Van Hove \cite{vanHove:1967zz} proposed a physically intuitive picture of a Reggeon by relating it to Feynman diagrams in the cross-channels. We will explore this picture of Reggeization with a simple model.
\subsection{Elementary $t$-channel exchanges}
Consider the amplitude corresponding to a particle with spin-$J$ and mass $m_J$ exchanged in the $t$-channel as:
%%
\begin{equation}
\label{eq:AJ}
A^J(s,t) = i \, g_J \, \left( q_1^{\mu_1} \dots q_1^{\muJ} \right) \frac{P^J{\mu_1\dots\mu_J,\nu_1\dots\nu_J}(k)}{mJ^2 - t} \left( q{\bar{2}}^{\nu1} \dots q{\bar{2}}^{\nu_J} \right)
\end{equation}
%%
where $gJ$ is a coupling constant with dimension $2-2J$ (i.e., $A^J(s,t)$ is dimensionless) and the projector of spin-$J$ is defined from the polarization tensor of rank-$J\geq1$ as
%%
\begin{equation}
P^J{\mu_1\dots\mu_J,\nu_1\dots\nuJ}(k) = \frac{(J+1)}{2} \, \sum{\lambda}
\epsilon^{\mu_1\dots\mu_J}(k,\lambda) \, \epsilon^{*\nu_1\dots\nu_J}(k,\lambda) ~.
\end{equation}
%%
Using the exchange momentum $k = q1 + q{\bar{3}} = q_1 - q_3$, calculate the amplitudes corresponding to $J=0,1,2$ exchanges in terms of $t= k^2$, the modulus of 3-momentum and cosine of scattering angle in the $t$-channel frame, $q_t$ and $\cos\thetat$ respectively. Use the explicit forms of the projectors:
%%
\begin{gather}
P^0(k^2) = 1 \
P^1{\mu\nu}(k^2) \equiv \tilde{g}{\mu\nu} = \frac{ k\mu \, k\nu}{ k^2} - g{\mu\nu}\
P^2{\mu\nu\alpha\beta}(k^2) = \frac{3}{4}\left(\tilde{g}{\mu\alpha} \, \tilde{g}{\nu\beta} + \tilde{g}{\mu\beta} \, \tilde{g}{\nu\alpha}\right) - \frac{1}{2} \, \tilde{g}{\mu\nu} \, \tilde{g}_{\alpha\beta} ~,
\end{gather}
%%
and conjecture a generalization of the amplitude for arbitrary integer $J$.
\noindent\textit{Hint: Show that in the $t$-channel frame, the exchange particle is at rest and therefore $\tilde{g}{\mu\nu}$ reduces to a $\delta{ij}$ with respect to only spacial momenta.}
\subsection{Unitarity vs Elementary exchanges}
Express the amplitude entirely in terms of invariants $s$ and $t$. Use the optical theorem to relate the elastic amplitude to a total hadronic cross section:
\begin{equation}
\sigma\text{tot} = \frac{1}{2q\sqrt{s}} \, \Im A^J(s,t=0)~.
\end{equation}
Unitarity (via the Froissart-Martin bound) prohibits $\sigma\text{tot}$ from growing faster than $\log^2 s$ as $s\to\infty$~. What is then the maximal spin a single elementary exchange can have while satisfying this bound? Why is this a problem?
\subsection{Van Hove Reggeon}
Consider an amplitude of the form
%%
\begin{equation}
\label{eq:vanhove}
A(s,t) = \sum_{J=0}^\infty g \, r^{2J} \, \frac{ (q_t^2 \, \cos\theta_t)^J}{J- \alpha(t)} ~.
\end{equation}
%%
Here $\alpha(t) = \alpha(0) + \alpha^\prime \, t$ is a real, linear Regge trajectory, $g$ is a dimensionless coupling constant and $r \sim 1$ fm is a range parameter.
Compare \cref{eq:vanhove} with \cref{eq:AJ}, write the mass of the $J$th pole, $m_J^2$, as a function of the Regge parameters $\alpha(0)$ and $\alpha^\prime$. Interpret the pole structure in terms of the spectrum of particles in the model.
\
\noindent If the sum is truncated to a finite $J_\text{max}$, and we take the $s\to \infty$ limit, what is the high energy behavior of the amplitude?
\subsection{Analytic continuation in $J$}
Show that if the summation is kept infinite, the amplitude can be re-summed to something that is entirely analytic in $s$, $t$, $u$, and $J$.
\noindent \textit{Hint: Use the Mellin transform
%%
\begin{equation}
\frac{1}{J-\alpha(t)} = \int_0^1 dx \, x^{J-\alpha(t) - 1} ~,
\end{equation}
%%
to express the amplitude in terms of the Gaussian hypergeometric function and the Euler Beta function
%%
\begin{equation}
B(b, c-b) \, _2F_1(1, b, c; z) = \int_0^1 dx \, \frac{ x^{b-1} \, (1-x)^{c-b-1}}{1-x \, z} ~.
\end{equation}
%%
}
\subsection{Unitarity vs Reggeized exchanges}
Revisit $\mathbf{b)}$ with the resummed amplitude. Take the $s\to\infty$ limit and set a limit on the maximal intercept $\alpha(0)$ which is allowed by unitarity.
\noindent \textit{Hint: Assume that $\alpha(0) > -1$ and use the asymptotic behavior of the hypergeometric function given by
%%
\begin{equation}
_2F_1(1,b,c;z) \to \frac{\Gamma(c)\,\Gamma(1-b)}{\Gamma(1) \,\Gamma(c-b)} \, (-z)^{-b} ~.
\end{equation}
%%
}
\subsection{The Reggeon ``propagator"}
Modify \cref{eq:AJ} to have a definite signature by defining
%%
\begin{equation}
A^{\pm}(s,t) = \frac{1}{2}\left[A(s,t) \pm A(u,t) \right]~.
\end{equation}
%%
Repeat $\textbf{d)}$ and $\textbf{e)}$ with this signatured amplitude. Compare with the canonical form of the Reggeon exchange:
%%
\begin{equation}
A^\pm_\mathbb{R}(s,t) = \beta(t) \, \frac{1}{2}\left[\pm1 + e^{-i\pi\alpha(t)}\right] \, \Gamma(-\alpha(t)) \, \left(\frac{s}{s_0}\right)^{\alpha(t)} ~.
\end{equation}
%%
Identify the Regge residue $\beta(t)$ and characteristic scale $s_0$ in terms of the parameters $g_0$ and $r$.
\section{Veneziano Amplitude}
The quintessential dual amplitude was first proposed by Veneziano for $\omega\to3\pi$ \cite{Veneziano:1968yb} and later applied to elastic $\pi\pi$ scattering by Shapiro \cite{Shapiro:1969km} and Lovelace \cite{Lovelace:1968kjy}. Consider the $\pi^+\pi^-$ scattering amplitude of the form
%%
\begin{equation}
\label{eq:Astu}
\mathcal{A}(s,t,u) = V(s,t) + V(s,u) - V(t,u) ~.
\end{equation}
%%
with each
%%
\begin{equation}
\label{eq:Fst}
V(s,t) = \frac{\Gamma(1-\alpha(s)) \, \Gamma(1-\alpha(t))}{\Gamma(1-\alpha(s) - \alpha(t))} ~,
\end{equation}
%%
where $\alpha(s) = \alpha(0) + \alpha^\prime s$ is a real, linear Regge trajectory with $\alpha^\prime > 0$.
\subsection{Duality}
Show that the function $V(s,t)$ is symmetric in $s\leftrightarrow t$ and dual, i.e., it can be written entirely as a sum of either $s$-channel poles OR $t$-channel poles but never both simultaneously. Compare with the Reggeized amplitude in the previous problem, was that amplitude dual?
