a note on the eigenvalue problem in crystal statistics. a note on

1

Progress of Theoretical Physics Vol. V, No.1, Jan.-Feb., 1950.

A Note on the Eigenvalue Problem in Crystal Statistics.

Yoichiro NAMBU

Department of Physics, Tok)'o Universit}'.

(Received September 1, 1949)

§ 1. Introduction.

In recent years there have been some remarkable developments in the mathematical theory of order-disorder transition. The problem of determining the partition function for a lattice with simple interaction model was found, by severai authors,!) to be reduced to an eigenvalue problem which is easier to handle with,and which, with suitable approximations, would reveal more refined details of the phase transition than the former theories. This method of attack culminated in the work of Onsager,2) who succeeded In giving the rigorous solution of the above mentioned eigenvalue problem for two-dimensional simple square Ising lattice. The mathematical tools involved, however, are so hopelessly complicated that one would quite simply lose sight in the jungle of hypercomplex numbers. A considerable improvement on this mathematical point has been attainp.d by Husimi and Syozi3) who used it to solve the eigenvalue problem for the honeycomb type lattice. The present author, independently of them,' has reached more or less similar ideas and considerations. which it is the purpos~ of the present paper to expose in brief detail. Though as yet no substantial applications has been attempted, nor anything physically new has been derived, it may be hoped that it will do some profit for those who are interested in such problems.

§ 2. Preliminary Considerations.

Let us take N identical particles 1, 2,···N, arranged in order, and consider the interchange p ..... +1 of two adjacent particles nand n+l (n=I,2,···1V). Such operators are met with in the theory o( ferromagnetism in which the eigenvalues of the operator (for one-dimensional case)

(1)

are asked. Here a.. means the spin matrix (a", a/J' az) at the nth site. As P C6mmutes with I::um =2S-N, wh~re S is the total number of spins oriented to .. +z direction, we can work within a subspace in which S is kept fixed. Then

1

Progress of Theoretical Physics Vol. V, No.1, Jan.-Feb., 1950.

A Note on the Eigenvalue Problem in Crystal Statistics.

Yoichiro NAMBU

Department of Physics, Tok)'o Universit}'.

(Received September 1, 1949)

§ 1. Introduction.

In recent years there have been some remarkable developments in the mathematical theory of order-disorder transition. The problem of determining the partition function for a lattice with simple interaction model was found, by severai authors,!) to be reduced to an eigenvalue problem which is easier to handle with,and which, with suitable approximations, would reveal more refined details of the phase transition than the former theories. This method of attack culminated in the work of Onsager,2) who succeeded In giving the rigorous solution of the above mentioned eigenvalue problem for two-dimensional simple square Ising lattice. The mathematical tools involved, however, are so hopelessly complicated that one would quite simply lose sight in the jungle of hypercomplex numbers. A considerable improvement on this mathematical point has been attainp.d by Husimi and Syozi3) who used it to solve the eigenvalue problem for the honeycomb type lattice. The present author, independently of them,' has reached more or less similar ideas and considerations. which it is the purpos~ of the present paper to expose in brief detail. Though as yet no substantial applications has been attempted, nor anything physically new has been derived, it may be hoped that it will do some profit for those who are interested in such problems.

§ 2. Preliminary Considerations.

Let us take N identical particles 1, 2,···N, arranged in order, and consider the interchange p ..... +1 of two adjacent particles nand n+l (n=I,2,···1V). Such operators are met with in the theory o( ferromagnetism in which the eigenvalues of the operator (for one-dimensional case)

(1)

are asked. Here a.. means the spin matrix (a", a/J' az) at the nth site. As P C6mmutes with I::um =2S-N, wh~re S is the total number of spins oriented to .. +z direction, we can work within a subspace in which S is kept fixed. Then

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2

P is sites. If we

Y. NAMBu

equivalent to an operator affecting the arrangement of S particles over N Each site is either occupied or not occupied by one of these particles.

introduce such operators, that change the unoccJlpied nth site· into occupied, and vice versa, by

+ (0 0) ( 0 a.,.= 1 0 ' a",= 0 ~ ), (2)

then P can be written in terms of them as follows:

(3)

The operators (2) remind one of the operators familiar in the second quantization, the only,- difference being the lack of anticommutativity between different sites. But we can easily see that the introduction of the sign functions necessary to make the a",'s and a~'s anticommute does not impair the relation (3), so that we are permitted to put

(4)

Now we make use of the Fourier transformation to rewrite (3) as

p,- L: (a';ama(n-m+ 1) -a~a ... a(n-m) +o~a,.a;:;ama(n-m+ 1) +COllj. n.m

+ ~a~a .. a;:;a, .. a(n-m ± 1) .. n.

(5)

where

(5')

There is a striking analogy to the ordinary quantum mechanics. P is just the Hamiltonian of interacting identical systems for which the .. kinetic energy" and the "interaction potential" are given respectively by

(6)

where k and z take discrete values. The kinetic terms alone lead to the well known Bloch spin waves whereas the interaction terms cause scattering of them. The latter can be neglected when the number of particles are small because the interaction force is of short range, which corresponds to the fact that the notion

2

P is sites. If we

Y. NAMBu

equivalent to an operator affecting the arrangement of S particles over N Each site is either occupied or not occupied by one of these particles.

introduce such operators, that change the unoccJlpied nth site· into occupied, and vice versa, by

+ (0 0) ( 0 a.,.= 1 0 ' a",= 0 ~ ), (2)

then P can be written in terms of them as follows:

(3)

The operators (2) remind one of the operators familiar in the second quantization, the only,- difference being the lack of anticommutativity between different sites. But we can easily see that the introduction of the sign functions necessary to make the a",'s and a~'s anticommute does not impair the relation (3), so that we are permitted to put

(4)

Now we make use of the Fourier transformation to rewrite (3) as

p,- L: (a';ama(n-m+ 1) -a~a ... a(n-m) +o~a,.a;:;ama(n-m+ 1) +COllj. n.m

+ ~a~a .. a;:;a, .. a(n-m ± 1) .. n.

