- The document presents new exact solutions to the one-dimensional, two-dimensional, and three-dimensional Ising models that show these models can display spontaneous magnetization. - It finds a mistake in the traditional solutions, which did not account for local spin order and incorrectly calculated the partition function and correlation function. - The new solutions show the one-dimensional Ising model has ferromagnetism, with magnetization decreasing asymptotically to zero with temperature. The two-dimensional and three-dimensional models also have non-zero magnetization that decreases to zero with increasing temperature.

1. Full Paper
Exact solutions of the one-dimensional, two-dimensional, and
three-dimensional Ising models
Igor A.Stepanov
Department of Materials Science and Engineering,University of Sheffield, SirRobert Hadfield
Building, Mappin Street,Sheffield, S1 3JD, (UK)
E–mail:i.stepanov@sheffield.ac.uk;igstepanov@yahoo.com
Received: 21st
February, 2012 ; Accepted: 21st
March, 2012
Magnetic materials;
Nanostructures;
Magnetic properties;
Magnetic structure;
Phase transitions.
KEYWORDSABSTRACT
According to the traditional solutions, there is no magnetisation in the
one-dimensional Ising model in the absence of external magnetic field. It is
shown that there is a mistake within them. In them, local order of spins was
not taken into account and the partition function and the correlation
function were calculated wrongly. In this paper, that mistake is improved
and a new solution is presented.According to it, there is ferromagnetism in
that model. Analogously, solutions of the two-dimensional and three-
dimensional Ising models are obtained.
 2012 Trade Science Inc. - INDIA
INTRODUCTION
There is a well-known result in physics of phase
transitions: the one-dimensional Ising model in the
absenceofexternalmagneticfieldhasnomagneticphase
transition. It has been a paradox since the predictions
ofWeiss’mean field theoryare independent of lattice
dimensionalityandhence the linear chainis predicted
toundergoaphasetransitionatnon-zerotemperature.[1-
3]
However, it is shown that there are mistakes in the
traditionalsolutionsinthecalculationofthecorrelation
functionandthepartitionfunction.Anothermethodfor
solvingthisproblemisproposed.Usingit,onecanshow
that the one-dimensional Ising model displays
spontaneous magnetisation which decreases
asymptoticallytozero with the temperature.The new
method is applied to the two-dimensional and three-
dimensionalproblemsandexactsolutionsareobtained.
An Indian Journal
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NSNTAIJ, 6(3), 2012 [118-122]
ISSN : 0974 - 7494
They also produce magnetisation which decreases
asymptoticallytozerowiththetemperature.
THEORY
Lest us considerthe one-dimensional Isingmodel
in the absence of external magnetic field.[1]
The
HamiltonianofachainofN atoms isoftheform




1N
1i
1iii
^
ssJÇ (1)
where the spins si
are one-dimensional unit vectors
assumingonlythediscretevalues+1and-1andJi
isthe
interactionenergybetweenspinssituatedonsitesiandi
+1.PositiveJi
favoursparallelandnegativeJi
antiparallel
alignment ofthe spins.Thepartitionfunction is
     
  


 






1
11s
1
12s
1
1Ns
1N
1i
1iii1N21NN ssJexp...J,...,J,JZZ (2)

2. Igor A.Stepanov 119
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Where
kT
1
 .
Forachainof2spins theHamiltonianwill be
211
^
ssJÇ  (3)
Thepartitionfunctionis
     1
1
11s
1
12s
211122 Jcosh4ssJexpJZZ   
 
(4)
Now we can calculate the nearest-neighbour cor-
relationfunctionforachainoftwospins:
   1
}s{
21121
1
2211 JtanhssJexpssZss)1(Ã  
(5)
where summation is over spins s1
and s2
. In Ref. 1,
usingamoretedious derivationit was shownthat fora
chain ofN spins
   kk Jtanh1Ã  (6)
Indeed, the expression for  1Ãk does not depend
onthenumberofspins inthechain.
In Ref. 1,the two-spin correlationfunctionfor achain
of N spins was calculated:
 
  











s
1N
1i
1iiirkk
1
Nrkkk ssJexpssZss)r(Ã (7)
where the symbol {s} denotes the N-fold summation
ofEq.(2).Hereristhedistancebetweenspins,inunits
of a latticeconstant. It equals
       
