Ice model (or 6-vertex model) and Yang-Baxter
Based on [Bax82, §8+9]
1
Description of the ice model
In an ice crystal, each of the two H-ion in H2 O forms a hydrogen bond with
another O-ion. The result is a lattice of O-ions with four nearest O-neighbours.
Each edge between two O-vertices is occupied by one H-ion which has a true bond
(close) with one O-vertex and a hydogen bond (distant) with the other O-vertex.
In 3D the O-vertices form alternating centres and vertices of regular tetrahedra.
In 2D we consider a planar M × N square lattice with oriented edges symbolising the displacement of the H-ion. According to the ice rule, every vertex has
2 incoming and 2 outgoing arrows, in total 6 possible vertex types:
- 6
6
1
?
?
- ?
?
2
?
6 - 6 ?
6
6
3
4
5
6
Assigning an energy ǫi to vertex type i, the partition function is given by
6
1 X
exp −
Z=
ǫi mi (C) ,
kB T i=1
C - ice configuration
X
(1)
where the sum is over all configurations C compatible with the ice rule and mi (C)
the number of vertices of type i in configuration C.
An alternative encoding of the vertices consists in drawing a line segment at a
vertex if the arrow points right or down. One concludes that in an infinite lattice,
lines never start, end, cross or branch. Assuming periodic boundary conditions,
or cylindrical compactifcation, we let n lines start on the bottom row (=circle).
The ice rule imples that these n lines run without intersection to the top row
(=circle). Every row has n incoming and n outgoing lines, necessary interlaced,
with compatible horizontal connections. The sum over C thus decomposes into
• a sum over the number n of lines,
• for each such n, and for every of the M rows, the sum over the positions
Xnj := (xj1 , . . . , xjn ) of the vertical line segments between rows j and j + 1.
There are Nn choices for Xnj for every j,
• for every row j, a sum over horizontal line configurations configurations
compatible with Xnj , Xnj+1.
1
We assume ǫ2i−1 = ǫ2i for i = 1, 2, 3, a symmetry which is interpreted as absence of
an electric field and define a := exp(− kBǫ1T ), b := exp(− kBǫ3T ) and c := exp(− kBǫ5T ).
This gives
Z=
N
X
X
Vn (Xn1 , Xn2 )Vn (Xn2 , Xn3 ) · · · Vn (XnM , Xn1 ) ,
(2)
1 ,...,X M
n=0 Xn
n
Vn (Xn , Yn ) =







0
X
am1 +m2 bm3 +m4 cm5 +m6
h-con
if 6 ∃ compatible horizontal connection between Xn and Yn
if horizontal connection h-con
contains mi vertices of type i.
There are zero, one or two horizontal connections compatible with Xn and Yn .
2
Bethe ansatz
All thermodynamical quantities are computable from log Z, which in the thermodynamic limit is dominated by the largest eigenvalue of {Vn }. The eigenvalues
can be identified by an ingeneous ansatz due to Bethe [Bet31].
The eigenvalue problem for Vn thus reads
X
N
Vn (Xn , Yn )u(Yn ) ,
(u(Yn )) ∈ C( n ) ,
(3)
Λn u(Xn ) =
Yn
where the sum is over the Nn possible choices for 1 ≤ y1 < y2 < · · · < yn ≤ N.
For n = 0 vertical lines, Y0 is the empty set, and a compatible horizontal line is
either empty (consisting of N vertice of type 1) or full (consisting of N vertices
of type 4). This means V0 (∅, ∅) = aN + bN , and the single eigenvalue is Λ0 =
aN + bN .
For n = 1 and Y1 = (y) we must distinguish the three cases
• y < x: a horizontal line from 1 to y and from x to N,
• y = x: either a full or an empty horizontal line,
• y > x: a horizontal line between x and y.
