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\begin{document}

\title[Lecture 11. Kondo problem: Bethe Ansatz]{Lecture 11\\Kondo problem: derivation of the Bethe Ansatz}
\date{}
\author{Michael Lashkevich}

\frame{\titlepage}

\begin{frame}
\frametitle{Kondo effect}
Beginning in the 1930s, a minimum was observed in the temperature dependence of the resistivity of some seemingly pure metals (Au, Ag, Cu) at low temperatures.
$$
\includegraphics[height=3cm]{Classickondo.png}
$$
\pause
Later it turned out that the anomaly is caused by the presence of a low concentration of impurity atoms of transition metals (Mn, Fe, Cr, Co, Ce, Y).
\pause
Jun Kond\=o (1964) explained this phenomenon by electron scattering on impurities described by the interaction ({\it $sd$ model\/})
\eq$$
V=J\sum_i\bsigma\bS_i\delta(\br-\bR_i)
\label{sdinteraction}
$$

\end{frame}

\begin{frame}
\frametitle{$sd$ Model}

In the first (Born) approximation the scattering amplitude is
$$
f^{(1)}_{\sigma'\sigma}\sim J(\bsigma\bS)_{\sigma'\sigma}.
$$
It is of the same order as the potential scattering and does not change the temperature behavior.
\pause

In the second order we have two processes:
$$
\begin{tikzpicture}[scale=.5,baseline=-2pt,>={Stealth[length=1mm]},decoration={markings,mark=at position .55 with {\arrow{>}}}]
\begin{scope}[color=blue]
\draw[-,postaction=decorate] (-3,-.5) node[left] {\black$\bp,\sigma$} -- node[above] {$e$} (-1,-.5);
  \draw[-,postaction=decorate] (-1,-.5) -- node[above,sloped] {$e$} (1,.5);
  \draw[-,postaction=decorate] (1,.5) -- node[above] {$e$} (3,.5) node[right] {\black$\bp',\sigma'$};
\end{scope}
\draw[fill] (-1,-.5) circle (2pt);
\draw[fill] (1,.5) circle (2pt);
\end{tikzpicture}
\quad\text{and}\quad
\begin{tikzpicture}[scale=.5,baseline=-2pt,>={Stealth[length=1mm]},decoration={markings,mark=at position .55 with {\arrow{>}}}]
\begin{scope}[color=blue]
\draw[-,postaction=decorate] (-3,.5) node[left] {\black$\bp,\sigma$} -- node[above] {$e$} (1,.5);
  \draw[-,postaction=decorate,color=red] (-1,-.5) -- node[above,sloped,pos=.3] {$h$} (1,.5);
  \draw[-,postaction=decorate] (-1,-.5) -- node[above] {$e$} (3,-.5) node[right] {\black$\bp',\sigma'$};
\end{scope}
\draw[fill] (-1,-.5) circle (2pt);
\draw[fill] (1,.5) circle (2pt);
\end{tikzpicture}
$$
\pause
The corresponding amplitude is $f^{(2)}_{\sigma'\sigma}\sim{}$
$$\hspace{-.5cm}
J^2\sum_{\sigma''}\int{d^3p''\over(2\pi)^3}\,{(\bsigma\bS)_{\sigma'\sigma''}(\bsigma\bS)_{\sigma''\sigma}(1-f(\bp''))\over\epsilon_\bp-\epsilon_{\bp''}}
-J^2\sum_{\sigma''}\int{d^3p''\over(2\pi)^3}\,{(\bsigma\bS)_{\sigma''\sigma}(\bsigma\bS)_{\sigma'\sigma''}f(\bp'')\over\epsilon_{\bp''}-\epsilon_{\bp'}}.
$$
\pause
Taking into account
$$
\Aligned{
\sigma^iS^i\sigma^jS^j
&=S(S+1)-\bsigma\bS,
\\
\sigma^iS^j\sigma^jS^i
&=S(S+1)+\bsigma\bS,
}
$$
\pause
we obtain the integral
$$
f^{(2)}_{\sigma'\sigma}\sim J^2\dashint{d^3p''\over(2\pi)^3}\,
\left({S(S+1)\delta_{\sigma'\sigma}\over\epsilon_\bp-\epsilon_{\bp''}}
+{2f(\bp'')-1\over\epsilon_\bp-\epsilon_{\bp''}}(\bsigma\bS)_{\sigma'\sigma}\right),
$$
The second term diverges on the Fermi surface at $T=0$.

