arXiv:1509.02305v1 [math-ph] 8 Sep 2015

Bethe’s Quantum Numbers And
Rigged Configurations
a,bAnatol N. Kirillov and cReiho Sakamotoa Research Institute for Mathematical Sciences,

Kyoto University, Sakyo-ku,
Kyoto, 606-8502, Japankirillov@kurims.kyoto-u.ac.jp

b The Kavli Institute for the Physics andMathematics of the Universe (IPMU),The University of Tokyo,
Kashiwa, Chiba, 277-8583, Japanc Department of Physics,

Tokyo University of Science, Kagurazaka,
Shinjuku, Tokyo, 162-8601, Japanreiho@rs.tus.ac.jp
reihosan@08.alumni.u-tokyo.ac.jp

Abstract
We propose a method to determine the quantum numbers, which we call the rigged
configurations, for the solutions to the Bethe ansatz equations for the spin-1/2 isotropic
Heisenberg model under the periodic boundary condition. Our method is based on theobservation that the sums of Bethe’s quantum numbers within each string behave
particularly nicely. We confirm our procedure for all solutions for length 12 chain(totally 923 solutions).

1

1

Introduction

Bethe’s seminal solution to the isotropic Heisenberg model under the periodic boundary
condition [B] in 1931 is one of the prototypical theories on the quantum integrable systems.
Basic procedure of the algebraic Bethe ansatz is as follows (see, e.g., [F, KBI]). First, we
solve the set of algebraic equations called the Bethe ansatz equations (see equation (12) inthe main text). Next, by using the solutions to the Bethe ansatz equations, we constructeigenstates of the Hamiltonian (see equation (11)). The main problem which we will consider
in the present paper is to show that the whole procedure involved in the Bethe ansatz method
is mathematically well-defined. We pursue the problem in the same setting as was treatedby Bethe.
For a long time it has been observed that there are several subtle points about theprocedure. One of such obstructions is the problem called the singular solutions (see equation(13)). Recently we have much progress on this issue (see [AV, BMSZ, NW, HNS, KS14a,
KS14b]) and now we attain fairly good understanding of the problem. However, the very
structure of the solutions to the Bethe ansatz equations itself still remains in rather foggy
situation.
In the analysis of the solutions to the Bethe ansatz equations, it has been customary toassume that roots of the solutions take a particular form called the strings. This idea wasalready apparent in the original Bethe’s paper. However, as we will explain shortly after, thenotion “string” has rather elusive nature and for a long time it has been a difficult problem
to understand how to utilize the string structure in a proper manner. One of the well-known
attempts on this problem is Takahashi’s theory [T]. Let us explain in more detail. Let N be
the length of the spin chain and let ℓ be the number of down spins. Then a main assumption
in the derivation of Takahashi’s quantum number is to suppose that the strings take exactly
the following form:1
a + bi, a + (b − 1)i, a + (b − 2)i, . . . , a − bi,(a ∈ R, b ∈ Z≥0 /2).

(1)

However, when ℓ ≥ 3, the assumption (1) has serious difficulty. Indeed, as we can easily
see in examples, the real part a as well as the intervals between successive roots are notunique if the lengths of the strings are larger than 2. As the result, Takahashi’s quantum
numbers are not uniquely defined and also they are not half-integers. In a previous paper[S15], we proposed to seek an alternative to Takahashi’s quantum numbers. In particular,
in that paper we showed that the correct quantum number for the exceptional real solution[EKS] is different from the one derived by Takahashi’s quantum numbers.
In the present paper, as a continuation of a previous work [S15], we propose a method
to assign quantum numbers to strings of roots. For this purpose, we start from the so-called
Bethe’s quantum numbers (see Section 3). Here, roughly speaking, Bethe’s quantum numbers
appear as phase factors after taking logarithm of the Bethe ansatz equations. By definition,
Bethe’s quantum numbers are uniquely defined and exactly integers or half-integers. Thenour basic observation is that the sum of the Bethe’s quantum numbers associated with all
1

We want to point out that the assumption of the validity of the string shape (1) allows to guess some
explicit expressions for q-weight multiplicities (Kostka–Foulkas polynomials) appearing in the representation theory of Lie algebras of type A. All these formulas has been proven rigorously in [KR] and lead toapplications in combinatorics, representation theory and discrete integrable systems.

