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\title{NEGATIVE CONTRIBUTIONS TO S FROM MAJORANA PARTICLES}
\author{Evalyn Gates and John Terning\\
Department of Physics, Yale University, New Haven, CT 06520}
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\begin{abstract}
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In this letter we study oblique corrections from heavy fermions.
For certain mass ranges we find negative contributions to the $S$ and $T$
parameters from Majorana particles.
\end{abstract}
Recently a number of authors \cite{STGB}~-~\cite{Sirlin} have focused attention
on the fact that heavy
particles do not necessarily decouple \cite{Appelquist}
in theories with broken gauge symmetries.
If new heavy particles exist and couple only weakly to light fermions,
then their effect on low-energy precision electroweak
measurements can only arise through contributions to gauge boson self energies.
This type of radiative correction is termed an oblique correction
\cite{Lynn,Peskin}.
If the masses of the new particles are much larger than the mass of the $Z$
boson, $M_Z$, then the effects of the oblique corrections can be
described by three parameters: $S$, $T$, and $U$
\cite{ST}.
These parameters are
defined in terms of the gauge boson self energies:
\begin{eqnarray}
S &=& 16\pi{d\over{dq^2}}(\Pi_{33}(q^2)-\Pi_{3Q}(q^2))|_{q^2=0} ~~,
\\
\alpha T & =& \Delta \rho_{*} = {e^2\over{s^2 M_W^2}}
(\Pi_{11}(0)-\Pi_{33}(0)) ~~,\nonumber \\
U &=& 16\pi{d\over{dq^2}}(\Pi_{11}(q^2)-\Pi_{33}(q^2))|_{q^2=0} ~~,
\nonumber
\label{STU}
\end{eqnarray}
where we have adopted the notation of Peskin and Takeuchi
\cite{ST,Takeuchi,Peskin}, and
$M_W$ is the mass of the $W$, $e$ is the electromagnetic charge,
and $s^2 \equiv \sin^2(\theta_W)$.
The indices 1 and 3 refer to $SU(2)_L$ currents,
while $Q$ refers to the electromagnetic current. So, for example, in this
notation the $W$ self energy is $i({e^2}/{s^2})\Pi_{11}(q^2)$.
The parameter $S$ is sometimes refered to as
$S_Z$, and $S_W$ is defined by $S_W \equiv S + U$.
A notation based on chiral Lagrangian conventions uses $L_{10} = -S/(16\pi)$
\cite{ST}.
(Note that $S$, $T$, and $U$ do not include standard model corrections to gauge
boson
self energies except for contributions from particles much heavier than the
$Z$, eg. the top quark or Higgs boson.)
Precision electroweak measurements are beginning to constrain these parameters.
While QCD-like technicolor theories predict $S$ and $T$ to be of order $1$
\cite{ST}, and SUSY theories predict $S$ and $T$ to be very small \cite{Haber},
recent global fits to data \cite{KL},
while not ruling out positive values, give central values of order $-1$.
If experiments continue to favor negative values of $S$ and $T$ while the errors
shrink, then it will be useful to have models that can accomodate such
values. In
this letter we point out that, for this purpose, models using heavy Majorana
particles are
viable, while those containing only heavy Dirac particles are not. (For an
alternative scenario for obtaining negative contributions to $S$
see ref.~\cite{Holdom}.)
To begin we will discuss the case of heavy degenerate multiplets of Dirac
particles. Contributions to $S$ will come from diagrams like those in Fig. 1.
By dimensional analysis, such contributions to $S$ must be independent of the
fermion mass, since $S$ is dimensionless and there is no other dimensionful
parameter aside from the fermion mass. (There is no cutoff dependence since $S$
is
finite, see ref. \cite{Peskin}.) Thus for the case of degenerate Dirac
multiplets,
we can discuss contributions from all heavy fermions simultaneously. To
simplify the discussion we assemble all the heavy fermion fields into a
column vector, and make use of the charge, weak isospin, and hypercharge
operators $Q$, $I_3^L$, and $Y^L$. These operators are diagonal matrices with
eigenvalues corresponding to the electromagnetic charge, third component of
weak isospin, and hypercharge of the left-handed fermions.
To be completely general we will consider the possibility that some of the
right-handed fields may also belong to arbitrary $SU(2)_L$ multiplets.
Thus we will also
make use of the operators which give the charges of the right-handed fields:
$I_3^R$ and $Y^R$.
