Download Chapter 2 Phase Transition and Critical Phenomena

Chapter 2
Phase Transition and Critical
Phenomena
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2.1 Introduction
A phase is a state of matter in thermodynamic equilibrium. The same matter (or
system) could be in several different states or phases depending upon the
macroscopic condition (Temperature, Pressure, etc.) of the system. Different
phases of water are our everyday experience. Ice, water and steam are the
different states or phases of a collection of large number of H 2 O molecules.
Given a macroscopic condition, the system spontaneously goes to a particular
phase corresponding to lowest free energy as shown in Figure 2.1. For a closed
system which exchange only energy with the surroundings has lowest Helmholtz
free energy ( F ) at equilibrium and for an open system which exchange energy as
well as mass with the surroundings the equilibrium corresponds to lowest Gibb's
free energy ( G ). The free energy of a system is the sum of various energies
associated with a collection of large number of atoms or molecules. Beside the
kinetic energies of the particles, the potential energies due to inter atomic (or
molecular) interactions contribute mostly to the free energy. Essentially, the
different phases of matter are a consequence of interaction among a large number
of atoms or molecules at a given thermodynamic condition.
Figure 2.1: Variation of Helmholtz free energy F against temperature T at a constant pressure for a
system in solid, liquid and gaseous phases. After , the system spontaneosly goes to the liquid phase.
There is a wide variety of phase transitions starting from liquid to gas,
paramagnet to ferromagnet, normal to super conductor, liquid to liquid crystal
and many other transitions. Most phase transitions belong to one of the two types
- first order and second order phase transition. As per Ehrenfest's criteria, nth
order phase transition corresponds to the discontinuity of the n th derivative of
the free energy functions. Thus, in a first order transition, the first derivative of
the free energy becomes discontinuous whereas in the second order phase
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transition, the second derivative of the free energy becomes discontinuous or
diverges at the transition point. First and second order phase transitions then can
be understood qualitatively in terms of discontinuity of free energy functions.
2.2 Thermodynamic Stability, positive response function and
convexity of free energy:
Free energy functions are often found a concave or a convex function of
thermodynamic parameters. A function f (x) is called a convex function of x if
 x + x  f ( x1 ) + f ( x2 )
f  1 2 ≤
2
 2 
for all x1 and x2 .
That is to say the chord joining the points f ( x1 ) and f ( x2 ) lies above or on the
curve f (x) for all x in the interval x1 < x < x2 for a convex function. Similarly,
a function f (x) is called a concave function of x if
 x + x  f ( x1 ) + f ( x2 )
f  1 2 ≥
2
 2 
for all x1 and x2 .
Thus, for a concave function, the chord joining the points f ( x1 ) and f ( x2 ) lies
below or on the curve f (x) for all x in the interval x1 < x < x2 . If the function is
differentiable and the derivative f ′(x ) exists, then a tangent to a convex function
always lies below the function except at the point of tangent whereas for a
concave function it always lies above the function except at the point of tangency.
If the second derivative exists, then for a convex function f ′′( x) ≥ 0 and for a
concave function f ′′( x) ≤ 0 for all x
On the other hand, thermodynamic response functions such as specific heat,
compressibility, susceptibility (for ferromagnetic systems) are found to be
positive and the positive values of the response function implies the convexity
properties of the free energy functions such as F = E − TS or G = F + PV .
The positive response function is a direct consequence of Le Chatelier's principle
for stable equilibrium. The principle says, if a system is in thermal equilibrium
any small spontaneous fluctuation in the system parameter, the system gives rise
to certain processes that tends to restore the system back to equilibrium. Suppose
there was a spontaneous temperature fluctuation in which the temperature of the
system increases from T to T ′ . In order to maintain the stability, the system
should absorb certain amount of heat ΔQ and as a consequence the specific heat
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C = ΔQ/ΔT must be positive since both ΔQ and ΔT are positive. If there occurs
a spontaneous pressure fluctuation, P → P′ and P′ > P , then the system will
reduce its volume by certain amount ΔV to maintain the stability. As a
consequence the compressibility κ = −ΔV/ΔP is also to be positive since ΔP is
positive but ΔV is negative. Thus, for thermally and mechanically stable fluid
system, the specific heat and compressibility should be positive for all T .
