\section{Free field representation: eight-vertex model via SOS model}

At the end of the last lecture we introduced the operator
$\Lambda(u_0)^{m'n'}_{mn}=T(u_0)^{m'n'}T(u_0)_{mn}$. We said that,
having a bosonization of the SOS model, this operator is the only thing
to be bosonized. To obtain bosonization for the $\Lambda$ operator let us
consider its commutation relations with the vertex operators. From the
commutation relations of the operators $T(u_0)^{mn}$ and $T(u_0)_{mn}$
we find
\Align{
\Lambda(u_0)^{m'}_m{}^{n'}_s\Phi(u)^s_n
&=\sum_{s'}\L(s,n;s',n'|u-u_0)\Phi(u)_{s'}^{n'}
\Lambda(u_0)^{m'}_m{}^{s'}_n,
\label{PhiLambdaCommut}
\\
\Psi^*(u)_s^{m'}\Lambda(u_0)^s_m{}^{n'}_n
&=\sum_{s'}\Lambda(u_0)^{m'}_{s'}{}^{n'}_n\Psi^*(u)_m^{s'}
\tL(s',m;s,m'|u-u_0-\Delta u_0),
\label{PsiLambdaCommut}
}
where
\eq{
\L(n_1,n_2;n_3,n_4|u)
=\sum_\ve t^*_\ve(u)^{n_2}_{n_1} t_\ve(u)^{n_4}_{n_3},
\qquad
\tL(m_1,m_2;m_3,m_4|u)
=\sum_\ve\tdt^*_\ve(u)^{m_2}_{m_1}\tdt_\ve(u)^{m_4}_{m_3}
=\left.\L(m_1,m_2;m_3,m_4|u)\right|_{r\to r-1}.
\label{LwDef}
}
Explicitly, we have
\eq{
\Aligned{
\L(n,n\pm1;n'\pm1,n'|u)
&={[u\pm{1\over2}(n-n')][{1\over2}(n+n')]\over [u][n]},
\\
\L(n,n\pm1;n'\mp1,n'|u)
&={[u\pm{1\over2}(n+n')][{1\over2}(n-n')]\over [u][n]}
}\label{LExplicit}
}
for $n'-n\in2\Z$. Note, that the number of type~I vertex operators and
that of type~II vertex operators in any meaningful trace is even. This
means that considering the $\Lambda$ operator for $n'-n\in2\Z$ and
$m'-m\in2\Z$ is natural.

Evidently,
\eq{
\L(n,n';n'',n|u)=\delta_{n'n''}.
\label{LDelta}
}
As the $\Lambda$ operator is a `half infinite product' of $L$s, we
easily conclude that
\eq{
\Lambda(u)^{m'n}_{mn}=\delta_{m'm}.
\label{LambdaDelta}
}
Besides,
\eq{
\Lambda(u_0)^{m'}_m{}^{n'}_n=0
\text{\quad if\quad $m'<m$, $n'>n$ or $m'>m$, $n'<n$}.
\label{LambdaGreaterLess}
}
Indeed, if, for example $m'\le m$, $n'>n$ for nonzero
$\Lambda(u)^{m'n'}_{mn}$
, there must be a point $j$ at the
paths where $n(j)=n'(j)$. But, due to (\ref{LDelta}), it means that
$n(j+1)=n'(j+1)$ and, by induction $n(k)=n'(k)$ for any $k>j$.
Therefore, $m'=m$.

