\ In extending the minimal standard model of quarks and leptons to include supersymmetry, the conservation of baryon and lepton numbers is no longer automatic. I show how the latter may be achieved with a new U(1) gauge symmetry and new supermultiplets at the TeV scale. Neutrino masses and a solution of the $\mu$ problem are essential features of this proposed extension.
hep-ph0201083.tex
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UCRHEP-T328\\ 
January 2002\
\end{flushright}
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\begin{center}
{\LARGE	\bf New U(1) Gauge Extension of the\\
Supersymmetric Standard Model\\}
\vspace{1.5in}
{\bf Ernest Ma\\}
\vspace{0.2in}
{Physics Department, University of California, Riverside, 
California 92521, USA\\}
\vspace{1.5in}
\end{center}
\begin{abstract}\
In extending the minimal standard model of quarks and leptons to include 
supersymmetry, the conservation of baryon and lepton numbers is no longer 
automatic.  I show how the latter may be achieved with a new U(1) gauge 
symmetry and new supermultiplets at the TeV scale.  Neutrino masses and 
a solution of the $\mu$ problem are essential features of this proposed 
extension.
\end{abstract}

\newpage
\baselineskip 24pt

It is well-known that the minimal standard model of quarks and leptons 
conserves both baryon number $B$ and lepton number $L$ automatically 
(as the consequence of the assumed $SU(3)_C \times SU(2)_L \times U(1)_Y$ 
gauge symmetry and its representation content).  It is also well-known 
that this is not true any more once it is extended to include 
supersymmetry.  Thus any such extension must be supplemented by a new 
symmetry which forbids the violation of $B$ or $L$ or both.  There are 
many ways to do this; the most direct is to impose the conservation of 
an odd-even discrete symmetry, i.e. $R \equiv (-1)^{2j+3B+L}$, which is 
of course the defining hypothesis of the Minimal Supersymmetric 
Standard Model (MSSM).

There are two additional features of the MSSM which are often called into 
question.  One is the absence of neutrino masses.  This is, however, 
easily remedied by the addition of three neutral singlet lepton superfields 
(analogs of the three right-handed singlet neutrinos of the nonsupersymmetric 
standard model).  The other is the presence of the so-called $\mu$ term in the 
MSSM superpotential, i.e. $\mu \hat \phi_1 \hat \phi_2$, where $\hat 
\phi_{1,2}$ are the two Higgs superfields which spontaneously break the 
electroweak gauge symmetry.  Since this term is allowed by the gauge symmetry 
and the supersymmetry, there is no understanding of why $\mu$ should be the 
order of the electroweak breaking scale, rather than some very large 
unification scale.

Whereas there are piecemeal solutions of all the above three problems of the 
MSSM, it is clearly desirable to have a single principle which works for all 
three at the same time.  In this paper I show how a new simple U(1) gauge 
extension of the MSSM may be used exactly for this purpose \cite{prev}.

Consider the gauge group $SU(3)_C \times SU(2)_L \times U(1)_Y \times U(1)_X$. 
The usual quark and lepton (left-handed) chiral superfields transform as 
follows:
\begin{eqnarray}
&& (\hat u, \hat d) \sim (3,2,1/6;n_1), ~~ \hat u^c \sim (3^*,1,-2/3;n_2), 
~~ \hat d^c \sim (3^*,1,1/3;n_3), \\ 
&& (\hat \nu, \hat e) \sim (1,2,-1/2;n_4), ~~ \hat e^c \sim (1,1,1;n_5), 
~~ \hat N^c \sim (1,1,0;n_6).
\end{eqnarray}
They are supplemented by the two Higgs doublet superfields
\begin{equation}
\hat \phi_1 \sim (1,2,-1/2;-n_1-n_3), ~~~ \hat \phi_2 \sim (1,2,1/2;-n_1-n_2),
\end{equation}
with
\begin{equation}
n_1 + n_3 = n_4 + n_5, ~~~ n_1 + n_2 = n_4 + n_6,
\end{equation}
as in the MSSM.  However, the $\mu$ term is replaced by the trilinear 
interaction $\hat \chi \hat \phi_1 \hat \phi_2$, where $\hat \chi$ is a 
Higgs singlet superfield transforming as
\begin{equation}
\hat \chi \sim (1,1,0;2n_1+n_2+n_3).
\end{equation}
Thus
\begin{equation}
2n_1 + n_2 + n_3 \neq 0
\end{equation}
is required so that the effective $\mu$ parameter of this model is determined 
by the $U(1)_X$ breaking scale, i.e. $\langle \hat \chi \rangle$.

To complete this model, I add two copies of the singlet quark superfields
\begin{equation}
\hat U \sim (3,1,2/3;n_7), ~~~ \hat U^c \sim (3^*,1,-2/3;n_8),
\end{equation}
and one copy of
\begin{equation}
\hat D \sim (3,1,-1/3;n_7), ~~~ \hat D^c \sim (3^*,1,1/3;n_8),
\end{equation}
with
\begin{equation}
n_7 + n_8 = -2n_1 - n_2 - n_3,
\end{equation}
so that their masses are also determined by the $U(1)_X$ breaking scale.

