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\begin{document}

\title[Lecture 1. Heisenberg spin chain]{Lecture 1\\Heisenberg spin chain}
\subtitle{A mini-course ``Solvable lattice models and Bethe Ansatz''\\(Ariel University, spring 2021)}
\author{Michael Lashkevich}
\institute{Landau Institute for Theoretical Physics,\\
Kharkevich Institute for Information Transmission Problems}
\date{}

\frame{\titlepage}

\begin{frame}
\frametitle{Single spin $1/2$}
\unpause

Consider a spin $S=1/2$.
\unpause
It means that any state $|\psi\rangle$ is a linear combination of `spin up' $|+\rangle=|\mathord\uparrow\rangle$ and `spin down' $|-\rangle=|\mathord\downarrow\rangle$ vectors:
$$
|\psi\rangle =\psi_+|\mathord\uparrow\rangle+\psi_-|\mathord\downarrow\rangle
=\sum_{\varepsilon=\pm\tikz[overlay] \node (varepsilonF) {};}\psi_\varepsilon|\varepsilon\rangle,
\quad
\psi_\pm\in\C.
\tikz[overlay,red,right,yshift=-.5cm] {\node[draw] (varepsilonI) {$\varepsilon=2s^z=\pm1=\pm$};
  \draw[->] (varepsilonI.west) to[out=180,in=-10] (varepsilonF);}
$$
\unpause
The space of these states is
$$
\cH_1=\C|\mathord\uparrow\rangle\oplus\C|\mathord\downarrow\rangle\cong\C^2.
$$
\unpause
In this space of states in the basis $(+-)$ act \emph{the Pauli matrices}:
\eq$$
\sigma^x=\pMatrix{0&1\\1&0},
\quad
\sigma^y=\pMatrix{0&-\i\\\i&0},
\quad
\sigma^z=\pMatrix{1&0\\0&-1}.
\label{sigmaxyz}
$$
\unpause
In other words, they act on the basic states as
\eq$$
\Aligned{
\sigma^x|\mathord\uparrow\rangle
&=|\mathord\downarrow\rangle,
&\sigma^y|\mathord\uparrow\rangle
&=\i|\mathord\downarrow\rangle,
&\sigma^z|\mathord\uparrow\rangle
&=|\mathord\uparrow\rangle,
\\
\sigma^x|\mathord\downarrow\rangle
&=|\mathord\uparrow\rangle,
&\sigma^y|\mathord\downarrow\rangle
&=-\i|\mathord\uparrow\rangle,
&\sigma^z|\mathord\downarrow\rangle
&=-|\mathord\downarrow\rangle.
}\label{sigmaxyz-ops}
$$
\unpause
It is convenient to introduce `spin raise' and `spin lower' operators:
\eq$$
\sigma^+={\sigma^x+\i\sigma^y\over2}=\pMatrix{0&1\\0&0},
\qquad
\sigma^-={\sigma^x-\i\sigma^y\over2}=\pMatrix{0&0\\1&0}
\label{sigmapm}
$$
\unpause
or
\eq$$
\Aligned{
\sigma^+|\mathord\uparrow\rangle
&=0,
&\sigma^-|\mathord\uparrow\rangle
&=|\mathord\downarrow\rangle,
\\
\sigma^+|\mathord\downarrow\rangle
&=|\mathord\uparrow\rangle,
&\sigma^-|\mathord\downarrow\rangle
&=0.
}\label{sigmapm-ops}
$$

\end{frame}

\begin{frame}
\frametitle{Spin system}
\unpause

Now consider a chain of $N$ spins $S=1/2$.
\unpause
Any state $|\psi\rangle$ is a combination
$$
\psi=\sum_{\varepsilon_1,\ldots,\varepsilon_N=\pm}\psi_{\varepsilon_1\ldots\varepsilon_N}|\varepsilon_1\ldots\varepsilon_N\rangle
\unpause
=\sum_{\varepsilon_1,\ldots,\varepsilon_N=\pm}\psi_{\varepsilon_1\ldots\varepsilon_N}|\varepsilon_1\rangle\otimes\cdots\otimes|\varepsilon_N\rangle.
$$
\unpause
Thus, the space of states is a \emph{tensor product}:
\eq$$
\cH_N=\underbrace{\C^2\otimes\cdots\otimes\C^2}_N,
\label{HNspace}
$$
\unpause
Define the Pauli matrices acting on the $n$th spin
\eq$$
\sigma^\alpha_n=1\otimes\cdots\otimes1\otimes\underbrace{\sigma^\alpha}_n\otimes1\otimes\cdots\otimes1.
\label{sigma_n-def}
$$
\unpause
Define the translation operator $T$ such that
\eq$$
T|\varepsilon_1\varepsilon_2\ldots\varepsilon_N\rangle=|\varepsilon_2\ldots\varepsilon_N\varepsilon_1\rangle.
\label{T-translation-def}
$$
\unpause
It satisfies the condition
\eq$$
T\sigma^\alpha_n=\sigma^\alpha_{n-1}T.
\label{T-translation-commut}
$$