 





r
1i
1ik
1rk1kkk
Jtanh
Jtanh...JtanhJtanhrÃ
(8)
Now we can find the temperature at which long-
range order sets in, i.e. the temperature at which the
two-spincorrelationfunctionfallsoffsufficientlyslowly
with theinterspindistancer thatthemagnetisationbe-
comes non-zero. The squared normalized zero-field
magnetisationis:[1]
 
 
 rÃlim
0H,0TM
0H,TM
k
r2
2
2




 (9)
From Eq. (8) we see that the hyperbolic tangents are
less than oneand hence the products approach zero in
thelimitofinfiniter.Thelinearchainthereforedisplays
zerospontaneousmagnetisationforallnon–zerovalues
of thetemperature. Nevertheless, there is a mistake in
the proof given byEqs. (8) and (9). Consider a chain
consistingof100spins,Jk
=Jforallk,and   99.01Ãk  .
Its squared magnetisation from Eqs. (8) and (9) is 2
˜
0.37. Now let us assume that J50
= 0. Then from Eqs.
(8) and (9), 2
= 0. It is a wrong result. In reality the
chain will split into two chains with 2
˜ 0.61 each. It
proves that theprevious solutionis wrong.
There is a flaw in the proof given byEqs. (8) and
(9). Let us assume theoreticallythat the correct solu-
tionoftheIsingmodelexistsand,accordingtoit,inthe
chain 90% of the spins are up and 10% are down.
(StanleyinRef.1presentedthesameargument: “Sup-
posethecontrary,namelythatthereexistsaphasetran-
sitionwithTc
>0.ThenforsomeT <Tc
letus‘flip’half
the chain to assume an opposite polarity.”)Then Eqs.
(8) and (9) predict that magnetisation of this chain is
zero.
Eq. (8) is of the form
  ...ssssssrà 433221k  (10)
whereeveryfactorislessthanunity.Imagineaninfinite
chainofvectors whichis constructedbythefollowing
rule: orientation of a vector depends on that of its left
neighbour:iftheneighbourisupthenwiththeprobabil-
ity90%the vectoris up.If the neighbour is down then
with the probability 9 0% the vector is up. Eq. (10)
predictsthatthelongrangecorrelationfunctionforthis
chainiszero,whichis wrong.
ConsideraninfiniteIsingchainwithastrongcorre-
lation,forexample,   99.01Ãk  .Itmeansthatalmostall
neighbour spins are aligned so as the central one, k.
Twocasesarepossible.Thefirstone:achainwithabout
99% ofthespins turnedupandonlyabout 1%ofthem
turneddown.Suchchainwillpossessmagnetisationac-
cordingtothedefinitionofthecorrelationfunction,al-
thoughtheprevioussolutionresultsinnomagnetisation
init.Thesecondcaseis:thechainconsists ofdomains,
eachoneconsistsinaverageof50spins.Insomeofthe
domains thespins areturned up,andinothers theyare
turned down. The net magnetisation is zero. If
  999.01Ãk  thenthechainconsistsofdomainsofabout
500 spins, etc. It is alreadya new result. If there is no
magnetisationatall,thensuchchainhasequal number
of spins upanddown, and theyareoriented randomly,
thatisitscorrelation   01Ãk  .Itcontradictsthedefini-
tion of the problem:   99.01Ãk  and, from Eq. (6),
  01Ãk  only at Jk
= 0 or T  .

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One can supposethat the first casetakes place be-
cause  1Ãk is identical for everyk, andthe directions
ofspins must bealsoidenticalforallkj(j istheindex
of spins directed down, theyare about 1% of the total
spinnumber)due tosymmetryofthe problem.
ThemagnetisationofanIsingchainwillbeequalto
thatofanaveragedelementarycell.Theelementarycell
consists of a spin and its nearest neighbour. The
magnetisationofthechainis,
 
 






i
1N
1i
i
2
Jtanh
1N
Jtanh
(11)
here < Ji
> is an average interaction energy.In the uni-
form case, Ji
= J, the squared normalized zero-field
magnetisationis
 Jtanh2
 (12)
And
mulaisvalidforalldimensions.Meanfieldtheorypro-
duces thefollowingequation:
  Jztanh (15)
Ifwedenotethemagnetisationofanelementarycellof
the crystal lattice byMi
then the magnetisation of the
two-andthree-dimensionalIsinglatticesis
cells
cellsN
1i
i
N
M
M


 (16)
whereNcells
isthenumberoftheelementarycells.Inthe
uniformcaseofequalJ,thenormalizedmagnetisation
is equal to that of one elementarycell. For that case, 
for hexagonal lattice, square lattice and simple cubic
lattice canbe written as
5.0
3
1k
k
3
)1(
