To ensure the absence of the axial-vector anomaly \cite{ava}, the following 
conditions are considered \cite{gema}. 
\begin{equation}
[SU(3)]^2 U(1)_X ~:~ 2n_1 + n_2 + n_3 + n_7 + n_8 = 0,
\end{equation}
\begin{eqnarray} 
[SU(2)]^2 U(1)_X ~:~ 3 (3n_1 + n_4) + (-n_1-n_3) + (-n_1-n_2) = 0,
\end{eqnarray}
\begin{eqnarray}
[U(1)_Y]^2 U(1)_X &:& 3 \left[ 6 \left( {1 \over 6} \right)^2 n_1 + 3 \left( 
-{2 \over 3} \right)^2 n_2 + 3 \left( {1 \over 3} \right)^2 n_3 
+ 2 \left( -{1 \over 2} \right)^2 n_4 + n_5 \right] \nonumber \\ 
&& + 2 \left[ 3 \left( {2 \over 3} \right)^2 n_7 + 3 \left( -{2 \over 3} 
\right)^2 n_8 \right] + 3 \left( -{1 \over 3} \right)^2 n_7 + 3 \left( 
{1 \over 3} \right)^2 n_8 \nonumber \\ && + 2 \left( -{1 \over 2} \right)^2 
(-n_1-n_3) + 2 \left( {1 \over 2} \right)^2 (-n_1-n_2) = 0,
\end{eqnarray}
\begin{eqnarray} 
U(1)_Y [U(1)_X]^2 &:& 3 \left[ 6 \left( {1 \over 6} \right) n_1^2 + 3 \left( 
-{2 \over 3} \right) n_2^2 + 3 \left( {1 \over 3} \right) n_3^2 
+ 2 \left( -{1 \over 2} \right) n_4^2 + n_5^2 \right] \nonumber \\ 
&& + 2 \left[ 3 \left( {2 \over 3} \right) n_7^2 + 3 \left( -{2 \over 3} 
\right) n_8^2 \right] + 3 \left( -{1 \over 3} \right) n_7^2 + 3 \left( 
{1 \over 3} \right) n_8^2 \nonumber \\ 
&& + 2 \left( -{1 \over 2} \right) (-n_1-n_3)^2 + 2 \left( {1 \over 2} 
\right) (-n_1-n_2)^2 = 0,
\end{eqnarray}
\begin{eqnarray}
[U(1)_X]^3 &:& 3 \left[ 6 n_1^3 + 3 n_2^3 + 3 n_3^3 + 2 n_4^3 + n_5^3 + n_6^3 
\right] + 3 (3n_7^3 + 3 n_8^3) \nonumber \\ && + 2 (-n_1-n_3)^3 + 
2 (-n_1-n_2)^3 + (2n_1+n_2+n_3)^3 = 0.
\end{eqnarray}

Using Eq.~(9), it is clear that Eq.~(10) is automatically satisfied.  Using 
Eqs.~(4) and (9), it is easily shown that both Eqs.~(11) and (12) are 
satisfied by the single condition
\begin{equation}
n_2 + n_3 = 7 n_1 + 3 n_4.
\end{equation}
Using Eqs.~(4), (9), and (15), it is then simple to show that Eq.~(13) becomes
\begin{equation}
6(3n_1+n_4)(2n_1-4n_2-3n_7) = 0.
\end{equation}
Using Eq.~(15), it is clear that $3n_1+n_4 = 0$ contradicts Eq.~(6).  Hence 
only the condition
\begin{equation}
2n_1 - 4n_2 - 3n_7 = 0
\end{equation}
will be considered from here on.

At this point, the eight parameters ($n_1$ to $n_8$) are constrained by the 
five conditions given by Eqs.~(4), (9), (15), and (17).  Consider $n_1$, 
$n_4$, and $n_6$ as the independent parameters.  The others are then given by
\begin{eqnarray}
n_2 &=& -n_1 + n_4 + n_6, \\ 
n_3 &=& 8n_1 + 2n_4 - n_6, \\ 
n_5 &=& 9n_1 + n_4 - n_6, \\ 
n_7 &=& 2n_1 - {4 \over 3} n_4 - {4 \over 3} n_6, \\ 
n_8 &=& -11 n_1 - {5 \over 3} n_4 + {4 \over 3} n_6.
\end{eqnarray}
It is now straightforward to simplify Eq.~(14) to read
\begin{equation}
-36 (3n_1+n_4)(9n_1+n_4-2n_6)(6n_1-n_4-n_6) = 0.
\end{equation}
Whereas one factor, i.e. $3n_1+n_4$, must be nonzero, there remain two 
possible solutions, i.e.
\begin{eqnarray}
({\rm A}) && n_6 = {1 \over 2} (9n_1+n_4), \\ 
({\rm B}) && n_6 = 6n_1 - n_4,
\end{eqnarray}
which render $U(1)_X$ free of the axial-vector anomaly.  This exact factoring 
of the sum of eleven cubic terms is certainly not a trivial result \cite{eg}.

