\section{Free field representation: SOS model}

The bosonization or free field representation appeared in conformal
field theory in the works by Feigin and Fuchs~\cite{FF} and by Dotsenko
and Fateev~\cite{DF} in 1983. It is no use to recall these papers for
our purposes. The most important outcome of these papers for us is that
some linear equations can be solved by representing the solution in
terms of expectation values of some quantum operators. The trace form
of the functions we want to obtain prompt us that it must be a thermal
average. If the Hamiltonian $H$ is quadratic in the bosonic field and
the operators $\Phi(u)^{n'}_n$ is expressed as exponentials of this
field, the problem will be solvable.

Let me first formally introduce the construction by Lukyanov and
Pugai~\cite{LP96} and then to explain how it can be obtained.

Consider a Heisenberg algebra of operators $a_k$ ($k\in\Z\setminus\{0\}$)
and a pair of `zero-mode' operators $\cP$ and $\cQ$ with the commutation
relations
\eq{
[\cP,\cQ]=-\i,\qquad
[a_k,a_l]=k{\[k\]_x\[(r-1)k\]_x\over\[2k\]_x\[rk\]_x}\delta_{k+l,0}
\quad\hbox{with}\quad
\[u\]_x={x^u-x^{-u}\over x-x^{-1}}.
\label{HeisAlg}
}
The `$q$-number' $\[u\]_x$ here should not be confused with the `elliptic
$q$-numbers' $[u]_i$ ($i=1,\ldots,4$). It is also useful to introduce
the operators
\eq{
\tilde a_k={\[rk\]_x\over\[(r-1)k\]_x}a_k.
\label{atilde}
}
The normal ordering operation $\lcolon\ldots\rcolon$
places $\cP$ to the right of $\cQ$ and $a_k$ with positive $k$
to the right of $a_{-k}$. It will be convenient to assign
\eq{
\alpha_+=\sqrt{a_+}=\sqrt{r\over r-1},
\qquad
\alpha_-=-\sqrt{a_-}=-\sqrt{r-1\over r},
\qquad
2\alpha_0=\alpha_++\alpha_-={1\over\sqrt{r(r-1)}}.
\label{alphapmdef}
}
Now introduce the fields
\eq{
\Aligned{
\varphi(z)
&={\alpha_-\over\sqrt2}(\cQ-\i\cP\log z)
-\sum_{k\neq0}{a_k\over\i k}z^{-k}.
\\
\tilde\varphi(z)
&={\alpha_+\over\sqrt2}(\cQ-\i\cP\log z)
+\sum_{k\neq0}{\tilde a_k\over\i k}z^{-k}.
}\label{varphidef}
}
These fields enter the exponential operators
\eq{
\Aligned{
V(u)
&=z^{(r-1)/4r}\lcolon e^{\i\varphi(z)}\rcolon,
&\qquad
\bV(u)
&=z^{(r-1)/r}\lcolon e^{-\i\varphi(x^{-1}z)-\i\varphi(xz)}\rcolon,
\\
\tV(u)
&=z^{r/4(r-1)}\lcolon e^{\i\tilde\varphi(z)}\rcolon,
&\qquad
\tbV(u)
&=z^{r/(r-1)}\lcolon e^{-\i\tilde\varphi(x^{-1}z)-\i\tilde\varphi(xz)}\rcolon,
}\label{VbarVdef}
}
and Lukyanov's screening operators
\eq{
\Aligned{
x(u,C)
&={\epsilon\over\eta}\int_C{dv\over\i\pi}\,
\bV(v){[v-u+{1\over2}-\sqrt{2r(r-1)}\,\cP]
\over[v-u-{1\over2}]},
\\
\tilde x(u,C)
&={\epsilon\over\eta'}\int_C{dv\over\i\pi}\,
\tbV(v){[v-u-{1\over2}+\sqrt{\textstyle2r(r-1)}\,\cP]'
\overtext[v-u+{1\over2}]'}.
}\label{screening}
}
The constants $\eta$, $\eta'$ will be fixed as
\eq{
\Aligned{
\eta^{-1}
&=\i[1]\,x^{r-1\over2r}
{(x^2;x^{2r})_\infty\over(x^{2r-2};x^{2r})_\infty}
{(x^6;x^4,x^{2r})_\infty(x^{2r+2};x^4,x^{2r})_\infty
\over
(x^4;x^4,x^{2r})_\infty(x^{2r+4};x^4,x^{2r})_\infty},
\\
\eta^{\prime-1}
&=-{2\epsilon\over\pi}[1]'x^{-{r\over2(r-1)}}
{(x^{2r-2};x^{2r-2})^2_\infty\over(x^{2r};x^{2r-2})^2_\infty}
{(x^4;x^4,x^{2r-2})_\infty(x^{2r+2};x^4,x^{2r-2})_\infty
\over
(x^2;x^4,x^{2r-2})_\infty(x^{2r+4};x^4,x^{2r-2})_\infty}.
}\label{etadef}
}
Now let us fix the contours. Let $C^-_u$ and $C^+_u$ go from
$u-{\i\pi\over2\epsilon}$ to $u+{\i\pi\over2\epsilon}$ to the left
and to the right of $u$ respectively.

