Theoretical and Mathematical Physics, 199(3): 909–921 (2019)

EIGENVALUES OF THE TRANSFER MATRIX OF THE

THREE-DIMENSIONAL ISING MODEL IN THE PARTICULAR CASE

n = m = 2

I. M. Ratner∗

The 16th-order transfer matrix of the three-dimensional Ising model in the particular case n = m = 2

(n×m is number of spins in a layer) is specified by the interaction parameters of three basis vectors. The

matrix eigenvectors are divided into two classes, even and odd. Using the symmetry of the eigenvectors,

we find their corresponding eigenvalues in general form. Eight of the sixteen eigenvalues related to odd

eigenvectors are found from quadratic equations. Four eigenvalues related to even eigenvectors are found

from a fourth-degree equation with symmetric coefficients. Each of the remaining four eigenvalues is equal

to unity.

Keywords: partition function, three-dimensional Ising model, spin, transfer matrix, eigenvalue, eigenvec-tor, symmetry

DOI: 10.1134/S0040577919060102

1. Introduction

The one-dimensional model of interacting spins was solved by Ising [1], and the two-dimensional modelwithout a magnetic field was solved by Onsager [2] and Kaufman [3]. To find the partition function of atwo-dimensional lattice, Onsager introduced the matrix (transfer matrix) V of the order 2n, where n isnumber of spins in a row. The order of the matrix is equal to the number of possible states of all the spinsof the row: the number of possible states of each spin is 2, and the number of spins is n.

If the number of rows in the lattice is m, then the partition function Z is equal to the trace of the mthpower of the matrix V :

Z = SpV m =N∑

i=1

λmi , (1)

where λi are eigenvalues of V and N = 2n. Formula (1) holds for cyclic boundary conditions: the (m+1)throw is identified with the first row. If the (n+1)th spin in a row is identified with the first spin, thenthe two-dimensional lattice closes into a torus (cyclic boundary conditions vertically and horizontally). Inthe thermodynamic limit n → ∞, m → ∞, the boundary conditions do not affect the thermodynamicproperties of the lattice.

Onsager introduced the transfer matrix V in the form of a product of two noncommuting matrices V 1

and V 2, where V 1 (diagonal) describes the interaction between spins in a row and V 2 (the direct product

∗Institute of Information Technology and Telecommunications, North-Caucasus Federal University, Stavropol,

Russia, e-mail: ratner.ilya@mail.ru.

Translated from Teoreticheskaya i Matematicheskaya Fizika, Vol. 199, No. 3, pp. 497–510, June, 2019. Received

June 22, 2018. Revised June 22, 2018. Accepted January 29, 2019.

0040-5779/19/1993-0909 c© 2019 Pleiades Publishing, Ltd. 909

of n identical second-order matrices) is the interaction between rows.

We can express the elements of V explicitly:

[s1]...

[sn]

∣∣∣∣∣∣∣∣V

∣∣∣∣∣∣∣∣

[s′1]...

[s′n]

=n∏

k=1

eβI1sksk+1eβI2sks′k , (2)

where β is the inverse temperature in energy units, I1 and I2 are the respective interaction parameters ina row and between rows, si = ±1 are spin variables, and n + 1 ≡ 1. We bring the factor (2 sinh(2βI2))n/2

outside V and obtain the matrix V ′ whose determinant is equal to unity:

V = (2 sinh(2βI2))n/2V ′. (3)

The matrix V ′ has the order 2n, but using the matrix symmetry, which reflects the translation symmetryof the lattice, Onsager computed the eigenvalues of V ′

λ = e±γ0±γ2±···±γ2n−2 (4)

or

λ = e±γ1±γ3±···±γ2n−1 (5)

with the hyperbolic angles γk found from the characteristic equation

cosh 2γk = cosh 2ϕ cosh2θ − cosπk

nsinh 2ϕ sinh 2θ, (6)

where ϕ = βI1, tanh θ = e−2βI2 , k = 0, 1, . . . , 2n − 1, and an even number of minus signs is taken in (4)and (5). Each of formulas (4) and (5) gives half of the eigenvalues of V ′ and together, the full spectrum.

