spacing=nonfrench

00footnotetext: Submitted in fulfillment of the requirements for a M.Sc.
degree in Theoretical Physics.

# Chiral Corrections to the Gell-Mann-Oakes-Renner Relation from
QCD Sum Rules

Preshin Moodley

Supervisor: Emeritus Professor C.A. Dominguez
Co-Supervisor: Professor A. Peshier

Department of Physics, University of Cape Town

(Submitted 12 October 2010)

### ABSTRACT

We calculate the next-to-leading order corrections to the $SU(2)\otimes SU(2)$
and $SU(3)\otimes SU(3)$ Gell-Mann-Oakes-Renner relations. We use a
pseudoscalar correlator calculated from Perturbative QCD up to five loops and
use the QCD Finite Energy Sum Rules with integration kernels tuned to suppress
the importance of the hadronic resonances. This leads to a substantial
reduction in the systematic uncertainties from the experimentally unknown
resonance spectral function. We use the method of Fixed Order and Fixed
Renormalization Scale Perturbation Theory to compute the integrals. We
calculate these corrections to be $\delta_{\pi}=0.060\pm 0.014$ and
$\delta_{K}=0.64\pm 0.24$. As a result of these new values we predict the
value of the light quark condensate
$\left\langle{0|\bar{q}q|0}\right\rangle=-\left({266\pm 5{\text{
MeV}}}\right)^{3}$ and the Chiral Perturbation Theory low energy constant
$H_{2}^{r}=-\left({4.9\pm 1.8}\right)\times 10^{-3}$. Results from this work
have been published as: J. Bordes, C.A. Dominguez, P. Moodley, J.
Pe$\widetilde{\text{n}}$arrocha and K. Schilcher, J. High Ener. Phys. 05
(2010) 064.

###### Contents

  1. 1 Introduction
  2. 2 A Low Energy Theorem
    1. 2.1 Hadronic Part
    2. 2.2 QCD Part
    3. 2.3 The Gell-Mann-Oakes-Renner Relation
  3. 3 The QCD Finite Energy Sum Rule
  4. 4 Integration in the Complex Plane
    1. 4.1 Fixed Order Perturbation Theory
    2. 4.2 Fixed Renormalization Scale Perturbation Theory
  5. 5 The Resonance Contribution $\left.{\delta_{5}\left(R\right)}\right|_{RES}$
  6. 6 Results
    1. 6.1 Frozen Order Results
      1. 6.1.1 $SU(2)\otimes SU(2)$
    2. 6.2 FOPT Results
      1. 6.2.1 $SU(2)\otimes SU(2)$
      2. 6.2.2 $SU(3)\otimes SU(3)$
    3. 6.3 Discussion of Results
  7. 7 Conclusion
  8. A Miscellaneous Material for the Gell-Mann-Oakes-Renner Relation
    1. A.1 A Step Towards Covariance
    2. A.2 Time Ordered Products
    3. A.3 Preparatory Work I
    4. A.4 Preparatory Work II
    5. A.5 Operator Identity
  9. B Perturbative QCD Calculations
    1. B.1 $\Pi_{0}(z)$ Calculations
      1. B.1.1 One Loop Calculation
      2. B.1.2 Two Loop Calculation
      3. B.1.3 Three Loop Calculation
      4. B.1.4 Four Loop Calculation
      5. B.1.5 Five Loop Calculation
    2. B.2 $\Pi_{2}(z)$ Calculations
    3. B.3 Cu Calculation
    4. B.4 Cj Calculations
  10. C Renormalization Group Equations for the coupling and mass
  11. D A comment about the CIPT
  12. E Coefficient fixing of $\Delta(z)$
  13. F Perturbative QCD Integrals
  14. G The Resonance Integrals
  15. H Exact solution of the RGE for $\alpha_{s}$ up to fifth order

###### List of Figures

  1. 1 Integration contour used for correlator. X indicates a pole and the broken line indicates a branch cut in the complex plane along the positive real axis corresponding to the resonances
  2. 2 Results for $\psi_{5}(0)$(GeV4) [30] as a function of the energy $s$ in FSPT with $\mu^{2}=4$GeV2. a, b and c correspond to polynomial orders 4, 5 and 6 respectively.
  3. 3 Results for $\psi_{5}(0)$(GeV4) as a function of the energy $s$ in FOPT. We have neglected the two resonances. Starting from the top we have $\psi_{5}(s,0.365)$ and $\psi_{5}(s,0.397)$ respectively
  4. 4 Results for $\delta_{\pi}$ as a function of the energy $s$ in FOPT. We have neglected the two resonances. Starting from the top we have $\delta_{\pi}(s,0.397)$ and $\delta_{\pi}(s,0.365)$ respectively
  5. 5 Results for $\psi_{5}(0)$(GeV4) as a function of the energy $s$ in FOPT. We have included the two resonances with the weighting factor $\kappa=0.5$. Starting from the top we have $\psi_{5}(s,0.365)$ and $\psi_{5}(s,0.397)$ respectively
  6. 6 Results for $\delta_{\pi}$ as a function of the energy $s$ in FOPT. We have included the two resonances with the weighting factor $\kappa=0.5$. Starting from the top we have $\delta_{\pi}(s,0.397)$ and $\delta_{\pi}(s,0.365)$ respectively
  7. 7 Plot of the various contributors to $\delta_{\pi}$ for $\Lambda_{QCD}=0.365$ GeV. Starting from the top we have $\ln(\delta_{\pi}^{\pi 0})$, $\ln(\delta_{\pi}^{C_{j}})$, $\ln(\delta_{\pi}^{C_{u}})$ and $\ln(\delta_{\pi}^{\pi 2})$ respectively
  8. 8 Plot of the various contributors to $\delta_{\pi}$ for $\Lambda_{QCD}=0.397$ GeV. Starting from the top we have $\ln(\delta_{\pi}^{\pi 0})$, $\ln(\delta_{\pi}^{C_{j}})$, $\ln(\delta_{\pi}^{C_{u}})$ and $\ln(\delta_{\pi}^{\pi 2})$ respectively
  9. 9 Results for $\psi_{5}(0)$(GeV4) as a function of the energy $s$ in FOPT. We have neglected the two resonances. Starting from the top we have $\psi_{5}(s,0.365)$ and $\psi_{5}(s,0.397)$ respectively
  10. 10 Results for $\delta_{\pi}$ as a function of the energy $s$ in FOPT. We have neglected the two resonances. Starting from the top we have $\delta_{K}(s,0.397)$ and $\delta_{K}(s,0.365)$ respectively
  11. 11 Results for $\psi_{5}(0)$(GeV4) as a function of the energy $s$ in FOPT. We have included the two resonances with the weighting factor $\kappa=0.5$. Starting from the top we have $\psi_{5}(s,0.365)$ and $\psi_{5}(s,0.397)$ respectively
  12. 12 Results for $\delta_{\pi}$ as a function of the energy $s$ in FOPT. We have included the two resonances with the weighting factor $\kappa=0.5$. Starting from the top we have $\delta_{K}(s,0.397)$ and $\delta_{K}(s,0.365)$ respectively
  13. 13 Plot of the various contributors to $\delta_{K}$ for $\Lambda_{QCD}=0.365$ GeV. Starting from the top we have $\ln(\delta_{K}^{\pi 0})$, $\ln(\delta_{K}^{\pi 2})$, $\ln(\delta_{K}^{C_{j}})$ and $\ln(\delta_{K}^{C_{u}})$ respectively
  14. 14 Plot of the various contributors to $\delta_{K}$ for $\Lambda_{QCD}=0.397$ GeV. Starting from the top we have $\ln(\delta_{K}^{\pi 0})$, $\ln(\delta_{K}^{\pi 2})$, $\ln(\delta_{K}^{C_{j}})$ and $\ln(\delta_{K}^{C_{u}})$ respectively
  15. 15 Numerical solution to the RGE for $\alpha_{s}(0,0.397)=0.11266$, with the top curve the real part and the bottom the imaginary part of $\alpha_{s}$

###### List of Tables

  1. 1 Tabulated values extracted from the plots 3 \- 5 which are used to estimate $\delta_{\pi}$.
  2. 2 Tabulated values extracted from the plots 9 \- 12 which are used to estimate $\delta_{K}$.

##  1 Introduction

In recent years it has become generally accepted that the correct physical
theory for the description of the strongly interacting particles is Quantum
Chromodynamics (QCD). Initially there had been strong criticism against QCD
since the fundamental constituents of the theory, the quarks and gluons, had
never been observed in nature or in the laboratory. This is still the case
today, but there have been many predictions from QCD about the bound states of
these quarks (hadrons) which can be tested in the laboratory. Since there has
been no analytic solution to QCD as yet, these predictions have been made
possible by exploiting one of the key features of QCD, which is Asymptotic
freedom and its closely related counterpart Confinement. These interesting
features of QCD were discovered by Gross, Wilcek and Politzer [1, 2].
Confinement provides a theoretical explanation for why there are no hadronic
colour multiplicity states observed in nature. The reason for this is, as the
quarks move further away from each other, their interaction starts to increase
i.e. the value of $\alpha_{s}$, the strong coupling (which sets the scale of
the interaction), increases up to the point where there is so much energy in
the two quark system that it is energetically more economical to bring two
more quarks into existence from the vacuum and decrease the total energy of
the system. For shorter distances between quarks the strong coupling goes to
zero.

This allows us to use the tools from Perturbation theory to make an expansion
of the equations of motion, which cannot be solved analytically (as yet), in
terms of this small parameter $\alpha_{s}$. In this way we can extract
information from QCD to get an approximate understanding of the behavior of
the quarks and gluons.

The two main ways to compute hadronic parameters from QCD have been with the
use of Lattice QCD were numerical solutions to the equations of motion are
found. This is good since no approximations are made and solutions obtained
are nearly exact, but since the computer is a black box we have no
understanding about the subtle mechanisms at work in the theory. This is an
undesirable feature of the results from Lattice QCD. The other way to extract
information from the QCD equations of motion is the use of the QCD sum rules.

The idea of the QCD sum rules was developed in the late 1970’s by Shifman _et
al_ [3, 4] and has now become one of the work horses of analytic methods to
extract information from QCD. QCD sum rules assume that confinement exists in
QCD and tries to parameterize this effect. QCD sum rules are based on two
ideas, the Operator Product Expansion (OPE) which separates the short and long
distance quark-gluon interactions, and the application of Cauchy’s Theorem in
the complex energy (squared) plane. The short distance interactions are
calculated using Perturbative QCD and the long distance interaction, which are
non-perturbative effects, are parameterized in terms of the vacuum
condensates.

In this project we use the method of QCD sum rules to make a direct estimate
of the breaking of the $SU(2)\otimes SU(2)$ and $SU(3)\otimes SU(3)$ Gell-
Mann-Oakes-Renner relations. There have been previous attempts to calculate
this symmetry breaking parameter [5, 6], but these have been indirect methods
to estimate the breaking parameter $\delta_{\pi}$, with large uncertainties
associated with them. The symmetry breaking parameter is also of great
interest to the field of Chiral Perturbation Theory where $\delta_{\pi}$ is
related to two of the theory’s low energy constants $L_{8}^{r}$ and
$H_{2}^{r}$.

##  2 A Low Energy Theorem

The Gell-Mann, Oakes, Renner Relation (GMOR) is a low energy relation which
comes from the global
$SU(N)\otimes SU(N)$ chiral symmetry which is present in the theory of
strongly interacting particles. We give a proof for the relation below. For
simplicity, we shall consider the case of the pion $\pi$ and hence $n=2$

###  2.1 Hadronic Part

We start with the hadronic pseudoscalar correlator which is the time-ordered
product of the hadronic currents

$\displaystyle\left.{\psi_{5}\left({q^{2}}\right)}\right|_{{\text{HAD}}}=i\int{e^{iq\cdot
x}\left\langle{0\left|{T\left[{\partial^{\mu}A_{\mu}(x)\partial^{\nu}A_{\nu}^{\dagger}(0)}\right]}\right|0}\right\rangle
dx}$ (2.1.1)

since the Fock space has the property of completeness we can insert a state
between the two currents without changing the correlator

$\displaystyle\left.{\psi_{5}\left({q^{2}}\right)}\right|_{{\text{HAD}}}=i\sum\limits_{n}{\int{e^{iq\cdot
x}\left\langle{0\left|{T\left[{\partial^{\mu}A_{\mu}(x)\left|n\right\rangle\left\langle
n\right|\partial^{\nu}A_{\nu}^{\dagger}(0)}\right]}\right|0}\right\rangle}dx}$
(2.1.2)

we will insert only the pion state and then truncate the series. From the
Partial Conserved Axial Current(PCAC) [7, 8, 9] we have

$\displaystyle\partial^{\mu}A_{\mu}\left(x\right)=\sqrt{2}f_{\pi}m_{\pi}^{2}\hat{\varphi}\left(x\right)$
(2.1.3)

were $m_{\pi}$ and $f_{\pi}$ are the pion mass and the pion decay constant
respectively. Inserting (2.1.3) into the correlator (2.1.1) we obtain

$\displaystyle\left.{\psi_{5}\left({q^{2}}\right)}\right|_{{\text{HAD}}}=2f_{\pi}^{2}m_{\pi}^{4}i\int{e^{iq\cdot
x}\left\langle{0\left|{T\left[{\hat{\varphi}\left(x\right)\hat{\varphi}^{\dagger}\left(0\right)}\right]}\right|0}\right\rangle
d^{4}x}$ (2.1.4)

The right-hand-side of (2.1.4) is just the Fourier transform of the vacuum-
expectation-value(vev) of some operator. This operator is the Green’s
function/propagator for the pion and can then be written as

$\displaystyle
i\Delta\left({x-y}\right)=\left\langle{0\left|{T\left[{\hat{\varphi}\left(x\right)\hat{\varphi}^{\dagger}\left(y\right)}\right]}\right|0}\right\rangle$
(2.1.5)

were $x$ and $y$ are arbitrary spacetime points. We can then split the time
ordered product (2.1.5) into two terms which depend on which event occurred
first.

$\displaystyle
i\Delta\left({x-y}\right)=\Theta\left({x_{0}-y_{0}}\right)\left\langle{0\left|{\hat{\varphi}\left(x\right)\hat{\varphi}\left(y\right)}\right|0}\right\rangle+\Theta\left({y_{0}-x_{0}}\right)\left\langle{0\left|{\hat{\varphi}\left(y\right)\hat{\varphi}\left(x\right)}\right|0}\right\rangle\hfill$
(2.1.6)

were $\Theta(x)$ is the Heaviside/Step function. The pion field can be written
as a sum over the positive and negative energy states with appropriate
creation and annihilation operators

$\displaystyle\hat{\varphi}\left(x\right)=\sum\limits_{q}{\left[{\hat{a}_{q}\varphi_{\underset{\raise
2.10002pt\hbox{$\smash{\scriptscriptstyle-}$}}{q}}^{(+)}\left(x\right)+\hat{a}^{\dagger}_{q}\varphi_{-\underset{\raise
2.10002pt\hbox{$\smash{\scriptscriptstyle-}$}}{q}}^{(-)}\left(x\right)}\right]}$
(2.1.7)

The $\varphi_{\underset{\raise
2.10002pt\hbox{$\smash{\scriptscriptstyle-}$}}{q}}^{(+)}\left(x\right),\varphi_{-\underset{\raise
2.10002pt\hbox{$\smash{\scriptscriptstyle-}$}}{q}}^{(-)}\left(x\right)$ are
the positive and negative energy plane wave solutions. Inserting the $\pi$
fields(2.1.7) into the propagator(2.1.6).

$\displaystyle
i\Delta\left({x-y}\right)=\Theta\left({{x_{0}}-{y_{0}}}\right)\left\langle{0\left|{\sum\limits_{q}{\left[{{{\hat{a}}_{\underline{q}}}\varphi_{\underline{q}}^{(+)}\left(x\right)+\hat{a}_{\underline{q}}^{\dagger}\varphi_{-\underline{q}}^{(-)}\left(x\right)}\right]}\sum\limits_{q^{\prime}}{\left[{{{\hat{a}}_{\underline{q}^{\prime}}}\varphi_{\underline{q}^{\prime}}^{(+)}\left(y\right)+\hat{a}_{\underline{q}^{\prime}}^{\dagger}\varphi_{-\underline{q}^{\prime}}^{(-)}\left(y\right)}\right]}}\right|0}\right\rangle+$
$\displaystyle+\Theta\left({{y_{0}}-{x_{0}}}\right)\left\langle{0\left|{\sum\limits_{q^{\prime}}{\left[{{{\hat{a}}_{\underline{q}^{\prime}}}\varphi_{\underline{q}^{\prime}}^{(+)}\left(y\right)+\hat{a}_{\underline{q}^{\prime}}^{\dagger}\varphi_{-\underline{q}^{\prime}}^{(-)}\left(y\right)}\right]}\sum\limits_{q}{\left[{{{\hat{a}}_{\underline{q}}}\varphi_{\underline{q}}^{(+)}\left(x\right)+\hat{a}_{\underline{q}}^{\dagger}\varphi_{\underline{-q}}^{(-)}\left(x\right)}\right]}}\right|0}\right\rangle$
(2.1.8)

since there is a product of sums and the sums are over different momenta we
can collect the summations together and proceed with normal multiplication.

$\displaystyle i\Delta\left({x-y}\right)$ $\displaystyle=$
$\displaystyle\Theta\left({{x_{0}}-{y_{0}}}\right)\left\langle{0\left|{\sum\limits_{q}{\sum\limits_{q^{\prime}}{\left[{\varphi_{\underline{q}}^{(+)}\left(x\right)\varphi_{\underline{q}^{\prime}}^{(+)}\left(y\right){{\hat{a}}_{\underline{q}}}{{\hat{a}}_{\underline{q}^{\prime}}}+\underline{\varphi_{\underline{q}}^{(+)}\left(x\right)\varphi_{-\underline{q}^{\prime}}^{(-)}\left(y\right){{\hat{a}}_{\underline{q}}}\hat{a}_{\underline{q}^{\prime}}^{\dagger}}}\right]}}}\right|0}\right\rangle+$
(2.1.9)
$\displaystyle\Theta\left({{x_{0}}-{y_{0}}}\right)\left\langle{0\left|{\sum\limits_{q}{\sum\limits_{q^{\prime}}{\left[{\varphi_{-\underline{q}}^{(-)}\left(x\right)\varphi_{\underline{q}^{\prime}}^{(+)}\left(y\right)\hat{a}_{\underline{q}}^{\dagger}{{\hat{a}}_{\underline{q}^{\prime}}}+\varphi_{-\underline{q}}^{(-)}\left(x\right)\varphi_{-\underline{q}^{\prime}}^{(-)}\left(y\right)\hat{a}_{\underline{q}}^{\dagger}\hat{a}_{\underline{q}^{\prime}}^{\dagger}}\right]}}}\right|0}\right\rangle+$
$\displaystyle\Theta\left({{y_{0}}-{x_{0}}}\right)\left\langle{0\left|{\sum\limits_{q}{\sum\limits_{q^{\prime}}{\left[{\varphi_{\underline{q}}^{(+)}\left(x\right)\varphi_{\underline{q}^{\prime}}^{(+)}\left(y\right){{\hat{a}}_{\underline{q}^{\prime}}}{{\hat{a}}_{\underline{q}}}+\underline{\varphi_{\underline{q}^{\prime}}^{(+)}\left(y\right)\varphi_{-\underline{q}}^{(-)}\left(x\right){{\hat{a}}_{\underline{q}^{\prime}}}\hat{a}_{\underline{q}}^{\dagger}}}\right]}}}\right|0}\right\rangle+$
$\displaystyle\Theta\left({{y_{0}}-{x_{0}}}\right)\left\langle{0\left|{\sum\limits_{q}{\sum\limits_{q^{\prime}}{\left[{\varphi_{-\underline{q}^{\prime}}^{(-)}\left(y\right)\varphi_{\underline{q}}^{(+)}\left(x\right)\hat{a}_{\underline{q}}^{\prime{\dagger}}{{\hat{a}}_{\underline{q}}}+\varphi_{-\underline{q}^{\prime}}^{(-)}\left(y\right)\varphi_{-\underline{q}}^{(-)}\left(x\right)\hat{a}_{\underline{q}^{\prime}}^{\dagger}\hat{a}_{\underline{q}}^{\dagger}}\right]}}}\right|0}\right\rangle$

all terms in the above expansion are zero except the underlined ones. These
are the only terms which contain the right combination of creation and
annihilation operators which connect the vacuum to the vacuum. This can be
seen if we use the commutation relation [10]

$\left[{{{\hat{a}}_{\underline{q}}};\hat{a}_{\underline{q}^{\prime}}^{\dagger}}\right]={\delta_{qq^{\prime}}}$

The propagator is then

$\displaystyle
i\Delta\left({x-y}\right)=\Theta\left({x_{0}-y_{0}}\right)\sum\limits_{q}{\left[{\varphi_{\underset{\raise
2.10002pt\hbox{$\smash{\scriptscriptstyle-}$}}{q}}^{(+)}\left(x\right)\varphi_{-\underline{q}}^{(-)}\left(y\right)}\right]}+\Theta\left({y_{0}-x_{0}}\right)\sum\limits_{q}{\left[{\varphi_{\underline{q}}^{(+)}\left(y\right)\varphi_{-\underline{q}}^{(-)}\left(x\right)}\right]}$
(2.1.10)

One of the sums over the momenta has been ‘killed’ by the delta function which
picked for us only one momentum state $q^{\prime}=q$. We now proceed by using
the plane wave solutions

$\displaystyle\varphi_{\pm\underline{q}}^{(\pm)}\left(x\right)=C_{\pm\underline{q}}e^{\mp
iq^{\mu}\cdot x_{\mu}}$ (2.1.11)

with the coefficient

$\displaystyle C_{\pm\underline{q}}=\frac{1}{{\left({2\pi}\right)^{{\raise
1.50694pt\hbox{$\scriptstyle 3$}\kern-0.70004pt/\kern-1.04996pt\lower
0.75346pt\hbox{$\scriptstyle 2$}}}}}\frac{1}{{\sqrt{2E_{\pm\underline{q}}}}}$
(2.1.12)

and the energy $E_{\pm\underline{q}}=\sqrt{\underline{q}^{2}+m^{2}}$, since
the energy is invariant of the sign of the momentum, the coefficient is in
turn also invariant and thus only the exponential carries the sign which
distinguishes the positive and negative plane waves. We will go from a
discrete set of $q$ to the continuum and the sum over the $q$ states will
change to an integral over the continuous momenta $q$. We shall also
henceforth drop the sign in the energy and the coefficient. Using (2.1.11) in
the propagator (2.1.10) we obtain

$\displaystyle i\Delta\left({x-y}\right)$
$\displaystyle=\Theta\left({x_{0}-y_{0}}\right)\int{C_{\underline{q}}^{2}e^{-iq^{\mu}\cdot\left({x_{\mu}-y_{\mu}}\right)}d^{3}q}+\Theta\left({y_{0}-x_{0}}\right)\int{C_{\underline{q}}^{2}e^{-iq^{\mu}\cdot\left({y_{\mu}-x_{\mu}}\right)}d^{3}q}$
$\displaystyle=\int{\Theta\left({x_{0}-y_{0}}\right)C_{\underline{q}}^{2}e^{-iE_{\underline{q}}\cdot\left({x_{0}-y_{0}}\right)}e^{i\underline{q}\cdot\left({\underline{x}-\underline{y}}\right)}d^{3}q}+\int{\Theta\left({y_{0}-x_{0}}\right)C_{\underline{q}}^{2}e^{-iE_{\underline{q}}\cdot\left({y_{0}-x_{0}}\right)}e^{i\underline{q}\cdot\left({\underline{y}-\underline{x}}\right)}d^{3}q}$
$\displaystyle=\int{\Theta\left({x_{0}-y_{0}}\right)C_{\underline{q}}^{2}e^{-iE_{\underline{q}}\cdot\left({x_{0}-y_{0}}\right)}e^{-i\underline{q}\cdot\left({\underline{y}-\underline{x}}\right)}d^{3}q}+\underline{\int{\Theta\left({y_{0}-x_{0}}\right)C_{\underline{q}}^{2}e^{-iE_{\underline{q}}\cdot\left({y_{0}-x_{0}}\right)}e^{i\underline{q}\cdot\left({\underline{y}-\underline{x}}\right)}d^{3}q}}$
(2.1.13)

making the change of variables $q\to-q$ for the second term which is
underlined in (2.1.13) and noting the earlier comment of the invariance of the
energy and coefficients under a change in sign in the momenta we arrive at

$\displaystyle i\Delta\left({x-y}\right)$ $\displaystyle=$
$\displaystyle\int{\Theta\left({x_{0}-y_{0}}\right)C_{\underline{q}}^{2}e^{-iE_{\underline{q}}\cdot\left({x_{0}-y_{0}}\right)}e^{-i\underline{q}\cdot\left({\underline{y}-\underline{x}}\right)}d^{3}q}+^{*}\int{\Theta\left({y_{0}-x_{0}}\right)C_{\underline{q}}^{2}e^{-iE_{\underline{q}}\cdot\left({y_{0}-x_{0}}\right)}e^{-i\underline{q}\cdot\left({\underline{y}-\underline{x}}\right)}d^{3}q}$
(2.1.14) $\displaystyle=$
$\displaystyle\int{C_{\underline{q}}^{2}e^{-i\underline{q}\cdot\left({\underline{y}-\underline{x}}\right)}\left[{\Theta\left({x_{0}-y_{0}}\right)e^{-iE_{\underline{q}}\cdot\left({x_{0}-y_{0}}\right)}+\Theta\left({y_{0}-x_{0}}\right)e^{-iE_{\underline{q}}\cdot\left({y_{0}-x_{0}}\right)}}\right]d^{3}q}$

* A subtle point to note about the change of variables here: even though there will be an overall negative sign coming from the differential $-d^{3}q$, we suppress the integrals signs denoting only one instead of a multitude, in this case there are three integrals that need to be done but with the change of variables we also swapped the three integration bounds which in turn gives $(-1)^{3}$ and with the negative sign from the differential leads to an overall positive sign.