\end{frame}

\begin{frame}
\frametitle{$sd$ Model}

Integration gives the amplitude proportional to
\eq$$
f_{\sigma'\sigma}\sim J(\bsigma\bS)_{\sigma'\sigma}\left(1+J\rho(\eF)\log{\eF\over\max(|\epsilon_\bp-\eF|,T)}\right).
\label{secondorder}
$$
\pause
For the resistivity it means
$$
\rho=\rho_v+\rho^{(0)}_J\left(1+2J\rho(\eF)\log{\eF\over T}\right).
$$
where $\rho_v$ is the potential contribution.
\pause

This formula contains an energy scale called the \emph{Kondo temperature}:
\eq$$
T_K\sim\eF e^{-1/J\rho(\eF)}.
\label{KondoTemperature}
$$
This is the only characteristic energy scale of the Kondo effect.
\pause

The formula (\ref{secondorder}) is applicable, if $T\gg T_K$. How to improve the result?
\pause

Abrikosov (1965) and Suhl (1965) independently summed up the diagrams that contain powers of logarithm and found
$$
\rho=\rho_v+{\rho^{(0)}_J\over
\left(1-J\rho(\eF)\log{\eF\over T}\right)^2}.
$$
\pause
This formula makes it possible to approach $T_K$ closer, but it has a singularity at $T=T_K$. We need a nonperturbative approach at $T\lesssim T_K$.

\end{frame}

\begin{frame}
\frametitle{Kondo effect. Characteristic features}

Not only the resistivity has anomalies due to the Kondo effect, but also thermodynamic quantities.
\pause

Experimental data shows that for $T\gg T_K$ the impurity contribution reads
$$
\Aligned{
\rho_{\text{imp}}(T)
&\simeq{\const\over\log^2{T\over T_K}},
\\
C_{\text{imp}}(T)
&\simeq{\const\over\log^2{T\over T_K}},
\\
\chi_{\text{imp}}(T)
&\simeq{\const\over T\log{T\over T_K}}.
}
$$
\pause
For $T\ll T_K$ it reads
$$
\Aligned{
\rho_{\text{imp}}(T)
&=\rho_{\text{imp}}(0)\biggl(
1-\kappa_R\left(T\over T_K\right)^2+\ldots
\biggr),
\\
C_{\text{imp}}(T)
&=\gamma{T\over T_K}\biggl(
1-\kappa_C\left(T\over T_K\right)^2+\ldots
\biggr),
\\
\chi_{\text{imp}}(T)
&=\chi_0\biggl(
1-\kappa_\chi\left(T\over T_K\right)^2+\ldots
\biggr),
}
$$
where $\kappa_R$, $\kappa_C$, $\kappa_\chi$ are quantities of order one.

\end{frame}

\begin{frame}
\frametitle{$sd$ Model. Simplifications}

I will describe a solution to the $sd$ model, which is correct under rather restrictive assumptions, but exact. In fact, finite temperatures nor transport properties will not be discussed here in any detail.
\pause

Consider an impurity in a metal, such that
\begin{itemize}
\item the metal is described by an isotropic Fermi gas of electrons;
\pause
\item the impurity is described by a point\-/like spin with isotropic exchange interaction and no potential interaction with the electron gas;
\pause
\item $J\rho(\epsilon_F)\ll1$.
\end{itemize}
\pause
The Hamiltonian:
\eq$$
H=H_0+J\bsigma(0)\bS,
\qquad
\label{sdHamiltonian}
$$
\pause
where
\eq$$
\Aligned{
H_0
&=\sum_\sigma\int d^3x\,\psi^+_\sigma(x)\epsilon(\nabla)\psi_\sigma(x)
=\sum_{\bp\sigma}\epsilon_\bp c^+_{\bp\sigma}c_{\bp\sigma},
\\
\bsigma(0)
&=\sum_{\sigma'\sigma}\psi^+_{\sigma'}(0)\bsigma_{\sigma'\sigma}\psi_\sigma(0)
=\sum_{\bp'\sigma',\bp\sigma}
c^+_{\bp'\sigma'}\bsigma_{\sigma'\sigma}c_{\bp\sigma}.
}\label{sdHam-comps}
$$
\pause
We will also assume that
\begin{itemize}
\item the spectrum is nearly linear: $\epsilon_\bp=\eF+v_F(p-p_F)$.
\end{itemize}