(17)
5.0
4
1k
k
4
)1(
















(18)
and
5.0
12
1k
k
12
)1(
















(19)
respectively.Tocalculate  1Ãk ,onemust findthepar--
titionfunctionoftheelementarycell.Theenergyofthe
hexagonallatticeis
...
2
ssJssJssJ
2
ssJssJssJ
...EEEÇ
232345424133322211
321
^





(20)
Herethefirsttermontherighthandsideisthecon-
tributionofthefirst elementarycell,thesecond term is
that of the adjacent cell, etc. The Hamiltonian of one
cell in a lattice differs from that of a single triangle by
division by 2 because every term in the numerator is
taken twice ( 322 ssJ and 232 ssJ in Eq. (20)). For the cu-
bic lattice, it differs from that of a single cubebydivi-
sionby4.Thepartitionfunctionofthehexagonallattice
Figure 1: Dependence of the spontaneousmagnetisation per
spin ontemperatureTfortheone-dimensionalIsing model,
Eq. (13), J = 1.
   5.0
Jtanh  (13)
The plot of this function is given in Figure 1. The
magnetisation decreases from unity at T = 0 and as-
ymptoticallytendstozero withthetemperature.Mean
fieldtheorypredictsthattheIsingmodelacquiresmag-
netism at cTT  [1, 3]
where Tc
can be found from
zJkTc  (14)
Here z is the number of nearest neighbours. This for-

4. Igor A.Stepanov 121
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intheuniformcaseis:
   ...ZZZ...EEEexpZ 321
}s{
321hex   (21)
where Z1
, Z2
, Z3
, ... are the partition functions of the
elementarycells.Thepartitionfunction of acell inthe
hexagonallatticeintheuniformcaseis:
)J5.0exp(6)J5.1exp(2Zi  (22)
Nowletusfindthenearest-neighbourcorrelationfunc-
tionbetweenspin1andspin2,  1Ã1 ,itequalsall other
nearest-neighbourcorrelationfunctions:
 
  
1
}s{
121
321
}s{ }s{
3
}s{
2121
hex
}s{
32121
1
Z
)Eexp(ss
...ZZZ
...)Eexp()Eexp()Eexp(ss
Z
...EEEexpss
1Ã

 







(23)
where E1
, E2
, E3
, ... are defined in Eq. (20). So, this
correlationfunctionis
   
)J5.0exp(6)J5.1exp(2
)J5.0exp(2)J5.1exp(2
1Ã1Ã k1


 (24)
andthemagnetisationis:
5.0
)J5.0exp(3)J5.1exp(
)J5.0exp()J5.1exp(








 (25)
Theplotofthisfunctionis giveninFigure2.
UsingthismethodinitiallydevelopedinRef.1,one
caneasyshowthatthelongrangecorrelationfunctionis
    r
1k rÃrà  (26)
It tends tozero with r and, therefore, accordingto
Ref.1,thehexagonaltwo-dimensionallatticepossesses
no spontaneous magnetisation.This result contradicts
thetraditionalsolutionforthetwo-dimensionalcaseand
is another evidence that the proof proposed in Ref. 1
(Eqs. (8), (9)) is untenable.
Analogously,onecanshowthatthepartitionfunc-
tionofacell in thesquarelatticeintheuniform caseis:
12)J2cosh(4Zi  (27)
the nearest-neighbour correlation function is
 
12)J2cosh(4
)J2sinh(4
1Ãk

 (28)
andthemagnetisationis:
5.0
3)J2cosh(
)J2sinh(







 (29)
Again,thelongrangecorrelationfunctionis   r
1 rÃ
and the magnetisation of this lattice must be zero ac-
cordingtoRef.1.TheplotofthisisshowninFigure2.
Figure2 : Dependence of thespontaneousmagnetisationper
spin  on temperature T for the two-dimensional hexagonal
lattice(solidline)and two-dimensionalsquarelattice(dashed
line) Ising models, Eqs. (25) and (29), J = 1. The both curves
almost coincide
It practically coincides with that of Eq.(25). The
magnetisationdecreasesasymptoticallytozerowiththe
temperature from the maximum value at T=0.
Usingthesame method, one canfind thepartition
functionofacell inthesimplecubiclatticefortheuni-
formcase:
64)J5.0cosh(96)Jcosh(60)J5.1cosh(32)J3cosh(4
Zi


