- #1
Maybe_Memorie
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I'm studying the SU(3) invariant XXX chain as part of my Bachelor's thesis.
The monodromy matrix of this system can be written as a 3x3 matrix. We perform a 2x2 decomposition of it and write is as ##T(\mu)=\left(
\begin{array}{cc}
A(\mu) & B(\mu) \\
C(\mu) & D (\mu)
\end{array} \right)##
For a system to be integrable it has an R-matrix which satisfies the Yang-Baxter relation. I don't know how to decompose the R-matrix though.
Kulish/Resithiken writes it as, for the GL(N) case,
##R(\mu)=\left( \begin{array}{cccc}
\mu & 0 & 0 & 0 \\
0 & \mu I & I & 0 \\
0 & I & \mu I & 0 \\
0 & 0 & 0 & S(\mu)
\end{array} \right) ##
where I is the 2x2 identity matrix and S(u) is the SU(2) R-matrix.
The main reason this is confusing me is because when we write the Identity matrix and S(u) in the blocks we don't get the correct R-matrix, so i suspect there's a change of basis going on or something.
Any help is appreciated.
The monodromy matrix of this system can be written as a 3x3 matrix. We perform a 2x2 decomposition of it and write is as ##T(\mu)=\left(
\begin{array}{cc}
A(\mu) & B(\mu) \\
C(\mu) & D (\mu)
\end{array} \right)##
For a system to be integrable it has an R-matrix which satisfies the Yang-Baxter relation. I don't know how to decompose the R-matrix though.
Kulish/Resithiken writes it as, for the GL(N) case,
##R(\mu)=\left( \begin{array}{cccc}
\mu & 0 & 0 & 0 \\
0 & \mu I & I & 0 \\
0 & I & \mu I & 0 \\
0 & 0 & 0 & S(\mu)
\end{array} \right) ##
where I is the 2x2 identity matrix and S(u) is the SU(2) R-matrix.
The main reason this is confusing me is because when we write the Identity matrix and S(u) in the blocks we don't get the correct R-matrix, so i suspect there's a change of basis going on or something.
Any help is appreciated.