Kaufman [3] significantly simplified the Onsager solution using the theory of spinors in multidimensionalspaces. Rumer [4] established the correspondence between matrices of the orders 2n and 2n and found theeigenvalues of the 2nth-order matrices as the angles γk in formulas (4) and (5) and also analyzed the natureof the spectrum of V ′ at the critical temperature and temperatures above and below critical.

Kac and Ward [5] justified the combinatorial method for solving the Ising problem, an alternative to thematrix method. In the combinatorial method, which reduces to the problem of enumerating graphs, eachterm in the partition function is associated with a closed graph on the lattice. By counting the numbers ofclosed graphs with different numbers of vertical and horizontal lines, we can obtain the partition function.Various ways to count the number of closed graphs were presented in [6]–[9].

Zinoviev [10] proved the Onsager formula for the partition function of a two-dimensional lattice withcyclic boundary conditions. The thermodynamic limit of the free energy of the two-dimensional Ising modelwith free boundary conditions was calculated in [11], and the spontaneous magnetization of this model wascalculated in [12].

It was shown in [13] that a two-dimensional Ising model can be defined by setting the number n ofspins in a row, the number m of rows, and a four-index matrix of the second order (a four-matrix) with

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elements (written slightly modified)

[s3][s1] T [s2]

[s4]= δs1s3e

θ1s1s2eθ2s3s4 , (7)

where θ1 = βI1, θ2 = βI2, and si = ±1. The Kronecker symbol in (7),

δs1s3 =

⎧⎪⎨

⎪⎩

1 if s1 = s3,

0 otherwise,(8)

reflects the fact that the same spin interacts with horizontally and vertically neighboring spins. If this factoris omitted in (7), then formula (7) corresponds to a lattice with two independent spins at each site, one spininvolved in the horizontal interaction and the other, in the vertical. In this case, the lattice splits into n

vertical and m horizontal decoupled one-dimensional chains, and the partition function, as a multiplicativequantity, is equal to

Z = Zm1 Zn

2 , (9)

where Z1 and Z2 are the partition functions of the respective vertical and horizontal chains.Onsager matrix (2) can be obtained by vertically multiplying n four-matrices T given by (7) and closing

the obtained product into a ring:

[s1]

[s2]

...

[sn]

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

V

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

[s′1]

[s′2]

...

[s′n]

=

[i]

[s1] T [s′1]

[j]

•

[j]

[s2] T [s′2]

[p]

•...

[q]

[sn] T [s′n]

[i]

•

(10)

A dot between identical indices facing each other means summation over the two possible values of theseindices. A dot below the last index means summation over the lowest and the highest indices, i.e., latticeclosure into a ring (cyclic boundary conditions).

Omitting the index designations (empty square brackets) and letting a dot denote summation over thesame indices facing each other, we can represent the partition function Z of a two-dimensional lattice as

911

the trace of degree n × m of the four-matrix T given by (7):

Z =

[ ][ ] T [ ]•

[ ]•

[ ][ ] T [ ]•

[ ]•...

[ ][ ] T [ ]•

[ ]•

[ ][ ] T [ ]•

[ ]•

[ ][ ] T [ ]•

[ ]•...

[ ][ ] T [ ]•

[ ]•

· · ·

[ ][ ] T [ ]•

[ ]•

[ ][ ] T [ ]•

[ ]•...

[ ][ ] T [ ]•

[ ]•

︸ ︷︷ ︸m

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

n = SpT n×m. (11)

Dots below and to the right mean the closure of a lattice into a torus. From (11), we can see that thefour-matrix T given by (7) is characterized by essentially flat multiplication.

In [14], formula (2) for the Onsager matrix was written as

〈s1, . . . , sn|V |s′1, . . . , s′n〉 =n∏

k=1

eβI1sksk+1eβI2sks′k ,

i.e., in the Dirac bracket notation for matrix elements. Here, the sets of spin variables si and s′i label rowsand columns, for example, in binary system. We enclose the indices in square brackets, as is customary inprogramming languages, and arrange the indices such that the matrix multiplication is clear.

