text:
\section{Introduction}
\label{sec:intro}

The study of possible modifications of hadron properties in the nuclear medium
is one of the central problems of contemporary nuclear physics. In principle,
these and related phenomena in nuclear physics are governed by the
fundamental theory of the strong interactions, quantum chromodynamics (QCD).
However, although QCD has been very successful in explaining a large
class of hadronic processes at high energy and large momentum transfer, typical
nuclear phenomena at lower energies cannot be derived from QCD with the
theoretical tools presently available. The difficulty of using QCD for
phenomena at the nuclear scale is related to the nonperturbative nature
of these. Due to the asymptotic freedom property of QCD, high energy
processes are calculable by perturbative techniques in the quark-gluon
coupling constant. On the other hand, since there are no reliable systematic
approximation schemes in field theory for performing nonperturbative
calculations, the construction of models is an important aspect of low energy
QCD. While there is considerable optimism that eventually one will be able to
solve QCD numerically on the lattice using supercomputers, the development of
analytical approximation methods are in urgent need to make contact with the
wealth of data on nonperturbative phenomena, presently available, or that will
be available when the new experimental facilities under construction start
operating. The $\delta$ expansion~\cite{original} is an example of a method
recently developed with the aim of studying nonperturbative phenomena in field
theory.

The idea of the $\delta$ expansion is to perturb the original theory by the
introduction of an artificial expansion parameter $\delta$, absent in the
original theory. The parameter $\delta$ is introduced in such a way that it
interpolates between the theory one wants to solve and another theory that
one knows how to solve. The $\delta$ expansion can be formulated in two
different forms, the logarithmic $\delta$ expansion~\cite{original} and the
linear $\delta$ expansion~\cite{linear}-\cite{su}. In this paper we consider
the linear form. Specifically, let $S$ be the action of the theory one wants to
solve, and $S_0$ the action of the soluble theory. Then, the interpolating
action $S(\delta)$ is defined as
\begin{equation}
S(\delta)=(1-\delta)S_0 + S \delta ,
\label{sdelta}
\end{equation}
so that $S(0)=S_0$ and $S(1)=S$. The next step involves the evaluation of
desired physical quantities as a perturbation series in powers of $\delta$,
which is then set equal to $1$ at the end. A crucial aspect of the method is
the recognition that $S_0$ involves arbitrary unknown (dimensionful and/or
dimensionless) parameters. If one were able to solve the new theory to all
orders in $\delta$, the unknown parameters would not play any role, since no
physical observable would depend on them. However, since we will be able to
solve the interpolated theory only to a finite order in $\delta$, there will
remain a residual dependence of the results on the parameters of $S_0$. These
arbitrary parameters must therefore be determined according to some criterion
and in fact there are many ways in which this can be
done~\cite{linear}-\cite{QM}. One physically appealing way to fix the unknown
parameters, which is the one adopted here, is the principle of minimal
sensitivity (PMS) introduced in Ref.~\cite{PMS}. This principle amounts to the
requirement that physical quantities should be at least locally independent of
the parameters. In the original applications of the method, the unknown
parameters were set to be equal to unity. The $\delta$ expansion, together
with the criterion of the PMS of physical observables, is known as the
{\em optimized} $\delta$ {\em expansion}. The convergence of the optimized
$\delta$ expansion has been proved in Ref.~\cite{ian}.

The different forms of the $\delta$ expansion have been successfully applied
to many different problems in quantum mechanics \cite{QM}, particle
theory~\cite{part,mar}, statistical physics~\cite{stat} and lattice
field theory~\cite{du,hugh}. Motivated by these successes, in this
paper we investigate the possibility of employing the linear $\delta$
expansion to study medium effects in hadron and nuclear physics using
typical field theoretic models. In the next section, we consider the use
of the $\delta$ expansion  in the study of density effects on the chiral
symmetry breaking in the Nambu--Jona-Lasinio (NJL) model. The PMS criterion
for the typical chiral quantities is applied following previous experience
with the method. We also investigate an alternative way of fixing the
parameters by applying the PMS to the energy density of the system. In
Section~\ref{sec:Walecka} we consider the Walecka model and study the effective
nucleon mass in nuclear matter. Both problems can be treated by obtaining the
propagators of the fields involved. Traditionally, the propagators are obtained
in the Hartree-Fock (HF) approximation. In the case of the NJL model this
approximation amounts to neglecting the corrections to the four-point vertex.
In the Walecka model, the HF approximation consists in obtaining self-
consistently the nucleon propagator with bare meson propagators, and neglecting
corrections to the meson-nucleon vertices. An additional approximation in the
Walecka model is the neglection of vacuum effects in the nucleon propagator.
With the purpose of comparing the $\delta$ expansion method with the
traditional HF approximation, we also neglect vertex corrections in both
models, and neglect the vacuum in the Walecka model. Conclusions and
discussions of the perspectives for future calculations appear in
Section~\ref{sec:concl}.

\section{Nambu--Jona-Lasinio model}
\label{sec:NJL}

In the limit of zero current quark masses, the two-flavor Lagrangian
density of the Nambu--Jona-Lasinio (NJL) model~\cite{njl} is given by
\begin{equation}
{\cal L}_{\rm NJL} = \bar q(i \partial \raise.3ex \hbox {\kern -.55 em/})q+G\left[\left(\bar q q\right)^2-
\left(\bar q\gamma_5\mbox{\boldmath$\tau$} q\right)^2\right]\;,
\label{NJLag}
\end{equation}
where the quark field operators $q=q(x)$ represent the doublet of $u$ and $d$
quarks. Since the model is non-renormalizable, one has a cutoff $\Lambda$ as
an extra parameter of the model, besides $G$.

According to Eq.~(\ref{sdelta}), one needs to introduce a Lagrangian
density ${\cal L}_0$ such that
\begin{equation}
{\cal L}_{\rm NJL}(\delta)=(1-\delta){\cal L}_0+\delta {\cal L}_{\rm NJL}\;,
\label{NJLinterp}
\end{equation}
where the equations of motion derived from ${\cal L}_0$ can be solved as
exactly as possible. Since we are looking for solutions which break chiral
symmetry, the natural choice for ${\cal L}_0$ is
\begin{equation}
{\cal L}_0=\bar q (i \partial \raise.3ex \hbox {\kern -.55 em/} - \mu)q\;,
\label{L0}
\end{equation}
where $\mu$ is an arbitrary mass parameter introduced for dimensional reasons.
Therefore, the interpolated NJL Lagrangian density can be written as
\begin{eqnarray}
{\cal L}_{\rm NJL}(\delta)&=&(1-\delta)\left[\bar q (i \partial \raise.3ex \hbox {\kern -.55 em/}-\mu)q\right]+
\delta \left\{\bar q (i \partial \raise.3ex \hbox {\kern -.55 em/})q+G\left[\left(\bar q q\right)^2-
\left(\bar q\gamma_5\mbox{\boldmath$\tau$} q\right)^2\right]\right\} \nonumber \\
&=&\bar q (i \partial \raise.3ex \hbox {\kern -.55 em/} - \mu) q+\delta\left\{G\left[\left(\bar q q\right)^2-
\left(\bar q\gamma_5\mbox{\boldmath$\tau$} q\right)^2\right]+\mu \bar q q\right\}\;.
\label{Linterp}
\end{eqnarray}

The evaluation of physical quantities is performed using perturbation theory
in the parameter $\delta$. The physical quantities of interest, whose values
characterize the chiral symmetry breaking, are the constituent quark mass
$M_q$, the quark condensate $<\bar q q>$ and the pion decay constant $f_\pi$.
The quark condensate for a given flavor is given by
\begin{equation}
<\bar q q>=-i \int \frac{d^4 p}{(2\pi)^4} {\rm Tr}\left[S(p)\right]\;,
\label{qbq}
\end{equation}
where the trace is over spinor and color indices. The pion decay
constant can be evaluated using the Pagels-Stokar~\cite{PS} formula:
\begin{equation}
if_{\pi}q_{\mu}\delta^{ab} = g_{\pi qq} \int \frac{d^4p}{(2\pi)^4}
{\rm Tr}
\left[\gamma_{\mu}\gamma_5 \tau^a S(q)\gamma_5 \frac{\tau^b}{2}
S(q+p)\right]\;,
\label{fpi}
\end{equation}
where the trace now is over spinor, flavor and color indices. The quark-pion
coupling is obtained from the Goldberger-Treiman relation.

