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\begin{document}

\centerline{\bf Lecture~2}
\centerline{\bf Bosonization of the Thirring model}
\bigskip
Consider the {\it massive Thirring model\/} in the Minkowski space:
\eq$$
S^{MT}[\psi,\bpsi]
=\int d^2x\,
\left(\bpsi(i\hat\d-m)\psi-{g\over2}(\bpsi\gamma^\mu\psi)^2\right).
\label{MT}
$$
Here $\psi(x)$, $\bpsi(x)$ are the fermion field and its Dirac conjugate, $\gamma^\mu$ are the Dirac matrices, and $\hat\d=\gamma^\mu\d_\mu$. The Dirac matrices satisfy the standard relations
$$
\gamma^\mu\gamma^\nu+\gamma^\nu\gamma^\mu=2g^{\mu\nu},
\qquad
\gamma^{\mu+}=\gamma^0\gamma^\mu\gamma^0.
$$
In the two\-/dimensional case the gamma\-/matrices can be written as:
\eq$$
\gamma^0=\pMatrix{&-i\\i&},
\qquad
\gamma^1=\pMatrix{&i\\i&},
\qquad
\gamma^3=\gamma^0\gamma^1=\pMatrix{1&\\&-1}.
\label{gamma-matrices}
$$
The model has a conserved current
\eq$$
j^\mu=\bpsi\gamma^\mu\psi.
\label{MTcurrent}
$$
When $m=0$ there is another conserved current
\eq$$
j_3^\mu=\bpsi\gamma^3\gamma^\mu\psi=-\epsilon^{\mu\nu}j_\nu.
\label{MT3current}
$$

In the previous lecture we considered the sine\-/Gordon model:
\eq$$
S^{SG}[\phi]=\int d^2x\,\left({(\d_\mu\phi)^2\over8\pi}+\mu\cos\beta\phi\right).
\label{SG}
$$
This model has a topological number
\eq$$
q={\beta\over2\pi}(\phi(t,+\infty)-\phi(t,-\infty)),
\label{topnumber}
$$
which takes integer values. It can be written as
\eq$$
q={\beta\over2\pi}\int^\infty_{-\infty} dx\,\d_1\phi(t,x).
$$
This allows us to define a current responsible for the topological charge:
\eq$$
j^\mu_\text{top}=-{\beta\over2\pi}\epsilon^{\mu\nu}\d_\nu\phi.
\label{topcurrent}
$$
This current satisfies the continuity equation $\d_\mu j^\mu_\text{top}=0$ due to the antisymmetry of the symbol $\epsilon^{\mu\nu}$ and the commutativity of derivatives.

In the present lecture we will make sure that the massive Thirring model and the sine\-/Gordon model are equivalent\cite{Coleman,Mandelstam}, while the parameters of the two models are related according to
\Align$$
g
&=\pi(\beta^{-2}-1),
\label{gbeta}
\\
\mu
&\sim mr_0^{\beta^2-1},
\label{mum}
$$
and the Thirring current coincides with the topological one:
\eq$$
j^\mu=j^\mu_\text{top}.
\label{MT-SG-cur}
$$
This is an extremely important correspondence called {\it bosonization}. The equation (\ref{MT-SG-cur}) plays a key role in the bosonization.