The four-matrix T given by (7) can be subjected to the similarity transformation

[ p ]A−1[ i ] •

[ r ]B−1

[ k ]•

[ k ][ i ] T [ j ]•

[ l ]•

[ l ]B

[ s ]

[ j ]A[ q ] =[ r ]

[ p ] T ′ [ q ][ s ]

, (12)

where A, A−1 and B, B−1 are mutually inverse two-matrices (ordinary matrices) of the second order.If we replace the four-matrix T everywhere in (11) with the transformed four-matrix T ′, then the trace

of the product, i.e., the partition function does not change, because the matrix product is associative. Asa result of replacing T in (11) with T ′, the products AA−1 and BB−1 appear, and they can be replacedwith unit matrices and excluded from consideration.

The simplest formula for the partition function Z is obtained if similarity transformation (12) is usedto bring the original four-matrix T to a diagonal form in at least one pair of conjugate indices (either

912

vertical or horizontal). This is possible in some special cases, but four-matrix (7) of the two-dimensionalIsing model is not diagonalizable. Indeed, when diagonalizing by horizontal indices with n = 2, for example,two of the four eigenvalues must be equal, but all four eigenvalues in fact differ.

If we eliminate the Kronecker symbol δs1s3 in formula (7) for the four-matrix T , then we can diagonalizethe four-matrix T along both pairs of conjugate indices and represent it as a direct product of the diagonalmatrices of one-dimensional lattices horizontally and vertically.

But using similarity transform (12), we can reduce the four-matrix T given by (7) to a form allowingfurther combinatorial calculations. Choosing

A = A−1 = B = B−1 =1√2

(1 1

1 −1

), (13)

we bring the four-matrix T to the form

T ′ = 2

⎛

⎜⎜⎜⎜⎝

cosh θ1 cosh θ2 0 0 sinh θ1 sinh θ2

0 cosh θ1 sinh θ2 sinh θ1 cosh θ2 0

0 cosh θ1 sinh θ2 sinh θ1 cosh θ2 0cosh θ1 cosh θ2 0 0 sinh θ1 sinh θ2

⎞

⎟⎟⎟⎟⎠, . (14)

Here, we use the notation for the elements of multi-index matrices [15]. The arrows indicate the directionin which the corresponding indices increase.

With each term of sum (11) with T ′ replacing T , we can associate a specific graph containing thickand thin horizontal and vertical lines connecting the individual lattice sites. The combination of thick andthin horizontal and vertical lines fills all the connections between the lattice sites such that each site isconnected to its four neighbors with thick or thin lines, and the lattice is assumed to be closed into a torus.Such a graph is called a vertex-complete graph of thin and thick lines. A vertex-complete graph has fourlines adjacent to each vertex: either four thick, or two thick and two thin, or four thin. Because each vertexis adjacent to an even number of lines of the same type, the sets of graphs of thick and thin lines must beclosed separately.1

We assign factors to the lines: sinh θ1 to thick horizontal lines, cosh θ1 to thin horizontal lines, sinh θ2

to thick vertical lines, and cosh θ2 to thin vertical lines.If we bring the factor cosh θ1 cosh θ2 outside the four-matrix T ′ given by (14), then we can sum only

graphs with thick lines:

T ′ = 2 cosh θ1 cosh θ2

⎛

⎜⎜⎜⎜⎝

1 0 0 tanh θ1 tanh θ2

0 tanh θ2 tanh θ1 0

0 tanh θ2 tanh θ1 01 0 0 tanh θ1 tanh θ2

⎞

⎟⎟⎟⎟⎠, . (15)

In this case, thick horizontal lines correspond to the factor tanh θ1, and thick vertical lines correspond tothe factor tanh θ2. Factors equal to unity correspond to thin horizontal and vertical lines according toformula (15). Therefore, a graph with thin lines that complements a graph with thick lines to a vertex-complete graph has the factor unity, which can be omitted. It is therefore possible to consider only the sumover graphs with thick lines.