(5)

where

(5')

There is a striking analogy to the ordinary quantum mechanics. P is just the Hamiltonian of interacting identical systems for which the .. kinetic energy" and the "interaction potential" are given respectively by

(6)

where k and z take discrete values. The kinetic terms alone lead to the well known Bloch spin waves whereas the interaction terms cause scattering of them. The latter can be neglected when the number of particles are small because the interaction force is of short range, which corresponds to the fact that the notion


A Note on the Eigenvalue Problem irt Cr),stal Statistics. 3

of spin wave is a good" approximation when the magnetization is nearly complete (low temperature).

As the above example shows, the introduction of a set' of anticommuting operators is particularly convenient because of its orthogonal property4). Instead of the creation and annihilation operators a: and a", we can also use the quantities a'; +a" and i(a,; -a,.) , or in other words, a set of anticommuting quantities with the commutation relations

(7)

According to the theory of spinors, such quantities can be considered as making up an orthogonal basis for a 2N-dimentional vectors space5). A product of two vectors X'~. commutes with X', if t~r, s. Thus we have a means for expressing commutable quantities in terms of ~nticommutable quantities.

As a next example we shall take the eigenvalue problem for the operator N

p' = ~ (0',...0,,+1.:0 + 0'''1/0'11+1.11) = ~ (A,. + B .. ), (8) ,.=1

which is a part of the operator P defined in (1). The commutation relations for A and Bare

[AR ,· A",]_=[B .. , B",]_=O,

[A .. , 8",]_=0 if n=m or I n-m I ~1, (9)

[A .. , B":l:t3+=O, A .. 2=B .. 2=1.

That is, A .. and En anticommute when they are adjacent, otherwise simply commuting. Such quantities can equally be composed of totally anticommuting operators. Indeed, let us introduce 2N basic vectors X'1> X'2" 'X'2N with the commutatio~ relations (7), and define (Fig. 1), for N even, .

.2n-l 2n .2n+l .2n+.2 0---0 0 0 0---0

All-I An Au+l . n=l, 2,···N; 2N+1=1. (10) 0------0

Bn

Fig. 1. Obviously these quantities satisfy all the relations. (8). The only difference is as

follows. From (8) we have N N Il A .. = If B .. =1,

n=l tI=1

while from (10), on the other hand,

llA,.=iJ!.'X'2X"s '" X'2#1= -iNX'IX'2 ••• X'2N=.x'= ±l,

llB .. =iNX'IX"4·X's%O·X'r.Zs ... X'2N-IZ2

(11)

(11')

A Note on the Eigenvalue Problem irt Cr),stal Statistics. 3

of spin wave is a good" approximation when the magnetization is nearly complete (low temperature).

As the above example shows, the introduction of a set' of anticommuting operators is particularly convenient because of its orthogonal property4). Instead of the creation and annihilation operators a: and a", we can also use the quantities a'; +a" and i(a,; -a,.) , or in other words, a set of anticommuting quantities with the commutation relations

(7)

According to the theory of spinors, such quantities can be considered as making up an orthogonal basis for a 2N-dimentional vectors space5). A product of two vectors X'~. commutes with X', if t~r, s. Thus we have a means for expressing commutable quantities in terms of ~nticommutable quantities.

As a next example we shall take the eigenvalue problem for the operator N

p' = ~ (0',...0,,+1.:0 + 0'''1/0'11+1.11) = ~ (A,. + B .. ), (8) ,.=1

which is a part of the operator P defined in (1). The commutation relations for A and Bare

[AR ,· A",]_=[B .. , B",]_=O,

[A .. , 8",]_=0 if n=m or I n-m I ~1, (9)

[A .. , B":l:t3+=O, A .. 2=B .. 2=1.

That is, A .. and En anticommute when they are adjacent, otherwise simply commuting. Such quantities can equally be composed of totally anticommuting operators. Indeed, let us introduce 2N basic vectors X'1> X'2" 'X'2N with the commutatio~ relations (7), and define (Fig. 1), for N even, .

.2n-l 2n .2n+l .2n+.2 0---0 0 0 0---0

All-I An Au+l . n=l, 2,···N; 2N+1=1. (10) 0------0

Bn

Fig. 1. Obviously these quantities satisfy all the relations. (8). The only difference is as

follows. From (8) we have N N Il A .. = If B .. =1,

n=l tI=1

while from (10), on the other hand,

llA,.=iJ!.'X'2X"s '" X'2#1= -iNX'IX'2 ••• X'2N=.x'= ±l,

llB .. =iNX'IX"4·X's%O·X'r.Zs ... X'2N-IZ2

(11)

(11')


4 Y. NAMBu

so that the operators in (8) cover only a subspace of the operator domains for

(10). We have deliberately to select those solutions which lie in the subspace

~=l.

Now (8) can be written as

N N P' = i2J (%~2n+l + %2n-1J;2n+2) = z1J (.X"~2m+l + X2n-1X2m+2)~"'"

n=1 R~l

where

[Xk, X-IJ+=[Yk, Y-I]+=~kl' k, I -N, - (N-1), ... +N, (12')

[Zk, 3,]_=0, [X k , ..1'~']+=O.

Thus we see that the eigenvalue of P' isa sum of the contributions from individual

terms with Zk' which have four eigenvalues 0, 0, ±1 as shown by matrix repre

sentation:

Then

Xk=(Ol)X(l_l)' %-k=(l O)x(l -1)'

..1'k=( 1 1 ) x ( 0 1 ), ..1'-k=( 1 1 ) x ( 1 0 ),

, NI2 • 27l'k P =4~Sk stn --, 8k=0, 0, 1 or -1.

k=1 N

~( eN

o ) .

The restriction %=1 becomes as follows. We write

N

tI3)

(14)

O(cos O+sin OX2nX2n+l) =0 exp (O%2n~2rl+l)=exp [O~X~2n+J] (15) n-l

4 Y. NAMBu

so that the operators in (8) cover only a subspace of the operator domains for

(10). We have deliberately to select those solutions which lie in the subspace

~=l.