Using the relation
\begin{equation}
Q = I_3^L + {{Y^L}\over 2} = I_3^R + {{Y^R}\over 2} ~~,
\label{Q}
\end{equation}
we find
\begin{equation}
\Pi_{33}(q^2)-\Pi_{3Q}(q^2) = -{{1}\over 2} \Pi_{3Y}(q^2) ~~,
\label{Pi3Y}
\end{equation}
so we see that it is sufficient to calculate the (momentum dependent)
$W_3-B$ gauge field mixing (Fig.~1)
in order to determine $S$. Now, the graph in
Fig.~1a is proportional to Tr($I_3^L Y^L$), which vanishes within in
each multiplet, and the graph in Fig.~1b is proportional to Tr($I_3^L Y^R$).
Using Eq.~\ref{Q} we find
\begin{equation}
{\rm Tr}(I_3^L Y^R) = {\rm Tr}(I_3^L (2I_3^L - 2I_3^R + Y^L))
= 2 {\rm Tr}(I_3^L I_3^L - I_3^L I_3^R) ~~.
\label{Tr}
\end{equation}
If some of the right handed fields are not singlets, then there will be other
contributions to $S$ like those in Fig.~1b, but with L and R interchanged. Thus
the full contribution to $S$ from degenerate multiplets of Dirac fermions is
\begin{equation}
S_{\rm Deg. Dirac} =
{1\over{3\pi}} {\rm Tr}(I_3^L I_3^L - 2 I_3^L I_3^R + I_3^R I_3^R)
= {1\over{3\pi}} {\rm Tr}(I_3^L - I_3^R)^2 ~~,
\label{SDD}
\end{equation}
which is positive semi-definite.
We now turn to the case of non-degenerate Dirac particles. For simplicity
we will consider the example of a generic doublet
$\left(\begin{array}{c}U\\D\end{array}\right)_L$ with right-handed
singlets $U_R$ and $D_R$. Evaluating the graphs in Fig.~1 we find a
contribution of the form
\begin{equation}
S_{\rm Dirac} = {1\over{6\pi}} \left[Y \ln\left({{M_U^2}\over{M_D^2}}\right) + 1
\right] ~~.
\label{SD}
\end{equation}
We see that the logarithmic term, which comes from the graph in Fig.~1a,
may have either sign depending on the masses and hypercharge of the particles.
However, it is well known that (in the absence of flavor-changing neutral
currents) the contribution of Dirac particles to $T$
(or $\Delta \rho_*$) is
positive semi-definite \cite{delrho}:
\begin{equation}
T_{\rm Dirac} = {1\over{16\pi s^2 M_W^2}} \left[
M_U^2+M_D^2-{{2M_U^2M_D^2}\over{M_U^2-M_D^2}}\ln\left({{M_U^2}\over{M_D^2}}
\right) \right] ~~.
\label{TD}
\end{equation}
Thus, although mass splittings may give rise to negative contributions to $S$,
they also give positive contributions to $T$. Since experiment favors a negative
value for $T$, and the top quark is also expected to give a positive
contribution
to $T$ (like that in Eq.~\ref{TD} multiplied by the number of colors),
producing a negative value only for $S$ seems unpromising.
A way around these difficulties is suggested by the result of Bertolini and
Sirlin \cite{Sirlin}. These authors found that Majorana fermions can
give negative contributions to $T$. Thus we turn to the calculation of $S$ for
this case. For completeness we will also present a somewhat simplified
discussion of the calculation of $T$.
We consider a model containing two Majorana (self-conjugate) fields
$N = \left(\begin{array}{c}N_1\\N_2\end{array}\right)$ with a mass term
\begin{equation}
{\cal L}_M =
-{1\over 2} \overline{N} \left(\begin {array}{cc}M_1 & 0 \\0 & M_2\end{array}
\right) N ~~.
\label{LM}
\end{equation}
The left-handed weak eigenstates are given by \cite{Sirlin}
\begin{equation}
\left(\begin {array}{c}\tilde{N}_L\\ \tilde{N}_R^c\end{array}\right)
= \left(\begin {array}{cc}ic_\theta & s_\theta \\-is_\theta & c_\theta
\end{array}\right) {1\over 2}(1-\gamma_5) N ~~,
\label{lhwe}
\end{equation}
where $\tilde{N}_R^c \equiv C(\tilde{N}_R)^T$ is a left-handed charge-conjugate
field, and
$c_\theta$ and $s_\theta$ denote
$\cos(\theta)$ and $\sin(\theta)$, with
\begin{eqnarray}
\tan(2\theta)&=&{{2 \sqrt{M_1 M_2}}\over{M_2 - M_1}} ~~, \\
\cos^2(\theta)&=&{M_2\over{M_1+M_2}} ~~,\nonumber \\
\sin^2(\theta)&=&{M_1\over{M_1+M_2}} ~~. \nonumber
\label{sc}
\end{eqnarray}
In terms of the $\tilde{N}$ fields, Eq.~\ref{LM} corresponds to a Dirac
mass equal to $\sqrt{M_1 M_2}$ for both left and right-handed weak eigenstates,
and a Majorana mass
equal to $M_2-M_1$ for the right-handed fields \cite{Sirlin}.