\end{frame}

\begin{frame}
\frametitle{Reduction to a one\-/dimensional model}

Decompose the creation\-/annihilation operators into spherical functions:
\eq$$
c^+_{\bp\sigma}=\sum_{lm}Y_{lm}(\bp/p)c^+_{plm\sigma}.
\label{cpangexp}
$$
\pause
Let us make the substitution ($N$ being the number of electrons)
$$
H\to H+N\epsilon_F,
\quad
\epsilon_p\to\epsilon_p+\epsilon_F,
\quad
p\to p+p_F,
$$
and choose measure units so that $v_F=1$.
\pause
Then
\eq$$
H=\sum_{plm\sigma}pc^+_{plm\sigma}c_{plm\sigma}
+J\smashoperator{\sum_{p'\sigma',p\sigma}}c^+_{p'00\sigma'}c_{p00\sigma}
\bsigma_{\sigma'\sigma}\bS.
\label{Hangexp}
$$
\pause
Notice, that electrons with nonzero angular momenta do not interact with the impurity and, therefore, do not contribute the Kondo effect. Get rid of them:
\eq$$
H=\sum_{p\sigma}pc^+_{p\sigma}c_{p\sigma}
+J\smashoperator{\sum_{p'p\sigma'\sigma}}c^+_{p'\sigma'}c_{p\sigma}
\bsigma_{\sigma'\sigma}\bS,
\label{Hl0}
$$
\pause
Make a Fourier transform
\eq$$
c(x)=\pMatrix{c_+(x)\\c_-(x)}
=\sum_pe^{ipx}\pMatrix{c_{p+}\\c_{p-}}.
\label{c(x)-def}
$$
\pause
Then
\eq$$
H=\int dx\,(-ic^+(x)\d_xc(x)
+Jc^+(x)(\bsigma\bS)c(x)\delta(x)).
\label{Hcoord}
$$
We have a \emph{one\-/dimensional} Hamiltonian.
\pause
Particles move with the same velocity from left to right.
\pause
The $x<0$ semiaxis corresponds to the falling waves, while the $x>0$ semiaxis corresponds to the diverging waves.

\end{frame}

\begin{frame}
\frametitle{Many\-/particle states}

Since the Hamiltonian preserves the number of particles $N$ (now it will be $s$ electrons only), define the $N$\-/particle states.
\pause
First define the vacuum
\eq$$
c_\sigma(x)|\Omega\rangle=S^+|\Omega\rangle=0,
\quad
S^\pm=S^x\pm iS^y.
\label{Omegadef}
$$
\pause
Then define the states
\vspace{-8pt}
\Multline$$
|\Psi_N\rangle
=\int dx_1\ldots dx_N\,\sum_{\sigma_1\ldots\sigma_N}\sum^S_{s=-S}
\Psi^{\sigma_1\ldots\sigma_N,s}(x_1,\ldots,x_N)
\\
\times c^+_{\sigma_1}(x_1)\ldots c^+_{\sigma_N}(x_N)(S^-)^{S-s}|\Omega\rangle.
\label{Psidef}
$$
\pause
The action of the Hamiltonian on the wave function is
\eq$$
\hat H_N\Psi^{\sigma_1\ldots\sigma_N,s}
=-i\sum^N_{j=1}\d_{x_j}\Psi^{\sigma_1\ldots\sigma_N,s}
+J\sum^N_{j=1}\sum_{\sigma'_j,s'}
\delta(x_j)\bsigma_{\sigma_j\sigma'_j}\bS_{ss'}
\Psi^{\sigma_1\ldots\sigma'_j\ldots\sigma_N,s'}
\label{HcoordN}
$$
\vspace{-12pt}

\pause
Consider the case $N=1$. Look for the wave function in the form
\eq$$
\Psi^{\sigma,s}_p(x)
=\Cases{
A^{\sigma,s}_pe^{ipx},
&x<0,
\\
B^{\sigma,s}_pe^{ipx},
&x>0.
}
\label{Psi1}
$$
\pause
Substituting it to the Schrödinger equation, we obtain
\eq$$
A^{\sigma,s}_p=\sum_{\sigma',s'}R^{\sigma s}_{\sigma's'}B^{\sigma',s'},
\qquad
R=e^{iJ\bsigma\bS}.
\label{Rmat1}
$$

\end{frame}

\begin{frame}
\frametitle{$N=2$ states}

Consider now the case $N=2$.

