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\begin{document}

\centerline{\bf Lecture~11}
\centerline{\bf Solving massive Thirring model by Bethe Ansatz: thermodynamic limit}
\bigskip
On the last lecture we derived the Bethe equation for $N$ pseudoparticles
\eq$$
p_0(\lambda_k)L+\sum^N_{l=1}\Phi(\lambda_k-\lambda_l)=2\pi n_k,
\qquad
n_k\in\Z+{\delta\over2}.
\label{BetheEquation-log}
$$
Today we will use it to describe the true vacuum of the system and excitations above the vacuum. The Bethe equation is a system of nonlinear algebraic equations so that it is probable impossible to solve it analytically. For the Thirring model there is no need to solve them for arbitrary finite $N$, since we are only interested in the limit $N\to\infty$. In principle, this limit is achieved by sending to infinity the cut-off scale~$\Lambda$, and we may consider arbitrary values of~$L$. In fact, here we will only discuss the case of large enough $L$, namely $L\gg m^{-1}$, where $m$ is the physical mass of excitations.

When $g=0$ the vacuum had the following structure. The parameters $\lambda_k$ lied on the axis $\R+\i\pi$, so that $\lambda_k=\i\pi+\beta_k$. It corresponded to the negative energy `excitations'. Besides, the negative energy band was full, so that we could write $n_k=-k+\const$. In the limit $L\to\infty$ the roots $\beta_k$ became dense. Note that in this dense case the answer is not sensible (in the leading approximation) to the periodicity condition (NS or~R). Suppose that this structure is preserved for small enough values of~$g$. Then for the \emph{vacuum solution} we have
$$
p_0(\beta_k)L=\sum^N_{l=1}\Phi(\beta_k-\beta_l)+2\pi k+\const.
$$
Let us take a difference of two these equations with neighboring values of~$k$:
$$
p_0(\beta_{k+1})-p_0(\beta_k)={1\over L}\sum^N_{l=1}(\Phi(\beta_{k+1}-\beta_l)-\Phi(\beta_k-\beta_l)+{2\pi\over L}.
$$
Define the \emph{density of roots} for the vacuum solution:
$$
\rho(\beta_k)={1\over L(\beta_{k+1}-\beta_k)}.
$$
If roots are dense, we may write $f(\lambda_{k+1})-f(\lambda_k)=f'(\lambda_k)(\lambda_{k+1}-\lambda_k)$ for any slowly changing function~$f$. Hence,
$$
p'(\beta_k)={1\over L}\sum^N_{l=1}\Phi'(\beta_k-\beta_l)+2\pi\rho(\beta_k).
$$
Finally substitute the sum by and integral with the measure $dl=L\rho(\beta_l)\,d\beta_l$:
\eq$$
p'(\beta)=2\pi\rho(\beta)+\int^{\Theta}_{-\Theta} d\gamma\,\Phi'(\beta-\gamma)\rho(\gamma)
\quad\text{for $-\Theta\le\beta\le\Theta$.}
\label{BetheEquation-continuous-vac}
$$
The parameter $\beta_0$ is related to the cut-off parameter $\Lambda$:
\eq$$
\Lambda={m_0\over2}\e^{\Theta}.
\label{Lambda-Theta-rel}
$$
On the other hand, it is related to the spacial density of pseudoparticles in the vacuum state:
\eq$$
{N\over L}=\int^\Theta_{-\Theta}d\beta\,\rho(\beta).
\label{N-Theta-rel}
$$
We do not intend to solve the equation (\ref{BetheEquation-continuous-vac}) since it only gives the vacuum energy, which we are not interested in. It can be shown that
\eq$$
\rho(\beta)\propto\exp{\beta\over1+g},
\label{rho-fin}
$$
but it will not help us. We will be interested in excitations. There is a large zoo of excitations in this theory, but we will consider the simplest kind of them: holes. Consider the state in which $\lambda_k=\i\pi+\beta_k$, $\beta_k\in\R$, as before, but the values $n_k$ do not form a solid segment, but have some omissions.

Return for a moment to the general equation (\ref{BetheEquation-log}). Let $\{\lambda_k\}$ be a given solution, and $\{n_k\}$ is the corresponding set of integers (or half-integers). Then define the function $\lambda(n)$ as a solution to the equation
\eq$$
\pi(\lambda(n))L+\sum^N_{l=1}\Phi(\lambda(n)-\lambda_l)=2\pi n.
\label{lambda(n)-def}
$$
If $n=n_k$ the value of the function coincides with the corresponding root: $\lambda(n_k)=\lambda_k$.