(We assume that the contours $C^\pm_u$ go to the left of all
poles in the `main rectangle' related to the operators that are to the
right of the screening operator and to the right of all poles related
to the operators placed to the left of the screening operators. The
`main rectangle' is understood as a rectangle with sides $r$ along the
real axis and $\pi\over\epsilon$ along the imaginary axis that contains
all points $u_i$, $v_i$ {\it etc.} It is well defined for large enough
$r$ and for points $u_i$, $v_i$,\dots\ close enough to each other. In
the general case the operator products are considered as analytic
continuation from this region.)

Then
\eq{
\Aligned{
X(u)
&=x(u,C^-_{u+1/2}),
&\qquad
Y(u)
&=x(u-1,C^+_{u-1/2}),
\\
\tX(u)
&=\tilde x(u,C^-_{u-1/2}),
&\qquad
\tY(u)
&=\tilde x(u+1,C^+_{u+1/2}).
}\label{XYscreening}
}
These operators satisfy the equations
\eq{
Y(u)V(u)=V(u)X(u),
\qquad
\tY(u)\tV(u)=\tV(u)\tX(u).
\label{YXcorresp}
}

Define the Fock spaces $\cF_{mn}$ generated by the operators $a_{-k}$
($k>0$) from the highest weight vectors $|P_{mn}\rangle$ such that
\eq{
a_k|P_{mn}\rangle=0\quad (k>0),
\qquad
\cP|P_{mn}\rangle=P_{mn}|P_{mn}\rangle,
\qquad
P_{mn}={1\over\sqrt2}(\alpha_+m+\alpha_-n).
\label{Pvacdef}
}
There are strong evidences that $\cF_{mn}$ can be identified with
$\cH_{mn}$ for generic $r$.

The vertex operators are defined on $\cF_{mn}$ as follows:
\eq{
\Aligned{
\Phi(u)^{n+1}_n
&={\i^{m-n}\over[n]}V(u),
\\
\Phi(u)^{n-1}_n
&=-{\i^{m-n}\over[n]}V(u)X(u),
\\
\Psi^*(u)^{m+1}_m
&=\tV(u),
\\
\Psi^*(u)^{m-1}_m
&=(-1)^{m-n}\tY(u)\tV(u).
}\label{bosvertex}
}
The corner Hamiltonian $H_{mn}$ is the restriction to $\cF_{mn}$ of the
operator
\eq{
H={\cP^2\over2}+\sum_{k=1}^\infty
{\[2k\]_x\[rk\]_x\over\[k\]_x\[(r-1)k\]_x}a_{-k}a_k.
\label{bosH}
}
Ooh! This is the end at last!

