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\begin{document}

\title[Lecture 11. Kondo Problem: Solving Bethe Equations]{Lecture 12\\Kondo Problem: Solving Bethe Equations}
\date{}
\author{Michael Lashkevich}

\frame{\titlepage}

\begin{frame}
\frametitle{Bethe equations}
Recall the Bethe equations for the $sd$ model:
\Gather$$
e^{ip_aL}=e^{iJS}\prod^n_{i=1}{v_i+i/2\over v_i-i/2},
\label{cyclicfin}
\\
\uncover<2->{\left(v_i+i/2\over v_i-i/2\right)^N{v_i+iS+1/g\over v_i-iS+1/g}=-\prod^n_{j=1}{v_i-v_j+i\over v_i-v_j-i},}
\label{Bethefin}
\\
a=1,\ldots,N,
\qquad
\uncover<2->{i,j=1,\ldots,n,}
\nonumber
$$
\pause\pause
while
\eq$$
g={1\over S+1/2}\tg J(S+1/2).
\label{gdef}
$$
\pause
The energy of the system is
\eq$$
E=\sum^N_{a=1}p_a.
\label{energy}
$$
\pause
Now we will study these equations in the thermodynamic limit $L\to\infty$, $N\to\infty$.

\end{frame}

\begin{frame}
\frametitle{Logarithm of the Bethe equations}

Take logarithm of the Bethe equations:
\Gather$$
p_aL=2\pi I_a+JS-\sum^n_{i=1}(\pi+p(v_i)),
\label{b1}
\\
\uncover<2->{Np(v_i)+\delta_S(v_i)
=}
\uncover<3->{2\pi J_i+\sum^n_{j=1}\Phi(v_i-v_j),}
\label{b2}
\\
p(v)=2\arctg2v,
\qquad
\uncover<2->{\delta_S(v)=p((v+1/g)/2S),}
\qquad
\uncover<3->{\Phi(v)=p(v/2),}
\label{pfuncdef}
\\
I_a\in\Z,
\qquad
\uncover<3->{J_i\in\Cases{\Z,&\text{$N-n\in2\Z$},\\
\Z+1/2.&\text{$N-n\in2\Z+1$}.}}
\label{IJdef}
$$
\pause\pause\pause
Besides, all the numbers $J_i$ should be pairwise distinct, and all the numbers $I_a$ too.
\pause

The total energy:
\Align$$
E&=\Ech+\Esp,
\label{Esplit}
\\
\uncover<6->{\Ech}
&\uncover<6->{={2\pi\over L}\sum^N_{a=1}I_a{\only<7>{\red}-{\pi N^2\over2L}},}
\label{Echdef}
\\
\uncover<7->{\Esp}
&\uncover<7->{={\only<7>{\red}\only<8>{\blue}{\pi N^2\over2L}}+{NJS\over L}-{N\over L}\sum^n_{i=1}({\only<8>{\blue}\pi}+p(v_i))}
\uncover<9>{
\tikz[overlay,color=red]\draw (current page.south east) node[rectangle,draw,left=.5cm,yshift=2cm] (sumPhi) {$\displaystyle\sum^N_{i,j=1}\Phi(v_i-v_j)=0$};
}
\nonumber
\\
&\uncover<8->{=-{2\pi\over L}\sum^n_{i=1}J_i+{\only<8>{\blue}{\pi\over L}N\left({N\over2}-n\right)}
+{NJS\over L}+{1\over L}\sum^n_{i=1}\delta_S(v_i),\tikz[overlay,baseline=-2pt] \node (sumPhiTo) {};
\uncover<9>{\tikz[overlay] \draw[->,color=red] (sumPhi) to[out=-90,in=0] (sumPhiTo);}}
\label{Espdef}
$$

\end{frame}

\begin{frame}
\frametitle{Charge energy for $J=0$. Usual description}

Let $J=0$: the case of free electrons. We have two pictures:


{\bf1.~Usual description:}
$$
p_aL=2\pi I_a
\quad
I_a\in\Z,
$$
where pairs of $I_a$ may coincide, but if $I_a=I_b$ ($a\ne b$), then $\forall c\ne a,b:\ I_c\ne I_a$.
\pause