$\displaystyle X\equiv\Theta\left({x_{0}-y_{0}}\right)e^{-iE_{\underset{\raise
1.50002pt\hbox{$\smash{\scriptscriptstyle-}$}}{q}}\cdot\left({x_{0}-y_{0}}\right)}+\Theta\left({y_{0}-x_{0}}\right)e^{-iE_{\underset{\raise
1.50002pt\hbox{$\smash{\scriptscriptstyle-}$}}{q}}\cdot\left({y_{0}-x_{0}}\right)}$
(2.1.15)

leading to

$\displaystyle
i\Delta\left({x-y}\right)=\int{C_{\underline{q}}^{2}e^{-i\underline{q}\cdot\left({\underline{y}-\underline{x}}\right)}Xd^{3}q}$
(2.1.16)

We need (2.1.16) to be a Lorentz covariant object111Lorentz Covariance is an
extremely powerful idea which has come to be accepted as an imperative
component of any modern Quantum Field Theory. This elegant idea was developed
by Einstein for use in the Special Theory of Relativity. There was earlier
mention of the idea by Mach but no mathematical formulation was developed. so
we will need to go to the complex plane ($\mathbb{C}$) to accomplish this
task. We can write $X$ as

$\displaystyle X=\frac{{2E}}{{2\pi
i}}\int\limits_{-\infty}^{\infty}{\frac{{e^{-i\left({x_{0}-y_{0}}\right)q_{0}}}}{{m_{\pi}^{2}-q^{2}-i\varepsilon^{\prime}}}dq_{0}}$
(2.1.17)

See Appendix A.1 for details. Substituting (2.1.17) into (2.1.16) we obtain

$\displaystyle i\Delta\left({x-y}\right)$ $\displaystyle=$
$\displaystyle\int{\frac{{2E}}{{2\pi
i}}C_{\underline{q}}^{2}\frac{{e^{-i\left({x_{0}-y_{0}}\right)q_{0}}e^{i\underline{q}\cdot\left({\underline{x}-\underline{y}}\right)}}}{{m_{\pi}^{2}-q^{2}-i\varepsilon^{\prime}}}dq_{0}d^{3}q}$
(2.1.18) $\displaystyle=$ $\displaystyle\int{\frac{{2E}}{{2\pi
i}}C_{\underline{q}}^{2}\frac{{e^{-i\left[{q_{0}\left({x_{0}-y_{0}}\right)-\underline{q}\cdot\left({\underline{x}-\underline{y}}\right)}\right]}}}{{m_{\pi}^{2}-q^{2}-i\varepsilon^{\prime}}}dq_{0}d^{3}q}$
$\displaystyle=$ $\displaystyle\int{\frac{{2E}}{{2\pi
i}}C_{\underline{q}}^{2}\frac{{e^{-iq_{\mu}\left({x^{\mu}-y^{\mu}}\right)}}}{{m_{\pi}^{2}-q^{2}-i\varepsilon^{\prime}}}d^{4}q}$
$\displaystyle=$ $\displaystyle\int{\frac{{2E}}{{2\pi
i}}C_{\underline{q}}^{2}\frac{{e^{-iq\cdot\left({x-y}\right)}}}{{m_{\pi}^{2}-q^{2}-i\varepsilon^{\prime}}}d^{4}q}$

inserting the coefficient (2.1.12) into (2.1.18) leads to

$\displaystyle i\Delta\left({x-y}\right)$ $\displaystyle=$
$\displaystyle\int{\frac{{2E}}{{2\pi
i}}\frac{1}{{\left({2\pi}\right)^{3}}}\frac{1}{{2E}}\frac{{e^{-iq\cdot\left({x-y}\right)}}}{{m_{\pi}^{2}-q^{2}-i\varepsilon^{\prime}}}d^{4}q}$
(2.1.19) $\displaystyle=$
$\displaystyle\frac{1}{{\left({2\pi}\right)^{4}}}\int{e^{-iq\cdot\left({x-y}\right)}\frac{{-i}}{{m_{\pi}^{2}-q^{2}-i\varepsilon^{\prime}}}d^{4}q}$
$\displaystyle=$
$\displaystyle\frac{1}{{\left({2\pi}\right)^{4}}}\int{e^{-iq\cdot\left({x-y}\right)}\frac{i}{{q^{2}-m_{\pi}^{2}+i\varepsilon^{\prime}}}d^{4}q}$

(2.1.19) is now a Lorentz covariant object and retains its form in all
inertial reference frames. We note now that the right-hand-side of (2.1.19) is
just the Fourier transform of the propagator from momentum space to coordinate
space. The propagator in momentum is then simply

$\displaystyle\Delta\left(q\right)=\frac{1}{{q^{2}-m_{\pi}^{2}+i\varepsilon^{\prime}}}$
(2.1.20)

Now looking back at the correlator (2.1.4) and the definition of the
propagator (2.1.5) which is shown below

$\displaystyle\left.{\psi_{5}\left({q^{2}}\right)}\right|_{{\text{HAD}}}=2f_{\pi}^{2}m_{\pi}^{4}i\int{e^{iq\cdot
x}\left\langle{0\left|{T\left[{\hat{\varphi}\left(x\right)\hat{\varphi}^{\dagger}\left(0\right)}\right]}\right|0}\right\rangle
d^{4}x}$ (2.1.21) $\displaystyle
i\Delta\left({x-y}\right)=\left\langle{0\left|{T\left[{\hat{\varphi}\left(x\right)\hat{\varphi}^{\dagger}\left(y\right)}\right]}\right|0}\right\rangle$
(2.1.22)

substituting (2.1.22) into (2.1.21) we obtain

$\displaystyle\left.{\psi_{5}\left({q^{2}}\right)}\right|_{{\text{HAD}}}=-2f_{\pi}^{2}m_{\pi}^{4}\int{e^{iq\cdot
x}\Delta\left(x-y\right)d^{4}x}$ (2.1.23)

we note now that the right-hand-side of (2.1.23) is just the Fourier transform
of the propagator from coordinate space to momentum space but we know what
this transform is in momentum space, it is (2.1.20) for the pion $\pi^{0}$,
which leads to

$\displaystyle\left.{\psi_{5}\left({q^{2}}\right)}\right|_{{\text{HAD}}}=-2f_{\pi}^{2}m_{\pi}^{4}\frac{1}{{q^{2}-m_{\pi}^{2}+i\varepsilon^{\prime}}}$
(2.1.24)

and finally in the low energy limit and $\varepsilon^{\prime}\rightarrow 0$

$\displaystyle\left.{\psi_{5}\left({0}\right)}\right|_{{\text{HAD}}}=2f_{\pi}^{2}m_{\pi}^{2}$
(2.1.25)

###  2.2 QCD Part

We now turn to the QCD correlator. Starting from the axial two point
correlating function of quark currents

$\displaystyle\Pi_{\mu\nu}^{5}=i\int{e^{iq\cdot
x}\left\langle{0\left|{T\left[{A_{\mu}\left(x\right)A_{\nu}^{\dagger}\left(0\right)}\right]}\right|0}\right\rangle
d^{4}x}$ (2.2.1)

the time ordered product can be expressed differently as

$\displaystyle
T\left[{A_{\mu}\left(x\right)A_{\nu}^{\dagger}\left(0\right)}\right]=\left[{A_{\mu}\left(x\right);A_{\nu}^{\dagger}\left(0\right)}\right]\Theta\left({x_{0}}\right)+A_{\nu}^{\dagger}\left(0\right)A_{\mu}\left(x\right)$
(2.2.2)

(See Appendix A.2 for proof.) Inserting the time ordered product (2.2.2) into
(2.2.1) we obtain

$\displaystyle\Pi_{\mu\nu}^{5}=i\int{e^{iq\cdot
x}\left\langle{0\left|\left[{A_{\mu}\left(x\right);A_{\nu}^{\dagger}\left(0\right)}\right]\Theta\left({x_{0}}\right)+A_{\nu}^{\dagger}\left(0\right)A_{\mu}\left(x\right)\right|0}\right\rangle
d^{4}x}$ (2.2.3)

we now apply the momentum operator

$\displaystyle q^{\mu}=-i\frac{\partial}{{\partial x_{\mu}}}=i\partial^{\mu}$
(2.2.4)

to (2.2.3)

$\displaystyle q^{\mu}\Pi_{\mu\nu}^{5}=-i^{2}\partial^{\mu}\int{e^{iq\cdot
x}\left\langle{0\left|\left[{A_{\mu}\left(x\right);A_{\nu}^{\dagger}\left(0\right)}\right]\Theta\left({x_{0}}\right)+A_{\nu}^{\dagger}\left(0\right)A_{\mu}\left(x\right)\right|0}\right\rangle
d^{4}x}$ (2.2.5)

since the fields are local the integral in (2.2.5) yields a function of $q$,
acting on it with the momentum operator gives zero which leads to

$\displaystyle 0$ $\displaystyle=$
$\displaystyle-{i^{2}}\int{{\partial^{\mu}}\left\\{{{e^{iq\cdot
x}}\left\langle{0\left|{\left[{{A_{\mu}}\left(x\right);A_{\nu}^{\dagger}\left(0\right)}\right]\Theta\left({{x_{0}}}\right)+A_{\nu}^{\dagger}\left(0\right){A_{\mu}}\left(x\right)}\right|0}\right\rangle}\right\\}{d^{4}}x}$
(2.2.6) $\displaystyle=$ $\displaystyle{q^{\mu}}\underline{i\int{{e^{iq\cdot
x}}\left\langle{0\left|{\left[{{A_{\mu}}\left(x\right);A_{\nu}^{\dagger}\left(0\right)}\right]\Theta\left({{x_{0}}}\right)+A_{\nu}^{\dagger}\left(0\right){A_{\mu}}\left(x\right)}\right|0}\right\rangle{d^{4}}x}}$
$\displaystyle+$ $\displaystyle\int{{e^{iq\cdot
x}}\left\langle{0\left|{\underline{\left[{{\partial^{\mu}}{A_{\mu}}\left(x\right);A_{\nu}^{\dagger}\left(0\right)}\right]\Theta\left({{x_{0}}}\right)}+\left[{{A_{\mu}}\left(x\right);A_{\nu}^{\dagger}\left(0\right)}\right]{\partial^{\mu}}\Theta\left({{x_{0}}}\right)+\underline{A_{\nu}^{\dagger}\left(0\right){\partial^{\mu}}{A_{\mu}}\left(x\right)}}\right|0}\right\rangle{d^{4}}x}$

we note that the first underlined term in (2.2.6 ) is $\Pi_{\mu\nu}^{5}$ in
(2.2.3) and interestingly the second and third underlined terms together form
a time-ordered product

$\displaystyle\left[{{\partial^{\mu}}{A_{\mu}}\left(x\right);A_{\nu}^{\dagger}\left(0\right)}\right]\Theta\left({{x_{0}}}\right)+A_{\nu}^{\dagger}\left(0\right){\partial^{\mu}}{A_{\mu}}\left(x\right)=T\left[{{\partial^{\mu}}{A_{\mu}}\left(x\right)A_{\nu}^{\dagger}\left(0\right)}\right]$
(2.2.7)

which leads to

$\displaystyle 0$
$\displaystyle={q^{\mu}}\Pi_{{}_{\mu\nu}}^{5}+\int{{e^{iq\cdot
x}}\left\langle{0\left|{T\left[{{\partial^{\mu}}{A_{\mu}}\left(x\right)A_{\nu}^{\dagger}\left(0\right)}\right]}\right|0}\right\rangle{d^{4}}x}+\int{{e^{iq\cdot
x}}\left\langle{0\left|{\left[{{A_{\mu}}\left(x\right);A_{\nu}^{\dagger}\left(0\right)}\right]{\partial^{\mu}}\Theta\left({{x_{0}}}\right)}\right|0}\right\rangle{d^{4}}x}$
$\displaystyle={q^{\mu}}\Pi_{{}_{\mu\nu}}^{5}+\int{{e^{iq\cdot
x}}\left\langle{0\left|{T\left[{{\partial^{\mu}}{A_{\mu}}\left(x\right)A_{\nu}^{\dagger}\left(0\right)}\right]}\right|0}\right\rangle{d^{4}}x}+\int{{e^{iq\cdot
x}}\left\langle{0\left|{\left[{{A_{0}}\left(x\right);A_{\nu}^{\dagger}\left(0\right)}\right]{\partial^{0}}\Theta\left({{x_{0}}}\right)}\right|0}\right\rangle{d^{4}}x}$
(2.2.8)

but the derivative of the Heaviside function leads to the delta function i.e.
${\partial^{0}}\Theta\left({{x_{0}}}\right)=\delta\left({{x_{0}}}\right)$ and
defining the following functions:

$\displaystyle\Pi_{{}_{\nu}}^{5}$ $\displaystyle\equiv$
$\displaystyle-\int{{e^{iq\cdot
x}}\left\langle{0\left|{T\left[{{\partial^{\mu}}{A_{\mu}}\left(x\right)A_{\nu}^{\dagger}\left(0\right)}\right]}\right|0}\right\rangle{d^{4}}x}$
(2.2.9) $\displaystyle\Delta_{{}_{\nu}}^{5}$ $\displaystyle\equiv$
$\displaystyle\int{{e^{iq\cdot
x}}\left\langle{0\left|{\left[{{A_{0}}\left(x\right);A_{\nu}^{\dagger}\left(0\right)}\right]\delta\left({{x_{0}}}\right)}\right|0}\right\rangle{d^{4}}x}$
(2.2.10)

and substituting into (2.2.8) gives us the Ward(I) identity222the ‘I’ we have
attached to this Ward identity serves the purpose of distinguishing it from
the Ward(II) identity which will appear later. It is not a convention and will
not be found in other articles on the topic

$\displaystyle
0={q^{\mu}}\Pi_{{}_{\mu\nu}}^{5}-\Pi_{{}_{\nu}}^{5}+\Delta_{{}_{\nu}}^{5}$
(2.2.11)

We can express (2.2.9) as

$\displaystyle\Pi_{{}_{\nu}}^{5}=\int{{e^{-iq\cdot
x}}\left\langle{0\left|{T\left[{A_{\nu}^{\dagger}\left(x\right){\partial^{\mu}}{A_{\mu}}\left(0\right)}\right]}\right|0}\right\rangle{d^{4}}x}$
(2.2.12)

See Appendix A.3 for details. We now apply the momentum operator (2.2.4) to
(2.2.12). Since the fields are local the derivative yields zero

$\displaystyle 0$ $\displaystyle=$
$\displaystyle-i\partial^{\nu}\int{e^{-iq\cdot
x}\left\langle{0\left|{T\left[{A_{\nu}^{\dagger}\left(x\right)\partial^{\mu}A_{\mu}\left(0\right)}\right]}\right|0}\right\rangle
d^{4}x}$ (2.2.13) $\displaystyle=$
$\displaystyle-i\int{\partial^{\nu}\left\\{{e^{-iq\cdot
x}\left\langle{0\left|{T\left[{A_{\nu}^{\dagger}\left(x\right)\partial^{\mu}A_{\mu}\left(0\right)}\right]}\right|0}\right\rangle}\right\\}d^{4}x}$
$\displaystyle=$ $\displaystyle-i\int{\left\\{{-iq^{\nu}e^{-iq\cdot
x}\left\langle{0\left|{T\left[{A_{\nu}^{\dagger}\left(x\right)\partial^{\mu}A_{\mu}\left(0\right)}\right]}\right|0}\right\rangle+e^{-iq\cdot
x}\left\langle{0\left|{\partial^{\nu}T\left[{A_{\nu}^{\dagger}\left(x\right)\partial^{\mu}A_{\mu}\left(0\right)}\right]}\right|0}\right\rangle}\right\\}d^{4}x}$
$\displaystyle=$ $\displaystyle-q^{\nu}\underline{\int{e^{-iq\cdot
x}\left\langle{0\left|{T\left[{A_{\nu}^{\dagger}\left(x\right)\partial^{\mu}A_{\mu}\left(0\right)}\right]}\right|0}\right\rangle
d^{4}x}}-i\int{e^{-iq\cdot
x}\left\langle{0\left|{\partial^{\nu}T\left[{A_{\nu}^{\dagger}\left(x\right)\partial^{\mu}A_{\mu}\left(0\right)}\right]}\right|0}\right\rangle
d^{4}x}$

we first note that the underlined term in (2.2.13) is the $\Pi_{{}_{\nu}}^{5}$
in (2.2.12) and the time-ordered product can be expressed as

$\displaystyle
T\left[{A_{\nu}^{\dagger}\left(x\right)\partial^{\mu}A_{\mu}\left(0\right)}\right]=\left[{A_{\nu}^{\dagger}\left(x\right);\partial^{\mu}A_{\mu}\left(0\right)}\right]\Theta\left({x_{0}}\right)+\partial^{\mu}A_{\mu}\left(0\right)A_{\nu}^{\dagger}\left(x\right)$
(2.2.14)

now acting on (2.2.14) by the partial derivative operator we obtain

$\displaystyle\partial^{\nu}T\left[{A_{\nu}^{\dagger}\left(x\right)\partial^{\mu}A_{\mu}\left(0\right)}\right]$
$\displaystyle=$
$\displaystyle\left[{\partial^{\nu}A_{\nu}^{\dagger}\left(x\right);\partial^{\mu}A_{\mu}\left(0\right)}\right]\Theta\left({x_{0}}\right)+\left[{A_{\nu}^{\dagger}\left(x\right);\partial^{\mu}A_{\mu}\left(0\right)}\right]\partial^{\nu}\Theta\left({x_{0}}\right)+\partial^{\mu}A_{\mu}\left(0\right)\partial^{\nu}A_{\nu}^{\dagger}\left(x\right)$
$\displaystyle=$
$\displaystyle\left[{\partial^{\nu}A_{\nu}^{\dagger}\left(x\right);\partial^{\mu}A_{\mu}\left(0\right)}\right]\Theta\left({x_{0}}\right)+\left[{A_{0}^{\dagger}\left(x\right);\partial^{\mu}A_{\mu}\left(0\right)}\right]\partial^{0}\Theta\left({x_{0}}\right)+\partial^{\mu}A_{\mu}\left(0\right)\partial^{\nu}A_{\nu}^{\dagger}\left(x\right)$
$\displaystyle=$
$\displaystyle\underline{\left[{\partial^{\nu}A_{\nu}^{\dagger}\left(x\right);\partial^{\mu}A_{\mu}\left(0\right)}\right]\Theta\left({x_{0}}\right)}+\left[{A_{0}^{\dagger}\left(x\right);\partial^{\mu}A_{\mu}\left(0\right)}\right]\delta\left({x_{0}}\right)+\underline{\partial^{\mu}A_{\mu}\left(0\right)\partial^{\nu}A_{\nu}^{\dagger}\left(x\right)}$

the underlined terms of (LABEL:ptop) forms a new time-ordered product

$\displaystyle\left[{\partial^{\nu}A_{\nu}^{\dagger}\left(x\right);\partial^{\mu}A_{\mu}\left(0\right)}\right]\Theta\left({x_{0}}\right)+\partial^{\mu}A_{\mu}\left(0\right)\partial^{\nu}A_{\nu}^{\dagger}\left(x\right)=T\left[{\partial^{\nu}A_{\nu}^{\dagger}\left(x\right)\partial^{\mu}A_{\mu}\left(0\right)}\right]$
(2.2.16)

so the time-ordered product (LABEL:ptop) is

$\displaystyle\partial^{\nu}T\left[{A_{\nu}^{\dagger}\left(x\right)\partial^{\mu}A_{\mu}\left(0\right)}\right]=T\left[{\partial^{\nu}A_{\nu}^{\dagger}\left(x\right)\partial^{\mu}A_{\mu}\left(0\right)}\right]+\left[{A_{0}^{\dagger}\left(x\right);\partial^{\mu}A_{\mu}\left(0\right)}\right]\delta\left({x_{0}}\right)$
(2.2.17)

substituting (2.2.17) into (2.2.13) we obtain

$\displaystyle 0=-q^{\nu}\Pi_{\nu}^{5}-i\int{e^{-iq\cdot
x}\left\langle{0\left|{T\left[{\partial^{\nu}A_{\nu}^{\dagger}\left(x\right)\partial^{\mu}A_{\mu}\left(0\right)}\right]}\right|0}\right\rangle
d^{4}x}-i\int{e^{-iq\cdot
x}\left\langle{0\left|{\left[{A_{0}^{\dagger}\left(x\right);\partial^{\mu}A_{\mu}\left(0\right)}\right]\delta\left({x_{0}}\right)}\right|0}\right\rangle
d^{4}x}$ (2.2.18)

we now define a new function

$\displaystyle\Delta_{1}^{5}\equiv-i\int{e^{-iq\cdot
x}\left\langle{0\left|{\left[{A_{0}^{\dagger}\left(x\right);\partial^{\mu}A_{\mu}\left(0\right)}\right]\delta\left({x_{0}}\right)}\right|0}\right\rangle
d^{4}x}$ (2.2.19)

and then we arrive at

$\displaystyle 0=-q^{\nu}\Pi_{\nu}^{5}-i\int{e^{-iq\cdot
x}\left\langle{0\left|{T\left[{\partial^{\nu}A_{\nu}^{\dagger}\left(x\right)\partial^{\mu}A_{\mu}\left(0\right)}\right]}\right|0}\right\rangle
d^{4}x}+\Delta_{1}^{5}$ (2.2.20)

to proceed further with the analysis we need to cast second term of (2.2.20)
into a recognizable form. We can express (2.2.20) as

