Theoretical and Mathematical Physics, 204(3): 1195–1200 (2020)
WRONSKIAN-TYPE FORMULA FOR INHOMOGENEOUS TQ
EQUATIONS
Rafael I. Nepomechie
It is known that the transfer-matrix eigenvalues of the isotropic open Heisenberg quantum spin-1/2 chain
with nondiagonal boundary magnetic fields satisfy a TQ equation with an inhomogeneous term. We
derive a discrete Wronskian-type formula relating a solution of this inhomogeneous TQ equation to the
corresponding solution of a dual inhomogeneous TQ equation.
Keywords: Bethe ansatz, TQ equation, discrete Wronskian, boundary integrability
DOI: 10.1134/S004057792009007X
1. Introduction and summary of results
We consider the famous Baxter TQ equation for the closed periodic XXX spin chain of length N:
T (u)Q(u)=(u
+
)
N
Q
−−
(u)+(u
)
N
Q
++
(u). (1.1)
Here and hereafter, we use the brief notation f
±
(u)=f(u ±i/2) and f
±±
(u)=f(u ±i). It is known that
for a given transfer-matrix eigenvalue T (u), Eq. (1.1) can be regarded as a second-order finite-difference
equation for Q(u). The eigenvalue T (u) is necessarily a polynomial in u (of degree N ) because the model
is integrable. It is well known that Eq. (1.1) has two independent polynomial solutions [1]. One of them is
a polynomial Q(u)ofdegreeM N/2oftheform
Q(u)=
M
k=1
(u u
k
), (1.2)
whose zeros {u
k
} are solutions of the Bethe equations that follow directly from (1.1)
u
j
+ i/2
u
j
i/2
N
=
M
k=1,
k=j
u
j
u
k
+ i
u
j
u
k
i
,j=1,...,M. (1.3)
The other is a polynomial P (u)ofdegreeN M +1>N/2 corresponding to Bethe roots “on the other
side of the equator.” These two solutions are related by a discrete Wronskian (or Casoratian) formula
P
+
(u)Q
(u) P
(u)Q
+
(u) u
N
, (1.4)
Physics Department, University of Miami, Coral Gables, Florida, USA, e-mail: nep[email protected].
This research was supported in part by a Cooper fellowship.
Prepared from an English manuscript submitted by the author; for the Russian version, see Teoreticheskaya
i Matematicheskaya Fizika, Vol. 204, No. 3, pp. 430–435, September, 2020. Received January 17, 2020. Revised
March 23, 2020. Accepted March 23, 2020.
0040-5779/20/2043-1195
c
2020 Pleiades Publishing, Ltd.
1195
where denotes equality up to a multiplicative constant. The existence of a second polynomial solution of
the TQ equation is equivalent to the admissibility of the Bethe roots [2], [3]. Using the Wronskian formula,
we can succinctly reformulate the Q-system for this model [4] (which provides an efficient way to compute
the admissible Bethe roots) in terms of Q and P [5], [6].
A generalization of Wronskian formula (1.4)fortheopenXXX spin chain with diagonal boundary
fields was recently obtained [7]:
g(u)P
+
(u)Q
(u) f (u)P
(u)Q
+
(u) u
2N+1
, (1.5)
where Q(u)andP (u) are the respective polynomial solutions of a TQand a dual TQequation (see Eqs. (2.7)
and (2.12) below). Moreover, the functions f(u)andg(u) are given by (diagonal case)
f(u)=(u )(u + ),g(u)=f(u)=(u + )(u ), (1.6)
where α and β are boundary parameters. This result was used in [7]toformulateaQ-system for the model.
Our main result is a further generalization of the Wronskian formula to the case of nondiagonal bound-
ary fields:
g(u)P
+
(u)Q
(u) f (u)P
(u)Q
+
(u)=μ(u)u
2N+1
, (1.7)
where Q(u)andP (u) are the respective polynomial solutions of TQ equation (2.7) and dual TQ equa-
tion (2.12), f(u)andg(u)arenowgivenby(2.9), and, most importantly, μ(u) is a polynomial that satisfies
the remarkably simple relation
1
μ
+
(u) μ
(u)=γu(Q(u) P (u)). (1.8)
In other words, μ(u)isthediscrete integral of γu(Q(u) P (u)). In the diagonal case, γ = 0; it then follows
from (1.8)thatμ(u)=const,and(1.7) hence reduces to (1.5). The appearance of the nontrivial factor μ(u)
in Wronskian-type formulas (1.7)and(1.8) is due to the presence of an inhomogeneous term in the TQ
equation for the model [9]–[11]. We expect that this Wronskian-type formula will be useful for formulating
a Q-system for this model, but this remains a challenge.