Let us start derivation from the case $m'=m$. This case is sufficient
for calculation of correlation functions. Consider the commutation
relation (\ref{PhiLambdaCommut}) in the limit $u\to u_0$. In this limit
$L(s,n;s',n|u-u_0)\to\infty$ so that
$$
L(n\pm1,n;n+1,n|u-u_0)/L(n\pm1,n;n-1,n|u-u_0)\to1.
$$
We obtain the relation (we omit the indices $m$ for simplicity)
$$
\Phi(u)^{n'}_{n'-1}\Lambda(u)^{n'-1}_n
=-\Phi(u)^{n'}_{n'+1}\Lambda(u)^{n'+1}_n
$$
which amounts
$$
V(u)\Lambda(u)^{n'-1}_n
={[n'-1]\over[n'+1]}V(u)X(u)\Lambda(u)^{n'+1}_n.
$$
We can conjecture that
\eq{
\Lambda(u)^{m\>n-2l}_{m\>n}
={[n-2l]\over[n]}X^l(u)\quad\text{for}\quad l\geq0.
\label{DiagLambdaBos}
}
If we substitute this solution to the commutation relation
(\ref{PhiLambdaCommut}) in its general form, we can make sure that this
is indeed the solution to these commutation relations $n'\le n$.

But what to do in the case $n'>n$? Look at the weights of the SOS model.
They are invariant with respect to the reflection
$$
m\to-m,\qquad n\to-n.
$$
It means that we can identify with $\cH_{mn}$ not only $\cF_{mn}$, but
also $\cF_{-m,-n}$. With this identification we have an alternate
bosonization for the vertex operators of the SOS model:
\eq{
\Aligned{
\Phi(u)^{n+1}_n
&={\i^{n-m}\over[n]}V(u)X(u),
\\
\Phi(u)^{n-1}_n
&=-{\i^{n-m}\over[n]}V(u),
\\
\Psi^*(u)^{m+1}_m
&=(-1)^{m-n}\tY(u)\tV(u),
\\
\Psi^*(u)^{m-1}_m
&=\tV(u).
}\label{VOBosAlternate}
}
In this alternate bosonization we have
\eq{
\Lambda(u)^{m\>n+2l}_{m\>n}
={[n+2l]\over[n]}X^l(u)\quad\text{for}\quad l\geq0.
\label{DiagLambdaAltBos}
}
We use different free field representation in different cases. This
fact must not embarrass you, because the $\Lambda$ operator enters the
trace once. Hence, we simply need to use different bosonization for
different traces.

This is sufficient for calculation of correlation functions. But the
construction contains two free parameters $u_0$ and $m$. If there would
be a rigorous proof of our construction, we would be sure that the
answer is independent of these parameters. But from the mathematical
point of view our construction is nothing but a conjecture. As a test we
must check the $u_0$ and $m$ independence. The $u_0$ independence can be
proven from periodicity properties: it turns out that the correlation
function constructed from this bosonization procedure must be periodic
in $u_0$ with two periods, $2$ and $2r$. As $r$ is, generically,
irrational, the answer is a constant. We have no proof of $m$
independence, but the simplest examples of the one- and two-point
functions demonstrate this independence. The answer for the one-point
function is nothing but the famous Baxter's staggered spontaneous
polarization:
$$
\langle\sigma^z\rangle^{(i)}
\equiv P^{(i)}{}^+_+-P^{(i)}{}^-_-
=(-)^i{(x^2;x^2)_\infty^2(-x^{2r};x^{2r})_\infty^2
\over(-x^2;x^2)_\infty^2(x^{2r};x^{2r})_\infty^2}.
$$