In order to calculate these quantities one needs the quark propagator $S(p)$,
which can be obtained using Dyson's equation. Expressed in terms of the
self-energy $\Sigma(p)$ the quark propagator reads
\begin{equation}
S^{-1}(p)=S^{-1}_0(p)-\Sigma(p),
\label{Dyson}
\end{equation}
where $S^{-1}_0(p)$ is the inverse of the quark propagator corresponding to
${\cal L}_0$:
\begin{equation}
S^{-1}_0(p)={\not\!p}-\mu.
\label{freeS}
\end{equation}
$\Sigma(p)$ is calculated as a power series in $\delta$, and $S(p)$ is then
obtained by inverting Eq.~(\ref{Dyson}). Since the self-energy is calculated
perturbatively, this is still a perturbative scheme. It is the application of
the PMS to physical quantities which furnishes the nonperturbative character
to the $\delta$ expansion.

In zeroth order in $\delta$, one has
\begin{equation}
\Sigma^{(0)}(p)=0\;.
\label{Sigma0}
\end{equation}
Therefore, one can invert Eq.~(\ref{Dyson}) by using the well-known modified
Feynman $i\epsilon$ prescription for particles in a Fermi sea~\cite{hs}.
The in-medium quark propagator is then given by:
\begin{equation}
S^0(p)=\frac{{\not\!p}+\mu}{p^2-\mu^2+i\epsilon}+2\pi i
\frac{{\not\!p}+\mu}
{2E_0(p)}\delta\left(p^0-E_0(p)\right)\theta\left(P_F-\absp
\right)\;,
\label{MediumS0}
\end{equation}
where $E_0(p)=\left(\bp^2+\mu^2\right)^{{1 \over 2}}$, and $P_F$ is the Fermi
momentum, which relates to the density $\rho$ via $P_F=(\pi^2 \rho/2)
^{1/3}$.

At this zeroth order in $\delta$, no dynamical content from the model has been
used. The dynamics of the model starts to show up at order $\delta$. At first
order in $\delta$, the self-energy $\Sigma(p)$ is given by:
\begin{eqnarray}
\Sigma^{(1)}(p)=&-&\delta \mu\nonumber\\
&+& 2i\delta G \int\frac{d^4q}{(2\pi)^4}\left\{{\rm Tr}\left[S_0(q)
\right]
- S_0(q)-\gamma_5 \tau^a {\rm Tr}\left[\tau^a S_0(q)\gamma_5\right] +
\gamma_5 \tau^a S_0(q) \tau^a \gamma_5 \right\}\;,
\label{SDENJL}
\end{eqnarray}
where a sum over the isospin index $a$ is implied.

Substituting Eq.~(\ref{MediumS0}) into this equation, we obtain for
$\Sigma^{(1)}$ the following expression:
\begin{equation}
\Sigma^{(1)}(p)=- \delta \mu+M_1-\gamma_0 \Sigma_0\;,
\label{sigma1}
\end{equation}
where
\begin{equation}
M_1=\delta{G\over \pi^2}\mu\left( N_cN_f+{1\over 2} \right )\left
\{ \Lambda\left(\Lambda^2+
\mu^2\right)^{{1 \over 2}}-P_F\left(P_F^2+\mu^2\right)^{{1 \over 2}} -\mu^2
\ln\left[{\Lambda + \left(\Lambda^2+\mu^2\right)^{{1 \over 2}} \over P_F +
\left(P_F^2+\mu^2\right)^{{1 \over 2}}}\right]\right\}\;,
\label{defM1}
\end{equation}
and
\begin{equation}
\Sigma_0= - 4 \delta G \int \frac{d^3q}{(2\pi)^3}\theta(P_F-\absq) .
\label{defSig0}
\end{equation}
Since the effect of $\Sigma_0$ is just to shift the chemical
potential~\cite{klev} one may write the constituent quark mass to $O(\delta)$
as
\begin{equation}
M_q= \mu - \delta \mu + M_1 .
\label{defMq1}
\end{equation}

Substituting Eq.~(\ref{MediumS0}) into Eq.~(\ref{qbq}) and Eq.~(\ref{fpi}), one
gets for the order parameter per flavor and for the pion decay constant the
following lowest order expressions:
\begin{equation}
<\bar q q>=-{2N_c\mu \over (2 \pi)^2}\left \{ \Lambda\left(\Lambda^2+
\mu^2\right)^{{1 \over 2}}-P_F\left(P_F^2+\mu^2\right)^{{1 \over 2}} -\mu^2
\ln\left[{\Lambda + \left(\Lambda^2+\mu^2\right)^{{1 \over 2}}
\over P_F +
\left(P_F^2+\mu^2\right)^{{1 \over 2}}}\right]\right\}\;,
\label{qbq0}
\end{equation}
and
\begin{equation}
f_{\pi}^2={N_cN_f\mu^2 \over (2 \pi)^2}\left \{ \ln \left[
{\Lambda + \left(\Lambda^2+\mu^2\right)^{{1 \over 2}} \over
P_F+(P_F^2+\mu^2)^{{1 \over 2}} } \right ] - \left( 1 + {\mu^2
\over \Lambda^2}\right )^{-{1 \over 2}} +\left(1+{\mu^2 \over P_F^2}
\right)^{-{1 \over 2}} \right\}\;,
\label{fpi0}
\end{equation}
where the lowest order Goldberger-Treiman relation,
$g_{\pi qq}=\mu/f_{\pi}$,
has been used.

The next step in the process is to fix $\mu$. In a previous work~\cite{mar}
this arbitrary parameter was determined by requiring $f_{\pi}$, which is
a fundamental quantity in the study of chiral symmetry breaking, to be
stationary with respect to variations in $\mu$. This is also a convenient
choice since, apart from having a well known empirical value, $f_{\pi}$ is the
only one of the studied quantities in Ref.~\cite{mar} which has a well defined
stationary point for finite values of $\mu$. To fix the noncovariant cutoff
$\Lambda$ one uses the empirical value $f_{\pi}$ at zero density. If one
applies the criterion of stationarity to $f_\pi$ in the vacuum, one obtains
(for $N_c=3$ and $N_f=2$)
\begin{equation}
\mu = 0.97 \times \Lambda\;.
\end{equation}
With the input $f_{\pi}= 93~~{\rm MeV}$, one finds $\Lambda=571$ MeV,
which implies $\mu=553$ MeV and $<\bar q q>= - (250~~{\rm MeV})^3$. Using
$G\Lambda^2=2.89$ as in Ref.~\cite{mar} and setting $\delta=1$ in
Eq.~(\ref{defMq1}) one finds the constituent quark mass to $O(\delta)$ to be
$M_q=574$ MeV. Figure 1 shows $\mu$ as a function of the density and has been
obtained by applying the PMS to $f_\pi$ for different values of $P_F$.
The results obtained for $M_q$ and $-<\bar q q>^{1/3}$, for different values
of $P_F$ are shown in Figure 2 (solid and dashed lines respectively). In Figure
3 the solid line shows the $P_F$ dependence of $f_{\pi}$. Contrary to what
happens at finite temperature~\cite{mar} we find that our result are
insensitive to whether $\mu(P_F)$ (as in Figure 1) or
$\mu(0) = 533~~{\rm MeV}$ is used.

A natural question which arises at this point is the uniqueness of the
of the value of $\mu$. If one were to use other physical quantities to fix
$\mu$, as for example the masses of the vector mesons, it is very  likely
that one would obtain a different value for $\mu$. In such a case, one
would have a different quark propagator for each observable. Of course this
would not be catastrophic if the spread of the values of $\mu$ determined with
different observables is not too large. In order to avoid such potential
uncertainties, we propose to fix $\mu$ by demanding that the energy density of
the system be stationary with respect to variations of $\mu$. The energy
density can always be written in terms of the propagators of the theory, and
then it is natural to demand stationarity of the energy with respect to the
unknown parameters of the propagators. Then, all physical observables are
determined from the same quark propagator.