Solution (A) is thus given by
\begin{eqnarray}
&& n_2 = n_3 = {1 \over 2} (7n_1+3n_4), ~~~ n_5 = n_6 = {1 \over 2} 
(9n_1 + n_4), \\ && n_7 = -4n_1 - 2n_4, ~~~ n_8 = -5n_1 - n_4.
\end{eqnarray}
In the MSSM, $\hat L$ and $\hat \phi_1$ transform identically under 
$SU(3)_C \times SU(2)_L \times U(1)_Y$.  Here $\hat L$ and $\hat \phi_1$ are 
distinguished by $U(1)_X$ if
\begin{equation}
9n_1 + 5n_4 \neq 0.
\end{equation}
Hence the lepton number $L$ may be automatically conserved as in the 
nonsupersymmetric standard model.

In the MSSM, the term $\hat u^c \hat d^c \hat d^c$ is allowed in the 
superpotential.  Here it is forbidden if
\begin{equation}
7n_1 + 3n_4 \neq 0.
\end{equation}
Hence the baryon number $B$ may be automatically conserved as well.

Solution (B) has
\begin{eqnarray}
&& n_2 = 5n_1, ~~ n_3 = 2n_1 + 3n_4, ~~ n_5 = 3n_1 + 2n_4, \\ 
&& n_6 = 6n_1 - n_4, ~~ n_7 = -6n_1, ~~ n_8 = -3n_1 - 3n_4.
\end{eqnarray}
Hence $L$ is automatically conserved if
\begin{equation}
3n_1 + 4n_4 \neq 0,
\end{equation}
and $B$ is automatically conserved if 
\begin{equation}
3n_1 + 2n_4 \neq 0.
\end{equation}
Note that solutions (A) and (B) are identical if $n_4 = n_1$.  This turns 
out to be also the condition \cite{coc} for $U(1)_X$ to be orthogonal to 
$U(1)_Y$, i.e.
\begin{eqnarray} 
&& 3 \left[ 6 \left( {1 \over 6} \right) n_1 + 3 \left( 
-{2 \over 3} \right) n_2 + 3 \left( {1 \over 3} \right) n_3 
+ 2 \left( -{1 \over 2} \right) n_4 + n_5 \right] \nonumber \\ 
&& + 2 \left[ 3 \left( {2 \over 3} \right) n_7 + 3 \left( -{2 \over 3} 
\right) n_8 \right] + 3 \left( -{1 \over 3} \right) n_7 + 3 \left( 
{1 \over 3} \right) n_8 \nonumber \\ 
&& + 2 \left( -{1 \over 2} \right) (-n_1-n_3) + 2 \left( {1 \over 2} 
\right) (-n_1-n_2) = 0.
\end{eqnarray}


There are two more anomalies to consider.  The global SU(2) chiral gauge 
anomaly \cite{witten} is absent because the number of $SU(2)_L$ doublets 
is even.  The mixed gravitational-gauge anomaly \cite{mixed} is proportional 
to the sum of $U(1)_X$ charges, i.e.
\begin{eqnarray}
&& 3(6n_1 + 3n_2 + 3n_3 + 2n_4 + n_5 + n_6) + 3(3n_7 + 3n_8) \nonumber \\ 
&& + 2(-n_1-n_3) + 2(-n_1-n_2) + (2n_1 + n_2 + n_3) = 6(3n_1 + n_4),
\end{eqnarray}
which is not zero.  This anomaly may be tolerated if gravity is neglected. 
On the other hand, it may be rendered zero by adding $U(1)_X$ 
supermultiplets as follows: one with charge $3(3n_1+n_4)$, three with charge 
$-2(3n_1+n_4)$, and three with charge $-(3n_1+n_4)$.  Hence they contribute 
$3 + 3(-2-1) = -6$ (in units of $3n_1+n_4$) to Eq.~(35), but $27 + 3(-8-1) 
= 0$ to Eq.~(14).

The allowed terms in the superpotential of either solution (A) or (B) 
consist of the usual allowed terms of the MSSM with $\mu \hat \phi_1 
\hat \phi_2$ replaced by $\hat \chi \hat \phi_1 \hat \phi_2$.  In (A), 
the usual $R$-parity violating terms are forbidden by Eqs.~(28) and (29). 
As for the interactions of the exotic quark singlets of Eqs.~(7) and (8), 
$n_1 + n_4 \neq 0$ forbids $\hat U^c \hat d^c \hat d^c$, $\hat u^c \hat d^c 
\hat D^c$, and $(\hat u_i \hat d_j - \hat d_i \hat u_j) \hat D$; 
$13n_1 + n_4 \neq 0$ forbids $\hat U^c \hat d^c \hat D^c$; and $n_1 \neq 0$ 
forbids $\hat e^c \hat u^c \hat D$, $(\hat \nu \hat d - \hat e \hat u) 
\hat D^c$, $\hat N^c \hat u^c \hat U$, and $\hat N^c \hat d^c \hat D$. 
This means that if $n_4 = -n_1$, then $\hat U^c$ and $\hat D^c$ are diquark 
superfields, and if $n_1 = 0$, then $\hat U$ and $\hat D$ are leptoquark 
superfields.