2

roots of a given string behaves particularly simple manner. This is a remarkable property
since individual Bethe’s quantum numbers behave in a rather complicated way. Based on
this observation we propose a method to determine complete set of the quantum numbers,
which we call the rigged configurations, from Bethe’s quantum numbers. We confirmed theseobservations for all solutions of N = 12 case by using the numerical data given in [GD, HNS].
As we noted previously, we do not have thorough understanding of the string patternwhich appear in the solutions to the Bethe ansatz equations. However, our main aim in thepresent paper is to propose a method which could be made mathematically rigorous once weobtain sound understanding of the string structure. Our expectation relies on the fact that
Bethe’s numbers are uniquely defined and exactly half-integers.
To our best knowledge, Bethe’s quantum numbers had been introduced and studied in theoriginal paper by Bethe [B], formulas (37a) and (37b), and has been studied more recently
in [LR] in their study of the set of real solutions to the Bethe ansatz equations for ℓ = 2,
and in [HC] concerning the deviation of solutions to the Bethe ansatz equations from having
exact string structure with numerical evidence for N = 10.
The outline of the present paper is as follows. In Section 2, we provide necessary facts
about the Bethe ansatz method and the rigged configurations. In Section 3 we collectnecessary facts about Bethe’s quantum numbers. In Section 4 we provide the main algorithm.
We provide examples in Section 5. Finally we conclude in Section 6. Besides the main article,
there are supplementary tables (see remarks after Conjecture 2).22

Algebraic Bethe ansatz

In this section we briefly overview some basic facts concerning the spin-1/2 isotropic Heisenberg model, also known as the XXX model. The space of states of the spin-1/2 XXX chainNof length N is the complex space C2 , which one can identify with the tensor product of N
copies of C2 , namely
HN =

N
O

Vj ,

V j ≃ C2 .

(2)

j=1

The Hamiltonian describing the interaction of particles has the formN

JX x xyz
HN =(σk σk+1 + σky σk+1
+ σkz σk+1
− IN ),
4

(3)

k=1

a
a
where we assume σN+1 = σ1 , a ∈ {x, y, z}, and




0 1
0 −i
x
y
σ =
,
σ =
,
1 0
i 0
2

z

σ =



1 0
0 −1



(4)

After completing the present work, we notice that the paper [DG15] appears. In this paper the authors
also study Bethe’s quantum numbers. However our work is significantly different from theirs since our main
motivation is to understand the situation which appear when ℓ ≥ 3 whereas their paper considers the caseℓ = 2 exclusively. Note that by a result of [V] the real part “a” of the string (1) is uniquely defined whenℓ = 2 (except for the exceptional real solutions).

3

stand for the Pauli matrices, andσka = I ⊗ · · · ⊗ |{z}σ a ⊗ · · · ⊗ I.

(5)k

Let us introduce the vectors

 
 
0
1
and v− =
v+ =
1
0

(6)

and the global vacuum (ground state) as|0iN = v+ ⊗ · · · ⊗ v+ ∈ HN .
Let us consider the local transfer matrixi −λIN + 2i σkzσk2
Lk (λ) =i +σλIN − 2i σkz
2 k
where σk± = σkx ± iσky , and consider the transfer matrixAN (λ) BN (λ).
TN (λ) = LN (λ)LN −1 (λ) · · · L1 (λ) =CN (λ) DN (λ)

(7)

(8)

(9)

It follows from the fact that the local transfer matrices satisfy the relation R12 L1 L2 =L2 L1 R12 for some matrix R ∈ C2 ⊗ C2 where R satisfies the quantum Yang–Baxter equation
R12 R13 R23 = R23 R13 R12 .

(10)

The quantum Yang–Baxter equation (10) implies that the transfer matrices TN (λ) and TN (µ)
commute for different values of λ and µ. As a consequence of the relation (10), we can deducethat [BN (λ), BN (µ)] = 0 for different values of the parameters λ and µ. The key observation
due to Bethe is that the vectorΨN (λ1 , . . . , λℓ ) = BN (λ1 ) · · · BN (λℓ )|0iN(11)

is an eigenstates of the Hamiltonian HN if and only if the parameters λ1 , . . . , λℓ satisfy thefollowing system of algebraic equations, known as the Bethe ansatz equations (BAE(ℓ) for
short)
λk + 2iλk − 2iN=ℓY
λk − λj + i
j=1
j6=k

λk − λj − i,(k = 1, · · · , ℓ).