 From the Lagrangian density, Eq.~(\ref{NJLag}), we have that the
energy-momentum
tensor is given by:
\begin{eqnarray}
T^{\mu\nu}_{\rm NJL}&=&i\bar q \gamma^{\mu} \partial^\nu q
-g^{\mu\nu}{\cal L}_{\rm NJL}\nonumber\\
&=i&\bar q \gamma^{\mu} \partial^\nu q
-g^{\mu\nu}\left\{\bar q(i \partial \raise.3ex \hbox {\kern -.55 em/})q + G\left[\left(\bar q q\right)^2-
\left(\bar q\gamma_5\mbox{\boldmath$\tau$} q\right)^2\right]\right\}\;.
\label{Tmunu}
\end{eqnarray}
Note that we have not used the equation of motion for the quark field operator.
The energy density is the volume integral of the expectation value of
$T^{00}$ in the many-quark state. The expectation value of the field operators
can be evaluated using the usual Wick contraction technique. This leads
to
\begin{eqnarray}
{\cal E_{\rm NJL}}&=& \frac{1}{V}\int d^3x\;<T^{00}>\nonumber\\
&=&-i\int\frac{d^4q}{(2\pi)^4}q^0{\rm Tr}\left[\gamma^0S(q)\right]+
i\int\frac{d^4q}{(2\pi)^4}{\rm Tr}\left[{\not\!q}S(q)\right]-
G\left\{-\left[\int\frac{d^4q}{(2\pi)^4}{\rm Tr}\left[S(q)\right]
\right]^2 \right. \nonumber\\
&&+\int\frac{d^4q}{(2\pi)^4}\frac{d^4k}{(2\pi)^4}{\rm Tr}
\left[S(q)S(k) \right] +\left[\int\frac{d^4q}{(2\pi)^4}{\rm Tr}
\left[\mbox{\boldmath$\tau$}\gamma_5S(q)\right]\right]^2\nonumber\\
&&-\left. \int\frac{d^4q}{(2\pi)^4}\frac{d^4k}{(2\pi)^4}{\rm Tr}
\left[\gamma_5\tau^aS(q)\gamma_5\tau^a S(k)\right]\right\}\;.
\label{ENJL}
\end{eqnarray}

Substituting Eq.~(\ref{MediumS0}) into the expression above, we obtain:
\begin{equation}
{\cal E}^{(0)}_{\rm NJL}=-2N_cN_f\int_{P_F}^\Lambda
\frac{d^3q}{(2\pi)^3}\frac{\bq^2}{E_0(q)}-
2GN_cN_f(2N_cN_f+1) \left[\int_{P_F}^\Lambda
\frac{d^3q}{(2\pi)^3} \frac{\mu}{E_0(q)}
\right]^2\;.
\label{ENJLM0}
\end{equation}

The requirement that ${\cal E}$ be stationary with respect to variations in
$\mu$ leads to
\begin{equation}
\mu=4G\left(N_cN_f+\frac{1}{2}\right) \int_{P_F}^\Lambda \frac{d^3q}{(2\pi)^3}
\frac{\mu}{E_0(q)}\;.
\label{mustat}
\end{equation}
This is is the familiar Hartree-Fock gap equation of the model, where
$\mu$ has the interpretation of the dynamically generated mass.

Next, we consider the first order self-energy. By inverting Dyson's equation,
Eq.~(\ref{Dyson}), one obtains the quark propagator:
\begin{equation}
S^{(1)}(p)=\frac{{\not\!p}_1+M_1}{p^{2}_1-M^{2}_1+i\epsilon}+
2\pi i\frac{{\not\!p}_1+M_1}
{2E_1(p)}\delta\left(p^{0}_1-E_1(p)\right)\theta\left(P_F-\absp
\right)\;,
\label{MediumS1}
\end{equation}
where
\begin{eqnarray}
&&p^{\mu}_1=(p^{0}_1, {\bp})=(p^0+\Sigma_0, \bp)\;,\\
&&E_1(p)=\left[\bp^2 + (M_1)^2 \right]^{{1 \over 2}}\;.
\label{aux1}
\end{eqnarray}
The superscript $(1)$ in $S^{(1)}$ indicates that the propagator has been
obtained with a self-energy calculated to first order in $\delta$. Note that
we are not expanding the propagator in powers of $\delta$. The process of
obtaining the propagator by inverting Dyson's equation with a self-energy
calculated in perturbation theory is known as the chain approximation.

Substituting Eq.~(\ref{MediumS1}) into Eq.~(\ref{ENJL}), we obtain:
\begin{equation}
{\cal E}^{(1)}_{\rm NJL}=-2N_cN_f\int_{P_F}^\Lambda
\frac{d^3q}{(2\pi)^3}\frac{\bq^2}{E_1(q)}-
2GN_cN_f(2N_cN_f+1) \left[\int_{P_F}^\Lambda
\frac{d^3q}{(2\pi)^3} \frac{M_1}
{E_1(q)} \right]^2\;,
\label{ENJLM1}
\end{equation}
where $M_1$ is given by Eq.~(\ref{defM1}) and $E_1(q)$ is defined in
Eq.~(\ref{aux1}).

Application of the PMS to ${\cal E}^{(1)}_{\rm NJL}$,
\begin{equation}
\frac{d{\cal E}^{(1)}_{\rm NJL}}{d\mu}=\frac{d{\cal E}^{(1)}_
{\rm NJL}}{dM_1}
\frac{dM_1}{d\mu}=0\;,
\end{equation}
leads to
\begin{equation}
M_1=4G\left(N_cN_f+\frac{1}{2}\right) \int_{P_F}^\Lambda
\frac{d^3q}{(2\pi)^3}
\frac{M_1}{E_1(q)}\;.
\label{M1stat}
\end{equation}
Again, we have obtained the familiar Hartree-Fock gap equation
for the dynamically generated mass.

If one proceeds to higher orders in $\delta$, in the scheme of neglecting
vertex corrections, the higher order quark propagator will always be of the
form of Eq.~(\ref{MediumS1}), with $M_1$ replaced by another constant, say $M$,
which is a function of $\mu$. However, because of the PMS condition on
${\cal E}$, $M$ at each order will always be given by the same value.
This value is the one that satisfies the usual gap equation:
\begin{equation}
M=4G\left(N_cN_f+\frac{1}{2}\right) \int_{P_F}^\Lambda
 \frac{d^3q}{(2\pi)^3}
\frac{M}{E(q)}\;,
\label{Mstat}
\end{equation}
where
\begin{equation}
E(q)=\left(\bq^2 + M^2\right)^{{1 \over 2}}\;.
\label{E}
\end{equation}
Therefore, the PMS condition on the energy density is equivalent
to the usual Hartree-Fock solution for the dynamically generated mass, in the
approximation of neglecting vertex corrections.

This result should be compared to the one presented in Ref.~\cite{su} where,
in the context of the effective potential, it was found that the $\delta$
expansion and the $1/N$ expansion are identical in the large $N$ limit.

In Figure 2 we compare the results obtained for the quark mass and
the quark condensate, when the two above described ways of applying
the PMS are used. We call PMS1 the results obtained by imposing $f_\pi$
to be stationary with respect to $\mu$, and PMS2 the results obtained
when the PMS is imposed to the energy density. The solid and dashed
lines give, respectively, $M_q$ and $-<\bar q q>^{1/3}$ obtained with PMS1,
and the dotted and dot-dashed lines give, respectively, $M_q$ and
$-<\bar q q>^{1/3}$ obtained with PMS2. In both cases we used the same set of
parameters: $\Lambda=571$ MeV and $G\Lambda^2=2.89$.

In Figure 3 we show the results obtained for $f_\pi$ with PMS1 (solid
line), PMS2 (dotted line), both with the parameters given above, and with
PMS2 (dashed line) with a new set of parameters: $\Lambda=653$ MeV and
$G\Lambda^2=1.98$. The last set of parameters was fixed by renormalizing
$f_\pi$ and $<\bar q q>$ at $P_F=0$ to their experimental values and requiring
$M_q$ to be roughly one-third of the nucleon mass. The curves for the
quark mass and condensate obtained by PMS2 with the renormalized parameters
are not shown because their behavior is analogous to $f_\pi$: they go to zero
at $P_F= 1.6$ fm$^{-1}$. At $P_F=0$ we have $M_q=314$ MeV. These are the usual
HF results and the renormalized parameters are the same as used in
Ref.~\cite{klev} for two flavors and a three momentum cutoff.

 From these figures we see that the results change appreciably when different
criteria are used. The main difference is related with the density
dependence of the quantities: while with PMS1 the quantities smoothly approach
zero at some critical density, they go to zero through a first order phase
transition with PMS2 (or Hartree-Fock)~\cite{klev}.

\section{Walecka model}
\label{sec:Walecka}

In this section we consider the Walecka model~\cite{wal} for nuclear matter.
The Lagrangian density of the model is given by
\begin{equation}
{\cal L}_{\rm W}=\bar \psi\left[\gamma_\mu(i\partial^{\mu}-g_\omega V^{\mu})-
(M-g_\sigma\phi)\right]\psi+\frac{1}{2}(\partial_{\mu}\phi\partial^{\mu}
\phi-m_\sigma^2 \phi^2)-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\frac{1}{2}
m_\omega^2V_{\mu}V^{\mu}\;,
\label{LW}
\end{equation}
where $\psi$ represents the nucleon field operators, $\phi$ and $V_{\mu}$ are
respectively the field operators of the scalar and vector mesons, and
$F_{\mu\nu}=\partial_{\mu}V_{\nu}-\partial_{\nu}V_{\mu}$.