This means
Λ1 u(x) =
x−1
X
y
by−1 cax−y−1 cbN −x u(y)
x
y=1
y
x
+ bx−1 abN −x u(x) + ax−1 baN −x u(x)
+
N
X
y
x
y
ax−1 cby−x−1 caN −y u(y)
x
y=x+1
2
The Bethe ansatz consists in trying u(x) = z x for some complex number z and
gives
Λ1 z x =
2
2
bN −x ax−1 c2 z
(a2 − c2 ) − abz x
N ab + (c − b )z x
(1 − z N ) + bN
z
+
a
z .
a − bz
ab − b2 z
a2 − abz
Choosing z N = 1 we extract N pairs (Λ1 , u(x)) of eigenvalues and eigenfunctions:
Λ1 = aN L(z) + bN M(z) ,
u(x) = z x ,
where
ab + (c2 − b2 )z
ab + (c2 − a2 )z −1
,
M(z)
=
.
a2 − abz
b2 − abz −1
An important observation is that all eigenvectors are independent of a, b, c.
For n ≥ 2 the Bethe ansatz is
X
Ap1 ...pN zpx11 zpx22 · · · zpxnn .
u(x1 , . . . , xn ) =
L(z) =
(p1 ,...,pn )∈Sn
A similar calculation leads to eigenvalues
Λn = aN L(z1 ) · · · L(zn ) + bN M(z1 ) · · · M(zn ) .
(4)
All other contributions are killed by an appropriate choice of (AP , zi ) which turns
out to be
zjN
= (−1)
n−1
n
Y
slj (zl , zj )
,
sjl (zj , zl )
Y
= sign(p1 , . . . , pn )
spj pi (zpj , zpi )
(5)
l=1
Ap1 ...pn
1≤i<j≤n
where
a2 + b2 − c2
.
(6)
2ab
The next step consists in solving the system of equations (5) for zi . The solutions
are inserted into (4) to identify the largest of the Nn eigenvalues of Λn , and
then one maximises over 0 ≤ n ≤ N. This complicated programme was first
completed by [Lie67]. It turns our that only the case ∆ < 1 is interesting. For
∆ ≥ 1 the largest eigenvalue is Λ0 = aN + bN , hence kB T log Z is independent of
T , corresponding to a frozen system.
For ∆ < 1 one can reparametrise
sij (zi , zj ) = 1 − 2∆zj + zi zj ,
a = ρ sinh
λ−ν
,
2
b = ρ sinh
λ+ν
,
2
∆ :=
c = ρ sinh λ
3
⇒
∆ = − cosh λ , (7)
where ρ, λ, ν are possibly complex variables. The equations for z can be solved
with the ansatz
zj =
sinh
sinh
λ−νj
2
λ+νj
2
⇒
sjk =
L(zj ) =
2λ+νj −νk
2
λ+νj
λ+νk
sinh 2 sinh 2
ν−νj +2λ
sinh
2
,
−
ν−ν
sinh 2 j
sinh λ sinh
,
M(zj ) = −
ν−νj −2λ
2
ν−ν
sinh 2 j
sinh
.
Inserted into (4) one gets
φ(λ − ν)qn (ν + 2λ − 2πi) + φ(λ + ν)qn (ν − 2λ + 2πi)
qn (ν)
n
Y
µ − νl
N µ
N
sinh
,
qn (µ) :=
.
where φ(µ) := ρ sinh
2
2
l=1
Λn (ρ, λ, ν) =
(8)
It remains to satisfy (5), but this is equivalent to the requirement that Λn is an
entire function of ν. Since qn (νi ) = 0, Λn being an entire function of ν enforces
qn (νj − 2λ + 2πi)
φ(λ − νj )
=−
φ(λ + νj )
qn (νj + 2λ − 2πi)
∀j = 1, . . . , n ,
(9)
and this is precisely
(5). The equations (9) have multiple solutions; to show that
N
there are n of them requires work. These solutions νj (λ) depend only on λ but
not on ρ and ν.