However, for a magnetic system such arguments that the susceptibility χ and
specific heat C both are positive cannot be made. It is known that for
diamagnetic materials, χ < 0 . The ferromagnetic materials on the other hand
have positive χ . It can be shown that such systems are described by the
Hamiltonian =
− ⃗ ∙ ⃗.
The response functions are not all independent. One could show for a fluid
system,
CP − CV = TV α P2 /κ T and κT − κ S = TV α P2 /C P
where α P = (∂V/∂T ) P /V , the thermal expansion coefficient and similarly for a
magnetic system,
CH − CM = T α H2 /χT and χT − χ S = T α H2 /CH
where α H = (∂M/∂T ) H . Since, the specific heat and compressibility are positive, it
could be shown from the above relations that C P ≥ CV and κ T ≥ κ S . The equality
holds either at T = 0 or at α = 0 , for example, α = 0 for water at 4 oC .
From thermodynamic relations, it is already known that
 ∂ 2G 
CP
 2 =−
T
 ∂T  P
and
 ∂2 F 
CV
.
 2 =−
T
T
∂

V
Since the specific heats are positive, these second derivatives are negative and as
a consequence G and F both are concave functions of temperature T . It is also
known that
 ∂ 2G 
 ∂2F 
1
=
and
=
.
V
κ
−
 2

T
2 
κ
P
V
V
∂
∂

T

T
T
Since the compressibility is a positive quantity, G is a concave function of P
whereas F is a convex function of V .
It can be shown that G (T , H ) is a concave function of both T and H whereas
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F (T , M ) is a concave function of T but a convex function of M for magnetic
=
systems described by the Hamiltonian
− ⃗ ∙ ⃗.
(a)
(c)
(b) S = −(∂F/∂T ) P
(d) V = (∂G/∂P)T
Figure 2.2: Variation of free energies and their first derivatives with respect to the respective
parameters around a first order transition. (a ) Plot of Helmholtz free energy F against
temperature T at a constant pressure. (b) Plot of the entropy S , first derivative of F with
respect to T , against T . (c ) Plot of Gibb's free energy G against pressure P at a constant
temperature. (d ) Plot of the volume V , first derivative of G with respect to P , against P .
Discontinuities in entropy S (latent heat) as well as in volume V are marks of a first order
transition.
2.3 First and second order transitions
Both the Helmholtz free energy F and the Gibb's free energy G are concave
function of temperature T and pressure P. However, F is a decreasing
function of temperature T and G is an increasing function of pressure P.
Variation of the Helmholtz free energy ( F ) with temperature T and the Gibb's
free energy ( G ) with pressure P are shown in Fig.2.2 for a fluid system.
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Around a first order transition point (T * , P* ) , the free energy curves of the two
phases meet with difference in slopes and both stable and metastable states exist
for some region of temperature and pressure. At the transition temperature ∗ ,
the tangent to the curve F (T ) versus T changes discontinuously and similarly
for G (P ) versus P , the tangent changes discontinuously at the transition
pressure P* . The change in the slope of F with respect to T corresponds to
entropy S = −(∂F/∂T ) P discontinuity. The change in the slope of G with respect
to P corresponds to discontinuity in the volume V = (∂G/∂P)T . The transition is
called first order because both entropy S and volume V are the first derivatives
of the free energy functions and exhibit discontinuities. First order transitions are
generally abrupt and are associated with an emission (or absorption) of the latent
heat L = T *ΔS . Latent heat is released when the material cools through an
infinitesimally small temperature change around the transition temperature. Most
crystallization and solidification are first order transitions. For example, the latent
heat L = 334 Jg −1 comes out when water becomes ice. This happens sharply at 0o
C temperature under the atmospheric pressure, when the H 2 O molecules which
wander around in the water phase gets packed in FCC ice structure releasing the
excess energy as the latent heat. It can also be noted that in a first order phase
transition there is generally a radical change in the structure of the material.
