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Correct typo with regex

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Signed-off-by: Riccardo Finotello <riccardo.finotello@gmail.com>
This commit is contained in:
Riccardo
2020-10-30 12:25:29 +01:00
parent 1c0d765856
commit 6f2efba172
2 changed files with 106 additions and 101 deletions
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207
sec/part2/divergences.tex
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@@ -287,14 +287,14 @@ We therefore have
\end{equation}
Using Fourier transforms it follows that the eigenmodes are
\begin{equation}
\phi_{k_-kr}\qty(u,\, v,\, z,\, \vec{x})
\phi_{\kmkr}\qty(u,\, v,\, z,\, \vec{x})
=
e^{i k_+ v + i l z + i \vec{k} \cdot \vec{x}}\,
\tphi_{k_-kr}(u),
\tphi_{\kmkr}(u),
\end{equation}
with
\begin{equation}
\tphi_{k_-kr}(u)
\tphi_{\kmkr}(u)
=
\frac{1}{\sqrt{\qty( 2 \pi )^D~ \abs{2 \Delta k_+\, u}}}
e^{
@@ -304,7 +304,7 @@ with
\end{equation}
and
\begin{equation}
\phi^*_{k_-kr}\qty(u,\, v,\, z,\, \vec{x})
\phi^*_{\kmkr}\qty(u,\, v,\, z,\, \vec{x})
=
\phi_{\mkmkr}\qty(u,\, v,\, z,\, \vec{x}).
\end{equation}
@@ -312,7 +312,7 @@ We chose the numeric factor in order to get a canonical normalisation:
\begin{equation}
\begin{split}
&
\qty( \phi_{k_-krN{1}},\, \phi_{k_-krN{2}} )
\qty( \phi_{\kmkrN{1}},\, \phi_{\kmkrN{2}} )
\\
= &
\int\limits_{\R^{D-3}} \dd[D-3]{\vec{x}}\,
@@ -320,7 +320,7 @@ We chose the numeric factor in order to get a canonical normalisation:
\infinfint{v}\,
\finiteint{z}{0}{2\pi}
\abs{\Delta u}\,
\phi_{k_-krN{1}}\, \phi_{k_-krN{2}}
\phi_{\kmkrN{1}}\, \phi_{\kmkrN{2}}
\\
= &
\delta^{D-3}( \vec{k}_{\qty(1)} + \vec{k}_{\qty(2)})\,
@@ -337,8 +337,8 @@ We can then perform the off-shell expansion
\infinfint{k_+}
\infinfint{r}
\infinfsum{l}
\cA_{k_-kr}\,
\phi_{k_-kr}\qty(u,\, v,\, z,\, \vec{x}),
\cA_{\kmkr}\,
\phi_{\kmkr}\qty(u,\, v,\, z,\, \vec{x}),
\end{equation}
such that the scalar kinetic term becomes
\begin{equation}
@@ -349,8 +349,8 @@ such that the scalar kinetic term becomes
\infinfint{r}
\infinfsum{l}
\qty(r - M^2)\,
\cA_{k_-kr}\,
\cA_{k_-kr}^*.
\cA_{\kmkr}\,
\cA_{\kmkr}^*.