3
Algebraic interpretation
We put the
N
n
eigenvalues Λn into a diagonal matrix and arrange these diagonal
N
matrices to a big 2N × 2N -matrix VD . Viewing the eigenvectors u(Xn ) ∈ C( n )
N
as U(Xn ) ∈ C2 such that VD U(Xn ) = Λn U(Xn ), and putting the resulting 2N
different U(Xn ) next to each other in a 2N × 2N -matrix U, the diagonalisation of
V reads
V = UVD U −1 .
The matrices V, VD depend on (ρ, λ, ν), but U is only dependent on λ. We will
focus on the ν-dependence and thus write V (ν) and VD (ν), whereas U is independent of ν. Consequently, matrices V (ν) of different ν (but same λ) commute,
V (ν)V (ν ′ ) = V (ν ′ )V (ν ′ ). We could also admit different ρ, ρ′ since ρ only appears
as a global scaling factor in φ.
Similarly we can put the Nn numbers qn (µ) after corresponding n-arrangement
into a 2N × 2N -matrix QD (µ). Defining Q(µ) := UQD (µ)U −1 , the relation (8)
becomes (φ is a scalar)
V (ν)Q(ν) = φ(λ − ν)Q(ν + 2λ − 2πi) + φ(λ + ν)Q(ν − 2λ + 2πi) .
4
(10)
By construction, Q(ν) commutes with any Q(ν ′ ) and V (ν ′ ).
Conversely, one can check that the solution of the ice model follows from the
following algebraic requirements:
1. Given a, b, c in terms of ρ, λ, ν, then for fixed λ, ρ any two transfer matrices
V (ν), V (ν ′ ) commute.
2. There are matrices Q(ν) 6= 0 which commute with all Q(ν ′ ), V (ν ′ ) and
satisfy (10) for some scalar function φ.
3. V (ν) and Q(ν) arePentire functions of ν, and the eigenvalues of Q(ν) have a
representation as r Dr exp( νr
), for r either an integer or half-integer with
2
−N ≤ r ≤ N.
Indeed the mutual commutation implies simultaneous diagonalisability V (ν) =
U diag(Λ(ν)) U −1 and Q(ν) = U diag(q(ν)) U −1 with U independent of ν. The
diagonal entries of (10) then lead
at least for some ν, and 3. implies a
Qn to (8) ν−ν
polynomial factorisation q(ν) = l=1 sinh 2 l with 0 ≤ n ≤ N and certain zeros
νl . Then (9) follows from entireness of V (ν), and the solution of these equations
gives the eigenvalues Λn .
4
Yang-Baxter equation
One can show that 1. is solved by the Yang-Baxter relation. For that consider
a row of vertices of the ice model with attached lower/upper vertical lines with
orientations α = (αi ) and β = (βi ). The transfer matrix reads
β
Vαβ =
X
wξ1 α1 |β1 ξ2 wξ2 α2 |β2 ξ3 · · · wξN αN |βN ξ1 ,
wξα|βξ′ =
ξ
ξ
w
ξ′
(11)
α
with summation over the possible orientations ξ compatible with the ice rule.
Writing up or right as +, down or left as −, the only non-vanishing weights are
w++|++ = w−−|−− == a ,
w+−,−+ = w−+|+− = b ,
w+−|+− = w−+|−+ = c .
(12)
The product of two transfer matrices reads
X
X
Vαγ Vγβ =
Sξ1 η1 |ξ2 η2 |α1 β1 Sξ2 η2 |ξ3 η3 |α2 β2 · · · SξN ηN |ξ1 η1 |αN βN
(V V ′ )αβ =
γ
(13)
ξ,η
η
β
η′
w′
Sξη|ξ′ η′ |αβ =
X
′
wξα|γξ′ wηγ|βη
′ =
X
γ
w
γ=±
γ=±
ξ
5
(14)
α
ξ′
If W is the 2-dimensional vector space with basis (+, −), the we can collect for
fixed α, β the Sξη|ξ′ η′ |αβ to a linear operator
S(α, β) : W ⊗ W → W ⊗ W ,
S(α, β)(x, y)
ξ,η
=:
X
Sξη|ξ′ η′ |αβ xξ′ yη′ .
