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\begin{document}

\title[Lecture 8. Heisenberg spin chain]{Lecture 8\\Heisenberg spin chain and its scaling limit}
\date{}

\frame{\titlepage}

\begin{frame}
\frametitle{Heisenberg spin chain}

Consider a chain of $N$ spins $S=1/2$, that is, the space
\eq$$
\cH_N=\underbrace{\C^2\otimes\cdots\otimes\C^2}_N,
\label{HNspace}
$$
\pause
The Hamiltonian
\eq$$
H_{\it XYZ}=-{1\over2}\sum^N_{n=1}
\left(J_x\sigma^x_n\sigma^x_{n+1}
+J_y\sigma^y_n\sigma^y_{n+1}
+J_z\sigma^z_n\sigma^z_{n+1}\right)
\label{HXYZgen}
$$
with the periodic boundary condition $\sigma^i_{N+1}=\sigma^i_1$.
\pause
\begin{itemize}
\item XYZ chain: generic $J_i$;
\item XXZ chain: $|J_x|=|J_y|$;
\item XXX chain: $|J_x|=|J_y|=|J_z|$;
\item \emph{XY chain}: $J_z=0$. Today we will mostly consider this case.
\end{itemize}
\pause

Let $J_x=1$, $J_y=\Gamma$, $J_z=\Delta$ with $|\Gamma|\le1$. Then
\pause
\Align$$
H_{\it XYZ}
&=-{1\over2}\sum^{\smash N}_{n=1}
\Bigl(
(1+\Gamma)(\sigma^+_n\sigma^-_{n+1}+\sigma^-_n\sigma^+_{n+1})
+(1-\Gamma)(\sigma^+_n\sigma^+_{n+1}+\sigma^-_n\sigma^-_{n+1})
\notag
\\
&\quad+\Delta\sigma^z_n\sigma^z_{n+1}
\Bigr),
\label{HXYZ}
$$
where
\eq$$
\sigma^+={\sigma^x+i\sigma^y\over2}=\pMatrix{0&1\\0&0},
\qquad
\sigma^-={\sigma^x-i\sigma^y\over2}=\pMatrix{0&0\\1&0}
\label{sigmapm}
$$

\end{frame}

\begin{frame}
\frametitle{Jordan\--Wigner transformation}

The matrices $\sigma^\pm$ behave like fermions
$$
\sigma^+\sigma^-+\sigma^-\sigma^+=1,
\qquad
(\sigma^+)^2=(\sigma^-)^2=0.
$$
The operator
$$
\sigma^+\sigma^-={\sigma^z+1\over2}
$$
behaves like the fermion number operator.
\pause

\emph{But} the operators from different nodes commute: $\sigma^i_m\sigma^j_n=\sigma^j_n\sigma^i_m$. We cannot completely fermionize the model in this way.
\pause

Introduce the non\-/local operators (\emph{Jordan\--Wigner transformation})
\eq$$
a_n=\sigma^-_n\prod^{n-1}_{j=1}(-\sigma^z_j),
\qquad
a^+_n=\sigma^+_n\prod^{n-1}_{j=1}(-\sigma^z_j).
\label{JWinv}
$$
\pause
Then
$$
a^+_ma_n+a_na^+_m=\delta_{mn},
\qquad
a_ma_n+a_na_m=a^+_ma^+_n+a^+_na^+_m=0.
$$
\pause
It is invertible
\eq$$
\sigma^z_n=2a^+_na_n-1,
\qquad
\sigma^+_n=a^+_n\exp\biggl(i\pi\sum^{n-1}_{j=1}a^+_ja_j\biggr),
\qquad
\sigma^-_n=a_n\exp\biggl(-i\pi\sum^{n-1}_{j=1}a^+_ja_j\biggr).
\label{JW}
$$


\end{frame}

\begin{frame}
\frametitle{Jordan\--Wigner transformation 2}

The price payed for the transformation is a small change of the boundary condition:
\eq$$
a_{N+1}=a_1(-1)^M,
\qquad
M=\sum^N_{k=1}a_k^+a_k=S^z+{N\over2},
\label{a-boundary}
$$
which depends on the total number of fermion particles.
\pause

The Hamiltonian:
\Align$$
H_{\it XYZ}
&=-\sum^N_{n=1}
\biggl({1+\Gamma\over2}(a^+_{n+1}a_n+a^+_na_{n+1})
+{1-\Gamma\over2}(a^+_na^+_{n+1}-a_na_{n+1})
\nonumber
\\
&\quad
+2\Delta(a^+_na_na^+_{n+1}a_{n+1}-a^+_na_n)\biggr).
\label{HXYZfermion}
$$
When is this Hamiltonian solvable?
\pause
Evidently, for $\Delta=0$.