Now (8) can be written as

N N P' = i2J (%~2n+l + %2n-1J;2n+2) = z1J (.X"~2m+l + X2n-1X2m+2)~"'"

n=1 R~l

where

[Xk, X-IJ+=[Yk, Y-I]+=~kl' k, I -N, - (N-1), ... +N, (12')

[Zk, 3,]_=0, [X k , ..1'~']+=O.

Thus we see that the eigenvalue of P' isa sum of the contributions from individual

terms with Zk' which have four eigenvalues 0, 0, ±1 as shown by matrix repre

sentation:

Then

Xk=(Ol)X(l_l)' %-k=(l O)x(l -1)'

..1'k=( 1 1 ) x ( 0 1 ), ..1'-k=( 1 1 ) x ( 1 0 ),

, NI2 • 27l'k P =4~Sk stn --, 8k=0, 0, 1 or -1.

k=1 N

~( eN

o ) .

The restriction %=1 becomes as follows. We write

N

tI3)

(14)

O(cos O+sin OX2nX2n+l) =0 exp (O%2n~2rl+l)=exp [O~X~2n+J] (15) n-l


A N9te on the Eigenvalue Problem in Crystal Statistics.

=D{ 1 +R,/(cos 20-1) +i sin 20R,I:},

since

Putting 0=1t/2, we have

(15')

(16)

(17)

R,I: turns out to commute with Z,I: and has the eigenvalues e,l:2. Hence we get the condition

(IS)

which determines the number of non-zero e,l:.

Thus we see that the use of anticommuting quantities is successful for the operator (S). The problem of diagonalizing the operator has been reduced to diagonalizing a quadratic form. However, this ceases to be the case when we take instead of (S) the operator P defined in (1), which calls for a quartic form rather than a quadratic.

§ 3. Onsager's Problem.

Now let us take up the case of Onsager), that is, a plane square lattice with interaction according to Ising model:

(19)

where s, represents the spin (= ± 1) of the ith site on a certain row and s/ that of the neighboring one on the next row. The summation extends over all the sites in the lattice. According to Onsager, the problem of obtaining the partition function F Tr exp (-E/kT) is reduced to asking for the largest eigenvalue of the operator

(20)

where

[s", s ... ]_=[s .. , c ... ]_=[c ... c ... ]_=O. n~m.

We shall not occupy ourselves with the deduction of the formula (20). but suppose it as given. and try to solve the eigenvalue problem for H. Writing (20) simply as

S,. == S,.S"+l' C,. :5! C,. •

(21)

A N9te on the Eigenvalue Problem in Crystal Statistics.

=D{ 1 +R,/(cos 20-1) +i sin 20R,I:},

since

Putting 0=1t/2, we have

(15')

(16)

(17)

R,I: turns out to commute with Z,I: and has the eigenvalues e,l:2. Hence we get the condition

(IS)

which determines the number of non-zero e,l:.

Thus we see that the use of anticommuting quantities is successful for the operator (S). The problem of diagonalizing the operator has been reduced to diagonalizing a quadratic form. However, this ceases to be the case when we take instead of (S) the operator P defined in (1), which calls for a quartic form rather than a quadratic.

§ 3. Onsager's Problem.

Now let us take up the case of Onsager), that is, a plane square lattice with interaction according to Ising model:

(19)

where s, represents the spin (= ± 1) of the ith site on a certain row and s/ that of the neighboring one on the next row. The summation extends over all the sites in the lattice. According to Onsager, the problem of obtaining the partition function F Tr exp (-E/kT) is reduced to asking for the largest eigenvalue of the operator

(20)

where

[s", s ... ]_=[s .. , c ... ]_=[c ... c ... ]_=O. n~m.

We shall not occupy ourselves with the deduction of the formula (20). but suppose it as given. and try to solve the eigenvalue problem for H. Writing (20) simply as

S,. == S,.S"+l' C,. :5! C,. •

(21)


6 Y. NAMBU

we see that S,. and e" have the properties

[S", C":l:l] + =0, otherwise all commuting, and

S:=SIS2 ' .. SN=l, C= C1 C2 ••• CN = ±1. (21')

In view of the relations (20) and (21'), a natural idea which suggests itself is to express H in terms of the operators of the second quantization as was illustrated in the last section. Indeed, we can put

with

S .. S"+1 = i(a: -a,,) (a,t+1 +a"+1) , c .. =2a,ta .. -1,

a,i +a,,=c1c2 ••• C"_1S,, , i(a,t -a,.) =C1C2 ••• eftS",

[a,., a!]+ = a ......

(22)

Here Cn plays just the role of the sign function. Such a procedure. has been independently devised by Husimi. Instead of (22) we may directly start from the anticommuting orthogonal basis {x,,}. Thus we replace S" and C" by

S .. =iX2nX2n+l' C .. = iXl"_lX2n ,

[xr' x.]+=2a ... , r, s=l, 2, ... 2N,

s=ns,,=iNx~3 ... X2NX1 = -X,. (23)

C nC .. =i",j,·lX'l ... X2N==X,

This substitution corresponds to the case C = -1 if we confine ourselves to the subspace X=l. The case C=l obtains when one of the defining equations, say for C;, is substituted by

C;= -iX1X 2• (23')

However, such minor modifications do not affect the final result for large N except some delicate problems like the boundary tension, so that we shall look apart from the eigenvalueo; of Sand C in order to avoid unnecessary complexity. With the choice (20), the operator H can now be written. as

(24)

As mentioned before, the quantities Xu X 2, ••• .x2N may be regarded as orthogonal basic vectors in a 2N~imensional space. In this space, a rotation of angle 0 in the plane spanned by Xr and x, is expressed by the operator

Z .. _U-1X"U,

U=exp [(O/2)x~s]=cos (O/2)+sin (O/2)x~,.