The energy-momentum tensor density corresponding to this Lagrangian density is
given by:
\begin{equation}
T^{\mu\nu}_{\rm W}=i\bar\psi\gamma^\mu \partial^\nu\psi+\partial^
\mu\phi
\partial^\nu\phi+\partial^\nu V_{\lambda}F^{\lambda\mu}-
g^{\mu\nu}{\cal L}_{\rm W}\;.
\label{TmunuW}
\end{equation}
Note that we have not used the nucleon equation of motion. Next, we eliminate
the meson field operators in favor of the nucleon field operators. The
Euler-Lagrange equations yield the meson field equations:
\begin{eqnarray}
\left(\partial_\mu\partial^\mu+m_\sigma^2\right)\phi&=&g_\sigma
\bar\psi\psi\;,\label{eqsmes}
\\
\left(\partial_\mu\partial^\mu+m_\omega^2\right)V^\nu&=&
g_\omega\bar\psi\gamma^\nu\psi\;.
\label{eqsmev}
\end{eqnarray}
In obtaining the second equation above we have used baryon current
conservation, which implies that $\partial_\mu V^\mu=0$. These
equations can formally be integrated as:
\begin{eqnarray}
\phi(x)&=&-g_\sigma\int d^4y\Delta_\sigma(x-y)\bar\psi(y)\psi(y)\;,
\label{phisol}
\\
V_{\mu}(x)&=&-g_\omega\int d^4y\Delta_\omega(x-y)\bar\psi(y)
\gamma_\mu\psi(y)\;,
\label{omegsol}
\end{eqnarray}
where $\Delta_i(x)$, $i=\sigma, \omega$, is given by:
\begin{equation}
\Delta_i(x)=\int \frac{d^4q}{(2\pi)^4}\frac{1}{q^2-m_i^2+i\epsilon}e^{-iqx} \;.
\label{mesprop}
\end{equation}
In Eq.~(\ref{omegsol}) above, because of baryon current conservation, we
have neglected the term proportional to $p^\mu p^\nu/m_\omega^2$ in the vector
meson propagator.

Using the expressions above for $\phi$ and $V^{\mu}$ in Eq.~(\ref{TmunuW}),
taking the expectation value of the resulting expression in the
many-nucleon state, and evaluating this with the help of Wick's contraction
technique, we obtain:
\begin{equation}
\langle T^{\mu\nu} \rangle=-i\int\frac{d^4q}{(2\pi)^4}\left\{{\rm Tr}
\left[\gamma^\mu q^\nu-g^{\mu\nu}({\not\!q}-M)\right]S(q)\right\}+
\langle T^{\mu\nu} \rangle_\sigma+\langle T^{\mu\nu} \rangle_\omega\;,
\end{equation}
with
\begin{eqnarray}
\langle T^{\mu\nu} \rangle_\sigma &=&\frac{1}{2}\frac{g_\sigma^2}{m_\sigma^2}
\left[\int\frac{d^4q}{(2\pi)^4}{\rm Tr}S(q)\right]^2 g^{\mu\nu}-g_\sigma^2
\int\frac{d^4q}{(2\pi)^4} \frac{d^4k}{(2\pi)^4}{\rm Tr}\left[S(q+k)S(k)\right]
\Delta_\sigma(q^2) \nonumber\\
&\times&\left\{ \left[\frac{1}{2}(q^2-m_\sigma^2)\Delta_\sigma(q^2)-1\right]
g^{\mu\nu}-q^{\mu}q^{\nu}\Delta_\sigma(q^2)\right\},
\label{Tsigma}
\end{eqnarray}
and
\begin{eqnarray}
\langle T^{\mu\nu} \rangle_\omega&=&
- \frac{1}{2}\frac{g_\omega^2}{m_\omega^2}\left[\int\frac{d^4q}{(2\pi)^4}
{\rm Tr}\gamma_\mu S(q)\right]\left[\int\frac{d^4q}{(2\pi)^4}{\rm Tr}
\gamma^\mu S(q)\right] g^{\mu\nu}\nonumber\\
&+& g_\omega^2\int\frac{d^4q}{(2\pi)^4}\frac{d^4k}{(2\pi)^4}
{\rm Tr}\left[\gamma_\lambda S(q+k)\gamma^\lambda S(k)\right]
\Delta_\omega(q^2)\nonumber\\
&\times&\left\{
\left[\frac{1}{2}(q^2-m_\omega^2)\Delta_\omega(q^2)-1\right]g^{\mu\nu}
- q^{\mu}q^{\nu}\Delta_\omega(q^2)\right\}\;.
\label{Tomega}
\end{eqnarray}

In the same way as in the NJL model, the nucleon propagator is obtained by
inverting Dyson's equation
\begin{equation}
S^{-1}(p)=S_0^{-1}(p)-\Sigma(p)\;,
\label{DysonN}
\end{equation}
where $S_0$ is the propagator corresponding to ${\cal L}_0$ in
Eq.~(\ref{Walinter}) below,
\begin{equation}
S_0^{-1}(p)={\not\!p}-M_0\;,
\end{equation}
with the self-energy $\Sigma(p)$ calculated as a perturbation expansion in
powers of $\delta$.

For infinite nuclear matter, because of the translational, rotational, parity
and time reversal invariances, $\Sigma(p)$ can be generally written in terms of
the unit matrix and the Dirac $\gamma_\mu$ matrices as follows~\cite{hs}:
\begin{eqnarray}
\Sigma(p)&=&\Sigma^s(p)-\gamma_\mu \Sigma^\mu(p) \nonumber\\
&=&\Sigma^s(p^0,\absp)-\gamma^0\Sigma^0(p^0,\absp)+\mbox{\boldmath{$\gamma$}}\bcdot
\bp \Sigma^v(p^0, \absp) \;.
\label{genSig}
\end{eqnarray}
Defining the following auxiliary quantities \cite{hs}:
\begin{eqnarray}
M^*(p)    &\equiv & M_0+\Sigma^s(p)\;,\nonumber\\
{\bp}^*   &\equiv & {\bp}\left[1+\Sigma^{v}(p)\right]\;,\nonumber\\
E^*(p)    &\equiv & \left[{\bp}^{*2}+M^{*2}(p)\right]^{{1 \over 2}}\;,\\
\label{auxW*}
p^{* \mu} & =     & p^\mu + \Sigma^\mu (p)=\left[p^0+\Sigma^0(p),
{\bp}^*\right]\;, \nonumber
\end{eqnarray}
we can invert Eq.~(\ref{DysonN}) and write the nucleon propagator in the
compact form:
\begin{eqnarray}
S(p)&=&S_F(p)+S_D(p)\\
\label{SFSD}
S_F(p)&=&\left[\gamma^\mu p^*_\mu+M^*(p)\right]\frac{1}{p^{*\mu}p^*_{\mu}-
M^{*2}(p)+i\epsilon}\\
\label{SF}
S_D(p)&=&\left[\gamma^\mu p^*_\mu+M^*(p)\right]\frac{i\pi}{E^*(p)}
\delta\left(p^0-E(p)\right)\theta\left(P_F-\absp\right)\;,
\label{SD}
\end{eqnarray}
where $E(p)$ is the single-particle energy, which is the solution of the
transcendental equation:
\begin{eqnarray}
E(p)&=&\left[E^*(p)-\Sigma^0(p)\right]_{p_0=E(p)}\nonumber\\
&=&\left\{\bp^2\left[1+\Sigma^v(\absp, E(p))\right]^2+\left[M+
\Sigma^s(\absp,E(p))\right]^2\right\}^{{1 \over 2}}-\Sigma^0(\absp, E(p))\;.
\label{selfE}
\end{eqnarray}
Note that we have assumed that the nucleon propagator has simple
poles with unit residue. Within the approximation scheme we are working in
this paper, this assumption is satisfied, as can be seen below.