$\displaystyle-i\int{e^{-iq\cdot
x}\left\langle{0\left|{T\left[{\partial^{\nu}A_{\nu}^{\dagger}\left(x\right)\partial^{\mu}A_{\mu}\left(0\right)}\right]}\right|0}\right\rangle
d^{4}x}=i\int{e^{iq\cdot
x}\left\langle{0\left|{T\left[{\partial^{\mu}A_{\mu}\left(x\right)\partial^{\nu}A_{\nu}^{\dagger}\left(0\right)}\right]}\right|0}\right\rangle
d^{4}x}$ (2.2.21)

See Appendix A.4 for details, the right-hand-side of (2.2.21) is a vaguely
familiar function and if we refer back to (2.1.1) which is shown below

$\displaystyle\left.{\psi_{5}\left({q^{2}}\right)}\right|_{{\text{HAD}}}=i\int{e^{iq\cdot
x}\left\langle{0\left|{T\left[{\partial^{\mu}A_{\mu}(x)\partial^{\nu}A_{\nu}^{\dagger}(0)}\right]}\right|0}\right\rangle
dx}$ (2.2.22)

so

$\displaystyle-i\int{e^{-iq\cdot
x}\left\langle{0\left|{T\left[{\partial^{\nu}A_{\nu}^{\dagger}\left(x\right)\partial^{\mu}A_{\mu}\left(0\right)}\right]}\right|0}\right\rangle
d^{4}x}=\psi_{5}\left(q^{2}\right)$ (2.2.23)

now going back and substituting (2.2.23) into (2.2.20) we arrive at the
Ward(II) identity

$\displaystyle 0=-q^{\nu}\Pi_{\nu}^{5}+\psi_{5}+\Delta_{1}^{5}$ (2.2.24)

###  2.3 The Gell-Mann-Oakes-Renner Relation

We have now set the stage and all the pieces are ready to be put together to
demonstrate the Gell-Mann Oakes Renner relation. We start with the Ward(I)
identity (2.2.11)

$\displaystyle
0={q^{\mu}}\Pi_{{}_{\mu\nu}}^{5}-\Pi_{{}_{\nu}}^{5}+\Delta_{{}_{\nu}}^{5}$
(2.3.1)

Multiplying both sides of the (2.3.1) by $q^{\nu}$

$\displaystyle
0=q^{\nu}q^{\mu}\Pi_{\mu\nu}^{5}-q^{\nu}\Pi_{\nu}^{5}+q^{\nu}\Delta_{\nu}^{5}$
(2.3.2)

now substituting the Ward(II) identity (2.2.24) into (2.3.2)

$\displaystyle
0=q^{\nu}q^{\mu}\Pi_{\mu\nu}^{5}-\psi_{5}-\Delta_{1}^{5}+q^{\nu}\Delta_{\nu}^{5}$
(2.3.3)

and rearranging we obtain

$\displaystyle
q^{\nu}\left({q^{\mu}\Pi_{\mu\nu}^{5}+\Delta_{\nu}^{5}}\right)-\psi_{5}=\Delta_{1}^{5}$
(2.3.4)

If we take the limit on both sides of (2.3.4) as $q^{\nu}\to 0$

$\displaystyle\mathop{\lim}\limits_{q^{\nu}\to
0}\left[{q^{\nu}\left({q^{\mu}\Pi_{\mu\nu}^{5}+\Delta_{\nu}^{5}}\right)-\psi_{5}}\right]=\mathop{\lim}\limits_{q^{\nu}\to
0}\Delta_{1}^{5}$ (2.3.5)

Since the terms in the round brackets are analytic functions this just gives

$\displaystyle\mathop{-\lim}\limits_{q^{\nu}\to
0}\psi_{5}=\mathop{\lim}\limits_{q^{\nu}\to 0}\Delta_{1}^{5}$ (2.3.6)

but the limit of the right-hand-side

$\displaystyle\mathop{\lim}\limits_{q^{\nu}\to 0}\Delta_{1}^{5}$
$\displaystyle=$ $\displaystyle-i\mathop{\lim}\limits_{q^{\nu}\to
0}\int{e^{-iq\cdot
x}\left\langle{0\left|{\left[{A_{0}^{\dagger}\left(x\right);\partial^{\mu}A_{\mu}\left(0\right)}\right]\delta\left({x_{0}}\right)}\right|0}\right\rangle
d^{4}x}$ (2.3.7) $\displaystyle=$
$\displaystyle-i\int{\left\langle{0\left|{\left[{A_{0}^{\dagger}\left({0,\underline{x}}\right);\partial^{\mu}A_{\mu}\left(0,\underline{0}\right)}\right]}\right|0}\right\rangle
d^{3}x}$

So we can now express the currents and their derivatives as a product of quark
fields

$\displaystyle\partial^{\mu}A_{\mu}\left({0;\underline{0}}\right)$
$\displaystyle=$ $\displaystyle
i\left({m_{i}+m_{j}}\right)\overline{\psi}_{j}\left(0\right)\gamma_{5}\psi_{i}\left(0\right)$
(2.3.8) $\displaystyle A_{\mu}\left({0,\underline{x}}\right)$ $\displaystyle=$
$\displaystyle\overline{\psi}_{j}\left(0\right)\gamma_{\mu}\gamma_{5}\psi_{i}\left(0\right)$
(2.3.9)

The $\mu=0$ component of (2.3.9) is :

$\displaystyle
A_{0}\left({0,\underline{x}}\right)=\overline{\psi}_{j}\left(0\right)\gamma_{0}\gamma_{5}\psi_{i}\left(0\right)$
(2.3.10)

and the conjugate field can be expressed as :

$\displaystyle\overline{\psi}_{j}=\psi_{j}^{\dagger}\gamma_{0}$ (2.3.11)

Some useful properties of the gamma matrices are listed below

$\displaystyle\gamma_{0}\gamma_{0}$ $\displaystyle=$ $\displaystyle 1$
(2.3.12) $\displaystyle\gamma_{5}\gamma_{5}$ $\displaystyle=$ $\displaystyle
1$ (2.3.13) $\displaystyle\gamma_{5}^{\dagger}$ $\displaystyle=$
$\displaystyle\gamma_{5}$ (2.3.14)
$\displaystyle\left[{\gamma_{0},\gamma_{5}}\right]$ $\displaystyle=$
$\displaystyle 0$ (2.3.15)

Using properties (2.3.11-2.3.14) we obtain

$\displaystyle A_{0}$ $\displaystyle=$
$\displaystyle\psi_{j}^{\dagger}\gamma_{0}\gamma_{0}\gamma_{5}\psi_{i}=\psi_{j}^{\dagger}\gamma_{5}\psi_{i}$
$\displaystyle A_{0}^{\dagger}$ $\displaystyle=$
$\displaystyle\left({\psi_{j}^{\dagger}\gamma_{5}\psi_{i}}\right)^{\dagger}=\psi_{i}^{\dagger}\gamma_{5}^{\dagger}\left({\psi_{j}^{\dagger}}\right)^{\dagger}=\psi_{i}^{\dagger}\gamma_{5}\psi_{j}$
(2.3.16) $\displaystyle\partial^{\mu}A_{\mu}$ $\displaystyle=$ $\displaystyle
i\left({m_{i}+m_{j}}\right)\psi_{j}^{\dagger}\gamma_{0}\gamma_{5}\psi_{i}$
(2.3.17)

and substituting (2.3.16) and (2.3.17) into the commutator of (2.3.7), we get

$\displaystyle\left[{A_{0}^{\dagger}\left({0,\underline{x}}\right);\partial^{\mu}A_{\mu}\left({0;\underline{0}}\right)}\right]$
$\displaystyle=$
$\displaystyle\left[{\psi_{i}^{\dagger}\gamma_{5}\psi_{j};i\left({m_{i}+m_{j}}\right)\psi_{j}^{\dagger}\gamma_{0}\gamma_{5}\psi_{i}}\right]$
(2.3.18) $\displaystyle=$ $\displaystyle
i\left({m_{i}+m_{j}}\right)\left[{\psi_{i}^{\dagger}\gamma_{5}\psi_{j};\psi_{j}^{\dagger}\gamma_{0}\gamma_{5}\psi_{i}}\right]$

We can move the gamma matrices around, but being careful not to move them past
each other since they do not commute with each other, we arrive at

$\displaystyle\left[{\psi_{i}^{\dagger}\gamma_{5}\psi_{j};\psi_{j}^{\dagger}\gamma_{0}\gamma_{5}\psi_{i}}\right]$
$\displaystyle=$
$\displaystyle\psi_{i}^{\dagger}\gamma_{5}\psi_{j}\psi_{j}^{\dagger}\gamma_{0}\psi_{i}\gamma_{5}-\psi_{j}^{\dagger}\gamma_{0}\psi_{i}\psi_{i}^{\dagger}\gamma_{5}\psi_{j}\gamma_{5}$
(2.3.19) $\displaystyle=$
$\displaystyle\left({\psi_{i}^{\dagger}\gamma_{5}\psi_{j}\psi_{j}^{\dagger}\gamma_{0}\psi_{i}-\psi_{j}^{\dagger}\gamma_{0}\psi_{i}\psi_{i}^{\dagger}\gamma_{5}\psi_{j}}\right)\gamma_{5}$
$\displaystyle=$
$\displaystyle\left[{\psi_{i}^{\dagger}\psi_{j}\gamma_{5};\psi_{j}^{\dagger}\psi_{i}\gamma_{0}}\right]\gamma_{5}$

We now use an operator identity

$\displaystyle\left[{A^{\dagger}B;C^{\dagger}D}\right]$ $\displaystyle=$
$\displaystyle
A^{\dagger}\left\\{{B;C^{\dagger}}\right\\}D-C^{\dagger}\left\\{{D;A^{\dagger}}\right\\}B-\left\\{{A^{\dagger};C^{\dagger}}\right\\}BD+C^{\dagger}A^{\dagger}\left\\{{B;D}\right\\}$
(2.3.20)

were $A,B,C,D$ are Grassman [11] or Fermion fields. We can express the right-
hand-side of (2.3.20) as

$\displaystyle\left[{A^{\dagger}B;C^{\dagger}D}\right]=A^{\dagger}D\delta_{BC}-C^{\dagger}B\delta_{AD}-BD\delta_{AC}+C^{\dagger}A^{\dagger}\delta_{BD}$
(2.3.21)

and considering $B=C$ and $A=D$ $\Rightarrow$ $\delta_{AC}=\delta_{BD}=0$ so

$\displaystyle\left[{A^{\dagger}B;C^{\dagger}D}\right]=A^{\dagger}D\delta_{BC}-C^{\dagger}B\delta_{AD}$
(2.3.22)

Now comparing the terms in the commutator of (2.3.19) and (2.3.22) we see that

$\displaystyle A^{\dagger}$ $\displaystyle=$
$\displaystyle\psi_{i}^{\dagger}\hfill$ (2.3.23) $\displaystyle B$
$\displaystyle=$ $\displaystyle\psi_{j}\gamma_{5}\hfill$ (2.3.24)
$\displaystyle C^{\dagger}$ $\displaystyle=$
$\displaystyle\psi_{j}^{\dagger}\hfill$ (2.3.25) $\displaystyle D$
$\displaystyle=$ $\displaystyle\psi_{i}\gamma_{0}$ (2.3.26)

together with

$\displaystyle\delta_{BC}$ $\displaystyle=$
$\displaystyle\delta\left({\underline{x}}\right)\gamma_{5}\hfill$ (2.3.27)
$\displaystyle\delta_{AD}$ $\displaystyle=$
$\displaystyle\delta\left({\underline{x}}\right)\gamma_{0}$ (2.3.28)

We obtain

$\displaystyle\left[{\psi_{i}^{\dagger}\psi_{j}\gamma_{5};\psi_{j}^{\dagger}\psi_{i}\gamma_{0}}\right]$
$\displaystyle=$
$\displaystyle\psi_{i}^{\dagger}\psi_{i}\gamma_{0}\delta\left({\underline{x}}\right)\gamma_{5}-\psi_{j}^{\dagger}\psi_{j}\gamma_{5}\delta\left({\underline{x}}\right)\gamma_{0}$
(2.3.29) $\displaystyle=$
$\displaystyle\left[{\psi_{i}^{\dagger}\psi_{i}\gamma_{0}\gamma_{5}-\psi_{j}^{\dagger}\psi_{j}\gamma_{5}\gamma_{0}}\right]\delta\left({\underline{x}}\right)$
$\displaystyle=$
$\displaystyle\left[{\psi_{i}^{\dagger}\gamma_{0}\psi_{i}\gamma_{5}+\psi_{j}^{\dagger}\gamma_{0}\psi_{j}\gamma_{5}}\right]\delta\left({\underline{x}}\right)$
$\displaystyle=$
$\displaystyle\left[{\bar{\psi}_{i}\psi_{i}+\bar{\psi}_{j}\psi_{j}}\right]\delta\left({\underline{x}}\right)\gamma_{5}$

Substituting (2.3.29) into (2.3.19) and noting property (2.3.13), which is in
turn substituted into (2.3.18) and finally substituting back into (2.3.7), we
obtain

$\displaystyle\mathop{\lim}\limits_{q^{\nu}\to 0}\Delta_{1}^{5}$
$\displaystyle=$
$\displaystyle-i^{2}\left({m_{i}+m_{j}}\right)\int{\left\langle{0\left|{\bar{\psi}_{i}\psi_{i}+\bar{\psi}_{j}\psi_{j}}\right|0}\right\rangle\delta\left({\underline{x}}\right)d^{3}x}$
(2.3.30) $\displaystyle=$
$\displaystyle\left({m_{i}+m_{j}}\right)\left\langle{0\left|{\bar{\psi}_{i}\psi_{i}+\bar{\psi}_{j}\psi_{j}}\right|0}\right\rangle$

and in the case of the $SU(2)\otimes SU(2)$ symmetry with just the up(u) and
down(d) quarks we have

$\displaystyle\mathop{\lim}\limits_{q^{\nu}\to
0}\Delta_{1}^{5}=\left({m_{u}+m_{d}}\right)\left\langle{0\left|{\bar{u}u+\bar{d}d}\right|0}\right\rangle$
(2.3.31)

so referring back to (2.3.6) we have

$\displaystyle\mathop{-\lim}\limits_{q^{\nu}\to
0}\psi_{5}=\left({m_{u}+m_{d}}\right)\left\langle{0\left|{\bar{u}u+\bar{d}d}\right|0}\right\rangle$
(2.3.32)

but the left-hand-side of (2.3.32) also has a low energy limit which was
worked out in (2.1.25) so we finally arrive at the Gell-Mann Oakes Renner low
energy relation

$\displaystyle-2f_{\pi}^{2}m_{\pi}^{2}=\left({m_{u}+m_{d}}\right)\left\langle{0\left|{\bar{u}u+\bar{d}d}\right|0}\right\rangle$
(2.3.33)

In the chiral limit the up and down quarks have zero mass and we expect that
there would be three massless Goldstone boson’s when this symmetry is broken.
Indeed we do find these three boson’s the $\pi^{-},\pi^{0},\pi^{+}$ but their
masses are not zero. This is an indication to us that the Chiral symmetry is
only an approximate symmetry and that the quarks are not massless. Looking at
(2.3.33) we see that in the limit of $m_{u}=m_{d}=0$ the symmetry is exact and
the pion’s have zero mass. This provides an explanation for the existence of
the light mass mesons we observe from experiment.

##  3 The QCD Finite Energy Sum Rule

In this project we will estimate how much the Gell-Mann-Oakes-Renner relation
in (2.3.33) is broken with the inclusion of chiral corrections to (2.3.33).
The amount the symmetry is broken by is measured by the parameter
$\delta_{\pi}$ which is given by the relation

$\displaystyle\psi_{5}\left(0\right)\equiv-\left({m_{u}+m_{d}}\right)\left\langle{0|\bar{u}u+\bar{d}d|0}\right\rangle=2f_{\pi}^{2}M_{\pi}^{2}\left({1-\delta_{\pi}}\right)$
(3.1)

We calculate $\psi_{5}\left(0\right)$ using the QCD Sum rules and perturbation
theory and thus make a direct estimate of $\delta_{\pi}$

We will derive the Finite Energy Sum Rule(FESR) [12] in this section. We start
by using the Cauchy-Goursat Theorem on the pseudoscalar correlator
$\psi_{5}(z)$ which has a branch cut along the positive real axis in the
complex plane We will choose the integration contour suitably so that
$\psi_{5}(z)$ is an _analytic_ function in the region of integration which
will be as shown in figure 1. The integration region is to the left of the
bounding contour if we follow the contour in an anti-clockwise manner. In this
way we avoid the poles which are on the positive real axis and we can consider
$\psi_{5}(z)$ to be _analytic_

Figure 1: Integration contour used for correlator. X indicates a pole and the
broken line indicates a branch cut in the complex plane along the positive
real axis corresponding to the resonances

We will also introduce a kernel $\Delta(z)$ which is an arbitrary complete(the
zeroth and first order terms are included) quadratic polynomial of $z$. We
introduce $\Delta(z)$ to reduce the systematic uncertainties which are
associated to the resonances, by constraining $\Delta(z)=0$ at the two
resonance peaks. This fixes for us the coefficients of the kernel uniquely.

$\displaystyle\psi_{5}\left({z_{0}}\right)\Delta\left({z_{0}}\right)$
$\displaystyle=\frac{1}{{2\pi
i}}\mathop{\oint\mkern-0.8mu}\limits_{C}{\frac{{\psi_{5}\left(z\right)\Delta\left(z\right)}}{{z-z_{0}}}dz}$
$\displaystyle=\mathop{\lim}\limits_{\varepsilon\to 0}\frac{1}{{2\pi
i}}\int\limits_{z_{th}}^{R}{\frac{{\psi_{5}\left({z+i\varepsilon}\right)\Delta\left({z+i\varepsilon}\right)}}{{z-z_{0}}}dz}+\frac{1}{{2\pi
i}}\mathop{\oint\mkern-0.8mu}\limits_{C\left(R\right)}{\frac{{\psi_{5}\left(z\right)\Delta\left(z\right)}}{{z-z_{0}}}dz}+\mathop{\lim}\limits_{\varepsilon\to
0}\frac{1}{{2\pi
i}}\int\limits_{R}^{z_{th}}{\frac{{\psi_{5}\left({z-i\varepsilon}\right)\Delta\left({z-i\varepsilon}\right)}}{{z-z_{0}}}dz}$
$\displaystyle=\frac{1}{{2\pi
i}}\mathop{\oint\mkern-0.8mu}\limits_{C\left(R\right)}{\frac{{\psi_{5}\left(z\right)\Delta\left(z\right)}}{{z-z_{0}}}dz}+\mathop{\lim}\limits_{\varepsilon\to
0}\frac{1}{{2\pi
i}}\int\limits_{z_{th}}^{R}{\frac{{\psi_{5}\left({z+i\varepsilon}\right)\Delta\left({z+i\varepsilon}\right)}}{{z-z_{0}}}dz}-\mathop{\lim}\limits_{\varepsilon\to
0}\frac{1}{{2\pi
i}}\int\limits_{z_{th}}^{R}{\frac{{\psi_{5}\left({z-i\varepsilon}\right)\Delta\left({z-i\varepsilon}\right)}}{{z-z_{0}}}dz}$
$\displaystyle=\frac{1}{{2\pi
i}}\mathop{\oint\mkern-0.8mu}\limits_{C\left(R\right)}{\frac{{\psi_{5}\left(z\right)\Delta\left(z\right)}}{{z-z_{0}}}dz}+\mathop{\lim}\limits_{\varepsilon\to
0}\frac{1}{{2\pi
i}}\int\limits_{z_{th}}^{R}{\frac{{\psi_{5}\left({z+i\varepsilon}\right)\Delta\left({z+i\varepsilon}\right)-\psi_{5}\left({z-i\varepsilon}\right)\Delta\left({z-i\varepsilon}\right)}}{{z-z_{0}}}dz}$
(3.2)

The residue from the pole is included in the hadronic spectral function. Since
$\Delta(z)$ is a single valued analytic function, in the limit

$\mathop{\lim}\limits_{\varepsilon\to
0}\Delta\left({z+i\varepsilon}\right)=\mathop{\lim}\limits_{\varepsilon\to
0}\Delta\left({z-i\varepsilon}\right)=\Delta\left(z\right)$

${\psi_{5}\left({z+i\varepsilon}\right)}$ and
${\psi_{5}\left({z-i\varepsilon}\right)}$ are the functions evaluated in the
two different branches or Riemann sheets of the function and we have that

$\mathop{\lim}\limits_{\varepsilon\to
0}\psi_{5}\left({z+i\varepsilon}\right)\neq\mathop{\lim}\limits_{\varepsilon\to
0}\psi_{5}\left({z-i\varepsilon}\right)$

we can now write the (3.2) as

$\displaystyle\psi_{5}\left({z_{0}}\right)\Delta\left({z_{0}}\right)=\frac{1}{{2\pi
i}}\mathop{\oint\mkern-0.8mu}\limits_{C\left(R\right)}{\frac{{\psi_{5}\left(z\right)\Delta\left(z\right)}}{{z-z_{0}}}dz}+\frac{1}{{2\pi
i}}\int\limits_{z_{th}}^{R}{\frac{{\Delta\left(z\right)}}{{z-z_{0}}}\mathop{\lim}\limits_{\varepsilon\to
0}\left[{\psi_{5}\left({z+i\varepsilon}\right)-\psi_{5}\left({z-i\varepsilon}\right)}\right]dz}$
(3.3)

let us now consider an analytic function $f(z+i\varepsilon)$, we can Laurent
expand this function around $z$ as