(30)
Thenearest-neighbourcorrelationfunctionis
 
16)J5.0cosh(24)Jcosh(15)J5.1cosh(8)J3cosh(
)J5.0sinh(4)Jsinh(5)J5.1sinh(4)J3sinh(
1Ãk



(31)
andthemagnetisationis:
5.0
16)J5.0cosh(24)Jcosh(15)J5.1cosh(8)J3cosh(
)J5.0sinh(4)Jsinh(5)J5.1sinh(4)J3sinh(









(32)

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Figure 3 : Dependence of thespontaneous magnetisation per
spinontemperatureTforthethree-dimensionalIsingmodel,
Eq. (32), J = 1.
The plot of this result is shown in Figure 3. rations, and, therefore, are wrong.
CONCLUSIONS
Earlier it has been thought that the Ising model,
unlikethemeanfieldtheory,predictsnophasetransition
for a linear chain system in the absence of external
magnetic field.It is shownthat thistraditional result is
wrongbecauseitisbasedonawrongcalculationofthe
correlation function or taking redundant terms in the
partition function.Anothersolutionis obtained which
does predict magnetisation. It acquires its maximum
value at T = 0 and decreases asymptotically to zero
withthetemperature.Ithasbeenproventhatthesolution
oftheone-dimensionalcaseisexact.Usingthismethod,
exact solutions of the two-dimensional and three-
dimensionalIsingmodelsareobtained.Theyalsoacquire
their maximum values at T = 0 and decrease
asymptoticallytozerowiththetemperature.Itisshown
that there is a mistake in the former solutions of the
two-andthree-dimensional Isingmodels.
REFERENCES
[1] H.E.Stanley; ‘Introduction to PhaseTransitions and
Critical Phenomena’, Oxford UniversityPress, New
York, Oxford, (1987).
[2] K.Binder, Ising Model; in M.Hazewinkel, (Ed),
‘Encyclopaedia of Mathematics’, Kluwer Aca
demic Publishers, London, 279-281 (2000).
[3] J.M.Yeomans; ‘Statistical Mechanics of Phase
Transitions’, Oxford University Press, Oxford,
(1993).
[4] E.Ising; Z.Phys., 31, 253 (1925).
[5] R.Becker,W.Doring; ‘Ferromagnetismus’, Springer
Verlag, Berlin, (1939).
[6] S.V.Vonsovskij, Ya.S.Shur; ‘Ferromagnetizm’,
Gostehizdat, Leningrad, 161-167 (1948).
[7] G.F.Newell, E.W.Montroll; Rev.Mod.Phys., 25, 159
(1953).
[8] G.F.Newell, E.W.Montroll; Rev.Mod.Phys., 25, 353
(1953).
[9] C.Domb; Adv.Phys., 9, 149 (1960).
[10] A.Vindigni, A.Rettori, M.G.Pini, C.Carbone,
P.Gambardella; Appl.Phys.A, 82, 385 (2006).
[11] Z.-D.Zhang; Phil.Mag., 87, 5309 (2007).
[12] V.Matveev, R.Shrock; Phys.A: Math.Theor., 41,
135002 (2008).
Isingproposedanothersolutionoftheone-dimen-
sional Isingmodel.[4]
He considereda chainofNspins
and usedthe expression
Zln
H
kT


 (33)
whereHistheexternal magneticfieldandZ isthepar-
titionfunction.Inthepartitionfunction,thesummation
is over all configurations with N1
spins up and N – N1
spins down, where N1
changes from 0 to N. However,
if   99.01Ãk  and the case 1 described above takes
place, then N – N1
is of the order 1% and configura-
tions with N – N1
>> 1% are impossible and may not
be taken into account.
If one knows  1Ãk one can calculate exactly the
numberofspinsupanddownintheinfinitechain.Inthe
calculationofthepartitionfunctiononlytheconfigura-
tions with that number of spins up and down must be
taken.Asthedirectionsupanddownhaveequalrights,
onemusttakealsotheconfigurationswithN1
spinsdown
and N – N1
spins up. Such partition function will be
exact and calculations with it will produce correct re-
sult. Otherconfigurations are impossible andmaynot
betakenintoaccountincalculation ofZ.InRefs.1,3-
12,othersolutionsoftheone-dimensional,two-dimen-
sional,andthree-dimensionalIsingmodelsweregiven.
TheyalsouseEq.(33)andsummationoverallconfigu-