Rewrite the action of the Thirring model by using the explicit form of the gamma\-/matrices:
$$
S^{MT}[\psi,\bpsi]
=\int d^2x\,(i\psi^+_1(\d_0+\d_1)\psi_1+i\psi^+_2(\d_0-\d_1)\psi_2+im(\psi^+_1\psi_2-\psi^+_2\psi_1)-2g\psi^+_1\psi^+_2\psi_2\psi_1).
$$
Substituting $z=x^1-x^0$, $\bz=x^1+x^0$, we obtain
$$
S^{MT}[\psi,\bpsi]
=\int d^2x\,(2i\psi^+_1\bd\psi_1-2i\psi^+_2\d\psi_2+im(\psi^+_1\psi_2-\psi^+_2\psi_1)-2g\psi^+_1\psi^+_2\psi_2\psi_1).
$$
In these components the Thirring current has the form:
\eq$$
j_z=-\psi^+_1\psi_1,
\qquad
j_\bz=\psi^+_2\psi_2.
\label{j-psipsi}
$$
Consider the case $m=0$, which admits an exact solution~\cite{Thirring}. Start with solving classical equations of motion
\eq$$
\Aligned{
\bd\psi_1
&=-ig\psi^+_2\psi_2\psi_1\equiv-igj_\bz\psi_1,
\\
\d\psi_2
&=ig\psi^+_1\psi_1\psi_2\equiv-igj_z\psi_2.
}\label{Teqmotion}
$$
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$$
j_\mu=-{\beta\over2\pi}\d_\mu\tphi.
\label{Ttphi}
$$
It is convenient to consider the field $\tphi$ as a dual to another field $\phi$, as it was described in the last lecture. Both satisfy the d'Alembert equation:
$$
\d_\mu\d^\mu\phi=\d_\mu\d^\mu\tphi=0.
$$
The general solution to these equations can be written as
\eq$$
\Aligned{
\phi(x)
&=\varphi(z)+\bvarphi(\bz),
\\
\tphi(x)
&=\varphi(z)-\bvarphi(\bz),
}\label{phi-tphi-varphi}
$$
where $\varphi(z)$ and $\bvarphi(\bz)$ are arbitrary functions of the only $z$ and $\bz$ variables respectively. The coefficient in (\ref{Ttphi}) is arbitrary. We choose it in such a way that the relation~(\ref{MT-SG-cur}) is formally satisfied, if we identify the field $\phi$ with that in the sine\-/Gordon model.

We see that the massless Thirring model is equivalent to the free massless boson model. From relation (\ref{Ttphi}) we have
\eq$$
{\beta\over2\pi}\d\varphi=\psi^+_1\psi_1,
\qquad
{\beta\over2\pi}\bd\bvarphi=\psi^+_2\psi_2.
\label{Tcureq}
$$
To continue searching for a classical solution, we have to substitute these functions into the equations (\ref{Teqmotion}). By solving the last one we obtain
\eq$$
\psi_1(z,\bz)=F_1(z)e^{-i{g\beta\over2\pi}\bar\varphi(\bz)},
\qquad
\psi_2(z,\bz)=F_2(\bz)e^{i{g\beta\over2\pi}\varphi(z)}
\label{psibpsi-classical}
$$
with arbitrary functions $F_i$. By substituting them back into (\ref{Tcureq}), we have
\eq$$
{\beta\over2\pi}\d\varphi(z)=F_1(z)F_1^*(z),
\qquad
{\beta\over2\pi}\bd\bar\varphi(\bz)=F_2(\bz)F_2^*(\bz),
\label{Tcur-classical}
$$
where the star denotes complex conjugation under the assumption of real arguments. It is left to integrate these equations and substitute the result into (\ref{psibpsi-classical}). As a result the fields $\psi_i$ are expressed in terms of two functions $F_i$ and two integration constants.

