476 V.V. Bazhanov et al. / Nuclear Physics B 927 (2018) 468–515
where the indices ,
take N
+
different values, whereas the index J take N different values.
The parameters p
1
, p
2
are defined as
p
±
=p
2
±p
1
. (2.29)
Different solutions of (2.28) correspond to different eigenvalues of the transfer matrix (2.26).
Next, in order to satisfy the boundary conditions (2.23) the rapidities θ
j
should satisfy (2.25).
Together with (2.27) these conditions take the form
−1 = e
2πi(p
1
−p
2
)
e
iM
0
R sinh θ
J
sinh
θ
J
−u
−
iπ
2
δ
sinh
θ
J
−u
+
iπ
2
δ
. (2.30)
Thus, we have obtained the system of coupled BAE (2.28) and (2.30) for the set of rapidities {u
}
and {θ
J
}. Once they are solved, the energy and momentum of the state can be calculated from
(2.13). To complete the construction the eigenvectors (2.12) one needs to find the corresponding
eigenvector φ of the transfer matrix (2.26) of the six-vertex model. This part of analysis is well
known and can be found in [27–30], as well as in the original BL paper [1]. We do not reproduce
these results here since e
xplicit expressions for the eigenvectors are not used in this paper.
It is wo
rth noting again that the parameter M
0
used in this section is the bare mass parameter.
Its relationship with the physical fermion mass M in (1.1) follows from the requirement that the
scaling function (1.3), calculated from BAE, at large distances should decay as ∝ exp(−MR).
As shown in [2] this is achieved if one sets
M
0
=M cos
πδ
2
. (2.31)
In what follows this relation will always be assumed.
2.3. The vacuum state
It is useful to rewrite the above equations (2.28), (2.30) in the logarithmic form
m
J
=
1
2
+p
1
−p
2
+
1
2π
r cos(πδ/2) sinh(θ
J
) +
φ
2δ
(θ
J
−u
) (2.32a)
m
=
1
2
−2p
1
−
φ
4δ
(u
−u
) +
J
φ
2δ
(u
−θ
J
), (2.32b)
where r =MR,
φ
α
(θ) =
1
2πi
log
sinh
iπ
4
α −θ
sinh
iπ
4
α +θ
(2.33)
with the phase of the logarithm is chosen such that φ
α
(0) = 0. The integer phases {m
J
} and {m
}
play the rôle of quantum numbers, which uniquely characterize solutions of the BAE. Different
solutions define different eigenstates of the Hamiltonian. The energy of the corresponding state
reads
E =−M cos(π δ/2)
J
cosh(θ
J
). (2.34)
For the vacuum state the location of zeroes and the associated phases in (2.32) were analyzed in
[1] and [2]. First note that the vacuum is flavor neutral. Thus, we set N
+
= N
−
and N = 2N
+
.
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