\noindent \textit{Hint: Relate $V(s,t)$ to the Euler Beta function
\begin{equation}
B(x,y) = \frac{\Gamma(x) \, \Gamma(y)}{\Gamma(x+y)}
\end{equation}
and use the identities $B(x,y) = B(y,x)$ and
%%
\begin{equation}
B(p-x, q-y) = \sum_{J=1}^\infty \, \frac{\Gamma(J-p+1 +x)}{\Gamma(J) \, \Gamma(-p + 1 +x)} \, \frac{1}{J-1+q-y} ~.
\end{equation}
%%
}
\subsection{Isospin basis}
Define the $s$-channel isospin basis through
%%
\begin{equation}
\begin{pmatrix}
\mathcal{A}^{(0)}(s,t,u) \
\mathcal{A}^{(1)}(s,t,u) \
\mathcal{A}^{(2)}(s,t,u)
\end{pmatrix}
=
\frac{1}{2}\begin{pmatrix}
3 & 1 & 1 \
0 & 1 & -1 \
0 & 1 & 1
\end{pmatrix}
\begin{pmatrix}
\mathcal{A}(s,t,u) \
\mathcal{A}(t,s,u) \
\mathcal{A}(u,t,s)
\end{pmatrix}
~.
\end{equation}
%%
Write down the definite-isospin amplitudes in terms of $V$'s. Comment on the symmetry properties of each isospin amplitude with respect to $t \leftrightarrow u$.
\subsection{Chew-Frautshi plot}
Locate where each $\mathcal{A}^{(I)}(s,t,u)$ will have poles in the $s$-channel physical region. What is their residue? Draw a schematic Chew-Frautschi plot of the resonance spectrum in each isospin channel.
\subsection{Regge limit}
Now consider the limit $t \to \infty$ and $u \to - \infty$ with $s \leq 0$ is fixed. What is the asymptotic behavior of $V(s,t)$ and $V(s,u)$? Assume that $V(t,u)$ vanishes faster than any power of $s$ in this limit. What is the resulting behavior of the isospin amplitudes $\mathcal{A}^{(I)}(s,t,u)$ in this limit?
\noindent \textit{ Hint: Use the Sterling approximation of the $\Gamma$ function., i.e. as $|x|\to \infty$
\begin{equation}
\Gamma(x) \to \sqrt{\frac{2\pi}{x}} \left( \frac{x}{e} \right)^x ~.
\end{equation}
}
\subsection{Ancestors and Strings}
Consider the model now with a complex trajectory $\alpha(s) = a_0 + \alpha^\prime \, s + i \,\Gamma$ with $\Gamma > 0$ to move the poles off the real axis. Reexamine the the Chew-Frautshi plot for the $I=1$ amplitude using this trajectory, why is the resulting spectrum problematic? Try a real but non-linear trajectory, say $\alpha(s) = \alpha_0 + \alpha^\prime \, s + \alpha^{\prime\prime} \, s^2$, what is the spectrum like now?
Compare the requirements of the trajectory for $V(s,t)$ to make sense with the energy levels of a rotating relativistic string with a string tension $T$:
%%
\begin{equation}
E_J^2 = \frac{1}{2\pi \, T} \, J ~.
\end{equation}
%%
What is a possible microscopic picture of hadrons if the Veneziano amplitude is believed?
\section{Sommerfeld-Watson Transform}
\subsection{Geometric series}
Prove the well known resummation of the geometric series:
%%
\begin{equation}\label{eq:SW_0}
1 + x + x^2 + x^3 + \dots = \frac{1}{1-x} \quad \text{for } |x| < 1
\end{equation}
%%
can be analytically continued to $|x| \geq 1$ with the Sommerfeld-Watson Transform.
Assume that $|x| > 1$ and show that the summation can be written as an integral over the complex plane
%%
\begin{equation}
\label{eq:SW_1}
\int \frac{d\ell}{2i} \frac{(-x)^\ell}{\sin \pi \ell} = 1 + x + x^2 + \dots ~.
\end{equation}
%%
Draw the contour around which the above integration should be taken (careful with orientations and signs).
Deform the contour such that you can relate \cref{eq:SW_1} to the series
%%
\begin{equation}
\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + \dots = \frac{1}{1- \frac{1}{x}}
\end{equation}
%%
and arrive at \cref{eq:SW_0}.
\subsection{Van Hove Reggeon}
Revisit the Regge behavior of \cref{eq:vanhove} using the S-W transform. How does the inclusion of poles at $\alpha(s) = \ell$ change the contour of integration and the leading contribution to the asymptotic behavior?
\section{Finite Energy Sum Rules}
Consider $z$ a complex variable and $\alpha$ a real fixed parameter. What is the analytic structure of the function $z^\alpha$? What is the discontinuity across the cut?
Write a Cauchy contour $C$ surrounding the cut and closing it with a circle of radius $\Lambda$ in the complex $z$ plane, and check that
\begin{align}
\oint_{C} z^\alpha \mathrm d z & = 0
\end{align}
You can start with the simple case $\alpha = 1/2$, {\it i.e.} $\sqrt{z}$, then generalize to any real $\alpha$.
% \section{Crossing and $t$-channel kinematics}
%
\noindent In this sheet we will consider the elastic scattering of two identical, spinless particles with 4-momentum $q_{i}$ and with mass $q_i^2 = m^2$.
We define the usual Mandelstam variables
%%
\begin{gather}
s = (q_1 + q_2)^2 = (q_3 + q_4)^2 \nonumber\
t = (q_1 - q_3)^2 = (q_4 - q_2)^2 \nonumber \
u = (q_1 - q_4)^2 = (q_2 - q_3)^2 \nonumber
\end{gather}
%%
We refer to the $s$-channel as the physical region describing the process
%%
\begin{equation}
1 \, (q_1) + 2 \, (q_2) \to 3 \, (q_3) + 4 \, (q_4) ~, \nonumber
\end{equation}
%%
while in the $t$-channel we consider
\begin{equation}
1 \, (q1) + \bar{3} \, (q{\bar{3}})
\to \bar{2} \, (q_{\bar{2}}) + 4 \, (q_4) ~. \nonumber
\end{equation}
%%
\section{Reggeons and Unitarity}
\subsection{Elementary $t$-channel exchanges}
The $J=0$ is trivial
\begin{equation}
A^0(s,t) = i \, g_0 \, \frac{1}{m_0^2 - t} = i \, g_0 \, \frac{P_0(\cos\theta_t)}{m_0^2 - t} ~.
\end{equation}
For $J=1$ use $q_1 = (\sqrt{t}/2, qt \, \hat{z})$ and $q{\bar{3}} = (\sqrt{t}/2, - q_t \, \hat{z} )$.
In the $t$-channel CM frame we have $k = (q_1 - q_3) = (q1 + q{\bar{3}}) = (\sqrt{t}, \vec{0})$ and
%%
\begin{equation}
-\tilde{g}{\mu\nu} = - g{\mu\nu} + \frac{k\mu \, k\nu}{t}
= -\left[\delta{\mu0} \, \delta{\nu0} - \delta{ij} \right] + \frac{\sqrt{t}^2}{t} \, \delta{\mu0} \, \delta{\nu0} = +\delta{ij} ~.
