2}.

Now, consider the monopole background (\ref{ccon}). 
Obviously it is invariant under the color reflection (\ref{cri}).
Indeed not only $\vec C_\mu$ but also the background 
$\hat A_\mu$ itself is invariant 
under the color reflection (because the electric 
potential $A_\mu$ is uniquely defined only up to the signature).
Clearly we must honor this gauge invariance in our calculation
of the functional determinant (\ref{gdet1}). Unfortunately we did
not. To see this consider  
${\rm Det} (\hD^2+2iaN)(\hD^2-2iaN)$ 
in (\ref{gdet2}) first. This is made of two determinants, 
${\rm Det} (\hD^2+2iaN)={\rm Det} (\tilde D^2+2a)^2$ 
and ${\rm Det} (\hD^2-2iaN)={\rm Det} (\tilde D^2-2a)^2$,
and we have to solve the eigenvalue problem for 
each determinant separately. At first sight it appears that 
only the second determinant contains the tachyonic eigenstates.
However, observe that $\hD^2$ is invariant under (\ref{cri}),
but $N_{ab}$ changes the signature. So the two determinants are 
the color reflection counterpart of each other. This means that, 
after the color reflection, it becomes the first determinant 
which contains the tachyonic eigenstates.
From this it must become clear that actually both determinants
contain the tachyonic eigenstates. {\it More importantly, 
in both determinants, the tachyonic eigenstates are precisely those
which do not remain invariant
under the color reflection. Furthermore, the color 
reflection invariance tells that the two determinants
must have identical eigenvalues. This means
that the physical eigenstates of  
${\rm Det} (\hD^2+2iaN)(\hD^2-2iaN)$ must be given by
${\rm Det} (\tilde D^2+2a)^2(\tilde D^2+2a)^2$, not by
${\rm Det} (\tilde D^2+2a)^2(\tilde D^2-2a)^2$, and should not
contain any tachyonic states}.

Exactly the same argument applies to  
${\rm Det} (\hD^2+2bN)(\hD^2-2bN)$ in (\ref{gdet2}),
made of two determinants ${\rm Det} (\hD^2+2bN)={\rm Det} (-\tilde D^2-2ib)^2$ 
and ${\rm Det} (\hD^2-2bN)={\rm Det} (-\tilde D^2+2ib)^2$.
In this case only the second determinant appears to have acausal eigenstates 
propagating backward in time when $k^2<b$, 
where $k$ is the momentum of eigenstate in the direction of 
the background color electric field. But again, the above argument
tells that the two determinants are the color reflection
counterpart of each other. So, under the color reflection
the role of the first and second determinants are interchanged,
and only the causal eigenstates which become invariant
under the color reflection qualify as the physical eigenstates.
From this we conclude that ${\rm Det} (\hD^2+2bN)(\hD^2-2bN)$ 
should be written as ${\rm Det} (-\tilde D^2-2ib)^2(-\tilde D^2-2ib)^2$,
not as ${\rm Det} (-\tilde D^2-2ib)^2(-\tilde D^2+2ib)^2$.
So, excluding the unphysical eigenstates we have
\begin{widetext}
\bea
& \ln {\rm Det} K = \ln {\rm Det} \Big[(\hD^2+2igaN)(\hD^2+2igaN) \Big]_{ab}
+\ln {\rm Det} \Big[(\hD^2+2gbN)(\hD^2+2gbN)\Big]_{ab} \nn\\
&= 2 \ln {\rm Det} (-\tilde D^2+2a)(-\tilde D^2+2a)
+2 \ln {\rm Det} (-\tilde D^2-2ib)(-\tilde D^2-2ib).
\label{gdeto}
\eea
From this we finally have 
\bea
\Delta S = i \ln {\rm Det} [(-\tilde D^2+2a)(-\tilde D^2+2a)]
+ i \ln {\rm Det} [(-\tilde D^2-2ib)(-\tilde D^2-2ib)]
- 2i \ln {\rm Det}(-\tilde D^2),
\label{fdabo}
\eea
and
\bea
&\Delta {\cal L} =  \dfrac{}{} \lim_{\epsilon\rightarrow0}
\dfrac{1}{16 \pi^2}  \int_{0}^{\infty}
\dfrac{dt}{t^{3-\epsilon}} \dfrac{ a b t^2 / \mu^4}{\sinh (at/\mu^2)
\sin (bt/\mu^2)} \Big[ \exp(-2at/\mu^2)+\exp(-2at/\mu^2) \nn\\
&+\exp(+2ibt/\mu^2)+\exp(+2ibt/\mu^2)-2 \Big].
\label{eaabo}
\eea
\end{widetext}
Obviously this has no infra-red divergence.
This tells that, only when one mistakenly includes the
unphysical eigenstates in the calculation of the functional
determinant, one has to worry about which infra-red 
regularization is the correct one.
 