\end{frame}

\begin{frame}
\frametitle{XY model. Bogoliubov transform: XX case}

Set $\Delta=0$. Let us pass to the momentum space
\eq$$
a_n={1\over N^{1/2}}\sum_{-\pi<k\le\pi}a_ke^{ikn},
\quad
{kN\over2\pi}\in\Cases{\Z,&M\text{ even;}\\\Z+{1\over2},&M\text{ odd.}}
\label{fourier}
$$
\pause
The Hamiltonian is
\eq$$
H_{\it XY}=-\sum_{-\pi<k\le\pi}\left(
(1+\Gamma)\cos k\cdot a^+_ka_k
+i{1-\Gamma\over2}\sin k\cdot(a^+_ka^+_{-k}+a_ka_{-k})
\right).
\label{HXYak}
$$
\pause
Let $\Gamma=1$: XX model. Then the spectrum $\epsilon^{(0)}_k=-2\cos k$ divides in two regions:
\begin{itemize}
\item for $|k|>k_F={\pi\over2}$ we have $\epsilon^{(0)}_k>0$;
\item for $|k|<k_F={\pi\over2}$ we have $\epsilon^{(0)}_k<0$.
\end{itemize}
The last region must be filled up in the ground state (`Dirac sea').
\pause

Since the space of states is finite\-/dimensional, we may redefine the oscillators:
\eq$$
b_k=a_{k-\pi},
\quad
b^+_k=a^+_{k-\pi},
\quad
b'_k=ia^+_k,
\quad
b^{\prime+}_k=-ia_k,
\quad
{\blue-{\pi\over2}<k\le{\pi\over2}},
\label{Boglolyubov-Gamma=1}
$$
This is the simplest \emph{Bogoliubov transform}.
\pause
Then
$$
H_{\it XX}=\sum_{-\pi/2<k\le\pi/2}\epsilon_k(b^+_kb_k+b^{\prime+}_kb'_k),
\quad
\epsilon_k=\cos k\ge0.
$$
But there are \emph{particles} $b_k$ and \emph{antiparticles} $b'_k$.


\end{frame}

\begin{frame}
\frametitle{XY model. Bogoliubov transform: general case}

Now return to generic $\Gamma$. Consider the \emph{Bogoliubov transform}
\eq$$
\Gathered{
a_{k-\pi}=\alpha_kb_k+\beta_kb^+_{-k},
\quad
a^+_{k-\pi}=\beta^{\prime*}_kb_{-k}+\alpha^{\prime*}_kb^+_k,
\\
a_k=\alpha'_kb'_k+\beta'_kb^{\prime+}_{-k},
\quad
a^+_k=\beta^{\prime*}_kb'_{-k}+\alpha^{\prime*}_kb^{\prime+}_k,
}\label{Bogoliubov}
$$
where
$$
\Gathered{
|\alpha_k|^2+|\beta_k|^2=1,
\quad
\alpha_k\beta_{-k}+\alpha_{-k}\beta_k=0,
\\
|\alpha'_k|^2+|\beta'_k|^2=1,
\quad
\alpha'_k\beta'_{-k}+\alpha'_{-k}\beta'_k=0.
}
$$
\pause
Now we should adjust the coefficients $\alpha_k,\ldots$ to eliminate terms of the form $bb$, $b^+b^+$, $b'b'$, $b^{\prime+}b^{\prime+}$ in the Hamiltonian. We obtain
\eq$$
\Aligned{
\alpha_k
&=\cos{\kappa\over2},
&\beta_k
&=i\sin{\kappa\over2},
\\
\alpha'_k
&=-\sin{\kappa\over2},
&\beta'_k
&=i\cos{\kappa\over2},
}\qquad
\tg\kappa={1-\Gamma\over1+\Gamma}\tg k.
\label{bgsolution}
$$
\pause
The Hamiltonian takes the form
\eq$$
H_{\it XY}=\sum_{-\pi/2<k\le\pi/2}\epsilon_k(b^+_kb_k+b^{\prime+}_kb'_k),
\quad
\epsilon_k=\sqrt{(1+\Gamma)^2\cos^2k+(1-\Gamma)^2\sin^2k}.
\label{HXYbk}
$$