To make the total energy finite, we need a cutoff at negative momenta: $p_a\ge-\eF$.
\pause
The density of states in the momentum space is $2{L\over2\pi}$.
\pause
Thus
\eq$$
N={L\eF\over\pi}.
\label{N-eF-rel}
$$
\pause
We will be interested in the thermodynamic limit
$$
L\to\infty,
\qquad
N\to\infty,
\qquad
{N\over L}={\eF\over\pi}=\const,
$$
\pause
The ground state is defined by $-{N\over2}\le I_a\le0$, and the energy is equal to
$$
E_0=-{\pi N^2\over2L}=-{\eF N\over2}.
$$

\end{frame}

\begin{frame}
\frametitle{Charge energy for $J=0$. Bethe description}

{\bf2.~Bethe description:}
$$
p_aL=2\pi I_a-\sum^n_{i=1}(\pi+p(v_i)),
\quad
I_a\in\Z,
$$
where $I_a$ are all different: $I_a\ne I_b$ if $a\ne b$.
\pause

In the ground state the total spin is zero: $n=N/2$.
\pause
From the XXZ model we know that the total momentum in the spin space is equal to zero:
$$
\sum^{N/2}_{i=1}p(v_i)=0.
$$
\pause
Hence
$$
p_a={2\pi\over L}I_a-{\pi N\over2L}={2\pi\over L}I_a-{\eF\over2}.
$$
\pause
The ground state energy is
$$
E_0=-{\eF N\over2}+\sum^N_{a=1}I_a
\quad\Rightarrow\quad
\sum^N_{a=1}I_a=0.
$$
\pause
All other energies must be larger. Therefore, we obtain the admissibility condition for solutions to the Bethe Ansatz equations
$$
\boxed{I_a\ge-{N/2}.}
$$

\end{frame}

\begin{frame}
\frametitle{Two descriptions: comparison}

$$
\begin{tikzpicture}
\draw[->] (0,-2.2) -- (0,3) node[right] {$\epsilon$};
\draw[-] (0,-1) -- (.1,-1) node[right] {$-\eF$};
\draw[-] (0,1) -- (.1,1) node[right] {$0$};
\foreach \y in {-1,-.8,...,.8} {
  \draw[-] (0,\y) node[left] {\tiny$\uparrow\downarrow$};
}
\end{tikzpicture}
\quad
\begin{tikzpicture}
\draw[->] (0,-2.2) -- (0,3) node[right] {$\epsilon$};
\draw[-] (0,-2) -- (.1,-2) node[right] {$-{3\over2}\eF$};
\draw[-] (0,-1) -- (.1,-1) node[right] {$-\eF$};
\draw[-] (0,1) -- (.1,1) node[right] {$0$};
\draw[-] (0,2) -- (.1,2) node[right] {${1\over2}\eF$};
\foreach \y in {-2,-1.8,...,1.8} {
  \draw[-] (0,\y) node[left] {\tiny$\bullet$};
}
\end{tikzpicture}
\qquad
\pause
\qquad
\begin{tikzpicture}
\draw[->] (0,-2.2) -- (0,3) node[right] {$\epsilon$};
\draw[-] (0,-1) -- (.1,-1) node[right] {$-\eF$};
\draw[-] (0,1) -- (.1,1) node[right] {$0$};
\foreach \y in {-1,-.8,...,0,.4,.6,.8} {
  \draw[-] (0,\y) node[left] {\tiny$\uparrow\downarrow$};
}
\draw[-] (0,.2) node[left] (spinhole)
  {\tiny\only<2,4>{\blue$\uparrow\mathclose{\phantom{\downarrow}}$}\only<3,5>{\red$\mathopen{\phantom{\uparrow}}\downarrow$}};
\draw[-] (0,1.4) node[left] (spinparticle)
  {\tiny\only<2,5>{\red$\mathopen{\phantom{\uparrow}}\downarrow$}\only<3,4>{\blue$\uparrow\mathclose{\phantom{\downarrow}}$}};
\draw[->,overlay] (spinhole.west) to[out=135,in=-135] (spinparticle.west);
\end{tikzpicture}
\quad
\begin{tikzpicture}
\draw[->] (0,-2.2) -- (0,3) node[right] {$\epsilon$};
\draw[-] (0,-2) -- (.1,-2) node[right] {$-{3\over2}\eF$};
\draw[-] (0,-1) -- (.1,-1) node[right] {$-\eF$};
\draw[-] (0,1) -- (.1,1) node[right] {$0$};
\draw[-] (0,2) -- (.1,2) node[right] {${1\over2}\eF$};
\foreach \y in {-2,-1.8,...,1,1.4,1.6,1.8} {
  \draw[-] (0,\y) node[left] {\tiny$\bullet$};
}
\draw[-] (0,1.2) node[left] (hole) {\tiny$\phantom{\bullet}$};
\only<2,3>{\draw[-] (0,2.4) node[left] (particle) {\tiny$\bullet$};
\draw[->,overlay] (hole.west) to[out=135,in=-135] (particle.west);}
\only<4>{\draw[-] (0,2.2) node[left] (particle) {\tiny$\bullet$};
\draw[->,overlay] (hole.west) to[out=135,in=-135] (particle.west);}
\only<5->{\draw[-] (0,2.6) node[left] (particle) {\tiny$\bullet$};
\draw[->,overlay] (hole.west) to[out=135,in=-135] (particle.west);}