Return to our dense states. If we order the set of $n_k$ in such a way that $n_{k+1}<n_k$, we may define the \emph{density of states} $\rho(\beta)$ and the \emph{density of roots} $\rho_r(\beta)$:
\eq$$
\rho(\beta(n))={1\over L(\beta(n+1)-\beta(n))},
\qquad
\rho_r(\beta(n_k))={1\over L(\beta_{k+1}-\beta_k)}.
\label{rho-rhor-def}
$$
We also may define the \emph{density of holes}
\eq$$
\rho_h(\beta)=\rho(\beta)-\rho_r(\beta).
\label{rhoh-def}
$$
It is easy to write the integral equation for such a state
\eq$$
p'(\beta)=2\pi\rho(\beta)+\int^{\Theta}_{-\Theta} d\gamma\,\Phi'(\beta-\gamma)(\rho(\gamma)-\rho_h(\gamma))
\quad\text{for $-\Theta\le\beta\le\Theta$.}
\label{BetheAnsatz-continuous-holes}
$$
It is important that $\rho(\beta)$ is the unknown, while $\rho_h(\beta)$ is a free parameter. We may choose any function $\rho_h(\beta)$ if only the solution will satisfy $\rho(\beta)\ge\rho_h(\beta)$. For example, the choice $\rho_h(\beta)=L^{-1}\delta(\beta-\beta_0)$ corresponds to the state with a single hole of rapidity~$\beta_0$.

Let $\rho_0(\beta)$ be the vacuum density of states, i.e.\ the solution to the equation~(\ref{BetheEquation-continuous-vac}) and $\rho(\beta)=\rho_0(\beta)+\delta\rho(\beta)$ be the solution to the equation~(\ref{BetheAnsatz-continuous-holes}) with a given density of holes~$\rho_h(\beta)$. By taking the difference of these two equations we obtain
\eq$$
2\pi\delta\rho(\beta)+\int^\Theta_{-\Theta}d\gamma\,\Phi'(\beta-\gamma)\delta\rho(\gamma)
=\int^\Theta_{-\Theta}d\gamma\,\Phi'(\beta-\gamma)\rho_h(\gamma).
\label{deltarho-eq}
$$
Note that $\delta\rho(\beta)$ is not supposed to be small.

Now let us solve the equation. It turns out that it is sufficient to solve it for $\Theta=\infty$. In this case we may apply the Fourier method. Let
\Gather$$
\tilde\Phi'(\omega)=\int^\infty_{-\infty}d\beta\,\Phi'(\beta)\e^{\i\omega\beta}
=-2\pi{\sh\pi g\omega\over\sh\pi\omega},
\label{Phi'-Fourier}
\\
\delta\tilde\rho(\omega)=\int^\infty_{-\infty}d\beta\,\delta\rho(\beta)\e^{\i\omega\beta},
\qquad
\tilde\rho_h(\omega)=\int^\infty_{-\infty}d\beta\,\rho_h(\beta)\e^{\i\omega\beta}.
\label{rho-Fourier}
$$
After the Fourier transform equation (\ref{deltarho-eq}) reads
\eq$$
2\pi\delta\tilde\rho(\omega)+\tilde\Phi'(\omega)\delta\tilde\rho(\omega)=\tilde\Phi'(\omega)\tilde\rho_h(\omega).
\label{deltarho-eq-Fourier}
$$
It solution is given by
\eq$$
\delta\tilde\rho(\omega)
=-{\sh\pi g\omega\over2\sh{\pi(1-g)\omega\over2}\ch{\pi(1+g)\omega\over2}}\tilde\rho_h(\omega).
\label{deltarho-solution}
$$
Suppose that $\tilde\rho_h(\omega)$ does not have poles in a large enough strip around the real axis. In particular, for a finite number of holes this function has no poles at all. Compute the energy (comparing to the vacuum energy) and momentum of the state:
\eq$$
\Aligned{
E
&=L\int^\Theta_{-\Theta}d\beta\,\epsilon_0(\beta)(\rho_h(\beta)-\delta\rho(\beta)),
\\
P
&=L\int^\Theta_{-\Theta}d\beta\,p_0(\beta)(\rho_h(\beta)-\delta\rho(\beta))\sh\beta.
}\label{energy-momentum-rhoh}
$$
Consider the case $g<0$. In this case these integrals are convergent as $\Theta\to\infty$, and we easily obtain that for $\Theta=\infty$
$$
E+P=m_0L(\tilde\rho_h(\i)-\delta\tilde\rho(\i))=0,
\qquad
E-P=m_0L(\tilde\rho_h(-\i)-\delta\tilde\rho(-\i))=0.
$$
In fact, surely, the energy and momentum are nonzero due to finiteness of $\Theta$. It can be shown that
\eq$$
E=L\int^\Theta_{-\Theta}d\beta\,\rho_h(\beta)\epsilon\left(\beta\over1+g\right),
\qquad
P=L\int^\Theta_{-\Theta}d\beta\,\rho_h(\beta)p\left(\beta\over1+g\right),
\label{EP-fin}
$$
where $\epsilon(\theta),p(\theta)$ are energy and momentum of physical excitations:
\Gather$$
\epsilon(\theta)=m\ch\theta,
\qquad
p(\theta)=m\sh\theta,
\label{ep-phys}
\\
m={m_0\over g}\ctg\left({\pi\over2}{1-g\over1+g}\right)\e^{{g\over1+g}\Theta}\sim m_0\left(m_0\over\Lambda\right)^{-{g\over1+g}}.
\label{m-phys}
$$
Comparing with the case $g=0$ we conclude that this excitation is an antifermion%
%
\footnote{Note that if we try to obtain the fermion as a pseudoparticle with real $\lambda$, we will fail. For indefinitely small though nonzero $g$ the pseudoparticle repulse states in the vicinity of $\lambda+\i\pi$, and we obtain a fermion\-/antifermion bound state instead. The construction of the fermion is more involved.}
%
We see that the rapidity is renormalized in consistency with the density of states~(\ref{rho-fin}).