Now turn our attention to the case $m'\ne m$. Due to the selection rule
(\ref{LambdaGreaterLess}), we have to use the first bosonization in the
case $n'\le n$, $m'\le m$ and the alternative bosonization in the case
$n'\ge n$, $m'\ge m$. First of all, let us try to find
$\Lambda(u)^{m-2}_m{\>}^{n-2}_n$. Take the commutation
relation~(\ref{PsiLambdaCommut}) for $s=m$, $m'=m-1$. In this case it
becomes
$$
\Psi^*(u)^{m-1}_m\Lambda(u_0)^m_m{}^{n'}_n
=\Lambda(u_0)^{m-1}_{m-1}{\>}^{n'}_n\Psi^*(u)^{m-1}_m
+\Lambda(u_0)^{m-1}_{m+1}{\>}^{n'}_n\Psi^*(u)^{m+1}_m
{[u-u_0-\Delta u_0-m]'[1]'\overtext[u-u_0-\Delta u_0]'[m+1]'}.
$$
For $n'=n-2$ we have
$$
{[n-2]\over[n]}\tY(u)X(u_0)\tV(u)
={[n-2]\over[n]}X(u_0)\tY(u)\tV(u)
+(-1)^{n-m}\Lambda(u_0)^{m-1}_{m-1}{\>}^{n-2}_n\tV(u)
{[u-u_0-\Delta u_0-m]'[1]'\overtext[u-u_0-\Delta u_0]'[m+1]'}.
$$
Erasing $\tV(u)$ we obtain
\eq{
\Lambda(u)^{m-2}_m{\>}^{n-2}_n\tV(u_0)
=(-)^{n-m}{[u-u_0-\Delta u_0-m+1]'[1]'\overtext[u-u_0-\Delta u_0]'[m]'}
{[n-2]\over[n]}[\tY(u),X(u_0)]\tV(u).
\label{LambdatV}
}
What is the commutator $[\tY(u),X(u_0)]$? If the contours of the
commutator would not catch poles it would be zero. But it can be proven
that this commutator is a simple combinations of exponentials:
\Align{
[\tY(u'),X(u)]|_{\cF_{mn}}
&={\epsilon\over\eta\eta'\sh\epsilon}\biggl(
{[m-1]'\overtext\partial[0]'}
{[u'-u+{1\over2}-n]\over[u'-u-{1\over2}]}
\,W_+(u')
\notag
\\
&\quad
+{[n-1]\over\partial[0]}
{[u'-u+{3\over2}-m]'\overtext[u'-u+{1\over2}]'}
\,W_-(u)\biggr).
\label{YXCommut}
}
Here $\partial[0]=d[u]/du|_{u=0}$,
$\partial[0]'=d[u]'/du|_{u=0}$ and
\eq{
W_+(u)=W(u+{\textstyle{r\over2}}),
\qquad
W_-(u)=W(u-{\textstyle{r-1\over2}})
\label{WpmDef}
}
with
\eq{
\Aligned{
W(u)
&=z^{1/r(r-1)}\lcolon e^{\i\varphi_0(z)}\rcolon,
\\
\varphi_0(z)
&=-2\sqrt2\alpha_0(\cQ-\i\cP\log z)
-\sum_{k\ne0}{\[2k\]_x\over\[(r-1)k\]_x}{a_k\over\i k}z^{-k}.
}\label{WDef}
}
It can be checked that
$$
W_+(u)\tV(u)=0.
$$
Therefore the first term in the commutator does not affect the operator
$\Lambda(u)$. But the second exponential $W_-$ has the argument of the
$X$ screening operator. The difference $u-u_0$ drop out
of~(\ref{LambdatV}) with these substitution, if we accept that
\eq{
\Delta u_0=-1/2.
\label{Deltau0}
}
Finally, we obtain
\eq{
\Lambda(u)^{m-2}_m{\>}^{n-2}_n
=(-)^{n-m}{\epsilon\over\eta\eta'\sh\epsilon}
{[m]'\overtext[1]'}{[n-1][n-2]\over\partial[0]\,[n]}
\,W_-(u).
\label{LambdaSimplest}
}
We see that $\Lambda(u)^{m-2}_m{\>}^{n-2}_n$ is proportional to
$W_-(u)$. If we return to the commutation relation $[\tY(u'),X(u)]$ we 
shall see that at the point $u'=u-1/2$ the second term with $W_-(u)$ has
a pole, while the first one remains finite. Therefore,
$$
\lim_{u'\to u-1/2}[u'-u+1/2]'[\tY(u'),X(u)]\sim W_-(u).
$$
One can check that this pole is appears in the product $\tY(u')X(u)$,
while the product $X(u)\tY(u')$ is regular at that point. It means that
$$
\Lambda(u)^{m-2}_m{\>}^{n-2}_n
\sim\lim_{u'\to u}[u'-u]'\tY(u'-1/2)X(u).
$$
We may conjecture that
\eq{
\Lambda(u)^{m-2k}_m{\>}^{n-2l}_n
\sim\lim_{u'\to u}[u'-u]'\tY^k(u'-1/2)X^l(u)
\quad\text{for $k,l>0$}.
\label{LambdaGeneralConjecture}
}
It can be checked that the product in the right hand side has just a
simple pole and that
$$
\lim_{u'\to u}[u'-u]'\tY^k(u'-1/2)X^l(u)
\sim\tY^{k-1}(u-1/2)W_-(u)X^{l-1}(u).
$$
The coefficients in (\ref{LambdaGeneralConjecture}) can be obtained from
the intertwining relation (\ref{PsiLambdaCommut}) in the limit $u\to
u_0-1/2$ just in the same way as we obtained the coefficient
in~(\ref{DiagLambdaBos}) using the equation
$$
\Psi^*(u)^{m-2k+1}_{m-2k}\Lambda(u_0)^{m-2k}_m{\,}^{n-2l}_n
=\Psi^*(u)^{m-2k+1}_{m-2k+2}\Lambda(u_0)^{m-2k+2}_m{\,}^{n-2l}_n+O(1)
\quad\text{as}\quad
u\to u_0-{\textstyle{1\over2}}
$$
and one more equation that follows from (\ref{PhiLambdaCommut}) in the
limit  $u\to u_0$ (but written in the form $\Phi(u)\Lambda(u_0)=\ldots$).