\end{frame}

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

The Hamiltonian
\eq$$
H_{\rm XYZ}=-{1\over2}\sum^N_{n=1}
\left(J_x\sigma^x_n\sigma^x_{n+1}
+J_y\sigma^y_n\sigma^y_{n+1}
+J_z\sigma^z_n\sigma^z_{n+1}\right)
\label{HXYZgen}
$$
defines a \emph{Heisenberg spin chain}.
\unpause
The Hamiltonian is underdefined. Indeed, what is $\sigma^\alpha_{N+1}$?
\unpause
We will impose the \emph{periodic boundary condition} by assuming $\sigma^\alpha_{N+1}=\sigma^\alpha_1$.
\unpause

There are three main cases:
\begin{itemize}
\item<+->XYZ chain: generic $J_\alpha$.
\unpause
It is exactly solvable, but its solution is too tricky for introductory lectures.
\item<+->\emph{XXZ chain}: $|J_x|=|J_y|$.
\unpause
Its solutions is much simpler, but the solution is rich enough and reflects the main ideas well. It is the main subject of these lectures.
\item<+->XXX chain: $|J_x|=|J_y|=|J_z|$.
\unpause
It is too degenerate, so its direct solution has subtleties. It is simpler to consider it as a limit from the XXZ case.
\end{itemize}

With the cyclic boundary condition the Hamiltonian is translationally invariant.
\unpause
It commutes with the translation operator:
\eq$$
[H_{\rm XYZ},T]=0.
\label{H-translation-commut}
$$


\end{frame}

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

Let $J_x=1$, $J_y=\Gamma$, $J_z=\Delta$ with $|\Gamma|\le1$. Then
\unpause
\Align$$
H_{\rm XYZ}
&=-{1\over2}\sum^{\smash N}_{n=1}
\Bigl(
(1+\Gamma)(\sigma^+_n\sigma^-_{n+1}+\sigma^-_n\sigma^+_{n+1})
+(1-\Gamma)(\sigma^+_n\sigma^+_{n+1}+\sigma^-_n\sigma^-_{n+1})
\notag
\\
&\quad+\Delta\sigma^z_n\sigma^z_{n+1}
\Bigr).
\label{HXYZ}
$$
\unpause
If $\Gamma=1$, the Hamiltonian is
\eq$$
H_{\rm XXZ}=-{1\over2}\sum^{\smash N}_{n=1}\left(2(\sigma^+_n\sigma^-_{n+1}+\sigma^-_n\sigma^+_{n+1})+\Delta\sigma^z_n\sigma^z_{n+1}\right).
\label{HXXZ}
$$
\unpause
Each term either does not change the spins or flips one spin up and another spin down. Hence
\eq$$
[H_{\rm XXZ},S^z]=0,
\quad
S^z={1\over2}\sum^N_{n=1}\sigma^z_n.
\label{HXXZ-Sz-commut}
$$
\unpause
Therefore we can split the space of states into the subspaces of spin eigenvectors:
\eq$$
\cH_N=\bigoplus^N_{k=0}(\cH_N)_k,
\quad
(\cH_n)_k=\left\{|\psi\rangle\>\big|\>S^z|\psi\rangle=(N/2-k)|\psi\rangle\right\}.
\label{cHN-split}
$$
\unpause
The Hamiltonian $H_{\rm XXZ}$ acts as an operator on each of these subspaces.

