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\begin{document}

\title[Lecture 2. Bosonization of the Thirring model]{Lecture 2.\\Bosonization of the Thirring model}
\date{}
\author{Michael Lashkevich}

\frame{\titlepage}

\begin{frame}
\frametitle{Thirring model}
\unpause

The {\it massive Thirring model\/} in the Minkowski space:
\eq$$
S^{MT}[\psi,\bpsi]
=\int d^2x\,
\left(\bpsi(i\gamma^\mu\d_\mu-m)\psi-{g\over2}(\bpsi\gamma^\mu\psi)^2\right).
\label{MT}
$$
\unpause
Here $\psi(x)$, $\bpsi(x)$ are the Dirac fermion field and its Dirac conjugate. The $\gamma$ matrices satisfy the relations
$$
\gamma^\mu\gamma^\nu+\gamma^\nu\gamma^\mu=2g^{\mu\nu},
\qquad
\gamma^{\mu+}=\gamma^0\gamma^\mu\gamma^0.
$$
\unpause
They can be chosen as
\eq$$
\gamma^0=\pMatrix{&-i\\i&},
\qquad
\gamma^1=\pMatrix{&i\\i&},
\qquad
\gamma^3=\gamma^0\gamma^1=\pMatrix{1&\\&-1}.
\label{gamma-matrices}
$$
\unpause
The model has a conserved current
\eq$$
j^\mu=\bpsi\gamma^\mu\psi.
\label{MTcurrent}
$$
\unpause
When $m=0$ there is another conserved current
\eq$$
j_3^\mu=\bpsi\gamma^3\gamma^\mu\psi=-\epsilon^{\mu\nu}j_\nu,
\qquad
\epsilon^{01}=-\epsilon^{10}=-1.
\label{MT3current}
$$
\unpause
Excitations: fermion, antifermion, and for $g>0$ neutral boson bound states.

\end{frame}

\begin{frame}
\frametitle{Sine\-/Gordon model}
\unpause

The sine\-/Gordon model:
\eq$$
S^{SG}[\phi]=\int d^2x\,\left({(\d_\mu\phi)^2\over8\pi}+\mu\cos\beta\phi\right).
\label{SG}
$$
\unpause
This model has a \emph{topological number}
\eq$$
q={\beta\over2\pi}(\phi(t,+\infty)-\phi(t,-\infty))\in\Z.
\label{topnumber}
$$
\unpause
It conserves on solutions and can be written as
\eq$$
q={\beta\over2\pi}\int^\infty_{-\infty}dx\,\d_1\phi(t,x)=\int^\infty_{-\infty}df_\mu\,j^\mu_\text{top},
$$
where $df_\mu=\epsilon_{\mu\nu}dx^\nu$ is the one\-/dimensional surface element
\unpause
and $j^\mu_\text{top}$ is the \emph{topological current}:
\eq$$
j^\mu_\text{top}=-{\beta\over2\pi}\epsilon^{\mu\nu}\d_\nu\phi,
\quad
\d_\mu j^\mu_{\rm top}=-{\beta\over2\pi}\epsilon^{\mu\nu}\d_\mu\d_\nu\phi=0.
\label{topcurrent}
$$

\end{frame}

\begin{frame}
\frametitle{Digression: Stationary solutions of field equations}
\unpause

Consider a two\-/dimensional model of one scalar field with the action
$$
S[\phi]=\int d^2x\,\left({(\d_\mu\phi)^2\over2}-U(\phi)\right).
$$
Suppose that the potential $U(\phi)$ possesses a set of degenerate absolute minima $\phi_i$. Let us order these minima so that $\phi_i<\phi_{i+1}$.
\unpause
We say that a solution $\phi(x)$ with finite action has \emph{topological charge} $q$, if
\begin{itemize}
\item$\phi(x)\to\phi_i$ as $x^1\to-\infty$,
\item$\phi(x)\to\phi_{i+q}$ as $x^1\to+\infty$.
\end{itemize}
\unpause

\begin{block}{Theorem}

Subject to these conditions any nontrivial \emph{static} solution $\phi(x)=\varphi(x^1)$ has topological charge $q=\pm1$.