\end{equation}
@@ -461,7 +461,7 @@ We proceed hierarchically: first we solve for $a_v$ and $a_i$ whose equations ar
We get the solutions:
\begin{equation}
\begin{split}
\norm{\tildea_{k_-kr\, \alpha}(u)}
\norm{\tildea_{\kmkr\, \alpha}(u)}
\,=
\mqty(%
\tildea_u
@@ -479,7 +479,7 @@ We get the solutions:
\qty{ \underu, \underv, \underz,\underi }
}
\pol{\alpha}
\norm{\tildea^{\underline{\alpha}}_{k_-kr\, \alpha}(u)}
\norm{\tildea^{\underline{\alpha}}_{\kmkr\, \alpha}(u)}
\\
& =
\pol{u}
@@ -492,7 +492,7 @@ We get the solutions:
\\
0
)\,
\tphi_{k_-kr}(u)
\tphi_{\kmkr}(u)
\\
& +
\pol{v}
@@ -507,7 +507,7 @@ We get the solutions:
\\
0
)\,
\tphi_{k_-kr}(u)
\tphi_{\kmkr}(u)
\\
& +
\pol{z}
@@ -520,7 +520,7 @@ We get the solutions:
\\
0
)\,
\tphi_{k_-kr}(u)
\tphi_{\kmkr}(u)
\\
& +
\pol{j}
@@ -533,7 +533,7 @@ We get the solutions:
\\
\delta_{\underline{ij}}
)\,
\tphi_{k_-kr}(u),
\tphi_{\kmkr}(u),
\label{eq:Orbifold_spin1_pol}
\end{split}
\end{equation}
@@ -549,13 +549,13 @@ then we can expand the off-shell fields as
}
\infinfsum{l}
\pol{\alpha}\,
{a}^{\underline{\alpha}}_{k_-kr\, \alpha}\qty(u,\, v,\, z,\, \vec{x} ),
{a}^{\underline{\alpha}}_{\kmkr\, \alpha}\qty(u,\, v,\, z,\, \vec{x} ),
\end{equation}
where
\begin{equation}
a^{\underline{\alpha}}_{k_-kr\, \alpha}\qty(u,\, v,\, z,\, \vec{x})
a^{\underline{\alpha}}_{\kmkr\, \alpha}\qty(u,\, v,\, z,\, \vec{x})
=
\tildea^{\underline{\alpha}}_{k_-kr\, \alpha}(u)\,
\tildea^{\underline{\alpha}}_{\kmkr\, \alpha}(u)\,
e^{i\, \qty( k_+ v + l z + \vec{k} \cdot \vec{x})}
\end{equation}
and $\int \ccD k = \int\limits_{\R^{D-3}} \dd[D-3]{\vec{k}} \infinfint{k_+} \infinfint{r}$.
@@ -573,7 +573,7 @@ We can also compute the normalisation as
\\
& \times
g^{\alpha\beta}\,
a_{k_-krN{1}\, \alpha}\, a_{k_-krN{2}\, \beta}
a_{\kmkrN{1}\, \alpha}\, a_{\kmkrN{2}\, \beta}
\\
& =
\genpolN{1} \circ \genpolN{2}
@@ -624,9 +624,9 @@ The photon kinetic term becomes
\infinfint{r}
\infinfsum{l}\,
\frac{r}{2}\,
\cE_{k_-kr}\,
\cE_{\kmkr}\,
\circ
\cE_{k_-kr}^*.
\cE_{\kmkr}^*.
\end{equation}
@@ -654,7 +654,7 @@ Its computation involves integrals such as
\int \dd{u}\,
\abs{\Delta u}\,
\qty(\frac{l}{u})^2
\finiteprod{i}{1}{3} \tphi_{k_-krN{i}}
\finiteprod{i}{1}{3} \tphi_{\kmkrN{i}}
\sim
\int\limits_{u \sim 0} \dd{u}\,
\qty(\frac{l^2}{\abs{u}^{\frac{5}{2}}})
@@ -668,7 +668,7 @@ and
\int \dd{u}\,
\abs{\Delta u}\,
\qty(\frac{1}{u})
\finiteprod{i}{1}{3} \tphi_{k_-krN{i}}
\finiteprod{i}{1}{3} \tphi_{\kmkrN{i}}
\sim
\int\limits_{u \sim 0} \dd{u}\,
\qty(\frac{1}{u\, \abs{u}^{\frac{1}{2}}})
@@ -720,7 +720,7 @@ and we get:\footnotemark{}
\delta_{\finitesum{i}{1}{3} l_{\qty(i)},\, 0}\,
\\
& \times
\qty(\cA_{\mkmkrN{2}})^*\, \cA_{k_-krN{3}}
\qty(\cA_{\mkmkrN{2}})^*\, \cA_{\kmkrN{3}}
\\
& \times
\Biggl\lbrace
@@ -782,19 +782,19 @@ In the previous expressions we also defined for future use:
\infinfint{u}\,
\abs{\Delta u}\, u^{\nu}\,
\finiteprod{i}{1}{N}
\tphi_{k_-krN{i}}
\tphi_{\kmkrN{i}}
\\
\cJ_{\qty{N}}^{\qty[\nu]}
& = &
\infinfint{u}\,
\abs{\Delta}\, \abs{u}^{1 + \nu}
\finiteprod{i}{1}{N} \tphi_{k_-krN{i}}.
\finiteprod{i}{1}{N} \tphi_{\kmkrN{i}}.