(12)

We are interested in the so-called physical solutions to BAE(ℓ), i.e., the solutions suchthat ΨN (λ) 6= 0. The set of physical solutions is divided into the regular solutions, thatis, the solutions (λ1 , . . . , λℓ ) which do not contain a pair (λα , λβ ) = ( 2i , − 2i ), and the set ofphysical singular solutions of the form


i
i
λ=
, − , λ3 , . . . , λℓ .
(13)
2 2
4

In the case of the regular solutions, one hasℓ

JX 1
Eλ1 ,...,λℓ := −.
2 j=1 λ2j + 14

HN ΨN (λ1 , . . . , λℓ ) = Eλ1 ,...,λℓ ΨN (λ1 , . . . , λℓ ),(14)

However for singular solutions, the energy Eλ1 ,...,λℓ does not be defined. It was suggested to“resolve” singularity of Eλ1 ,...,λℓ by using a deformed form of singular solutions, namely,

and consider the limit

e2 = − i + ǫ,λ2e1 = i + ǫ + c ǫN ,λ21e1 )BN (λe2 )BN (λ3 ) · · · BN (λℓ ) = Ψe λ.
lim N BN (λǫ→0 ǫe λ 6= 0 if and only if
It was conjecture by [NW], that Ψ−ℓY
λj + i2

λ − 2i
j=3 j

!N

= 1,

(15)

(16)

(17)

where
N +1

c = 2iℓY
λj + 3i2

λ − 2i
j=3 j

.

(18)

We call the singular solutions which satisfy the condition (17) the physical singular solution.
The corresponding energy is [KS14b]ℓ

Eλ = −J −JX 1.
2 j=3 λ2j + 41

It is also conjectured [HNS] that the set of physical singular and regular solutions to BAE(ℓ)
exhausts the set of all physical solutions to BAE(ℓ).
Finally, let us define the rigged configurations. The rigged configurations (ν, I) are comprised of a partition ν = (ν1 , . . . , νl ), called the configuration, and a sequence of non-negative
integers Ij , called the riggings, satisfying the following condition. Define the vacancy number
Pk (ν) by the following formula
Pk (ν) = N − 2(ν1′ + · · · + νk′ ) = N − 2lXmin(k, νj )

(19)

j=1

where νj′ represent components of the transformed partition ν ′ . If we regard the rigged
configuration as the set of pairs (ν, I) = {(ν1 , I1 ), . . . , (νl , Il )}, then the riggings must satisfy
0 ≤ Ij ≤ Pνj (ν)5(20)

for all j = 1, . . . , l. We note that in the rigged configuration theory, the order of the riggings
associated with rows of the configuration ν with the same length is not essential. Therefore
if we have the subset {(νk , Ik,1 ), . . . , (νk , Ik,m)} within (ν, I), we assume that Ik,1 ≤ · · · ≤ Ik,mto erase ambiguity.
Now the basic observation in our earlier paper [KS14a] is that we have an one to one
correspondence between the set of physical solutions to BAE(ℓ) and the set of the rigged
configurations (ν, I) of type ν ⊢ ℓ. Let us define the flip operator on the set of physicalsolutions to BAE(ℓ) as follows
κ : {λ1 , λ2 , . . . , λℓ } 7−→ {−λ1 , −λ2 , . . . , −λℓ }.

(21)Then another conjecture in [KS14a] states that the corresponding rigged configuration areconnected by the following operation on the rigged configuration (ν, I). Namely, we replaceeach element (νj , Ij ) by (νj , Pνj (ν) − Ij ) and reorder the riggings if necessary. This property
is useful in the following discussion.

3

Bethe’s quantum numbers

In what follows, we use function Arctan(z), defined as an analytic continuation of the function
arctan(x) (x ∈ R, arctan(0) = 0) to all complex plane with branch cuts (i, +i∞) and(−i, −i∞) along with the imaginary axis. For example, Arctan(xi) = i arctanh(x) and
Arctan(−xi) = −i arctanh(x) for x > 1. If we go across the branch cuts, we have
Arctan(xi) − lim Arctan(xi − ǫ) = π,ǫ→0Arctan(−xi) − lim Arctan(−xi + ǫ) = −πǫ→0for x > 1 and ǫ > 0.
We use the following formulalog

z−i= 2i Arctan(z) + (2n + 1)πiz+i(n ∈ Z).