\end{equation}
%%
thus $q1^\mu \, \tilde{g}{\mu\nu} q_{\bar{2}}^\nu = \vec{q}1 \, \cdot \, \vec{q}{\bar{2}} = q_t^2 \, \cos\theta_t$. Similarly $q1^\mu \, \tilde{g}{\mu\nu} q{1}^\nu = q{\bar{2}}^\mu \, \tilde{g}{\mu\nu} q{\bar{2}}^\nu = q_t^2$ and we have
%
\begin{equation}
A^1(s,t) = i g_1 \, q_t^2 \frac{ \cos\theta_t}{m_1^2 - t} = i g_1 \, q_t^2 \frac{P_1(\cos\theta_t)}{m_1^2 - t} ~,
\end{equation}
%%
and finally also
%%
\begin{equation}
A^2(s,t) = i g_2 \, q_t^4 \,\frac{\frac{1}{2}(3 \cos\theta_t - 1)}{m_2^2 - t} = i g_2 \, q_t^4 \, \frac{ P_2(\cos\theta_t)}{m_2^2 - t} ~.
\end{equation}
%%
The generalization to arbitrary $J$ is
%%
\begin{equation}
A^J(s,t) = ig_J \, q_t^{2J} \, \frac{ P_J(\cos\theta_t)}{m_J^2 - t} ~.
\end{equation}
%%
\subsection{Unitarity vs Elementary exchanges}
We have
%%
\begin{equation}
\sigma_\text{tot} = \left. \frac{1}{2q\sqrt{s}} \, \frac{g_J}{m_J^2} \, q_t^{2J} \, P_J(\cos\thetat) \right |{t=0} ~.
\end{equation}
%%
We have $q_t^2 \, \cos\thetat = (s-u)/ 4$ so that as $s\to \infty$, we have:
%%
\begin{equation}
\sigma\text{tot} \sim s^{J-1} ~.
\end{equation}
%%
To satisfy the Froissart bound, the maximally allowed spin then is $J=1$.
\subsection{Van Hove Reggeon}
We can write
%%
\begin{equation}
J-\alpha(t) = J - \alpha(0) - \alpha^\prime \, t = \alpha^\prime \left( (J-\alpha(0))/\alpha' - t\right)
\end{equation}
%5
and thus we have $m_J^2 = (J-\alpha(0))/\alpha^\prime$.
We can use
%%
\begin{align}
(\cos\thetat)^J &= \sum{J+J^\prime \text{ even}}^J \frac{(J+1)!}{(J-J')!! \, (J+J'+1)!!} \, P_{J'}(\cos\thetat) \
&= \sum{J+J^\prime \text{ even}}^J \mu{JJ'} \, P{J'}(\cos\theta_t)
\end{align}
%%
to write
%%
\begin{align}
A(s,t) &= \sumJ \, \sum{J'=0}^J \left(\frac{g \, r^{2J} \mu_{JJ'}}{\alpha'}\right) qt^{2J} \, \frac{P{J'}(\cos\theta_t)}{m_J^2 - t} ~,
\
&= \sumJ \, \sum{J'=0}^J g_{JJ'} \, qt^{2J} \, \frac{P{J'}(\cos\theta_t)}{m_J^2 - t} ~,
\end{align}
%%
Comparing with the form of our elementary exchanges, this amplitude is an infinite sum of particles with spin-$J$ and mass $m_J^2$ but also all same parity daughters at the same mass.
If the sum is truncated at $J\text{max}$ the $s\to\infty$ limit is dominated by the largest spin exchange and we have $A\text{trunc}(s,t) \propto s^{J_\text{max}}$.
\subsection{Analytic continuation in $J$}
Go back to the original form in terms of monomials, we can write
%%
\begin{equation}
A(s,t) = \sum_{J=0} \int_0^1 dx \, g \, r^{2J} (q_t^2 \, \cos\theta_t)^J \, x^{J-\alpha(t)-1} ~.
\end{equation}
%%
Collecting all things with powers of $J$, we notice a geometric series which can be summed analytically
%%
\begin{equation}
A(s,t) = g \,\int_0^1 dx \, \frac{x^{-\alpha(t) -1}}{1 - r^2 \, q_t^2 \, \cos\theta_t \, x} ~.
\end{equation}
%%
Comparing with the definition of the hypergeometric function, we can identify $z = r^2 \, q_t^2 \, \cos\theta_t$ and $b = -\alpha(t)$. Since there is no $(1-x)$ term we require $c = b+1= 1-\alpha(t)$. Thus we have
%5
\begin{align}
A(s,t) &= \frac{ \Gamma(-\alpha(t))}{\Gamma(1-\alpha(t))} \, _2F_1\left(1,-\alpha(t), 1-\alpha(t), (q_t \,r)^2 \, \cos\theta_t \right)
\
&= \Gamma(-\alpha(t)) \, _2\tilde{F}_1\left(1,-\alpha(t), 1-\alpha(t), (q_t \,r)^2 \, \cos\theta_t \right) ~.
\end{align}
%%
\subsection{Unitarity vs Reggeized exchanges}
From the hypergeometric form we can take $s\to \infty$ which takes $q_t^2 \, \cos\theta_t = (s-u)/4 \to \infty$ and we can write
%%
\begin{equation}
A(s,t) = g0 \, \Gamma(-\alpha(t)) \, \Gamma(1+\alpha(t)) \, \left(\frac{u-s}{4r^{-2}}\right)^{\alpha(t)} ~.
\end{equation}
%%
So, we have
%%
\begin{equation}
\Im A(s,0) \propto \Im (-s)^{\alpha(0)} \propto \sin \pi \alpha(0) \, s^{\alpha(0)}
\end{equation}
%%
so $\sigma\text{tot} \sim s^{\alpha(0)-1}$ and unitarity requires $\alpha(0) \leq 1$.
\subsection{The Reggeon ``propagator"}
As we see above, switching $s\leftrightarrow u$ introduces a minus sign and we have
%%
\begin{equation}
A^\pm(s,t) = g_0 \, \Gamma(1+\alpha(t)) \, \frac{1}{2}[\pm1 + e^{-i\pi\alpha(t)}] \, \Gamma(-\alpha(t)) \left(\frac{s-u}{4r^{-2}}\right)^{\alpha(t)} ~,
\end{equation}
%%
and we can read off $\beta(t) = g_0 \, \Gamma(1+\alpha(t))$. Using $s \sim -u$ we also see $s_0 = 2 \, r^{-2}$
\section{Veneziano Amplitude}
\subsection{Duality}
We can write:
%%
\begin{equation}
V(s,t) = (1-\alpha(s) - \alpha(t)) \, B(1-\alpha(s), 1-\alpha(t)) ~.
\end{equation}
%%
Then using the expansion of the Beta function on its first argument, we have
%%
\begin{equation}
V(s,t) = (1-\alpha(s) - \alpha(t)) \, \sum_{J=1}^\infty \, \frac{\Gamma(J-1+\alpha(t))}{\Gamma(J) \, \Gamma(\alpha(t))} \, \frac{1}{J-\alpha(s)}
\end{equation}
%%
which only has poles in $\alpha(s)$. Because of the $s\leftrightarrow t$ symmetry, we can write the exact same expression with only poles in $\alpha(t)$.
\subsection{Isospin basis}
We have:
%%
\begin{gather}
\mathcal{A}^{(0)}(s,t,u) = \frac{1}{2} \left[ 3 \, V(s,t) + 3\,V(s,u) - V(t,u)\right] \
\mathcal{A}^{(1)}(s,t,u) = V(s,t) - V(s,u) \
\mathcal{A}^{(2)}(s,t,u) = V(t,u) ~.
\end{gather}
%%
$I=0,2$ are symmetric in $t\leftrightarrow u$ while $I=1$ is anti-symmetric as required by Bose symmetry.