\end{eqnarray}
For the sake of brevity from now on we use
\begin{eqnarray}
\tphi_{\qty(i)} & = & \tphi_{k_-krN{i}},
\tphi_{\qty(i)} & = & \tphi_{\kmkrN{i}},
\\
\tphi_{\qty(i)} & = & \tphi_{k_-krN{i}}
\tphi_{\qty(i)} & = & \tphi_{\kmkrN{i}}
\end{eqnarray}
when not causing confusion.
@@ -835,7 +835,7 @@ which can be expressed using the modes as:
& \times
\Biggl\lbrace
e^2\,
\qty(\cA_{\mkmkrN{3}})^* \cA_{k_-krN{4}}
\qty(\cA_{\mkmkrN{3}})^* \cA_{\kmkrN{4}}
\\
& \times
\Biggl[
@@ -879,8 +879,8 @@ where
\qty(\cA_{\mkmkrN{2}})^*\,
\\
& \times
\cA_{k_-krN{3}}\,
\cA_{k_-krN{4}}.
\cA_{\kmkrN{3}}\,
\cA_{\kmkrN{4}}.
\end{split}
\end{equation}
When setting $l_{\qty(*)} = 0$ all the surviving terms are divergent.
@@ -896,12 +896,12 @@ From the discussion in the previous section the origin of the divergences is the
When $l = 0$ the highest order singularity of the Fourier transformed d'Alembertian equation vanishes.
Explicitly we have:
\begin{equation}
A\, \ipd{u} \tphi_{k_-kr}
A\, \ipd{u} \tphi_{\kmkr}
+
B(u)\, \tphi_{k_-kr}
B(u)\, \tphi_{\kmkr}
=
A\, e^{-\int^u \frac{B(u)}{A} du}\,
\ipd{u} \qty[ e^{+\int^u \frac{B(u)}{A} du} \tphi_{k_-kr} ]
\ipd{u} \qty[ e^{+\int^u \frac{B(u)}{A} du} \tphi_{\kmkr} ]
=
0,
\end{equation}
@@ -1269,7 +1269,7 @@ to finally get:\footnotemark{}
& =
\cN\,
\infinfsum{l}
\phi_{k_-kr}\qty(u,\, v,\, z,\, \vec{x})
\phi_{\kmkr}\qty(u,\, v,\, z,\, \vec{x})
e^{i\, l\, \frac{k_2}{\Delta k_+}},
\end{split}
\label{eq:Psi_phi}
@@ -1284,7 +1284,7 @@ when $k_+ \neq 0$ and where
The fact that $\Psi$ depends only on the equivalence class $\qty[k_+\, k_-\, k_2\, k]$ allows us to restrict $0 \le \frac{k_2}{\Delta\, \abs{k_+}} < 2 \pi$ so that we can invert the previous expression and get:
\begin{equation}
\begin{split}
\phi_{k_-kr}\qty(u,\, v,\, z,\, \vec{x})
\phi_{\kmkr}\qty(u,\, v,\, z,\, \vec{x})
& =
\frac{1}{\cN}\,
\frac{1}{2 \pi \Delta \abs{k_+}}
@@ -1336,7 +1336,7 @@ The explicit expression for the eigenfunction with constant $\epsilon_+$, $\epsi
}
\\
& =
\cN
\cN\,
\psi^{[1]}_{k_+\, k_-\, k_2;\, \epsilon_+\, \epsilon_-\, \epsilon_2}\qty(u,\, v,\, z),
\end{split}
\end{equation}
@@ -1370,7 +1370,7 @@ Building the corresponding function on the orbifold amounts to summing the image
\psi_{k}\qty( \cK^n x)
=
\infinfsum{n}
\cK^{-n}
\cK^{-n}\,
\vec{\epsilon} \cdot \dd{x}~
\psi_{\cK^{-n} k}\qty(x).
\end{split}
@@ -1508,7 +1508,7 @@ Then it follows that
\label{eq:a_uvz_from_covering}
\end{equation}
Many coefficients of $\Psi$ or its derivatives contain $k_2$.
They cannot be expressed using the quantum numbers $k_-kr$ of the orbifold but are invariant on it.
They cannot be expressed using the quantum numbers $\kmkr$ of the orbifold but are invariant on it.
They are new orbifold quantities we interpret as orbifold polarisations.