In solution (B), the usual $R$-parity violating terms are forbidden by 
Eqs.~(32) and (33).  Furthermore, $n_1+3n_4 \neq 0$ forbids $\hat U^c 
\hat d^c \hat d^c$; $n_1 \neq 0$ forbids $\hat u^c \hat d^c \hat D^c$ and 
$(\hat u_i \hat d_j - \hat d_i \hat u_j) \hat D$; $4n_1 + 3n_4 \neq 0$ 
forbids $\hat U^c \hat d^c \hat D^c$; $n_1+n_4 \neq 0$ forbids $\hat e^c 
\hat u^c \hat D$, $(\hat \nu \hat d - \hat e \hat u) \hat D^c$, and 
$\hat N^c \hat d^c \hat D$; and $5n_1-n_4 \neq 0$ forbids $\hat N^c 
\hat u^c \hat U$.  This means that $\hat U^c$ is a diquark if $n_4 = -n_1/3$, 
and $\hat U$ is a leptoquark if $n_4 = 5n_1$; whereas $\hat D^c$ is a 
diquark if $n_1 = 0$, and $\hat D$ is a leptoquark if $n_4 = -n_1$.

Even with the imposition of $R$-parity, there are higher-dimensional operators 
in the MSSM which may induce proton decay, i.e. $\hat q \hat q \hat q 
\hat l$ and $\hat u^c \hat u^c \hat d^c \hat e^c$.  In the nonsupersymmetric 
standard model, since quarks and leptons are fermions, these operators have 
dimension six, but in the MSSM they have dimension-five pieces.  Hence 
proton decay may not be sufficiently suppressed, which is a well-known 
problem of the MSSM.  In this model, these terms are forbidden [in both 
solutions (A) and (B)] by $3n_1+n_4 \neq 0$, i.e. the same condition that 
forbids the $\mu$ term.

As it stands, this model pairs $\nu$ with $N^c$ to form a Dirac neutrino 
with mass proportional to $\langle \phi_2 \rangle$.  This would require 
extremely small Yukawa couplings and is generally considered to be very 
unnatural.  On the other hand, if $n_6 = 3n_1 + n_4$ is assumed [i.e. 
$n_4 = 3n_1$ in solution (A) or $n_4 = 3n_1/2$ in solution (B)], then the 
extra singlets used to cancel the mixed gravitational-gauge anomaly of 
Eq.~(35) are exactly of the right number and structure to allow neutrinos 
to acquire naturally small seesaw Dirac masses, as shown below.

In addition to the 3 singlets $N^c$ of $U(1)_X$ charge $n_6$, there are now 
also 3 singlets $N$ of charge $-n_6$ and 3 singlets $S^c$ of charge $-2n_6$. 
The $12 \times 12$ mass matrix spanning $(\nu, S^c, N, N^c)$ is then of the 
form
\begin{equation}
{\cal M} = \left[ \begin{array} {c@{\quad}c@{\quad}c@{\quad}c} 0 & 0 & 0 & m_1 
\\ 0 & 0 & m_2 & 0 \\ 0 & m_2 & 0 & M \\ m_1 & 0 & M & 0 \end{array} \right],
\end{equation}
where $m_1$ comes from $\nu N^c \phi_2^0$ with $\langle \phi_2^0 \rangle 
\neq 0$, $m_2$ comes from $N S^c \chi$ with $\langle \chi \rangle \neq 0$, 
and $M$ is an allowed invaraint mass.  Thus $m_1 \sim 10^2$ GeV, $m_2 \sim 
10^3$ GeV, and $M \sim 10^{16}$ GeV are expected.  In the reduced $(\nu, S^c)$ 
sector, the effective $6 \times 6$ mass matrix is still exactly of the Dirac 
form, i.e.
\begin{equation}
{\cal M}_\nu = \left[ \begin{array} {c@{\quad}c} 0 & -m_1 m_2/M \\ -m_1 m_2/M 
& 0 \end{array} \right],
\end{equation}
and $m_1 m_2/M \sim 10^{-2}$ eV is the right order of magnitude for 
realistic neutrino masses.


In conclusion, a remarkable new U(1) gauge symmetry has been identified in a 
simple extension of the supersymmetric standard model which is capable of 
enforcing $B$ or $L$ conservation or both, as well as the absence of the $\mu$ 
term and the presence of neutrino masses.  Two solutions have been obtained 
[from the exact factoring of Eq.~(14) to become Eq.~(23)] with many possible 
variations regarding new interactions beyond the MSSM, as summarized in 
Tables 1 and 2.  The origin of this new U(1) gauge symmetry is unknown at 
present; it has no obvious fit into any simple model of grand unification 
or string theory.