$\displaystyle
f\left({z+i\varepsilon}\right)=\sum\limits_{n=0}^{\infty}{\frac{{f^{\left(n\right)}\left({i\varepsilon}\right)}}{{n!}}z^{n}}$
(3.4)

were $f^{\left(n\right)}\left({i\varepsilon}\right)$ is the nth derivative of
the function $f(z)$ evaluated at the point $z=i\varepsilon$. We can do the
same for the function $f(z-i\varepsilon)$ which is

$\displaystyle
f\left({z-i\varepsilon}\right)=\sum\limits_{n=0}^{\infty}{\frac{{f^{\left(n\right)}\left({-i\varepsilon}\right)}}{{n!}}z^{n}}$
(3.5)

We can now evaluate the difference of these functions

$\displaystyle f\left({z+i\varepsilon}\right)-f\left({z-i\varepsilon}\right)$
$\displaystyle=$
$\displaystyle\sum\limits_{n=0}^{\infty}{\frac{{f^{\left(n\right)}\left({i\varepsilon}\right)}}{{n!}}z^{n}}-\sum\limits_{n=0}^{\infty}{\frac{{f^{\left(n\right)}\left({-i\varepsilon}\right)}}{{n!}}z^{n}}$
$\displaystyle=$
$\displaystyle\sum\limits_{n=0}^{\infty}{\frac{{z^{n}}}{{n!}}}\left[{f^{\left(n\right)}\left({i\varepsilon}\right)-f^{\left(n\right)}\left({-i\varepsilon}\right)}\right]$
$\displaystyle=$
$\displaystyle\sum\limits_{n=0}^{\infty}{\frac{{z^{n}}}{{n!}}}\frac{{d^{n}}}{{dz^{n}}}\left[{\left.{f\left(z\right)}\right|_{z=i\varepsilon}-\left.{f\left(z\right)}\right|_{z=-i\varepsilon}}\right]$
$\displaystyle=$
$\displaystyle\sum\limits_{n=0}^{\infty}{\frac{{z^{n}}}{{n!}}\frac{{d^{n}}}{{dz^{n}}}\left.{\left[{f\left(z\right)-f\left({\bar{z}}\right)}\right]}\right|_{z=i\varepsilon}}$
$\displaystyle=$
$\displaystyle\sum\limits_{n=0}^{\infty}{\frac{{z^{n}}}{{n!}}\frac{{d^{n}}}{{dz^{n}}}\left.{\left[{f\left(z\right)-\overline{f\left(z\right)}}\right]}\right|_{z=i\varepsilon}}$
$\displaystyle=$
$\displaystyle\sum\limits_{n=0}^{\infty}{\frac{{z^{n}}}{{n!}}\frac{{d^{n}}}{{dz^{n}}}\left.{2i\operatorname{Im}\left[{f\left(z\right)}\right]}\right|_{z=i\varepsilon}}$
$\displaystyle=$ $\displaystyle
2i\operatorname{Im}\left[{\sum\limits_{n=0}^{\infty}{\frac{{z^{n}}}{{n!}}\left.{\frac{{d^{n}}}{{dz^{n}}}f\left(z\right)}\right|_{z=i\varepsilon}}}\right]$
$\displaystyle=$ $\displaystyle
2i\operatorname{Im}\left[{\sum\limits_{n=0}^{\infty}{\frac{{z^{n}}}{{n!}}f^{\left(n\right)}\left({i\varepsilon}\right)}}\right]$
$\displaystyle=$ $\displaystyle
2i\operatorname{Im}\left[{f\left({z+i\varepsilon}\right)}\right]$ (3.7)

We have used the Schwartz Reflection Principle [13, 14, 15] in (3) to
analytically continue the correlator onto the other side of the branch cut. We
now take the limit $\varepsilon\Rightarrow 0$ to obtain

$\displaystyle\mathop{\lim}\limits_{\varepsilon\to
0}\left[{f\left({z+i\varepsilon}\right)-f\left({z-i\varepsilon}\right)}\right]=\mathop{\lim}\limits_{\varepsilon\to
0}2i\operatorname{Im}\left[{f\left({z+i\varepsilon}\right)}\right]=2i\operatorname{Im}\left[{f\left(z\right)}\right]$
(3.8)

Now analogously we can do the same for the $\psi_{5}(z)$ to get

$\displaystyle\mathop{\lim}\limits_{\varepsilon\to
0}\left[{\psi_{5}\left({z+i\varepsilon}\right)-\psi_{5}\left({z-i\varepsilon}\right)}\right]=2i\operatorname{Im}\left[{\psi_{5}\left(z\right)}\right]$
(3.9)

and substituting (3.9) into (3.3) we obtain

$\psi_{5}\left({z_{0}}\right)\Delta\left({z_{0}}\right)=\frac{1}{{2\pi
i}}\mathop{\oint\mkern-0.8mu}\limits_{C\left(R\right)}{\frac{{\psi_{5}\left(z\right)\Delta\left(z\right)}}{{z-z_{0}}}dz}+\frac{1}{\pi}\int\limits_{z_{th}}^{R}{\frac{{\Delta\left(z\right)}}{{z-z_{0}}}\operatorname{Im}\left[{\psi_{5}\left(z\right)}\right]dz}$

and in the low energy limit we get

$\displaystyle\psi_{5}\left(0\right)\Delta\left(0\right)=\frac{1}{{2\pi
i}}\mathop{\oint\mkern-0.8mu}\limits_{C\left(R\right)}{\frac{{\Delta\left(z\right)}}{z}\psi_{5}\left(z\right)dz}+\frac{1}{\pi}\int\limits_{z_{th}}^{R}{\frac{{\Delta\left(z\right)}}{z}\operatorname{Im}\left[{\psi_{5}\left(z\right)}\right]dz}$
(3.10)

Equation (3.10) is referred to as the Finite Energy Sum Rule. The size of the
radius of the circle used in the contour integration in (3.10) hints to us
which correlating function will be used, if we choose the radius to be large
enough then we can use the Perturbative QCD (pQCD) correlator for high
energies for $\psi_{5}(z)$ in the contour integral on the circle [16, 17]. For
the second term in (3.10) we shall use the Hadronic spectral function.

$\displaystyle\psi_{5}\left(0\right)\Delta\left(0\right)=\frac{1}{{2\pi
i}}\mathop{\oint\mkern-0.8mu}\limits_{C\left(R\right)}{\frac{{\Delta\left(z\right)}}{z}\left.{\psi_{5}\left(z\right)}\right|_{PQCD}dz}+\frac{1}{\pi}\int\limits_{z_{th}}^{R}{\frac{{\Delta\left(z\right)}}{z}\left.{\operatorname{Im}\left[{\psi_{5}\left(z\right)}\right]}\right|_{HAD}dz}$
(3.11)

For the case of the pion, the hadronic spectral function is the sum of the
pion pole term and the resonance terms i.e.

$\displaystyle\frac{1}{\pi}\left.{\operatorname{Im}\left[{\psi_{5}\left(z\right)}\right]}\right|_{HAD}=2f_{\pi}^{2}M_{\pi}^{4}\delta\left({z-M_{\pi}^{2}}\right)+\frac{1}{\pi}\left.{\operatorname{Im}\left[{\psi_{5}\left(z\right)}\right]}\right|_{RES}$
(3.12)

substituting (3.12) into the sum rule (3.11) to get

$\displaystyle\psi_{5}\left(0\right)\Delta\left(0\right)=2f_{\pi}^{2}M_{\pi}^{2}\Delta\left({M_{\pi}^{2}}\right)+\frac{1}{\pi}\int\limits_{z_{th}}^{R}{\frac{1}{z}\Delta\left(z\right)\left.{\operatorname{Im}\left[{\psi_{5}\left(z\right)}\right]}\right|_{RES}dz}+\frac{1}{{2\pi
i}}\mathop{\oint\mkern-0.8mu}\limits_{C\left(R\right)}{\frac{1}{z}\Delta\left(z\right)\left.{\psi_{5}\left(z\right)}\right|_{PQCD}dz}$
(3.13)

For convenience we define two new functions to simplify the work

$\displaystyle\left.{\delta_{5}\left(R\right)}\right|_{RES}$
$\displaystyle\equiv$
$\displaystyle\frac{1}{\pi}\int\limits_{z_{th}}^{R}{\frac{1}{z}\Delta\left(z\right)\left.{\operatorname{Im}\left[{\psi_{5}\left(z\right)}\right]}\right|_{RES}dz}$
(3.14) $\displaystyle\left.{\delta_{5}\left(R\right)}\right|_{PQCD}$
$\displaystyle\equiv$ $\displaystyle\frac{1}{{2\pi
i}}\mathop{\oint\mkern-0.8mu}\limits_{C\left(R\right)}{\frac{1}{z}\Delta\left(z\right)\left.{\psi_{5}\left(z\right)}\right|_{PQCD}dz}$
(3.15)

the FESR which we will use in the subsequent analysis is then

$\displaystyle\psi_{5}\left(0\right)\Delta\left(0\right)=2f_{\pi}^{2}M_{\pi}^{2}\Delta\left({M_{\pi}^{2}}\right)+\left.{\delta_{5}\left(R\right)}\right|_{RES}+\left.{\delta_{5}\left(R\right)}\right|_{PQCD}$
(3.16)

##  4 Integration in the Complex Plane

We will use three methods to evaluate the
$\left.{\delta_{5}\left(R\right)}\right|_{PQCD}$ of (3.15). We will use the
method of Fixed Order Perturbation Theory (FOPT) and the method of Fixed
Renormalization Scale/Frozen Order Perturbation Theory (FSPT).

The corresponding Perturbative QCD expression for (3.15) can be improved with
the help of the Renormalization Group (RG) improvement. In the course of
calculating (3.15) we will also need to perform the contour integration over
the circle. We now have a choice of which order the integral and the
renormalization group improvement will be performed.

###  4.1 Fixed Order Perturbation Theory

Fixed Order Perturbation Theory [18, 19, 20, 21] refers to first evaluating
the contour integral and then performing the renormalization group
improvement. The integrals are performed in the complex energy squared plane
(z-plane) on a contour which is a circle with fixed energy (constant radius).
Since the strong coupling $\alpha_{s}\left(z\right)$ and the masses of the
quarks $m_{q}\left(z\right)$ depend only on the magnitude of the complex
variable $z$ i.e. depend only on $R$ which is the squared energy in the
complex plane, these quantities can be taken out of the contour integral. Only
after the integrals are done, we then implement the renormalization group
improvement by setting the renormalization scale equal to the radius of the
circle i.e. $\mu^{2}=R$, so that all the logarithmic terms in the form
$\ln(R/\mu^{2})$ vanish.

The Perturbative QCD correlator given in [19, 22, 23, 24, 25, 26] is

$\displaystyle\left.{\psi_{5}\left({q^{2}}\right)}\right|_{PQCD}=m_{q}^{2}\left({\left|q\right|^{2}}\right)\left[{-q^{2}\Pi_{0}\left({q^{2}}\right)+m_{q}\left({\left|q\right|^{2}}\right)\Pi_{2}\left({q^{2}}\right)+\frac{{C_{u}}}{{-q^{2}}}\left.{\left\langle{m_{q}\bar{u}u}\right\rangle}\right|_{\mu_{0}}+\sum\limits_{j=1}^{3}{\frac{{C_{j}}}{{-q^{2}}}\left\langle{\hat{O}_{j}}\right\rangle}+\ldots}\right]$
(4.1.1)

were $\mu_{0}$ is the scale at which the condensate is measured.

$\displaystyle\Pi_{0}\left({q^{2}}\right)$ $\displaystyle=$
$\displaystyle\frac{1}{{16\pi^{2}}}\left[{-12+6\ln\left({\frac{{-q^{2}}}{{\mu^{2}}}}\right)+\left({\frac{{\alpha_{s}\left({\left|q\right|^{2}}\right)}}{\pi}}\right)\left({24\zeta\left(3\right)-\frac{{131}}{2}+34\ln\left({\frac{{-q^{2}}}{{\mu^{2}}}}\right)-6\ln^{2}\left({\frac{{-q^{2}}}{{\mu^{2}}}}\right)}\right)}\right]$
(4.1.2) $\displaystyle+$
$\displaystyle\frac{1}{{16\pi^{2}}}\left({\frac{{\alpha_{s}\left({\left|q\right|^{2}}\right)}}{\pi}}\right)^{2}\left[{{\text{A}_{3}}\ln\left({\frac{{-q^{2}}}{{\mu^{2}}}}\right)+{\text{B}_{3}}\ln^{2}\left({\frac{{-q^{2}}}{{\mu^{2}}}}\right)+{\text{C}_{3}}\ln^{3}\left({\frac{{-q^{2}}}{{\mu^{2}}}}\right)}\right]$
$\displaystyle+$
$\displaystyle\frac{1}{{16\pi^{2}}}\left({\frac{{\alpha_{s}\left({\left|q\right|^{2}}\right)}}{\pi}}\right)^{3}\left[{A\ln\left({\frac{{-q^{2}}}{{\mu^{2}}}}\right)+B\ln^{2}\left({\frac{{-q^{2}}}{{\mu^{2}}}}\right)+C\ln^{3}\left({\frac{{-q^{2}}}{{\mu^{2}}}}\right)+D\ln^{4}\left({\frac{{-q^{2}}}{{\mu^{2}}}}\right)}\right]$
$\displaystyle+$
$\displaystyle\frac{1}{{16\pi^{2}}}\left({\frac{{\alpha_{s}\left({\left|q\right|^{2}}\right)}}{\pi}}\right)^{4}\sum\limits_{i=1}^{5}{H_{i}L^{i}}$

with $H_{i}$ are a set of real numbers and $L^{i}=\rm{ln^{i}}(-q^{2}/\mu^{2})$
and

$\displaystyle\Pi_{2}\left({q^{2}}\right)=\frac{1}{{16\pi^{2}}}\left\\{{-12+12\ln\left({\frac{{-q^{2}}}{{\mu^{2}}}}\right)+\frac{{\alpha_{s}\left({\left|{q^{2}}\right|}\right)}}{\pi}\left[{-1000+64\ln\left({\frac{{-q^{2}}}{{\mu^{2}}}}\right)-24\left[{\ln\left({\frac{{-q^{2}}}{{\mu^{2}}}}\right)}\right]^{2}+48\zeta\left(3\right)}\right]}\right\\}$
(4.1.3)

substituting (4.1.2) and (4.1.3) into (4.1.1) and using
$\Delta\left(z\right)=1-a_{0}z-a_{1}z^{2}$ as the integration kernel in (3.15)
we obtain

$\displaystyle\left.{\delta_{5}\left(R\right)}\right|_{PQCD}=\left.{\delta_{5}\left(R\right)}\right|_{\Pi_{0}}+\left.{\delta_{5}\left(R\right)}\right|_{\Pi_{2}}+\left.{\delta_{5}\left(R\right)}\right|_{C_{u}}+\left.{\delta_{5}\left(R\right)}\right|_{C_{j}}$
(4.1.4)

with

$\displaystyle\left.{\delta_{5}\left(R\right)}\right|_{\Pi_{0}}=\left.{\delta_{5}}\right|_{1{\text{loop}}}+\left.{\delta_{5}}\right|_{{\text{2loop}}}+\left.{\delta_{5}}\right|_{{\text{3loop}}}+\left.{\delta_{5}}\right|_{{\text{4loop}}}+\left.{\delta_{5}}\right|_{{\text{5loop}}}$
(4.1.5)

See explicit forms of these functions which have been calculated in the
Appendix B.

With the definitions given below we can calculate numerically the behavior of
$\left.{\delta_{5}\left(R\right)}\right|_{PQCD}$ as a function of the energy
$R$.

The coefficients of the QCD Beta function [25] series for $n$ quark types:

$\displaystyle\beta_{0}$ $\displaystyle=$ $\displaystyle
0.25\left({11-\frac{2}{3}n}\right)$ $\displaystyle\beta_{1}$ $\displaystyle=$
$\displaystyle\frac{1}{{16}}\left({102.0-\frac{{38}}{3}n}\right)$
$\displaystyle\beta_{2}$ $\displaystyle=$
$\displaystyle\frac{1}{{64}}\left({\frac{{2857.0}}{2}-\frac{{5033}}{{18}}n+\frac{{325}}{{54}}n^{2}}\right)$
$\displaystyle\beta_{3}$ $\displaystyle=$
$\displaystyle\frac{1}{{256.0}}\left({\frac{{149753}}{{6.0}}+3564\zeta(3)-\left({\frac{{1078361}}{{162.0}}+\frac{{6508}}{{27}}\zeta(3)}\right)n+\left({\frac{{50065}}{{162.0}}+\frac{{6472}}{{81}}\zeta(3)}\right)n^{2}+\frac{{1093}}{{729.0}}n^{3}}\right)$

$\Lambda\equiv\Lambda_{\rm{QCD}}$ for the definitions found in [19, 26]:

$\displaystyle L[z,\Lambda]$ $\displaystyle=$
$\displaystyle\ln\left[{\frac{z}{{\Lambda^{2}}}}\right]$ $\displaystyle
a1[z,\Lambda]$ $\displaystyle=$
$\displaystyle-\frac{{\beta_{1}}}{{\beta_{0}^{2}}}\ln[L[z,\Lambda]]$
$\displaystyle a2[z,\Lambda]$ $\displaystyle=$
$\displaystyle\frac{1}{{\beta_{0}^{2}}}\left({\frac{{\beta_{1}^{2}}}{{\beta_{0}^{2}}}\left({\ln[L[z,\Lambda]]^{2}-\ln[L[z,\Lambda]]-1}\right)+\frac{{\beta_{2}}}{{\beta_{0}}}}\right)$
$\displaystyle a3[z,\Lambda]$ $\displaystyle=$
$\displaystyle\frac{1}{{\beta_{0}^{3}}}\left({\frac{{\beta_{1}^{3}}}{{\beta_{0}^{3}}}\left({-\ln[L[z,\Lambda]]^{3}+2.5\ln[L[z,\Lambda]]^{2}+2\ln[L[z,\Lambda]]-0.5}\right)-3\frac{{\beta_{1}\beta_{2}}}{{\beta_{0}^{2}}}\ln[L[z,\Lambda]]+0.5\frac{{\beta_{3}}}{{\beta_{0}}}}\right)$
$\displaystyle\alpha[z,\Lambda]$ $\displaystyle=$
$\displaystyle\frac{\pi}{{\beta_{0}L[z,\Lambda]}}\left({1+\frac{{a1[z,\Lambda]}}{{L[z,\Lambda]}}+\frac{{a2[z,\Lambda]}}{{L[z,\Lambda]^{2}}}+\frac{{a3[z,\Lambda]}}{{L[z,\Lambda]^{3}}}}\right)$
(4.1.6)

The coefficients of the QCD Gamma function series for quark masses:

$\displaystyle\gamma_{0}$ $\displaystyle=$ $\displaystyle 1$
$\displaystyle\gamma_{1}$ $\displaystyle=$
$\displaystyle\frac{1}{{16}}\left({\frac{{202}}{3}-\frac{{20}}{9}n}\right)$
$\displaystyle\gamma_{2}$ $\displaystyle=$
$\displaystyle\frac{1}{{64}}\left[{1249.0-\left({\frac{{2216}}{{27.0}}+\frac{{160}}{3}\zeta\left(3\right)}\right)n-\frac{{140}}{{81.0}}n^{2}}\right]$
$\displaystyle\gamma_{3}$ $\displaystyle=$
$\displaystyle\frac{1}{{256}}\left[{\left({\frac{{4603055}}{{162.0}}+\frac{{135680}}{{27}}\zeta\left(3\right)-8800\zeta\left(5\right)}\right)+\left({-\frac{{91723}}{{27}}-\frac{{34192}}{9}\zeta\left(3\right)+880\zeta\left(4\right)+\frac{{18400}}{9}\zeta\left(5\right)}\right)n}\right]$
$\displaystyle+$
$\displaystyle\frac{1}{{256}}\left[{\left({\frac{{5242}}{{243}}+\frac{{800}}{9}\zeta\left(3\right)-\frac{{160}}{3}\zeta\left(3\right)}\right)n^{2}+\left({-\frac{{332}}{{243}}+\frac{{64}}{{27}}\zeta\left(3\right)}\right)n^{3}}\right]$

$\displaystyle d1[z,\Lambda]$ $\displaystyle=$
$\displaystyle-\frac{{\gamma_{0}\beta_{1}}}{{\beta_{0}^{3}}}{\text{ln}}[L[z,\Lambda]]$
$\displaystyle d2[z,\Lambda]$ $\displaystyle=$
$\displaystyle\frac{{\gamma_{0}}}{{\beta_{0}^{3}}}\left({\frac{\beta_{1}^{2}}{{\beta_{0}^{2}}}\left({{\text{ln}}[L[z,\Lambda]]^{2}-{\text{ln}}[L[z,\Lambda]]-1}\right)+\frac{\beta_{2}}{{\beta_{0}}}}\right)+0.5\frac{{\gamma_{0}}}{{\beta_{0}}}\left({\frac{{\gamma_{0}}}{{\beta_{0}}}-1}\right)\frac{{{\beta_{1}}^{2}}}{{\beta_{0}^{4}}}{\text{ln}}[L[z,\Lambda]]^{2}$
$\displaystyle d3[z,\Lambda]$ $\displaystyle=$
$\displaystyle\frac{{\gamma_{0}}}{{\beta_{0}^{4}}}\left({\frac{{{\beta_{1}}^{3}}}{{\beta_{0}^{3}}}\left({-{\text{ln}}[L[z,\Lambda]]^{3}+2.5{\text{ln}}[L[z,\Lambda]]^{2}+2{\text{ln}}[L[z,\Lambda]]-0.5}\right)-3\frac{{\beta_{1}\beta_{2}}}{{\beta_{0}^{2}}}{\text{ln}}[L[z,\Lambda]]+0.5\frac{{\beta_{3}}}{{\beta_{0}}}}\right)$
$\displaystyle+$ $\displaystyle
0.5\frac{{\gamma_{0}}}{{\beta_{0}}}\left({\frac{{\gamma_{0}}}{{\beta_{0}}}-1}\right)\left({-2\frac{{\beta_{1}}}{{\beta_{0}^{4}}}}\right)\left({\frac{{\beta_{1}^{2}}}{{\beta_{0}^{2}}}\left({{\text{ln}}[L[z,\Lambda]]^{2}-{\text{ln}}[L[z,\Lambda]]-1}\right)+\frac{{\beta_{2}}}{{\beta_{0}}}}\right){\text{ln}}[L[z,\Lambda]]$
$\displaystyle-$
$\displaystyle\frac{1}{6}\frac{{\gamma_{0}}}{{\beta_{0}}}\left({\frac{{\gamma_{0}}}{{\beta_{0}}}-1}\right)\left({\frac{{\gamma_{0}}}{{\beta_{0}}}-2}\right)\frac{{\beta_{1}^{3}}}{{\beta_{0}^{6}}}{\text{ln}}[L[z,\Lambda]]^{3}$