The operators $H$ and $\Phi(u)^{n'}_n$ satisfy the necessary algebra of
commutation relations. Besides, the operators $\Psi^*(u)^{n'}_n$ satisfy
a similar algebra
\Align{
\span
\Psi^*(u)^{m'}_mx^{2vH_{mn}}=x^{2vH_{m'n}}\Psi^*(u-v)^{m'}_m,
\label{SOSAPsiCommut}
\\
\span
\sum_{s'}\tW(s',m;s,m'|u_1-u_2)
\Psi^*(u_1)^{m'}_{s'}\Psi^*(u_2)^{s'}_m
=\Psi^*(u_2)^{m'}_s\Psi^*(u_1)^s_m,
\label{SOSPsiPsiCommut}
\\
\span
\Psi(u')^{m'}_{m''}\Psi^*(u)^{m''}_m={1\over\pi}{\delta^{m'}_m\over u'-u}
+O(1),
\qquad
\Psi(u)^{m'}_m={1\over[m]'}\Psi^*(u-1)^{m'}_m.
\label{SOSPsiNorm}
}
Here
$$
\tW(m_1,m_2;m_3,m_4|u)=-\left.\W(m_1,m_2;m_3,m_4|u)\right|_{r\to r-1}.
$$
The operators $\Phi(u)$ are called the {\it type~I\/} vertex operators,
while $\Psi^*(u)$ are called the {\it type~II\/} vertex operators. The
difference between these operators is in their physical meaning. The
type~I operators are, as we already said, the half transfer matrices,
while the type~II vertex operators represent one-particle excitation
states. Both types of operators satisfy the relation
\eq{
\Phi(u_1)^{n'}_n\Psi^*(u_2)^{m'}_m
=\tau(u_1-u_2)\Psi^*(u_2)^{m'}_m\Phi(u_1)^{n'}_n,
\qquad
\tau(u)
=\i{\theta_1\left({1\over4}-{u\over2};{\i\pi\over2\epsilon}\right)
\overtext
\theta_1\left({1\over4}+{u\over2};{\i\pi\over2\epsilon}\right)}.
\label{SOSPhiPsiCommut}
}
The function $\prod^l_{j=1}\tau(u-v_j)$ is the eigenvalue of the
transfer matrix $T_{\rm SOS}(u)$ on the excited states of $l$ particles.
The functions $\tW(m_1,m_2;m_3,m_4|v_1-v_2)$ provide the scattering
matrix of two excitations. A trace function with a product of
$\Psi^*(v)$ and $\Psi(v)$ inserted represents a matrix element of a
local operator described by $\Phi$s instead of its vacuum expectation
value. Namely,
\Multline{
\biggl\langle\Matrix{mm'_1\ldots m'_{l'-1}m'\\v'_1,\ldots,v'_{l'}}\biggr|
\cO\Bigl(n\Matrix{n_1\ldots n_{k-1}\\n'_1\ldots n'_{k-1}}n'\Bigr)
\biggl|\Matrix{mm_1\ldots m_{l-1}m'\\v_1,\ldots,v_l}\biggr\rangle
\\
={1\over\chi_m}
\Tr_{\cF_{mn}}(\Psi(v'_1)^m_{m'_1}\ldots\Psi(v'_{l'})^{m'_{l'-1}}_{m'}
\Psi^*(v_l)^{m'}_{m_{l-1}}\ldots\Psi^*(v_1)^{m_1}_m
\\
\times
\Phi^*(u_1)^n_{n'_1}\ldots\Phi^*(u_k)^{n'_{k-1}}_{n'}
\Phi(u_k)^{n'}_{n_{k-1}}\ldots\Phi(u_1)^{n_1}_nx^{4H}),
\notag
}
where $\cO\Bigl(n\Matrix{n_1\ldots n_{k-1}\\n'_1\ldots
n'_{k-1}}n'\Bigr)$ is the operator corresponding to the picture in the
last lecture. The corresponding vacuum expectation values are just the
probabilities $P\Bigl(n\Matrix{n_1\ldots n_{k-1}\\n'_1\ldots
n'_{k-1}}n'\Bigr)$.