This work was supported in part by the U.~S.~Department of Energy under 
Grant No.~DE-FG03-94ER40837.


\begin{table}[htb]
\caption{Solutions (A) and (B) where $n_i = a n_1 + b n_4$.}
\begin{center}
\begin{tabular}{||c|c|c||c|c||}
\hline \hline
& \multicolumn{2}{c||}{(A)} & \multicolumn{2}{c||}{(B)} \\ 
\cline{2-5}
 & $a$ & $b$ & $a$ & $b$ \\ 
\hline
$n_2$ & 7/2 & 3/2 & 5 & 0 \\ 
$n_3$ & 7/2 & 3/2 & 2 & 3 \\ 
$n_5$ & 9/2 & 1/2 & 3 & 2 \\ 
$n_6$ & 9/2 & 1/2 & 6 & --1 \\ 
$n_7$ & --4 & --2 & --6 & 0 \\ 
$n_8$ & --5 & --1 & --3 & --3 \\ 
\hline
$-n_1-n_3$ & --9/2 & --3/2 & --3 & --3 \\ 
$-n_1-n_2$ & --9/2 & --3/2 & --6 & 0 \\ 
$2n_1+n_2+n_3$ & 9 & 3 & 9 & 3 \\ 
\hline \hline
\end{tabular}
\end{center}
\end{table}


\begin{table}[htb]
\caption{Conditions on $n_1$ and $n_4$ in (A) and (B).}
\begin{center}
\begin{tabular}{||c|c||c|c|l||}
\hline \hline
\multicolumn{2}{||c||}{(A)} & \multicolumn{2} {c|}{(B)} & \\ 
\hline
$c$ & $d$ & $c$ & $d$ & $cn_1+dn_4 \neq 0$ forbids \\ 
\hline
3 & 1 & 3 & 1 & $\mu$ term \\ 
9 & 5 & 3 & 4 & $L$ violation \\ 
7 & 3 & 3 & 2 & $B$ violation \\ 
1 & 1 & 1 & 3 & $U^c$ as diquark \\ 
1 & 1 & 1 & 0 & $D^c$ as diquark \\ 
1 & 0 & 5 & --1 & $U$ as leptoquark \\ 
1 & 0 & 1 & 1 & $D$ as leptoquark \\ 
13 & 1 & 4 & 3 & $U^c$, $D^c$ as semiquarks \\ 
\hline \hline
\end{tabular}
\end{center}
\end{table}

%\newpage
\bibliographystyle{unsrt}
\begin{thebibliography}{99}
\bibitem{prev} For previous studies, see for example H.-C. Cheng, B. A. 
Dobrescu, and K. T. Matchev, Phys. Lett. {\bf B439}, 301 (1998); Nucl. Phys. 
{\bf B543}, 47 (1999); J. Erler, Nucl. Phys. {\bf B586}, 73 (2000).
\bibitem{ava} S. L. Adler, Phys. Rev. {\bf 177}, 2426 (1969); J. S. Bell and 
R. Jackiw, Nuovo Cimento {\bf A60}, 47 (1969); W. A. Bardeen, Phys. Rev. 
{\bf 184}, 1848 (1969).
\bibitem{gema} C. Q. Geng and R. E. Marshak, Phys. Rev. {\bf D39}, 693 (1989); 
X.-G. He, G. C. Joshi, and R. R. Volkas, Phys. Rev. {\bf D41}, 278 (1990).
\bibitem{eg} Consider for example $n_1=2/3$, $n_4=-1$ in solution (A), then 
Eq.~(14) is realized as $18(2/3)^3 + 18(5/6)^3 + 6(-1)^3 + 6(5/2)^3 + 
9(-2/3)^3 + 9(-7/3)^3 + 4(-3/2)^3 + (3)^3 = 0$.
\bibitem{coc} W. Loinaz and T. Takeuchi, Phys. Rev. {\bf D60}, 115008 (1999).
\bibitem{witten} E. Witten, Phys. Lett. {\bf 117B}, 324 (1982).
\bibitem{mixed} R. Delbourgo and A. Salam, Phys. Lett. {\bf 40B}, 381 (1972); 
T. Eguchi, and P. G. O. Freund, Phys. Rev. Lett. {\bf 37}, 1251 (1976); 
L. Alvarez-Gaume and E. Witten, Nucl. Phys. {\bf B234}, 269 (1984).
\end{thebibliography}
\end{document}






To obtain nonzero neutrino masses so as to explain the observed atmospheric 
\cite{atm} and solar \cite{sol} neutrino oscillations, the minimal standard 
model of particle interactions is often extended to include three neutral 
fermion singlets, often referred to as right-handed singlet neutrinos.  If 
they have large Majorana masses, then the famous seesaw mechanism 
\cite{seesaw} allows 
the observed neutrinos to acquire naturally small Majorana masses. 