\end{block}
\unpause
\emph{Proof.} The static solution satisfies the equation
$$
\d_1^2\varphi=U'(\varphi).
$$
\unpause
This is the Newton equation with the potential $-U(\varphi)$. The points $\phi_i$ correspond to \emph{maxima} of this potential. The field $\varphi$, which starts its `movement' at the point $\phi_i$ may only finish it at the points $\phi_{i\pm1}$.
\unpause

If there is a static solution, we may define a family of solutions moving with any velocity $-1<v<1$:
$$
\phi(x)=\varphi\left(x^1-vx^0\over\sqrt{1-v^2}\right).
$$

\end{frame}

\begin{frame}
\frametitle{Digression: Classical solutions of the sine\-/Gordon equation}
\unpause

There are \emph{kink} ($q=1$) and \emph{antikink} ($q=-1$) solutions:
$$
\phi(x)=\pm{4\over\beta}\arctg\exp{m(x^1-vx^0-x^1_0)\over\sqrt{1-v^2}}
$$
\unpause
The \emph{breather} solution
$$
\phi(x)=4\arctg{\sqrt{1-\omega^2}\cos\omega\tau\over\omega\cos\sqrt{1-\omega^2}\xi},
\quad
\tau={x^0-vx^1-x^0_0\over\sqrt{1-v^2}},
\quad
\xi={x^1-vx^0-x^1_0\over\sqrt{1-v^2}},
$$
depends on the parameter $\omega$ ($0\le\omega<1$).
\unpause
It is not stationary and can be considered as a kink\--antikink bound state.
\unpause

Kinks, antikinks and breathers are the only \emph{solitary waves} of the sine\-/Gordon equation.
\unpause
It can be checked that any solution decays to a system of solitary waves with different velocities as $x^0\to\pm\infty$.
\unpause
Moreover, the composition and velocities of the solitary waves in far past and far future coincide, so that the scattering of solitary wave reduces to shifts of the parameters $x_0^0,x_0^1$ as a result of interaction.
\unpause

One may expect the quantum spectrum to consist of a kink ($q=1$), antikink ($q=-1$) and a discrete series of breathers. The density of the mass spectrum of breathers decreases with increasing $\beta$.

\end{frame}

\begin{frame}
\frametitle{Equivalence of the two models}
\unpause

We will see that these two models are equivalent subject to
\eq$$
g=\pi(\beta^{-2}-1),
\label{gbeta}
$$
\unpause
The massive term in the Thirring model corresponds to the cosine term in the sine\-/Gordon model, so that
\eq$$
\mu\sim mr_0^{\beta^2-1},
\label{mum}
$$
\unpause
The conserved currents also coincide
\eq$$
j^\mu=j^\mu_{\rm top},
\label{MT-SG-cur}
$$
so that the fermion number in the Thirring model coincides with the topological number in the sine\-/Gordon model.

\end{frame}

\begin{frame}
\frametitle{Massless Thirring model}
\unpause

Rewrite the action in terms of the light cone variables:
$$
S^{MT}[\psi,\bpsi]
=\int d^2x\,(2i\psi^+_1\bd\psi_1-2i\psi^+_2\d\psi_2+im(\psi^+_1\psi_2-\psi^+_2\psi_1)-2g\psi^+_1\psi^+_2\psi_2\psi_1).
$$
\unpause
The components of the currents are
\eq$$
j_z=-\psi^+_1\psi_1,
\qquad
j_\bz=\psi^+_2\psi_2.
\label{j-psipsi}
$$
\unpause
In the case \emph{$m=0$} the equations of motion read
\eq$$
\Aligned{
\bd\psi_1
&=-ig\psi^+_2\psi_2\psi_1\equiv-igj_\bz\psi_1,
\\
\d\psi_2
&=ig\psi^+_1\psi_1\psi_2\equiv-igj_z\psi_2.
}\label{Teqmotion}
$$
\unpause
Since $\epsilon^{\mu\nu}\d_\mu j_\nu=\d_\mu j_3^\mu=0$ the current $j_\mu$ is a gradient of a free field:
\eq$$
j_\mu=-{\beta\over2\pi}\d_\mu\tphi\uncover<.(1)->{{}=-{\beta\over2\pi}\epsilon_{\mu\nu}\d^\nu\phi}.
\label{Ttphi}
$$
\unpause
We will think of $\tphi$ as of the dual of another field $\phi$.
\unpause
Both satisfy the d'Alembert equation:
$$
\d_\mu\d^\mu\phi=\d_\mu\d^\mu\tphi=0.
$$
\unpause
The solution to these equations reads
\eq$$
\phi(x)=\varphi(z)+\bvarphi(\bz),
\quad
\tphi(x)=\varphi(z)-\bvarphi(\bz),
\label{phi-tphi-varphi}
$$
\unpause
Thus we have
\eq$$
j_z=-{\beta\over2\pi}\d\varphi,
\quad
j_\bz={\beta\over2\pi}\bd\bvarphi.
\label{j-varphi}
$$