(22)

We verify the formula as follows. We see that the differentiations of the both sides give z 22i+1 .
In order to determine the remaining constant, we compare the values of the both sides at
z = 0 as follows;
0+i= log(−1) = log(eπi · e2nπi ) = (2n + 1)πi
0−i
2i Arctan(0) = 0.
log

6(n ∈ Z),

3.1

Regular caseFirst let us consider a regular solution. By using the formula (22), we take the logarithm of
the Bethe ansatz equations (12);ℓi N Yλk − λj − i  λk + 2log i
λk − λj + i
λk − 2j=1
j6=k= −N · 2i Arctan(2λk ) +ℓX

2i Arctan(λk − λj ) + (−N + ℓ − 1)πi + 2nπi (n ∈ Z).

j=1
j6=k

By the Bethe ansatz equations, this should coincide with log(1) = 2mπi for m ∈ Z. Writing
2Jk = 2n − 2m − N + ℓ − 1, we obtain


ℓ
N 
2 X

Jk =
Arctan(λk − λj ) .
(23)
2 Arctan(2λk ) −
2π
N j=1
j6=k

We call the half integers Jk the Bethe’s quantum numbers.
Here we describe the basic property of the Bethe’s quantum numbers in relation with themultiplication by (−1).
Proposition 1. Suppose that we have the following two solutions of the Bethe ansatz equations{λ1 , λ2 , . . . , λℓ },e1 , e{λλ2 , . . . , eλℓ } := {−λ1 , −λ2 , . . . , −λℓ }.

eα ). Then we
Let Jα (resp. Jeα ) be the Bethe’s quantum number corresponding to λα (resp. λhave Jα = −Jeα .
Proof. Since Arctan(−z) = − Arctan(z), we obtain the result by (23).

3.2

Singular solutions caseNow let us consider the singular solution (13) when N is even. The computation for Jk(k ≥ 3) is same with the previous case. For the computations of J1 and J2 , we should be
reminded that the function Arctan(z) has the logarithmic singularities at z = ±i. Recall
that we assume ǫ ∈ R in (15). We introduce the deformation of singular solutionsi+ (reiθ ) + c(reiθ )N ,2i
λ2 (θ) = − + (reiθ )2

λ1 (θ) =7

where r ∈ R>0 and study the limits ofN2
b1 (θ) =2 Arctan(2λ1 (θ)) − Arctan(λ1 (θ) − λ2 (θ)) ,
2π
N


N
2
b2 (θ) =2 Arctan(2λ2 (θ)) − Arctan(λ2 (θ) − λ1 (θ))
2πNwhen r → 0. As a result we obtain the Bethe numbers J1 (θ) and J2 (θ) which depend on
the choice of θ. The following two graphs show J1 (θ) (left) and J2 (θ) (right) for the solution{i/2, −i/2} of N = 12.

We note that the sum J1 (θ) + J2 (θ) is constant for − π2 < θ < π2 .

4Main algorithm

We start from a given physical solution λ = (λ1 , . . . , λℓ ) to the Bethe ansatz equations. We
divide λ into subsets called strings. Here we say that S = {η1 , . . . , ηl }, (ηi = aj + ibj ) forms
a string if
aj − aj+1 ≈ 0,
bj − bj+1 ≈ 1(j = 1, . . . , l − 1).
By a result of Vladimirov [V] one can assume that η1 = η l if l > 1. We define the real partof the string by Re(S) = Re(η1 ).
We take a component of the solution λ to the Bethe ansatz equations with maximal
imaginary part and starting from that component we form a string of maximal length.
To proceed, we erase the string created, and apply the above procedure for the remaining
components of the solution λ. As a result we partition all component of the solution λ intoa collection of distinct strings.
One of the main ingredients of our algorithm is the sum of the Bethe numbers corresponding to a string S
X
J(S) =Jη .
η∈S

A striking feature of the quantity J(S) is that it exhibits very stable behavior even if individual Bethe numbers Jη behave in a complicated way.

8

4.1

Configuration

Starting from the solution λ, we divide it into strings S1 , . . . , Sp . Let the lengths of Sk be
lk . Then the partition ν which appears in the rigged configuration (ν, I) is ν = (l1 , . . . , lp )+ .
Here the symbol (l1 , . . . , lp )+ denotes the decreasing order of numbers (l1 , . . . , lp ).