\subsection{Chew-Frautshi plot}
A single $V(s,t)$ will have poles at all $\alpha(s) = J\geq 1$ and all possible daughters. The residues are
%%
\begin{equation}
(J-1+\alpha(t)) \, \frac{\Gamma(J-1+\alpha(t))}{\Gamma(J) \, \Gamma(\alpha(t))} = \frac{-1}{\Gamma(J)} \frac{\Gamma(J+\alpha(t))}{\Gamma(\alpha(t))} = - \frac{(\alpha(t))_J}{(J-1)!}.
\end{equation}
%%
For a linear trajectory this is a order $J$ polynomial in $t$ and therefore in $z_s$.
The symmetry factors in $I=0,1$ will remove all odd (even) $J$ daughters. The $I=2$ amplitude has no $s$ dependence and therefore no isospin-2 poles at all.
\subsection{Ancestors}
If we allow $\alpha(t)$ to be complex, then at a pole $\alpha(s) \to J + i \Gamma$ and the residue we calculated
\begin{equation}
\frac{\Gamma(J+i \,\Gamma + \alpha(t))}{\Gamma(\alpha(t))} ~,
\end{equation}
is no longer a fixed order polynomial in $t$. It will thus give contributions to ALL spins at each pole, i.e. introduce an infinite number of ancestors. Similarly if $\alpha(s)$ is non-linear, we will have finitely many ancestors but still unphysical poles nonetheless.
This means the Veneziano amplitude \textit{only} gives a physical picture for real and linear trajectories. This means we require $J \propto s \sim m^2$ which mimics the spectrum of states in a relativistic rotating string. This gives rise to the stringy picture of a $q\bar{q}$ pair connected by a gluon flux tube and later the entire field of string theories.
\section{Sommerfeld-Watson Transform}
\subsection{Geometric series}
We want to show that we can analytically continue the geometric series to $|x|>1$ using the Sommerfeld-Watson Transform.
First, we will prove that the sum can be written as an integral over the complex plane:
\begin{equation}
\intC\frac{d\ell}{2i}\frac{(-x)^\ell}{\sin\pi\ell} \ ,
\end{equation}
where the function $1/\sin\pi\ell$ has poles at integer values of $\ell=\ldots, -2,-1,0,1,2,\ldots$
We use the Cauchy Residue Theorem:
\begin{equation}
\oint f(z) dz = \pm 2\pi i \sum{k}\textrm{Res}_{z=z_k}f(z) \ ,
\end{equation}
with sign $+$ for a counterclockwise contour and $-$ for a clockwise contour around the pole at $zk$.
The residue of $f(\ell)=(-x)^\ell/\sin\pi\ell$ at $\ell=k$ is given by
\begin{equation}
\textrm{Res}{\ell=k}\frac{(-x)^\ell}{\sin\pi\ell}=\lim{\ell\to k}(\ell-k)\frac{(-x)^\ell}{\sin\pi\ell}=\lim{\ell\to k}\frac{(-x)^\ell}{\pi\cos\pi\ell}=\frac{x^k}{\pi}
\end{equation}
Therefore, if we encircle all the poles at $\ell=k$ for $k\geq 0$ with counterclockwise contours $Ck$, which we can combine to a single counterclockwise contour $C$ encircling all the poles at $0$ and positive integers, we obtain the geometric series:
\begin{equation}
\sum{k=0}^\infty\int_{C_k}\frac{d\ell}{2i}\frac{(-x)^\ell}{\sin\pi\ell}=\intC\frac{d\ell}{2i}\frac{(-x)^\ell}{\sin\pi\ell}=\sum{k=0}^\infty x^k=1+x+x^2+\cdots
\end{equation}
Next, we deform the contour to a vertical line from $\sigma+i\infty$ to $\sigma-i\infty$, with $-1<\sigma<0$, and further deform it to enclose all negative integers in a clockwise contour $C'$, that we can split in individual clockwise contours $C{k'}$:
\begin{align}
\int{C'}\frac{d\ell}{2i}\frac{(-x)^\ell}{\sin\pi\ell}&=\sum{C{k'}}\int{C{k'}}\frac{d\ell}{2i}\frac{(-x)^\ell}{\sin\pi\ell}=-\frac{1}{x}-\frac{1}{x^2}-\frac{1}{x^3}-\cdots \
& =-\frac{1}{x}\left(1+\frac{1}{x}+\frac{1}{x^2}+\cdots\right) \nonumber
\end{align}
and given the assumption $|x|>0$, we have $|1/x|<1$, and we can sum the geometric series:
\begin{equation}
-\frac{1}{x}\left(1+\frac{1}{x}+\frac{1}{x^2}+\cdots\right) = -\frac{1}{x}\frac{1}{1-\frac{1}{x}}=\frac{1}{1-x} \ .
\end{equation}
\subsection{Van Hove Reggeon}
If we include a Regge pole $[\ell-\alpha(t)]^{-1}$, in the process of deforming the contour from $C$ to $C'$ we have to pick the residue of the pole at $\ell=\alpha(t)$, with $\textrm{Re}\,\alpha(t)<0$. This means we have an extra clockwise contour $C_\alpha$, from which we obtain an extra contribution $\propto x^{\alpha(t)}$. Since $x \sim q_t^2 \, z_t \sim s$ this yields the $s^{\alpha(t)}$ behavior.
\section{Finite Energy Sum Rules}
The function $z^\alpha$ has a branch cut for $z\in [-\infty, 0]$. The Cauchy contour enclose the cut and the discontinuity across that cut is
\bsub\begin{align}
(z+i\epsilon)^\alpha - (z-i\epsilon)^\alpha & = (|z| e^{i \pi})^\alpha- (|z| e^{-i \pi})^\alpha & \text{for real negative} z
\
&= |z|^\alpha \left( e^{i\pi \alpha} - e^{-i\pi \alpha} \right)
\
& = 2 i |z|^\alpha \sin \pi \alpha
\end{align}\esub
The Cauchy contour is then, with $C\Lambda$ being the circle of radius $\Lambda$ in the positive sense,
\bsub\begin{align}
\oint z^\alpha \mathrm d z & = \int{-\Lambda}^0 (z+i \epsilon)^\alpha \mathrm d z + \int^{-\Lambda}0 (z-i \epsilon)^\alpha \mathrm d z + \oint{C\Lambda} z^\alpha \mathrm d z
\
& = 2i \sin \pi \alpha \int{-\Lambda}^0 |z|^\alpha \mathrm d z + \oint{C\Lambda} z^\alpha \mathrm d z
\end{align}\esub
The first integral is easily done
\begin{align}
2i \sin \pi \alpha \int{-\Lambda}^0 |z|^\alpha \mathrm d z & = \frac{2i \Lambda^{\alpha+1}}{\alpha+1} \sin \pi \alpha.
\end{align}
For the second integral, we need the change of variable $z = \Lambda \exp(i \theta)$, with $\theta \in [-\pi,\pi]$. We obtain
\bsub\begin{align}
\oint{C\Lambda} z^\alpha \mathrm d z & =
i\Lambda^{\alpha+1} \int{-\pi}^\pi e^{i \theta(\alpha+1)} \mathrm d \theta = \frac{i\Lambda^{\alpha+1}}{\alpha+1}
\left(e^{i \pi(\alpha+1)} -e^{-i \pi(\alpha+1)} \right)
\
& = - \frac{2i \Lambda^{\alpha+1}}{\alpha+1} \sin \pi \alpha
\end{align}\esub
We used $\exp(i\pi) = -1$.
mmikhasenko
commented
4 months ago
\documentclass[a4paper,11pt]{article} \usepackage{style/preamble} % \titleformat{\subsection}[runin]{}{}{}{}[] \renewcommand{\thesection}{\Roman{section}} \renewcommand{\thesubsection}{\alph{subsection})} \renewcommand*{\thesubsubsection}{\roman{subsubsection}}
\begin{document}
\section{Reggeization and Unitarity}
Van Hove \cite{vanHove:1967zz} proposed a physically intuitive picture of a Reggeon by relating it to Feynman diagrams in the cross-channels. We will explore this picture of Reggeization with a simple model.