Let us turn to the quantum case. The situation is simpler just in the quantum case. Let us look for a solution to the equations (\ref{Tcureq}) in the form
\eq$$
\psi_i(x)=\eta_i\sqrt{N_i\over2\pi}e^{i\alpha_i\varphi (z)+i\beta_i\bvarphi(\bz)},
\qquad
\psi^+_i(x)=\eta_i^{-1}\sqrt{N_i\over2\pi}e^{-i\alpha_i\varphi(z)-i\beta_i\bvarphi(\bz)},
\label{psibpsiansatz}
$$
where $\eta_i$ are the algebraic factors necessary to ensure the fermionic behavior of the fields~$\psi_i$. It turns out that solutions can be found, if we assume
\eq$$
\eta_1\eta_2=-\eta_2\eta_1.
\label{algfactors}
$$
First of all, demand that the fields $\psi_i(x)$ behave like fermions. Consider the product
\eq$$
\psi_i(x')\psi_j(x)
=\eta_i\eta_j{\sqrt{N_iN_j}\over2\pi}
(z'-z)^{\alpha_i\alpha_j}
(\bz'-\bz)^{\beta_i\beta_j}
e^{i\alpha_i\varphi(z')+i\beta_i\bvarphi(\bz')
+i\alpha_j\varphi(z)+i\beta_j\bvarphi(\bz)}.
\label{psipsi}
$$
This expression continues well into the Euclidean region. From the anticommutativity requirement it is easy to obtain that
\eq$$
\alpha_i^2-\beta_i^2\in2\Z+1,
\qquad
\alpha_1\alpha_2-\beta_1\beta_2\in2\Z.
\label{fermion-limitation}
$$
From (\ref{psipsi}) it can be seen that products like $\psi^+_1\psi_1$ are poorly defined. Let us {\it define\/} these products as follows. Consider another product:
\eq$$
\psi^+_1(x')\psi_1(x)
={N_1\over2\pi}(z'-z)^{-\alpha_1^2}\left(\bz'-\bz\right)^{-\beta_1^2}\left(1-i\alpha_1(z'-z)\,\d\phi(x)-i\beta_1(\bz'-\bz)\,\bd\phi(x)+\cdots\right).
\label{bpsi1psi1}
$$
Take an average of this product over the circle $|z'-z|^2=r_0^2$ and assume that $r_0$ is small. The leading term in the expansion in $r_0$ will be assumed for $\psi^+_1(x)\psi_1(x)$. Suppose that
\eq$$
\alpha_1^2-\beta_1^2=1.
\label{eq1}
$$
Then the first and third terms in the expansion (\ref{bpsi1psi1}) vanish after averaging. The leading nonzero term is the second:
$$
N_1r_0^{-2\beta_1^2}\left(-i\alpha_1\d\varphi\over2\pi\right).
$$
It is just what we will identify with~$\psi^+_1\psi_1$. The coefficient $N_1$ must be imaginary for consistency with the Hermicity of $\psi^+\psi$. Comparing with (\ref{Tcureq}), we obtain
\eq$$
\beta=-ir_0^{-2\beta_1^2}N_1\alpha_1.
\label{betaN1}
$$
Similarly, assuming
\eq$$
\alpha_2^2-\beta_2^2=-1,
\label{eq2}
$$
we obtain
\eq$$
\beta=-ir_0^{-2\alpha_2^2}N_2\beta_2.
\label{betaN2}
$$
Now consider the equations of motion (\ref{Teqmotion}). Substituting (\ref{Tcureq}) and (\ref{psibpsiansatz}), we obtain
$$
\Aligned{
i\beta_1\,\bd\bvarphi\,e^{i\alpha_1\varphi+i\beta_1\bvarphi}
&=-ig{\beta\over2\pi}\bd\bvarphi\,e^{i\alpha_1\varphi+i\beta_1\bvarphi},
\\
i\alpha_2\,\d\varphi\,e^{i\alpha_2\varphi+i\beta_2\bvarphi}
&=ig{\beta\over2\pi}\d\varphi\,e^{i\alpha_2\varphi+i\beta_2\bvarphi}.
}
$$
From this we have
\eq$$
\alpha_2=-\beta_1={g\beta\over2\pi},
\label{eq3}
$$
which is surely consistent with the classical solution.

To fix the coefficients $\alpha_i$, $\beta_i$, we need to define the mass term consistently in such a way that it commute with the fermion charge
\eq$$
Q=\int df_\mu\,j^\mu,
\label{Qcharge-def}
$$
where $df_\mu=\epsilon_{\mu\nu}\,dx^\nu$ is the one\-/dimensional surface element. Consider the expansion
$$
\psi^+_2(x')\psi_1(x)
=-\eta_1\eta_2^{-1}{\sqrt{N_1N_2}\over2\pi}
(z'-z)^{-\alpha_1\alpha_2}(\bz'-\bz)^{-\beta_1\beta_2}
\left(
e^{i(\alpha_1-\alpha_2)\varphi(z)+i(\beta_1-\beta_2)\bvarphi(\bz)}+\cdots
\right).
$$
The first term survives under averaging over the corners if
\eq$$
\alpha_1\alpha_2=\beta_1\beta_2,
\label{eq4}
$$
which is consistent with (\ref{eq1}--\ref{eq3}) and yields
\eq$$
\alpha_1=-\beta_2.
\label{eq4a}
$$
By taking the angular average we obtain the definition of the products
\eq$$
\Aligned{
\psi^+_2\psi_1
&=-\eta_1\eta_2^{-1}{\sqrt{N_1N_2}\over2\pi}r_0^{-2\alpha_1\alpha_2}e^{i(\alpha_1-\alpha_2)\phi},
\\
\psi^+_1\psi_1
&=-\eta_2\eta_1^{-1}{\sqrt{N_1N_2}\over2\pi}r_0^{-2\alpha_1\alpha_2}e^{-i(\alpha_1-\alpha_2)\phi}.
}\label{massterms-intermediate}
$$