We are assuming that the right-handed field $\tilde{N}_R$ is an $SU(2)_L$
singlet, and that the left-handed field is in a doublet with a negatively
charged partner, $E$, with mass $M_E$,
i.e. $\left(\begin{array}{c}\tilde{N} \\ E \end{array}\right)
_L$. Our results can be easily modified for the case of a positively charged
partner, i.e. $\left(\begin{array}{c}E \\ \tilde{N} \end{array}\right)
_L$. We will not consider any mixing between the new heavy particles and the
light generations.
With these assumptions (and making use of the self-conjugacy
properties of the
Majorana fields $N$), the left-handed neutral current is
\begin{equation}
\overline{\tilde{N}_L}\gamma^\mu\tilde{N}_L =
-{{c_\theta^2}\over 2}\overline{N_1}\gamma^\mu\gamma_5N_1
-{{s_\theta^2}\over 2}\overline{N_2}\gamma^\mu\gamma_5N_2
+is_\theta c_\theta \overline{N_2}\gamma^\mu N_1 ~~.
\label{lhnc}
\end{equation}
Similarly, the charged current is
\begin{equation}
\overline{\tilde{N}_L}\gamma^\mu E_L + h.c. =
-ic_\theta\overline{N_{1L}}\gamma^\mu E_L
+s_\theta\overline{N_{2L}}\gamma^\mu E_L + h.c.
\label{lhcc}
\end{equation}
We can now calculate the $S$, $T$, and $U$ parameters for the case of
heavy Majorana fermions.
The Feynman rules for Majorana particles are well known
\cite{Gates} and will not be reviewed here.
It is convenient to express the gauge boson self energies in terms of
vacuum polarizations of left- and
right-handed currents. Using an ultra-violet cut-off $\Lambda$
we have \cite{Peskin}:
\begin{eqnarray}
\lefteqn{\Pi_{LL}({m_1}^2,{m_2}^2,q^2) = \Pi_{RR}({m_1}^2,{m_2}^2,q^2) =}
\label{PiLL} \\
& & {-{4\over{{(4\pi)}^2}}\int_0^1 dx \ln\left[{\Lambda^2\over{M^2-x(1-x)q^2}}
\right]\left(x(1-x)q^2 - {1\over 2}M^2\right),} \nonumber \\
& & \nonumber \\
\lefteqn{\Pi_{LR}({m_1}^2,{m_2}^2,q^2) =}
\label{PiLR} \\
& & {-{4\over{{(4\pi)}^2}}\int_0^1 dx \ln\left[{\Lambda^2\over{M^2-x(1-x)q^2}}
\right]\left({1\over 2}m_1m_2\right)} ~, \nonumber
\end{eqnarray}
where where $m_1$ and $m_2$ are the masses of the fermions in the loop, and
$M^2 = xm_1^2 + (1-x)m_2^2$.