Using~\eqref{eq:Psi_phi} we can finally write
\begin{equation}
@@ -1516,7 +1516,7 @@ Using~\eqref{eq:Psi_phi} we can finally write
\Psi^{[1]}_{[k,\, \epsilon]}\qty(\qty[x])
& =
\infinfsum{l}
\phi_{k_-kr}\qty(u,\, v,\, z,\, \vec{x})
\phi_{\kmkr}\qty(u,\, v,\, z,\, \vec{x})
e^{i\, l \frac{k_2}{\Delta k_+}}
\\
& \times
@@ -1931,7 +1931,7 @@ The final expression for the orbifold symmetric tensor is
\Psi^{[2]}_{\qty[\vec{k},\, S]}\qty(\qty[x])
& =
\infinfsum{l}
\phi_{k_-kr}\qty(u,\, v,\, z,\, \vec{x})
\phi_{\kmkr}\qty(u,\, v,\, z,\, \vec{x})
e^{i\, l \frac{k_2}{\Delta k_+}}
\\
& \times
@@ -2870,26 +2870,29 @@ We therefore need solve:
\end{equation}
To this purpose, we introduce a Fourier transformation over $v,\, w,\, z,\, \vec{x}$:
\begin{equation}
\phi_r\qty( u,\, v,\, w,\, z,\, \vec{x})
=
\infinfsum{l}\,
\int\limits_{\R^{D-4}} \dd[D-4]{\vec{k}}
\infinfint{k_+}
\infinfint{p}\,
e^{i\, \qty( k_+ v + p w + l z + \vec{k} \cdot \vec{x} )}
\tphi_{k_-krgen}(u),
\begin{split}
& \phi_r\qty( u,\, v,\, w,\, z,\, \vec{x})
\\
& =
\infinfsum{l}\,
\int\limits_{\R^{D-4}} \dd[D-4]{\vec{k}}
\infinfint{k_+}
\infinfint{p}\,
e^{i\, \qty( k_+ v + p w + l z + \vec{k} \cdot \vec{x} )}
\tphi_{\kmkrgen}(u),
\end{split}
\end{equation}
where we defined $k_+,\, p,\, l,\, \vec{k}$ as associated momenta to $v,\, w,\, z,\, \vec{x}$ respectively.
We find:
\begin{equation}
\phi_{k_-krgen}\qty( u,\, v,\, w,\, z,\, \vec{x} )
\phi_{\kmkrgen}\qty( u,\, v,\, w,\, z,\, \vec{x} )
=
e^{i\, \qty( k_+ v + p w + l z + \vec{k} \cdot \vec{x} )}
\tphi_{k_-krgen}( u ).
\tphi_{\kmkrgen}( u ).
\end{equation}
where
\begin{equation}
\tphi_{k_-krgen}( u )
\tphi_{\kmkrgen}( u )
=
\frac{1}{2 \sqrt{\qty(2 \pi)^D \abs{\Delta_2 \Delta_3 k_+}}}\,
\frac{1}{\abs{u}}
@@ -2906,7 +2909,7 @@ where
These solutions present the right normalisation, as we can verify through the product:
\begin{equation}
\begin{split}
& \left( \phi_{k_-krgenN{1}},\, \phi_{k_-krgenN{2}} \right)
& \left( \phi_{\kmkrgenN{1}},\, \phi_{\kmkrgenN{2}} \right)
\\
& =
\int\limits_{\R^{D-4}} \dd[D-4]{\vec{x}}
@@ -2917,8 +2920,8 @@ These solutions present the right normalisation, as we can verify through the pr
2 \abs{\Delta_2 \Delta_3} u^2
\\
& \times
\phi_{k_-krgenN{1}}~
\phi_{k_-krgenN{2}}
\phi_{\kmkrgenN{1}}~
\phi_{\kmkrgenN{2}}
\\
& =
\delta^{D - 4}\qty( \vec{k}_{\qty(1)} + \vec{k}_{\qty(2)} )\,
@@ -2942,7 +2945,7 @@ Then we have the off-shell expansion:
\infinfint{r}
\\
& \times
\frac{\cA_{k_-krgen}}{\abs{u}}
\frac{\cA_{\kmkrgen}}{\abs{u}}
e^{%
i\, \qty(%
k_+ v + p w + l z + \vec{k} \cdot \vec{x}
@@ -3115,15 +3118,15 @@ These equations can be solved using standard techniques through a Fourier transf
\\
& \times
e^{i\, \qty( k_+ v + p w + l z + \vec{k} \cdot \vec{x} )}
\tildea_{k_-krgen\, \alpha}(u).