\end{frame}

\begin{frame}
\frametitle{XY model. Spectrum}

$$
\epsilon_k=\sqrt{(1+\Gamma)^2\cos^2k+(1-\Gamma)^2\sin^2k}
$$

$$
\hphantom{1-\Gamma=.3}\begin{tikzpicture}[scale=2]
\draw[->] (-pi/2,0) node[below] {$-{\pi\over2}$} -- (0,0) -- node[below] {$k$} (pi/2,0) node[below] {$\pi\over2$};
\draw[-] (-pi/2,0) -- (-pi/2,2) -- (pi/2,2) -- (pi/2,0);
\draw[->] (0,0) node[below] {$0$} -- (0,2) node[below left] {$\epsilon_k$};
\draw[-,thick,domain=-pi/2:pi/2,samples=200] plot function{sqrt(2.56*cos(x)**2+.16*sin(x)**2)};
\draw (0,1.6) node[above right] {$1+\Gamma$};
\draw[dashed] (pi/2,.4) -- (0,.4) node[left] {$1-\Gamma$};
\uncover<2->{\draw[-,domain=-pi/2:pi/2,samples=200,color=red] plot function{sqrt(2.89*cos(x)**2+.09*sin(x)**2)} node[right] {\small$1-\Gamma=.3$};}
\uncover<3->{\draw[-,domain=-pi/2:pi/2,samples=200,color=orange] plot function{sqrt(3.24*cos(x)**2+.04*sin(x)**2)} node[right] {\small$1-\Gamma=.2$};}
\uncover<4->{\draw[-,domain=-pi/2:pi/2,samples=200,color=green] plot function{sqrt(3.61*cos(x)**2+.01*sin(x)**2)} node[right] {\small$1-\Gamma=.1$};}
\uncover<5->{\draw[dashed,domain=-pi/2:pi/2,samples=200,color=blue] plot function{2*cos(x)} node[right] {\small$1-\Gamma=0$};}
\end{tikzpicture}
$$

\uncover<6->{
The minimum of $\epsilon_k$ is equal to
\begin{itemize}
\item $1-\Gamma$ and achieved at $k={\pi\over2}$ for $\Gamma>0$;
\item $1+\Gamma$ and achieved at $k=0$ for $\Gamma<0$.
\end{itemize}
}
\uncover<7->{Hence, the spectrum has a mass gap of $2\min(1+\Gamma,1-\Gamma)$. In the limits $\Gamma\to\pm1$ the gap disappears and the system admits scaling limit. Without loss of generality we will consider the limit $\Gamma\to1$.}


\end{frame}

\begin{frame}
\frametitle{XY model. Scaling limit}

Let $1-\Gamma\ll1$. Consider low\-/lying excitations. Let
$$
pa=\Cases{{\pi\over2}-k,&k>0;\\-{\pi\over2}-k,&k<0.}
$$
Take the limit $a\to0$ with finite $p$.
\pause

Then
\eq$$
\epsilon(p)=2a\sqrt{m^2+p^2},
\qquad m={1-\Gamma\over2a}.
\label{FFscaling}
$$
Up to a factor $2a$ this is the spectrum of a free relativistic particle. So define
$$
H_{FF}={1\over2a}H_{\it XY}.
$$
\pause

The parameter of the Bogoliubov transform
$$
\ctg\kappa={|p|\over m},
$$


\end{frame}

\begin{frame}
\frametitle{Scaling XX model}

Consider the case $m=0$ ($\Gamma=1$ or XX model). The Bogoliubov transform becomes trivial
\eq$$
\Aligned{
a_{\pi/2-pa}
&=ib^{\prime+}_{-\pi/2+pa},
&\qquad
a_{\pi/2+pa}
&=b_{-\pi/2+pa},
\\
a_{-\pi/2+pa}
&=ib^{\prime+}_{\pi/2-pa},
&\qquad
a_{-\pi/2-pa}
&=b_{\pi/2-pa}.
}\label{bglsollim}
$$
\pause
Introduce the operators
\eq$$
\psi_\pm(x)=\int^\infty_{-\infty}{dp\over2\pi}\,a_\pm(p)e^{ipx}
\qquad
a_\pm(p)=(Na)^{1/2}a_{\pm\pi/2+pa},
\label{psi_pm-def}
$$
\pause
It is easy to check that
$$
\psi(x)=\pMatrix{\psi_-(x)\\-i\psi_+(x)}
$$
satisfies the massless Dirac equation: $\hat\d\psi=0$ and satisfies the correct anticommutation equations.
\pause

The initial fermions $a_n$ are expressed in terms of them as
$$
a_n=a^{1/2}(i^n\psi_+(an)+i^{-n}\psi_-(an)).
$$