In Sec. 2, we briefly review the considered model and its TQ equations and then obtain a dual TQ
equation. In Sec. 3, we derive the Wronskian-type formulas (1.7)and(1.8).
2. The model and its TQ equations
We consider the isotropic (XXX) open Heisenberg quantum spin-1/2 chain of length N with boundary
magnetic fields, whose Hamiltonian is given by
H =
N1
k=1
σ
k
·σ
k+1
ξ
β
σ
x
1
1
β
σ
z
1
+
1
α
σ
z
N
, (2.1)
where σ =(σ
x
y
z
) are the usual Pauli matrices and α, β,andξ are arbitrary real parameters. This
model is not U (1)-invariant in the nondiagonal case ξ =0.
To construct the corresponding transfer matrix, we use the R-matrix (solution of the Yang–Baxter
equation) given by the 4×4matrix
R(u)=
u
i
2
I + iP (2.2)
1
After completing this work, we learned of a similar result for the closed XXX spin chain with a nondiagonal twist (see
Theorem 4.10 in [8].
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(where P is the permutation matrix and I is the identity matrix) and the K-matrices (solutions of boundary
Yang–Baxter equations) given by the 2×2 matrices [12], [13]
K
R
(u)=
i(α 1/2) + u 0
0 i(α +1/2) u
,
K
L
(u)=
i(β 1/2) u ξ(u + i/2)
ξ(u + i/2) i(β +1/2) + u
,
(2.3)
which depend on the boundary parameters α, β,andξ.
The transfer matrix T(u)=T(u; α, β, ξ)isgivenby[14]
T(u)=tr
0
K
L
0
(u)M
0
(u)K
R
0
(u)
M
0
(u), (2.4)
where M and
M are monodromy matrices given by
M
0
(u)=R
01
(u)R
02
(u) ···R
0N
(u),
M
0
(u)=R
0N
(u) ···R
02
(u)R
01
(u). (2.5)
The transfer matrix is constructed to satisfy the fundamental commutativity condition
[T(u), T(v)] = 0 (2.6)
and the relation T(u)=T(u). Hamiltonian (2.1) is proportional to (dT(u)/du)
u=i/2
up to an additive
constant.
The eigenvalues T (u) of the transfer matrix T(u)arepolynomialsinu (as a consequence of (2.6)) and
satisfy the TQ equation [9]–[11]
uT (u)Q(u)=g
(u)(u
+
)
2N+1
Q
−−
(u)+f
+
(u)(u
)
2N+1
Q
++
(u) γu(u
u
+
)
2N+1
, (2.7)
where Q(u)isanevenpolynomialofdegree2N
Q(u)=
N
k=1
(u u
k
)(u + u
k
), (2.8)
the functions f (u)andg(u)aregivenby
f(u)=(u )(u
1+ξ
2
+ ),g(u)=f(u)=(u + )(u
1+ξ
2
), (2.9)
and γ is defined by
γ = 2(1
1+ξ
2
). (2.10)
We note the presence of an inhomogeneous term (proportional to γ)inTQ equation (2.7). In the diagonal
case ξ =0,from(2.10), we see that γ = 0 and the inhomogeneous term hence disappears. In this case, the
functions f (u)andg(u)in(2.9) reduce to (1.6).