$\displaystyle c1$ $\displaystyle=$
$\displaystyle\frac{{\gamma_{1}}}{{\beta_{0}}}-\frac{{\gamma_{0}\beta_{1}}}{{\beta_{0}^{2}}}$
$\displaystyle c2$ $\displaystyle=$ $\displaystyle
0.5\left({\frac{{\gamma_{2}}}{{\beta_{0}}}-\frac{{\gamma_{1}\beta_{1}}}{{\beta_{0}^{2}}}+\frac{{\gamma_{0}}}{{\beta_{0}^{2}}}\left({\frac{{\beta_{1}^{2}}}{{\beta_{0}}}-\beta_{2}}\right)}\right)$
$\displaystyle c3$ $\displaystyle=$
$\displaystyle\frac{1}{3}\left({\frac{{\gamma_{0}}}{{\beta_{0}^{2}}}\left({\frac{{\beta_{1}\beta_{2}}}{{\beta_{0}}}-\frac{{\beta_{1}}}{{\beta_{0}}}\left({\frac{{\beta_{1}^{2}}}{{\beta_{0}}}-\beta_{2}}\right)-\beta_{3}}\right)+\frac{{\gamma_{1}}}{{\beta_{0}^{2}}}\left({\frac{{\beta_{1}^{2}}}{{\beta_{0}}}-\beta_{2}}\right)-\frac{{\beta_{1}\gamma_{2}}}{{\beta_{0}^{2}}}+\frac{{\gamma_{3}}}{{\beta_{0}}}}\right)$

For convenience we have split the right-hand-side of the quark mass function
into a product of the invariant quark mass $\widehat{m}_{\Lambda}$ and the
functional dependence on the energy $z$ and the $\Lambda_{\rm{QCD}}$ scale in
the $U\left({z,\Lambda}\right)$

$\displaystyle U\left({z,\Lambda}\right)=$
$\displaystyle\frac{1}{{\left({\frac{1}{2}L\left[{z,\Lambda}\right]}\right)^{\frac{{\gamma_{0}}}{{\beta_{0}}}}}}\Biggl{\\{}1+\left({\frac{{c1}}{{\beta_{0}}}+d1\left[{z,\Lambda}\right]}\right)\frac{1}{{L[z,\Lambda]}}+$
$\displaystyle+\left[{\left({c1\frac{{a1\left[{z,\Lambda}\right]}}{{\beta_{0}}}+\frac{{0.5c1^{2}+c2}}{{\beta_{0}^{2}}}}\right)+d2\left[{z,\Lambda}\right]+c1\frac{{d1\left[{z,\Lambda}\right]}}{{\beta_{0}}}}\right]\frac{1}{{L\left[{z,\Lambda}\right]^{2}}}$
$\displaystyle+$
$\displaystyle\left[{c1\frac{{a2\left[{z,\Lambda}\right]}}{{\beta_{0}}}+\frac{{\left({c2+0.5c1^{2}}\right)}}{{\beta_{0}^{2}}}\left({d1\left[{z,\Lambda}\right]-2\frac{{\beta_{1}}}{{\beta_{0}^{2}}}\ln\left[{L\left[{z,\Lambda}\right]}\right]}\right)+\frac{{c3+c1c2+\frac{1}{6}c1^{3}}}{{\beta_{0}^{3}}}}\right]\frac{1}{{L\left[{z,\Lambda}\right]^{3}}}$
$\displaystyle+$
$\displaystyle\left[{c1\frac{{d1\left[{z,\Lambda}\right]a1\left[{z,\Lambda}\right]}}{{\beta_{0}}}+c1\frac{{d2\left[{z,\Lambda}\right]}}{{\beta_{0}}}+d3\left[{z,\Lambda}\right]}\right]\frac{1}{{L\left[{z,\Lambda}\right]^{3}}}\Biggr{\\}}$
(4.1.7)

were

$\displaystyle
m\left({z,\Lambda}\right)=\widehat{m}_{\Lambda}U\left({z,\Lambda}\right)$
(4.1.8)

###  4.2 Fixed Renormalization Scale Perturbation Theory

The method of Fixed Renormalization Scale Perturbation Theory333work done in
collaboration with J.M. Bordes and J. Pe$\widetilde{\text{n}}$arrocha at the
Departamento de Fisica Teorica, Universitat de Valencia, and Instituto de
Fisica Corpuscular, Centro Mixto Universitat de Valencia-CSIC between November
2009 - March 2010 [27, 28, 29] is very similar to the method of Fixed Order
Perturbation Theory. In this case we generalize the integration kernel from a
simply second order polynomial to a $n^{th}$ order polynomial and instead of
choosing the renormalization scale to be radius of the circle in the complex
energy plane we will instead choose a fixed scale which will be varied over a
large interval of $\mu^{2}=2-50$ GeV2. The method is implemented by using the
Legendre polynomials [30] and making use of their orthogonality property. We
first generalize the kernel to

$\displaystyle\tilde{\Delta}_{n}\left(z\right)\equiv\frac{{P_{n}\left[{x\left(z\right)}\right]}}{{P_{n}\left[{x\left({M_{\pi}^{2}}\right)}\right]}}=\sum\limits_{j=0}^{n}{a_{j}z^{j}}$
(4.2.1)

were $P_{n}$ is the Legendre polynomial which can be generated by the
Rodriguez Formula

$\displaystyle
P_{n}\left(x\right)=\frac{1}{{2^{n}n!}}\left({\frac{d}{{dx}}}\right)^{n}\left({x^{2}-1}\right)^{n}$
(4.2.2)

We use the orthogonality of the polynomials i.e.

$\int\limits_{-1}^{1}{P_{n}\left(x\right)P_{m}\left(x\right)dx}=0$

for $n\neq m$ by rescaling the domain of integration as

$x\left(z\right)\equiv\frac{{2z-\left({z_{0}-z_{th}}\right)}}{{z_{th}-z_{0}}}$

with $z\in\left[{z_{th};z_{0}}\right]$ yields the following constraints on
(4.2.1)

$\displaystyle\tilde{\Delta}_{n}\left({M_{\pi}^{2}}\right)$ $\displaystyle=$
$\displaystyle 1$ (4.2.3)
$\displaystyle\int\limits_{z_{th}}^{z_{0}}{z^{k}\tilde{\Delta}_{n}\left(z\right)dz}$
$\displaystyle=$ $\displaystyle 0{\text{ }}k\in\mathbb{N},k<n$ (4.2.4)

Imposing the orthogonality condition (4.2.4) forces the generalized polynomial
$\tilde{\Delta}_{n}\left(z\right)$ to minimize the contributions of the
continuum in the range $z\in\left[{z_{th};z_{0}}\right]$.

We then substitute (4.2.1) into (3.15) and (3.14) and choose allow $\mu^{2}$
to vary in the range $2-50$ GeV2. In our case we found the best(largest stable
domain) results for $\mu^{2}=4$ GeV2. We varied the value of $\mu^{2}$ in the
range $2-50$ GeV2 to obtain an uncertainty in the value of the $\delta_{\pi}$

##  5 The Resonance Contribution
$\left.{\delta_{5}\left(R\right)}\right|_{RES}$

To evaluate the resonance term (3.14) we require information about the
hadronic spectral function beyond the pion pole. At the current time this
information is not available from experiment and we only have information
about the mass and width of the first two excited states of the pion i.e. the
$\Pi(1300)$ [31] and the $\Pi(1800)$ excitations. To proceed further we shall
have to model the hadronic spectral function by using threshold constraints
from Chiral Perturbation Theory ($\chi$-PT) which are imposed on a linear
combination of Breit-Wigner profiles to describe the two excitations of the
pion. Alas by introducing this model we will also introduce systematic
uncertainties and solutions obtained thereafter are dependent on the model
used. However, we can reduce these systematic uncertainties by introducing
analytic integration kernels that are forced to be zero at the resonance
points. This will reduce the effect of the systematic uncertainties in the
neighborhood of the resonances. We have introduced the quadratic polynomial
$\Delta(z)$ earlier precisely because of this!

We shall start by modeling the hadronic spectral function as

$\displaystyle\frac{1}{\pi}\left.{\operatorname{Im}\left[{\psi_{5}\left(z\right)}\right]}\right|_{RES}=\frac{1}{\pi}\left.{\operatorname{Im}\left[{\psi_{5}\left(z\right)}\right]}\right|_{3\pi}BW\left(z\right)$
(5.1)

were

$\displaystyle\frac{1}{\pi}\left.{\operatorname{Im}\left[{\psi_{5}\left(z\right)}\right]}\right|_{3\pi}=\frac{1}{{3\cdot
2^{8}\pi^{4}}}\frac{{M_{\pi}^{4}}}{{f_{\pi}^{2}}}z\Theta\left(z\right)$ (5.2)

is the constraint from $\chi$-PT [32]. The Breit-Wigner profile BW(z) will be
in the form

$\displaystyle BW\left(z\right)=\frac{{BW_{1}\left(z\right)+\kappa
BW_{2}\left(z\right)}}{{1+\kappa}}$ (5.3)

the $\kappa$ is just a weighting parameter which controls the relative weight
of the resonances with $\kappa\in\mathbb{R}^{+}$. $\kappa=0.5$ was chosen for
the computational work since the width of the $\Pi(1300)$ is twice as broad as
the width of the $\Pi(1812)$ resonance and thus the $\Pi(1812)$ resonance
would only contribute half as much as the $\Pi(1300)$ resonance. The
$BW_{1}\left(z\right)$ and $BW_{2}\left(z\right)$ are isolated Breit-Wigner
resonances. We normalize the Breit-Wigner profile at zero to be $BW(0)=1$,
this implies that $BW_{1}(0)=1$ and $BW_{2}(0)=1$. This sets the form of the
Breit-Wigner resonances we will have to use

$\displaystyle
BW_{k}\left(z\right)=\frac{{M_{k}^{2}\left({M_{k}^{2}+\Gamma_{k}^{2}}\right)}}{{\left({z-M_{k}^{2}}\right)^{2}+M_{k}^{2}\Gamma_{k}^{2}}}$
(5.4)

we can now proceed with the calculation, substituting (5.2) and (5.3) into
(5.1) we obtain

$\displaystyle\frac{1}{\pi}\left.{\operatorname{Im}\left[{\psi_{5}\left(z\right)}\right]}\right|_{RES}=\frac{1}{{3\cdot
2^{8}\pi^{4}}}\frac{{M_{\pi}^{4}}}{{f_{\pi}^{2}}}\frac{1}{{1+\kappa}}z\Theta\left(z\right)\left[{BW_{1}\left(z\right)+\kappa
BW_{2}\left(z\right)}\right]$ (5.5)

for convenience we define a new parameter $\lambda$ which is just a collection
of all the constants

$\lambda\equiv\frac{1}{{3\cdot
2^{8}\pi^{4}}}\frac{{M_{\pi}^{4}}}{{f_{\pi}^{2}}}\frac{1}{{1+\kappa}}$

and

$\displaystyle\frac{1}{\pi}\left.{\operatorname{Im}\left[{\psi_{5}\left(z\right)}\right]}\right|_{RES}=\lambda
z\Theta\left(z\right)\left[{BW_{1}\left(z\right)+\kappa
BW_{2}\left(z\right)}\right]$ (5.6)

we can now calculate the resonance contribution in this model by substituting
(5.6) into (3.14) to get

$\displaystyle\left.{\delta_{5}\left(R\right)}\right|_{RES}$ $\displaystyle=$
$\displaystyle\int\limits_{z_{th}}^{R}{\frac{1}{z}\Delta\left(z\right)\lambda
z\Theta\left(z\right)\left[{BW_{1}\left(z\right)+\kappa
BW_{2}\left(z\right)}\right]dz}$ (5.7) $\displaystyle=$
$\displaystyle\lambda\int\limits_{9M_{\pi}^{2}}^{R}{\frac{1}{z}\Delta\left(z\right)z\Theta\left(z\right)BW_{1}\left(z\right)dz}+\lambda\kappa\int\limits_{9M_{\pi}^{2}}^{R}{\frac{1}{z}\Delta\left(z\right)z\Theta\left(z\right)BW_{2}\left(z\right)dz}$
$\displaystyle=$
$\displaystyle\lambda\left[{G_{1}\left(R\right)-G_{1}\left({9M_{\pi}^{2}}\right)}\right]+\lambda\kappa\left[{G_{2}\left(R\right)-G_{2}\left({9M_{\pi}^{2}}\right)}\right]$

were the function $G_{k}\left(x\right)$ is

$\displaystyle
G_{k}\left(z\right)=M_{k}^{2}\left({M_{k}^{2}+\Gamma_{k}^{2}}\right)\Biggl{\\{}\frac{1}{{M_{k}\Gamma_{k}}}$
$\displaystyle\left[{1-a_{0}M_{k}^{2}-a_{1}\left({M_{k}^{4}-M_{k}^{2}\Gamma_{k}^{2}}\right)}\right]\arctan\left({\frac{{z-M_{k}^{2}}}{{M_{k}\Gamma_{k}}}}\right)+$
$\displaystyle-\left({\frac{1}{2}a_{0}+a_{1}M_{k}^{2}}\right)\ln\left[{\left({z-M_{k}^{2}}\right)^{2}+M_{k}^{2}\Gamma_{k}^{2}}\right]-a_{1}\left({z-M_{k}^{2}}\right)\Biggr{\\}}$

which is derived in Appendix G. The inclusion of the analytic integration
kernel $\Delta(z)$ attempts to suppress the uncertainties from the resonance
regions which is poorly understood. Even with this suppression we cannot
eliminate entirely these uncertainties. To get a handle on how good the
suppression is we make a Taylor expansion of the $\delta_{5}$ or equivalently
the $G_{k}(z)$ around the resonance peak. For convenience we shall choose
$k=1$ and keeping only the first order terms in the expansion we obtain to
leading order

$\displaystyle\delta_{5}\left(R\right)\approx\lambda
M_{1}^{2}\left({M_{1}^{2}+\Gamma_{1}^{2}}\right)\frac{{M_{1}^{2}-M_{2}^{2}}}{{2M_{1}^{2}M_{2}^{2}}}\left\\{{\left({\frac{{R-M_{1}^{2}}}{{M_{1}\Gamma_{1}}}}\right)+\ln\left({M_{1}\Gamma_{1}}\right)^{2}}\right\\}$
(5.9)

The coefficient of the braces is the gradient of this line and sets the size
on the contribution from the resonances. The magnitude of this coefficient is

$\lambda
M_{1}^{2}\left({M_{1}^{2}+\Gamma_{1}^{2}}\right)\frac{{M_{1}^{2}-M_{2}^{2}}}{{2M_{1}^{2}M_{2}^{2}}}=1.2\times
10^{-7}$

We see that not only in the immediate neighborhood ($R\approx M_{1}^{2}$) is
the suppression effective but due to the tiny coefficient, the contribution
from the resonances is effectively eliminated!

##  6 Results

We consider two cases,

  * •

$SU(2)\otimes SU(2)$ for the Pions

  * •

$SU(3)\otimes SU(3)$ case for the Kaons

of which the FOPT results will be of main interest.We use the particle data
from the Particle Data Group [33, 34] for the $SU(2)\otimes SU(2)$ Pion case

  * •

$m_{\pi}=139.642400\pm 0.00035$ MeV

  * •

$f_{\pi}=92.21\pm 0.14$ MeV

  * •

$m_{\pi}^{*}=1300\pm 100$ MeV

  * •

$\Gamma_{\pi}^{*}=400\pm 200$ MeV

  * •

$m_{\pi}^{**}=1812\pm 14$ MeV

  * •

$\Gamma_{\pi}^{**}=207\pm 13$ MeV

  * •

$\overline{m}_{u}(2\rm{GeV})=2.9\pm 0.2$ MeV

  * •

$\overline{m}_{d}(2\rm{GeV})=5.3\pm 0.4$ MeV

  * •

$\hat{m}_{u}(2\rm{GeV})=3.8-3.9$ MeV

  * •

$\hat{m}_{d}(2\rm{GeV})=6.4-7.2$ MeV

$SU(3)\otimes SU(3)$ Kaon[30, 34] case

  * •

$m_{K}=493.677\pm 0.005$ MeV

  * •

$m_{K}^{*}=1460$ MeV

  * •

$\Gamma_{K}^{*}=250$ MeV

  * •

$m_{K}^{**}=1830$ MeV

  * •

$\Gamma_{K}^{**}=250$ MeV

  * •

$\overline{m}_{u}(2\rm{GeV})=2.9\pm 0.2$ MeV

  * •

$\hat{m}_{u}(2\rm{GeV})=3.8-3.9$ MeV

  * •

$\hat{m}_{s}(2\rm{GeV})=130-150$ MeV

We note here that for the plots presented below we will be searching for the
so-called stable region. This is the domain on the plot where the function
varies the slowest. This domain then defines the stability region. Once we
have identified the stability region we read of the average value of the
function over this domain. We shall use either $\Lambda_{\rm{QCD}}=0.365$ GeV
or $\Lambda_{\rm{QCD}}=0.397$ GeV. It has become standard practice to make
predictions about hadronic parameters at these scales.

###  6.1 Frozen Order Results

####  6.1.1 $SU(2)\otimes SU(2)$

Figure 2: Results for $\psi_{5}(0)$(GeV4) [30] as a function of the energy $s$
in FSPT with $\mu^{2}=4$GeV2. a, b and c correspond to polynomial orders 4, 5
and 6 respectively.

From the Fixed Scale Perturbation Theory we see that the stability is over a
small interval instead of the whole integration range. From the stable
intervals we extract a value of $\delta_{\pi}=0.056\pm 0.011$

###  6.2 FOPT Results

####  6.2.1 $SU(2)\otimes SU(2)$

Figure 3: Results for $\psi_{5}(0)$(GeV4) as a function of the energy $s$ in
FOPT. We have neglected the two resonances. Starting from the top we have
$\psi_{5}(s,0.365)$ and $\psi_{5}(s,0.397)$ respectively

Figure 4: Results for $\delta_{\pi}$ as a function of the energy $s$ in FOPT.
We have neglected the two resonances. Starting from the top we have
$\delta_{\pi}(s,0.397)$ and $\delta_{\pi}(s,0.365)$ respectively

Figure 5: Results for $\psi_{5}(0)$(GeV4) as a function of the energy $s$ in
FOPT. We have included the two resonances with the weighting factor
$\kappa=0.5$. Starting from the top we have $\psi_{5}(s,0.365)$ and
$\psi_{5}(s,0.397)$ respectively

Figure 6: Results for $\delta_{\pi}$ as a function of the energy $s$ in FOPT.
We have included the two resonances with the weighting factor $\kappa=0.5$.
Starting from the top we have $\delta_{\pi}(s,0.397)$ and
$\delta_{\pi}(s,0.365)$ respectively

Figure 7: Plot of the various contributors to $\delta_{\pi}$ for
$\Lambda_{QCD}=0.365$ GeV. Starting from the top we have
$\ln(\delta_{\pi}^{\pi 0})$, $\ln(\delta_{\pi}^{C_{j}})$,
$\ln(\delta_{\pi}^{C_{u}})$ and $\ln(\delta_{\pi}^{\pi 2})$ respectively

The various contributions are defined as

$\displaystyle\psi_{5}$ $\displaystyle=$ $\displaystyle
2f_{\pi}^{2}M_{\pi}^{2}\Delta\left({M_{\pi}^{2}}\right)+\delta_{5PQCD}+\delta_{5RES}$
$\displaystyle\delta_{\pi}$ $\displaystyle=$ $\displaystyle
1-\frac{{\psi_{5}}}{{2f_{\pi}^{2}M_{\pi}^{2}}}=1-\frac{{2f_{\pi}^{2}M_{\pi}^{2}\Delta\left({M_{\pi}^{2}}\right)+\delta_{5PQCD}+\delta_{5RES}}}{{2f_{\pi}^{2}M_{\pi}^{2}}}$
(6.1) $\displaystyle=$
$\displaystyle\underbrace{1-\frac{{2f_{\pi}^{2}M_{\pi}^{2}\Delta\left({M_{\pi}^{2}}\right)+\delta_{5RES}}}{{2f_{\pi}^{2}M_{\pi}^{2}}}}_{=\delta_{\pi}^{0}}-\underbrace{\frac{{\left.{\delta_{5}}\right|_{\Pi
0}}}{{2f_{\pi}^{2}M_{\pi}^{2}}}}_{=\delta_{\pi}^{\Pi
0}}-\underbrace{\frac{{\left.{\delta_{5}}\right|_{\Pi
2}}}{{2f_{\pi}^{2}M_{\pi}^{2}}}}_{=\delta_{\pi}^{\Pi
2}}-\underbrace{\frac{{\left.{\delta_{5}}\right|_{Cu}}}{{2f_{\pi}^{2}M_{\pi}^{2}}}}_{=\delta_{\pi}^{Cu}}-\underbrace{\frac{{\left.{\delta_{5}}\right|_{Cj}}}{{2f_{\pi}^{2}M_{\pi}^{2}}}}_{=\delta_{\pi}^{Cj}}$
$\displaystyle=$ $\displaystyle\delta_{\pi}^{0}-\delta_{\pi}^{\Pi
0}-\delta_{\pi}^{\Pi 2}-\delta_{\pi}^{Cu}-\delta_{\pi}^{Cj}$

Figure 8: Plot of the various contributors to $\delta_{\pi}$ for
$\Lambda_{QCD}=0.397$ GeV. Starting from the top we have
$\ln(\delta_{\pi}^{\pi 0})$, $\ln(\delta_{\pi}^{C_{j}})$,
$\ln(\delta_{\pi}^{C_{u}})$ and $\ln(\delta_{\pi}^{\pi 2})$ respectively

In the case of FOPT the results show good stability, looking at figures 3 \- 5
in the range $3\leq s\leq 5$ $\times 10^{-4}\rm{GeV}^{2}$ there is very little
change in $\psi_{5}(s)$ and $\delta_{\pi}(s)$. With the inclusion of the
resonances the functions still display good stability. From the data used to
generate the above plots, the result are tabulated below

|  | Pion data |  |
---|---|---|---|---
| No Resonances | With Resonances
$\Lambda_{QCD}$(GeV) | $\psi_{5}(0)(\rm{GeV}^{4})\times 10^{-4}$ | $\delta_{\pi}$ | $\psi_{5}(0)(\rm{GeV}^{4})\times 10^{-4}$ | $\delta_{\pi}$
0.365 | 3.12-3.14 | 0.059-0.062 | 3.12-3.16 | 0.052-0.054
0.397 | 3.04-3.09 | 0.072-0.073 | 3.04-3.09 | 0.068-0.071
Table 1: Tabulated values extracted from the plots 3 \- 5 which are used to
estimate $\delta_{\pi}$.

from values listed in table 1, we estimate the value of $\delta_{\pi}=0.064\pm
0.008$.