Following the scheme of neglecting the Feynman part of the nucleon propagator,
Eq.~(\ref{SFSD}), we obtain for the energy density of nuclear matter the
following expression:
\begin{eqnarray}
{\cal E}_{\rm W}&=&\frac{1}{V}\int d^3x T^{00}_{\rm W}-{\rm V.E.V.}
\nonumber\\
&=&\gamma\int_0^{P_F}\frac{d^3q}{(2\pi)^3}\frac{\bq\cdot\bq^*+
MM^*(q)}{E^*(q)}+{\cal E}^D_{\rm W}+{\cal E}^E_{\rm W}\;,
\label{EW}
\end{eqnarray}
where ${\rm V.E.V.}$ means the vacuum expectation value of $T^{00}_{\rm W}$,
and ${\cal E}^D_{\rm W}$ and ${\cal E}^E_{\rm W}$ are the direct and
exchange contributions, given by:
\begin{eqnarray}
{\cal E}^D_{\rm W}&=& \left(\frac{1}{2}-1\right)\frac{g_\sigma^2}
{m_\sigma^2}\left[\gamma\int_0^{P_F}\frac{d^3q}{(2\pi)^3}\frac{M^*(q)}
{E^*(q)}\right]^2-\left(\frac{1}{2}-1\right)\frac{g_\omega^2}{m_\omega^2}
\left[\gamma\int_0^{P_F}\frac{d^3q}{(2\pi)^3}\right]^2\;,
\label{EWD}
\end{eqnarray}
\begin{eqnarray}
{\cal E}^E_{\rm W}&=&\frac{1}{2}\gamma \int_0^{P_F}\frac{d^3q}
{(2\pi)^3E^*(q)}\int_0^{P_F}\frac{d^3k}{(2\pi)^3E^*(k)}\left\{g_\sigma^2
\Delta_\sigma(q-k) \right.\nonumber\\
&&\times\left[\left(\frac{1}{2}-1\right)-[E(q)-E(k)]^2
\Delta_\sigma(q-k)\right]\left[q^{*\mu}k^*_{\mu}+M^*(q)M^*(k)\right]\nonumber\\
&&+2g_\omega^2  \Delta_\omega(q-k)\left[\left(\frac{1}{2}-1\right)-
[E(q)-E(k)]^2\Delta_\omega(q-k)\right]\nonumber\\
&&\times\left.\left[q^{*\mu}k^*_{\mu}-2M^*(q)M^*(k)\right]
\right\}\;.
\label{EWE}
\end{eqnarray}
These expressions are very similar to the ones obtained in the Hartree-Fock
approximation. Differences are contained in the fermion kinetic energy,
the first term in Eq.~(\ref{EW}), and in the factors $\left(\frac{1}{2}
- 1\right)$ in Eqs. (\ref{EWD}) and (\ref{EWE}). These differences arise
because we are not using the nucleon field equation of motion.

To implement the $\delta$ expansion, we need to specify ${\cal L}_0$.
We choose
\begin{equation}
{\cal L}_0=\bar \psi\left(i\gamma_{\mu}\partial^{\mu}-M_0\right)\psi
+\frac{1}{2}(\partial_{\mu}\phi\partial^{\mu}\phi-m_\sigma^2\phi^2)-
\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\frac{1}{2}m_\omega^2V_{\mu}V^{\mu}\;,
\label{L0Wal}
\end{equation}
where
\begin{equation}
M_0 \equiv M+\mu\;.
\label{M0W}
\end{equation}
The interpolated Walecka model is then given by
\begin{equation}
{\cal L}_{\rm W}(\delta)={\cal L}_0+\delta\left(-g_\omega\bar\psi\gamma_{\mu}
V^{\mu}\psi+g_\sigma \bar\psi\phi\psi+\mu\bar\psi\psi\right)\;.
\label{Walinter}
\end{equation}
Notice that the $\delta$ expansion technique could have also been applied
to the meson fields explicitly. However, we have chosen to eliminate the
meson fields in favor of the nucleon fields by means of Eqs.~(\ref{phisol}) and
(\ref{omegsol}), therefore meson effects enter via the nucleon fields.

Next we obtain the self-energy in perturbation theory, always neglecting vertex
corrections and the Feynman part of the nucleon propagator. In zeroth order in
$\delta$, the nucleon self-energy, corresponding to the interpolated Lagrangian
Eq.~(\ref{Walinter}), is obviously zero:
\begin{equation}
\Sigma^{(0)}=0\;.
\end{equation}
Therefore, the auxiliary quantities to be used in Eq.~(\ref{auxW*})
become:
\begin{eqnarray}
M^*(p)    &\equiv & M_0 = M + \mu\;,\nonumber\\
{\bp}^*   &\equiv & {\bp}\;,\nonumber\\
E^*(p)    &\equiv & \left[{\bp}^{2}+M_0^2\right]^{{1 \over 2}}\;,\\
\label{auxW0}
p^{* \mu} & =     & p^\mu = (p^0,{\bp})\;.\nonumber
\end{eqnarray}
The single-particle energy is simply given by:
\begin{equation}
E(p)=E^*(p)=E_0(p)=\left[{\bp}^{2}+M_0^2\right]^{{1 \over 2}}\;.
\end{equation}
Using these in Eqs. (\ref{EWD}) and (\ref{EWE}), we obtain for the zeroth order
energy density the following expression:
\begin{equation}
{\cal E}^{(0)}_{\rm W}=\gamma\int_0^{P_F}\frac{d^3q}{(2\pi)^3}
\frac{\bq^2+MM_0}{E_0(q)}+{\cal E}_{\rm W}^{(0)D}+{\cal E}_
{\rm W}^
{(0)E}\;,
\label{EW0DE}
\end{equation}
with the direct and exchange contributions given by:
\begin{eqnarray}
{\cal E}_{\rm W}^{(0)D}&=&\left(\frac{1}{2}-1\right)\frac{g_\sigma^2}
{m_\sigma^2}\left[\gamma\int_0^{P_F} \frac{d^3q}{(2\pi)^3}\frac{M_0}
{E_0(q)}\right]^2-\left(\frac{1}{2}-1\right)\frac{g_\omega^2}{m_\omega^2}
\left[\gamma\int_0^{P_F}\frac{d^3q}{(2\pi)^3}\right]^2\;,\\
\label{EW0D}
{\cal E}_{\rm W}^{(0)E}&=& \frac{1}{2} \gamma\int_0^{P_F}\frac{d^3q}
{(2\pi)^3E_0(q)}\int_0^{P_F}\frac{d^3k}{(2\pi)^3E_0(k)}\left\{g_\sigma^2
\Delta_\sigma([E_0(q)-E_0(k)]^2-(\bq-\bk)^2) \right.\nonumber\\
&\times&\left[\left(\frac{1}{2}-1\right)-[E_0(q)-E_0(k)]^2
\Delta_\sigma([E_0(q)-E_0(k)]^2-(\bq-\bk)^2)\right] \left[E_0(q)E_0(k)
\right.\nonumber\\
&-&\left.\bq\cdot\bk +M_0^2 \right]+2g_\omega^2
\Delta_\omega([E_0(q)-E_0(k)]^2-(\bq-\bk)^2)\left[\left(\frac{1}{2}-1\right)-
[E_0(q)-E_0(k)]^2\right.\nonumber\\
&\times&\left.\left.\Delta_\omega([E_0(q)-E_0(k)]^2-(\bq-\bk)^2)\right]
\left[E_0(q)E_0(k)-\bq\cdot\bk-2M_0^2\right]\right\}\;.
\label{EW0E}
\end{eqnarray}

Let us consider the direct term first. Application of the PMS to it:
\begin{equation}
\frac{d{\cal E}^{(0)}_{\rm W}}{d\mu}=\frac{d{\cal E}^{(0)}_{\rm W}}{dM_0}
\frac{dM_0}{d\mu}=\frac{d{\cal E}^{(0)}_{\rm W}}{dM_0}=0\;,
\end{equation}
yields the following self-consistency condition for $M_0$:
\begin{equation}
M_0=M-\frac{g_\sigma^2 }{m_\sigma^2}\gamma\int_0^{P_F}
\frac{d^3q}
{(2\pi)^3}
\frac{M_0}{E_0(q)}\;.
\label{M00}
\end{equation}
This is exactly the same self-consistency condition for the effective nucleon
mass obtained by means of the Hartree, or mean-field, approximation.