\end{frame}

\begin{frame}
\frametitle{Massless Thirring model: quantization}
\unpause

The equations of motion (\ref{Teqmotion}) have the solution
$$
\psi_1(z,\bz)=F_1(z)e^{-i{g\beta\over2\pi}\bvarphi(\bz)},
\quad
\psi_2(z,\bz)=F_2(\bz)e^{i{g\beta\over2\pi}\varphi(z)},
$$
\unpause
while equation (\ref{j-psipsi}) gives
\eq$$
{\beta\over2\pi}\d\varphi=\psi^+_1\psi_1=F_1(z)F_1^*(z),
\qquad
{\beta\over2\pi}\bd\bvarphi=\psi^+_2\psi_2=F_2(\bz)F_2^*(\bz).
\label{Tcureq}
$$
\unpause
This makes it possible to find a classical solution in terms of two analytic functions $F_i(z)$.
\unpause

What can be done in the quantum case?
\unpause
Conjecture the fields $\psi_i$ in the form
\eq$$
\psi_i(x)=\eta_i\sqrt{N_i\over2\pi}e^{i\alpha_i\varphi (z)+i\beta_i\bvarphi(\bz)},
\qquad
\psi^+_i(x)=\eta_i^{-1}\sqrt{N_i\over2\pi}e^{-i\alpha_i\varphi(z)-i\beta_i\bvarphi(\bz)},
\label{psibpsiansatz}
$$
\unpause
where $N_i$ are constants
\unpause
and $\eta_i$ are algebraic elements with the relations
\eq$$
\eta_1\eta_2=-\eta_2\eta_1.
\label{algfactors}
$$
\unpause
As in the classical case we have
\eq$$
\alpha_2=-\beta_1={g\beta\over2\pi},
\label{eq3}
$$
\unpause
But the exponential form of $F_i$ seems to be strange. To understand them consider the operator product expansions of~$\psi_i$.

\end{frame}

\begin{frame}
\frametitle{Massless Thirring model: quantization}
\unpause

We have
\Multline$$
\psi_i(x')\psi_j(x)
=\eta_i\eta_j{\sqrt{N_iN_j}\over2\pi}
(z'-z)^{\alpha_i\alpha_j}
(\bz'-\bz)^{\beta_i\beta_j}
\\
\times e^{i\alpha_i\varphi(z')+i\beta_i\bvarphi(\bz')
+i\alpha_j\varphi(z)+i\beta_j\bvarphi(\bz)}.
\label{psipsi}
$$
\unpause
Then $\psi_i(x')\psi_j(x)=-\psi_j(x)\psi_i(x')$, if
\eq$$
\alpha_i^2-\beta_i^2\in2\Z+1,
\qquad
\alpha_1\alpha_2-\beta_1\beta_2\in2\Z.
\label{fermion-limitation}
$$
\unpause
Now let us expand the product $\psi^+_1(x')\psi_1(x)$ in powers of $x'-x$:
\Multline$$
\psi^+_1(x')\psi_1(x)
={N_1\over2\pi}(z'-z)^{-\alpha_1^2}
\left(\bz'-\bz\right)^{-\beta_1^2}
\\
\times
\alt<+->{\left({\red\alt<.(3)->{\transparent1}{\transparent0}\xcancel{\transparent1\black1}}
-i\alpha_1(z'-z){\red\alt<.(5)->{\transparent1}{\transparent0}\boxed{\transparent1\black\d\phi(x)}}
-{\red\alt<.(3)->{\transparent1}{\transparent0}\xcancel{\transparent1\black i\beta_1(\bz'-\bz)\,\bd\phi(x)}}+
{\red\alt<.(4)->{\transparent1}{\transparent0}\xcancel{\transparent1\black\cdots{}}}
\right)}{e^{-i\alpha_1(\varphi(z')-\varphi(z))-i\beta_1(\bvarphi(\bz')-\bvarphi(\bz))}}.
\label{bpsi1psi1}
$$
\unpause
Define the current $j_z$ as follows:
$$
j_z(z,\bz)=-\int^{2\pi}_0{d\theta\over2\pi}\psi_1^+(z+r_0e^{i\theta},\bz+r_0e^{-i\theta})\psi_1(z,\bz)
$$
for some small~$r_0$.
\unpause
\eq$$
\alpha_1^2-\beta_1^2=1,
\label{eq1}
$$
\unpause
the first and the third terms give zero contribution due to the angular dependence
\unpause
and the terms from the forth on are negligible.
\unpause
The second term only contributes the current.