First order transition often (not always) ends up at a critical point where second
order transition takes place. In Fig.2.3, the characteristic behaviour of second
order phase transitions is shown for a fluid system. At a first order phase
transition the free energy curves of the two phases meet with a difference in
slopes whereas at a second order transition the two free energy curves meet
tangentially at the critical point ( Tc , Pc ). The slopes of the curves changes
continuously across the critical point. Therefore, there is no discontinuity either
in entropy or in volume. Since there is no entropy discontinuity in second order
transition, there is no emission (or absorption) of latent heat in this transition. It is
a continuous phase transition where the system goes continuously from one phase
to another without any supply of latent heat.
Not only the first derivatives but also the second derivatives of the free energy
show a drastic difference in their behaviour around the transition point in the first
and second order transitions. For example, the specific heat C P = −T (∂ 2G/∂T 2 ) P
diverges in the first order transition whereas in the second order transition
specific heat has a finite discontinuity or logarithmic divergence at the critical
point. Infinite specific heat in first order transition can be easily visualized by
considering boiling of water. Any heat absorbed by the system will drive the
transition ( 100o C water to 100o C steam) rather than increasing the temperature of
the system. There is then an infinite capacity of absorption of heat by the system.
In the second order phase transition, the response functions, the second
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derivatives of the free energy functions, are expected to diverge at the critical
1  ∂ 2G 
point. For example, the isothermal compressibility κ T = −  2  of a fluid
V  ∂P T
system diverges at the critical point.
(a)
(c)
(b) S = −(∂F/∂T ) P
(d) V = (∂G/∂P)T
Figure 2.3: Variation of free energies and their first derivatives with respect to the respective
parameters around a second order transition. (a ) Plot of Helmholtz free energy F against
temperature T at a constant pressure. (b) Plot of the entropy S , the first derivative of F with
respect to T , against T . (c ) Plot of Gibb's free energy G against pressure P at a constant
temperature. (d ) Plot of the volume V , the first derivative of G with respect to P , against P .
No discontinuity is present in either entropy S or volume V .
2.4. Continuous phase transition or critical phenomena
Critical phenomena are the characteristic features that accompany the second
order phase transition at a critical point. The critical point is reached by tuning
thermodynamic parameters (for example temperature T or pressure P or both).
A critical phenomenon is seen as T ( P ) approaches the critical point Tc ( Pc ) . In
order to understand the characteristic features appearing at the critical point, one
must study the macroscopic properties of the system at the critical point. In
principle, all macroscopic properties can be obtained from the free energy or the
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partition function of a given system. However, since the critical phenomena, a
second order phase transition or a continuous phase transitions, involve
discontinuities in the response functions (which are second derivatives of the free
energy function) at the critical point there must be singularities in the free energy
at the critical point. On the other hand, the canonical partition function of a finite
number of particles is always analytic. The critical phenomena then can only be
associated with infinitely many particles, i.e. in the ``thermodynamic limit'', and
to their cooperative behaviour. The study of critical phenomena is thus essentially
related to finding the origin of various singularities in the free energy and
characterizing them.
Let us consider more carefully the two classic examples of second order phase
transition involving condensation of gas into liquid and transformation of
paramagnet to ferromagnet. In the liquid-gas or fluid system the thermodynamic
parameters are ( P, V , T ) and in magnetic system the corresponding
thermodynamic parameters are ( H , M , T ) . One may note the correspondence
between the thermodynamic parameters of fluid and magnetic systems as:
V → − M and P → H . In the case of fluid, instead of volume V we will be
considering the density ρ as a parameter. The equations of states in these
systems are then given by f ( P, ρ , T ) = 0 and f ( H , M , T ) = 0 respectively. A
second order phase transition is a qualitative change in the system behaviour at a
sharply defined parameter value, the critical point, when the parameter changes
continuously. The critical points are usually denoted by ( Pc , ρ c , Tc ) and
( H c , M c , Tc ) . Commonly, phase transitions are studied varying the temperature T
of the system and a phase transition occurs at T = Tc . We will be describing the
features appearing at the critical point by considering different phase diagrams
such as PT , HT ; Pρ , HM ; ρ T , MT of the full three dimensional phase space
of ( P, ρ , T ) or ( H , M , T ) .