Now, application of the PMS to the full energy density, which includes both
direct and exchange contributions, leads to a nonlinear equation for $\mu$, or
equivalently for $M_0$, which is more complicated than the one of
Eq.~(\ref{M00}). We do not present the expression here because it is
rather lengthy and not very instructive. To investigate the size of the
exchange corrections we carry out two sets of comparisons. In Figure~4 we
compare the nucleon binding energy, obtained by using only the first and direct
terms in Eq.~(\ref{EW0DE}) (solid line) and coupling constants fixed by
fitting the binding energy and density of equilibrium nuclear matter, with
the full binding energy, keeping the same coupling constants (dotted line).
The value of the coupling constants are $g_s^2=91.64$ and $g_v^2=136.2$.
The masses used in all calculations are $M=939$ MeV, $m_v=783$ MeV and
$m_\sigma=550$ MeV. We find that the exchange corrections coincide with those
obtained in a relativistic Hartree-Fock calculation~\cite{wal,hs} which
we also show for comparison (long-dashed line). The dashed line shows the
results obtained for the total binding energy (including both direct and
exchange contributions) after renormalizing the model parameters to reproduce
the bulk saturation properties of nuclear matter: $g_s^2=83.11$ and
$g_v^2=108.05$. These coupling constants are the same used when renormalizing
the relativistic Hartree-Fock calculation of Ref.~\cite{hs}. Therefore, the
PMS condition on the zeroth order energy density of the Walecka model is also
equivalent to the usual Hartree-Fock solution.

In Figure~5 we show the results for $\mu$ as a function of the Fermi
momentum $P_F$ obtained with the application of the PMS to the zeroth order
energy density. The solid line corresponds to the first and direct terms only
and the
dashed one (almost unnoticeable) corresponds to the full energy density
with the
renormalized constants. In Figure~6 we compare the results for the effective
nucleon mass in nuclear matter as a function of $P_F$ obtained from $\mu$.
In both figures, it is clear that the results with the exchange terms
and renormalized constants coincide with the results obtained by using the
direct terms only.

We now consider the second-order contribution to the self-energy. The
self-energy to second-order in delta is is given by:
\begin{eqnarray}
\Sigma^{(2)}(p)=&-&\mu \delta+i\frac{g_\sigma^2 \delta^2}{m_\sigma^2}
\int\frac{d^4q}{(2\pi)^4}{\rm Tr}\left[S^{(0)}(q)\right]-
i\frac{g_\omega^2 \delta^2}{m_\omega^2}\int\frac{d^4q}{(2\pi)^4}
\gamma_\mu {\rm Tr}\left[\gamma^{\mu}S^{(0)}(q)\right]
\nonumber\\
&+&ig_{\sigma}^2 \delta^2 \int\frac{d^4 q}{(2\pi)^4}S^{(0)}(q)
\Delta_{\sigma}(p-q)-ig_{\omega}^2 \delta^2 \int\frac{d^4 q}{(2\pi)^4}
\gamma_\mu S^{(0)}(q)\Delta_{\omega}(p-q)\gamma_{\mu}\;.
\label{SigmaW2}
\end{eqnarray}
where $\Delta_{\sigma}$ and $\Delta_{\omega}$ are given in Eq.~(\ref{mesprop}),
and again we have made use of the baryon current conservation. We evaluate
this expression neglecting the Feynman part of the nucleon propagator, the term
$S_F(p)$ given by Eq.~(\ref{SFSD}). Because of this, all integrals in
Eq.~(\ref{SigmaW2}) are finite and can easily be evaluated; there is no
need for renormalization. The first term in Eq.~(\ref{SigmaW2}) comes from the
first order contribution in $\delta$ and must be kept at second order. The
results are very similar to the ones obtained with the Hartree-Fock
approximation~\cite{hs}. Since there are subtle differences, we write them
explicitly below. Each component of the self-energy, $\Sigma^s$, $\Sigma^0$,
and $\Sigma^v$, can be decomposed in a direct and an exchange part. The direct
components are given by:
\begin{eqnarray}
\Sigma^{s(2)}_D&=&-\delta \mu-\gamma \frac{g_\sigma^2 \delta^2}
{m_\sigma^2} \int_0^{P_F}
\frac{d^3q}{(2\pi)^3}\frac{M_0}{E_0(q)}\;,\\
\Sigma^{0(2)}_D&=&-\gamma\frac{g_\omega^2 \delta^2}
{m_\omega^2}\int_0^{P_F}\frac{d^3q}{(2\pi)^3}\;,\\
\Sigma^{v(2)}_D&=&0\;.
\label{sigmaD1}
\end{eqnarray}
Notice that they are independent of energy and momentum. The exchange terms are
given by:
\begin{eqnarray}
\Sigma^{s(2)}_E(p)&=&\frac{1}{4\pi^2p}\int_0^{P_F}dq\;q\;\frac{M_0}{E_0(q)}
\left[\frac{1}{4}g_\sigma^2 \delta^2 \Theta_\sigma(p,q)-g_\omega^2 \delta^2
\Theta_\omega(p,q)\right]\;,\\
\Sigma^{0(2)}_E(p)&=&-\frac{1}{4\pi^2p}\int_0^{P_F}dq\;q\;\left[\frac{1}{4}
g_\sigma^2 \delta^2 \Theta_\sigma(p,q)+\frac{1}{2} {g_\omega^2 \delta^2}
\Theta_\omega(p,q)\right]\;,\\
\Sigma^{v(2)}_E(p)&=&-\frac{1}{4\pi^2p^2}\int_0^{P_F}dq\;q\;\frac{q}{E_0(q)}
\left[\frac{1}{2}g_\sigma^2 \delta^2 \Phi_\sigma(p,q)+g_\omega^2 \delta^2
\Phi_\omega(p,q)\right]\;,
\label{sigmaE1}
\end{eqnarray}
where the functions $\Theta_i(p,q), \Phi_i(p,q), i=\sigma, \omega$, are
defined by:
\begin{eqnarray}
\Theta_i(p,q)&=&\ln \left| \frac{A_i(p,q)+2pq}{A_i(p,q)-2pq}\right|\;,\\
\Phi_i(p,q)&=&\frac{1}{4pq}A_i(p,q)\Theta_i(p,q)-1\;,
\label{ThetaPhi}
\end{eqnarray}
where
\begin{equation}
A_i(p,q)=\bp^2 + \bq^2 + m_i^2 -\left[E(p)-E_0(q)\right]^2\;.
\label{Apq}
\end{equation}

The auxiliary quantities to be substituted into Eq.~(\ref{auxW*})
are then given by:
\begin{eqnarray}
M^*(p) &\equiv & M_0+\left[\Sigma_D^{s(2)} + \Sigma_E^{s(2)}(p)\right]\;,\nonumber\\
\bp^* &\equiv & {\bp}\left[1+\Sigma_E^{v(2)}(p)\right]\;,\nonumber\\
E^*(p)    &\equiv & \left[\bp^{*2}+M^{*2}\right]^{{1 \over 2}}\;,\\
\label{auxW2}
p^{* \mu} & = & p^\mu + \Sigma^\mu (p)=\left[p^0+\Sigma_D^{0(2)}+
\Sigma_E^{0(2)}(p), {\bp}^*\right]\;,\nonumber
\end{eqnarray}
and the single-particle energy is the solution of:
\begin{eqnarray}
E(p)&=&\left[E^*(p)-\Sigma^{0(2)}(p)\right]_{p_0=E(p)}\nonumber\\
&=&\left\{\bp^2\left[1+\Sigma^{v(2)}(\absp, E(p))\right]^2+
\left[M+\Sigma^{s(2)}(\absp,E(p))\right]^2\right\}^{{1 \over 2}}-
\Sigma^{0(2)}(\absp, E(p))\;.
\label{selfE2}
\end{eqnarray}

We are in the position to calculate the energy density. Initially we consider
the direct terms only. The energy density is given by:
\begin{equation}
{\cal E}_W^{(2)D}=\gamma\int_0^{P_F}\frac{d^3q}{(2\pi)^3}\frac{\bq\cdot\bq^*+
MM^*}{E^*(q)} + \frac{g_\omega^2 }{2m_\omega}\left[\frac{2}
{3\pi^2}P_F^3\right]^2-\frac{g_\sigma^2 }{2m_\sigma^2}
\left[\gamma \int_0^{P_F}\frac{d^3q}{(2\pi)^3}
\frac{M^*}{E^*(q)}\right]^2\;.
\label{Hartree}
\end{equation}
Application of the PMS to this yields again the familiar Hartree
result, Eq. (\ref{M00}), with $M_0$ and $E_0(q)$ replaced respectively
by $M^*$ and $E^*(q)$. The exchange terms are shown in Eq.~(\ref{EWE}) and we
believe it is not necessary to rewrite them here. Also in this case, despite
numerical imprecisions, the exchange corrections coincide with the usual
Hartree-Fock solution, as can be seen in Figure~7. The behavior of $M^*$ as a
function of the Fermi momentum at this order does not show any noticeable
difference as compared with the zeroth order results. However, as can be seen
in Figure~8 the behavior of $\mu$ as a function of $P_F$ obtained with the
application of the PMS to the full second order energy density (dashed line)
is rather different from the one obtained when only the direct term is taken
into account (solid line). It is very interesting to notice that this different
behavior does not manifest itself either in the values of $M^*$ or of the
binding
energy. This is because the energy density $\cal E$ is a very flat function
of $\mu$, as can be seen in Figure~9, where the energy density is shown
for $P_F=1.19$ fm$^{-1}$. Recall that if one had an exact solution, the energy
density would be independent of $\mu$. The solid line is obtained without the
inclusion of the exchange term (the PMS solution in this case is given by
$\mu/M=-0.275$) and the dashed line gives the full second order density energy
(the PMS solution is $\mu/M=-0.35$). This stability in the value of the energy
density as a function of $\mu$ is very desirable and guarantees that even
big changes in the value of $\mu$ will not affect physical quantities, as the
binding energy for instance.