\end{frame}

\begin{frame}
\frametitle{Structure of subspaces $(\cH_N)_k$}
\unpause

The space $(\cH_N)_0$ is one\-/dimensional:
\eq$$
(\cH_N)_0=\C|\Omega_+\rangle,
\quad
|\Omega_+\rangle=|\mathord\uparrow\mathord\uparrow\ldots\mathord\uparrow\rangle.
\label{cH0-def}
$$
\unpause
This state can be also defined by the condition
\eq$$
\sigma^+_n|\Omega_+\rangle=0
\quad\forall n=1,\ldots,N.
\label{Omega-def}
$$
\unpause
The space $(\cH_N)_1$ is $N$\-/dimensional:
\eq$$
(\cH_N)_1=\bigoplus_{1\le n\le N}\C|n\rangle,
\quad
|n\rangle=\sigma^-_n|\Omega_+\rangle=|\mathord\uparrow\ldots\mathord\uparrow\mathord{\mathop\downarrow\limits_n}\mathord\uparrow\ldots\mathop\uparrow\rangle.
\label{cH1-def}
$$
\unpause
The space $(\cH_N)_2$ is $N(N-1)/2$\-/dimensional:
$$
\Gathered{
(\cH_N)_2=\bigoplus_{1\le n_1<n_2\le N}\C|n_1,n_2\rangle,
\\
|n_1,n_2\rangle=\sigma^-_{n_1}\sigma^-_{n_2}|\Omega_+\rangle
=|\mathord\uparrow\ldots\mathord\uparrow\mathord{\mathop\downarrow\limits_{n_1}}\mathord\uparrow\ldots\mathop\uparrow
\mathord{\mathop\downarrow\limits_{n_2}}\mathord\uparrow\ldots\mathop\uparrow\rangle.
}\label{cH2-def}
$$
\unpause
Generally we have
$$
\Gathered{
(\cH_N)_k=\bigoplus_{1\le n_1<\cdots<n_k\le N}\C|n_1,\ldots,n_k\rangle,
\\
|n_1,\ldots,n_k\rangle=\sigma^-_{n_1}\cdots\sigma^-_{n_k}|\Omega_+\rangle
=|\mathord\uparrow\ldots\mathord\uparrow\mathord{\mathop\downarrow\limits_{n_1}}\mathord\uparrow\ldots\mathop\uparrow
\mathord{\mathop\downarrow\limits_{n_k}}\mathord\uparrow\ldots\mathop\uparrow\rangle.
}\label{cHk-def}
$$
\unpause
This space is $N\choose k$\-/dimensional.

\end{frame}

\begin{frame}
\frametitle{Eigenvectors and eigenvalues: $k=0,1$}
\unpause

Now let us find eigenvectors.
\unpause
Since $(\cH_N)_0$ is one\-/dimensional the vector $|\Omega_+\rangle$ is an eigenvector of the Hamiltonian:
\eq$$
H_{\rm XXZ}|\Omega_+\rangle=-{N\Delta\over2}|\Omega_+\rangle.
\label{H-Omega-action}
$$
\unpause
The action of $H_{\rm XXZ}$ in $(\cH_N)_1$ is
$$
H_{\rm XXZ}|n\rangle=-|n-\rangle-|n+1\rangle-(N-4){\Delta\over2}|n\rangle.
$$
\unpause
Thus the vectors $|n\rangle$ are not eigenvectors.
\unpause
Make use of the translational invariance of $H_{\rm XXZ}$. Introduce the eigenvector $|z\rangle$ of the operator $T$:
\eq$$
T|z\rangle=z|z\rangle.
\label{z-vector}
$$
\unpause
Since $T|n\rangle=|n-1\rangle$ we have a wave solution\tikz[overlay,red] \node[draw,left=5pt,yshift=-.5cm] (zI) at (current page.east) {$z=e^{\i p}$, $p$ is the quasimomentum.};
\eq$$
|z\rangle=\sum^N_{n=1}\tikz[overlay] \node[above] (zF) {};z^n|n\rangle
\tikz[overlay,red] \draw[->] (zI.west) to[bend right] (zF);
\label{z-explicit}
$$
\unpause
subject to the condition
\eq$$
z^N=1
\uncover<+->{\quad\Leftrightarrow\quad
z=e^{2\pi\i{n\over N}},
\quad
n=0,1,\ldots,N-1.}
\label{zN=1}
$$
\unpause
The state $|z\rangle$ is an eigenvector of the Hamiltonian:
\eq$$
H_{\rm XXZ}|z\rangle
=\left(-{N\Delta\over2}+\epsilon(z)\right)|z\rangle,
\qquad
\epsilon(z)=2\Delta-z-z^{-1}.
\label{HPsi1}
$$