\end{frame}

\begin{frame}
\frametitle{Massless Thirring model: conditions for $\alpha_i,\beta_i,N_i$}
\unpause

We obtain
$$
j_z=-N_1r_0^{-2\beta_1^2}\left(-i\alpha_1\d\varphi\over2\pi\right).
$$
\unpause
Comparing with $j_z={\beta\over2\pi}\d\varphi$ we find
\eq$$
\beta=-ir_0^{-2\beta_1^2}N_1\alpha_1.
\label{betaN1}
$$
\unpause
Repeating the same for $\psi^+_2\psi_2$ and assuming
\eq$$
\alpha_2^2-\beta_2^2=-1,
\label{eq2}
$$
\unpause
we obtain
\eq$$
\beta=-ir_0^{-2\alpha_2^2}N_2\beta_2.
\label{betaN2}
$$
\unpause
We need one more equation. To find it, we have to analyze the mass term.

\end{frame}

\begin{frame}
\frametitle{Thirring model: mass term}
\unpause

Consider now the mass term $-m\bpsi\psi=im(\psi^+_1\psi_2-\psi^+_2\psi_1)$. Both terms must be \emph{well\-/defined} and be consistent with the \emph{conservancy of the fermion charge}.
\unpause
Consider the expansion
\Multline$$
\psi^+_2(x')\psi_1(x)
=-\eta_1\eta_2^{-1}{\sqrt{N_1N_2}\over2\pi}(z'-z)^{-\alpha_1\alpha_2}(\bz'-\bz)^{-\beta_1\beta_2}
\\
\times
\left(e^{i(\alpha_1-\alpha_2)\varphi(z)+i(\beta_1-\beta_2)\bvarphi(\bz)}+\cdots\right).
$$
\unpause
The first term survives in the averaged product defined similarly to $j_z$, if
\eq$$
\alpha_1\alpha_2=\beta_1\beta_2
\quad\Rightarrow\quad
\alpha_1=-\beta_2.
\label{eq4}
$$
\unpause
With this assumption we may define
$$
\psi^+_2\psi_1(x)=-\eta_1\eta_2^{-1}{\sqrt{N_1N_2}\over2\pi}r_0^{-2\alpha_1\alpha_2}e^{i(\alpha_1-\alpha_2)\varphi(z)+i(\beta_1-\beta_2)\bvarphi(\bz)}.
$$
\unpause
Now check the consistency with the conservancy of the fermion charge $Q$ means that it should be a function of $\phi$.
\unpause
Consider the commutator
\Multline*$$
[O(0),Q]=\oint df_\mu\,j^\mu(x)O(0)=\oint dx^\nu\,\epsilon_{\mu\nu}j^\mu(x)O(0)
=-{\beta\over2\pi}\oint dx^\nu\,\epsilon_{\mu\nu}\d^\mu\tphi(x)\,O(0)
\\
=-{\beta\over2\pi}\oint dx^\nu\,\epsilon_{\mu\nu}\epsilon^{\mu\lambda}\d_\lambda\phi(x)\,O(0)
={\beta\over2\pi}\oint dx^\lambda\,\d_\lambda\phi(x)\,O(0)={\beta\over2\pi}\Delta\phi(x)\,O(0),
$$
where $\Delta$ means the increment of the field while passing the closed contour.

\end{frame}

\begin{frame}
\frametitle{Thirring model: mass term}
\unpause

Let $O(x)=e^{\alpha\varphi(z)+\alpha'\bvarphi(\bz)}$. Then
\Multline*$$
[O(0),Q]={\beta\over2\pi}\Delta(\varphi(z)+\bvarphi(\bz))e^{i\alpha\varphi(z)+i\alpha'\bvarphi(\bz)}
\\
={i\beta\over2\pi}\Delta\left(\alpha\log{1\over z}+\alpha'\log{1\over\bz}\right)e^{i\alpha\varphi(0)+i\alpha'\bvarphi(0)}
=\beta(\alpha-\alpha')e^{i\alpha\varphi(0)+i\alpha'\bvarphi(0)}.
$$
\unpause
Hence, the operator $Q$ is commuting with $\psi^+_2\psi_1$, since $\alpha_1-\alpha_2=\beta_1-\beta_2$.
\unpause