\draw[overlay] (0,2.4) node[right] {\only<2-3>{$n={N\over2}$}\only<4>{\blue$n=\smash{N\over2}-1$}\only<5>{\red$\times$}};
\end{tikzpicture}
$$%
\vtop to 1.5cm{
\only<1-3>{$I'_a=I_a+L\Delta E/2\pi$.}%
\only<4>{$I'_a=I_a+L\Delta E/2\pi{\blue{}-1}$ since $\smash{\Delta\sum(-\pi-p(v_i))\simeq{2\pi\over N}}$.}%
\only<5->{$I'_a=I_a+L\Delta E/2\pi{\red{}+1}$? {\red BUT:} Calculation of state with $n>\smash{N\over2}$ ($S^z_\text{tot}<0$) is problematic within the Bethe Ansatz technique.}
\vfill
}

\end{frame}

\begin{frame}
\frametitle{States of nonzero spin}

Assume $J$ to be arbitrary. Return to the Bethe equations for $v_i$:
\eq$$
Np(v_i)+\delta_S(v_i)=2\pi J_i+\sum^n_{j=1}\Phi(v_i-v_j),
\quad
i=1,\ldots,n.
\tag{\ref{b2}}
$$
\pause
Since $p(v)$, $\delta_S(v)$ and $\Phi(v)$ are increasing odd functions and tend to $\pi$ as $v\to\infty$, we have
$$
J_i\to\pm\smash{N+2-n\over2}\text{ as $v_i\to\pm\infty$.}
$$
\pause
Hence
\eq$$
-{N+1-n\over2}\le J_i\le{N+1-n\over2}.
\label{Jirange}
$$
\pause
The minimum of energy corresponds to larger $v_i$s and, hence, to larger $J_i$s. Therefore, for the ground state in the spin space $J_i$s densely fill the region
\eq$$
J_\text{min}\le J_i\le{N+1-n\over2}.
\label{JiGSrange}
$$
with a certain $J_\text{min}$
\pause
This corresponds to the region
$$
-b\le v_i<\infty
$$
with a certain value $b$.

\end{frame}

\begin{frame}
\frametitle{Pauli paramagnetism}

It is easy to find the energy of these states in the case $J=0$. Indeed, evidently
$$
{N+2-n\over2}-J_\text{min}=n
$$
\pause
Hence
$$
J_\text{min}={N+2-3n\over2}.
$$
\pause
Now let us calculate the spin energy. Since $\delta_S(v)=\pi$, we have
$$
\Espel=-{2\pi\over L}\sum^{n-1}_{i=0}(J_\text{min}+i)+\eF\left({N\over2}-n+{n\over N}\right)
={2\eF\over N}\left({N\over2}-n\right)^2={2\eF\over N}(S_\el^z)^2.
$$
The superscript $\el$ means that later we will separate this contribution as the energy of the electron subsystem.
\pause

If we define external magnetic field $H$ as $\Espel(H)=\Espel-S_\el^zH$, and minimize this energy in $S^z_\el$, we obtain the \emph{Pauli paramagnetism} of the $s$ electrons:
\eq$$
H={4\eF\over N}S^z=4\eF M_\el,
\label{HSzrel}
$$
where $M_\el$ is the magnetization, i.e.\ spin per electron.
\pause
This formula will make it possible to express impurity energy and magnetization in terms of external magnetic field.