Thus, the graphs in Fig.~1 with the appropriate combinations of mass
eigenstates going around the loops can be summarized as:
\begin{eqnarray}
\Pi_{3Y}(q^2) & = & {{c_\theta^4}\over 2}\left[\Pi_{LR}({M_1}^2,{M_1}^2,q^2)-
\Pi_{LL}({M_1}^2,{M_1}^2,q^2)\right]
\label{3Y} \\
& &\mbox{}+{{s_\theta^4}\over 2}\left[\Pi_{LR}({M_2}^2,{M_2}^2,q^2)-
\Pi_{LL}({M_2}^2,{M_2}^2,q^2)\right]\nonumber \\
& &\mbox{}- s_\theta^2c_\theta^2\left[\Pi_{LR}({M_1}^2,{M_2}^2,q^2)+
\Pi_{LL}({M_1}^2,{M_2}^2,q^2)\right]\nonumber \\
& &\mbox{}+\Pi_{LR}({M_E}^2,{M_E}^2,q^2)+{1\over 2}
\Pi_{LL}({M_E}^2,{M_E}^2,q^2) ~~. \nonumber
\end{eqnarray}
Using Eqs.~\ref{PiLL}, ~\ref{PiLR}, and \ref{3Y}
we find the contribution to $S$ is given by
\begin{eqnarray}
\lefteqn{S_{M} = } \label{SM} \\
& & {1\over{6\pi}} \left\{
\begin{array}{l}\raisebox{0ex}[1ex][3ex]{}
c_\theta^2\ln\left({{\displaystyle \raisebox{0ex}[1ex][.8ex]{} M_1^2}\over
{\displaystyle \raisebox{0ex}[2ex][1ex]{} M_E^2}}\right)
+
s_\theta^2\ln\left({{\displaystyle \raisebox{0ex}[1ex][.8ex]{} M_2^2}\over
{\displaystyle \raisebox{0ex}[2ex][1ex]{} M_E^2}}\right)
+ {{\displaystyle 3}\over {\displaystyle 2}}
\nonumber \\
\mbox{}\raisebox{0ex}[1ex][3ex]{}
- s_\theta^2c_\theta^2\left[ {{\displaystyle 8} \over {\displaystyle 3}}
+f_1(M_1,M_2) +
f_2(M_1,M_2)
\ln\left({{\displaystyle \raisebox{0ex}[1ex][.8ex]{}
M_1^2}\over{\displaystyle \raisebox{0ex}[1.8ex][1ex]{} M_2^2}}\right) \right]
\end{array}
\right\}
\nonumber ~~,
\end{eqnarray}
where
\begin{eqnarray}
f_1(M_1,M_2) & = &
{{3M_1M_2^3+3M_1^3M_2-4M_1^2M_2^2}\over{\left(M_1^2-M_2^2\right)^2}} ~~,
\label{f1f2}\\
f_2(M_1,M_2) & = &
{{M_1^6-3M_1^4M_2^2+6M_1^3M_2^3-3M_1^2M_2^4+M_2^6}\over
{\left(M_1^2-M_2^2\right)^3}} ~~. \nonumber
\end{eqnarray}
The calculation of $T$ is summarized by
\begin{eqnarray}
\Pi_{11}(0)-\Pi_{33}(0) & = &
{{c_\theta^2}\over 2}\Pi_{LL}({M_1}^2,{M_E}^2,0)
+{{s_\theta^2}\over 2}\Pi_{LL}({M_2}^2,{M_E}^2,0)
\label{11m33} \\
& &\mbox{}+{{c_\theta^4}\over 4}\left[\Pi_{LR}({M_1}^2,{M_1}^2,0)-
\Pi_{LL}({M_1}^2,{M_1}^2,0)\right] \nonumber \\
& &\mbox{}+{{s_\theta^4}\over 4}\left[\Pi_{LR}({M_2}^2,{M_2}^2,0)-
\Pi_{LL}({M_2}^2,{M_2}^2,0)\right]\nonumber \\
& &\mbox{}- {{s_\theta^2c_\theta^2}\over 2}\left[\Pi_{LR}({M_1}^2,{M_2}^2,0)+
\Pi_{LL}({M_1}^2,{M_2}^2,0)\right]\nonumber \\
& &\mbox{}-{1\over 2}\Pi_{LL}({M_E}^2,{M_E}^2,0) ~~. \nonumber
\end{eqnarray}
For the contribution to $T$ we find
\begin{eqnarray}
\lefteqn{T_{M} =}
\label{TM} \\
& & {1\over{16\pi s^2 M_W^2}} \left\{ \begin{array}{l}
\raisebox{0ex}[1ex][3ex]{}c_\theta^2\left[
M_1^2+M_E^2-{{\displaystyle \raisebox{0ex}[1ex][.8ex]{} 2M_1^2M_E^2}\over
{\displaystyle \raisebox{0ex}[1.1ex][1ex]{} M_1^2-M_E^2}}
\ln\left({{\displaystyle \raisebox{0ex}[1ex][.8ex]{}
M_1^2}\over{\displaystyle \raisebox{0ex}[1.1ex][1ex]{}M_E^2}}\right)\right]
\nonumber \\
\mbox{}+\raisebox{0ex}[1ex][3ex]{} s_\theta^2\left[
M_2^2+M_E^2-{{\displaystyle \raisebox{0ex}[1ex][.8ex]{} 2M_2^2M_E^2}\over
{\displaystyle \raisebox{0ex}[1.1ex][1ex]{} M_2^2-M_E^2}}
\ln\left({{\displaystyle \raisebox{0ex}[1ex][.8ex]{} M_2^2}\over
{\displaystyle \raisebox{0ex}[1.1ex][1ex]{} M_E^2}}\right)\right]
\\
\mbox{} -\raisebox{0ex}[1ex][3ex]{} s_\theta^2c_\theta^2
\left[\begin{array}{l} \raisebox{0ex}[1ex][2ex]{}
M_1^2+M_2^2 - 4M_1M_2 \nonumber \\ \raisebox{0ex}[1ex][3ex]{}
\mbox{} +2{{\displaystyle \raisebox{0ex}[1ex][.8ex]{}
M_1^3M_2-M_1^2M_2^2+M_1M_2^3}\over
{\displaystyle \raisebox{0ex}[1.1ex][1ex]{} M_1^2-M_2^2}}
\ln\left({{\displaystyle \raisebox{0ex}[1ex][.8ex]{} M_1^2}\over
{\displaystyle \raisebox{0ex}[1.1ex][1ex]{} M_2^2}}\right)
\end{array}\right]\end{array} \right\} ~~.