Now fix the $\beta$ parameter. Set $O(x)=\psi_i(x)$. Since the operators $\psi_i$ have the fermion charge $-1$, we have
$$
\psi_i(0)=[\psi_i(0),Q]=\beta(\alpha_i-\beta_i)\psi_i(0)=\beta(\alpha_1+\alpha_2)\psi_i(0)
$$
\unpause
Hence
$$
\alpha_1+\alpha_2=\beta^{-1}
\quad\Rightarrow\quad
\alpha_1-\alpha_2=\beta.
$$
\unpause
Hence
\eq$$
\Aligned{
\alpha_1=-\beta_2
&={1\over2}\left({1\over\beta}+\beta\right),
\\
\alpha_2=-\beta_1
&={1\over2}\left({1\over\beta}-\beta\right).
}
$$
\unpause
Using $\alpha_2=g\beta/2\pi$ we obtain the relation between coupling constants:
\eq$$
g=\pi(\beta^{-2}-1).
\tag{\ref{gbeta}}
$$

\end{frame}

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

Substituting it we obtain
\eq$$
N_1=-N_2=ir_0^{{\beta^2\over2}+{1\over2\beta^2}-1}{2\beta^2\over\beta^2+1},
\label{N12}
$$
\unpause
Hence
$$
\Aligned{
-i\psi_2^+\psi_1
&={1\over\pi}{\beta^2\over\beta^2+1}r_0^{\beta^2-1}\left(i\eta_1\eta_2^{-1}\right)e^{i\beta\phi},
\\
i\psi_1^+\psi_2
&={1\over\pi}{\beta^2\over\beta^2+1}r_0^{\beta^2-1}\left(i\eta_1\eta_2^{-1}\right)^{-1}e^{-i\beta\phi}.
}
$$
\unpause
Consider the mass contribution
$$
im(\psi^+_2\psi_1-\psi^+_1\psi_2)\sim\left(i\eta_1\eta_2^{-1}\right)e^{i\beta\phi}+\left(i\eta_1\eta_2^{-1}\right)^{-1}e^{-i\beta\phi}
$$
as a perturbation.
\unpause
Due to the `neutrality' on the infinite plane, $\#(e^{i\beta\phi})-\#(e^{-i\beta\phi})$ is the same in each perturbation term.
\unpause
Hence the substitution
$$
\left(i\eta_1\eta_2^{-1}\right)^{\alpha/\beta}e^{i\alpha\phi}\to e^{i\alpha\phi}.
$$
makes it possible to get rid of the algebraic elements $\eta_i$.
\unpause
Then we have
\eq$$
i(\psi^+_1\psi_2-\psi^+_2\psi_1)={2\over\pi}{\beta^2\over\beta^2+1}r_0^{\beta^2-1}\cos\beta\phi,
\label{psicos}
$$
\unpause
from which we find
\eq$$
\mu\sim mr_0^{\beta^2-1},
\tag{\ref{mum}}
$$

\end{frame}

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

Correlation functions of the Thirring and the sine\-/Gordon models coincide in each order of the perturbation theory in the parameter $m$. This is a strong argument for their coincidence.
\unpause

The parameter $\mu$ of the sine\-/Gordon model is finite and measurable. It is related to the physical mass $m_\text{phys}$ of a particle:
$$
\mu\sim m_{\rm phys}^{2-\beta^2}.
$$
The coefficient is known exactly, but its calculation is far beyond the scope of these lectures.

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Hence the parameter $m$ is divergent:
\eq$$
m\sim m_{\rm phys}(m_{\rm phys}r_0)^{1-\beta^2}
=m_{\rm phys}(m_{\rm phys}r_0)^{g/\pi\over1+g/\pi},
\label{mphys}
$$
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Consider the operators from Lecture~1:
\eq$$
e^{iJ\varphi}=e^{{iJ\over2\beta}\tilde\phi},
\qquad
J\in\Z.
\label{exp-Jvarphi}
$$
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These operators change the topological charge by $J$: $q\to q+J$.
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For $J=\pm1$ they are \emph{boson} kink creation\-/annihilation operators.


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\frametitle{Seminar}
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