Let $\Psi^{\sigma_1\sigma_2,s}(x_1,x_2)=A^{\sigma_1\sigma_2,s}e^{ip_1x_1+ip_2x_2}-A^{\sigma_2\sigma_1,s}e^{ip_1x_2+ip_2x_1}$, $x_1,x_2<0$.
\pause
$$
\begin{tikzpicture}
\draw[->] (-3,0) -- (3,0) node[below] {$x_1$};
\draw[->] (0,-2) -- (0,3) node[left] {$x_2$};
\draw (-1,-1) node (Psi) {$\Psi_{120}$};
\uncover<3->{
  \draw (1.5,-1) node (R10Psi) {$R_{10}\Psi_{120}$};
  \draw[->,color=blue] (Psi) to[bend right] (R10Psi);
}
\uncover<4->{
  \draw (2,.5) node (R20R10Psi) {$R_{20}R_{10}\Psi_{120}$};
  \draw[->,color=blue] (R10Psi) to[bend right] (R20R10Psi);
}
\uncover<5->{
  \draw (-1,1.5) node (R20Psi) {$R_{20}\Psi_{120}$};
  \draw[->,color=green!50!gray] (Psi) to[bend left] (R20Psi);
}
\uncover<6->{
  \draw (1,2) node (R10R20Psi) {$R_{10}R_{20}\Psi_{120}$};
  \draw[->,color=green!50!gray] (R20Psi) to[bend left] (R10R20Psi);
}
\uncover<8->{
  \draw[dashed,color=red] (0,0) -- node[pos=.7] (cutline) {} (3,3);
  \draw (3,1.8) node (cut) {\rlap{discontinuity}};
  \draw[->] (cut) to[bend left] (cutline);
}
\uncover<12->{
  \draw[dashed,color=red] (-2,-2) -- (0,0);
}
\end{tikzpicture}
$$
\pause\pause\pause\pause\pause
But {\red$R_{20}R_{10}\ne R_{10}R_{20}$}!
\pause\pause
Is it possible?
\pause
\emph{Yes}. For $x_1,x_2\ne0$ the Schrödinger equation reads
\vspace{-8pt}
$$
E\Psi(x_1,x_2)=-i(\d_{x_1}+\d_{x_2})\Psi(x_1,x_2)
\pause\quad\Rightarrow\quad
\Psi(x_1,x_2)=e^{iE{x_1+x_2\over2}}f(x_1-x_2)
$$
with an \emph{arbitrary} function $f$, including \emph{piecewise smooth}.

\end{frame}

\begin{frame}
\frametitle{$N=2$ states}

Thus, we will search for the wave functions in the form
$$
\Psi^{\sigma_1\sigma_2,s}(x_1,x_2)=\Cases{
  A^{\sigma_1\sigma_2,s}_{12,\ve_1\ve_2}e^{ip_1x_1+ip_2x_2}-A^{\sigma_2\sigma_1,s}_{21,\ve_2\ve_1}e^{ip_2x_1+ip_1x_2},
  &x_1<x_2;
  \\
  A^{\sigma_1\sigma_2,s}_{21,\ve_1\ve_2}e^{ip_1x_1+ip_2x_2}-A^{\sigma_2\sigma_1,s}_{12,\ve_2\ve_1}e^{ip_2x_1+ip_1x_2},
  &x_1>x_2,
}
$$
where $\ve_i=\sign x_i$. There are six groups of coefficients: $A_{12,--}$, $A_{12,-+}$, $A_{12,++}$, $A_{21,--}$, $A_{21,+-}$, $A_{21,++}$.
\pause
The scattering on the impurity imposes the relations
$$
A^{\sigma_1\sigma_2,s}_{ij,-\,\ve_2}=\sum_{\sigma'_1,s'}R^{\sigma_1s}_{\sigma'_1s'}A^{\sigma'_1\sigma_2,s'}_{ij,+\,\ve_2},
\quad
A^{\sigma_1\sigma_2,s}_{ij,\ve_1\,-}=\sum_{\sigma'_2,s'}R^{\sigma_2s}_{\sigma'_2s'}A^{\sigma_1\sigma'_2,s'}_{ij,\ve_1\,+}.
$$
\pause
The scattering of electrons is very special. Due to this map from the 3D space to a line, the unit $S$ matrix in 3D maps to the transposition matrix~$P$ in~1D:
$$
A_{21,\ve\ve}=PA_{12,\ve\ve}
$$
\pause
or, with all superscripts,
$$
A^{\sigma_1\sigma_2,s}_{ij,\ve\ve}=A^{\sigma_2\sigma_1,s}_{ji,\ve\ve},\quad\ve=\pm.
$$
\pause