\end{eqnarray}
It is not surprising that the contribution to $T$ can be
negative, since Eq.~\ref{lhnc} contains a mass-eigenstate-changing neutral
current.
For completeness we also present the contribution to $U$:
\pagebreak
\begin{eqnarray}
\lefteqn{U_{M} = }
\label{UM} \\
& & {1\over{6\pi}} \left\{
\begin{array}{l}\raisebox{0ex}[1ex][3ex]{} c_\theta^2\left[
f_3(M_1,M_E)\ln\left({{\displaystyle \raisebox{0ex}[1.1ex][.8ex]{} M_1^2}\over
{\displaystyle \raisebox{0ex}[1ex][.8ex]{} M_E^2}}\right)
+{{\displaystyle \raisebox{0ex}[1ex][1.1ex]{}4M_1^2M_E^2}\over
{\left(\displaystyle \raisebox{0ex}[1.1ex][1ex]{} M_1^2-M_E^2\right)
\displaystyle^2}} \right] \\
\mbox{}\raisebox{0ex}[1ex][3ex]{} + s_\theta^2\left[
f_3(M_2,M_E)\ln\left({{\displaystyle \raisebox{0ex}[1ex][.8ex]{} M_2^2}\over
{\displaystyle \raisebox{0ex}[1.1ex][1ex]{} M_E^2}}\right)
+ {{\displaystyle \raisebox{0ex}[1ex][.8ex]{} 4M_2^2M_E^2}\over
{\left(\displaystyle \raisebox{0ex}[1.1ex][1ex]{} M_2^2-M_E^2\right)
\displaystyle^2}}
\right] - {{\displaystyle 13}\over{ \displaystyle 6}}
\nonumber \\
\mbox{}\raisebox{0ex}[1ex][3ex]{}
+ s_\theta^2c_\theta^2\left[\raisebox{0ex}[5ex][2ex]{}
{{\displaystyle 8} \over{\displaystyle 3}}+f_1(M_1,M_2) -
f_2(M_1,M_2)
\ln\left({{\displaystyle \raisebox{0ex}[1ex][.8ex]{} M_1^2}\over
{\displaystyle \raisebox{0ex}[1.1ex][1ex]{} M_2^2}}\right) \right]
\end{array}
\right\}
\nonumber ~~,
\end{eqnarray}
where
\begin{equation}
f_3(M_1,M_E) = {{M_1^6-3M_1^4M_E^2-3M_1^2M_E^4+M_E^6}\over
{\left(M_1^2-M_E^2\right)^3}}
~~.
\label{f3}
\end{equation}
It can easily be checked that Eqs.~\ref{SM} and \ref{TM} reduce to
Eqs.~\ref{SD} and \ref{TD} in the limit that $M_1 \rightarrow M_2$.
In order to simplify the discussion of our results we have plotted the curves
$S=0$ and $T=0$ (lines (a) and (b) respectively) in the $M_1$, $M_2-M_1$ plane
in Fig.~2. The region where $S<0$ lies to the left of curve (a), and the
region $T<0$ lies to the right of and above curve (b).
The region $U<0$ is not shown, but it lies entirely in the $T<0$ region
and does not overlap with the $S<0$ region.
Note that the familiar
case of Dirac fermions lies along the bottom edge of the graph where $M_1=M_2$.
For the Dirac case we see (as expected) that $S > 0$ for
$M_1/M_E > e^{-1/2}$ (where $e$ here is the base of the natural logarithm), and
$T \geq 0$ for all values of the masses. (The region $M_2