4.2

Riggings

We determine the rigging corresponding to a string in the following way. If a string S satisfies
Re(S) = 0, then the corresponding rigging is the half of the corresponding vacancy number.
Otherwise consider the set of all solutions with a configuration ν, denoted bySol(ν) = {λ(1) , λ(2) , . . .} = {λ(α) }.
To start with we compute all Bethe numbers for Sol(ν).
The main tool to construct bijection between the set of solutions to the Bethe ansatz
equations and rigged configurations is the following string crossing rule. Let us consider astring S satisfying Re(S) > 0 and length n. Let Si1 , . . . , Siq be strings such that• length of Sik = lk < n,
• Re(S) < Re(Sik )for k = 1, . . . , q. Similarly let Sj1 , . . . , Sjq′ be strings such that• length of Sjk = lk′ > n,
• Re(Sjk ) < Re(S)for k = 1, . . . , q ′ . Then we consider
Je(S) := J(S) −

q
X
k=1

′

∆(lk , n) +q
X∆(lk′ , n),∆(l, n) := l + n − 3.

(24)

k=1

(α)

Let us fix some νi of the partition ν and denote the multiplicity of νi by m. Let Siα,k be
the string of length νi within λ(α) . Here we assume that the strings of the same length νi
(α)
(α)
(α)
satisfy Re(Siα,1 ) > Re(Siα,2 ) > · · · > Re(Siα,m ). Define(α)

eMνi ,k = max{J(Siα ,k )}.
α(α)

Suppose that we have Re(Siα ,k ) > 0. Then the corresponding rigging is(α)

e (α) )}.
Pνi (ν) − {Mνi,k − J(Siα ,kIf we have Re(Siα ,k ) < 0, we can use the flip operation to obtain the rigging.
Then our main conjecture is the following.
9

Conjecture 2. The above procedure gives one-to-one correspondence between the set ofphysical solutions to the Bethe ansatz equations and the rigged configurations.
We confirm the conjecture for the length 12 chain based on the numerical data by [GD].
As supplementary data, we provide plots of the solutions to the Bethe ansatz equations for
N = 12 case (except for simpler ones). For each solution, we assign the numbers corresponding to the solution numbers in [GD].
Remark 3. Roughly speaking, the riggings represent positions of strings. Namely thelarger rigging corresponds to the larger real part Re(S). In [S15], we observed that in several
examples we can readily read off the riggings if we arrange the solutions according to thereal parts of the largest strings. We confirm that this algorithm works for all solutions oflength 12 if the number of the strings is at most 3. In other cases, the appearance of thesolutions in the ordered set of solutions becomes more complicated. Nevertheless we observe
there is a strong relation between the riggings and the positions of the strings even in such
case.

4.3

Explicit formulas for M

We expect that the numbers M depend only on the length N of the chain and the shape ofthe configuration ν. We make the following conjecture. If ν2 < 3, we haveMν1 ,1 =(N − ν1′ − max{1, ν2′ })ν1.
2

If ν1 > 1 and ν2 = 1, we haveM1,1 =

N + ν1 + ν2 − 2ν1′ − 3.
2

We checked these formulas for length 12 chain.

5

ExamplesWe consider the case of ν = (3, 2, 1) and N = 12. Below we give a list of the solutions
depicted on the complex plane which we arrange according to the real part of the 3-string.
To each root we provide the corresponding Bethe numbers. Following the diagrams we givee Note that thetables of the explicit numerical values which we need in the computation of J.
solution #11 contains a singular string. See also [S15, Section 5] for an alternative method
to determine the riggings.
Group 1.

10

#2

#1
✻

−7/2
t
t

−5/2 t
−5/2

#3
✻

t7/2

−7/2
t

t5/2✲

−3/2
t
−5/2

t

✻

t7/2

−7/2
t

t5/2✲

t

#4

t

−1/2
t
−5/2

t

#5
−7/2
t
t

3/2

t

−5/2

t

−5/2

✻ 9/2
t

−7/2
t

t3/2 ✲

3/2

t t 5/2✲

t

1/2

−5/2

t

9/2
✻t9/2
t3/2t ✲

5/2

t

−5/2

t

9/2

11/2

#1 0.54241927−0.45810568 + 0.50017785i#2 0.52708058−0.64127830 + 0.50335013i#3 0.49893578−0.69284388 + 0.50515234i#4 0.46564665−0.71976299 + 0.50614240i#5 0.42430010−0.73869344 + 0.50683156i#6 0.38490522 + 0.01906127i−0.75221326 + 0.50729383i#7 0.23056669−0.76056174 + 0.50745313i