####  6.2.2 $SU(3)\otimes SU(3)$

Figure 9: Results for $\psi_{5}(0)$(GeV4) as a function of the energy $s$ in
FOPT. We have neglected the two resonances. Starting from the top we have
$\psi_{5}(s,0.365)$ and $\psi_{5}(s,0.397)$ respectively

Figure 10: Results for $\delta_{\pi}$ as a function of the energy $s$ in FOPT.
We have neglected the two resonances. Starting from the top we have
$\delta_{K}(s,0.397)$ and $\delta_{K}(s,0.365)$ respectively

Figure 11: Results for $\psi_{5}(0)$(GeV4) as a function of the energy $s$ in
FOPT. We have included the two resonances with the weighting factor
$\kappa=0.5$. Starting from the top we have $\psi_{5}(s,0.365)$ and
$\psi_{5}(s,0.397)$ respectively

Figure 12: Results for $\delta_{\pi}$ as a function of the energy $s$ in FOPT.
We have included the two resonances with the weighting factor $\kappa=0.5$.
Starting from the top we have $\delta_{K}(s,0.397)$ and $\delta_{K}(s,0.365)$
respectively

Figure 13: Plot of the various contributors to $\delta_{K}$ for
$\Lambda_{QCD}=0.365$ GeV. Starting from the top we have $\ln(\delta_{K}^{\pi
0})$, $\ln(\delta_{K}^{\pi 2})$, $\ln(\delta_{K}^{C_{j}})$ and
$\ln(\delta_{K}^{C_{u}})$ respectively

Figure 14: Plot of the various contributors to $\delta_{K}$ for
$\Lambda_{QCD}=0.397$ GeV. Starting from the top we have $\ln(\delta_{K}^{\pi
0})$, $\ln(\delta_{K}^{\pi 2})$, $\ln(\delta_{K}^{C_{j}})$ and
$\ln(\delta_{K}^{C_{u}})$ respectively

Looking at figures 9 \- 12 in the range $4\leq s\leq 5$ $\times
10^{-4}\rm{GeV}^{2}$ there is very little change in $\psi_{5}(s)$ and
$\delta_{K}(s)$. With the inclusion of the resonances the functions still
display good stability. From the data used to generate the above plots, the
result are tabulated below:

Kaon data
---
| No Resonances | With Resonances
$\Lambda_{QCD}(\rm{GeV})$ | $\psi_{5}(0)(\rm{GeV}^{4})$ | $\delta_{K}$ | $\psi_{5}(0)(\rm{GeV}^{4})$ | $\delta_{K}$
0.365 | 0.0017-0.0019 | 0.6929-0.6963 | 0.0018-0.0019 | 0.6921-0.6959
0.397 | 0.0013-0.0015 | 0.7535-0.7542 | 0.0014-0.0015 | 0.7493-0.7578
Table 2: Tabulated values extracted from the plots 9 \- 12 which are used to
estimate $\delta_{K}$.

From values listed in table 2, we estimate the value of $\delta_{K}=0.64\pm
0.24$.

###  6.3 Discussion of Results

We have estimated the values of the $\delta_{\pi}$ and $\delta_{K}$ as (taking
averages and adding the errors in quadrature)

$\displaystyle\delta_{\pi}$ $\displaystyle=$ $\displaystyle 0.060\pm 0.014$
(6.1) $\displaystyle\delta_{K}$ $\displaystyle=$ $\displaystyle 0.64\pm 0.24$
(6.2)

The estimated value of $\delta_{\pi}$ is larger than the value estimated in
[35] which is based on perturbative QCD up to next-to-leading order. We can
use the value for $\delta_{\pi}$ to find the quark condensate by using the
current quark mass of $0.5\left({m_{u}+m_{d}}\right)=4.1\pm 0.2$ MeV and
substituting into (3.1) to get

$\displaystyle\left\langle{0|\bar{q}q|0}\right\rangle$ $\displaystyle=$
$\displaystyle-\frac{{f_{\pi}^{2}M_{\pi}^{2}\left({1-\delta_{\pi}}\right)}}{{2\cdot
0.5\left({m_{u}+m_{d}}\right)}}$ (6.3) $\displaystyle=$
$\displaystyle-\left({266\pm 5{\text{ MeV}}}\right)^{3}$

Comparing to the value calculated in [5] were

$\displaystyle\left\langle{0|\bar{q}q|0}\right\rangle=-\left({275\pm 15{\text{
MeV}}}\right)^{3}$

We see that the two values are in agreement within experimental uncertainty.
$\delta_{\pi}$ is also related to the low energy constants of chiral
perturbation theory by (6.4)

$\displaystyle\delta_{\pi}=4\frac{{M_{\pi}^{2}}}{{f_{\pi}^{2}}}\left({2L_{8}^{r}-H_{2}^{r}}\right)$
(6.4)

Using (6.4) and $L_{8}^{r}=\left({0.88\pm 0.24}\right)\times 10^{-3}$ [5] we
obtain

$\displaystyle H_{2}^{r}$ $\displaystyle=$ $\displaystyle
2L_{8}^{r}-\frac{{f_{\pi}^{2}}}{{4M_{\pi}^{2}}}\delta_{\pi}$ (6.5)
$\displaystyle=$ $\displaystyle-\left({4.9\pm 1.8}\right)\times 10^{-3}$

which is in agreement with the value quoted in [5]. Alternative comparisons
come from Lattice QCD [36] which predicts a value, using
$L_{8}^{r}=\left({0.58\pm 0.09}\right)\times 10^{-3}$ of

$\displaystyle H_{2}^{r}=-\left({5.5\pm 2.0}\right)\times 10^{-3}$ (6.6)

Using $L_{8}^{r}=\left({0.62\pm 0.20}\right)\times 10^{-3}$ [37] we calculate
a value of $H_{2}^{r}=-\left({5.4\pm 1.2}\right)\times 10^{-3}$, which is in
agreement with (6.5). The unphysical constant $H_{2}^{r}$ enters the next-to-
leading order definition of the quark condensate and leads to some ambiguity
in the definition of the condensate.

##  7 Conclusion

In this project we have made a direct estimate of $\delta_{\pi}$ and
$\delta_{K}$ which measures how much the $SU(2)\otimes SU(2)$ and
$SU(3)\otimes SU(3)$ symmetry groups are broken with the inclusion of chiral
corrections to their respective correlators. We have made use of the methods
of Fixed Order and Fixed Renormalization Scale Perturbation theory to
calculate the symmetry breaking parameters $\delta_{\pi}$ and $\delta_{K}$. We
also attempted to use the method of Contour Improved Perturbation theory, but
this analysis leads to $\delta_{5}\left(z\right)$ being dependent on both
$\psi_{5}\left(z\right)$ and $\psi^{{}^{\prime}}_{5}\left(z\right)$ which are
not physical quantities as they depend on the renormalization and the
regularization schemes used.

We have also performed our analysis using an integration kernel which is a
complete second order polynomial which was constrained, with an appropriate
choice of its coefficients, to vanish at the known two resonance points in the
hadronic region. This leads to drastic reduction of systematic uncertainties
that have plagued previous calculations of this nature. For $\delta_{\pi}$ and
$\delta_{K}$ we predict values of

$\displaystyle\delta_{\pi}$ $\displaystyle=$ $\displaystyle 0.060\pm 0.014$
$\displaystyle\delta_{K}$ $\displaystyle=$ $\displaystyle 0.64\pm 0.24$

and as a result of our new values for $\delta_{\pi}$ and $\delta_{K}$ we
calculated the light quark condensate to be

$\displaystyle\left\langle{0|\bar{q}q|0}\right\rangle=-\left({266\pm 5{\text{
MeV}}}\right)^{3}$

The chiral perturbation theory low energy constant

$\displaystyle H_{2}^{r}=-\left({4.9\pm 1.8}\right)\times 10^{-3}$

We have compared these predicted values of the symmetry breaking parameters,
the quark condensate and the low energy constant to those found in the
literature and the agreement is to within one standard deviation of the mean.

## Acknowledgment

I would like to thank my supervisor Professor Cesareo Dominguez who patiently
guided and helped me during the past two years in completing this project.
Prof. Dominguez has treated me kindly letting my mind run wild and only when I
had strayed too far, bringing me back on the right track! I have matured as a
theoretical physicist under his constant guidance. I would also like to thank
our collaborators, J.M. Bordes and J.A. Pe$\widetilde{\text{n}}$arrocha from
Spain, and Professor Karl Schilcher from Germany for all the discussions and
help they had given me for this project. Furthermore, I would also like to
thank Yingwen Zhang for helpful discussions on programming techniques and
optimization of numerical routines and Sebastian Bodenstein for technical
discussions about branch cuts and complex analysis in general and the
Renormalization Group.

Finally, I would like to thank the National Institute of Theoretical
Physics(NITheP) which provided financial support over the past two years for
this project.

##  Appendix A Miscellaneous Material for the Gell-Mann-Oakes-Renner Relation

###  A.1 A Step Towards Covariance

The Heaviside/Step function is represented piecewise as the following function

$\displaystyle\Theta\left(t\right)=\left\\{{\begin{array}[]{*{20}c}1&{,t\geq
0}\\\ 0&{,t\leq 0}\\\ \end{array}}\right.$ (A.1.3)

were $t\in\mathbb{R}$. In the complex plane the Heaviside function is
represented more compactly as a contour integral with the contour
$C=C^{\prime}\cup\mathbb{R}$ where $C^{\prime}$ is a positively oriented
semicircle and $\mathbb{R}$ is the real axis from $+\infty$ to $-\infty$

$\displaystyle\Theta\left(t\right)$ $\displaystyle=$
$\displaystyle\frac{1}{{2\pi
i}}\mathop{\oint\mkern-0.8mu}\limits_{C}{\frac{{e^{-izt}}}{{z+i\varepsilon}}dz}$
(A.1.4) $\displaystyle=$ $\displaystyle\frac{1}{{2\pi
i}}\int\limits_{\infty}^{-\infty}{\frac{{e^{-izt}}}{{z+i\varepsilon}}dz}+\frac{1}{{2\pi
i}}\int\limits_{C^{\prime}}{\frac{{e^{-izt}}}{{z+i\varepsilon}}dz}$
$\displaystyle=$ $\displaystyle-\frac{1}{{2\pi
i}}\int\limits_{-\infty}^{\infty}{\frac{{e^{-izt}}}{{z+i\varepsilon}}dz}+\underline{\frac{1}{{2\pi
i}}\int\limits_{C^{\prime}}{\frac{{e^{-izt}}}{{z+i\varepsilon}}dz}}$
$\displaystyle=$ $\displaystyle-\frac{1}{{2\pi
i}}\int\limits_{-\infty}^{\infty}{\frac{{e^{-izt}}}{{z+i\varepsilon}}dz}$
(A.1.5)

were $t\in\mathbb{R}$ and $z\in\mathbb{C}$ and we have used Cauchy’s Theorem
on the analytic function $e^{-izt}$ with a closed contour which is a semi-
circle in the lower half plane sent to infinity. The $i\varepsilon$ is added
to the denominator to prevent the denominator ever going to zero and the limit
is taken at the end of $i\varepsilon\to 0$. This is not a non-rigorous
procedure and we can perform the integral without the $i\varepsilon$, but in
this case we would have to resort to the method of the Principle Values to
extract the integral. Both methods lead to the same solution. The Principle
Value is though, aesthetically more pleasing a method, quite difficult to use
if there are repeated zeros. The second term which is underlined in (A.1.4)
does not contribute to the integral due to Jordan’s Lemma[13]

Consider now

$\displaystyle
e^{-iE\left({x_{0}-y_{0}}\right)}\Theta\left({x_{0}-y_{0}}\right)$
$\displaystyle=$ $\displaystyle-\frac{1}{{2\pi
i}}\int\limits_{-\infty}^{\infty}{\frac{{e^{-iz\left({x_{0}-y_{0}}\right)}e^{-iE\left({x_{0}-y_{0}}\right)}}}{{z+i\varepsilon}}dz}$
(A.1.6) $\displaystyle=$ $\displaystyle-\frac{1}{{2\pi
i}}\int\limits_{-\infty}^{\infty}{\frac{{e^{-i\left({x_{0}-y_{0}}\right)\left({z+E}\right)}}}{{z+i\varepsilon}}dz}$

if we make the following change of variables $q_{0}=z+E$ we get

$\displaystyle
e^{-iE\left({x_{0}-y_{0}}\right)}\Theta\left({x_{0}-y_{0}}\right)=\frac{1}{{2\pi
i}}\int\limits_{-\infty}^{\infty}{\frac{{e^{-i\left({x_{0}-y_{0}}\right)q_{0}}}}{{E-q_{0}-i\varepsilon}}dq_{0}}$
(A.1.7)

we perform the same procedure on the remaining term of (2.1.15) we obtain

$\displaystyle
e^{-iE\left({y_{0}-x_{0}}\right)}\Theta\left({y_{0}-x_{0}}\right)=\frac{1}{{2\pi
i}}\int\limits_{-\infty}^{\infty}{\frac{{e^{-i\left({y_{0}-x_{0}}\right)q_{0}}}}{{E-q_{0}-i\varepsilon}}dq_{0}}$
(A.1.8)

making a change of variables for (A.1.8) with $q_{0}\to-q_{o}$ which leads to

$\displaystyle
e^{-iE\left({y_{0}-x_{0}}\right)}\Theta\left({y_{0}-x_{0}}\right)=\frac{1}{{2\pi
i}}\int\limits_{-\infty}^{\infty}{\frac{{e^{-i\left({x_{0}-y_{0}}\right)q_{0}}}}{{E+q_{0}-i\varepsilon}}dq_{0}}$
(A.1.9)

$X$ can now be evaluated by combining (A.1.7) and (A.1.9)

$\displaystyle X$ $\displaystyle=$ $\displaystyle\frac{1}{{2\pi
i}}\int\limits_{-\infty}^{\infty}{\frac{{e^{-i\left({x_{0}-y_{0}}\right)q_{0}}}}{{E-q_{0}-i\varepsilon}}dq_{0}}+\frac{1}{{2\pi
i}}\int\limits_{-\infty}^{\infty}{\frac{{e^{-i\left({x_{0}-y_{0}}\right)q_{0}}}}{{E+q_{0}-i\varepsilon}}dq_{0}}$
(A.1.10) $\displaystyle=$ $\displaystyle\frac{1}{{2\pi
i}}\int\limits_{-\infty}^{\infty}{\frac{{e^{-i\left({x_{0}-y_{0}}\right)q_{0}}}}{{E-q_{0}-i\varepsilon}}+\frac{{e^{-i\left({x_{0}-y_{0}}\right)q_{0}}}}{{E+q_{0}-i\varepsilon}}dq_{0}}$
$\displaystyle=$ $\displaystyle\frac{1}{{2\pi
i}}\int\limits_{-\infty}^{\infty}{e^{-i\left({x_{0}-y_{0}}\right)q_{0}}\left[{\frac{1}{{E-q_{0}-i\varepsilon}}+\frac{1}{{E+q_{0}-i\varepsilon}}}\right]dq_{0}}$
$\displaystyle=$ $\displaystyle\frac{{2E}}{{2\pi
i}}\int\limits_{-\infty}^{\infty}{\frac{{e^{-i\left({x_{0}-y_{0}}\right)q_{0}}}}{{E^{2}-q_{0}^{2}-i\varepsilon^{\prime}}}dq_{0}}$

we note that the denominator

$E^{2}-q_{0}^{2}-i\varepsilon^{\prime}=m_{\pi}^{2}+\underline{q}^{2}-q_{0}^{2}-i\varepsilon^{\prime}=m_{\pi}^{2}-\left({-\underline{q}^{2}+q_{0}^{2}}\right)-i\varepsilon^{\prime}=m_{\pi}^{2}-q_{\mu}q^{\mu}-i\varepsilon^{\prime}=m_{\pi}^{2}-q^{2}-i\varepsilon^{\prime}$

leading to

$\displaystyle X=\frac{{2E}}{{2\pi
i}}\int\limits_{-\infty}^{\infty}{\frac{{e^{-i\left({x_{0}-y_{0}}\right)q_{0}}}}{{m_{\pi}^{2}-q^{2}-i\varepsilon^{\prime}}}dq_{0}}$
(A.1.11)

###  A.2 Time Ordered Products

$\displaystyle
T\left[{A_{\mu}\left(x\right)A_{\nu}^{\dagger}\left(0\right)}\right]=\left[{A_{\mu}\left(x\right);A_{\nu}^{\dagger}\left(0\right)}\right]\Theta\left({x_{0}}\right)+A_{\nu}^{\dagger}\left(0\right)A_{\mu}\left(x\right)$
(A.2.1)

we prove this as follows by starting with the right-hand-side(RHS)

$\displaystyle\rm{RHS}$ $\displaystyle=$
$\displaystyle\left[{A_{\mu}\left(x\right);A_{\nu}^{\dagger}\left(0\right)}\right]\Theta\left({x_{0}}\right)+A_{\nu}^{\dagger}\left(0\right)A_{\mu}\left(x\right)$
(A.2.2) $\displaystyle=$ $\displaystyle
A_{\mu}\left(x\right)A_{\nu}^{\dagger}\left(0\right)\Theta\left({x_{0}}\right)-A_{\nu}^{\dagger}\left(0\right)A_{\mu}\left(x\right)\Theta\left({x_{0}}\right)+A_{\nu}^{\dagger}\left(0\right)A_{\mu}\left(x\right)$
$\displaystyle=$ $\displaystyle
A_{\mu}\left(x\right)A_{\nu}^{\dagger}\left(0\right)\Theta\left({x_{0}}\right)-A_{\nu}^{\dagger}\left(0\right)A_{\mu}\left(x\right)\left[{\Theta\left({x_{0}}\right)-1}\right]$

but noting that

$\Theta\left({x_{0}}\right)-1=-\Theta\left({-x_{0}}\right)$

and substituting back into (A.2.2) which leads to

$\displaystyle\rm{RHS}$ $\displaystyle=$ $\displaystyle
A_{\mu}\left(x\right)A_{\nu}^{\dagger}\left(0\right)\Theta\left({x_{0}}\right)+A_{\nu}^{\dagger}\left(0\right)A_{\mu}\left(x\right)\Theta\left({-x_{0}}\right)$
$\displaystyle=$ $\displaystyle
T\left[{A_{\mu}\left(x\right)A_{\nu}^{\dagger}\left(0\right)}\right]$
$\displaystyle=$ $\displaystyle\rm{LHS}$

###  A.3 Preparatory Work I

We will now do some preparatory work on (2.2.9) before proceeding. In the
Heisenberg representation operators can be propagated through spacetime via a
product of unitary transformations. In our case

$\displaystyle{\partial^{\mu}}{A_{\mu}}\left(x\right)$ $\displaystyle=$
$\displaystyle{e^{ip\cdot
x}}{\partial^{\mu}}{A_{\mu}}\left(0\right){e^{-ip\cdot x}}$ (A.3.1)
$\displaystyle A_{\nu}^{\dagger}\left(0\right)$ $\displaystyle=$
$\displaystyle{e^{ip\cdot x}}A_{\nu}^{\dagger}\left({-x}\right){e^{-ip\cdot
x}}$ (A.3.2)

here we have taken the operators ${\partial^{\mu}}{A_{\mu}}\left(0\right)$ and
$A_{\nu}^{\dagger}\left({-x}\right)$ and propagated them from the spacetime
points $0$ and $-x$ to the spacetime points $x$ and $0$ respectively.
Substituting (A.3.1) and (A.3.2) into the time-order product (2.2.9)

$\displaystyle
T\left[{{\partial^{\mu}}{A_{\mu}}\left(x\right)A_{\nu}^{\dagger}\left(0\right)}\right]=$
$\displaystyle{e^{ip\cdot
x}}{\partial^{\mu}}{A_{\mu}}\left(0\right){e^{-ip\cdot x}}{e^{ip\cdot
x}}A_{\nu}^{\dagger}\left({-x}\right){e^{-ip\cdot
x}}\Theta\left({{x_{0}}}\right)+$ $\displaystyle+{e^{ip\cdot
x}}A_{\nu}^{\dagger}\left({-x}\right){e^{-ip\cdot x}}{e^{ip\cdot
x}}{\partial^{\mu}}{A_{\mu}}\left(0\right){e^{-ip\cdot
x}}\Theta\left({-{x_{0}}}\right)$ $\displaystyle={e^{ip\cdot
x}}\left[{{\partial^{\mu}}{A_{\mu}}\left(0\right)A_{\nu}^{\dagger}\left({-x}\right)\Theta\left({{x_{0}}}\right)+A_{\nu}^{\dagger}\left({-x}\right){\partial^{\mu}}{A_{\mu}}\left(0\right)\Theta\left({-{x_{0}}}\right)}\right]{e^{-ip\cdot
x}}$ (A.3.3)

substituting (A.3.3) into (2.2.9) we obtain

$\displaystyle\Pi_{{}_{\nu}}^{5}\equiv-\int{{e^{iq\cdot x}}{e^{ip\cdot
x}}\left\langle{0\left|{\left[{{\partial^{\mu}}{A_{\mu}}\left(0\right)A_{\nu}^{\dagger}\left({-x}\right)\Theta\left({{x_{0}}}\right)+A_{\nu}^{\dagger}\left({-x}\right){\partial^{\mu}}{A_{\mu}}\left(0\right)\Theta\left({-{x_{0}}}\right)}\right]}\right|0}\right\rangle{e^{-ip\cdot
x}}{d^{4}}x}$ (A.3.4)

we now make a change of variables with $x\mapsto x=-y$ and $d^{4}x=-d^{4}y$
and then

$\displaystyle\Pi_{{}_{\nu}}^{5}$ $\displaystyle\equiv$
$\displaystyle\int{{e^{-iq\cdot y}}{e^{-ip\cdot
y}}\left\langle{0\left|{\left[{{\partial^{\mu}}{A_{\mu}}\left(0\right)A_{\nu}^{\dagger}\left(y\right)\Theta\left({-{y_{0}}}\right)+A_{\nu}^{\dagger}\left(y\right){\partial^{\mu}}{A_{\mu}}\left(0\right)\Theta\left({{y_{0}}}\right)}\right]}\right|0}\right\rangle{e^{ip\cdot
y}}{d^{4}}y}$ (A.3.5) $\displaystyle=$ $\displaystyle\int{{e^{-iq\cdot
y}}\left\langle{0\left|{T\left[{A_{\nu}^{\dagger}\left(y\right){\partial^{\mu}}{A_{\mu}}\left(0\right)}\right]}\right|0}\right\rangle{d^{4}}y}$

and now just renaming $y\mapsto y=x$ we obtain

$\displaystyle\Pi_{{}_{\nu}}^{5}=\int{{e^{-iq\cdot
x}}\left\langle{0\left|{T\left[{A_{\nu}^{\dagger}\left(x\right){\partial^{\mu}}{A_{\mu}}\left(0\right)}\right]}\right|0}\right\rangle{d^{4}}x}$
(A.3.6)

###  A.4 Preparatory Work II

We first express the time-ordered product as

$\displaystyle
T\left[{\partial^{\nu}A_{\nu}^{\dagger}\left(x\right)\partial^{\mu}A_{\mu}\left(0\right)}\right]=\partial^{\nu}A_{\nu}^{\dagger}\left(x\right)\partial^{\mu}A_{\mu}\left(0\right)\Theta\left({x_{0}}\right)+\partial^{\mu}A_{\mu}\left(0\right)\partial^{\nu}A_{\nu}^{\dagger}\left(x\right)\Theta\left({-x_{0}}\right)$
(A.4.1)

and as previously stated operators expressed in the Heisenberg representation
can be propagated through spacetime via a product of unitary transformations
and in this case