\tildea_{\kmkrgen\, \alpha}(u).
\end{split}
\end{equation}
We first solve the equations for $\tildea_{k_-krgen\, v}$ and $\tildea_{k_-krgen\, i}$ since they are identical to the scalar equation~\eqref{eq:scalar_eom}.
We then insert their solutions as sources for the equations for $\tildea_{k_-krgen\, u}$, $\tildea_{k_-krgen\, w}$ and $\tildea_{k_-krgen\, z}$.
We first solve the equations for $\tildea_{\kmkrgen\, v}$ and $\tildea_{\kmkrgen\, i}$ since they are identical to the scalar equation~\eqref{eq:scalar_eom}.
We then insert their solutions as sources for the equations for $\tildea_{\kmkrgen\, u}$, $\tildea_{\kmkrgen\, w}$ and $\tildea_{\kmkrgen\, z}$.
The solutions can be written as the expansion:
\begin{equation}
\begin{split}
\norm{\tildea_{k_-krgen\, \alpha}(u)}
\norm{\tildea_{\kmkrgen\, \alpha}(u)}
& =
\mqty(%
\tildea_u
@@ -3139,16 +3142,16 @@ The solutions can be written as the expansion:
\\
& =
\sum\limits_{\ualpha \in \qty{\underu, \underv, \underw, \underz, \underi}}
\cE_{k_-krgen\, \ualpha}\,
\norm{\tildea^{\ualpha}_{k_-krgen\, \alpha}(u)}
\cE_{\kmkrgen\, \ualpha}\,
\norm{\tildea^{\ualpha}_{\kmkrgen\, \alpha}(u)}
\\
& =
\cE_{k_-krgen\, \underu}\,
\cE_{\kmkrgen\, \underu}\,
\mqty( 1 \\ 0 \\ 0 \\ 0 \\ 0 )\,
\tphi_{k_-krgen}
\tphi_{\kmkrgen}
\\
& +
\cE_{k_-krgen\, \underv}\,
\cE_{\kmkrgen\, \underv}\,
\mqty(%
\frac{i}{2 k_+ u}
+
@@ -3163,10 +3166,10 @@ The solutions can be written as the expansion:
\\
0
)\,
\tphi_{k_-krgen}
\tphi_{\kmkrgen}
\\
& +
\cE_{k_-krgen\, \underw}\,
\cE_{\kmkrgen\, \underw}\,
\mqty(
\frac{1}{4 k_+ \abs{u}}
\qty( \frac{l + p}{\Delta_2^2} - \frac{l - p}{\Delta_3^2} )
@@ -3179,10 +3182,10 @@ The solutions can be written as the expansion:
\\
0
)\,
\tphi_{k_-krgen}
\tphi_{\kmkrgen}
\\
& +
\cE_{k_-krgen\, \underz}\,
\cE_{\kmkrgen\, \underz}\,
\mqty(
\frac{1}{4 k_+ \abs{u}}
\qty( \frac{l + p}{\Delta_2^2} + \frac{l - p}{\Delta_3^2} )
@@ -3195,20 +3198,20 @@ The solutions can be written as the expansion:
\\
0
)\,
\tphi_{k_-krgen}
\tphi_{\kmkrgen}
\\
& +
\cE_{k_-krgen\, \underj}\,
\cE_{\kmkrgen\, \underj}\,
\mqty( 0 \\ 0 \\ 0 \\ 0 \\ \delta_{\underline{i j}} )\,
\tphi_{k_-krgen}
\tphi_{\kmkrgen}
\end{split}
\end{equation}
Consider the Fourier transformed functions:
\begin{equation}
a^{\ualpha}_{k_-krgen\, \alpha}\qty( u,\, v,\, w,\, z,\, \vec{x} )
a^{\ualpha}_{\kmkrgen\, \alpha}\qty( u,\, v,\, w,\, z,\, \vec{x} )
=
e^{i\, \qty(k_+ v + p w + l z + \vec{k} \cdot \vec{x})}
\tildea^{\ualpha}_{k_-krgen\, \alpha}( u ),
\tildea^{\ualpha}_{\kmkrgen\, \alpha}( u ),
\end{equation}
then we can expand the off shell fields as
\begin{equation}
@@ -3223,8 +3226,8 @@ then we can expand the off shell fields as
\\
& \times
\sum_{\ualpha \in \qty{\underu, \underv, \underw, \underz, \underi}}
\cE_{k_-krgen\, \alpha}\,
a^{\ualpha}_{k_-krgen\, \alpha}(x).