\end{frame}

\begin{frame}
\frametitle{Three regimes}
\unpause

We may think of the state $|\Omega_+\rangle$ as of some `vacuum' and of the state $|z\rangle$ as of an `excitation' with the energy $\epsilon(z)$.
\unpause
Consider three regimes:
\begin{itemize}
\item<+->$\Delta>1$: $\epsilon(z)>0$ $\forall z$. The state $|\Omega_+\rangle$ is the ground state. The excitation are physical excitations (\emph{magnons}).
\unpause
Due to the symmetry $\sigma^z_n\rightarrow-\sigma^z_n$, $\sigma^+_n\leftrightarrow\sigma^-_n$, there is another ground state $|\Omega_-\rangle\in(\cH_N)_N$, so that the ground state of the system is doubly degenerate.
\item<+->$\Delta<-1$: $\epsilon(z)<0$ $\forall z$. The states $|\Omega_\pm\rangle$ are states of the \emph{highest} energy. The ground state corresponds to $S^z=0$ or $\pm{1\over2}$, and excited states separated by an \emph{energy gap}.
\unpause
We shall speak of the states $|\Omega_\pm\rangle$ as of \emph{pseudovacuum}, and of the excitations like $|z\rangle$ as of \emph{pseudoparticles}.
\item<+->$-1<\Delta<1$: $\epsilon(z)$ does not have definite sign. The ground state corresponds to $S^z=0$ or $\pm{1\over2}$. There is \emph{no energy gap}.
\end{itemize}
\unpause
This conclusion would follow from the conjecture that the energies of excitations are additive $E=-{N\Delta\over2}+\sum_i\epsilon(z_i)$ with $|z_i|=1$, but this conjecture \emph{is not correct}.
\unpause
The energies are indeed additive, but $z_i$ are generally not on the unit circle. In fact, there are also bound states that complicate the situation. We will see later that nevertheless the boundaries between these three regimes are exact.


\end{frame}

\begin{frame}
\frametitle{Eigenvectors and eigenvalues: $k=2$}
\unpause

Consider the case $k=2$. Let us search for an eigenstate in the form of a combination of wave solutions:
\eq$$
|z_1,z_2\rangle
=\sum_{n_1<n_2}(A_{12}z_1^{n_1}z_2^{n_2}+A_{21}z_2^{n_1}z_1^{n_2})
|n_1,n_2\rangle.
\label{Psi2}
$$
\unpause
The action of the Hamiltonian moves $n_i$ by $0,\pm1$. Thus, the action on the contributions with $n_2-n_1>1$ does not differ from the action on the one\-/particle state.
\unpause
Hence, \emph{if} the state is an eigenstate, we have
$$
H_{\rm XXZ}|z_1,z_2\rangle
=\left(-{N\Delta\over2}+\epsilon(z_1)+\epsilon(z_2)\right)
|z_1,z_2\rangle.
$$
\unpause
When is it the case? First, check the action on the terms with $n_2-n_1=1$.
\unpause
After a calculation we obtain
\eq$$
{A_{21}\over A_{12}}=S(z_1,z_2)
\equiv-{1+z_1z_2-2\Delta z_2\over 1+z_1z_2-2\Delta z_1}.
\label{S12def}
$$
\unpause
Second, we have to impose the periodicity condition:
\eq$$
z_1^NS(z_1,z_2)=1,
\quad
z_2^NS(z_2,z_1)=1.
\label{k2betheeqs}
$$
\unpause
These two equations determine the allowed values of the pairs $(z_1,z_2)$. Thus, though the total energy looks additive, the sets of allowed quasimomenta of excitations are different in the cases of one\-/particle and of two\-/particle states.

\end{frame}

\begin{frame}
\frametitle{Eigenvectors: Bethe Ansatz and Bethe equations}
\unpause

Consider general $k$. The \emph{Bethe Ansatz} is
$$
|z_1,\ldots,z_k\rangle=\sum_{n_1<\ldots<n_k}\sum_{\sigma\in S_k}
A_{\sigma_1\ldots\sigma_k}\prod_{j=1}^k z_{\sigma_j}^{n_j}
|n_1,\ldots,n_k\rangle.
$$
\unpause
It is an eigenvector of the Hamiltonian, if (1)
\eq$$
{A_{\ldots ji\ldots}\over A_{\ldots ij\ldots}}=S(z_i,z_j)
\label{AjiAijratio}
$$
\unpause
and (2) the \emph{Bethe equations}
\eq$$
z_i^N\prod_{j,\ j\ne i}S(z_i,z_j)=1
\label{periodicity}
$$
are satisfied.
\unpause
The eigenvalue is given by
\eq$$
H_{\rm XXZ}|\Psi_k(z_1,\ldots,z_k)\rangle
=\left(-{N\Delta\over2}+\sum_{i=1}^k\epsilon(z_i)\right)
|\Psi_k(z_1,\ldots,z_k)\rangle,
\label{HPsik}
$$


\end{frame}


\end{document}