In fact it is easily verified that

U-1xrU=cos O.l:r+sin Ox"

U-1x.U=-sin Oxr+cos Ox ,

U-1x"U =.x .. , n!!rt r, s.

(25)

(26)

6 Y. NAMBU

we see that S,. and e" have the properties

[S", C":l:l] + =0, otherwise all commuting, and

S:=SIS2 ' .. SN=l, C= C1 C2 ••• CN = ±1. (21')

In view of the relations (20) and (21'), a natural idea which suggests itself is to express H in terms of the operators of the second quantization as was illustrated in the last section. Indeed, we can put

with

S .. S"+1 = i(a: -a,,) (a,t+1 +a"+1) , c .. =2a,ta .. -1,

a,i +a,,=c1c2 ••• C"_1S,, , i(a,t -a,.) =C1C2 ••• eftS",

[a,., a!]+ = a ......

(22)

Here Cn plays just the role of the sign function. Such a procedure. has been independently devised by Husimi. Instead of (22) we may directly start from the anticommuting orthogonal basis {x,,}. Thus we replace S" and C" by

S .. =iX2nX2n+l' C .. = iXl"_lX2n ,

[xr' x.]+=2a ... , r, s=l, 2, ... 2N,

s=ns,,=iNx~3 ... X2NX1 = -X,. (23)

C nC .. =i",j,·lX'l ... X2N==X,

This substitution corresponds to the case C = -1 if we confine ourselves to the subspace X=l. The case C=l obtains when one of the defining equations, say for C;, is substituted by

C;= -iX1X 2• (23')

However, such minor modifications do not affect the final result for large N except some delicate problems like the boundary tension, so that we shall look apart from the eigenvalueo; of Sand C in order to avoid unnecessary complexity. With the choice (20), the operator H can now be written. as

(24)

As mentioned before, the quantities Xu X 2, ••• .x2N may be regarded as orthogonal basic vectors in a 2N~imensional space. In this space, a rotation of angle 0 in the plane spanned by Xr and x, is expressed by the operator

Z .. _U-1X"U,

U=exp [(O/2)x~s]=cos (O/2)+sin (O/2)x~,.

In fact it is easily verified that

U-1xrU=cos O.l:r+sin Ox"

U-1x.U=-sin Oxr+cos Ox ,

U-1x"U =.x .. , n!!rt r, s.

(25)

(26)


A Note on the Eigenvalue Problem in Crystal Statistics. 7

Then (24) means nothing but a product of rotations, first by angle 2ai, in the planes Z2lI_1Z2n, and then by angle 2{1i, in the planes .x'2n.x'2n+1' Let us suppose that in a suitable coordinate system {.x' • .'} the rotation H is brought to Jordan's standard form, so that

(27)

Since I tx,.' .x',.+/ I = I, the eigenvalue of the operator H that operates on a eigenvector 1J! is found in the expression

H1J!=exp [~6nr,.]1J!, 6 .. = ±1. •

(28)

We have only to solve the problem in the 2N-dimensional subspace {x,.} of basic vectors, instead of the whole space of 1J! with 2N dimensions. By the first rotation H;., a vector (x .. ) is transformed into another vector er .. ), which, by the second rotation 9,., again changes into a third (8 .. ). Thus

or

Here ,,'=ai, {1'={1i,

In matrix form,

-6

C H=

)'271-1 = cos 2<'1;' .x'2n-1 + sin 2{1' X2n ,

)'271= -sin 2,,' ZlIII_1 + cos 2{1'.x'2lI ,

82r&=COS 2{1~2n+sin 2{1'y2n+l'

S2n+l= -sin 2)'Y2r&+cos 2{1':Y2n+l,

82,,= -6Z2n_1 +dx2n +a.x'2n+1 + CZ2r&+2;

S= -ax2n+dx2r&+1 +6X2n+2+C.x'2n_1.

a=cos 2,,' sin 2{1'~ 6=sin 2a' cos 2{1',

c=sin 2u.' sin 2{1', d=cos 2a' cos 2{1'.

d a c . . -a d b

-6 d a c .. C -a d 6

The eigenvalue equation for H becomes

or

-O.x'2n_l + (d·-l).x'2n+aX'2n+1 +CX2'H2=0,

C.x'2"_1-aX'2n + (d-l).~2n+1 +OZ2n+2=0,

aZ2n+1 +cz2n+2=OX2n_1 + (l-d)zin ,

(1-d)z2n+1:-o.x'2n+2=cZ2n_l-a£lln.

(28)

(28')

(28")

(28"')

···2n (29)

···2n+l

(30')

A Note on the Eigenvalue Problem in Crystal Statistics. 7

Then (24) means nothing but a product of rotations, first by angle 2ai, in the planes Z2lI_1Z2n, and then by angle 2{1i, in the planes .x'2n.x'2n+1' Let us suppose that in a suitable coordinate system {.x' • .'} the rotation H is brought to Jordan's standard form, so that

(27)

Since I tx,.' .x',.+/ I = I, the eigenvalue of the operator H that operates on a eigenvector 1J! is found in the expression

H1J!=exp [~6nr,.]1J!, 6 .. = ±1. •

(28)

We have only to solve the problem in the 2N-dimensional subspace {x,.} of basic vectors, instead of the whole space of 1J! with 2N dimensions. By the first rotation H;., a vector (x .. ) is transformed into another vector er .. ), which, by the second rotation 9,., again changes into a third (8 .. ). Thus

or

Here ,,'=ai, {1'={1i,

In matrix form,

-6

C H=

)'271-1 = cos 2<'1;' .x'2n-1 + sin 2{1' X2n ,

)'271= -sin 2,,' ZlIII_1 + cos 2{1'.x'2lI ,

82r&=COS 2{1~2n+sin 2{1'y2n+l'

S2n+l= -sin 2)'Y2r&+cos 2{1':Y2n+l,

82,,= -6Z2n_1 +dx2n +a.x'2n+1 + CZ2r&+2;

S= -ax2n+dx2r&+1 +6X2n+2+C.x'2n_1.

a=cos 2,,' sin 2{1'~ 6=sin 2a' cos 2{1',

c=sin 2u.' sin 2{1', d=cos 2a' cos 2{1'.

d a c . . -a d b

-6 d a c .. C -a d 6

The eigenvalue equation for H becomes

or

-O.x'2n_l + (d·-l).x'2n+aX'2n+1 +CX2'H2=0,

C.x'2"_1-aX'2n + (d-l).~2n+1 +OZ2n+2=0,

aZ2n+1 +cz2n+2=OX2n_1 + (l-d)zin ,

(1-d)z2n+1:-o.x'2n+2=cZ2n_l-a£lln.