\subsection{Elementary $t$-channel exchanges} Consider the amplitude corresponding to a particle with spin-$J$ and mass $m_J$ exchanged in the $t$-channel as: %% \begin{equation}
\label{eq:AJ} A^J(s,t) = i \, g_J \, \left( q_1^{\mu_1} \dots q_1^{\muJ} \right) \frac{P^J{\mu_1\dots\mu_J,\nu_1\dots\nu_J}(k)}{mJ^2 - t} \left( q{\bar{2}}^{\nu1} \dots q{\bar{2}}^{\nu_J} \right) \end{equation} %% where $gJ$ is a coupling constant with dimension $2-2J$ (i.e., $A^J(s,t)$ is dimensionless) and the projector of spin-$J$ is defined from the polarization tensor of rank-$J\geq1$ as %% \begin{equation} P^J{\mu_1\dots\mu_J,\nu_1\dots\nuJ}(k) = \frac{(J+1)}{2} \, \sum{\lambda} \epsilon^{\mu_1\dots\mu_J}(k,\lambda) \, \epsilon^{*\nu_1\dots\nu_J}(k,\lambda) ~. \end{equation} %%
Using the exchange momentum $k = q1 + q{\bar{3}} = q_1 - q_3$, calculate the amplitudes corresponding to $J=0,1,2$ exchanges in terms of $t= k^2$, the modulus of 3-momentum and cosine of scattering angle in the $t$-channel frame, $q_t$ and $\cos\thetat$ respectively. Use the explicit forms of the projectors: %% \begin{gather} P^0(k^2) = 1 \ P^1{\mu\nu}(k^2) \equiv \tilde{g}{\mu\nu} = \frac{ k\mu \, k\nu}{ k^2} - g{\mu\nu}\ P^2{\mu\nu\alpha\beta}(k^2) = \frac{3}{4}\left(\tilde{g}{\mu\alpha} \, \tilde{g}{\nu\beta} + \tilde{g}{\mu\beta} \, \tilde{g}{\nu\alpha}\right) - \frac{1}{2} \, \tilde{g}{\mu\nu} \, \tilde{g}_{\alpha\beta} ~, \end{gather} %% and conjecture a generalization of the amplitude for arbitrary integer $J$.
\noindent\textit{Hint: Show that in the $t$-channel frame, the exchange particle is at rest and therefore $\tilde{g}{\mu\nu}$ reduces to a $\delta{ij}$ with respect to only spacial momenta.}
\subsection{Unitarity vs Elementary exchanges} Express the amplitude entirely in terms of invariants $s$ and $t$. Use the optical theorem to relate the elastic amplitude to a total hadronic cross section: \begin{equation} \sigma\text{tot} = \frac{1}{2q\sqrt{s}} \, \Im A^J(s,t=0)~. \end{equation} Unitarity (via the Froissart-Martin bound) prohibits $\sigma\text{tot}$ from growing faster than $\log^2 s$ as $s\to\infty$~. What is then the maximal spin a single elementary exchange can have while satisfying this bound? Why is this a problem?
\subsection{Van Hove Reggeon}
Consider an amplitude of the form %% \begin{equation} \label{eq:vanhove} A(s,t) = \sum_{J=0}^\infty g \, r^{2J} \, \frac{ (q_t^2 \, \cos\theta_t)^J}{J- \alpha(t)} ~. \end{equation} %% Here $\alpha(t) = \alpha(0) + \alpha^\prime \, t$ is a real, linear Regge trajectory, $g$ is a dimensionless coupling constant and $r \sim 1$ fm is a range parameter. Compare \cref{eq:vanhove} with \cref{eq:AJ}, write the mass of the $J$th pole, $m_J^2$, as a function of the Regge parameters $\alpha(0)$ and $\alpha^\prime$. Interpret the pole structure in terms of the spectrum of particles in the model. \
\noindent If the sum is truncated to a finite $J_\text{max}$, and we take the $s\to \infty$ limit, what is the high energy behavior of the amplitude?
\subsection{Analytic continuation in $J$} Show that if the summation is kept infinite, the amplitude can be re-summed to something that is entirely analytic in $s$, $t$, $u$, and $J$.
\noindent \textit{Hint: Use the Mellin transform %% \begin{equation} \frac{1}{J-\alpha(t)} = \int_0^1 dx \, x^{J-\alpha(t) - 1} ~, \end{equation} %% to express the amplitude in terms of the Gaussian hypergeometric function and the Euler Beta function %% \begin{equation} B(b, c-b) \, _2F_1(1, b, c; z) = \int_0^1 dx \, \frac{ x^{b-1} \, (1-x)^{c-b-1}}{1-x \, z} ~. \end{equation} %% } \subsection{Unitarity vs Reggeized exchanges} Revisit $\mathbf{b)}$ with the resummed amplitude. Take the $s\to\infty$ limit and set a limit on the maximal intercept $\alpha(0)$ which is allowed by unitarity.
\noindent \textit{Hint: Assume that $\alpha(0) > -1$ and use the asymptotic behavior of the hypergeometric function given by %% \begin{equation} _2F_1(1,b,c;z) \to \frac{\Gamma(c)\,\Gamma(1-b)}{\Gamma(1) \,\Gamma(c-b)} \, (-z)^{-b} ~. \end{equation} %% }
\subsection{The Reggeon ``propagator"} Modify \cref{eq:AJ} to have a definite signature by defining %% \begin{equation} A^{\pm}(s,t) = \frac{1}{2}\left[A(s,t) \pm A(u,t) \right]~. \end{equation} %% Repeat $\textbf{d)}$ and $\textbf{e)}$ with this signatured amplitude. Compare with the canonical form of the Reggeon exchange: %% \begin{equation} A^\pm_\mathbb{R}(s,t) = \beta(t) \, \frac{1}{2}\left[\pm1 + e^{-i\pi\alpha(t)}\right] \, \Gamma(-\alpha(t)) \, \left(\frac{s}{s_0}\right)^{\alpha(t)} ~. \end{equation} %% Identify the Regge residue $\beta(t)$ and characteristic scale $s_0$ in terms of the parameters $g_0$ and $r$.
\section{Veneziano Amplitude}
The quintessential dual amplitude was first proposed by Veneziano for $\omega\to3\pi$ \cite{Veneziano:1968yb} and later applied to elastic $\pi\pi$ scattering by Shapiro \cite{Shapiro:1969km} and Lovelace \cite{Lovelace:1968kjy}. Consider the $\pi^+\pi^-$ scattering amplitude of the form %% \begin{equation} \label{eq:Astu} \mathcal{A}(s,t,u) = V(s,t) + V(s,u) - V(t,u) ~. \end{equation} %% with each %% \begin{equation}
\label{eq:Fst} V(s,t) = \frac{\Gamma(1-\alpha(s)) \, \Gamma(1-\alpha(t))}{\Gamma(1-\alpha(s) - \alpha(t))} ~, \end{equation} %% where $\alpha(s) = \alpha(0) + \alpha^\prime s$ is a real, linear Regge trajectory with $\alpha^\prime > 0$.