Furthermore, the absence of the mixed gravitational-gauge anomaly \cite{mixed} 
requires the sum of $U(1)_X$ charges to vanish, i.e.
\begin{equation}
U(1)_X ~:~ 6n_1 - 3n_2 - 3n_3 + 2n_4 - n_5 - 3 n_6 = 0. 
\end{equation}
Since the number of $SU(2)_L$ doublets remains even (it is in fact unchanged), 
the global SU(2) chiral gauge anomaly \cite{witten} is absent automatically.

Equations (4), (6), and (7) do not involve $n_6$.  Together they allow two 
solutions:
\begin{equation}
({\rm I}) ~ n_4 = -3n_1, ~~~ ({\rm II}) ~ n_2 = {1 \over 4} (7n_1 - 3n_4).
\end{equation}
Using Eq.~(5), solution (I) implies $n_6 = 0$, from which it can easily be 
seen that $U(1)_X$ is proportional to $U(1)_Y$.  In other words, no new 
gauge symmetry has been discovered.

Consider now solution (II).  Using Eqs.~(4) and (6), it implies
\begin{equation}
n_3 = {1 \over 4} (n_1+3n_4), ~~~ n_5 = {1 \over 4} (-9n_1+5n_4).
\end{equation}
Equations (5), (8), and (9) are then \underline {all} satisfied with
\begin{equation}
n_6 = {1 \over 4} (3n_1+n_4).
\end{equation}
This is a remarkable and highly nontrivial result.

The $U(1)_X$ charges of the possible Higgs doublets are:
\begin{equation}
n_1-n_3 = n_2-n_1 = n_6-n_4 = {3 \over 4} (n_1-n_4), ~~~ n_4-n_5 = {1 \over 4} 
(9n_1-n_4),
\end{equation}
which means that two distinct Higgs doublets are sufficient for all possible 
Dirac fermion masses in this model.  If $n_4 = -3n_1$ is chosen, then again 
$U(1)_X$ will be proportional to $U(1)_Y$.  However, for $n_4 \neq -3n_1$, 
a new class of models is now possible with $U(1)_X$ as a genuinely new gauge 
symmetry.

To summarize, the quarks and leptons transform under $U(1)_X$ as follows:
\begin{eqnarray}
&& (u,d)_L \sim n_1, ~~~ u_R \sim {1 \over 4} (7n_1 - 3n_4), 
~~~ d_R \sim {1 \over 4} (n_1 + 3n_4), \\ 
&& (\nu,e)_L \sim n_4, ~~~ e_R \sim {1 \over 4} (-9n_1 + 5n_4),  
~~~ \Sigma_R \sim {1 \over 4} (3n_1 + n_4).
\end{eqnarray}
The above charge assignments do not correspond to any existing model of quark 
and lepton interactions.  For example, if $n_4=n_1$ is assumed, then
\begin{equation} 
n_1 = n_2 = n_3 = n_4 = -n_5 = n_6,
\end{equation}
which means that $X$ couples 
vectorially to quarks, but its coupling to charged leptons is purely 
axial-vector.  On the other hand, if $n_4 = 9n_1$ is assumed, then
\begin{equation}
n_1 = 1, ~~ n_2 = -5, ~~ n_3 = 7, ~~ n_4 = 9, ~~ n_5 = 9, ~~ n_6 = 3 
\end{equation}
is a solution with $X$ coupling vectorially to charged leptons.

Consider $\nu q$ and $\bar \nu q$ deep inelastic scattering.  It has 
recently been reported \cite{nutev} by the NuTeV Collaboration that their 
measurement of the effective $\sin^2 \theta_W$, i.e. $0.2277 \pm 0.0013 \pm 
0.0009$, is about $3\sigma$ away from the standard-model prediction of 
$0.2227 \pm 0.00037$.  In this model, the $X$ gauge boson also contributes 
with
\begin{eqnarray}
J_X^\mu &=& n_1 \bar u \gamma^\mu \left( {1-\gamma_5 \over 2} \right) u + 
n_1 \bar d \gamma^\mu \left( {1-\gamma_5 \over 2} \right) d \nonumber \\ 
&& +~n_2 \bar u \gamma^\mu \left( {1+\gamma_5 \over 2} \right) u + 
n_3 \bar d \gamma^\mu \left( {1+\gamma_5 \over 2} \right) d + n_4 \bar \nu 
\gamma^\mu \left( {1-\gamma_5 \over 2} \right) \nu.
\end{eqnarray}
Assuming very small $X-Z$ mixing ($|\sin \theta| << 1$), the effective 
neutrino-quark interactions are then given by
\begin{equation}
{\cal H}_{int} = {G_F \over \sqrt 2} \bar \nu \gamma^\mu (1-\gamma_5) \nu 
[\epsilon_L^q \bar q \gamma_\mu (1-\gamma_5) q + \epsilon_R^q \bar q 
\gamma_\mu (1+\gamma_5) q],
\end{equation}
where
\begin{eqnarray}
\epsilon_L^u &=& (1-\xi)\left( {1 \over 2} - {2 \over 3} \sin^2 \theta_W 
\right) + n_1 \zeta, \\ 
\epsilon_L^d &=& (1-\xi)\left( -{1 \over 2} + {1 \over 3} \sin^2 \theta_W 
\right) + n_1 \zeta, \\ 
\epsilon_R^u &=& (1-\xi)\left( -{2 \over 3} \sin^2 \theta_W \right) 
+ n_2 \zeta, \\ 
\epsilon_R^d &=& (1-\xi)\left( {1 \over 3} \sin^2 \theta_W \right) 
+ n_3 \zeta,
\end{eqnarray}
with
\begin{eqnarray}
\xi &=& n_4 \sin \theta \left( 1 - {M_Z^2 \over M_X^2} \right) {g_X \over 
g_Z}, \\ 
\zeta &=& - \sin \theta \left( 1 - {M_Z^2 \over M_X^2} \right) {g_X \over g_Z} 
+ n_4 \left( {M_Z^2 \over M_X^2} \right) {g_X^2 \over g_Z^2}.
\end{eqnarray}