Then we have to substitute the answer into the relations
(\ref{PhiLambdaCommut}) and  (\ref{PsiLambdaCommut}) in the general form
and to check it. It is a very cumbersome calculation, but it was done.
Now the answer is
\Align{
\left.\Lambda(u)^{m-2k}_m{\,}^{n-2l}_n\right|_{\cF_{mn}}
&=C^{m-2k}_m{\,}^{n-2l}_n
\lim_{u'\to u}[u'-u]'\tY^k(u'-{\textstyle{1\over2}})X^l(u)
\notag
\\*
&=D^{m-2k}_m{\,}^{n-2l}_n\tX^{k-1}(u-{\textstyle{1\over2}})
W_-(u)Y^{l-1}(u),
\label{Lambdafin a}
\\
\left.\Lambda(u)^{m+2k}_m{\,}^{n+2l}_n\right|_{\cF_{-m\,-n}}
&=C^{m+2k}_m{\,}^{n+2l}_n
\lim_{u'\to u}[u'-u]'\tY^k(u'-{\textstyle{1\over2}})X^l(u)
\notag
\\*
&=D^{m+2k}_m{\,}^{n+2l}_n\tX^{k-1}(u-{\textstyle{1\over2}})
W_-(u)Y^{l-1}(u)
\label{Lambdafin b}
}
for $k,l>0$ with
\eq{
\Aligned{
D^{m-2k}_m{\,}^{n-2l}_n
&=(-)^{(m-n+1)k+l+1}{\epsilon\over\eta\eta'\sh\epsilon}
{[m]'\overtext[1]'}
{[l][n-l][n-2l]\over\partial[0]\,[1][n]}.
\\
C^{m-2k}_m{\,}^{n-2l}_n
&=(-)^{(m-n)k}{[m]'\overtext[k]'[m-k]'}
{[n-2l]\over [n]}.
}\label{Lambdacoeffs}
}

The two-particle form factors of any of operators $\sigma^a$ can be
obtained without integrations. Any of these calculations (for example
for $\sigma^z$) allows to fix the function $\tf(u)$ from the condition
of $u_0$ independence of the answer. The explicit (and readable)
answers for these quantities can be found in the paper by Lukyanov and
Terras~\cite{LT}.
