Student Seminar: Classical and Quantum Integrable Systems

This results into the following equation which describes how the components of the
monodromy transform under the global symmetry generators
[S α , T a (λ)] = 1 2 [T a(λ), σ α a ] .
Thus, we end up with three separate equations
)
,
[S 3 , T a (λ)] = 1 2 [T a(λ), σa] 3 = 1 [ ( A(λ) B(λ)
2 C(λ) D(λ)
[S + , T a (λ)] = 1 [ (
2 [T a(λ), σ a + A(λ) B(λ)
] =
C(λ) D(λ)
and
[S − , T a (λ)] = 1 [ (
2 [T a(λ), σa − A(λ) B(λ)
] =
C(λ) D(λ)
)
,
)
,
( ) 1 0 ]
=
0 − 1
( ) 0 1 ]
=
0 0
( ) 0 0 ] (
=
1 0
Essentially, we need the following commutation relations
[S 3 , B] = −B , [S + , B] = A − D .
( )
0 −B(λ)
,
C(λ) 0
( )
−C(λ) A(λ) − D(λ)
,
0 C(λ)
B(λ) 0
D(λ) − A(λ) −B(λ)
The action of the symmetry generators on the pseudo-vacuum have been already
derived
S + |0〉 = 0 , S 3 |0〉 = L 2 |0〉 .
So the state |0〉 is the highest weight state of the symmetry algebra. Further, we find
and
S 3 |λ 1 , · · · , λ M 〉 =
( L
2 − M )
|λ 1 , · · · , λ M 〉
)
.
S + |λ 1 , · · · , λ M 〉 = ∑ j
B(λ 1 ) . . . B(λ j−1 )(A(λ j ) − D(λ j ))B(λ j+1 ) . . . B(λ M )|0〉
= ∑ j
O j B(λ 1 ) . . . B(λ j−1 ) ˆB(λ j )B(λ j+1 ) . . . B(λ M )|0〉 .
The coefficients O j are unknown for the moment. To calculate O j we will use the
arguments similar to those for computing Wj
A and Wj D . The only contributions to
O j will come from
B(λ 1 ) . . . B(λ k−1 )(A(λ k ) − D(λ k ))B(λ k+1 ) . . . B(λ M )|0〉 with k ≤ j.
If k = j this contribution will be
M∏
k j +1
λ j − λ k − i
(λ j + i ) L ∏ M
λ j − λ k + i
−
(λ j − i ) L
λ j − λ k 2
λ j − λ k 2
k j +1
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