On the other hand, we may write a trivial identity
\eq$$
P_{12}R_{10}R_{20}=R_{20}R_{10}P_{12},
\label{YBconst}
$$
\pause
which have the form of the Yang\--Baxter equation.

\end{frame}

\begin{frame}
\frametitle{Formal Yang\--Baxter equation}

Indeed, rewrite the equation
\eq$$
P_{12}R_{10}R_{20}=R_{20}R_{10}P_{12}
\tag{\ref{YBconst}}
$$
graphically.
\pause
If we assume
$$
R_{10}=\tikz[baseline=-2pt,scale=.5]{
  \draw[->] (-.5,-.5) node[below left] {$1$} -- (.5,.5);
  \draw[->,dashed] (.5,-.5) node[below right] {$0$} -- (-.5,.5);
}
\quad
P_{12}=\tikz[baseline=-2pt,scale=.5]{
  \draw[->] (-.5,-.5) node[below left] {$1$} -- (.5,.5);
  \draw[->] (.5,-.5) node[below right] {$2$} -- (-.5,.5);
}
=\tikz[baseline=-2pt,scale=.5]{
  \draw[->] (-.5,-.5) node[below left] {$1$} -- (-.2,-.2) to[out=45,in=-45] (-.2,.2) -- (-.5,.5);
  \draw[->] (.5,-.5) node[below right] {$2$} -- (.2,-.2) to[out=135,in=-135] (.2,.2) -- (.5,.5);
}
$$
\pause
then eq.~(\ref{YBconst}) reads
$$
\begin{tikzpicture}[baseline=-2pt,scale=.5]
\draw[->] (-1,-.5) -- (1,.5);
\draw[->] (-1,.5) -- (1,-.5);
\draw[->,dashed] (.5,-1) -- (.5,1);
\end{tikzpicture}
\quad=\quad
\begin{tikzpicture}[baseline=-2pt,scale=.5]
\draw[->] (-1,-.5) -- (1,-.5);
\draw[->] (-1,.5) -- (1,.5);
\draw[->,dashed] (0,-1) -- (0,1);
\end{tikzpicture}
\quad=\quad
\begin{tikzpicture}[baseline=-2pt,scale=.5]
\draw[->] (-1,-.5) -- (1,.5);
\draw[->] (-1,.5) -- (1,-.5);
\draw[->,dashed] (-.5,-1) -- (-.5,1);
\end{tikzpicture}
$$

\end{frame}

\begin{frame}
\frametitle{Periodic boundary condition}

Impose the periodic boundary condition
\eq$$
\Psi(x_1,\ldots,x_j,\ldots,x_N)
=\Psi(x_1,\ldots,x_j+L,\ldots,x_N).
\label{cyclic}
$$
In the physical 3D space it means that we have a spherical mirror of the radius $L/2$ with an appropriate boundary condition.
\pause