\subsection{Duality} Show that the function $V(s,t)$ is symmetric in $s\leftrightarrow t$ and dual, i.e., it can be written entirely as a sum of either $s$-channel poles OR $t$-channel poles but never both simultaneously. Compare with the Reggeized amplitude in the previous problem, was that amplitude dual?
\noindent \textit{Hint: Relate $V(s,t)$ to the Euler Beta function \begin{equation} B(x,y) = \frac{\Gamma(x) \, \Gamma(y)}{\Gamma(x+y)} \end{equation} and use the identities $B(x,y) = B(y,x)$ and %% \begin{equation} B(p-x, q-y) = \sum_{J=1}^\infty \, \frac{\Gamma(J-p+1 +x)}{\Gamma(J) \, \Gamma(-p + 1 +x)} \, \frac{1}{J-1+q-y} ~. \end{equation} %% }
\subsection{Isospin basis} Define the $s$-channel isospin basis through %% \begin{equation} \begin{pmatrix} \mathcal{A}^{(0)}(s,t,u) \ \mathcal{A}^{(1)}(s,t,u) \ \mathcal{A}^{(2)}(s,t,u)
\end{pmatrix} = \frac{1}{2}\begin{pmatrix} 3 & 1 & 1 \ 0 & 1 & -1 \ 0 & 1 & 1 \end{pmatrix} \begin{pmatrix} \mathcal{A}(s,t,u) \ \mathcal{A}(t,s,u) \ \mathcal{A}(u,t,s)
\end{pmatrix} ~. \end{equation} %% Write down the definite-isospin amplitudes in terms of $V$'s. Comment on the symmetry properties of each isospin amplitude with respect to $t \leftrightarrow u$.
\subsection{Chew-Frautshi plot} Locate where each $\mathcal{A}^{(I)}(s,t,u)$ will have poles in the $s$-channel physical region. What is their residue? Draw a schematic Chew-Frautschi plot of the resonance spectrum in each isospin channel.
\subsection{Regge limit} Now consider the limit $t \to \infty$ and $u \to - \infty$ with $s \leq 0$ is fixed. What is the asymptotic behavior of $V(s,t)$ and $V(s,u)$? Assume that $V(t,u)$ vanishes faster than any power of $s$ in this limit. What is the resulting behavior of the isospin amplitudes $\mathcal{A}^{(I)}(s,t,u)$ in this limit?
\noindent \textit{ Hint: Use the Sterling approximation of the $\Gamma$ function., i.e. as $|x|\to \infty$ \begin{equation} \Gamma(x) \to \sqrt{\frac{2\pi}{x}} \left( \frac{x}{e} \right)^x ~. \end{equation} }
\subsection{Ancestors and Strings} Consider the model now with a complex trajectory $\alpha(s) = a_0 + \alpha^\prime \, s + i \,\Gamma$ with $\Gamma > 0$ to move the poles off the real axis. Reexamine the the Chew-Frautshi plot for the $I=1$ amplitude using this trajectory, why is the resulting spectrum problematic? Try a real but non-linear trajectory, say $\alpha(s) = \alpha_0 + \alpha^\prime \, s + \alpha^{\prime\prime} \, s^2$, what is the spectrum like now?
Compare the requirements of the trajectory for $V(s,t)$ to make sense with the energy levels of a rotating relativistic string with a string tension $T$: %% \begin{equation} E_J^2 = \frac{1}{2\pi \, T} \, J ~. \end{equation} %% What is a possible microscopic picture of hadrons if the Veneziano amplitude is believed?
\section{Sommerfeld-Watson Transform}
\subsection{Geometric series}
Prove the well known resummation of the geometric series: %% \begin{equation}\label{eq:SW_0} 1 + x + x^2 + x^3 + \dots = \frac{1}{1-x} \quad \text{for } |x| < 1 \end{equation} %% can be analytically continued to $|x| \geq 1$ with the Sommerfeld-Watson Transform.
Assume that $|x| > 1$ and show that the summation can be written as an integral over the complex plane %% \begin{equation} \label{eq:SW_1} \int \frac{d\ell}{2i} \frac{(-x)^\ell}{\sin \pi \ell} = 1 + x + x^2 + \dots ~. \end{equation} %% Draw the contour around which the above integration should be taken (careful with orientations and signs). Deform the contour such that you can relate \cref{eq:SW_1} to the series %% \begin{equation} \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + \dots = \frac{1}{1- \frac{1}{x}} \end{equation} %% and arrive at \cref{eq:SW_0}.
\subsection{Van Hove Reggeon} Revisit the Regge behavior of \cref{eq:vanhove} using the S-W transform. How does the inclusion of poles at $\alpha(s) = \ell$ change the contour of integration and the leading contribution to the asymptotic behavior?
\section{Finite Energy Sum Rules} Consider $z$ a complex variable and $\alpha$ a real fixed parameter. What is the analytic structure of the function $z^\alpha$? What is the discontinuity across the cut?
Write a Cauchy contour $C$ surrounding the cut and closing it with a circle of radius $\Lambda$ in the complex $z$ plane, and check that \begin{align} \oint_{C} z^\alpha \mathrm d z & = 0 \end{align}
You can start with the simple case $\alpha = 1/2$, {\it i.e.} $\sqrt{z}$, then generalize to any real $\alpha$.
\bibliographystyle{apsrev4-2.bst} \bibliography{biblio}
\newpage \begin{center} SOLUTIONS \end{center}
% \section{Crossing and $t$-channel kinematics} % \noindent In this sheet we will consider the elastic scattering of two identical, spinless particles with 4-momentum $q_{i}$ and with mass $q_i^2 = m^2$. We define the usual Mandelstam variables %% \begin{gather} s = (q_1 + q_2)^2 = (q_3 + q_4)^2 \nonumber\ t = (q_1 - q_3)^2 = (q_4 - q_2)^2 \nonumber \ u = (q_1 - q_4)^2 = (q_2 - q_3)^2 \nonumber \end{gather} %% We refer to the $s$-channel as the physical region describing the process %% \begin{equation} 1 \, (q_1) + 2 \, (q_2) \to 3 \, (q_3) + 4 \, (q_4) ~, \nonumber \end{equation} %% while in the $t$-channel we consider \begin{equation} 1 \, (q1) + \bar{3} \, (q{\bar{3}}) \to \bar{2} \, (q_{\bar{2}}) + 4 \, (q_4) ~. \nonumber \end{equation} %%
\section{Reggeons and Unitarity}
\subsection{Elementary $t$-channel exchanges} The $J=0$ is trivial \begin{equation} A^0(s,t) = i \, g_0 \, \frac{1}{m_0^2 - t} = i \, g_0 \, \frac{P_0(\cos\theta_t)}{m_0^2 - t} ~. \end{equation} For $J=1$ use $q_1 = (\sqrt{t}/2, qt \, \hat{z})$ and $q{\bar{3}} = (\sqrt{t}/2, - q_t \, \hat{z} )$. In the $t$-channel CM frame we have $k = (q_1 - q_3) = (q1 + q{\bar{3}}) = (\sqrt{t}, \vec{0})$ and %% \begin{equation} -\tilde{g}{\mu\nu} = - g{\mu\nu} + \frac{k\mu \, k\nu}{t} = -\left[\delta{\mu0} \, \delta{\nu0} - \delta{ij} \right] + \frac{\sqrt{t}^2}{t} \, \delta{\mu0} \, \delta{\nu0} = +\delta{ij} ~. \end{equation} %% thus $q1^\mu \, \tilde{g}{\mu\nu} q_{\bar{2}}^\nu = \vec{q}1 \, \cdot \, \vec{q}{\bar{2}} = q_t^2 \, \cos\theta_t$. Similarly $q1^\mu \, \tilde{g}{\mu\nu} q{1}^\nu = q{\bar{2}}^\mu \, \tilde{g}{\mu\nu} q{\bar{2}}^\nu = q_t^2$ and we have % \begin{equation} A^1(s,t) = i g_1 \, q_t^2 \frac{ \cos\theta_t}{m_1^2 - t} = i g_1 \, q_t^2 \frac{P_1(\cos\theta_t)}{m_1^2 - t} ~, \end{equation} %% and finally also %% \begin{equation} A^2(s,t) = i g_2 \, q_t^4 \,\frac{\frac{1}{2}(3 \cos\theta_t - 1)}{m_2^2 - t} = i g_2 \, q_t^4 \, \frac{ P_2(\cos\theta_t)}{m_2^2 - t} ~. \end{equation} %% The generalization to arbitrary $J$ is %% \begin{equation} A^J(s,t) = ig_J \, q_t^{2J} \, \frac{ P_J(\cos\theta_t)}{m_J^2 - t} ~. \end{equation} %%
\subsection{Unitarity vs Elementary exchanges} We have %% \begin{equation} \sigma_\text{tot} = \left. \frac{1}{2q\sqrt{s}} \, \frac{g_J}{m_J^2} \, q_t^{2J} \, P_J(\cos\thetat) \right |{t=0} ~. \end{equation} %% We have $q_t^2 \, \cos\thetat = (s-u)/ 4$ so that as $s\to \infty$, we have: %% \begin{equation} \sigma\text{tot} \sim s^{J-1} ~. \end{equation} %% To satisfy the Froissart bound, the maximally allowed spin then is $J=1$.