Figure 1 shows an example of a graph with thick and thin lines for the 3×3 lattice closed into a torus.We show the vertical and horizontal closing lines with arrows; of course, the graph remains undirected.

1If we follow the graph theory definitions strictly, then a vertex-complete graph is not a graph at all but splits into twoconjugate, possibly disconnected, closed graphs of thick and thin lines.

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Fig. 1. Closed graphs of thick and thin lines for the 3×3 lattice closed into a torus.

A graph of thick lines is closed, and a graph of thin lines is also closed if we consider it on a torus.We calculate the contribution to the partition function of the graph shown in Fig. 1. When calculating bythick and thin lines, we have four horizontal thick lines and four vertical thick lines, which corresponds tothe factor sinh4 θ1 sinh4 θ2. There are five thin horizontal lines including three closing lines and five thinvertical lines including three closing lines, and the factor from the thin lines is therefore cosh5 θ1 cosh5 θ2.The overall result is sinh4 θ1 sinh4 θ2 cosh5 θ1 cosh5 θ2. This value must also be multiplied by a factor of 2taken from the four-matrix raised to the power equal to the number of lattice sites, 3 × 3 = 9. Hence, thegraph contribution to the partition function is 29 sinh4 θ1 sinh4 θ2 cosh5 θ1 cosh5 θ2.

When calculating only along thick lines, we have the factor tanh4 θ1 tanh4 θ2. This value must alsobe multiplied by the factor 2 cosh θ1 cosh θ2 raised to the power equal to the number of lattice sites, 9.As a result, we have the contribution to the partition function: 29 cosh9 θ1 cosh9 θ2 tanh4 θ1 tanh4 θ2 =29 sinh4 θ1 sinh4 θ2 cosh5 θ1 cosh5 θ2.

As can be seen, calculations for thick and thin closed graphs and calculations for only thick closedgraphs give the same result. Of course, it is more convenient to carry out calculations only for graphs ofone type.

Specific calculations of the sum over the graphs of this type are the subject of the combinatorial methoddescribed in detail in [5]–[9].

2. Transfer matrix of the three-dimensional model

The three-dimensional Ising model has not yet been solved in the general case, although the phasetransition temperature has been found numerically with sufficient accuracy [16].

In [17], the eigenvalues of the transfer matrix of the three-dimensional Ising model (Onsager matrix)were calculated in some cases, which can be useful for establishing general regularities. Here, we find thegeneral form of the eigenvalues of the transfer matrix of the three-dimensional Ising model in the particularcase n = m = 2.

For the three-dimensional Ising model, we let I1, I2, and I3 denote the interaction parameters alongthree basis vectors. The “unit cell” is a six-index matrix (six-matrix) with indices arranged in pairs alongthe directions of the basis vectors X, Y, Z:

[s3] [s2]| �

[s5] R [s6]� |

[s1] [s4]

= δs1s3s5eθ1s1s2eθ2s3s4eθ3s5s6 , , (16)

where θ1 = βI1, θ2 = βI2, θ3 = βI3, and the three-index Kronecker symbol

δs1s3s5 =

⎧⎨

⎩1, if s1 = s3 = s5,

0 otherwise(17)

914

reflects the fact that the same spin interacts with neighboring spins in the directions of the three basisvectors of the lattice. Therefore, in six-matrix (16), only 16 of the 26=64 elements are nonzero.

Multiplying six-matrices (16) is essentially spatial. Multiplying n six-matrices in the X direction andthen the m obtained matrices in the Y direction and closing on the boundary indices, we obtain a matrixwith 2nm indices of which nm indices correspond to the rows and nm indices correspond to the columnsof the transfer matrix:

[s11] · · · [s1m]...

[sn1 · · · [snm]

∣∣∣∣∣∣∣V

∣∣∣∣∣∣∣

[s′11] · · · [s′1m]...