Nevertheless, here we have a problem. We have concluded in (\ref{m-phys}) that the physical mass is related to the bare mass as $m\sim m_0(m_0/\Lambda)^{g/(1+g)}\sim m_0(m_0r)^{g/(1+g)}$. But this contradicts the relation obtained by the boson\--fermion correspondence:
$$
m\sim m_0(m_0r)^{{1\over2(1-\beta^2)}-1}\sim m_0(m_0r)^{-{g\over1+2g}}.
$$
The formula (\ref{m-phys}) would be obtained if we assumed $g=1-2\beta^2$ instead of $g=(2\beta^2)^{-1}-1$. We may conjecture that the coupling constant $g$ in the Bethe Ansatz, being based on the anticommutation relation for fermions, differs with that of obtained from operator product expansion due renormalization of the fermion. If we denote the coupling constant in Bethe Ansatz by $g_{\text{BA}}$ and that used in the boson\--fermion correspondence by $g_{\text{OPT}}$, we get
\eq$$
g_{\text{BA}}={g_{\text{OPE}}\over1+g_{\text{OPE}}}.
\label{g-BA-OPE}
$$
In fact, after assuming this relation all results of the Bethe Ansatz and of the conformal perturbation theory become consistent.

Can we say anything about interaction of the antifermions? Yes. To extract this information, consider an auxiliary model of spinless fermions of mass $m$ with the wave function of the form
$$
\chi_{\theta_1\ldots\theta_N}(x_1,\ldots,x_N)
=\sum_{\tau\in S^N}(-1)^{\sigma\tau}A_\tau\prod^N_{k=1}\e^{\i mx_{\sigma_k}\sh\theta_{\tau_k}},
\quad\text{if $x_{\sigma_1}<\cdots<x_{\sigma_N}$.}
$$
The constants $A_\tau$ are related as
\eq$$
A_{\ldots ij\ldots}=S^{-1}(\theta_i-\theta_j)A_{\ldots ji\ldots},
\label{A=SA}
$$
where $S(\theta)$ is a given function, which play a role of the $S$ matrix of two particles. In fact, this model is an approximation to system of physical fermions in a strong external field, which suppress their antifermions. As we will see below, wave functions with such long-distance asymptotics are characteristic for integrable systems.