\cE_{\kmkrgen\, \alpha}\,
a^{\ualpha}_{\kmkrgen\, \alpha}(x).
\end{split}
\end{equation}
@@ -3241,7 +3244,7 @@ We can compute the normalisation as:
2 \abs{\Delta_2 \Delta_3} u^2
\\
& \times
\qty(g^{\alpha\beta}\, a_{k_-krgenN{1}\, \alpha}\, a_{k_-krgenN{2}\, \beta})
\qty(g^{\alpha\beta}\, a_{\kmkrgenN{1}\, \alpha}\, a_{\kmkrgenN{2}\, \beta})
\\
& =
\delta^{D-4}\qty( \vec{k}_{\qty(1)} + \vec{k}_{\qty(2)} )\,
@@ -3251,7 +3254,7 @@ We can compute the normalisation as:
\delta\qty( r_1 - r_2 )
\\
& \times
\cE_{k_-krgenN{1}} \circ \cE_{k_-krgenN{2}},
\cE_{\kmkrgenN{1}} \circ \cE_{\kmkrgenN{2}},
\end{split}
\end{equation}
where
@@ -3285,16 +3288,16 @@ where
is independent of the coordinates.
The Lorenz gauge now reads:
\begin{equation}
\eta^{i\underj}\, k_i \, \cE_{{k_-krgen} \underj}
\eta^{i\underj}\, k_i \, \cE_{{\kmkrgen} \underj}
-
k_+
\cE_{k_-krgen\, \underu}
\cE_{\kmkrgen\, \underu}
-
\frac{\vec{k}^2 + r}{2 k_+}
\cE_{k_-krgen\, \underv}
\cE_{\kmkrgen\, \underv}
= 0.
\end{equation}
As in the previous case, the constraint equation does not pose any condition on the transverse polarisations $\cE_{k_-krgen\, \underw}$ and $\cE_{k_-krgen\, \underz}$.
As in the previous case, the constraint equation does not pose any condition on the transverse polarisations $\cE_{\kmkrgen\, \underw}$ and $\cE_{\kmkrgen\, \underz}$.
\subsubsection{Cubic Interaction}
@@ -3331,16 +3334,16 @@ In the case of the \gnbo we find:
& \times
e~
\cA^*_{\mkmkrgenN{2}}
\cA_{k_-krgenN{3}}
\cA_{\kmkrgenN{3}}
\\
& \times
\Biggl\lbrace
\cE_{k_-krgenN{1}\, \underu}~
\cE_{\kmkrgenN{1}\, \underu}~
k_{\qty(2)\, +}~
\cI_{\qty{3}}^{\qty[0]}
\\
& +
\cE_{k_-krgenN{1}\, \underv}~
\cE_{\kmkrgenN{1}\, \underv}~
\Biggl[
\qty( \frac{\vec{k}_{\qty(2)}^2 + r_{(2)}}{2 k_{\qty(2)\, +}} )\,
\cI_{\qty{3}}^{\qty[0]}
@@ -3370,7 +3373,7 @@ In the case of the \gnbo we find:
\Biggr]
\\
& +
\qty( \cE_{k_-krgenN{1}\, \underw} - \cE_{k_-krgenN{1}\, \underz} )
\qty( \cE_{\kmkrgenN{1}\, \underw} - \cE_{\kmkrgenN{1}\, \underz} )
\\
& \times
\Biggl[
@@ -3402,18 +3405,18 @@ where we defined:
\cI_{\qty{N}}^{\qty[\nu]}
& =
\infinfint{u}
2\, \abs{\Delta_2 \Delta_3} u^2\, u^{\nu}\, \finiteprod{i}{1}{N} \tphi_{k_-krgenN{i}},
2\, \abs{\Delta_2 \Delta_3} u^2\, u^{\nu}\, \finiteprod{i}{1}{N} \tphi_{\kmkrgenN{i}},
\\
\cJ_{\qty{N}}^{\qty[\nu]}
& =
\infinfint{u}
2\, \abs{\Delta_2 \Delta_3} u^2\, \abs{u}^{\nu}\, \finiteprod{i}{1}{N} \tphi_{k_-krgenN{i}}.