ξ ′ ,η′
We can thus write (13) and similarly (V ′ V )αβ as
(V V ′ )αβ = trW ⊗W
N
Y
i=1
S(αi , βi ) ,
(V ′ V )αβ = trW ⊗W
N
Y
i=1
S ′ (αi , βi ) ,
′
′
′
where Sξη|ξ
′ η ′ |αβ =
γ=± wξα|γξ ′ wηγ|βη′ . Sufficient for commutativity of V, V is
the existence of a linear invertible operator T : W ⊗ W → W ⊗ W with S(α, β) =
′′
T S ′ (α, β)T −1 for all α, β. Let T ∈ Aut(W ⊗ W ) have matrix elements wηξ|η
′ ξ′ ,
′
then sufficient for commutativity is ST = T S or
X
X
′
′′
′′
′
(15)
wηξ|η
wξα|γξ′ wηγ|βη
′ wη ′ ξ ′ |η ′′ ξ ′′ =
′ ξ ′ wξ ′ α|γξ ′′ wη ′ γ|βη ′′ .
P
γ,ξ ′ ,η′ =±
γ,ξ ′ ,η′ =±
This identity has a graphical visualisation as transition between two trianges:
β
β
HH w′
η H
H
HHη ′
X
η ′′
HH
γ
HH
′′
w
γ,ξ ′ ,η′
H
H
′
ξ
ξ ′′
ξ w
η
′ X η
η
H
HH γ
H
γ,ξ ′ ,η′ w ′′ HH
ξ
H
ξ ′ HH
w′HH ξ
H
w
=
α
α
Equivalently, introducing linear operators Ri : W ⊗N → W ⊗N by
(Ri )αβ := δα1 β1 · · · δαi−1 βi−1 wαi βi |αi+1 βi+1 δαi+2 βi+2 · · · δαN βN ,
and similarly R′ , R′′ with weights w ′ , w ′′, then (15) reads
Ri+1 Ri′ Ri+1′′ = Ri′′ Ri+1′ Ri ,
Ri Rj = Rj Ri for |i − j| ≥ 2 .
(16)
For W the 2-dimensional vector space and the choice (12) of non-vanishing
weights, and similar for w ′, w ′′ , the 26 = 64 relations for external orientations
α, β, ξ, η, ξ ′, η ′ = ± reduce to only 3 equations
ac′ a′′ = bc′ b′′ + ca′ c′′ ,
ab′ c′′ = ba′ c′′ + cc′ b′′ ,
6
cb′ a′′ = ca′ b′′ + bc′ c′′ .
(17)
This system admits a non-trivial solution for (a′′ , b′′ , c′′ ) iff ∆ = ∆′ , thus reestablishing the results from the Bethe ansatz. Moreover, (17) is invariant under
exchange a′′ ↔ a, b′′ ↔ b, c′′ ↔ c so that ∆ = ∆′ = ∆′′ . It is then natural to
reparametrise in terms of (7) with ∆ = − cosh λ as before, and then (17) becomes sinh 41 (λ + ν − ν ′ − ν ′′ ) = 0, where without loss of generality the solution
is λ + ν − ν ′ − ν ′′ = 0. Introducing v = 21 (λ + ν) and similarly for v ′ , v ′′ , this
becomes v ′ = v + v ′′ , and we can write the original equation (16) in the form of
a parameter-dependent Yang-Baxter equation
Ri+1 (v)Ri (v + v ′′ )Ri+1 (v ′′ ) = Ri (v ′′ )Ri+1 (v + v ′′ )Ri (v)
(18)
for a single set of parametrised operators Ri (v) : W ⊗N → W ⊗N .