To account for the NuTeV result, i.e.
\begin{eqnarray}
(g_L^{eff})^2 &=& (\epsilon_L^u)^2 + (\epsilon_L^d)^2 = 0.3005 \pm 0.0014, \\ 
(g_R^{eff})^2 &=& (\epsilon_R^u)^2 + (\epsilon_R^d)^2 = 0.0310 \pm 0.0011,
\end{eqnarray}
against the standard-model prediction, i.e.
\begin{equation}
(g_L^{eff})^2_{SM} = 0.3042, ~~~ (g_R^{eff})^2_{SM} = 0.0301,
\end{equation}
consider the following specific model as an illustration:
\begin{equation}
n_1 = 0, ~~ n_2 = -{3 \over 4}, ~~ n_3 = {3 \over 4}, ~~ n_4 = 1, ~~ 
n_5 = {5 \over 4}, ~~ n_6 = {1 \over 4}.
\end{equation}
The central values of the NuTeV measurements are then obtained with
\begin{equation}
\xi = 0.0061, ~~~ \zeta = 0.0038,
\end{equation}
implying that
\begin{equation}
M_X \simeq 10 \left( {g_X \over g_Z} \right) M_Z, ~~~ \sin \theta \simeq 
0.006 \left( {g_Z \over g_X} \right).
\end{equation}
Whereas $M_X \sim 1$ TeV is certainly allowed by the present data, a smaller 
value of $\sin \theta$ is indicated by the precision measurements 
at the $Z$ pole.  A comprehensive numerical analysis of this and the more 
general case of $n_1 \neq 0$ will be given elsewhere \cite{maroy}.

In atomic parity nonconservation, the dominant effect comes from the 
axial-vector coupling of the electron.  In the model defined by Eq.~(29), 
this is given by $(n_4 - n_5)/2 = -1/8$; hence it is rather suppressed. 
Furthermore, the isoscalar vector coupling of the quarks in this model 
also vanishes.  Therefore, the contribution of $X$ is essentially 
negligible and there should be no observable deviation from the prediction 
\cite{apv} of the standard model, in agreement with the most recent data 
\cite{bewi}.

Consider now the anomalous magnetic moment of the muon.  A recent experimental 
result \cite{g-2}, after the latest theoretical corrections \cite{corr}, gives
its deviation from the standard model as
\begin{equation}
\Delta a_\mu = 2.5 \pm 1.6 \times 10^{-9},
\end{equation}
which is only an $1.6\sigma$ effect.  From the standpoint of the proposed 
$U(1)_X$ model, there are two possible contributions.  One comes from the 
$X$ boson which has a vector coupling, i.e. $(n_4 + n_5)/2$, to the muon. 
However, if $M_X \sim 1$ TeV, then this contribution is essentially 
negligible.  The other comes from the extended Higgs sector of this model. 
In particular, the coupling of $(\nu_\mu, \mu)_L$ to $\Sigma_R$ through the 
Higgs doublet with $X$ charge $3(n_1-n_4)/4$ provides two one-loop diagrams: 
one with $\Sigma^-$ and $\bar \phi^0$ as intermediate states, the other 
with $\Sigma^0$ and $\phi^-$.  If all these masses are equal, the former 
contributes with a coefficient of +2 and the latter with a coefficient of 
--1 to $\Delta a_\mu$.  Assuming masses of order 200 GeV, it is 
thus possible to account for Eq.~(32).