Now let us make some comments on the construction. Derivation of the
construction above starts from the following observation. Consider
first the commutation relation
$$
\Phi(u_1)^{n+2}_{n+1}\Phi(u_2)^{n+1}_n
=\W(n+1,n;n+1,n+2|u_1-u_2)\Phi(u_2)^{n+2}_{n+1}\Phi(u_1)^{n+1}_n.
$$
Since
$$
\W(n+1,n;n+1,n+2|u)=R_0(u)=z^{(r-1)/2r}{g(z^{-1})\over g(z)},
\qquad
z=x^{2u},
$$
we can rewrite it as
$$
(z_2/z_1)^{(r-1)/4r}g^{-1}(z_2/z_1)
\Phi(u_1)^{n+2}_{n+1}\Phi(u_2)^{n+1}_n
=(z_1/z_2)^{(r-1)/4r}g^{-1}(z_1/z_2)
\Phi(u_2)^{n+2}_{n+1}\Phi(u_1)^{n+1}_n
$$
This can be reproduced if $\Phi(u)^{n+1}_n\sim\lcolon
e^{\i\varphi(u)}\rcolon$ with $\varphi(u)$ has the form
(\ref{varphidef}). The pairs $(a_n,a_{-n})$ are supposed to form
independent Heisenberg algebra, but the normalization from
(\ref{HeisAlg}) is not supposed. It is known that
$$
\lcolon e^{\i\varphi_1}\rcolon\lcolon e^{\i\varphi_2}\rcolon
=e^{-\langle0|\varphi_1\varphi_2|0\rangle}
\lcolon e^{\i\varphi_1+\i\varphi_2}\rcolon,
$$
where $\varphi_1$, $\varphi_2$ are any linear combinations of $\cP$,
$\cQ$, $a_k$. It means that if
$$
\langle0|\varphi(u_1)\varphi(u_2)|0\rangle
=-\log((z_2/z_1)^{-(r-1)/4r}g(z_2/z_1)),
$$
we will be able to satisfy our equation.

Let us represent $\log g(z)$ in the form of a series in $z$. Namely, use
the identity
$$
\log(z;p_1,p_2)_\infty=\sum^\infty_{n_1,n_2=0}\log(1-zp_1^{n_1}p_2^{n_2})
=-\sum^\infty_{n_1,n_2=0}\sum^\infty_{m=1}
{z^mp_1^{mn_1}p_2^{mn_2}\over m}
=-\sum^\infty_{m=1}{z^m\over(1-p_1^m)(1-p_2^m)}.
$$
Applying it to the definition of $g(z)$, we obtain
$$
\log g(z)=-\sum^\infty_{m=1}
{(x^{2m}+x^{(2r+2)m}-x^{4m}-x^{2rm})z^m\over(1-x^{4m})(1-x^{2rm})}.
$$
This reproduces the normalizations in~(\ref{HeisAlg}).

Obtaining the other relations, e.g.
$$
\Phi(u_1)^n_{n-1}\Phi(u_2)^{n-1}_n
=(\text{something})\times\Phi(u_2)^n_{n-1}\Phi(u_1)^{n-1}_n
+(\text{something})\times\Phi(u_2)^n_{n+1}\Phi(u_1)^{n+1}_n
$$
Without the second term it could be reproduced by pure exponentials
(but with a wrong coefficient!), but the second term spoils everything.
From conformal field theory it is known that such commutation relations
can by obtained by use of the {\it screening operators}, which are
integral of exponentials. The particular form of the screening operator
(\ref{screening}) was guessed after long attempts to use a simpler form
without an elliptic function of the zero mode operator. The
normalization constant $\eta$ is extracted from the normalization
property for the vertex operators.

The $\Psi$ operators appeared as a natural generalizations of some
operators in the conformal field theory. Their meaning as
representatives of excited states was established by comparing with
similar construction for the six-vertex model by Jimbo and Miwa and with
Lukyanov's construction for form factors in quantum field theory.