\subsection{Van Hove Reggeon}
We can write %% \begin{equation} J-\alpha(t) = J - \alpha(0) - \alpha^\prime \, t = \alpha^\prime \left( (J-\alpha(0))/\alpha' - t\right) \end{equation} %5 and thus we have $m_J^2 = (J-\alpha(0))/\alpha^\prime$.
We can use %% \begin{align} (\cos\thetat)^J &= \sum{J+J^\prime \text{ even}}^J \frac{(J+1)!}{(J-J')!! \, (J+J'+1)!!} \, P_{J'}(\cos\thetat) \ &= \sum{J+J^\prime \text{ even}}^J \mu{JJ'} \, P{J'}(\cos\theta_t) \end{align} %% to write %% \begin{align} A(s,t) &= \sumJ \, \sum{J'=0}^J \left(\frac{g \, r^{2J} \mu_{JJ'}}{\alpha'}\right) qt^{2J} \, \frac{P{J'}(\cos\theta_t)}{m_J^2 - t} ~, \ &= \sumJ \, \sum{J'=0}^J g_{JJ'} \, qt^{2J} \, \frac{P{J'}(\cos\theta_t)}{m_J^2 - t} ~, \end{align} %% Comparing with the form of our elementary exchanges, this amplitude is an infinite sum of particles with spin-$J$ and mass $m_J^2$ but also all same parity daughters at the same mass.
If the sum is truncated at $J\text{max}$ the $s\to\infty$ limit is dominated by the largest spin exchange and we have $A\text{trunc}(s,t) \propto s^{J_\text{max}}$.
\subsection{Analytic continuation in $J$} Go back to the original form in terms of monomials, we can write %% \begin{equation} A(s,t) = \sum_{J=0} \int_0^1 dx \, g \, r^{2J} (q_t^2 \, \cos\theta_t)^J \, x^{J-\alpha(t)-1} ~. \end{equation} %%
Collecting all things with powers of $J$, we notice a geometric series which can be summed analytically %% \begin{equation} A(s,t) = g \,\int_0^1 dx \, \frac{x^{-\alpha(t) -1}}{1 - r^2 \, q_t^2 \, \cos\theta_t \, x} ~. \end{equation} %% Comparing with the definition of the hypergeometric function, we can identify $z = r^2 \, q_t^2 \, \cos\theta_t$ and $b = -\alpha(t)$. Since there is no $(1-x)$ term we require $c = b+1= 1-\alpha(t)$. Thus we have %5 \begin{align} A(s,t) &= \frac{ \Gamma(-\alpha(t))}{\Gamma(1-\alpha(t))} \, _2F_1\left(1,-\alpha(t), 1-\alpha(t), (q_t \,r)^2 \, \cos\theta_t \right) \ &= \Gamma(-\alpha(t)) \, _2\tilde{F}_1\left(1,-\alpha(t), 1-\alpha(t), (q_t \,r)^2 \, \cos\theta_t \right) ~. \end{align} %%
\subsection{Unitarity vs Reggeized exchanges} From the hypergeometric form we can take $s\to \infty$ which takes $q_t^2 \, \cos\theta_t = (s-u)/4 \to \infty$ and we can write %% \begin{equation} A(s,t) = g0 \, \Gamma(-\alpha(t)) \, \Gamma(1+\alpha(t)) \, \left(\frac{u-s}{4r^{-2}}\right)^{\alpha(t)} ~. \end{equation} %% So, we have %% \begin{equation} \Im A(s,0) \propto \Im (-s)^{\alpha(0)} \propto \sin \pi \alpha(0) \, s^{\alpha(0)} \end{equation} %% so $\sigma\text{tot} \sim s^{\alpha(0)-1}$ and unitarity requires $\alpha(0) \leq 1$.
\subsection{The Reggeon ``propagator"} As we see above, switching $s\leftrightarrow u$ introduces a minus sign and we have %% \begin{equation} A^\pm(s,t) = g_0 \, \Gamma(1+\alpha(t)) \, \frac{1}{2}[\pm1 + e^{-i\pi\alpha(t)}] \, \Gamma(-\alpha(t)) \left(\frac{s-u}{4r^{-2}}\right)^{\alpha(t)} ~, \end{equation} %% and we can read off $\beta(t) = g_0 \, \Gamma(1+\alpha(t))$. Using $s \sim -u$ we also see $s_0 = 2 \, r^{-2}$
\section{Veneziano Amplitude} \subsection{Duality} We can write: %% \begin{equation} V(s,t) = (1-\alpha(s) - \alpha(t)) \, B(1-\alpha(s), 1-\alpha(t)) ~. \end{equation} %% Then using the expansion of the Beta function on its first argument, we have %% \begin{equation} V(s,t) = (1-\alpha(s) - \alpha(t)) \, \sum_{J=1}^\infty \, \frac{\Gamma(J-1+\alpha(t))}{\Gamma(J) \, \Gamma(\alpha(t))} \, \frac{1}{J-\alpha(s)} \end{equation} %% which only has poles in $\alpha(s)$. Because of the $s\leftrightarrow t$ symmetry, we can write the exact same expression with only poles in $\alpha(t)$.
\subsection{Isospin basis} We have: %% \begin{gather} \mathcal{A}^{(0)}(s,t,u) = \frac{1}{2} \left[ 3 \, V(s,t) + 3\,V(s,u) - V(t,u)\right] \ \mathcal{A}^{(1)}(s,t,u) = V(s,t) - V(s,u) \ \mathcal{A}^{(2)}(s,t,u) = V(t,u) ~. \end{gather} %% $I=0,2$ are symmetric in $t\leftrightarrow u$ while $I=1$ is anti-symmetric as required by Bose symmetry.
\subsection{Chew-Frautshi plot} A single $V(s,t)$ will have poles at all $\alpha(s) = J\geq 1$ and all possible daughters. The residues are %%
\begin{equation}
The symmetry factors in $I=0,1$ will remove all odd (even) $J$ daughters. The $I=2$ amplitude has no $s$ dependence and therefore no isospin-2 poles at all.