Introduce the operator
\eq$$
T_j=P_{jj-1}\ldots P_{j1}R_{j0}P_{jN}\ldots P_{jj+1}.
\label{trmatdef}
$$
\pause
Then the periodic boundary condition reads
\eq$$
e^{ip_jL}\Psi=T_j\Psi.
\label{cyclictrmat}
$$
\pause
Example: $N=2$. Let $x_1<0<x_2<x_1+L$. Then
$$
\Aligned{
\Psi^{\sigma_1\sigma_2,s}(x_1,x_2)
&=A^{\sigma_1\sigma_2,s}_{12,-+}e^{ip_1x_1+ip_2x_2}-A^{\sigma_2\sigma_1,s}_{21,+-}e^{ip_2x_1+ip_1x_2},
\\
\Psi^{\sigma_1\sigma_2,s}(x_1+L,x_2)
&=e^{ip_1L}A^{\sigma_1\sigma_2,s}_{21,++}e^{ip_1x_1+ip_2x_2}-e^{ip_2L}A^{\sigma_2\sigma_1,s}_{12,++}e^{ip_2x_1+ip_1x_2}.
}
$$
\pause
By comparing the first terms we obtain
$$
e^{ip_1L}A^{\sigma_1\sigma_2,s}_{21,++}=A^{\sigma_1\sigma_2,s}_{12,-+}=R^{\sigma_1,s}_{\sigma'_1,s'}A^{\sigma'_1\sigma_2,s}_{12,++}
=R^{\sigma_1,s}_{\sigma'_1,s'}A^{\sigma_2\sigma'_1,s}_{21,++}=(R_{10}P_{12}A_{21,++})^{\sigma_1\sigma_2,s}.
$$
Comparing the second terms give the same result up to the permutation $1\leftrightarrow2$.

\end{frame}

\begin{frame}
\frametitle{$T_j$ operators}

Now rewrite the operators $T_j$. We have
\vspace{-8pt}
\Multline$$
T_j=
\begin{tikzpicture}[baseline=-2pt,scale=.5]
\draw[-] (0,-1) node[below] {$j$} -- (0,-.2) to[out=90,in=180] (.2,0) -- (1.5,0); \draw[dotted] (1.5,0) -- (2.5,0); \draw[-] (2.5,0) -- (5.5,0);
  \draw[dotted] (5.5,0) -- (6.5,0); \draw[->] (6.5,0) -- (7.8,0) to[out=0,in=-90] (8,.2) -- (8,1);
\draw[->] (1,-1) node[below] {$j+1$} -- (1,1);
\draw[->] (3,-1) node[below] {$N$} -- (3,1);
\draw[->,dashed] (4,-1) node[below] {$0$} -- (4,1);
\draw[->] (5,-1) node[below] {$1$} -- (5,1);
\draw[->] (7,-1) node[below] {$j-1$} -- (7,1);
\end{tikzpicture}
\ =\
\begin{tikzpicture}[baseline=-2pt,scale=.5]
\draw[-] (0,-1) node[below] {$j$} -- (0,-.2) to[out=90,in=180] (.2,0) -- (1.5,0); \draw[dotted] (1.5,0) -- (2.5,0); \draw[->] (2.5,0) -- (3.5,0);
\draw[->] (1,-1) node[below] {$j+1$} -- (1,1);
\draw[->] (3,-1) node[below] {$N$} -- (3,1);
\begin{scope}[xshift=-8cm]
  \draw[-] (3.5,0) -- (5.5,0); \draw[dotted] (5.5,0) -- (6.5,0); \draw[->] (6.5,0) -- (7.8,0) to[out=0,in=-90] (8,.2) -- (8,1);
  \draw[->,dashed] (4,-1) node[below] {$0$} -- (4,1);
  \draw[->] (5,-1) node[below] {$1$} -- (5,1);
  \draw[->] (7,-1) node[below] {$j-1$} -- (7,1);
\end{scope}
\end{tikzpicture}
\\
=\
\begin{tikzpicture}[baseline=-2pt,scale=.5]
\draw[-] (-4.5,0) -- (-2.5,0); \draw[dotted] (-2.5,0) -- (-1.5,0); \draw[-] (-1.5,0) -- (1.5,0); \draw[dotted] (1.5,0) -- (2.5,0);
  \draw[->] (2.5,0) -- (3.5,0);
\draw[->,dashed] (-4,-1) node[below] {$0$} -- (-4,1);
\draw[->] (-3,-1) node[below] {$1$} -- (-3,1);
\draw[->] (-1,-1) node[below] {$j-1$} -- (-1,1);
\draw[->] (0,-1) node[below] {$j$} -- (0,1);
\draw[->] (1,-1) node[below] {$j+1$} -- (1,1);
\draw[->] (3,-1) node[below] {$N$} -- (3,1);
\end{tikzpicture}
\ =T=\tr_\aux(P_{\aux N}\ldots P_{\aux1}R_{\aux0})
\label{trmat}
$$
\pause
How to diagonalize the matrix~$T$? We want to immerse it into a set of commuting transfer matrices $T(u)$, so that $T=T(0)$.
\pause

To do it let us recall the trivial identity
\eq$$
P_{12}R_{10}R_{20}=R_{20}R_{10}P_{12}
\tag{\ref{YBconst}}
$$
and try to deform it.