\end{frame}

\begin{frame}
\frametitle{Bethe equations in the thermodynamic limit}

Let us write down the Bethe equations for the ground state in the spin space in the thermodynamic limit:
\eq$$
\rho(v)=a_1(v)+{1\over N}a_{2S}(v+1/g)-\int^\infty_{-b}{dv'\over2\pi}\,a_2(v-v')\rho(v'),
\quad
-b\le v<\infty,
\label{b2int}
$$
where
\eq$$
\rho(v)={2\pi\over N}{dJ\over dv},
\qquad
a_t(v)={t\over v^2+t^2/4}.
\label{atdef}
$$
\pause
We have
\eq$$
S^z={N\over2}+S-n={N\over2}+S-N\int^\infty_{-b}{dv\over2\pi}\,\rho(v).
$$
\pause
and
\eq$$
\Esp={N\eF\over2}+{\eF JS\over\pi}-{N\eF\over\pi}\int^\infty_{-b}{dv\over2\pi}\,\rho(v)(\pi+p(v)).
\label{Espimexpr}
$$

\end{frame}

\begin{frame}
\frametitle{Splitting the integral equation}

Since the integral equation is linear, its solution can be exactly split into the sum
\eq$$
\rho(v)=\rho_0(v)+{\rho_1(v)\over N},
\label{rhovexp}
$$
\pause
where
\eq$$
\rho_0(v)=a_1(v)
-\int^\infty_{-b}{dv'\over2\pi}\,a_2(v-v')\rho_0(v'),
\qquad
-b\le v<\infty,
\label{b2int0}
$$
\pause
and
\eq$$
\rho_1(v)=a_{2S}(v+1/g)-\int^\infty_{-b}{dv'\over2\pi}\,a_2(v-v')\rho_1(v'),
\qquad
-b\le v<\infty.
\label{b2int1}
$$
\pause
Hence, $\rho_0(v)$ is the density of state for spins of free electrons, while $\rho_1(v)$ describes the impurity.
\pause
The magnetization and the energy split as well:
\eq$$
S^z=NM_\el+M_\im,
\pause
\quad
M_\el={1\over2}-\int^\infty_{-b}{dv\over2\pi}\,\rho_0(v),
\pause
\quad
M_\im=S-\int^\infty_{-b}{dv\over2\pi}\,\rho_1(v).
\label{Szexp}
$$
\pause
and
\Align$$
\Esp
&=\Espel+\Eim,
\label{Espexp}
\\
\uncover<9->{\Espel}
&\uncover<9->{={\eF N\over2}-{\eF\over\pi}\int^\infty_{-b}{dv\over2\pi}\,\rho_0(v)(\pi+p(v))}
\uncover<10->{=2\eF NM_\el^2,}
\label{Espeldef}
\\
\uncover<11->{\Eim}
&\uncover<11->{={\eF JS\over\pi}-{\eF\over\pi}\int^\infty_{-b}{dv\over2\pi}\,\rho_1(v)(\pi+p(v)).}
\label{Eimdef}
$$