It is well-known that given its particle content, the minimal standard model 
does not allow for the unification of gauge couplings.  The addition of 
$\Sigma_R$ in Eq.~(2) does not change the situation.  However, if 
gauge-coupling unification at $M_U \sim 10^{16}$ GeV is desired, one simple 
possibility is to add three charged lepton singlets with only vector 
interactions, i.e. $E_{L,R} \sim (1,1,-1;0)$, as well as an $SU(3)$ octet 
of neutral 
colored fermions, i.e. $\psi_{L,R} \sim (8,1,0;0)$.  It is clear that this 
model would still be anomaly-free, but the evolution equations of the gauge 
couplings would now change, assuming of course that all new particles have 
masses of order $10^2$ GeV.  Generically, the one-loop renormalization-group 
equations for the running of gauge couplings are given by
\begin{equation}
\alpha_i^{-1} (M_1) = \alpha_i^{-1}(M_2) - {b_i \over 2 \pi} 
\ln {M_1 \over M_2},
\end{equation}
where $\alpha_i \equiv g_i^2/4\pi$ and $b_i$ are constants determined by the 
particle content contributing to $\alpha_i$.  Here,
\begin{eqnarray}
&& b_3 = -11 + (3) {4 \over 3} + 4 = -3, \\ 
&& b_2 = -{22 \over 3} + (3) {4 \over 3} + (2) {1 \over 6} + (3) {4 \over 3} 
= 1, \\ 
&& b_Y = (3) {20 \over 9} + (2) {1 \over 6} + (3) {4 \over 3} = 11, \\ 
&& b_X = {1 \over 12} (585 n_1^2 - 282 n_1 n_4 + 177 n_4^2) = (40 ~{\rm if} 
~n_4 = n_1 = 1).
\end{eqnarray}

Using the precision measurements \cite{pdg}
\begin{equation}
\alpha^{-1} (M_Z) = 127.938 \pm 0.027, ~~~ \sin^2 \theta_W (M_Z) = 
0.23117 \pm 0.00016,
\end{equation}
and the relationships
\begin{equation}
\alpha_2^{-1} = \alpha^{-1} \sin^2 \theta_W, ~~~ \alpha_1^{-1} = {3 \over 5} 
\alpha_Y^{-1} = {3 \over 5} \alpha^{-1} \cos^2 \theta_W,
\end{equation}
I find from Eqs.~(33), (35), and (36) that
\begin{equation}
{M_U \over M_Z} = 2.223 \times 10^{14},
\end{equation}
from which $\alpha_3^{-1} (M_Z)$ is predicted by Eqs.~(33) and (34) to be 
8.544, in good agreement with the experimental value \cite{pdg} 
$\alpha_3 (M_Z) = 0.1192 \pm 0.0028$, i.e. $\alpha_3^{-1} = 8.39 
(+0.20/-0.19)$.

In the above, $U(1)_Y$ is normalized as in the standard model, but since the 
normalization of $U(1)_X$ is unknown, $g_X$ cannot be unified in analogy to 
$g_Y$.  This also means that a two-loop analysis of $\alpha_{1,2,3}$ would 
not be possible because it would involve $g_X$.  There is no obvious 
unification symmetry which includes the particle content of this model as 
an anomaly-free subset.

Instead of having one $\Sigma_R$ per family, consider the total of (A) one 
$\Sigma_R$, and (B) two $\Sigma_R$'s for the three families of quarks and 
leptons.  In either case, Eqs.~(4), (6), and (7) are unchanged.  Hence 
solution (II) of Eq.~(10) is still valid, together with Eq.~(11).  The analog 
of Eq.~(5) now implies
\begin{equation}
({\rm A})~~ n_6 = {3 \over 4} (3n_1 + n_4), ~~~ ({\rm B}) ~~n_6 = {3 \over 8} 
(3n_1 + n_4).
\end{equation}
Whereas the analog of Eq.~(9) is still automatically satisfied, that of 
Eq.~(8) is not.  On the other hand, if singlet $N_R$'s are added with $X$ 
charges given as follows:
\begin{eqnarray}
({\rm A}) &:& n_6, ~n_6, -{1 \over 3} n_6, -{5 \over 3} n_6, \\ 
({\rm B}) &:& n_6, {2 \over 3} n_6, -{5 \over 3} n_6,
\end{eqnarray}
the analogs of both Eqs.~(8) and (9) are again satisfied.  Note that in Case 
(A), there are two singlets with $X$ charge $n_6$, and in Case (B), there is 
one such singlet.  This means that the total number of triplets and singlets 
with $X$ charge $n_6$ is always three in each of the three models, thus 
allowing all three neutrinos to acquire small seesaw Majorana masses. 

To conclude, three anomaly-free $U(1)_X$ models have been discovered.  They 
are characterized by having fermions and Higgs bosons beyond those of the 
minimal standard model.  In the simplest case, each family of quarks and 
leptons is supplemented by a triplet of leptons.  In another case, i.e. (A),  
there is only one triplet for the three families, but there are four 
singlets with $X$ charges given by Eq.~(42).  In the third case, i.e. (B), 
there are two triplets and three singlets with $X$ charges given by Eq.~(43). 
If $U(1)_X$ is a relevant gauge symmetry at or near the electroweak breaking 
scale, then it may already be implicated in some recent experimental data 
which show possible deviations from the standard model, such as the NuTeV 
result \cite{nutev} and the muon $g-2$ measurement \cite{g-2}.  Of course, 
the main motivation for studying $U(1)_X$ is not predicated on these potential 
discrepancies, but rather on its fundamental theoretical appeal.  Details of 
other possible phenomenological consequences will be discussed elsewhere 
\cite{maroy}.

\end{document}














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