\end{frame}

\begin{frame}
\frametitle{Matrices $R(u)$ and $S(u)$: requirements}

Let us find the matrices $R(u)$ and $S(u)$, so that they satisfy the following requirements:
\pause
1.~The matrices $R(u)$ and $S(u)$ satisfy the Yang\--Baxter equation:
\subeq{
\Gather$$
S_{12}(u_1-u_2)R_{10}(u_1-u_0)R_{20}(u_2-u_0)
=R_{20}(u_2-u_0)R_{10}(u_1-u_0)S_{12}(u_1-u_2),
\label{SRR}
\\
S_{12}(u_1-u_2)S_{13}(u_1-u_3)S_{23}(u_2-u_3)
=S_{23}(u_2-u_3)S_{13}(u_1-u_3)S_{12}(u_1-u_2).
\label{SSS}
$$}
\pause

2.~At special points, the matrices $S(u)$ and $R(u)$ coincide with $S$ and $R$:
\eq$$
S(0)=P,
\qquad
R(1)=R=e^{iJ\bsigma\bS}.
\label{specialpoints}
$$
\pause

3.~$R$\-/matrices satisfy the unitarity condition:
\eq$$
S_{12}(u)S_{21}(-u)=1,
\qquad
R_{10}(u)R_{10}(-u)=1.
\label{unitarity}
$$
\pause

If we obtain such matrices, we will have a family of transfer matrices
\eq$$
T(u)=\tr_\aux L_\aux(u),
\qquad
L_\aux(u)=S_{\aux N}(u)\ldots S_{\aux 1}(u)R_{\aux0}(u+1),
\label{Tudef}
$$
such that
\eq$$
T(0)=T,
\qquad
[T(u),T(v)]=0.
\label{Tucommut}
$$

\end{frame}

\begin{frame}
\frametitle{Matrices $R(u)$ and $S(u)$: solution}

The solution can be represented as
\eq$$
\Aligned{
S_{12}(u)
&=w_0(u)+w(u)\bsigma_1\bsigma_2,
\\
R_{10}
&=w'_0(u)+2w'(u)\bsigma_1\bS_0.
}\label{SRform}
$$
\pause
It is convenient to introduce the notation
\eq$$
\Aligned{
a
&=w_0+w,
&\quad
b
&=w_0-w,
&\quad
c
&=2w,
\\
a'
&=w'_0+w',
&\quad
b'
&=w'_0-w',
&\quad
c'
&=2w'.
}\label{abcdef}
$$
\pause
In this case, the matrix $S(u)$ has the same form as the $R$\-/matrix of the XXZ model:
$$
S(u)=\pMatrix{a(u)&&&\\&b(u)&c(u)&\\&c(u)&b(u)&\\&&&a(u)}.
$$
\pause
By solving the Young\--Baxter equation, we find
\eq$$
\Aligned{
{b(u)\over a(u)}
&={b'(u)\over a'(u)}
={u\over u+ig},
\\
{c(u)\over a(u)}
&={c'(u)\over a'(u)}
={ig\over u+ig},
}\label{bacaratios}
$$
i.e.\ $S(u)$ is nothing but the $R$\-/matrix of the XXX model.

\end{frame}

\begin{frame}
\frametitle{Matrices $R(u)$ and $S(u)$: solution}

Impose the unitarity condition
\eq$$
a(u)a(-u)=1,
\qquad
a'(u)a'(-u)={g^2+u^2\over g^2(S+1/2)^2+u^2}.
\label{aunitdef}
$$
\pause
Finally, the condition (\ref{specialpoints}) gives
\eq$$
a(0)=1,\qquad
a'(1)={1+ig\over2}(e^{iJS}+e^{-iJ(S+1)})
\label{aspecial}
$$
and
\eq$$
g={1\over S+1/2}\tg J(S+1/2).
\label{gdef}
$$
Otherwise, $a(u)$, $a'(u)$ are arbitrary functions.