The transfer matrix transforms under charge conjugation by reflection (negation) of all the boundary
parameters:
CT(u; α, β, ξ)C = T(u; α, β,ξ), C =(σ
x
)
N
. (2.11)
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We thus obtain a dual TQequation from (2.7)asin[7] by changing Q(u) → P (u) and reflecting the boundary
parameters α →−α, β →−β,andξ →−ξ, which implies that f(u)andg(u) become interchanged,
uT (u)P (u)=f
(u)(u
+
)
2N+1
P
−−
(u)+g
+
(u)(u
)
2N+1
P
++
(u) γu(u
u
+
)
2N+1
, (2.12)
where P (u)isalsoanevenpolynomialofdegree2N,
P (u)=
N
k=1
(u ˜u
k
)(u u
k
), (2.13)
whose zeros can be regarded as dual Bethe roots. We emphasize that the same eigenvalue T (u)appearsin
both (2.7)and(2.12).
3. The Wronskian-type formula
We now consider the relation between Q(u)(asolutionofTQ equation (2.7) for some transfer-matrix
eigenvalue T (u)) and the corresponding P (u) (a solution of dual TQ equation (2.12) for the same transfer-
matrix eigenvalue T (u)). For this, we use the ansatz
g(u)P
+
(u)Q
(u) f (u)P
(u)Q
+
(u)=μ(u)u
2N+1
, (3.1)
where f(u)andg(u)aregivenby(2.9) and the function μ(u) is yet to be determined. This ansatz is
motivated by result (1.5) in the diagonal case.
We next define R(u), following the results in [1], as
R =
u
2N+1
Q
+
Q
=
1
μ
g
P
+
Q
+
f
P
Q
, (3.2)
where the second equality follows from ansatz (3.1). Dividing both sides of TQequation (2.7)byQQ
++
Q
−−
,
we obtain
uT
Q
++
Q
−−
= f
+
R
+ g
R
+
γuQR
+
R
. (3.3)
Substituting R in this equation using the second equality in (3.2) and then multiplying both sides of the
resulting equation by μ
+
μ
Q
++
Q
−−
,weobtain
uT μ
+
μ
= f
+
g
PQ
++
Q
−−
Q
(μ
+
μ
+ γuP) f
+
f
P
−−
Q
++
(μ
+
+ γuP)+
+ g
+
g
P
++
Q
−−
(μ
γuP)+f
g
+
P
−−
P
++
Q(γu). (3.4)
Similarly, we define S(u)as
S =
u
2N+1
P
+
P
=
1
μ
g
Q
P
f
Q
+
P
+
(3.5)
and divide both sides of dual TQ equation (2.12)byPP
++
P
−−
.Wethusobtain
uT
P
++
P
−−
= f
S
+
+ g
+
S
γuPS
+
S
. (3.6)
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Substituting the expression for S in this equation using the second equality in (3.5) and then multiplying
both sides by μ
+
μ
P
++
P
−−
,weobtain
uT μ
+
μ
= f
+
g
PQ
++
Q
−−
(γu) f
+
f
P
−−
Q
++
(μ
+ γuQ)+
+ g
+
g
P
++
Q
−−
(μ
+
γuQ) f
g
+
P
−−
P
++
Q
P
(μ
+
μ
γuQ). (3.7)
Equating the right-hand sides of (3.4)and(3.7), we obtain the constraint
[μ
+
μ
+ γu(P Q)]
1
QP
(gP
+
Q
fP
Q
+
)
+
(gP
+
Q
fP
Q
+
)
=0, (3.8)
which is obviously satisfied if we set
μ
+
μ
= γu(Q P ) (3.9)
as required by (1.8). For given polynomials Q(u)andP (u), Eq. (3.9) can be solved for a polynomial
function μ(u) up to an arbitrary additive constant.
Result (3.9) can in fact be obtained more directly.
2
Let
μ(u)=
1
u
2N+1
g(u)P
+
(u)Q
(u) f(u)P
(u)Q
+
(u)
, (3.10)
which is equivalent to (1.7). We multiply TQ equation (2.7)byP (u)/(u
u
+
)
2N+1
, multiply dual TQ
equation (2.12)byQ(u)/(u
u
+
)
2N+1
, and subtract the second equation from the first. The obtained result
writtenintermsofμ given by (3.10) exactly coincides with (3.9).
Acknowledgments
The author thanks the organizers of the CQIS-2019 workshop in St. Petersburg for the kind invitation.
Conflicts of interest. The author declares no conflicts of interest.
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