\end{frame}

\begin{frame}
\frametitle{Scaling XXZ model}

Now turn on $\Delta\ne0$, but we will think that $|\Delta|\ll1$.
\pause

Define
\eq$$
H_\Delta\equiv{1\over2a}(H_{\it XXZ}+\Delta N/2-2\Delta M)
=H_{FF}-{\Delta\over Na}\sum_q\rho_q\rho_{-q}\cos q,
\label{HXXZbk}
$$
where
\eq$$
\rho_q=\sum_ka^+_{k+q}a_{k'}
=\sum_na^+_na_n e^{iqn},
\qquad
M=\rho_0.
\label{rho_q-def}
$$
In the XXZ model the value of $M$ is conserved.
\pause
It can be shown that in the scaling limit
\eq$$
H_\Delta=H_{FF}-\Delta\int dx\,((\psi^+_+\psi_+)^2+(\psi^+_-\psi_-)^2+4\psi^+_+\psi_+\psi^+_-\psi_-).
\label{H_Delta-fin}
$$
\pause
The contributions $(\psi^+_\alpha\psi_\alpha)^2\sim(\d_x\varphi)^2$ renormalize the space coordinate~$x$. But the contribution $-4\Delta\int d^2x\,\psi^+_+\psi_+\psi^+_-\psi_-$ is of the Thirring model type. Therefore
\eq$$
g=-2\Delta
\qquad
(\Delta\ll1).
\label{gDeltapert}
$$
\pause
The exact relation for finite $\Delta$ is known from an exact solution of the XYZ model:
\eq$$
\beta^2={2\mu\over\pi},
\quad
{g\over\pi}={\pi/2-\mu\over\mu},
\quad
\Delta=-\cos\mu.
\label{gDelta}
$$


\end{frame}

\begin{frame}
\frametitle{XYZ model and sine\-/Gordon model}

$$
\boxed{\vcenter{\hbox{The \emph{XXZ}}\hbox{model}}}\ \xrightarrow{\vcenter{\hbox{scaling limit}}}\ \boxed{\vcenter{\hbox{\emph{massless}}\hbox{Thirring models}}}
\ \xleftrightarrow{\vcenter{\hbox{boson-fermion}\hbox{correspondence}}}\ \boxed{\vcenter{\hbox{free massless}\hbox{boson}}}
$$
\pause
Conjecture:
$$
\boxed{\vcenter{\hbox{The \emph{XYZ}}\hbox{model}}}\ \xrightarrow{\vcenter{\hbox{scaling limit}}}\ \boxed{\vcenter{\hbox{\emph{massive}}\hbox{Thirring models}}}
\ \xleftrightarrow{\vcenter{\hbox{boson-fermion}\hbox{correspondence}}}\ \boxed{\vcenter{\hbox{sine-Gordon}\hbox{model}}}
$$


\end{frame}

\begin{frame}
\frametitle{XYZ model and sine\-/Gordon model: symmetries}

Symmetry of the free massless boson:
$$
\varphi\to\varphi+\alpha,
\quad
\alpha\in\R.
$$
\pause
It is broken by the $\cos\beta\varphi$ interaction:
\eq$$
\alpha={2\pi n\over\beta},
\quad
n\in\Z.
\label{alphadisc}
$$
\pause
Symmetry of the XXZ model:
$$
\sigma^\pm\to e^{\pm i\lambda}\sigma^\pm,
\quad
\sigma^z\to\sigma^z,
\quad
\lambda\in\R.
$$
\pause
For $\Gamma<1$ there is a term
$$
\sum_n(\sigma^+_n\sigma^+_{n+1}+\sigma^-_n\sigma^-_{n+1}),
$$
which breaks the symmetry to
\eq$$
\lambda=\pi n,
\quad
n\in\Z.
\label{lambdadisc}
$$
\pause
Identification:
\eq$$
\alpha={2\lambda\over\beta}
$$


\end{frame}

\begin{frame}
\frametitle{XYZ model and sine\-/Gordon model: identification of operators}

Therefore
$$
\sigma^\pm_n\sigma^\pm_{n+1}\sim a^{\beta^2}e^{\pm i\beta\phi}.
$$
\pause

It can be conjectured that
$$
\sigma^\pm_n\sim a^{\beta^2/4}e^{\pm i{\beta\over2}\phi}.
$$
\pause

What is $\sigma^z$? A more accurate study gives
$$
\sigma^z_n=c_1a\d_t\phi+c_2(-1)^na^{1/\beta^2}\sin{i\tilde\phi\over\beta}.
$$

\end{frame}

\begin{frame}
\frametitle{Seminar}



\end{frame}




\end{document}