t 7/2

#7

−7/2
t

t t5/2✲

t 5/2✲

9/2#6✻ t7/2

−7/2
t

t

9/2

9/2

✻

t 7/2

t

11/2

0.54455699 + 0.99639165i−0.45810568 − 0.50017785i
0.52957875 + 0.99660493i−0.64127830 − 0.50335013i
0.50200196 + 0.99724969i−0.69284388 − 0.50515234i
0.46941736 + 0.99809739i−0.71976299 − 0.50614240i
0.42960641 + 0.99941330i−0.73869344 − 0.50683156i
0.36730804 + 0.99179719i−0.75221326 − 0.50729383i
0.23083274 + 0.99967059i−0.76056174 − 0.50745313i

0.54455699 − 0.99639165i−0.71532188
0.52957875 − 0.99660493i−0.30368149
0.50200196 − 0.99724969i−0.11725196
0.46941736 − 0.99809739i
0.035044606
0.42960641 − 0.99941330i
0.19387395
0.36730804 − 0.99179719i
0.38490522 − 0.01906127i
0.23083274 − 0.99967059i
0.82889133

Group 2.
#9#8
✻t 9/2
t

−5/2

t 7/2
t 1/2 ✲
t
t 7/2

11/2

#10
✻11/2
t
t 5/2
t t 1/2 ✲

#11
✻11/2
t
t 5/2
tt 1/2 ✲−3/2 t−1/2 t

t 5/2
13/2

t 5/2
13/2

11

✻9/2
t
t
t −1/2 ✲
t 1/2
t

−9/2

#12
−13/2 t✻−5/2 t−1/2 t t
t 1/2−5/2 t
−11/2

#8
#9
#10
#11
#12
#13
#14

✲

#13
−13/2 t✻−5/2 t−1/2 t t✲
3/2
t−5/2 t−11/2

0.20669577
0.10578435 + 0.50000000i
0.059726272
0.10847310 + 0.50000000i
0.010757119
0.06941354 + 0.50000000i
0.018539900i
0.50000000i−0.010757119−0.06941354 + 0.50000000i−0.059726272−0.10847310 + 0.50000000i−0.20669577−0.10578435 + 0.50000000i#14−11/2 t ✻−7/2 t−1/2 t

t ✲

5/2

t

−7/2 t−9/20.20597572 + 1.00038608i
0.10578435 − 0.50000000i
0.06007063 + 0.99927337i
0.10847310 − 0.50000000i
0.01249979 + 0.99958901i
0.06941354 − 0.50000000i
0.99377501i−0.50000000i−0.01249979 + 0.99958901i−0.06941354 − 0.50000000i−0.06007063 + 0.99927337i−0.10847310 − 0.50000000i−0.20597572 + 1.00038608i−0.10578435 − 0.50000000i

0.20597572 − 1.00038608i−0.83021590
0.06007063 − 0.99927337i−0.39681373
0.01249979 − 0.99958901i−0.17458378−0.99377501i−0.018539900i−0.01249979 − 0.99958901i
0.17458378−0.06007063 − 0.99927337i
0.39681373−0.20597572 − 1.00038608i
0.83021590