\section{Conclusions and Future Perspectives}
\label{sec:concl}

In this paper we have utilized the optimized $\delta$ expansion to study medium
effects in two commonly used models in hadron and nuclear physics: the
NJL model and the Walecka model. We have investigated an alternative way
of fixing the arbitrary parameters introduced by the $\delta$ expansion, by
applying the PMS to the energy density of the system.

The most important and concrete conclusion we can draw from this work
is that when applying the PMS to the energy density of the NJL model
we reproduce, at any order, the familiar Hartree-Fock solution for
the dynamically generated mass, in the approximation of neglecting
vertex corrections. In the case of Walecka model, we obtained results
quantitatively similar to
the ones of the usual Hartree-Fock approximation, although the analytical
expressions are not evidently equivalent. If one neglects the exchange term
in the energy density then clearly the mean-field solution is reproduced
at any order. It is also worth mentioning that, in the Walecka model, the
energy density is a very flat function of $\mu$ and this guarantees that
the PMS solution is indeed very stable.

On the basis of our results, we believe that the optimized $\delta$
expansion is a very robust nonperturbative approximation scheme. Compared with
the Hartree-Fock approximation, the $\delta$ expansion is very economical
because of its perturbative nature. Once the reliability of the scheme
has been established, one is ready to proceed to other interesting
applications. These include vertex and, obviously, vacuum effects. The
study of the vacuum in the Walecka model is an important issue since one
needs to know the limits of applicability of
such model to high densities and/or temperatures before quark and gluon
degrees of freedom have to be invoked. Of course, one has to face
renormalization problems when including the vacuum. Renormalization
in a Hartree-Fock scheme is very complicated~\cite{SerBiel} and one
expects that this will be facilitated within the $\delta$ expansion method.
\section{Acknowledgments}
This work was partially supported by CNPq and FAPESP.

redpajama_set_name: RedPajamaArXiv


At the age of 23, if anyone was smiling just before he was being hanged to death, it was Shaheed Bhagat Singh. He was born September 27, 1907 in the village Banga of Layalpur to Mata Vidyavati and Sardar Kishan Singh. His uncle, Sardar Ajit Singh, as well as his father, were great freedom fighters, so Bhagat Singh grew up in a patriotic atmosphere. Ajit Singh established the Indian Patriots' Association, along with Syed Haidar Raza, to organize the peasants against the Chenab Canal Colony Bill. He also established the secret organization, the Bharat Mata Society.
At an early age, Bhagat Singh started dreaming of uprooting the British empire. Never afraid of fighting during his childhood, he thought of "growing guns in the fields," so that he could fight against the British. The Ghadar Movement left a deep imprint on his mind. Kartar Sing Sarabha, hanged at the age of 19, became his hero. The massacre at Jallianwala Bagh on April 13, 1919 drove him to go to Amritsar, where he kissed the earth sanctified by the martyrs' blood and brought back home a little of the soaked soil. He studied in the D.A.V. School in Lahore. At the age of 16, he used to wonder why so many Indians could not drive away these fistful of invaders.
In search of revolutionary groups and ideas, he met Sukhdev and Rajguru. Bhagat Singh, along with the help of Chandrashekhar Azad, formed the Hindustan Socialist Republican Army (HSRA). The aim of this Indian revolutionary movement was now defined as not only to make India independent, but also to create "a socialist India."
During the Simon Commission, Sher-e-Punjab Lala Lajpat Rai was wounded and died later. To avenge his death, Bhagat Singh and Rajguru killed Mr. Saunders (one of the deputy officers in connection with the Simon Commission).
When the British government promulgated the two bills "Trade Union Dispute Bill" and "Public Safety Bill" which Bhagat Singh and his party thought were Black Laws aimed at curbing citizens' freedom and civil liberties, they decided to oppose these bills by throwing a bomb in the Central Assembly Hall (which is now Lok Sabha). However, things changed, and the Britishers arrested Bhagat Singh and his friends on April 8, 1929.
He and his friends wanted to be shot dead, since they were termed as prisoners of war. Their request was not fulfilled, and on March 23, 1931, Bhagat Singh, Shivram Rajguru, and Sukhdev were hanged to death.
This man's only mission in life was to see his country free from British rule. He did his best and when he was being led to the gallows, he was satisfied that he had lived up to his principles, irrespective of the consequences. The only thing that made him sad was that he couldn't do more for his country.
Bhagat Singh, September 27, 1907 – March 23, 1931) was an Indian freedom fighter, considered to be one of the most influential revolutionaries of the Indian independence movement. He is often referred to as Shaheed Bhagat Singh (the word shaheed means "martyr").
Bhagat Singh was born into a Jatt Sandhu family to Sardar Kishan Singh Sandhu and Vidyavati in the Khatkar Kalan village near Banga in the Lyallpur district of Punjab. Singh's given name of Bhagat means "devotee". He came from a patriotic Sikh family, some of whom had participated in movements supporting the independence of India and others who had served in Maharaja Ranjit Singh's army. His grandfather, Arjun Singh, was a follower of Swami Dayananda Saraswati's Hindu reformist movement, Arya Samaj, which would carry a heavy influence on Singh. His uncles, Ajit Singh and Swaran Singh, as well as his father were members of the Ghadar Party, led by Kartar Singh Sarabha Grewal and Har Dayal. Ajit Singh was forced to flee to Persia because of pending cases against him while Swaran Singh was hanged on December 19, 1927 for his involvement in the Kakori train robbery of 1925.
Unlike many Sikhs his age, Singh did not attend Khalsa High School in Lahore, because his grandfather did not approve of the school officials' loyalism to the British authorities. Instead, his father enrolled him in Dayanand Anglo Vedic High School, an Arya Samajist school. At age 13, Singh began to follow Mahatma Gandhi's Non-Cooperation Movement. At this point he had openly defied the British and had followed Gandhi's wishes by burning his government-school books and any British-imported clothing. Following Gandhi's withdrawal of the movement after the violent murders of policemen by villagers from Chauri Chaura, Uttar Pradesh, Singh, disgruntled with Gandhi's nonviolence action, joined the Young Revolutionary Movement and began advocating a violent movement against the British.
In 1923, Bhagat famously won an essay competition set by the Punjab Hindi Sahitya Sammelan. This grabbed the attention of members of the Punjab Hindi Sahitya Sammelan including its General Secretary Professor Bhim Sen Vidyalankar. At this age, he quoted famous Punjabi literature and discussed the Problems of the Punjab. He read a lot of poetry and literature which was written by Punjabi writers and his favourite poet was Allama Iqbal from Sialkot.
In his teenage years, Bhagat Singh started studying at the National College in Lahore, but ran away from home to escape early marriage, and became a member of the organization Naujawan Bharat Sabha ("Youth Society of India"). In the Naujawan Bharat Sabha, Singh and his fellow revolutionaries grew popular amongst the youth. He also joined the Hindustan Republican Association at the request of Professor Vidyalankar, which was then headed by Ram Prasad Bismil and Ashfaqulla Khan. It is believed that he had knowledge of the Kakori train robbery. He wrote for and edited Urdu and Punjabi newspapers published from Amritsar. In September 1928, a meeting of various revolutionaries from across India was called at Delhi under the banner of the Kirti Kissan Party. Bhagat Singh was the secretary of the meet. His later revolutionary activities were carried out as a leader of this association. The capture and hanging of the main HRA Leaders also allowed him to be quickly promoted to higher ranks in the party, along with his fellow revolutionary Sukhdev Thapar.
In the face of actions by the revolutionaries, the British government enacted the Defence of India Act to give more power to the police. The purpose of the Act was to combat revolutionaries like Bhagat Singh. The Act was defeated in the council by one vote. However, the Act was then passed under the ordinance that claimed that it was in the best interest of the public. In response to this act, the Hindustan Socialist Republican Association planned to explode a bomb in the Central Legislative Assembly where the ordinance was going to be passed. Originally, Chandrashekhar Azad, another prominent leader of the revolutionary movement attempted to stop Bhagat Singh from carrying out the bombing. However, the remainder of the party forced him to succumb to Singh's wishes. It was decided that Bhagat Singh and Batukeshwar Dutt, another revolutionary, would throw the bomb in the assembly.
On April 8, 1929, Singh and Dutt threw a bomb onto the corridors of the assembly and shouted "Inquilab Zindabad!" ("Long Live the Revolution!"). This was followed by a shower of leaflets stating that it takes a loud voice to make the deaf hear. The bomb neither killed nor injured anyone; Singh and Dutt claimed that this was deliberate on their part, a claim substantiated both by British forensics investigators who found that the bomb was not powerful enough to cause injury, and by the fact that the bomb was thrown away from people. Singh and Dutt gave themselves up for arrest after the bomb. He and Dutt were sentenced to 'Transportation for Life' for the bombing on June 12, 1929.
Bhagat Singh was known for his appreciation of martyrdom. His mentor as a young boy was Kartar Singh Sarabha. Singh is himself considered a martyr for acting to avenge the death of Lala Lajpat Rai, also considered a martyr. In the leaflet he threw in the Central Assembly on 9th April 1929, he stated that It is easy to kill individuals but you cannot kill the ideas. Great empires crumbled while the ideas survived. After engaging in studies on the Russian Revolution, he wanted to die so that his death would inspire the youth of India to unite and fight the British Empire.