[s′n1] · · · [s′nm]

=n∏

k=1

m∏

l=1

eθ1sklsk+1,leθ2sklsk,l+1eθ3skls′kl , (18)

where θ1 = βI1, θ2 = βI2, θ3 = βI3, skl = ±1, n + 1 ≡ 1, and m + 1 ≡ 1.To derive general laws, it is desirable to have an exact solution in some particular case that allows

calculating to the end. Such a particular case is the case n = m = 2. The number of spins in the layeris nm = 4, the number of possible configurations in the layer, N = 2nm = 16, is the order of the transfermatrix V , and its elements have the form

[s11] [s12][s21] [s22]

∣∣∣V∣∣∣[s′11] [s′12][s′21] [s′22]

=

= e2θ1s11s12+2θ1s21s22+2θ2s11s21+2θ2s12s22eθ3s11s′11+θ3s12s′

12+θ3s21s′21+θ3s22s′

22 . (19)

The first exponential describes the interaction in the layer, the second exponential describes the interactionbetween the layers, and the factor 2 in the first exponent reflects the cyclic boundary conditions in thelayer.

The partition function Z of lattices of q layers with cyclic boundary conditions (the qth layer interactswith the first) is represented in the form of the trace of the qth power of V :

Z = Sp V q =N∑

i=1

λqi , (20)

where λi are the eigenvalues of V and N = 2nm = 16.The transfer matrix V has the order 16, and the characteristic equation is therefore a 16th-degree

equation. We consider the possible 16 configurations in the order(− −− −

),

(− −− +

),

(− −+ −

),

(− −+ +

),

(− +

− −

),

(− +

− +

),

(− +

+ −

),

(− +

+ +

),

(+ −− −

),

(+ −− +

),

(+ −+ −

),

(+ −+ +

),

(+ +

− −

),

(+ +

− +

),

(+ +

+ −

),

(+ +

+ +

), (21)

and we number the rows and columns of the matrix V in the same order.It can be easily seen from formula (19) that the matrix V can be represented as the product of two

matrices V = V 1V 2, where V 1 is diagonal and expresses the interaction in the layer (first exponential)and V 2 (second exponential) expresses the interaction between the layers.

The 16 diagonal matrix elements of V 1 (in order (21)) are a1, 1, 1, a2, 1, a3, a4, 1, 1, a4, a3, 1, a2, 1, 1, a1,where

a1 = e4ϕ+4ψ, a2 = e4ϕ−4ψ,

a3 = e−4ϕ+4ψ, a4 = e−4ϕ−4ψ,ϕ = θ1, ψ = θ2. (22)

915

The matrix V 2 (off-diagonal) is a direct product of four second-order matrices,

T =

(eθ3 e−θ3

e−θ3 eθ3

). (23)

The determinant of V 1 is equal to unity because the matrix contains four pairs of mutually inverse elementsand eight ones. The determinant of V 2 is equal to the fourth power of the determinant of T :

detT = e2θ3 − e−2θ3 = 2 sinh 2θ3. (24)

Following Onsager and Kaufman, we take the denominator√

2 sinh 2θ3 outside each matrix T and thenhave

T =√

2 sinh 2θ3

(cosh θ sinh θ

sinh θ cosh θ

). (25)

The matrix V 2 is represented as the factor(√

2 sinh 2θ3

)4, and V ′2 is represented as the direct product of

four matrices T ′,

T ′ =

(cosh θ sinh θ

sinh θ cosh θ

). (26)

The determinant of V ′2 is equal to unity because the determinant of T ′ is equal to unity, and the value

θ can be found from the formula

θ =12

log1 + e−2θ3

1 − e−2θ3. (27)