Group 3.
#15
−11/2 t ✻
−3/2
t t
−5/2
t

#16
−11/2 t ✻5/2
t−3/2 t
−3/2

✲
t

7/2

t

−9/2

5/2
t
✲
t

−5/2 t
t

−7/2

5/2
t

t

✲

−5/2 t t
−3/2

7/2

−9/2
#19
−9/2 t ✻#17−9/2 t ✻

t

5/2
t
✲
t

−5/2 t

1/2t
7/2

t

5/2
t

t

✲

#21
−9/2 t ✻−5/2 t

5/2
t
t ✲
t 5/2

t

−7/2

12

t

−7/2

3/2t
7/2

−7/2

−5/2 t t
−1/2

7/2

−7/2
#20
−9/2 t ✻#18−9/2 t ✻

7/2

5/2
t
✲
t

7/2

#15 −0.23056669
0.76056174 + 0.50745313i#16 −0.38490522 + 0.01906127i
0.75221326 + 0.50729383i#17 −0.42430010
0.73869344 + 0.50683156i#18 −0.46564665
0.71976299 + 0.50614240i#19 −0.49893578
0.69284388 + 0.50515234i#20 −0.52708058
0.64127830 + 0.50335013i#21 −0.54241927
0.45810568 + 0.50017785i−0.23083274 + 0.99967059i
0.76056174 − 0.50745313i−0.36730804 + 0.99179719i
0.75221326 − 0.50729383i−0.42960641 + 0.99941330i
0.73869344 − 0.50683156i−0.46941736 + 0.99809739i
0.71976299 − 0.50614240i−0.50200196 + 0.99724969i
0.69284388 − 0.50515234i−0.52957875 + 0.99660493i
0.64127830 − 0.50335013i−0.54455699 + 0.99639165i
0.45810568 − 0.50017785i−0.23083274 − 0.99967059i−0.82889133−0.36730804 − 0.99179719i−0.38490522 − 0.01906127i−0.42960641 − 0.99941330i−0.19387395−0.46941736 − 0.99809739i−0.035044606−0.50200196 − 0.99724969i
0.11725196−0.52957875 − 0.99660493i
0.30368149−0.54455699 − 0.99639165i
0.71532188

Let us explain how our algorithm works for the examples listed.
3-strings. For solutions #1, . . . , #5, we have Je = 21/2 since there are no shorter stringson the right of the 3-string.
In solution #6, we have two roots 0.38 ± 0.02i. We classify one of them to 1-string andthe other to 3-string. In any case, 1-string is on the right of the 3-string. Thus we have
Je = 23/2 − ∆(3, 1) = 21/2. Similarly we have Je = 21/2 in solution #7.
In solution #8, we have J˜ = 21/2 since there are no shorter string on the right.
In solutions #9 and #10 we have Je = 25/2 − ∆(3, 2) = 21/2 since the 2-string is on theright of the 3-string.
In solution #11, the real part of the 3-string is 0. Thus the corresponding rigging is thehalf of the corresponding vacancy number P3 (3, 2, 1) = 0, that is, the rigging is 0.
In solutions #12, . . . , #21, we have Je < 0.
Summarizing, we have M3 = 21/2. Since we have Je = ±21/2 (except for the solution#11), we conclude that the corresponding rigging is always 0.
2-strings. In solutions #1, . . . , #7, #12, #13, and #14, the real part of the 2-strings arenegative.
In solution #8, we have J˜ = 14/2 since the 3-string is on the right and the 1-string is on
the left.
In solutions #9 and #10, we have Je = 10/2 + ∆(2, 3) = 14/2 since the 3-string is on theleft of the 2-string.
In solutions #15, . . . , #21, we have Je = 12/2 + ∆(2, 3) = 16/2 since the 3-string is on
the left of the 2-string. Note that the position of the 1-string does not affect Je since we have∆(2, 1) = 0.
To summarize, we have M2 = 16/2. Thus the riggings for the two strings are 0 for
solutions in Group 1, 1 for solutions in Group 2 and 2 for the solutions in Group 3.
13

1-strings. By a similar argument we see that M1 = 7/2. Then the riggings for each groupis 1, 2, . . . , 6 from the smaller solutions number to the larger solutions number.
The resulting rigged configurations agree with the ones obtained by a geometrical argument [S15] or by the computation of Takahashi’s quantum numbers [GD] (see also [DG14]).
Remark 4. The above example reminds us about the box-ball system (see [S12] for a review).
The box-ball system is a discrete soliton system where solitons are “crystal” analogue ofmagnons. One of the important properties is that the rigged configurations provide action
and angle variables of the box-ball system. In this picture each row νi of the configuration ν
represents a soliton of the same length. If we look at the above examples, it is tempting toconsider them as a propagation of the string of solutions where scattering of strings yields aphase shift ∆(m, n) = m+n−2. Note that the phase shift in the case of the box-ball system ismin(m, n) for the scattering of lengths m and n solitons. It will be interesting to remark thatin the box-ball system crystal analogue of the transfer matrices, which provide the quantum
integrability of the box-ball system, generate the corresponding rigged configurations which
are dynamical variables if we regard the box-ball system as classical integrable system [S07].

6

Conclusion

Summarizing, we describe an algorithm which associate collection of quantum numbers or
riggings to solutions to the Bethe ansatz equations for the spin-1/2 Heisenberg model, starting from the set of Bethe’s quantum numbers corresponding to those solutions and its string
structure for general N. We confirm our algorithm for all solutions to the Bethe ansatz
equations for N = 12.

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