(28)

(28')

(28")

(28"')

···2n (29)

···2n+l

(30')


8 Y. NAMBU

Let us write

and

A=( a, c) , B=( b, A-d). A-d, -b c, -a

Then (30') means

Aif/n+l=Bif/n, or rfu+l=A-1Brf" == Crfn,

hence

The periodicity condition requires that rfN+l=.rfl' or

(E-ClI')rfl=O, detl E-CNI =0.

(31)

(31')

(32)

(32')

(33)

If we denote the Nth roots of unity by ~.I:=exp (21tki/N)=exp (tpi), k=l, 2,.··

N, then

(33')

that is',

I ~.l:a-b, ~A:C-(A-d) I ~.I:(J.-d) -c, -~",b+a =0.

(33")

This gives a quadratic equation for J.:

~.(2_J. [2~d+~2C+C]+~(a2+b2+~+d2) -(1 +~2)(ab-cd) =0. (34)

Substituting (28"'),

a2+b2+~+d2=l, ab-cd=O,

.(2_A [2d+ (~+~-l)C] +1=0,

or A2_2J.(cos 2a' cos 2{1' + sin 2-d sin 2{1' cos tpi,) +1=0.

Putting the two roots as exp ( ±2r.t:' i).

(34')

(1/2)(e2T'1+e-2TIi)=cos 2r.t:'=cos 2, .. 1 cos 2{1' +sin 2a' sin 2{1' cos tp,.. (35)

Returning to the original constants a and {1, and writing r=r'i, we get just the relation given by Onsager:

cosh 2r.l:=cosh 2a cosh 2{1-sinh 2a sinh2{1 cos tp,.. tp,.=2'flk/ N. (35')

The largest eigenvalue of H becomes then, according to (28),

Hmttx=exp [:E I r", I] . (36)

Such a procedure also applies. mutatis mutandis, to certain variants of Onsager model, e.g. the honeycomb lattice treated by Husimi and Syozi. It fails.

8 Y. NAMBU

Let us write

and

A=( a, c) , B=( b, A-d). A-d, -b c, -a

Then (30') means

Aif/n+l=Bif/n, or rfu+l=A-1Brf" == Crfn,

hence

The periodicity condition requires that rfN+l=.rfl' or

(E-ClI')rfl=O, detl E-CNI =0.

(31)

(31')

(32)

(32')

(33)

If we denote the Nth roots of unity by ~.I:=exp (21tki/N)=exp (tpi), k=l, 2,.··

N, then

(33')

that is',

I ~.l:a-b, ~A:C-(A-d) I ~.I:(J.-d) -c, -~",b+a =0.

(33")

This gives a quadratic equation for J.:

~.(2_J. [2~d+~2C+C]+~(a2+b2+~+d2) -(1 +~2)(ab-cd) =0. (34)

Substituting (28"'),

a2+b2+~+d2=l, ab-cd=O,

.(2_A [2d+ (~+~-l)C] +1=0,

or A2_2J.(cos 2a' cos 2{1' + sin 2-d sin 2{1' cos tpi,) +1=0.

Putting the two roots as exp ( ±2r.t:' i).

(34')

(1/2)(e2T'1+e-2TIi)=cos 2r.t:'=cos 2, .. 1 cos 2{1' +sin 2a' sin 2{1' cos tp,.. (35)

Returning to the original constants a and {1, and writing r=r'i, we get just the relation given by Onsager:

cosh 2r.l:=cosh 2a cosh 2{1-sinh 2a sinh2{1 cos tp,.. tp,.=2'flk/ N. (35')

The largest eigenvalue of H becomes then, according to (28),

Hmttx=exp [:E I r", I] . (36)

Such a procedure also applies. mutatis mutandis, to certain variants of Onsager model, e.g. the honeycomb lattice treated by Husimi and Syozi. It fails.


A Note on tlte Eigenvalue Problem in Crystal Statistics. 9

however, in case of those lattices with too many nearest neighbours. in particular the three-dimensional lattices. For we cannot express a quantity which anticommutes with more than two neighbors by a product of two totally anticommuting vectors. In fact the three-dimensional model necessitates us to resort to products of four .%'s as illustrated below:

H exp [P2:!(s ..... s ... m+t+snmsr&+l .... ) exp [a,Ecll ... ]

=exp [iP2:!hrJ exp [al]k.] , (37)

h='%t'%a', .%4.%2", k='%t.%rs.%4' (See Fig. 2.)

Supplementary conditions are also to be consjdered. The occurence of quartic forms may look reasonable when we note the so<alled dual trans-formation. Such transformations do really exist 3

even in the three-dimensional case. In fact let us take a cubic lattice with atoms located at the middle point of each edge of the cell, and let each set of the four atoms lying on a side plane determine a four-body interaction energy PStS~sS4' s.= ±l. Then it is easy to show that this lattice is dual to a simple cubic lattice made up of the body center sites of the former with ordinary two-body interactions. Perhaps the four-body force may have some bearing on the quantity k. in (37).

4!

§ 4. Kramen-Wannier's method of. approach.

Fig. 2.