\end{frame}

\begin{frame}
\frametitle{The $b=\infty$ case}

In the $b=\infty$ case we may use the Fourier transform. Since
\eq$$
\tilde a_t(k)=\int^\infty_{-\infty}{dv\over2\pi}\,a_t(v)e^{ikv}
=e^{-t|k|/2},
\label{atFourier}
$$
\pause
we have
$$
\trho_0(k)=e^{-|k|/2}-e^{-|k|}\trho_0(k),
\qquad
\trho_1(k)=e^{-S|k|-ik/g}-e^{-|k|}\trho_1(k).
$$
\pause
The solution is
\eq$$
\trho_0(k)={1\over2\ch{k\over2}},
\qquad
\trho_1(k)={e^{-(S-1/2)k-ik/g}\over2\ch{k\over2}}.
\label{trhoksolutions}
$$
\pause
The point $k=0$ is of particular interest:
\eq$$
\trho_0(0)=\int{dv\over2\pi}\,\rho_0(v)={1\over2},
\qquad
\trho_1(0)=\int{dv\over2\pi}\,\rho_1(v)={1\over2}.
\label{trho0}
$$
\pause
From this we obtain
\subeq{
\Align$$
M_\el
&=1/2-\trho_0(0)=0,
\label{MelH0}
\\
M_\im
&=S-\trho_1(0)=S-1/2.
\label{MimH0}
$$}
\pause
The limit $b\to\infty$ corresponds to $H\to+0$. Therefore the total spin of the system is $S-1/2$ and, hence, \emph{the ground state is $2S$\-/fold degenerate}.

\end{frame}

\begin{frame}
\frametitle{Wiener\--Hopf method}

For finite $b$ both the equations for $\rho_0(v)$ and $\rho_1(v)$ have the form
\eq$$
f(x)+\int^\infty_0{dx'\over2\pi}\,K(x-x')f(x')=g(x),\qquad x>0.
\label{WHeq}
$$
\pause
These equations are solved by the Wiener\--Hopf method.
\pause
Arbitrarily continue the function $g(x)$ to negative values of~$x$. This continues the solution $f(x)$ to negative~$x$. Then perform the Fourier transform.
\pause
Let
\eq$$
\tilde f_+(k)=\int^\infty_0{dx\over2\pi}\,e^{ikx}f(x),
\quad
\tilde f_-(k)=\int^0_{-\infty}{dx\over2\pi}\,e^{ikx}f(x).
\label{WHfpm}
$$
The function $\tilde f_+(k)$ ($\tilde f_-(k)$) has no singularities in the upper (lower) half-plane. Here and below, such a property will be assumed for all functions with the $\pm$ subscripts.
\pause
We have
\eq$$
(1+\tilde K(k))\tilde f_+(k)+\tilde f_-(k)
=\tilde g(k).
\label{WHeqk}
$$
\pause
Represent the kernel $\tilde K(k)$ in the form
\eq$$
1+\tilde K(k)={\tilde K_+(k)\over\tilde K_-(k)}.
\label{WHfactor}
$$
\pause
Besides, we set
\eq$$
\tilde K_-(k)g(k)=\tilde q_+(k)+\tilde q_-(k).
\label{WHKgdecomp}
$$
\pause
Multiplying (\ref{WHeqk}) by $\tilde K_-(k)$, we obtain
\eq$$
\tilde K_+(k)\tilde f_+(k)+\tilde K_-(k)\tilde f_-(k)
=\tilde q_+(k)+\tilde q_-(k).
\label{HWeqdecomp}
$$

\end{frame}

\begin{frame}
\frametitle{Wiener\--Hopf method}

Thus
$$
\tilde K_+(k)\tilde f_+(k)-\tilde q_+(k)
=\tilde q_-(k)-\tilde K_-(k)\tilde f_-(k).
$$
\pause
The left-hand side has no singularities in the upper half-plane, and the right-hand side in the lower one. Thus, both sides of this equation have no singularities. Under some additional restrictions on the growth of the functions (which must be checked separately in each case), it follows that
\eq$$
\tilde K_+(k)\tilde f_+(k)=\tilde q_+(k),
\qquad
\tilde K_-(k)\tilde f_-(k)=\tilde q_-(k).
\label{WHeqsplitted}
$$
\pause
Finally,
\eq$$
f(x)=\int^\infty_{-\infty}dk\,{\tilde q_+(k)\over\tilde K_+(k)}e^{-ikx},
\qquad x>0.
\label{WHsolution}
$$