\end{frame}

\begin{frame}
\frametitle{Algebraic Bethe Ansatz}

Return to the definitions
\eq$$
T(u)=\tr_\aux L_\aux(u),
\qquad
L_\aux(u)=S_{\aux N}(u)\ldots S_{\aux 1}(u)R_{\aux0}(u+1),
\tag{\ref{Tudef}}
$$
\pause
The $L$ operator satisfy the relation
\eq$$
S_{\tilde1\tilde2}(u_1-u_2)L_{\tilde1}(u_1)L_{\tilde2}(u_2)
=L_{\tilde2}(u_2)L_{\tilde1}(u_1)S_{\tilde1\tilde2}(u_1-u_2).
\label{SLL}
$$
\pause
and has the matrix form in the auxiliary space $\aux$:
\eq$$
L(u)=\pMatrix{A(u)&B(u)\\C(u)&D(u)},
\label{ABCDdef}
$$
\pause
Hence, we can apply the algebraic Bethe Ansatz. Define the pseudovacuum $|\Omega_N\rangle$:
\eq$$
C(u)|\Omega_N\rangle=0.
\label{OmegaNdef}
$$
\pause
We have
$$
\Aligned{
A(u)|\Omega_N\rangle
&=\Lambda_A(u)|\Omega_N\rangle,
\\
D(u)|\Omega_N\rangle
&=\Lambda_D(u)|\Omega_N\rangle,
}
$$
\eq$$
\Aligned{
\Lambda_A(u)
&=((S+1/2)a'(u+1)-(S-1/2)b'(u+1))a^N(u),
\\
\Lambda_D(u)
&=((S+1/2)b'(u+1)-(S-1/2)a'(u+1))b^N(u).
}\label{LambdaADdef}
$$
\pause
The Bethe Ansatz has the form
\eq$$
|u_1,\ldots,u_n\rangle=B(u_1)\ldots B(u_n)|\Omega_N\rangle,
\qquad
S^z=N/2+S-n.
\label{ABA}
$$

\end{frame}

\begin{frame}
\frametitle{Bethe equations}

The Bethe equations are written in the standard form
\eq$$
{\Lambda_D(u_i)\over\Lambda_A(u_i)}
=\prod_{\scriptstyle j=1\atop\scriptstyle j\ne i}^n
{a(u_j-u_i)b(u_i-u_j)\over b(u_j-u_i)a(u_i-u_j)}.
\label{Betheeq}
$$
\pause
The eigenvalues of $T(u)$ are given by
\eq$$
\Lambda(u;u_1,\ldots,u_N)
=\Lambda_A(u)\prod_{i=1}^n{a(u_i-u)\over b(u_i-u)}
+\Lambda_D(u)\prod_{i=1}^n{a(u-u_i)\over b(u-u_i)}.
\label{Lambda}
$$
\pause
Taking $u=0$ we obtain
\vspace{-4pt}
\eq$$
e^{ip_jL}=\Lambda(0;u_1,\ldots,u_N)=\Lambda_A(0)\prod_{i=1}^n{a(u_i)\over b(u_i)}.
\label{cyclicLambda}
$$
\pause
It is convenient to use variables $v_j$
$$
u_j=g(v_j-i/2).
$$
\pause
Explicitly, the system of Bethe equations have the form
\eq$$
\left(v_i+i/2\over v_i-i/2\right)^N
{v_i+iS+1/g\over v_i-iS+1/g}
=-\prod^n_{j=1}{v_i-v_j+i\over v_i-v_j-i},
\label{Bethefin}
$$
\eq$$
e^{ip_jL}=e^{iJS}\prod^n_{i=1}{v_i+i/2\over v_i-i/2}.
\label{cyclicfin}
$$
This reduces the solution of the Kondo problem to the joint solution of the equations (\ref{Bethefin}) and~(\ref{cyclicfin}).

\end{frame}

\begin{frame}
\frametitle{Seminar}




\end{frame}


\end{document}