The matrix V ′2 in explicit form is

V ′2 =

⎛

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

w4

w3

w3

w2

w3

w2

w2

w1

w3

w2

w2

w1

w2

w1

w1

w0

w3

w4

w2

w3

w2

w3

w1

w2

w2

w3

w1

w2

w1

w2

w0

w1

w3

w2

w4

w3

w2

w1

w3

w2

w2

w1

w3

w2

w1

w0

w2

w1

w2

w3

w3

w4

w1

w2

w2

w3

w1

w2

w2

w3

w0

w1

w1

w2

w3

w2

w2

w1

w4

w3

w3

w2

w2

w1

w1

w0

w3

w2

w2

w1

w2

w3

w1

w2

w3

w4

w2

w3

w1

w2

w0

w1

w2

w3

w1

w2

w2

w1

w3

w2

w3

w2

w4

w3

w1

w0

w2

w1

w2

w1

w3

w2

w1

w2

w2

w3

w2

w3

w3

w4

w0

w1

w1

w2

w1

w2

w2

w3

w3

w2

w2

w1

w2

w1

w1

w0

w4

w3

w3

w2

w3

w2

w2

w1

w2

w3

w1

w2

w1

w2

w0

w1

w3

w4

w2

w3

w2

w3

w1

w2

w2

w1

w3

w2

w1

w0

w2

w1

w3

w2

w4

w3

w2

w1

w3

w2

w1

w2

w2

w3

w0

w1

w1

w2

w2

w3

w3

w4

w1

w2

w1

w3

w2

w1

w1

w0

w3

w2

w2

w1

w3

w2

w2

w1

w4

w3

w3

w2

w1

w2

w0

w1

w2

w3

w1

w2

w2

w3

w1

w2

w3

w4

w2

w3

w1

w0

w2

w1

w2

w1

w3

w2

w2

w1

w3

w2

w3

w2

w4

w3

w0

w1

w1

w2

w1

w2

w2

w3

w1

w2

w2

w3

w2

w3

w3

w4

⎞

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

, (28)

wherew0 = sinh4 θ, w1 = cosh θ sinh3 θ, w2 = cosh2 θ sinh2 θ,

w3 = cosh3 θ sinh θ, w4 = cosh4 θ.(29)

In what follows, we find the eigenvalues of

V ′ = V 1V′2, (30)

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whose determinant is equal to unity.The eigenvalues of the original matrix V can be found by multiplying them by the factor

(√2 sinh 2θ3

)4,which does not affect the thermodynamic properties of the lattice.

Although the characteristic equation for V ′ is a 16th-degree equation, the degree of the equationsfor the eigenvalues can be reduced using the matrix symmetry, which reflects the lattice symmetry: atranslation symmetry in two directions and an inversion symmetry.

The symmetry under inversion is because the energy of any spin configuration is unchanged if all spinvariables are replaced with their opposites. Mathematically, this means that V ′ commutes with U , whichis the direct product of four Pauli spin matrices (in the notation in [3])

C =

(0 1

1 0

)(31)

with units on the secondary diagonal and the other elements equal to zero.The square of U is equal to the identity matrix, and the eigenvalues of U are either 1 or −1 with an

equal number of positive and negative values. The eigenvectors of V ′, which commutes with U , are dividedinto two classes, even and odd.

We consider configuration sequence (21). Elements equidistant from the ends of the configuration haveopposite signs of all the spins; we say that such configurations are opposite. If we renumber configurationsfrom 0 to 15, then the sum of numbers of opposite configurations are equal to 15 (0 and 15, 1 and 14, etc.).Such numbers are also said to be opposite.

For even eigenvectors of V ′, components with opposite numbers are equal; for odd eigenvectors, com-ponents with opposite numbers have opposite signs. The numbers of even and odd eigenvectors are thesame: there are eight of them.

In addition to the inversion, which changes the configuration number to the opposite, the matrix V ′

also has other symmetry operations. For example, translation in a layer in the vertical direction replacesthe upper row of characters with the lower row and vice versa (the configuration

(− −− +

)goes into

(− +− −

),

and so on).

3. Odd eigenvectors

If in an odd eigenvector, some element remains unchanged in magnitude and sign under the actionof one of the symmetry operations except inversion and its sign changes under the action of the inversionoperation, then only the zero element can satisfy these conditions. Hence, some elements must be zero forodd eigenvectors. We therefore begin our consideration with odd eigenvectors.

For brevity, we write the elements of an eigenvector as a comma-separated string. Because the eigen-vectors are given up to an arbitrary factor, we take unity as one of the vector components; this makesfinding the eigenvalues easier.