\end{frame}

\begin{frame}
\frametitle{Wiener\--Hopf method. Application to $b<\infty$}

Let
$$
f_i(x)=\rho_i(x-b).
$$
\pause
Then
\eq$$
\tilde K(k)=e^{-|k|},
\qquad
\tilde g_0(k)=e^{ikb-|k|/2},
\qquad
\tilde g_1(k)=e^{ikb-ik/g-S|k|}.
\label{Kgik-def}
$$
\pause
We use a trick to obtain a few simple results. Rewrite equation (\ref{WHeqk}) in the form
\eq$$
\tilde f_{i+}(k)+{\tilde f_{i-}(k)\over1+\tilde K(k)}={\tilde g_i(k)\over1+\tilde K(k)}.
\label{WHeqk1}
$$
\pause
Perform the inverse Fourier transform:
\eq$$
f_i(x)+\int^0_{-\infty}{dx'\over2\pi}\,R(x-x')f_i(x')=h_i(x),
\label{WHeq1}
$$
\pause
where
\eq$$
\Gathered{
R(x)=\int^\infty_{-\infty}dk\,e^{-ikx}\left({1\over1+\tilde K(k)}-1\right)=-\int^\infty_{-\infty}dk\,{e^{-ikx}\over1+e^{|k|}},
\\
h_0(x)={\pi\over\ch\pi(x-b)},
\qquad
h_1(x)=\int^\infty_{-\infty}dk\,e^{-ik(x-b+1/g)}{e^{-(2S-1)|k|/2}\over2\ch{k\over2}}.
}\label{Rhi-def}
$$

\end{frame}

\begin{frame}
\frametitle{Application to $b<\infty$}

If $b\gg1$, for small enough $x$ we may approximate
\eq$$
h_0(x)\simeq2\pi e^{\pi(x-b)}.
\label{h0-approx}
$$
It works, if we want to calculate $\tilde f_{0-}(k)$. Thus we have $\tilde f_{0-}(k)\sim e^{-\pi b}$.
\pause

Take into account that $\tilde K(0)=\tilde g_i(0)=1$. Therefore
$$
2\tilde f_{i+}(0)+\tilde f_{i-}(0)=1
$$
\pause
and hence
$$
{H\over4\eF}=M_\el={1\over2}-\int^\infty_{-b}{dv\over2\pi}\,\rho_0(v)={1\over2}-\tilde f_{0+}(0)={1\over2}\tilde f_-(0)\sim e^{-\pi b}.
$$
\pause
More precisely (and it needs accurate solution of the integral equation)
\eq$$
{H\over2\eF}=e^{-\pi b}\left(2\over\pi e\right)^{1/2}.
\label{bHrel}
$$
\pause
For very large $b$ and for the impurity spin $S=1/2$ we may also write
\eq$$
h_1(x)\simeq2\pi e^{\pi(x-b)+\pi/g}.
\label{h1-approx}
$$
\pause
Hence ${\tilde f_{1-}(k)\over\tilde f_{0-}(k)}=e^{\pi/g}$, and we have a precise result for the susceptibility:
\eq$$
\chi_\im={M_\im\over H}={1\over4\eF}{M_\im\over M_\el}={e^{\pi/g}\over4\eF},
\quad
\text{if $S=1/2$.}
\label{chim-fin}
$$

\end{frame}

\begin{frame}
\frametitle{Magnetization for finite $H$}

An accurate calculation by the Wiener\--Hopf method gives the formula
\Multline$$
M_\im(H)=S-{1\over2}+
\\
+{i\over4\pi^{3/2}}\int^\infty_{-\infty}d\omega\,\left(H\over T_H\right)^{-2i\omega}{\Gamma(i\omega+1/2)\over\omega+i0}
\left(-i\omega+0\over e\right)^{-2iS\omega}\left(i\omega+0\over e\right)^{i(2S-1)\omega},
\label{MHrel}
$$
\pause
where
\eq$$
T_H=\left(2\pi\over e\right)^{1/2}{2\eF\over\pi}e^{-\pi/g}\sim T_K.
\label{THdef}
$$
\pause
It provide the asymptotics
$$
M_\im(H)=S\left(1-{1\over\log(H/T_K)^2}-{\log\log(H/T_K)^2\over\log^2(H/T_K)^2}+\cdots\right),
\quad
H\gg T_K,
$$
\pause
and
$$
\Gathered{
M_\im(H)=(S-1/2)\left(1+{1\over\log(T_K/H)^2}-{\log\log(T_K/H)^2\over\log^2(T_K/H)^2}+\cdots\right),
\quad
H\ll T_K,\quad S>1/2;
\\
M_\im(H)={2\over\sqrt\pi}\sum^\infty_{n=0}\left(n+{1\over2}\over e\right)^{n+{1\over2}}{(-1)^n\over n!(n+{1\over2})}\left(H\over T_H\right)^{2n+1},
%\sqrt{2\over\pi e}{H\over T_H}+\cdots,
\quad
S=1/2.
}
$$