\begin{figure*}
\includegraphics{qcd1bfig.eps}
\caption{\label{Fig. 1} The effective potential of SU(2) QCD in
the pure magnetic background. Here (a) is the effective potential
and (b) is the classical potential.}
\end{figure*}

Next, consider the Savvidy background (\ref{sb}).
Here again one can show that the gauge invariance 
excludes the tachyonic modes from the physical 
eigenstates. To implement the gauge invariance to the Savvidy background, 
notice that $\vec G_{\mu\nu}$
must be gauge covariant. So, one should be able 
to change $\vec G_{\mu\nu}$ to
$-\vec G_{\mu\nu}$ by a gauge transformation. 
{\it In terms of $\bar H_{\mu\nu}$, this means that $\bar H_{\mu\nu}$ 
and $-\bar H_{\mu\nu}$ must be gauge equivalent to each other
and thus indistinguishable. This means that
the gauge invariant magnetic background can not have 
a polarization direction, because the polarization direction
of $\bar H_{\mu\nu}$ is not a gauge invariant concept}.
Consequently the eigenvalue
equation (\ref{eveh}) must be invariant under
the transformation $\bar H_{\mu\nu}$ to $-\bar H_{\mu\nu}$. 
This means that the gauge invariant eigenvalues 
must be those which are invariant under
the transformation $H$ to $-H$. This, in turn, means that
only the eigenstates which are invariant under the spin flip
$(S_3 \rightarrow -S_3)$ of the valence gluon
can be treated as the physical states. 
This is nothing but 
the color reflection invariance. Now, it must be clear that 
the tachyonic states
are precisely the states which violate this color reflection 
invariance. Again, this tells that one must exclude
the tachyonic states in one's calculation of 
the effective action (\ref{fdhx}). 

If one does so, 
the effective action (\ref{fdhx}) changes to
\bea
&\Delta S = i\ln {\rm Det} \Big[(-\bD^2+2gH)(-\bD^2+2gH) \Big],
\label{fdho}
\eea
and we have
\bea
&\Delta{\cal L} = \dfrac{}{} \lim_{\epsilon \rightarrow 0}
\dfrac{1}{16 \pi^2}\int_{0}^{\infty}
\dfrac{dt}{t^{2-\epsilon}} \dfrac{gH/ \mu^2}{\sinh (gHt/\mu^2)} \nn\\
&\times \Big[\exp (-2gHt/\mu^2 )
+ \exp (-2gHt/\mu^2) \Big],
\label{eaho}
\eea
which has no infra-red divergence. 
This precludes the necessity to make any infra-red regularization.

The fact that $\bar H_{\mu\nu}$ does not change 
to $-\bar H_{\mu\nu}$ under the reflection
$\n_0$ to $-\n_0$ in (\ref{sb}) simply confirms that the Savvidy background
is not gauge invariant. This must be compared with
the monopole background (\ref{ccon}), which clearly
has the advantage that it is invariant under
the color reflection (\ref{cri}). 

At this point we must clarify the meaning of the gauge invariant
background. By this
we do not mean that the background $\vec G_{\mu\nu}$ must be 
gauge invariant. It must be gauge covariant. By a gauge invariant
background we mean $\vec G_{\mu\nu}^2$ of a gauge covariant
$\vec G_{\mu\nu}$. Exactly the same interpretation applies to 
the Lorentz invariance. In this sense our vacuum made of
the monopole condensation is clearly gauge invariant
as well as Lorentz invariant. Of course one could choose
any gauge one likes to calculate the functional determinant (\ref{fd}).
For example for the monopole background 
one can certainly choose a gauge where $\n$ becomes
$\n_0$. When one does that the monopole background 
becomes $H_{\mu\nu} \n_0$, which looks almost identical to 
the Savvidy background $\bar H_{\mu\nu} \n_0$. But we emphasize that 
even in this gauge it remains invariant under the color reflection,
because $H_{\mu\nu}$ transforms to $-H_{\mu\nu}$ 
under (\ref{cri}).

Now, one can integrate (\ref{eaabo}) and obtain
\bea
{\cal L}_{eff}=\left\{{-\dfrac{a^2}{2g^2} -\dfrac{11a^2}{48\pi^2}
(\ln \dfrac{a}{\mu^2}-c),~~~~~~~~~~b=0
\atop \dfrac{b^2}{2g^2} +\dfrac{11b^2}{48\pi^2}
(\ln \dfrac{b}{\mu^2}-c)-i\dfrac{11b^2}{96\pi},~~a=0}\right.
\label{ceaab}
\eea
which is identical to the effective action that we obtained 
with the infra-red regularization by causality.
It is really remarkable that two completely independent
principles, the gauge invariance and the causality,
produce exactly the same effective action in QCD.