Imposing the periodicity condition produces the Bethe equation
\eq$$
\e^{\i p(\lambda_k)L}=\pm\prod^N_{\substack{l=1\\(l\ne k)}}S(\theta_k-\theta_l).
\label{BE-S}
$$
Let
\eq$$
S(\theta)=\e^{\i\Psi(\theta)}.
\label{S-Psi}
$$
In the limit of dense roots we obtain
\eq$$
p'(\theta)+\int^\infty_{-\infty}d\vartheta\Psi'(\theta-\vartheta)\rho_p(\vartheta)=2\pi\rho_s(\vartheta).
\label{BetheEquation-continuous-S}
$$
Let $\delta\rho_s(\theta)=\rho_s(\theta)-{1\over2\pi}p'(\theta)$. Then we get
\eq$$
\delta\rho_s(\vartheta)={1\over2\pi}\int^\infty_{-\infty}d\vartheta\Psi'(\theta-\vartheta)\rho_p(\vartheta).
\label{deltarhos-S}
$$
After applying the Fourier transform
$$
\tilde\rho_p(t)=\int^\infty_{-\infty}d\theta\,\rho_p(\theta)\e^{\i t\theta}\quad\text{etc.}
$$
we obtain
$$
\delta\tilde\rho_s(t)={1\over2\pi}\tilde\Psi'(t)\tilde\rho_p(t).
$$

If we assume that these particles correspond to holes in the Bethe Ansatz solution to the Thirring model, then the density of such particles $\rho_p(\theta)$ is proportional to the density of holes $\rho_h(\beta)$, while the density of states $\rho_s(\theta)$ is proportional to the density of roots $\rho(\beta)$:
\eq$$
\rho_p(\theta)=(1+g)\rho_h((1+g)\theta),
\qquad
\rho_s(\theta)=(1+g)\rho((1+g)\theta).
\label{rho-rel}
$$
In terms of Fourier components we have
\eq$$
\tilde\rho_p(t)=\tilde\rho_h\left(t\over1+g\right),
\qquad
\delta\tilde\rho_s(t)=\delta\tilde\rho\left(t\over1+g\right).
\label{tilderho-rel}
$$
Hence,
$$
\delta\tilde\rho\left(t\over1+g\right)={1\over2\pi}\tilde\Psi'(t)\tilde\rho_h\left(t\over1+g\right)
$$
By substituting (\ref{deltarho-solution}) we obtain
\eq$$
\tilde\Psi'(t)=-2\pi{\sh\pi{\pi gt\over1+g}\over2\sh{\pi(1-g)t\over(1+g)}\ch{\pi t\over2}}.
\label{Psi'-solution}
$$
Substituting $g$ by $1-2\beta^2$ and using the parametrization
$$
\beta^2={p\over p+1},
$$
we obtain
\eq$$
S(\theta)=\exp\left(-\int{dt\over t}{\sh{\pi t\over2}\sh{\pi(p-1)t\over2}\over\sh\pi t\sh{\pi pt\over2}}\e^{-\i\theta t}\right)
=\exp\left(2\i\int{dt\over t}{\sh{\pi t\over2}\sh{\pi(p-1)t\over2}\over\sh\pi t\sh{\pi pt\over2}}\sin\theta t\right).
\label{S-fin}
$$
Note that it is the $S$ matrix, which describes the scattering of two antifermions into two antifermions $S(\theta)^{--}_{--}$. Due to the charge symmetry of the model it coincides with the scattering of two fermions int two fermions $S(\theta)^{++}_{--}$. Due to the crossing symmetry of amplitudes it also provides the matrix elements $S(\theta)^{+-}_{+-}$ and $S(\theta)^{-+}_{-+}$:
\eq$$
S(\theta)^{++}_{++}=S(\theta)^{--}_{--}=S(\theta),
\qquad
S(\theta)^{+-}_{+-}=S(\theta)^{-+}_{-+}=S(\i\pi-\theta)=S(\theta){\sh{\theta\over p}\over\sh{\i\pi-\theta\over p}}.
\label{4S-fin}
$$
To obtain the whole $S$ matrix we have to learn more about general properties of integrable systems.







%{\raggedright
%\bibliographystyle{mybib}
%\bibliography{main}}

\bigbreak
\leftline{\bf Bibliography}
\medskip
[1]~V.~E.~Korepin, N.~M.~Bogoliubov, A.~G.~Izergin, Quantum Inverse Scattering Method and Correlation Functions, Cambridge, 1997.

\bigbreak
\bigskip
\centerline{\bf Problems}
\medskip

{\bf1.}~Derive (\ref{EP-fin})\--(\ref{m-phys}).

{\bf2.}~Calculate the change of the number of states in the interval $[-\Theta,\Theta]$ under the influence of holes:
$$
\delta n=\int^\Theta_{-\Theta}d\beta\,\delta\rho(\beta).
$$




\end{document}