Check now that the operators defined in such a way commute with $Q$. Let $O(x)$ be a local operator. Calculate the commutator
\Multline$$
[O(0),Q]=\oint df_\mu\,j^\mu(x)\,O(0)=\oint dx^\nu\,\epsilon_{\mu\nu}j^\mu(x)O(0)
=-{\beta\over2\pi}\oint dx^\nu\,\epsilon_{\mu\nu}\,\d^\mu\tphi(x)\,O(0)
\\*
=-{\beta\over2\pi}\oint dx^\nu\,\epsilon_{\mu\nu}\epsilon^{\mu\lambda}\,\d_\lambda\phi(x)\,O(0)
={\beta\over2\pi}\oint dx^\lambda\,\d_\lambda\phi(x)\,O(0)={\beta\over2\pi}\Delta\phi(x)\,O(0),
\label{OQ-commut}
$$
Here $\Delta\phi(x)$ is the increment of the field $\phi(x)$ while $x$ goes around zero counterclockwise. Let us apply this formula to the operator $O(x)=e^{i\alpha\varphi(z)+i\alpha'\bvarphi(\bz)}$:
\Multline$$
[e^{i\alpha\varphi(0)+i\alpha'\bvarphi(0)},Q]={\beta\over2\pi}\Delta(\varphi(z)+\bvarphi(\bz))e^{i\alpha\varphi(0)+i\alpha'\bvarphi(0)}
\\*
={i\beta\over2\pi}\Delta\left(\alpha\log{1\over z}+\alpha'\log{1\over\bz}\right)e^{i\alpha\varphi(0)+i\alpha'\bvarphi(0)}
=\beta(\alpha-\alpha')e^{i\alpha\varphi(0)+i\alpha'\bvarphi(0)}.
\label{expQ-commut}
$$
The commutator $[Q,O(x)]=0$, if $\alpha=\alpha'$ a, therefore, $O(x)=e^{i\alpha\phi(x)}$. For $\alpha=\alpha_1-\alpha_2$, $\alpha'=\beta_1-\beta_2$, this condition is satisfied due to (\ref{eq3}), (\ref{eq4a}).

Now fix the $\beta$ parameter. To do it we set $O(x)=\psi_i(x)$ in (\ref{expQ-commut}). Since the operators $\psi_i$ have the fermion charge equal to $-1$, we have
$$
\psi_i(0)=[\psi_i(0),Q]=\beta(\alpha_i-\beta_i)\psi_i(0)=\beta(\alpha_1+\alpha_2)\psi_i(0)
$$
Hence,
\eq$$
\alpha_1+\alpha_2=\beta^{-1},
\label{eq5}
$$
and we immediately obtain
\eq$$
\alpha_1-\alpha_2=\beta
\label{eq6}
$$
and
\eq$$
\Aligned{
\alpha_1=-\beta_2
&={1\over2}\left({1\over\beta}+\beta\right),
\\
\alpha_2=-\beta_1
&={1\over2}\left({1\over\beta}-\beta\right).
}
$$
Substituting the answer into~(\ref{eq3}), we get~(\ref{gbeta}).

From (\ref{betaN1}), (\ref{betaN2}) we find
\eq$$
N_1=-N_2=ir_0^{{\beta^2\over2}+{1\over2\beta^2}-1}{2\beta^2\over\beta^2+1},
\label{N12}
$$
From this we obtain
$$
\Aligned{
-i\psi_2^+\psi_1
&={1\over\pi}{\beta^2\over\beta^2+1}r_0^{\beta^2-1}\left(i\eta_1\eta_2^{-1}\right)e^{i\beta\phi},
\\
i\psi_1^+\psi_2
&={1\over\pi}{\beta^2\over\beta^2+1}r_0^{\beta^2-1}\left(i\eta_1\eta_2^{-1}\right)^{-1}e^{-i\beta\phi}.
}
$$
Since on the infinite plane the total ``charge'' must be equal to zero, the operators $e^{i\beta\phi}$ and $e^{-i\beta\phi}$ must occur in equal numbers for correlation functions polynomial in $\varphi$,~$\bar\varphi$. Therefore, the factors $\left(i\eta_1\eta_2^{-1}\right)^{\pm1}$ will also cancel each other. In a more general case, they can be omitted by redefining the operators:
$$
\left(i\eta_1\eta_2^{-1}\right)^{\alpha/\beta}e^{i\alpha\phi}\to e^{i\alpha\phi}.
$$
Then we have
\eq$$
i(\psi^+_1\psi_2-\psi^+_2\psi_1)
={2\over\pi}{\beta^2\over\beta^2+1}r_0^{\beta^2-1}
\cos\beta\phi,
\label{psicos}
$$
from which we find~(\ref{mum}).