\end{frame}

\begin{frame}
\frametitle{String solutions}

The Bethe equations admit complex roots. For large values of $N$ these roots form the so called strings:
\eq$$
v^p_{j,k}=v^p_j+{i\over2}(p+1-2k)+O(e^{-\const N}),
\qquad
k=1,2,\ldots,p.
\label{strings}
$$
Real roots can be considered as $1$\-/strings.
\pause
As we already discussed at a seminar, the difference between roots in a string corresponds to a zero in the r.h.s.\ of the Bethe equation. The Bethe equations of the centers of strings are obtained by multiplying Bethe equations for all roots that enter the string.
\pause
For the Kondo problem we have
\Gather$$
e^{ip_aL}=e^{iJS}\prod^\infty_{p=1}\prod^{n_p}_{j=1}e_p(v^p_j),
\label{b1strings}
\\
(e_p(v^p_j))^Ne_{p,S}(v^p_j+1/g)
=\prod^\infty_{p'=1}\prod^{n_m}_{j'=1}
E_{pp'}(v^p_j-v^{p'}_{j'}),
\label{b2strings}
$$
\pause
where
$$
\Gathered{
e_p(v)=-e^{iP_p(v)}={v+ip/2\over v-ip/2},
\qquad
e_{p,S}(v)=-e^{i\Delta_{p,S}(v)}=\prod^p_{k=1}{v+{i\over2}(p+1-2k)+iS\over v+{i\over2}(p+1-2k)-iS},
\\
E_{pp'}(v)=e^{i\Phi_{pp'}(v)}=e_{|p-p'|}(v)e^2_{|p-p'|+2}(v)\ldots e^2_{p+p'-2}(v)e_{p+p'}(v).
}
$$

\end{frame}

\begin{frame}
\frametitle{Finite temperatures}

Bethe equations may be applied to finite temperatures. To do it, we need to introduce two types of densities: density of states $\rho_p(v)$ ($p$ means the type of a string) and density of particles $\rho^\bullet_p(v)$. The Bethe equations make it possible to express $\rho_p(v)$ in term of $\rho^\bullet_p(v)$. It is convenient to use also the density of holes $\rho^\circ_p(v)=\rho_p(v)-\rho^\bullet_p(v)$.
\pause

Introduce the entropy of a set of states described by these densities:
\Align$$
\cS
&=\log\prod_{p,v}
{(N\rho_p(v){dv\over2\pi})!\over(N\rho^\bullet_p(v){dv\over2\pi})!(N\rho^\circ_p(v){dv\over2\pi})!}
\nonumber
\\
&=N\sum^\infty_{p=1}\int{dv\over2\pi}\,
(\rho_p(v)\log\rho_p(v)-\rho^\bullet_p(v)\log\rho^\bullet_p(v)
-\rho^\circ_p(v)\log\rho^\circ_p(v)).
\label{entropy}
$$
\pause
Then we have to minimize the free energy
$$
F[\rho^\bullet]=E-T\cS-HS^z.
$$
\pause
This minimization leads to a set of nonlinear equations (the \emph{Yang\--Yang equations}) of the form
$$
\epsilon_p(v)+\sum_{p'}\int{dv'\over2\pi}\Phi_{pp'}(v-v')\log(1+e^{-\epsilon_{p'}(v')})={1\over T}\left(P_p(v)+{1\over N}\Delta_{p,S}(v)+pH\right),
$$
where
\vspace{-8pt}
$$
{\rho^\bullet_p(v)\over\rho_p(v)}={1\over e^{\epsilon_p(v)}+1}.
$$
All thermodynamic quantities are expressed in terms of the \emph{pseudoenergies} $\epsilon_p(v)$.

\end{frame}

\begin{frame}
\frametitle{Seminar}




\end{frame}



\end{document}