\subsection{Regge limit} Starting with \begin{equation} V(s,t) \to \Gamma(1-\alpha(s)) \, (-\alpha(t))^{\alpha(s)} \sim \Gamma(1-\alpha(s)) \, \left(-\alpha^\prime \, t\right)^{\alpha(s)} ~. \end{equation} Similarly \begin{equation} V(s,u) \to \Gamma(1-\alpha(s)) \, (-\alpha(u))^{\alpha(s)} \sim \Gamma(1-\alpha(s)) \, \left(-\alpha^\prime \, t\right)^{\alpha(s)} ~. \end{equation} Thus the combination %% \begin{align} V(s,t) \pm V(s,u) &\to \Gamma(1-\alpha(s)) \times \left[\left(-\alpha^\prime \, t\right)^{\alpha(s)} \pm \left(-\alpha^\prime \, u\right)^{\alpha(s)} \right] \ &= \Gamma(1-\alpha(s)) \times \left[1\pm e^{-i\pi\alpha(s)}\right] \, \left(\alpha^\prime \, t\right)^{\alpha(s)} \end{align} %%
\subsection{Ancestors} If we allow $\alpha(t)$ to be complex, then at a pole $\alpha(s) \to J + i \Gamma$ and the residue we calculated \begin{equation} \frac{\Gamma(J+i \,\Gamma + \alpha(t))}{\Gamma(\alpha(t))} ~, \end{equation} is no longer a fixed order polynomial in $t$. It will thus give contributions to ALL spins at each pole, i.e. introduce an infinite number of ancestors. Similarly if $\alpha(s)$ is non-linear, we will have finitely many ancestors but still unphysical poles nonetheless.
This means the Veneziano amplitude \textit{only} gives a physical picture for real and linear trajectories. This means we require $J \propto s \sim m^2$ which mimics the spectrum of states in a relativistic rotating string. This gives rise to the stringy picture of a $q\bar{q}$ pair connected by a gluon flux tube and later the entire field of string theories. \section{Sommerfeld-Watson Transform} \subsection{Geometric series} We want to show that we can analytically continue the geometric series to $|x|>1$ using the Sommerfeld-Watson Transform.
First, we will prove that the sum can be written as an integral over the complex plane: \begin{equation} \intC\frac{d\ell}{2i}\frac{(-x)^\ell}{\sin\pi\ell} \ , \end{equation} where the function $1/\sin\pi\ell$ has poles at integer values of $\ell=\ldots, -2,-1,0,1,2,\ldots$ We use the Cauchy Residue Theorem: \begin{equation} \oint f(z) dz = \pm 2\pi i \sum{k}\textrm{Res}_{z=z_k}f(z) \ , \end{equation} with sign $+$ for a counterclockwise contour and $-$ for a clockwise contour around the pole at $zk$. The residue of $f(\ell)=(-x)^\ell/\sin\pi\ell$ at $\ell=k$ is given by \begin{equation} \textrm{Res}{\ell=k}\frac{(-x)^\ell}{\sin\pi\ell}=\lim{\ell\to k}(\ell-k)\frac{(-x)^\ell}{\sin\pi\ell}=\lim{\ell\to k}\frac{(-x)^\ell}{\pi\cos\pi\ell}=\frac{x^k}{\pi} \end{equation} Therefore, if we encircle all the poles at $\ell=k$ for $k\geq 0$ with counterclockwise contours $Ck$, which we can combine to a single counterclockwise contour $C$ encircling all the poles at $0$ and positive integers, we obtain the geometric series: \begin{equation} \sum{k=0}^\infty\int_{C_k}\frac{d\ell}{2i}\frac{(-x)^\ell}{\sin\pi\ell}=\intC\frac{d\ell}{2i}\frac{(-x)^\ell}{\sin\pi\ell}=\sum{k=0}^\infty x^k=1+x+x^2+\cdots \end{equation}
Next, we deform the contour to a vertical line from $\sigma+i\infty$ to $\sigma-i\infty$, with $-1<\sigma<0$, and further deform it to enclose all negative integers in a clockwise contour $C'$, that we can split in individual clockwise contours $C{k'}$: \begin{align} \int{C'}\frac{d\ell}{2i}\frac{(-x)^\ell}{\sin\pi\ell}&=\sum{C{k'}}\int{C{k'}}\frac{d\ell}{2i}\frac{(-x)^\ell}{\sin\pi\ell}=-\frac{1}{x}-\frac{1}{x^2}-\frac{1}{x^3}-\cdots \ & =-\frac{1}{x}\left(1+\frac{1}{x}+\frac{1}{x^2}+\cdots\right) \nonumber \end{align} and given the assumption $|x|>0$, we have $|1/x|<1$, and we can sum the geometric series: \begin{equation} -\frac{1}{x}\left(1+\frac{1}{x}+\frac{1}{x^2}+\cdots\right) = -\frac{1}{x}\frac{1}{1-\frac{1}{x}}=\frac{1}{1-x} \ . \end{equation}
\subsection{Van Hove Reggeon} If we include a Regge pole $[\ell-\alpha(t)]^{-1}$, in the process of deforming the contour from $C$ to $C'$ we have to pick the residue of the pole at $\ell=\alpha(t)$, with $\textrm{Re}\,\alpha(t)<0$. This means we have an extra clockwise contour $C_\alpha$, from which we obtain an extra contribution $\propto x^{\alpha(t)}$. Since $x \sim q_t^2 \, z_t \sim s$ this yields the $s^{\alpha(t)}$ behavior.
\section{Finite Energy Sum Rules} The function $z^\alpha$ has a branch cut for $z\in [-\infty, 0]$. The Cauchy contour enclose the cut and the discontinuity across that cut is \bsub\begin{align} (z+i\epsilon)^\alpha - (z-i\epsilon)^\alpha & = (|z| e^{i \pi})^\alpha- (|z| e^{-i \pi})^\alpha & \text{for real negative} z \ &= |z|^\alpha \left( e^{i\pi \alpha} - e^{-i\pi \alpha} \right) \ & = 2 i |z|^\alpha \sin \pi \alpha \end{align}\esub The Cauchy contour is then, with $C\Lambda$ being the circle of radius $\Lambda$ in the positive sense, \bsub\begin{align} \oint z^\alpha \mathrm d z & = \int{-\Lambda}^0 (z+i \epsilon)^\alpha \mathrm d z + \int^{-\Lambda}0 (z-i \epsilon)^\alpha \mathrm d z + \oint{C\Lambda} z^\alpha \mathrm d z \ & = 2i \sin \pi \alpha \int{-\Lambda}^0 |z|^\alpha \mathrm d z + \oint{C\Lambda} z^\alpha \mathrm d z \end{align}\esub The first integral is easily done \begin{align} 2i \sin \pi \alpha \int{-\Lambda}^0 |z|^\alpha \mathrm d z & = \frac{2i \Lambda^{\alpha+1}}{\alpha+1} \sin \pi \alpha. \end{align} For the second integral, we need the change of variable $z = \Lambda \exp(i \theta)$, with $\theta \in [-\pi,\pi]$. We obtain \bsub\begin{align} \oint{C\Lambda} z^\alpha \mathrm d z & = i\Lambda^{\alpha+1} \int{-\pi}^\pi e^{i \theta(\alpha+1)} \mathrm d \theta = \frac{i\Lambda^{\alpha+1}}{\alpha+1} \left(e^{i \pi(\alpha+1)} -e^{-i \pi(\alpha+1)} \right) \ & = - \frac{2i \Lambda^{\alpha+1}}{\alpha+1} \sin \pi \alpha \end{align}\esub We used $\exp(i\pi) = -1$.
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