Similar operators $\Psi^*_\ve(u)$ must exist for the eight-vertex model.
The algebra of these operators was established by Foda, Iohara, Jimbo,
Miwa, and Yan in 1994 in the study of the elliptic algebra ${\cal
A}_{q,p}(\widehat{sl}_2)$~\cite{FIJKMY}. Let us write down the result:
\Align{
\span
\sum_{\ve'_1\ve'_2}\tR(u_1-u_2)^{\ve'_1\ve'_2}_{\ve_1\ve_2}
\Psi^*_{\ve'_1}(u_1)\Psi^*_{\ve'_2}(u_2)
=\Psi^*_{\ve_2}(u_2)\Psi^*_{\ve_1}(u_1),
\label{PsiPsiCommut}
\\
\span
\Psi_{\ve_1}(u_1)\Phi_{\ve_2}(u_2)
=\tau(u_1-u_2)\Phi_{\ve_2}(u_2)\Psi_{\ve_1}(u_1),
\label{PsiPhiCommut}
\\
\span
\Psi_{\ve_1}(u')\Psi^*_{\ve_2}(u)
={1\over\pi}{\delta_{\ve_1\ve_2}\over u'-u}+O(1),
\qquad
\Psi_\ve(u)=\Psi^*_{-\ve}(u-1).
\label{PsiNorm}
}
Here the $\tR(u)$ matrix is defined by the weights
\eq{
\tilde a(u)=-a(u)|_{r\to r-1},
\quad
\tilde b(u)=-b(u)|_{r\to r-1},
\quad
\tilde c(u)=-c(u)|_{r\to r-1},
\quad
\tilde d(u)=d(u)|_{r\to r-1}
\label{tRdef}
}
and provides the $S$ matrix of the eight-vertex model. This form of the
$S$ matrix was confirmed by Takebe by means of the Bethe
ansatz~\cite{Takebe95}.

The function $\tau(u)$ here is the same function as in the similar
commutation relation (\ref{SOSPhiPsiCommut}) for the SOS model. It is
not surprising, because we know that the spectra of transfer matrices
of both models coincide.

How to calculate anything with these bosonic fields? Denote by $\Tr_*$
the trace over oscillator modes and by $H^*$ the oscillator
contribution to $H_{mn}$. Besides, let
$$
\chi^*=\Tr_*(x^{4H^*})={1\over(x^4;x^4)_\infty}.
$$
Then, according to the Wick theorem,
$$
\Tr_{\cF_{mn}}(U_N(u_N)\ldots U_1(u_1))
=\langle P_{mn}|U^0_N(u_N)\ldots U^0_1(u_1)|P_{mn}\rangle\,
\chi^*\prod_{i=1}^N c_i\prod_{i<j}g_{ij}(u_i-u_j)
$$
with
$$
\Aligned{
\log c_i
&={1\over\chi^*}\Tr_*(\phi^+_i(0)\phi^-_i(0)x^{4H^*}),
\\
\log g_{ij}(u)
&={1\over\chi^*}\Tr_*(\phi_i(0)\phi_j(u)x^{4H^*}),
\qquad
\phi_i(u)=\phi^+_i(u)+\phi^-_i(u).
}
$$
The constants $c_i$ and functions $g_{ij}$ are expressed in terms of the
infinite products $(z;p_1,\ldots,p_k)_\infty$ defined above. The
resulting form factors are expressed in terms of integrations of the
products of these functions.