Our next investigations concern the screw lattice model adopted by Kramers and Wannier for solving the two-dimensional case. This model seems more convenient for general purposes than that used by Onsager. Now let us try to transform the Onsager model into the K-W model as follows. Rewrite (20) (after reversing the order of Dt and H,) as

H,.~=exp [ifUt.%2] exp [ifUaXJ···exp [iP.%r3Jexp[i~.%4.%5}··

=exp [ifU1.%2] exp [iazs.%4}exp UP.%rJ exp [ifUsZG] exp [iP.%~5}··

N-t

=exp [Uut'%21 n (exp [iUX2n+1.%2nH] exp [iP.%216.%27<+l] exp [iP.%~J, (38) .. ~t

which can be modified, without grave consequences, to

N

H = n exp [iu'x27<+t.%2n+2] exp [iP.%~!JI+1] = oH .. , "-1

(38')

A Note on tlte Eigenvalue Problem in Crystal Statistics. 9

however, in case of those lattices with too many nearest neighbours. in particular the three-dimensional lattices. For we cannot express a quantity which anticommutes with more than two neighbors by a product of two totally anticommuting vectors. In fact the three-dimensional model necessitates us to resort to products of four .%'s as illustrated below:

H exp [P2:!(s ..... s ... m+t+snmsr&+l .... ) exp [a,Ecll ... ]

=exp [iP2:!hrJ exp [al]k.] , (37)

h='%t'%a', .%4.%2", k='%t.%rs.%4' (See Fig. 2.)

Supplementary conditions are also to be consjdered. The occurence of quartic forms may look reasonable when we note the so<alled dual trans-formation. Such transformations do really exist 3

even in the three-dimensional case. In fact let us take a cubic lattice with atoms located at the middle point of each edge of the cell, and let each set of the four atoms lying on a side plane determine a four-body interaction energy PStS~sS4' s.= ±l. Then it is easy to show that this lattice is dual to a simple cubic lattice made up of the body center sites of the former with ordinary two-body interactions. Perhaps the four-body force may have some bearing on the quantity k. in (37).

4!

§ 4. Kramen-Wannier's method of. approach.

Fig. 2.

Our next investigations concern the screw lattice model adopted by Kramers and Wannier for solving the two-dimensional case. This model seems more convenient for general purposes than that used by Onsager. Now let us try to transform the Onsager model into the K-W model as follows. Rewrite (20) (after reversing the order of Dt and H,) as

H,.~=exp [ifUt.%2] exp [ifUaXJ···exp [iP.%r3Jexp[i~.%4.%5}··

=exp [ifU1.%2] exp [iazs.%4}exp UP.%rJ exp [ifUsZG] exp [iP.%~5}··

N-t

=exp [Uut'%21 n (exp [iUX2n+1.%2nH] exp [iP.%216.%27<+l] exp [iP.%~J, (38) .. ~t

which can be modified, without grave consequences, to

N

H = n exp [iu'x27<+t.%2n+2] exp [iP.%~!JI+1] = oH .. , "-1

(38')


10 Y. NAMBU

It turns out more convenient to consider an operator which has its component factors Hn in the reverse order to (51) (which is also equivalent to a renumbering of the order). Then define a displacement operator p. which changes x .. into

.i:"+1' and put . 7Jf n =H JI..-1···Dt 7Jf 0

with obvious relations

H .. =P"HoP~". Ho=exp [iPX2.\'XIJ exp [itu'1%2].

The relation (a Schrodinger equation for discrete time variable!)

is turned into

by putting

7Jf"+1 = H .. 7Jf ..

7Jf n+~ = p-1 Ho7Jf ,,' == A7Jf .. '

p-n7Jf .. =7Jf .. ' or 7Jf .. = P"7Jf,.'.

(39)

(39')

(40)

(40')

( 40")

But p-1Ho is' now independent of 1't. hence the eigenvalue problem for 7Jf ... is equivalen t to

).7Jf'=A7Jff =P-1/fo7Jf'. ).N=E, (H7Jfo=7Jf N=E7Jfo)' (41)

P can be expressed in terms of the basic vectors as

P-exp [ : %l%2N-l] exp [- : %IX2N_a} •• exp [ ± : X1%3]

(42)

because of the relations

Thus the operator A can again be regarded as a product of rotation operators. Let us look for an eigenvector of the form

N-l. %::;:: ~l"-l (aX2n_l + bX2n) + C%2N+ ).N-laX2N_t •

6'-1 (43)

Substituting in the equation A%A-l=Az, we get the following result:

10 Y. NAMBU

It turns out more convenient to consider an operator which has its component factors Hn in the reverse order to (51) (which is also equivalent to a renumbering of the order). Then define a displacement operator p. which changes x .. into

.i:"+1' and put . 7Jf n =H JI..-1···Dt 7Jf 0

with obvious relations

H .. =P"HoP~". Ho=exp [iPX2.\'XIJ exp [itu'1%2].

The relation (a Schrodinger equation for discrete time variable!)

is turned into

by putting

7Jf"+1 = H .. 7Jf ..

7Jf n+~ = p-1 Ho7Jf ,,' == A7Jf .. '

p-n7Jf .. =7Jf .. ' or 7Jf .. = P"7Jf,.'.

(39)

(39')

(40)

(40')

( 40")

But p-1Ho is' now independent of 1't. hence the eigenvalue problem for 7Jf ... is equivalen t to

).7Jf'=A7Jff =P-1/fo7Jf'. ).N=E, (H7Jfo=7Jf N=E7Jfo)' (41)

P can be expressed in terms of the basic vectors as

P-exp [ : %l%2N-l] exp [- : %IX2N_a} •• exp [ ± : X1%3]

(42)

because of the relations

Thus the operator A can again be regarded as a product of rotation operators. Let us look for an eigenvector of the form

N-l. %::;:: ~l"-l (aX2n_l + bX2n) + C%2N+ ).N-laX2N_t •

6'-1 (43)

Substituting in the equation A%A-l=Az, we get the following result:


or

A Note on tlte Eigenvalue Prob!em in Crystal Statistics.