We choose one of the irreducible representations of the symmetry group of V ′ and consider the oddeigenvector

(−k,−1,−1, 0,−1, 0, 0, 1,−1, 0, 0, 1, 0, 1, 1, k) (32)

related to this representation. Here, k is some number. Multiplying V ′ by vector (32), we obtain theequations for the eigenvalue λ corresponding to a given eigenvector. There are only two independentequations:

a1(−w4 + w0)k − a1(w3 + w3 + w3 + w3) + a1(w1 + w1 + 2w1) = −λk,

(−w3 + w1)k − w4 − w2 − w2 + w2 − w2 + w2 + w2 + w0 = −λ,(33)

917

ora1(−w4k + w0k − 4w3 + 4w1) = −λk,

− w3k + w1k − w4 + w0 = −λ.(34)

Eliminating k, we obtain a quadratic equation for λ:

λ2 + [(a1 + 1)(w0 − w4)]λ + a1[(w0 − w4)2 − 4(w1 − w3)2] = 0. (35)

Using the values w0, w1, w2, w3, and w4 given by (29), we obtain

(w0 − w4)2 − 4(w1 − w3)2 = 1, (36)

and the equation for λ hence becomes

λ2 + [(a1 + 1)(w0 − w4)]λ + a1 = 0. (37)

Substituting the values a1, w0, and w4 given by (22) and (29), we obtain

λ2 − [e4ϕ+4ψ + 1] cosh(2θ)λ + e4ϕ+4ψ = 0. (38)

We hence obtain two eigenvalues related to eigenvectors with the structure of eigenvectors (32), differingby k:

λ0,1 =(e4ϕ+4ψ + 1) cosh 2θ

2±

√(e4ϕ+4ψ + 1)2 cosh2 2θ

4− e4ϕ+4ψ. (39)

The odd eigenvectors(0,−1,−1,−k, 1, 0, 0,−1, 1, 0, 0,−1, k, 1, 1, 0),

(0,−1, 1, 0,−1,−k, 0,−1, 1, 0, k, 1, 0,−1, 1, 0),

(0,−1, 1, 0, 1, 0, k, 1,−1,−k, 0,−1, 0,−1, 1, 0)

(40)

similarly lead to another three respective eigenvalue pairs

λ2,3 =(e4ϕ−4ψ + 1) cosh 2θ

2±

√(e4ϕ−4ψ + 1)2 cosh2 2θ

4− e4ϕ−4ψ,

λ4,5 =(e−4ϕ+4ψ + 1) cosh 2θ

2±

√(e−4ϕ+4ψ + 1)2 cosh2 2θ

4− e−4ϕ+4ψ,

λ6,7 =(e−4ϕ−4ψ + 1) cosh 2θ

2±

√(e−4ϕ−4ψ + 1)2 cosh2 2θ

4− e−4ϕ−4ψ.

(41)

4. Even eigenvectors

The situation with even eigenvectors is more complicated. First, they lack zero elements (at least dueto symmetry) and have a much richer set of different elements (five different elements). If we increase thesymmetry of the lattice from orthogonal (ϕ �= ψ) to tetragonal (ϕ = ψ), then the set of different elementsreduces to four. We therefore consider a tetragonal lattice: ϕ = ψ, a1 = e8ϕ, a4 = e−8ϕ, and a2 = a3 = 1.

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An even eigenvector belonging to the identity irreducible representation of the symmetry group of V ′

has the structure(x, 1, 1, y, 1, y, z, 1, 1, z, y, 1, y, 1, 1, x), (42)

where x, y, and z are some yet unknown numbers. Multiplying V ′ by this eigenvector, we obtain theequations for the eigenvalue λ corresponding to this vector. There are only four independent equations:

r1x + r0 + r2 + r1y + r2 + r1y + r1z + r2 = λ,

r2x + r1 + r1 + r0y + r1 + r2y + r2z + r1 = λy,

a1(r0x + r1 + r1 + r2y + r1 + r2y + r2z + r1) = λx,

a4(r2x + r1 + r1 + r2y + r1 + r2y + r0z + r1) = λz,

(43)

wherer0 = w0 + w4, r1 = w1 + w3, r2 = 2w2. (44)

Reducing the five parameters w0, w1, w2, w3, and w4 to the three parameters r0, r1, and r2 is possiblebecause in each row of V ′

2 given by (28), the elements w0 and w4, w1 and w3 are located at equal distancesfrom the ends of the row and the elements w2 are arranged symmetrically about the center of the row.