Rigorously speaking, so far we have found the exact solution for the massless Thirring model only. However, it follows from our reasoning that the perturbation theory in the term $m\bpsi\psi$ for the Thirring model and the perturbation theory in $\cos\beta\phi$ for the sine\-/Gordon model coincide, which gives a strong foundation in favor of the coincidence of the theories\cite{Coleman,Mandelstam}. Note that the coupling constant $g$ in the Thirring model does not renormalize, while the ``mass'' $m$ is not a physical quantity and substantially renormalizes. This is because the mass term $\psi^+_1\psi_2-\psi^+_2\psi_1$ has a scale dimension~$\beta^2$ due to redefinition of the product of fields. The constant $\mu$ in the sine\-/Gordon model is measurable, and
\eq$$
\mu\sim m_{\rm phys}^{2-\beta^2},
\qquad
m\sim m_{\rm phys}(m_{\rm phys}r_0)^{1-\beta^2}
=m_{\rm phys}(m_{\rm phys}r_0)^{g/\pi\over1+g/\pi},
\label{mphys}
$$
where $m_{\rm phys}$ is the mass of physical excitations (for example, the Thirring fermions) in theory. The proportionality coefficient between the parameter $\mu$ and $m_{\rm phys}^{2-\beta^2}$ is known exactly~\cite{AlZamolodchikov}.

One more question remains: what does the Thirring fermions in the sine\-/Gordon model correspond to? From the equality between the topological and fermion currents, we can conclude that they correspond to kinks, which are nontrivial excitations with topological numbers $q=\pm1$. Simultaneously the kink excitation can be generated not only by fermion operators, but also by boson operators. Consider the operators
\eq$$
e^{iJ\varphi}=e^{{iJ\over2\beta}\tilde\phi},
\qquad
J\in\Z,
\label{exp-Jvarphi}
$$
which entered the correlation functions of the last lecture. These operator acting on a state change the topological number: $q\to q+J$. For $J=\pm1$ they can be considered as boson creation\-/annihilation operators of the kinks.

\def\refname{\normalsize References}
\begin{thebibliography}{9}

\bibitem{Thirring}W.~E.~Thirring, {\it Annals Phys.}\ {\bf 3} (1958) 91

\bibitem{Coleman}S.~Coleman, {\it Phys.\ Rev.}\ {\bf D11} (1975) 2088

\bibitem{Mandelstam}S.~Mandelstam, {\it Phys.\ Rev.}\ {\bf D11} (1975) 3026

\bibitem{AlZamolodchikov}Al.~B.~Zamolodchikov, {\it Int.\ J.\ Mod.\ Phys.}~{\bf A10} (1995) 1125

\end{thebibliography}

\bigbreak
\bigskip
\centerline{\bf Problems}
\medskip
{\bf1.}~Prove that the current (\ref{MT3current}) is conserved in the massless Thirring model. Find the divergence of the current for nonzero mass.

{\bf2.}~In the model of free massless Dirac fermions ($m=0$, $g=0$) find the pair correlation functions of the fermion fields $\langle\psi^+_i(x')\psi_j(x)\rangle$.

{\bf3.}~Obtain all classical solutions $\phi(t,x)$ of the sine\-/Gordon equation with finite energy that only depend on the combination $x-vt$ with some constant $v$, $|v|<1$. Find topological charges of these solutions.

{\bf4.}~Repeat the reasoning of the lecture in the special case of a free fermion ($g=0$). Check that in this case $m_{\rm phys}=m=\pi\mu$. Show that the bosonization reproduces the correct commutation relations for massless fermions.

{\bf5$^*$.}~Show that in the Thirring model, in consistency with (\ref{mphys}), in the one\-/loop approximation, the mass renormalizes as follows
$$
m_{\rm phys}=m\left(1+{g\over\pi}\log{\Lambda\over m}\right),
$$
where $\Lambda$ is the momentum cutoff parameter.

While deriving the diagrammatic technique, it is convenient to use the representation for the action of the Thirring model with an auxiliary field:
$$
S[\psi,\bar\psi,A^\mu]=\int d^Dx\,\left(\bar\psi(i\hat\d-\hat A-m)\psi+{1\over2g}A^\mu A_\mu\right).
$$



\end{document}