$\displaystyle\partial^{\nu}A_{\nu}^{\dagger}\left(x\right)$ $\displaystyle=$
$\displaystyle e^{ip\cdot
x}\partial^{\nu}A_{\nu}^{\dagger}\left(0\right)e^{-ip\cdot x}$ (A.4.2)
$\displaystyle\partial^{\mu}A_{\mu}\left(0\right)$ $\displaystyle=$
$\displaystyle e^{ip\cdot x}\partial^{\mu}A_{\mu}\left({-x}\right)e^{-ip\cdot
x}$ (A.4.3)

now substituting (A.4.2) and (A.4.3) into the time-ordered product (A.4.1) we
obtain

$\displaystyle
T\left[{\partial^{\nu}A_{\nu}^{\dagger}\left(x\right)\partial^{\mu}A_{\mu}\left(0\right)}\right]=e^{ip\cdot
x}\left[{\partial^{\nu}A_{\nu}^{\dagger}\left(0\right)\partial^{\mu}A_{\mu}\left({-x}\right)\Theta\left({x_{0}}\right)+\partial^{\mu}A_{\mu}\left({-x}\right)\partial^{\nu}A_{\nu}^{\dagger}\left(0\right)\Theta\left({-x_{0}}\right)}\right]e^{-ip\cdot
x}$ (A.4.4)

we are now ready to cast the second term of (2.2.20)

$\begin{gathered}-i\int{e^{-iq\cdot
x}\left\langle{0\left|{T\left[{\partial^{\nu}A_{\nu}^{\dagger}\left(x\right)\partial^{\mu}A_{\mu}\left(0\right)}\right]}\right|0}\right\rangle
d^{4}x}\hfill\\\ =-i\int{e^{-iq\cdot x}e^{ip\cdot
x}\left\langle{0\left|{\left[{\partial^{\nu}A_{\nu}^{\dagger}\left(0\right)\partial^{\mu}A_{\mu}\left({-x}\right)\Theta\left({x_{0}}\right)+\partial^{\mu}A_{\mu}\left({-x}\right)\partial^{\nu}A_{\nu}^{\dagger}\left(0\right)\Theta\left({-x_{0}}\right)}\right]}\right|0}\right\rangle
e^{-ip\cdot x}d^{4}x}\hfill\\\ \end{gathered}$

making the change of variables $x\mapsto x=-y$ so $d^{4}x=-d^{4}y$ leads to

$\begin{gathered}-i\int{e^{-iq\cdot
x}\left\langle{0\left|{T\left[{\partial^{\nu}A_{\nu}^{\dagger}\left(x\right)\partial^{\mu}A_{\mu}\left(0\right)}\right]}\right|0}\right\rangle
d^{4}x}\hfill\\\ =i\int{e^{iq\cdot y}e^{-ip\cdot
y}\left\langle{0\left|{\left[{\partial^{\nu}A_{\nu}^{\dagger}\left(0\right)\partial^{\mu}A_{\mu}\left(y\right)\Theta\left({-y_{0}}\right)+\partial^{\mu}A_{\mu}\left(y\right)\partial^{\nu}A_{\nu}^{\dagger}\left(0\right)\Theta\left({y_{0}}\right)}\right]}\right|0}\right\rangle
e^{ip\cdot y}d^{4}y}\hfill\\\ \end{gathered}$

but we now have a new time-ordered product with

$\displaystyle\partial^{\nu}A_{\nu}^{\dagger}\left(0\right)\partial^{\mu}A_{\mu}\left(y\right)\Theta\left({-y_{0}}\right)+\partial^{\mu}A_{\mu}\left(y\right)\partial^{\nu}A_{\nu}^{\dagger}\left(0\right)\Theta\left({y_{0}}\right)=T\left[{\partial^{\mu}A_{\mu}\left(y\right)\partial^{\nu}A_{\nu}^{\dagger}\left(0\right)}\right]$
(A.4.5)

and

$\displaystyle-i\int{e^{-iq\cdot
x}\left\langle{0\left|{T\left[{\partial^{\nu}A_{\nu}^{\dagger}\left(x\right)\partial^{\mu}A_{\mu}\left(0\right)}\right]}\right|0}\right\rangle
d^{4}x}=i\int{e^{iq\cdot
y}\left\langle{0\left|{T\left[{\partial^{\mu}A_{\mu}\left(y\right)\partial^{\nu}A_{\nu}^{\dagger}\left(0\right)}\right]}\right|0}\right\rangle
d^{4}y}$ (A.4.6)

we now make a final change of variables of $y\mapsto y=x$ and substituting
into (A.4.6)and arrive at

$\displaystyle-i\int{e^{-iq\cdot
x}\left\langle{0\left|{T\left[{\partial^{\nu}A_{\nu}^{\dagger}\left(x\right)\partial^{\mu}A_{\mu}\left(0\right)}\right]}\right|0}\right\rangle
d^{4}x}=i\int{e^{iq\cdot
x}\left\langle{0\left|{T\left[{\partial^{\mu}A_{\mu}\left(x\right)\partial^{\nu}A_{\nu}^{\dagger}\left(0\right)}\right]}\right|0}\right\rangle
d^{4}x}$ (A.4.7)

###  A.5 Operator Identity

We give a quick proof of (2.3.20) starting from the right-hand-side

$\displaystyle\rm{RHS}$ $\displaystyle=$ $\displaystyle
A^{\dagger}\left\\{{B;C^{\dagger}}\right\\}D-C^{\dagger}\left\\{{D;A^{\dagger}}\right\\}B-\left\\{{A^{\dagger};C^{\dagger}}\right\\}BD+C^{\dagger}A^{\dagger}\left\\{{B;D}\right\\}$
$\displaystyle=$ $\displaystyle
A^{\dagger}\left({BC^{\dagger}+C^{\dagger}B}\right)D-C^{\dagger}\left({DA^{\dagger}+A^{\dagger}D}\right)B-\left({A^{\dagger}C^{\dagger}+C^{\dagger}A^{\dagger}}\right)BD+C^{\dagger}A^{\dagger}\left({BD+DB}\right)$
$\displaystyle=$ $\displaystyle
A^{\dagger}BC^{\dagger}D+A^{\dagger}C^{\dagger}BD-C^{\dagger}DA^{\dagger}B-C^{\dagger}A^{\dagger}DB-A^{\dagger}C^{\dagger}BD-C^{\dagger}A^{\dagger}BD+C^{\dagger}A^{\dagger}BD+C^{\dagger}A^{\dagger}DB$
$\displaystyle=$ $\displaystyle
A^{\dagger}BC^{\dagger}D-C^{\dagger}DA^{\dagger}B$ $\displaystyle=$
$\displaystyle\left[{A^{\dagger}B;C^{\dagger}D}\right]$ $\displaystyle=$
$\displaystyle\rm{LHS}$

##  Appendix B Perturbative QCD Calculations

In this section we will explicitly derive
$\left.{\delta_{5}\left(R\right)}\right|_{PQCD}$ from a clean slate. The
pseudoscalar correlator(B.1) and the kernel(B.2) are shown below

$\displaystyle\left.{\psi_{5}\left({q^{2}}\right)}\right|_{PQCD}=m_{q}^{2}\left({\left|q\right|^{2}}\right)\left[{-q^{2}\Pi_{0}\left({q^{2}}\right)+m_{q}\left({\left|q\right|^{2}}\right)\Pi_{2}\left({q^{2}}\right)+\frac{{C_{u}}}{{-q^{2}}}\left.{\left\langle{m_{s}\bar{u}u}\right\rangle}\right|_{\mu_{0}}+\sum\limits_{j=1}^{3}{\frac{{C_{j}}}{{-q^{2}}}\left\langle{\hat{O}_{j}}\right\rangle}+\ldots}\right]$
(B.1)
$\displaystyle\Delta\left({{q^{2}}}\right)=1-{a_{0}}{q^{2}}-{a_{1}}{q^{4}}$
(B.2)

were $\Pi_{0}\left({q^{2}}\right)$ is used depending on which loop order we
want. The remaining terms in the correlator are the correction terms and are
usually small relative to the terms in $\Pi_{0}\left({q^{2}}\right)$. We
attempt to calculate $\delta_{5}(q^{2})$ which is

$\displaystyle{\delta_{5}}\left({{q^{2}}}\right)=\frac{1}{{2\pi
i}}\oint\limits{\frac{1}{{{q^{2}}}}\Delta\left({{q^{2}}}\right){\psi_{5}}\left({{q^{2}}}\right)d{q^{2}}}$
(B.3)

Since $q^{2}$ is a complex variable, we shall for convenience rename
$q^{2}\rightarrow z$, articles published on the method of the QCD Sum Rules
tend to use $s$ but we shall use the letter $z$ as it is traditionally a
complex variable.

###  B.1 $\Pi_{0}(z)$ Calculations

####  B.1.1 One Loop Calculation

For the one loop

$\displaystyle\Pi_{0}\left({q^{2}}\right)$ $\displaystyle=$
$\displaystyle\frac{1}{{16\pi^{2}}}\left[{-12+6\ln\left({\frac{{-q^{2}}}{{\mu^{2}}}}\right)+O\left({\alpha_{s}}\right)}\right]$
(B.1.1)

$\displaystyle\psi_{5}\left({q^{2}}\right)=-\frac{{m_{q}^{2}\left({\left|q\right|^{2}}\right)}}{{16\pi^{2}}}q^{2}\left[{-12+6\ln\left({\frac{{-q^{2}}}{{\mu^{2}}}}\right)}\right]$
(B.1.2)

Along the contour the radius is a constant, thus $|q|^{2}=|z|=R$, so we can
take the $m_{q}^{2}(|q|^{2})$ term out of the integral. We now have to
integrate the following

$\displaystyle\left.{\delta_{5}}\right|_{1\text{loop}}$ $\displaystyle=$
$\displaystyle\frac{{m_{q}^{2}\left(R\right)}}{{16\pi^{2}}}\frac{1}{{2\pi
i}}\oint\limits_{C(R)}{\left({1-a_{0}z-a_{1}z^{2}}\right)\left({-z}\right)\left[{-12+6\ln\left({\frac{{-z}}{{\mu^{2}}}}\right)}\right]\frac{1}{z}dz}$
(B.1.3) $\displaystyle=$
$\displaystyle\frac{{m_{q}^{2}\left(R\right)}}{{16\pi^{2}}}\frac{1}{{2\pi
i}}\oint\limits_{C(R)}{\left({1-a_{0}z-a_{1}z^{2}}\right)\left[{12-6\ln\left({\frac{{-z}}{{\mu^{2}}}}\right)}\right]dz}$
(B.1.4)

making the change of variables $z=\mu^{2}w$ and $dz=\mu^{2}dw$

$\displaystyle\left.{\delta_{5}}\right|_{1\text{loop}}$ $\displaystyle=$
$\displaystyle\frac{{m_{q}^{2}\left(R\right)}}{{16\pi^{2}}}\mu^{2}\frac{1}{{2\pi
i}}\oint\limits_{C(r)}{\left({1-a_{0}\mu^{2}w-a_{1}\mu^{4}w^{2}}\right)\left[{12-6\ln\left({-w}\right)}\right]dw}$
(B.1.5) $\displaystyle=$
$\displaystyle\frac{{m_{q}^{2}\left(R\right)}}{{16\pi^{2}}}\left({-6\mu^{2}}\right)\frac{1}{{2\pi
i}}\oint\limits_{C(r)}{\left({1-a_{0}\mu^{2}w-a_{1}\mu^{4}w^{2}}\right)\ln\left({-w}\right)dw}$
(B.1.6)

Using the integrals generated by using (F.1) we can integrate the above to

$\displaystyle\left.{\delta_{5}}\right|_{1\text{loop}}=\frac{{m_{q}^{2}\left(R\right)}}{{16\pi^{2}}}\left({-6\mu^{2}}\right)\left[{r-a_{0}\mu^{2}\frac{1}{2}r^{2}-a_{1}\mu^{4}\frac{1}{3}r^{3}}\right]$
(B.1.7)

since the change of variables were $z=\mu^{2}w$ $\Rightarrow$ $|z|=\mu^{2}|w|$
thus $R=\mu^{2}r$, substituting back,then setting $\mu^{2}=R$ and simplifying
we obtain

$\displaystyle\left.{\delta_{5}}\right|_{1\text{loop}}=-\frac{{m_{q}^{2}\left(R\right)}}{{16\pi^{2}}}\left[{6R-3a_{0}R^{2}-2a_{1}R^{3}}\right]$
(B.1.8)

####  B.1.2 Two Loop Calculation

For the two loop

$\displaystyle\Pi_{0}\left({q^{2}}\right)=\frac{1}{{16\pi^{2}}}\left[{\ldots+\left({\frac{{\alpha_{s}}}{\pi}}\right)\left({24\zeta\left(3\right)-\frac{{131}}{2}+34\ln\left({\frac{{-q^{2}}}{{\mu^{2}}}}\right)-6\ln^{2}\left({\frac{{-q^{2}}}{{\mu^{2}}}}\right)}\right)+O\left({\alpha_{s}^{2}}\right)}\right]$
(B.1.1)

now substituting (B.1.1) into (B.1) we arrive at

$\displaystyle\left.{\delta_{5}}\right|_{2\text{loop}}=-\frac{{m_{q}^{2}\left(R\right)}}{{16\pi^{2}}}\left[{\frac{{\alpha_{s}}}{\pi}}\right]\frac{1}{{2\pi
i}}\oint\limits_{C(R)}{\left({1-a_{0}z-a_{1}z^{2}}\right)\left[{\left({24\zeta\left(3\right)-\frac{{131}}{2}}\right)+34\ln\left({\frac{{-z}}{{\mu^{2}}}}\right)-6\ln^{2}\left({\frac{{-z}}{{\mu^{2}}}}\right)}\right]dz}$
(B.1.2)

using the same change of variables as shown in the one-loop calculations and
simplify we obtain

$\displaystyle\left.{\delta_{5}}\right|_{2\text{loop}}=-\frac{{m_{q}^{2}\left(R\right)}}{{16\pi^{2}}}\left[{\frac{{\alpha_{s}}}{\pi}}\right]\Biggl{\\{}$
$\displaystyle 34\mu^{2}\frac{1}{{2\pi
i}}\oint\limits_{C(r)}{\left({1-a_{0}\mu^{2}w-a_{1}\mu^{4}w^{2}}\right)\ln\left({-w}\right)dw}+$
$\displaystyle-\frac{6\mu^{2}}{{2\pi
i}}\oint\limits_{C(r)}{\left({1-a_{0}\mu^{2}w-a_{1}\mu^{4}w^{2}}\right)\ln^{2}\left({-w}\right)dw}\Biggr{\\}}$
(B.1.3)

Using the integrals generated by using (F.1) we can integrate the above to

$\displaystyle\left.{\delta_{5}}\right|_{2\text{loop}}=-\frac{{m_{q}^{2}\left(R\right)}}{{16\pi^{2}}}\left[{\frac{{\alpha_{s}}}{\pi}}\right]$
$\displaystyle\Biggl{\\{}34\mu^{2}\left[{r-a_{0}\mu^{2}\frac{{r^{2}}}{2}-a_{1}\mu^{4}\frac{{r^{3}}}{3}}\right]+$
$\displaystyle-6\mu^{2}\left[{2r\left({\ln
r-1}\right)-a_{0}\mu^{2}r^{2}\left({\ln
r-\frac{1}{2}}\right)-a_{1}\mu^{4}\frac{{r^{3}}}{9}\left({6\ln
r-2}\right)}\right]\Biggr{\\}}$ (B.1.4)

since the change of variables were $z=\mu^{2}w$ $\Rightarrow$ $|z|=\mu^{2}|w|$
thus $R=\mu^{2}r$, substituting back,then setting $\mu^{2}=R$ and simplifying
we obtain

$\displaystyle\left.{\delta_{5}}\right|_{2\text{loop}}=-\frac{{m_{q}^{2}\left(R\right)}}{{16\pi^{2}}}\left[{\frac{{\alpha_{s}}}{\pi}}\right]\left[{46R-20a_{0}R^{2}-\frac{{38}}{3}a_{1}R^{3}}\right]$
(B.1.5)

####  B.1.3 Three Loop Calculation

For the three loop

$\displaystyle\Pi_{0}\left({q^{2}}\right)=\frac{1}{{16\pi^{2}}}\left[{\ldots+\left({\frac{{\alpha_{s}}}{\pi}}\right)^{2}\left({{\rm
A_{3}}\ln\left({\frac{{-q^{2}}}{{\mu^{2}}}}\right)+{\rm
B_{3}}\ln^{2}\left({\frac{{-q^{2}}}{{\mu^{2}}}}\right)+{\rm
C_{3}}\ln^{3}\left({\frac{{-q^{2}}}{{\mu^{2}}}}\right)}\right)+O\left({\alpha_{s}^{3}}\right)}\right]$
(B.1.1)

were

$\displaystyle{\rm A_{3}}$ $\displaystyle=$ $\displaystyle
4n_{f}\zeta\left(3\right)-\frac{{65}}{4}n_{f}-117\zeta\left(3\right)+\frac{{10801}}{{24}}+34$
(B.1.2) $\displaystyle{\rm B_{3}}$ $\displaystyle=$
$\displaystyle\frac{{11}}{3}n_{f}-106$ (B.1.3) $\displaystyle{\rm C_{3}}$
$\displaystyle=$ $\displaystyle\frac{{19}}{2}-\frac{1}{3}n_{f}$ (B.1.4)

now substituting (B.1.1) into (B.1) and making the change of variables we
arrive at

$\displaystyle\left.{\delta_{5}}\right|_{3\text{loop}}\left[{-\frac{{m_{q}^{2}\left(R\right)}}{{16\pi^{2}}}\left[{\frac{{\alpha_{s}}}{\pi}}\right]^{2}}\right]^{-1}\\!\\!\\!\\!\\!\\!=$
$\displaystyle\frac{{\text{A}_{3}}}{{2\pi
i}}\oint\limits_{C(r)}{\left({1-a_{0}\mu^{2}w-a_{1}\mu^{4}w^{2}}\right)\ln\left({-w}\right)dw}+$
$\displaystyle+\frac{{\text{B}_{3}}}{{2\pi
i}}\oint\limits_{C(r)}{\left({1-a_{0}\mu^{2}w-a_{1}\mu^{4}w^{2}}\right)\ln^{2}\left({-w}\right)dw}$
$\displaystyle+\frac{{\text{C}_{3}}}{{2\pi
i}}\oint\limits_{C(r)}{\left({1-a_{0}\mu^{2}w-a_{1}\mu^{4}w^{2}}\right)\ln^{3}\left({-w}\right)dw}$
(B.1.5)

Using the integrals generated by using (F.1) we can integrate the above to

$\displaystyle\left.{\delta_{5}}\right|_{3\text{loop}}\left[{-\frac{{m_{q}^{2}\left(R\right)}}{{16\pi^{2}}}\left[{\frac{{\alpha_{s}}}{\pi}}\right]^{2}}\right]^{-1}$
$\displaystyle=$
$\displaystyle{\text{A}_{3}}\left[{r-a_{0}\mu^{2}\frac{{r^{2}}}{2}-a_{1}\mu^{4}\frac{{r^{3}}}{3}}\right]+{\text{B}_{3}}\left[{2r\left({\ln
r-1}\right)-a_{0}\mu^{2}r^{2}\left({\ln
r-\frac{1}{2}}\right)-a_{1}\mu^{4}\frac{{r^{3}}}{9}\left({6\ln
r-2}\right)}\right]$ $\displaystyle+$
$\displaystyle{\text{C}_{3}}\left[{r\left({3\ln^{2}r-6\ln
r+6-\pi^{2}}\right)-a_{0}\mu^{2}r^{2}\left({\frac{3}{2}\ln^{2}r-\frac{3}{2}\ln
r+\frac{3}{4}-\frac{{\pi^{2}}}{2}}\right)-a_{1}\mu^{4}r^{3}\left({\ln^{3}r-2\ln
r+\frac{2}{9}-\frac{{\pi^{2}}}{3}}\right)}\right]$

since the change of variables were $z=\mu^{2}w$ $\Rightarrow$ $|z|=\mu^{2}|w|$
thus $R=\mu^{2}r$, substituting back,then setting $\mu^{2}=R$ and simplifying
we obtain

$\displaystyle\left.{\delta_{5}}\right|_{3\text{loop}}\left[{-\frac{{m_{q}^{2}\left(R\right)}}{{16\pi^{2}}}\left[{\frac{{\alpha_{s}}}{\pi}}\right]^{2}}\right]^{-1}=\left[{A_{3}-2B_{3}+C_{3}\left({6-\pi^{2}}\right)}\right]R+\frac{{a_{0}}}{2}\left[{B_{3}-A_{3}-2C_{3}\left({\frac{3}{4}-\frac{{\pi^{2}}}{2}}\right)}\right]R^{2}+$
$\displaystyle+\frac{{a_{1}}}{3}\left[{\frac{{2B_{3}}}{3}-A_{3}-C_{3}\left({\frac{{2-3\pi^{2}}}{3}}\right)}\right]R^{3}$
(B.1.7)

using the A3, B3 and C3 defined above in (B.1.2),(B.1.3),(B.1.4) we find the
coefficients of $R$,$R^{2}$ and $R^{3}$ are

$\displaystyle A_{3}-2B_{3}+C_{3}\left({6-\pi^{2}}\right)$
$\displaystyle=\left({4n_{f}-117}\right)\zeta\left(3\right)+\frac{{15889-566n_{f}}}{{24}}+\left({2n_{f}-57}\right)\left({\frac{{\pi^{2}}}{6}-1}\right)$
(B.1.8)
$\displaystyle\frac{{a_{0}}}{2}\left[{B_{3}-A_{3}-2C_{3}\left({\frac{3}{4}-\frac{{\pi^{2}}}{2}}\right)}\right]$
$\displaystyle=\frac{{a_{0}}}{2}\left[{\left({117-4n_{f}}\right)\zeta\left(3\right)+\frac{{478n_{f}-13345}}{{24}}+\left({57-2n_{f}}\right)\left({\frac{{\pi^{2}}}{6}-\frac{1}{4}}\right)}\right]$
(B.1.9)
$\displaystyle\frac{{a_{1}}}{3}\left[{\frac{{2B_{3}}}{3}-A_{3}-C_{3}\left({\frac{{2-3\pi^{2}}}{3}}\right)}\right]$
$\displaystyle=\frac{{a_{1}}}{3}\left[{\left({117-4n_{f}}\right)\zeta\left(3\right)+\frac{{1346n_{f}-37491}}{{72}}+\left({57-2n_{f}}\right)\left({\frac{{\pi^{2}}}{6}-\frac{1}{9}}\right)}\right]$
(B.1.10)