Sequentially eliminating the unknown quantities x, y, and z from Eqs. (43), we obtain a fourth-degreeequation for λ:

λ4 + b1λ3 + b2λ

2 + b1λ + 1 = 0, (45)

whereb1 = −(a1 + a4 + 2)r0 − 4r2,

b2 = (a1 + a4)(2r02 − 4r1

2 + 4r0r2 − 2r22) + 2r0

2 − 8r12 + 4r0r2 + 2r2

2.(46)

The symmetry of the coefficients in (45) is because the eigenvalues form pairs of mutually inverse elements.Formulas (45) and (46) hold for the tetragonal lattice ϕ = ψ. For the orthogonal lattice ϕ �= ψ,

Eq. (45) holds but the values of the coefficients b1 and b2 should be generalized. In the final formula, thepairs a1, a4 and a2, a3 must appear symmetrically, which follows from the structure of V 1. Therefore, somecoefficients 2 in formulas (46) must be replaced with a2 + a3 such that the pairs a1, a4 and a2, a3 appearsymmetrically. There are only a few options here, and after a short search, we find

b1 = − (a1 + a4 + a2 + a3)r0 − 4r2,

b2 = (a1 + a4 + a2 + a3)(−4r12 + 4r0r2) +

+ (a1 + a4)(a2 + a3)(r02 − r2

2) + 2(r2 − r0)2.

(47)

The criterion in seeking formulas (47) is that the results coincide with the results in [17]. Substitutinga1, a2, a3, and a4 (given by 22), r0, r1, and r2 given by (44), and w0, w1, w2, w3, and w4 given by (29), weobtain

b1 = − (cosh(4ϕ + 4ψ) + cosh(4ϕ − 4ψ))(2 cosh2 2θ − sinh2 2θ) − 2 sinh2 2θ,

b2 = (cosh(4ϕ + 4ψ) + cosh(4ϕ − 4ψ))(

12

sinh2 4θ − 2 sinh2 2θ

)+

+ cosh(4ϕ + 4ψ) cosh(4ϕ − 4ψ)(4 cosh4 2θ − sinh2 4θ) + 2.

(48)

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Using the symmetry of the coefficients, it is easy to reduce (45) to a quadratic equation. Dividing allterms in the equation by λ2, we obtain

(λ2 +

1λ2

)+ b1

(λ +

1λ

)+ b2 = 0. (49)

Because the eigenvalues of V ′ form mutually inverse pairs, it is reasonable to represent them as pairsof exponentials

λi = e±γi . (50)

Then (λ +

1λ

)= 2 coshγ, λ2 +

1λ2

= 4 cosh2 γ − 2, (51)

and Eq. (49) becomes4 cosh2 γ + 2b1 cosh γ + (b2 − 2) = 0. (52)

We hence obtain two values of cosh γ:

cosh γ1,2 =−b1 ±

√b1

2 − 4(b2 − 2)

4, (53)

where b1 and b2 can be found by formulas (48).We obtain four eigenvalues of V ′ by formulas (50):

λ8 = eγ1 , λ9 = e−γ1 , λ10 = eγ2 , λ11 = e−γ2 , (54)

where γi can be found by formulas (53).Finally, each of the four remaining eigenvalues of V ′, as shown in [17], is equal to one:

λ12 = λ13 = λ14 = λ15 = 1. (55)

In conclusion, we note that all 16 eigenvalues form eight pairs of mutually inverse elements,

λ0λ7 = λ1λ6 = λ2λ5 = λ3λ4 = λ8λ9 = λ10λ11 = λ12λ13 = λ14λ15 = 1, (56)

and can be represented in form (50). Calculating the values γi is a separate problem.

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