cos 2a' cos 219' .;t' - sin 2a' cos 2p:r + sin 219' s =).N:c ,

sin 2a' .;t' + cos 20.')1 =;'s ,

-sin 219' cos 2a' .;t'+sin 2a' sin 2J3:r+cos 2P'Z=).N-ly ,

cosh 2a cosh 2{d:c+i sinh 2a cosh 2py-i sinh 2{dS=).N~ ,

-i sinh 2a:c+cosh 2ay=k,

i sinh 2{d cosh 2a.x - sinh 2a sinh 2Py + cosh 2pz = ).N-~ . Eliminating s by virture of the second equation,

(cosh 2a cosh 2{d-sinh 2a sinh 2p).-1):c+I(sinh 2a cosh 2{d

-sinh 219 cosh 2a)'-~=;'N~ ,

i(sinh 219 cosh 2a-sinh 2a cos 2{d).-1):c- (sinh 2a sinh 2{d

-cosh 219 cosh 2a).-ly=).N-ly ,

which determine ). according to

).2N-l + sinh 2a sinh 2p).N + ( - cosh 219 cosh 2a- cosh 2a cosh 2p»).N-l

+ sinh 2a sinh 2P).N-2 + (sinh! 2a sinh! 219 + cosh! 213 cosh! 2a

- sinh! 2a cosh! 2f1- sinh! 219 cosh! 2a).-1

(44)

(44')

(44")

+( -sinh 2a cosh 2a sinh 219 cosh 2p+sinh 2a cosh 2a sinh 219 cosh2p»).-2

- cosh 2a sinh 2a cosh 219 cosh 219 + sinh 2a cosh 2a sinh 219 cosh 219=0.

This in turn reduces to

).2N+ sinh 2a sinh 2{d().+).-1)-2cosh 2a cosh 2p+).-N=0. (45')

Putting ).=exp 2r,

cosh 2Nr=cosh 2a cosh 2p-sinh 2a sinh 219 cosh r. (46)

11

This is the equation for r. To solve this, we assume that 2r=2ro+wi, and substitute in (46), thereby making use of the relation

cosh (2ro+wz)=cosh 2ro cos w+i sinh 2ro sin w.

Comparison of real and imaginary parts yields the relations

cosh ~Nro cos Nw=cosh 2a cosh 2p-sinh 2a sinh 219 cosh ro cos ltJ,

sinh 2Nro sin Nw = sinh 2a sinh 2Jhinh ro sin w .

(47)

(48)

Choose w=k1T:/N, k=l, 2,···2N, so that sin NctJ=O, cosNw=(-I)k. In the limit N _ 00 and Nro - r=finite, we have ro=O, and both. equations become satisfied by this choice of w. Thus

±cosh 2r-cosh 2a cosh 2p-sinh 2a sinh 219 cos w , (49)

or

A Note on tlte Eigenvalue Prob!em in Crystal Statistics.

cos 2a' cos 219' .;t' - sin 2a' cos 2p:r + sin 219' s =).N:c ,

sin 2a' .;t' + cos 20.')1 =;'s ,

-sin 219' cos 2a' .;t'+sin 2a' sin 2J3:r+cos 2P'Z=).N-ly ,

cosh 2a cosh 2{d:c+i sinh 2a cosh 2py-i sinh 2{dS=).N~ ,

-i sinh 2a:c+cosh 2ay=k,

i sinh 2{d cosh 2a.x - sinh 2a sinh 2Py + cosh 2pz = ).N-~ . Eliminating s by virture of the second equation,

(cosh 2a cosh 2{d-sinh 2a sinh 2p).-1):c+I(sinh 2a cosh 2{d

-sinh 219 cosh 2a)'-~=;'N~ ,

i(sinh 219 cosh 2a-sinh 2a cos 2{d).-1):c- (sinh 2a sinh 2{d

-cosh 219 cosh 2a).-ly=).N-ly ,

which determine ). according to

).2N-l + sinh 2a sinh 2p).N + ( - cosh 219 cosh 2a- cosh 2a cosh 2p»).N-l

+ sinh 2a sinh 2P).N-2 + (sinh! 2a sinh! 219 + cosh! 213 cosh! 2a

- sinh! 2a cosh! 2f1- sinh! 219 cosh! 2a).-1

(44)

(44')

(44")

+( -sinh 2a cosh 2a sinh 219 cosh 2p+sinh 2a cosh 2a sinh 219 cosh2p»).-2

- cosh 2a sinh 2a cosh 219 cosh 219 + sinh 2a cosh 2a sinh 219 cosh 219=0.

This in turn reduces to

).2N+ sinh 2a sinh 2{d().+).-1)-2cosh 2a cosh 2p+).-N=0. (45')

Putting ).=exp 2r,

cosh 2Nr=cosh 2a cosh 2p-sinh 2a sinh 219 cosh r. (46)

11

This is the equation for r. To solve this, we assume that 2r=2ro+wi, and substitute in (46), thereby making use of the relation

cosh (2ro+wz)=cosh 2ro cos w+i sinh 2ro sin w.

Comparison of real and imaginary parts yields the relations

cosh ~Nro cos Nw=cosh 2a cosh 2p-sinh 2a sinh 219 cosh ro cos ltJ,

sinh 2Nro sin Nw = sinh 2a sinh 2Jhinh ro sin w .

(47)

(48)

Choose w=k1T:/N, k=l, 2,···2N, so that sin NctJ=O, cosNw=(-I)k. In the limit N _ 00 and Nro - r=finite, we have ro=O, and both. equations become satisfied by this choice of w. Thus

±cosh 2r-cosh 2a cosh 2p-sinh 2a sinh 219 cos w , (49)


12 Y. NAMBU

with w=br/N.

But the minus sign is imposSible, hence the allowed values for ware

w=2k1L/N. k=l, 2 ... ·~V.

(49')

(49")

This completely agrees with the previous results. The case of the threedimensional lattice can be treated analogously. but we shall not give here the details.

§ 5. Additional Remarks.