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Dragos Bozdog earned his Ph.D. in Financial Engineering and M.S. in Financial Engineering from Stevens Institute of Technology, Ph.D. in Mechanical Engineering from The University of Toledo, and B.S. in Mechanical Engineering from Polytechnic University of Bucharest.
Doctor of Philosophy. , Stevens Institute of Technology, Hoboken, USA.
Master of Science, Stevens Institute of Technology,Hoboken, USA.
Post-Doctoral Fellow, The State University of New Jersey, USA.
Doctor of Philosophy, The University of Toledo, OH, USA.
Bachelor of Science , Polytechnic University of Bucharest, Romania.
Stevens Institute of Technology, Department of Financial Engineering, U.S.A.
Research Analyst Consultant - Commodity Futures Trading Commission (CFTC). May 2011 - Aug 2011.
Cohen Capital Group, New York, NY. Aug 2009 - Dec 2010.
Genesis Securities, USA. Summer 2009.
His research interests include mathematics of finance, algorithms and optimization, multivariate methods for rare events analysis of high-frequency financial data, development of early warning systems and threat assessment in financial sector.


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According to journalist and author Malcolm Gladwell, it takes around 10,000 hours to develop expertise in a skill or behavior, but the constant demands on leaders today makes it hard for them to carve out time to spend on self-development.
Researchers also indicate that you can create a habit within 21 days, which has become a format for various development activities (e.g. I recently did a 21-day meditation series from Oprah Winfrey).
Lastly, we know that the best way to develop your behaviors is on the job.
With these three things in mind—10,000 hours for mastery, 21 days for a habit to form, and development best on the job—I wrote 25 bite-sized questions to help leaders look at their role and responsibilities from various perspectives, and shared them on LinkedIn and Twitter, one question a day, for 25 days. The response to the series was incredible. I particularly loved how some readers used these daily prompts as a series within their organizations, sending them to leaders directly.
Want to be a more inspirational #leader? Here are 25 Questions to Help You Be a Better Leader in 2019.
Frequently during times of constant change, leaders struggle with demonstrating strength while remaining humble. How do you show your humility in the workplace?
Hard work and intellect aren't enough to be a leader—you need EQ, too, whether in an office or on a race track. How emotionally aware (of yourself or of others) are you?
Feedback is a gift, but remember it is only well received when well delivered. Feedback should be timely, balanced, and specific. Miss one of these, and your feedback may miss the mark. Do you have feedback to give to someone today?
Peter Drucker says, "Only three things happen naturally in organizations: friction, confusion, and underperformance. Everything else requires leadership." Which of these three elements is most prevalent in your team now and how will you address it?
How do you want to be described as a leader? What do you want your leadership legacy to be? Write a sentence or two to answer these questions right now—and then live that intention.
How can you show vulnerability today? Spend 20 minutes on your personal development so you can share who you really are with your teams. Need a hint? Check out this TED Talk by Brené Brown on the power of vulnerability.
Are you developing a possible successor? According to the Global Leadership Forecast 2018, only 14 percent of CEOs have the leadership talent to execute their strategy. How can you ensure your business has a sustainable leadership pipeline?
If you haven't seen Randy Pausch's "The Last Lecture," then check out YouTube for a humbling experience. One quote that really stood out to me: "When you're screwing up and nobody says anything to you anymore, that means they've given up on you. You may not want to hear it, but your critics are often the ones telling you they still love you and care about you and want to make you better." Are you giving feedback to your team or giving up on their growth? When did you last seek feedback?
What is your "Why"? Never mind the "What" or "Where" or "How." "Why" is the most important model for inspirational leadership.
Imagine that your work day finishes in three hours—what do you really need to achieve before you leave your desk? Do that first.
Author Stephen Covey wrote, "Seek first to understand, then to be understood." How well do you understand those you lead? Or do you treat all team members as one and miss their unique skills, motivations, and abilities?
In today's stressful world, leaders are under pressure to achieve results quickly; which means they often focus on the practical needs of the conversations but miss the personal needs of the person they are conversing with. You can address the practical side of the conversation, but completely miss the other person's need to feel valued or respected. Is this a pitfall into which you fall? How can you address both the practical and personal needs of the people on your team?
Michael Jordan said, "I've missed more than 9,000 shots in my career. I've lost almost 300 games. Twenty-six times, I've been trusted to take the game winning shot and missed. I've failed over and over and over again in my life. And that's why I succeed." What do you learn from the errors or mistakes that you have made? Do you make failing fun within your team or is it something that people fear?
Are you unlocking the hidden potential in your company by leveraging diversity and inclusion? Check out DDI's popular eBook "Unleash Hidden Potential: Build Your Competitive Edge Through Diversity and Inclusion" to learn more.
Speaker and author Dave Ulrich says that, "Leaders who listen to their employees as much as talking to them learn to communicate in ways that matter to employees." When you interact with your team members, what is the balance of you listening and you speaking?
According to LinkedIn CEO Jeff Weiner, the biggest U.S. skills gap is not coding, but soft skills. When it comes to leadership development it is so easy to focus on technical skills and miss the "soft" skills (which are actually the hard skills!) How are you developing your interaction skills?
The Wall Street Journal recently identified empathy as one of the most critical skills for leaders—and according to adventurer Bear Grylls, this applies to other extreme leadership situations too: "If you are stuck up a mountain for three months with someone, then the most important skill is not climbing. It's empathy." How effective are you at demonstrating empathy?
If you haven't seen Sheryl Sandberg's TED talk on women leaders, stop what you're doing and watch it. It's essential viewing for leaders who want to flourish through team diversity. One of my favorite quotes: "We need women at all levels, including the top, to change the dynamic, reshape the conversation, and make sure women's voices are heard and heeded, not overlooked and ignored." How can you apply this to your team today?
Are you effectively demonstrating empathy to bridge the gap between interacting with your team members and driving job performance? Read more on the topic here.
Do you embrace the misfits and rebels, or do you fear them? The famous Apple "think different" ad campaign heralded these people as changemakers and revolutionaries. How can you leverage differences further within your team to drive change?
Collective leadership is one of today's hottest topics. How are you encouraging that collective spirit within your team?
How are you using data to drive leadership development and growth within your organization? Check out this article about how you can do so—and do it well.
How do you manage uncertainty and complexity? Do you have the leadership skills to stay on course during the storms?
How will you continue your self-development, and how does your organization develop its leaders? Get some inspiration by attending our webinar on furthering leadership development and best practices on this topic. This webinar, "Ignore Frontline Leaders at Your Peril," provides more details on alternative leadership development insights and approaches, and I hope you can join us for this event.
Wishing you all good habits, mastery development, on-the-job application—Oh, and a Happy New Year!
Reserve your spot in the "Ignore Frontline Leaders at Your Peril," webinar on January 31st.


The PDF for the catalogue is now online and as usual the powerful Godolphin operation has supported the auction with a quality line up from group performers to unraced prospects.
Among the Godolphin entries are proven stakes performers Marsupial and Tessera.
Also joining the catalogue are impressive drafts from Aquis Farms and Jadeskye Racing.
The Autumn Racehorse Sale will commence following the sale of the last yearling on Tuesday.