The effective action (\ref{ceaab}) generates the much desired
phase transition in QCD, the phenomenon
Coleman and Weinberg first observed in massless scalar QED \cite{cole}.
To demonstrate this we first renormalize the effective
action. Notice that the effective action (\ref{ceaab})
provides the following effective potential
\bea
V=\dfrac{a^2}{2g^2}
\Big[1+\dfrac{11 g^2}{24 \pi^2}(\ln\dfrac{a}{\mu^2}-c)\Big].
\eea
So we define the running coupling $\bar g$ by \cite{savv,cho3}
\bea
\frac{\partial^2V}{\partial a^2}\Big|_{a=\bar \mu^2}
=\frac{1}{ \bar g^2}.
\eea
With the definition we find
\bea
\frac{1}{\bar g^2} =
\frac{1}{g^2}+\frac{11}{24 \pi^2}( \ln\frac{{\bar\mu}^2}{\mu^2}
- c + \dfrac{3}{2}),
\eea
from which we obtain the following $\beta$-function,
\bea
\beta(\bar\mu)= \bar\mu \dfrac{\partial \bar g}{\partial \bar\mu}
= -\frac{11 \bar g^3}{24\pi^2}.
\eea
This is exactly the same $\beta$-function that one obtained
from the perturbative QCD to prove the asymptotic freedom
\cite{wil}. The fact that the $\beta$-function obtained from
the effective action becomes identical to the one obtained by
the perturbative calculation is remarkable, because
this is not always the case. In fact in QED it has been
demonstrated that the running coupling and the $\beta$-function
obtained from the effective action is different from those
obtained from the perturbative method \cite{cho01,cho5}.

In terms of the running coupling the renormalized potential is given by
\bea
V_{\rm ren}=\dfrac{a^2}{2\bar g^2}
\Big[1+\dfrac{11 \bar g^2}{24 \pi^2 }
(\ln\dfrac{a}{\bar\mu^2}-\dfrac{3}{2})\Big],
\eea
which generates a non-trivial local minimum at
\bea
<a>=\bar \mu^2 \exp\Big(-\frac{24\pi^2}{11\bar g^2}+ 1\Big).
\eea
Notice that with ${\bar \alpha}_s = 1$ we have
\bea
\dfrac{<a>}{{\bar \mu}^2} = 0.48988... .
\eea
This is nothing but the desired magnetic condensation.
The corresponding effective potential is plotted in Fig. 1,
where we have assumed $\bar \alpha_s = 1$ and $~\bar \mu =1$.

Notice that Nelsen and Olesen have suggested that the existence
of the unstable tachyonic modes are closely related with
the asymptotic freedom in QCD \cite{niel}.  Our analysis tells 
that this is not true. Obviously our asymptotic freedom 
follows from a stable monopole condensation.

\section{Discussion}

To eastablish the monopole condensation with the effective action in QCD
has been extremely difficult to attain. The central issue here 
has been the stability of the monopole
condensation.  The earlier attempts
to prove the monopole condensation
have produced a negative result.
The effective action of QCD in the presence of
pure magnetic background has first been calculated
by Savvidy and subsequently by Nielsen and Olesen \cite{savv,niel}.
They obtained a non-trivial magnetic condensation which was unstable,
and correctly pointed out that the instability 
was due to the tachyonic eigenstates 
of the functional determinant of the gluon loop.
They knew that this had to do with the fact that 
the magnetic background they have used was 
not gauge invariant. Unfortunately they did not know that 
the tachyonic eigenstates were exactly the byproduct of this defect,
and have included the tachyonic modes
in their calculation of the effective action. This 
destablized the SNO vacuum.

But recently there has been a new calculation of the 
effective action with a gauge invariant monopole 
background \cite{cho3,cho4}. In the new calculation
the authors have pointed out that the tachyonic modes
should be excluded in the calculation of the effective action,
because they are not physical. In particular, they showed how to remove 
the tachyonic modes with the infra-red regularization by causality.
With this infra-red regularization by causality they 
were able to prove the existence of a stable monopole condensation.
Given the importance of the issue here, however, one need 
a further justification of the infra-red regularization by causality.