We know the vertex--face correspondence for the weights. Is it possible
to relate somehow the vertex operator algebras? Surely, it is, but we
need again a kind of `physical reasoning'. You remember the relation
$$
[m']\Tr_{\cH^{(i)}}(x^{2H})
=\sum_{n\in2\Z+m+i}\Tr_{\cH_{mn}}x^{4H_{mn}}.
$$
Analysis of the low temperature expansion really indicates the relation
between $i$th ground state of the eight vertex model and the
$n-m=i\pmod2$ ground states of the SOS model. So, suppose that there
exist the operators
\eq{
T(u_0)_{mn}\>:\>\cH_{mn}\to\cH^{(i)},
\qquad
T(u_0)^{mn}\>:\>\cH^{(i)}\to\cH_{mn},
\qquad
i=n-m\pmod2,
\label{Tdef}
}
such that
\eq{
[m]'x^{2H^{(i)}}=\sum_{n\in2\Z+m+i}T(u_0)_{mn}x^{4H_{mn}}T(u_0)^{mn}
\label{CTMIntert}
}
and
\eq{
\Aligned{
\sum_n t_\ve(u-u_0)^{n'}_n\Phi(u)^{n'}_nT(u_0)^{mn}
&=T(u_0)^{mn'}\Phi_\ve(u)
\\
\sum_\ve t^*_\ve(u-u_0)_{n'}^n\Phi_\ve(u)T(u_0)_{mn}
&=T(u_0)_{mn'}\Phi(u)_n^{n'},
}\label{PhiIntert}
}
The operators $T(u_0)^{mn}$ and $T(u_0)_{mn}$ can be considered as a
half of the $\tau(u_0)$ transfer matrix and its `conjugate'
$\tau^*(u_0)$ in the infinite volume limit:
$$
T(u_0)^{mn}
=\vcenter{\beginpicture
%
\setcoordinatesystem units <1cm,1cm> point at 0 0
\trl 0.0 -0.25 , 0.0 1.25
\setdashes
\trl 0.0 1.25 , 0.0 2.25
\setsolid
\ln 0.0 -0.1 , -0.1 0.0 \ln -0.1 0.0 , 0.0 0.1
\rl -0.1 0.0 , -0.7 0.0 \ar -0.68 0.0 , -0.7 0.0
\ln 0.0 0.4 , -0.1 0.5 \ln -0.1 0.5 , 0.0 0.6
\rl -0.1 0.5 , -0.7 0.5 \ar -0.68 0.5 , -0.7 0.5
\ln 0.0 0.9 , -0.1 1.0 \ln -0.1 1.0 , 0.0 1.1
\rl -0.1 1.0 , -0.7 1.0 \ar -0.68 1.0 , -0.7 1.0
\put{$\ve_1$} [rB] <-2pt,-2pt> at -0.7 0.0
\put{$\ve_2$} [rB] <-2pt,-2pt> at -0.7 0.5
\put{$\ve_3$} [rB] <-2pt,-2pt> at -0.7 1.0
\put{$n_1$} [lB] <2pt,-2pt> at 0.0 -0.25
\put{$n_2$} [lB] <2pt,-2pt> at 0.0 0.25
\put{$n_3$} [lB] <2pt,-2pt> at 0.0 0.75
\put{$n_4$} [lB] <2pt,-2pt> at 0.0 1.25
%
\endpicture
}
\qquad\qquad
T(u_0)_{mn}
=\vcenter{\beginpicture
%
\setcoordinatesystem units <1cm,1cm> point at 0 0
\trl 0.0 -0.25 , 0.0 1.25
\setdashes
\trl 0.0 1.25 , 0.0 2.25
\setsolid
\ln 0.0 -0.1 , 0.1 0.0 \ln 0.1 0.0 , 0.0 0.1
\rl 0.1 0.0 , 0.5 0.0
\ln 0.0 0.4 , 0.1 0.5 \ln 0.1 0.5 , 0.0 0.6
\rl 0.1 0.5 , 0.5 0.5
\ln 0.0 0.9 , 0.1 1.0 \ln 0.1 1.0 , 0.0 1.1
\rl 0.1 1.0 , 0.5 1.0
\put{$\ve_1$} [lB] <2pt,-2pt> at 0.5 0.0
\put{$\ve_2$} [lB] <2pt,-2pt> at 0.5 0.5
\put{$\ve_3$} [lB] <2pt,-2pt> at 0.5 1.0
\put{$n_1$} [rB] <-2pt,-2pt> at 0.0 -0.25
\put{$n_2$} [rB] <-2pt,-2pt> at 0.0 0.25
\put{$n_3$} [rB] <-2pt,-2pt> at 0.0 0.75
\put{$n_4$} [rB] <-2pt,-2pt> at 0.0 1.25
%
\endpicture
}
$$
How the operators $T(u_0)^{mn}$ and $T(u_0)_{mn}$ intertwine the
type~II operators? In the spirit of this algebraic approach, we can
suggest that the commutation relation look like
\eq{
\Aligned{
\Psi^*_\ve(u)T(u_0)_{mn}
&=\sum_{m'} T(u_0)_{m'n}\Psi^*(u)^{m'}_m
\tdt^*_\ve(u-u_0-\Delta u_0)^m_{m'},
\\
\Psi^*(u)^{m'}_mT(u_0)^{mn}
&=\sum_\ve T(u_0)^{m'n}\Psi^*_\ve(u)\tdt_\ve(u-u_0-\Delta u_0)^{m'}_m,
}\label{PsiIntert}
}
where $\Delta u_0$ is some shift and $\tdt_\ve(u)^{m'}_m$ are
intertwining functions after the substitution $r\to r-1$ and erasing
$e^{\pm\i\pi/4}$:
\eq{
\Aligned{
\tdt_+(u)^{n'}_n
&=\tf(u)
\theta_3\!\left({(n'-n)u+n'\over2r'};\i{\pi\over2\epsilon r'}\right),
\\
\tdt_-(u)^{n'}_n
&=\tf(u)
\theta_4\!\left({(n'-n)u+n'\over2r'};\i{\pi\over2\epsilon r'}\right),
\qquad r'=r-1.
}\label{TildeIntertwining}
}
The normalization conditions fix the overall factors $f(u)$ and $\tf(u)$
to be solutions of the equations
$$
\Aligned{{}
[u]f(u)f(u-1)
&=C
\equiv{[0]_4^2\over2\theta_3(0;\i\pi/2\epsilon r)
\theta_4(0;\i\pi/2\epsilon r)},
\\
[u]'\tf(u)\tf(u-1)
&=C'
\equiv{[0]_4^{\prime2}\over2\theta_3(0;\i\pi/2\epsilon r')
\theta_4(0;\i\pi/2\epsilon r')}.
}
$$
The function $f(u)$ is not essential for the answers since the type~I
vertex operators $\Phi$ are always accompanied by the
`conjugates'~$\Phi^*$. But the particular form of $\tf(u)$ is essential
and will be fixed on the basis of the bosonization procedure as well as
the value of the shift $\Delta u_0$.