5
The eigenvalue equation
Consider a vector yα = gα1 1 ⊗ gα2 2 ⊗ · · · ⊗ gαNN ∈ W ⊗N . Then (11) gives
X
X
Vαβ yβ = trW G1α1 G2α2 · · · GN
wξα|βξ′ gβi ,
(Giα )ξξ′ =
(V y)α =
αN ) ,
β
Gi+ =
β=±
i
ag+
0
i
i
cg− bg+
,
Gi− =
i
i
bg−
cg+
i
0 ag−
.
i+1 −1
i
i i
) can be transformed into an upper
Now we ask whether G
α = P Hα (P
i′
i′′′
gα gα
for both α = ±. This would give
triangular matrix Hαi =
0 gαi′′
(V y)α = trX Hα11 Hα22 · · · HαNN ) =
N
Y
i=1
gαi′i +
N
Y
gαi′′i .
(19)
i=1
If pi ∈ W is the first column of P i , this condition Giα P i+1 = P i Hαi means
X
i i′
Ri (· · · ⊗ g i ⊗ pi+1 ⊗ . . . ) = · · · ⊗ pi ⊗ g i′ ⊗ . . . .
wξα|βξ′ gβi pi+1
ξ ′ = pξ hα ,
β,ξ ′
(20)
i
i′
These are four independent equations per i for g±
and g±
(corresponding to
ξ, α = ±1) which have a non-trivial solution if
pi−
1 r i+1
ri i
∆=
+ i+1 ,
r := i ,
⇒ r i+1 = −r i exp(±λ) .
i
2 r
r
p+
The signs σ i = ±1 are independent, r i = (−1)i r exp(λ(σ 1 + · · · + σ i−i )) with
P
N
i
i
i
i=1 σ = 0 due to cyclicity. Taking p+ = g+ = 1 without loss of generality, the
solution of (20) turns out to be (all vectors in W )
g i ≡ hi (ν) ,
g i′ = ahi (ν + 2λ − 2iπ) ,
7
g i′′ = bhi (ν − 2λ + 2iπ) ,
1
where h (ν) =
1
i
r exp( 2 (λ + ν)σ i )
equation (19) becomes
i
. Thus choosing y(ν) = h1 (ν) ⊗ · · · hN (ν),
V (ν)y(ν) = aN y(ν + 2λ − 2πi) + bN y(ν − 2λ + 2πi) .
Constructing a 2N × 2N -matrix Q(ν) out of ν-independent linear combinations of
y(ν) which differ in the 2N choices of (σ i ) we have proved (10). With some more
work it is possible to show that Q(ν) can be achieved invertible for some ν0 and
commuting with Q(ν ′ ) and V (ν ′ ). The growth rate 3. is a consequence of hi (ν).
In summary the Yang-Baxter equation (15) and the flip (20) permit a purely
algebraic solution of the ice model. The solution relies heavily on the assignments
a, b, c of vertex weights.
One can try to employ this method to other models. A famous success is the
8-vertex model where two additional vertices, sink and source,
-
?
6
?
6
7
8
are added with vertex weight d. The previous steps generalise [Bax72] to this
2
2 −c2 −d2
case, with ∆ = a +b
and a new function Γ = ab−cd
. Hyperbolic functions
2(ab+cd)
ab+cd
are now replaced by Jacobi elliptic functions.
References
[Bax82] R. J. Baxter, “Exactly Solved Models in Statistical Mechanics,” Academic Press, 1982.
[Bax72] R. J. Baxter, Annals Phys. 70 (1972) 193–228 [Annals Phys. 281 (2000)
187–222].
[Bet31] H. Bethe, “Zur Theorie der Metalle. I. Eigenwerte und Eigenfunktionen
der linearen Atomkette,” Z. Phys. 71 (1931) 205–226.
[Lie67] E. H. Lieb, “Residual Entropy of Square Ice,” Phys. Rev. 162 (1967)
162–172.
8