The mathematical tricks employed in the preceding sections seem to allow a rather natural explanation from a somewhat more general point of view. The essential point is that we consider. instead of an ordinary eigenvalue problem. a different one of the form

[H, XJ_=.tX (50)

for the U eigenoperator" X 6). Important consequences drawn from this equation are: (a) A is the difference of two eigenvalues of H: A=En-E",; (b) X transforms certain eigenvector 1Jl '" of H with Em into another one 1Jl n with eigenvalue En=E:" +A; and (c) Product of two eigenoperators ~Xl is again an eigenoperator, with eigenvalue A=A1 +A2• transforming 1Jl ... into a third eigenvector 1Jl,• If we integrate the. relation (50) according to

d ds X-rHo X]. (~1)

we arrive at the equation

(51)

The equation (51) just corresponds to the problem of obtaining the eigenoperator { z .. '} for the rotation H encountered in Section 3 and 4. When the operator H has a simple structure. a general eigenoperator X will be factorized into a product of prime eigenoperators:

with

X=X1X 2 .. ·Xk ,

e1'=tA1+A.+ ... Ak • (52)

The largest A is the difference of the largest and smallest eigenvalues of sH in (52). If the eigenvalues E1 are symmetrically distributed around zero. Emax will itself be given by Emax= Amax/2, a fact which casts a sidelight on the procedure used in the previous sections.

The main part of the present work had been completed nearly two years ago. It is through the kindness of Professor Husimi a d Mr. Syozi of Osaka University that the author enjoys the opportunity of publishing this note. He wishes to express on this occasion his cordial thanks to them and also to Professor Kubo of Tokyo University for helpful discussions.

12 Y. NAMBU

with w=br/N.

But the minus sign is imposSible, hence the allowed values for ware

w=2k1L/N. k=l, 2 ... ·~V.

(49')

(49")

This completely agrees with the previous results. The case of the threedimensional lattice can be treated analogously. but we shall not give here the details.

§ 5. Additional Remarks.

The mathematical tricks employed in the preceding sections seem to allow a rather natural explanation from a somewhat more general point of view. The essential point is that we consider. instead of an ordinary eigenvalue problem. a different one of the form

[H, XJ_=.tX (50)

for the U eigenoperator" X 6). Important consequences drawn from this equation are: (a) A is the difference of two eigenvalues of H: A=En-E",; (b) X transforms certain eigenvector 1Jl '" of H with Em into another one 1Jl n with eigenvalue En=E:" +A; and (c) Product of two eigenoperators ~Xl is again an eigenoperator, with eigenvalue A=A1 +A2• transforming 1Jl ... into a third eigenvector 1Jl,• If we integrate the. relation (50) according to

d ds X-rHo X]. (~1)

we arrive at the equation

(51)

The equation (51) just corresponds to the problem of obtaining the eigenoperator { z .. '} for the rotation H encountered in Section 3 and 4. When the operator H has a simple structure. a general eigenoperator X will be factorized into a product of prime eigenoperators:

with

X=X1X 2 .. ·Xk ,

e1'=tA1+A.+ ... Ak • (52)

The largest A is the difference of the largest and smallest eigenvalues of sH in (52). If the eigenvalues E1 are symmetrically distributed around zero. Emax will itself be given by Emax= Amax/2, a fact which casts a sidelight on the procedure used in the previous sections.

The main part of the present work had been completed nearly two years ago. It is through the kindness of Professor Husimi a d Mr. Syozi of Osaka University that the author enjoys the opportunity of publishing this note. He wishes to express on this occasion his cordial thanks to them and also to Professor Kubo of Tokyo University for helpful discussions.


A Note on .the EigemJalue Problem i1l Crystal Statistics.

References.

(1) H. A. Kramers and G .. II. Wannier, Phys. Rev. no (1941), 252, 268. E. N. Lassetre and J. P. Howe, J. Chern. Phys. 9 (1941), 747 and 801. E. W. Montroll and J. E. Mayer, J. Chern. Phys. 9 (1941), .626.

(2) L. Onsager, Phys. Rev. 65 (1944), 117.

13

(3) Husimi and Sy6zi, Lectures at the Annual Meeting of the Phys. Soc., 19'48 and 1949; Husimi and Syozi, Prog. Theor. Phys. in press.

(4) An alternative way for solving (3) is to use commutative, rather than anticommutative, operators which corresponds to Bose statistics, together with some supplementary conditions. This leads to Bethe's rigorous solution for one-dimensional model (ZS. t Phys. 71 (1931), 205; Handbuck d. Pkys. XXIV/2, 604). We shall not here enter into details about it.

(5) Cf. Cartan: Le{ons sur la tkeone des sjJineut'S, 1938. (6) A deeper research on this subject will be found in: Y. Nambu, On tke metkod of tke tkird quan

tization, Prog. Theor. Phys. 4 (1949), 331 and 399.

A Note on .the EigemJalue Problem i1l Crystal Statistics.

References.

(1) H. A. Kramers and G .. II. Wannier, Phys. Rev. no (1941), 252, 268. E. N. Lassetre and J. P. Howe, J. Chern. Phys. 9 (1941), 747 and 801. E. W. Montroll and J. E. Mayer, J. Chern. Phys. 9 (1941), .626.

(2) L. Onsager, Phys. Rev. 65 (1944), 117.

13

(3) Husimi and Sy6zi, Lectures at the Annual Meeting of the Phys. Soc., 19'48 and 1949; Husimi and Syozi, Prog. Theor. Phys. in press.

(4) An alternative way for solving (3) is to use commutative, rather than anticommutative, operators which corresponds to Bose statistics, together with some supplementary conditions. This leads to Bethe's rigorous solution for one-dimensional model (ZS. t Phys. 71 (1931), 205; Handbuck d. Pkys. XXIV/2, 604). We shall not here enter into details about it.

(5) Cf. Cartan: Le{ons sur la tkeone des sjJineut'S, 1938. (6) A deeper research on this subject will be found in: Y. Nambu, On tke metkod of tke tkird quan

tization, Prog. Theor. Phys. 4 (1949), 331 and 399.


a note on the eigenvalue problem in crystal statistics. a note on

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