Consider any form factor from the eight-vertex model
$$
{1\over\chi^{(i)}}\Tr_{\cH^{(i)}}(
\Phi^*_{\ve'_1}(u_1)\ldots\Phi^*_{\ve'_k}(u_k)
\Phi_{\ve_k}(u_k)\ldots\Phi_{\ve_1}(u_1)
\Psi^*_{\alpha_l}(v_l)\ldots\Psi^*_{\alpha_1}(v_1)x^{2H^{(i)}})
$$
We can represent the operator $x^{2H^{(i)}}$ according to
(\ref{CTMIntert}) and push $T(u_0)_{mn}$ to the left and $T(u_0)^{mn}$
to the right by use of the intertwining relations (\ref{PhiIntert})
and~(\ref{PsiIntert}). We obtain an infinite liner combination of the
traces
$$
{[n]\over[m']'\chi^{(i)}}\Tr_{\cH_{m'n}}(
\Phi^*(u_1)^n_{n'_1}\ldots\Phi^*(u_k)^{n'_{k-1}}_{n''}
\Phi(u_k)^{n''}_{n_{k-1}}\ldots\Phi(u_1)^{n_1}_{n'}
\Lambda(u_0)^{m'n'}_{mn}
\Psi(u_l)^m_{m_{l-1}}\ldots\Psi(u_1)^{m_1}_{m'}x^{4H_{m'n}})
$$
with
\eq{
\Lambda(u_0)^{m'n'}_{mn}=T(u_0)^{m'n'}T(u_0)_{mn}
\>:\>\cH_{mn}\to\cH_{m'n'}.
\label{LambdaDef}
}
All ingredients in this trace are known from the free field
representation except the operators $\Lambda(u_0)^{m'n'}_{mn}$. In the
next lecture we fix the bosonic form of these operators using the
commutation relations of this operator with the vertex operators.