One can justify the infra-red regularization by causality
with the perturbative calculation
of the imaginary part of the one-loop effective action \cite{sch,cho4}.
This is made possible, because in QCD (and in massless QED)
the imaginary part of the one-loop effective action is
of the order $g^2$. This assures us that one can make
a perturbative expansion for the imaginary part of the effective action.
Remarkably the perturbative calculation
of the imaginary part does endorse the stability of the monopole condensation.
The justification of the infra-red regularization by causality
with the perturbative calculation
of the imaginary part of the one-loop effective action
may not be so surprising, because the perturbative calculation
is also based on causality. Since both rely on the same principle,
namely the causality, one might like to have another 
confirmation based on a completely independent principle.

In this paper we have provided 
an independent justification of the infra-red regularization 
by causality based on the gauge invariance. 
The gauge invariance clearly shows that the tachyonic modes 
are the gauge artifact which should not be treated
as physical. They are simply not gauge invariant, which
disqualifies them as physical states. This again endorses 
the infra-red regularization by causality.

As we have pointed out, there are two ways to exclude the 
unphysical states in the calculation of the effective action.
One could either exclude them when one calculates the functional
determinant (\ref{fd}). If one does this, the integral expression 
of effective action (\ref{eaabo}) no longer has any infra-red 
divergence and thus one does not need any infra-red regularization.
Or one could include them at this stage, and remove them later.
If one choose to do so, one obtains the integral expression 
(\ref{eaabx}) of effective action which has a severe
infra-red divergence. In this case one must 
exclude them with the infra-red regularization by causality,
not with the $\zeta$-function regularization.
The reason is not that $\zeta$-function regularization
has any intrinsic deficiency. On the contrary the
$\zeta$-function regularization turns out to be too
faithful to the functional determinant. It simply does not remove
any states included in the determinant (\ref{fdabx}).

One might like to think that the tachyonic eigenstates are an essential
characteristic, a sacred feature, of QCD. 
This is not so. Actually it is not rare 
for us to encounter tachyonic states in physics.
The tachyons appear when one does something wrong in physics.
Consider a spontaneously broken Abelian gauge theory 
coupled to a charged scalar field. In this case tachyons 
appear when one chooses a wrong vacuum, but they disappear when one
chooses the correct vacuum. In QCD we have exactly the same
situation. Our analysis tells that the tachyonic eigenstates
appear because we have not implemented the gauge
invariance properly. With a proper implementation 
of the gauge invariance, they disappear. So there is nothing 
misterious about the tachyons in QCD.

{\it To summarize, we have presented three independent arguments
which support the stability of the monopole condensation in QCD, 
the infra-red regularization of effective action by causality,
the perturbative calculation of the imaginary part of effective action,
and the gauge invariance of the vacuum, all of which
endorse the stability of the monopole condensation.
Furthermore all these calculations have been shown to be 
consistent with duality, a fundamental symmetry of 
effective actions in gauge theory}. 
This should be enough to settle the controversy on
the stability of the monopole condensation in QCD once and for all.
With this we can conclude that the quantum fluctuation does 
create a phase transition in QCD, 
which is triggered by the monopole condensation.

It must be emphasized that Nielsen and Olesen have
correctly understood that the instability of the SNO vacuum
originates from the fact that it was not gauge invariant.
To cure this defect they have introduced the Copenhagen 
vacuum. In this paper we have shown that there is much simpler 
and more natural way to cure this defect, to impose 
the color reflection invariance
to the SNO vacuum.

We have neglected the quarks in this paper. We simply remark
that the quarks, just like in asymptotic freedom \cite{wil},
tend to destabilize the monopole condensation. In fact the stability
puts exactly the same constraint on the number of quarks as
the asymptotic freedom \cite{cho6}.
Furthermore here we have considered only the pure magnetic
or pure electric background. So, to be precise,
the above result only proves the existence of a stable
monopole condensation for a pure magnetic background. To show that
this is the true vacuum of QCD, one must calculate the effective action
with an arbitrary background in the presence of the
quarks and show that the monopole condensation
remains a true minimum of the effective potential. Fortunately, one
can actually calculate the effective action with an arbitrary
constant background, and show that indeed the monopole condensation
becomes the true vacuum of $SU(2)$ QCD, at least at one-loop
level \cite{cho6}.

It is truly  remarkable (and surprising) that the principles of
quantum field theory allow us to demonstrate confinement
within the framework of QCD. There has been a proof of
monopole condensation in a supersymmetric QCD 
\cite {witt}. Our analysis shows that one can indeed establish 
the existence of the confinement phase within the conventional
QCD, with the existing
principles of quantum field theory.
This should be interpreted as a
most spectacular triumph of quantum field theory itself.

{\bf Acknowledgements}

~~~We thank Professor S. Adler and Professor F. Dyson
for the fruitful discussions, and Professor C. N. Yang for
the encouragements. 

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\end{document}