####  B.1.4 Four Loop Calculation

For the three loop

$\displaystyle\Pi\left({q^{2}}\right)=\frac{1}{{16\pi^{2}}}\left[{\ldots+\left({\frac{{\alpha_{s}}}{\pi}}\right)^{3}\left({A\ln\left({\frac{{-q^{2}}}{{\mu^{2}}}}\right)+B\ln^{2}\left({\frac{{-q^{2}}}{{\mu^{2}}}}\right)+C\ln^{3}\left({\frac{{-q^{2}}}{{\mu^{2}}}}\right)+D\ln^{4}\left({\frac{{-q^{2}}}{{\mu^{2}}}}\right)}\right)+O\left({\alpha_{{}_{s}}^{4}}\right)}\right]$
(B.1.1)

were

$\displaystyle A$ $\displaystyle=$ $\displaystyle A_{1}$ (B.1.2)
$\displaystyle B$ $\displaystyle=$
$\displaystyle-6\left[{\frac{{4781}}{{18}}-\frac{{475}}{8}\zeta\left(3\right)}\right]$
(B.1.3) $\displaystyle C$ $\displaystyle=$ $\displaystyle 229$ (B.1.4)
$\displaystyle D$ $\displaystyle=$ $\displaystyle-\frac{{221}}{{16}}$ (B.1.5)

now substituting (B.1.1) into (B.1) and making the change of variables we
arrive at

$\displaystyle\left.\delta\right|_{4\text{loop}}\left[{-\mu^{2}\frac{{m_{q}^{2}\left(R\right)}}{{16\pi^{2}}}\left[{\frac{{\alpha_{s}}}{\pi}}\right]^{3}}\right]^{-1}$
$\displaystyle=$ $\displaystyle\frac{A}{{2\pi
i}}\oint\limits_{C(r)}{\left({1-a_{0}\mu^{2}w-a_{1}\mu^{4}w^{2}}\right)\ln\left({-w}\right)dw}+\frac{B}{{2\pi
i}}\oint\limits_{C(r)}{\left({1-a_{0}\mu^{2}w-a_{1}\mu^{4}w^{2}}\right)\ln^{2}\left({-w}\right)dw}$
$\displaystyle+$ $\displaystyle\frac{C}{{2\pi
i}}\oint\limits_{C(r)}{\left({1-a_{0}\mu^{2}w-a_{1}\mu^{4}w^{2}}\right)\ln^{3}\left({-w}\right)dw}+\frac{D}{{2\pi
i}}\oint\limits_{C(r)}{\left({1-a_{0}\mu^{2}w-a_{1}\mu^{4}w^{2}}\right)\ln^{4}\left({-w}\right)dw}$

Using the integrals generated by using (F.1) we can integrate the above to

$\displaystyle\left.\delta\right|_{4\text{loop}}\left[{-\mu^{2}\frac{{m_{q}^{2}\left(R\right)}}{{16\pi^{2}}}\left[{\frac{{\alpha_{s}}}{\pi}}\right]^{3}}\right]^{-1}$
$\displaystyle=$ $\displaystyle
A\left[{r-a_{0}\mu^{2}\frac{{r^{2}}}{2}-a_{1}\mu^{4}\frac{{r^{3}}}{3}}\right]+B\left[{2r\left({\ln
r-1}\right)-a_{0}\mu^{2}r^{2}\left({\ln
r-\frac{1}{2}}\right)-a_{1}\mu^{4}\frac{{r^{3}}}{9}\left({6\ln
r-2}\right)}\right]$ $\displaystyle+$ $\displaystyle
C\left[{r\left({3\ln^{2}r-6\ln
r+6-\pi^{2}}\right)-a_{0}\mu^{2}r^{2}\left({\frac{3}{2}\left({\ln^{2}r-\ln
r}\right)+\frac{3}{4}-\frac{{\pi^{2}}}{2}}\right)-a_{1}\mu^{4}r^{3}\left({\ln^{3}r-2\ln
r+\frac{2}{9}-\frac{{\pi^{2}}}{3}}\right)}\right]$ $\displaystyle+$
$\displaystyle
D\left[{r\left({4\ln^{3}r-12\ln^{2}r+4\left({6-\pi^{2}}\right)\ln
r+4\pi^{2}-24}\right)-a_{0}\mu^{2}r^{2}\left({2\ln^{3}r-3\ln^{2}r+\left({3-2\pi^{2}}\right)\ln
r+\pi^{2}-\frac{3}{2}}\right)}\right]$ $\displaystyle+$ $\displaystyle
D\left[{-a_{1}\mu^{4}r^{3}\left({\frac{4}{3}\left({\ln^{3}r-\ln^{2}r+\left({\frac{2}{3}-\pi^{2}}\right)\ln
r}\right)+\frac{4}{{3^{2}}}\left({\pi^{2}-\frac{2}{3}}\right)}\right)}\right]$
(B.1.6)

since the change of variables were $z=\mu^{2}w$ $\Rightarrow$ $|z|=\mu^{2}|w|$
thus $R=\mu^{2}r$, substituting back,then setting $\mu^{2}=R$ and simplifying
we obtain

$\displaystyle\left.\delta\right|_{4\text{loop}}\left[{\frac{{-m_{q}^{2}\left(R\right)}}{{16\pi^{2}}}\left[{\frac{{\alpha_{s}}}{\pi}}\right]^{3}}\right]^{-1}$
$\displaystyle=$
$\displaystyle\left[{A-2B+\left({6-\pi^{2}}\right)C+4\left({\pi^{2}-6}\right)D}\right]R+\frac{{a_{0}}}{2}\left[{-A+B-2\left({\frac{3}{4}-\frac{{\pi^{2}}}{2}}\right)C-2\left({\pi^{2}-\frac{3}{2}}\right)D}\right]R^{2}$
$\displaystyle+$
$\displaystyle\frac{{a_{1}}}{3}\left[{-A-\frac{2}{3}B-\left({\frac{2}{3}-\pi^{2}}\right)C-\frac{4}{3}\left({\pi^{2}-\frac{2}{3}}\right)D}\right]R^{3}$
(B.1.7)

using the A, B, C and D defined above in (B.1.2-B.1.5) we find the
coefficients of $R$,$R^{2}$ and $R^{3}$ are

$\displaystyle A-2B+\left({6-\pi^{2}}\right)C+4\left({\pi^{2}-6}\right)D$
$\displaystyle=$ $\displaystyle{\rm A}_{\rm
1}+12\left[{\frac{{4781}}{{18}}-\frac{{475}}{8}\zeta\left(3\right)}\right]-\frac{{3411}}{2}\left({\frac{{\pi^{2}}}{6}-1}\right)$
(B.1.8)
$\displaystyle-A+B-2\left({\frac{3}{4}-\frac{{\pi^{2}}}{2}}\right)C-2\left({\pi^{2}-\frac{3}{2}}\right)D$
$\displaystyle=$ $\displaystyle-{\rm A}_{\rm
1}-6\left[{\frac{{4781}}{{18}}-\frac{{475}}{8}\zeta\left(3\right)}\right]+\frac{{6159}}{4}\left({\frac{{\pi^{2}}}{6}-\frac{1}{4}}\right)$
(B.1.9)
$\displaystyle-A-\frac{2}{3}B-\left({\frac{2}{3}-\pi^{2}}\right)C-\frac{4}{3}\left({\pi^{2}-\frac{2}{3}}\right)D$
$\displaystyle=$ $\displaystyle-{\rm A}_{\rm
1}+4\left[{\frac{{4781}}{{18}}-\frac{{475}}{8}\zeta\left(3\right)}\right]+\frac{{2969}}{2}\left({\frac{{\pi^{2}}}{6}-\frac{1}{4}}\right)$
(B.1.10)

####  B.1.5 Five Loop Calculation

For the three loop

$\displaystyle\Pi_{\rm
0}\left({q^{2}}\right)=\frac{1}{{16\pi^{2}}}\left[{\ldots+\left({\frac{{\alpha_{s}}}{\pi}}\right)^{4}\sum\limits_{i=1}^{5}{H_{i}L^{i}}+O\left({\alpha_{s}^{5}}\right)}\right]$
(B.1.1)

where

$L^{i}=\ln^{i}\left({\frac{{-q^{2}}}{{\mu^{2}}}}\right)$

and

$H_{i}\in\mathbb{R}$

now substituting (B.1.1) into (B.1) and making the change of variables we
arrive at

$\displaystyle\left.{\delta_{5}}\right|_{5\text{loop}}\left[{-\mu^{2}\frac{{m_{q}^{2}\left(R\right)}}{{16\pi^{2}}}\left[{\frac{{\alpha_{s}}}{\pi}}\right]^{4}}\right]^{-1}$
$\displaystyle=$ $\displaystyle\frac{{H_{1}}}{{2\pi
i}}\oint\limits_{C(r)}{\left({1-a_{0}\mu^{2}w-a_{1}\mu^{4}w^{2}}\right)\ln\left({-w}\right)dw}+\frac{{H_{2}}}{{2\pi
i}}\oint\limits_{C(r)}{\left({1-a_{0}\mu^{2}w-a_{1}\mu^{4}w^{2}}\right)\ln^{2}\left({-w}\right)dw}$
$\displaystyle+$ $\displaystyle\frac{{H_{3}}}{{2\pi
i}}\oint\limits_{C(r)}{\left({1-a_{0}\mu^{2}w-a_{1}\mu^{4}w^{2}}\right)\ln^{3}\left({-w}\right)dw}+\frac{{H_{4}}}{{2\pi
i}}\oint\limits_{C(r)}{\left({1-a_{0}\mu^{2}w-a_{1}\mu^{4}w^{2}}\right)\ln^{4}\left({-w}\right)dw}$
$\displaystyle+$ $\displaystyle\frac{{H_{5}}}{{2\pi
i}}\oint\limits_{C(r)}{\left({1-a_{0}\mu^{2}w-a_{1}\mu^{4}w^{2}}\right)\ln^{5}\left({-w}\right)dw}$
(B.1.2)

Using the integrals generated by using (F.1) we can integrate the above to

$\displaystyle\left.{\left[{-\frac{{m_{q}^{2}\mu^{2}}}{{16\pi^{2}}}\left[{\frac{{\alpha_{s}}}{\pi}}\right]^{4}}\right]^{-1}\delta_{5}}\right|_{5\text{loop}}$
$\displaystyle=$ $\displaystyle
H_{1}\left({r-\frac{{a_{0}\mu^{2}}}{2}r^{2}-\frac{{a_{1}\mu^{4}}}{4}r^{3}}\right)+H_{2}\left({2r\left({\ln
r-1}\right)-a_{0}\mu^{2}r^{2}\left({\ln
r-\frac{1}{2}}\right)-\frac{{a_{1}\mu^{4}}}{9}r^{3}\left({6\ln
r-2}\right)}\right)+$ $\displaystyle+$ $\displaystyle
H_{3}\left({r\left({3\ln^{2}r-6\ln
r+6-\pi^{2}}\right)-a_{0}\mu^{2}r^{2}\left({\frac{3}{2}\ln^{2}r-\frac{3}{2}\ln
r+\frac{3}{4}-\frac{{\pi^{2}}}{2}}\right)}\right)+$ $\displaystyle-$
$\displaystyle a_{1}\mu^{4}r^{3}H_{3}\left({\ln^{3}r-2\ln
r+\frac{2}{9}-\frac{{\pi^{2}}}{3}}\right)+$ $\displaystyle+$ $\displaystyle
rH_{4}\left({4\ln^{3}r-12\ln^{2}r+4\ln
r\left({6-\pi^{2}}\right)+4\left({\pi^{2}-6}\right)}\right)$ $\displaystyle-$
$\displaystyle a_{0}\mu^{2}r^{2}H_{4}\left({2\ln^{3}r-3\ln^{2}r+\ln
r\left({3-2\pi^{2}}\right)+\pi^{2}-\frac{3}{2}}\right)+$ $\displaystyle-$
$\displaystyle
a_{1}\mu^{4}r^{3}H_{4}\left({\frac{4}{3}\left({\ln^{3}r-\ln^{2}r+\ln
r\left({\frac{2}{3}-\pi^{2}}\right)}\right)+\frac{4}{9}\left({\pi^{2}-\frac{2}{3}}\right)}\right)+$
$\displaystyle+$ $\displaystyle rH_{5}\left({5\ln^{4}r-20\ln^{3}r+10\left({\ln
r-2}\right)\left({6-\pi^{2}}\right)\ln r+\pi^{4}-20\pi^{2}+120}\right)+$
$\displaystyle-$ $\displaystyle
a_{0}\mu^{2}r^{2}H_{5}\left({\frac{5}{2}\ln^{4}r-5\ln^{3}r+5\left({\ln
r-1}\right)\left({\frac{3}{2}-\pi^{2}}\right)\ln
r+\frac{{\pi^{4}}}{2}-\frac{5}{2}\pi^{2}+\frac{{15}}{4}}\right)+$
$\displaystyle-$ $\displaystyle
a_{1}\mu^{4}r^{3}H_{5}\left({\frac{5}{3}\ln^{4}r-\frac{{20}}{9}\ln^{3}r+\frac{{10}}{9}\left({\ln
r-\frac{2}{3}}\right)\left({\frac{6}{9}-\pi^{2}}\right)\ln
r+\frac{{\pi^{4}}}{3}-\frac{{20}}{{27}}\pi^{2}+\frac{{40}}{{81}}}\right)$
(B.1.3)

since the change of variables were $z=\mu^{2}w$ $\Rightarrow$ $|z|=\mu^{2}|w|$
thus $R=\mu^{2}r$, substituting back,then setting $\mu^{2}=R$ and simplifying
we obtain

$\displaystyle\left.{\delta_{5}}\right|_{5\text{loop}}=-\frac{{m_{q}^{2}}}{{16\pi^{2}}}\left[{\frac{{\alpha_{s}}}{\pi}}\right]^{4}\left[{\alpha
R+\beta R^{2}+\gamma R^{3}}\right]$ (B.1.4)

where

$\displaystyle\alpha$ $\displaystyle=$ $\displaystyle
H_{1}-2H_{2}+\left({4H_{4}-H_{3}}\right)\left({\pi^{2}-6}\right)+\left({\pi^{4}-20\pi^{2}+120}\right)H_{5}$
(B.1.5) $\displaystyle\beta$ $\displaystyle=$ $\displaystyle-
H_{1}+H_{2}+\left({H_{3}-2H_{4}}\right)\left({\pi^{2}-\frac{3}{2}}\right)-\left({\pi^{4}-5\pi^{2}+\frac{{15}}{2}}\right)H_{5}$
(B.1.6) $\displaystyle\gamma$ $\displaystyle=$ $\displaystyle-
H_{1}+\frac{2}{3}H_{2}+\left({H_{3}-\frac{4}{3}H_{4}}\right)\left({\pi^{2}-\frac{2}{3}}\right)-\left({\pi^{4}-\frac{{20}}{9}\pi^{2}+\frac{{40}}{{27}}}\right)H_{5}$
(B.1.7)

###  B.2 $\Pi_{2}(z)$ Calculations

$\displaystyle\Pi_{2}\left(z\right)=\frac{1}{{16\pi^{2}}}\left\\{{-12+12\ln\left({\frac{{-z}}{{\mu^{2}}}}\right)+{\frac{\alpha_{s}\left({\left|z\right|}\right)}{\pi}}\left[{-1000+64\ln\left({\frac{{-z}}{{\mu^{2}}}}\right)-24\left[{\ln\left({\frac{{-z}}{{\mu^{2}}}}\right)}\right]^{2}+48\zeta\left(3\right)}\right]}\right\\}$
(B.2.1)

Now substituting (B.2.1) into the correlator (B.1) and focusing on this term
in the correlator we get

$\displaystyle\psi_{5}\left(z\right)=m_{q}^{4}\left({\left|z\right|}\right)\Pi_{2}\left(z\right)$
(B.2.2)

we can now calculate the $\delta_{5}$ by using (B.3). We will break the
calculation up into to manageable pieces so

$\displaystyle\delta_{5}$ $\displaystyle=\frac{1}{{2\pi
i}}\oint\limits_{C\left(R\right)}{\frac{1}{z}\left({1-a_{0}z-a_{1}z^{2}}\right)\frac{{m_{q}^{4}\left({\left|z\right|}\right)}}{{16\pi^{2}}}\left({-12+12\ln\left({\frac{{-z}}{{\mu^{2}}}}\right)}\right)dz}$
$\displaystyle=\frac{1}{{2\pi
i}}\oint\limits_{C\left(R\right)}{\frac{1}{z}\left({1-a_{0}z-a_{1}z^{2}}\right)\frac{{-12m_{q}^{4}\left({\left|z\right|}\right)}}{{16\pi^{2}}}dz}+\frac{1}{{2\pi
i}}\oint\limits_{C\left(R\right)}{\frac{1}{z}\left({1-a_{0}z-a_{1}z^{2}}\right)\frac{{m_{q}^{4}\left({\left|z\right|^{2}}\right)}}{{16\pi^{2}}}12\ln\left({\frac{{-z}}{{\mu^{2}}}}\right)dz}$
(B.2.3)
$\displaystyle=\frac{{-12m_{q}^{4}\left(R\right)}}{{16\pi^{2}}}\frac{1}{{2\pi
i}}\oint\limits_{C\left(R\right)}{\frac{1}{z}\left({1-a_{0}z-a_{1}z^{2}}\right)dz}+\underline{\frac{{12m_{q}^{4}\left(R\right)}}{{16\pi^{2}}}\frac{1}{{2\pi
i}}\oint\limits_{C\left(R\right)}{\frac{1}{z}\left({1-a_{0}z-a_{1}z^{2}}\right)\ln\left({\frac{{-z}}{{\mu^{2}}}}\right)dz}}$
(B.2.4)

in (B.2.3) we are integrating on a circle of radius $R$ so the only change is
in the angle on the loop. Since the mass only depends on the magnitude of the
complex variable, we can take the mass term out of the integral since $|z|=R$.
Making a change of variables with $z=\mu^{2}w$ $\Rightarrow$ $dz=\mu^{2}dw$ on
the underlined term in (B.2.4) we obtain

$\displaystyle\delta_{5}=$
$\displaystyle\frac{{-12m_{q}^{4}\left(R\right)}}{{16\pi^{2}}}\frac{1}{{2\pi
i}}\oint\limits_{C\left(R\right)}{\left({\frac{1}{z}-a_{0}-a_{1}z}\right)dz}+$
$\displaystyle+\frac{{12m_{q}^{4}\left(R\right)}}{{16\pi^{2}}}\frac{1}{{2\pi
i}}\oint\limits_{C\left(r\right)}{\left({\frac{1}{w}\ln\left({-w}\right)-a_{0}\mu^{2}\ln\left({-w}\right)-a_{1}\mu^{4}w\ln\left({-w}\right)}\right)dw}$
(B.2.5)

were $C(r)$ is a new contour of integration. Now using the integrals generated
by (F.1) we can integrate (B.2.5) to

$\displaystyle\delta_{5}$ $\displaystyle=$
$\displaystyle\frac{{-12m_{q}^{4}\left(R\right)}}{{16\pi^{2}}}+\frac{{12m_{q}^{4}\left(R\right)}}{{16\pi^{2}}}\left[{\ln
r-a_{0}\mu^{2}r-\frac{1}{2}a_{1}\mu^{4}r^{2}}\right]$ (B.2.6) $\displaystyle=$
$\displaystyle-\frac{{12m_{q}^{4}\left(R\right)}}{{16\pi^{2}}}\left[{1-\ln
r+a_{0}\mu^{2}r+\frac{1}{2}a_{1}\mu^{4}r^{2}}\right]$

since $|z|=\mu^{2}|w|$ $\Rightarrow$ $R=\mu^{2}r$ now substituting for $r$
into (B.2.6) we obtain

$\displaystyle\delta_{5}=-\frac{{12m_{q}^{4}\left(R\right)}}{{16\pi^{2}}}\left[{1-\ln\left({\frac{R}{{\mu^{2}}}}\right)+a_{0}\mu^{2}\left({\frac{R}{{\mu^{2}}}}\right)+\frac{1}{2}a_{1}\mu^{4}\left({\frac{R}{{\mu^{2}}}}\right)^{2}}\right]$
(B.2.7)

now taking the limit as $\mu^{2}\to R$ we obtain

$\displaystyle\delta_{5}=-\frac{{12m_{q}^{4}\left(R\right)}}{{16\pi^{2}}}\left[{1+a_{0}R+\frac{1}{2}a_{1}R^{2}}\right]$
(B.2.8)

We now turn to the second term of (B.2.1)

$\displaystyle\Pi_{2}\left(z\right)=\frac{1}{{16\pi^{2}}}\left\\{{\cdots+\frac{{\alpha_{s}\left({\left|z\right|}\right)}}{\pi}\left[{-1000+64\ln\left({\frac{{-z}}{{\mu^{2}}}}\right)-24\left[{\ln\left({\frac{{-z}}{{\mu^{2}}}}\right)}\right]^{2}+48\zeta\left(3\right)}\right]}\right\\}$
(B.2.9)

now substituting (B.2.9) into the correlator and then into (B.3) we obtain

$\displaystyle\delta_{5}$ $\displaystyle=\frac{1}{{2\pi
i}}\oint\limits_{C\left(R\right)}{\frac{1}{z}\left({1-a_{0}z-a_{1}z^{2}}\right)\frac{{m_{q}^{4}\left({\left|z\right|}\right)}}{{16\pi^{2}}}\frac{{\alpha_{s}\left({\left|z\right|}\right)}}{\pi}\left[{\left({48\zeta\left(3\right)-1000}\right)+64\ln\left({\frac{{-z}}{{\mu^{2}}}}\right)-24\left[{\ln\left({\frac{{-z}}{{\mu^{2}}}}\right)}\right]^{2}}\right]dz}$
$\displaystyle=\frac{{m_{q}^{4}\left(R\right)}}{{16\pi^{2}}}\frac{{\alpha_{s}\left(R\right)}}{\pi}\left[{48\zeta\left(3\right)-1000}\right]\frac{1}{{2\pi
i}}\oint\limits_{C\left(R\right)}{\frac{1}{z}\left({1-a_{0}z-a_{1}z^{2}}\right)dz}+$
$\displaystyle+\underline{64\frac{{m_{q}^{4}\left(R\right)}}{{16\pi^{2}}}\frac{{\alpha_{s}\left(R\right)}}{\pi}\frac{1}{{2\pi
i}}\oint\limits_{C\left(R\right)}{\frac{1}{z}\left({1-a_{0}z-a_{1}z^{2}}\right)\ln\left({\frac{{-z}}{{\mu^{2}}}}\right)dz}}$
$\displaystyle-\underline{24\frac{{m_{q}^{4}\left(R\right)}}{{16\pi^{2}}}\frac{{\alpha_{s}\left(R\right)}}{\pi}\frac{1}{{2\pi
i}}\oint\limits_{C\left(R\right)}{\frac{1}{z}\left({1-a_{0}z-a_{1}z^{2}}\right)\left[{\ln\left({\frac{{-z}}{{\mu^{2}}}}\right)}\right]^{2}dz}}$