2\, \abs{\Delta_2 \Delta_3} u^2\, \abs{u}^{\nu}\, \finiteprod{i}{1}{N} \tphi_{\kmkrgenN{i}}.
\end{split}
\end{equation}
While in the \nbo case we need to regularise the integrals at least taking their principal part when all $l_{(*)} = 0$ in~\eqref{eq:nbo_div_integral}, the \gnbo does not need any specific manipulation.
In fact the form of $\tphi_{k_-krgenN{i}}$ in~\eqref{eq:GNBO_reg_wave_functions} prevents the formation of isolated zeros in the phase factor proportional to $u^{-1}$: the presence of the continuous momentum $p$, contrary to the \nbo where all momenta are discrete, gives the integrals a distributional interpretation, similar to a derivative of a Dirac $\delta$ function.
In fact the form of $\tphi_{\kmkrgenN{i}}$ in~\eqref{eq:GNBO_reg_wave_functions} prevents the formation of isolated zeros in the phase factor proportional to $u^{-1}$: the presence of the continuous momentum $p$, contrary to the \nbo where all momenta are discrete, gives the integrals a distributional interpretation, similar to a derivative of a Dirac $\delta$ function.
\subsubsection{Quartic Interactions}
@@ -3454,16 +3457,16 @@ As for the \nbo, we consider the quartic interaction for the scalar \qed action:
\Biggl\lbrace
e^2
\cA^*_{\mkmkrgenN{3}}
\cA_{k_-krgenN{4}}
\cA_{\kmkrgenN{4}}
\\
& \times
\Biggl[
\cE_{k_-krgenN{1}} \circ \cE_{k_-krgenN{2}}\,
\cE_{\kmkrgenN{1}} \circ \cE_{\kmkrgenN{2}}\,
\cI_{\qty{4}}^{\qty[0]}
\\
& - i
\cE_{k_-krgenN{1}\, \underv}\,
\cE_{k_-krgenN{2}\, \underv}
\cE_{\kmkrgenN{1}\, \underv}\,
\cE_{\kmkrgenN{2}\, \underv}
\\
& \times
\Biggl(
@@ -3497,8 +3500,8 @@ As for the \nbo, we consider the quartic interaction for the scalar \qed action:
\cA^*_{\mkmkrgenN{2}}
\\
& \times
\cA_{k_-krgenN{3}}
\cA_{k_-krgenN{4}}
\cA_{\kmkrgenN{3}}
\cA_{\kmkrgenN{4}}
\cI_{\qty{4}}^{\qty[0]}
\Biggr\rbrace,
\end{split}
@@ -3514,16 +3517,16 @@ where we defined:
\\
\tilde{\cE}_{\pm\,\left( a, b \right)}
& =
\cE_{k_-krgenN{a}\, \underv}
\cE_{\kmkrgenN{a}\, \underv}
\\
& \times
\qty( \cE_{k_-krgenN{b}\, \underw} \pm \cE_{k_-krgenN{b}\, \underz} )
\qty( \cE_{\kmkrgenN{b}\, \underw} \pm \cE_{\kmkrgenN{b}\, \underz} )
\\
& -
\cE_{k_-krgenN{b}\, \underv}
\cE_{\kmkrgenN{b}\, \underv}
\\
& \times
\qty( \cE_{k_-krgenN{a}\, \underw} \pm \cE_{k_-krgenN{a}\, \underz} )
\qty( \cE_{\kmkrgenN{a}\, \underw} \pm \cE_{\kmkrgenN{a}\, \underz} )
\end{split}
\end{equation}
for simplicity.
@@ -3931,6 +3934,8 @@ Now we use the basic trick used in Poisson resummation
\abs{\frac{k_+ x^+}{k_- x^-}}^{-i \frac{l}{2 \Delta} }\,
\infinfint{s}
e^{i\, 2 \pi\, l\, s}
\\
& \times
e^{i\, \sgn(k_+\, x^+) \sqrt{\abs{k_+ k_- x^+ x^-}}
\qty{
\Lambda^s
@@ -3973,7 +3978,7 @@ Now we use the basic trick used in Poisson resummation
\end{equation}
where the last line represents the change of quantum number from $m\, \beta$ to $m\, l$ and allows us to identify
\begin{equation}
\cN_{\text{BO}}
\cN_{\text{BO}}\,
\tphi_{\lsi}(\tau)
=
\frac{1}{2\pi}\,

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