PT and HT diagrams: In Fig.2.4, PT and HT diagrams are shown. It can be
seen that the first order transition line (the vapour pressure curve for the fluid
system and H − T line for the magnetic system) terminates in a critical point at
T = Tc . This means that the liquid can be converted to gas continuously without
crossing the first order transition line following a path shown by a curved dotted
line. Similarly, in the case of magnetic system a continuous change from up spin
region to down spin region is also possible.
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Fluid
Magnet
Figure 2.4: Schematic plot of pressure P versus temperature T for a fluid in a gas-liquid
transition and plot of magnetic field (H ) versus temperature T for an Ising ferromagnet.
The solid line is the first order transition line which ends at a critical point Tc .
Fluid
Magnet
Figure 2.5: Schematic plot of pressure P versus density ρ isotherms for a fluid system
and of H versus M for a magnetic system.
Pρ and HM diagrams: These phase diagrams shown in Fig.2.5. The most
striking feature in this phase diagram is the change in shape of the isotherms as
the critical point is approached. At high temperature (T >> Tc ) , the isotherms are
expected to be the straight lines given either by the ideal gas equation of state
P = ρk BT/m or by the Curie law M = cH/T , where m is the mass of a molecule
and c is a constant. As the temperature decreases toward the critical temperature
Tc , the isotherms develop curvature. At T = Tc the isotherms are just flat and one
have ∂P/∂ρ = 0 and ∂H/∂M = 0 . As a consequence, the response functions,
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isothermal compressibility κ T =
1
∂ρ/∂P
ρ
and the isothermal susceptibility
χT = ∂M/∂H , diverge as T → Tc . These response functions are second derivative
1
of the respective free energy function: κ T = − (∂ 2G/∂P 2 )T and χT = −(∂ 2 F/∂H 2 )T .
V
We see that the second derivatives of the free energy are singular (the first
derivatives are continuous) as we expect in second order phase transitions.
ρT and MT diagrams: ρT and MT diagrams are shown in Fig.2.6. From
these diagrams as well as from Fig.3, it can be seen that there is a large difference
in densities in the liquid and gas phases of a fluid at low temperature. In the
magnetic system, there is a large difference in spontaneous magnetization below
Tc . As Tc is approached from below, the density difference Δρ = ρ L − ρ G of a
fluid system and the spontaneous magnetization M of a magnetic system tend to
zero. A quantity which is non-zero below Tc and zero above Tc is called the
order parameter of the transition. Thus, Δρ and M serve as order parameter of
the fluid and magnetic system respectively. Note that below Tc the order
parameter is multivalued where as it is single valued (zero) above Tc . Thus, the
order parameter has a branch point singularity at T = Tc .
Fluid
Magnet
Figure 2.6: Schematic plot of density ρ for fluid system and spontaneous magnetization
M for magnetic system against temperature T .
The critical point (Tc , Pc , ρc ) at which the transition occurs is found to be
dependent on the details of interatomic interactions or underlying lattice structure
and varies from material to material. Critical temperatures of different materials
are listed in table 2.1.
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Fluids
Water
Alcohol
CO 2
Argon
Tc (K)
Pc (atm)
ρ c (g/cm 3 )
647.5
516.6
304.2
150.8
218.50
63.10
72.80
48.34
0.325
0.280
0.460
0.530
Magnets
Fe
Ni
CrBr 3
EuS
Tc (K)
1043.00
627.20
32.56
16.50
Table 2.1: Values of the critical parameters for different fluid and magnetic systems are listed. For
magnetic systems, other critical parameters are spontaneous magnetization (M ) and external
magnetic field (H ) . Both M and H are zero at the critical point.
2.5 Morphology, fluctuation and correlation
The isotherms, P ρ or HM curves in the respective phase diagrams (Fig.4) develop
curvature as the system approaches the critical temperature Tc from above. The
curvature in the isotherms is the manifestation of the long range correlation of the
molecules in the fluid or spins in magnets. At high temperature, the gas molecules
move randomly or the magnetic moments flip their orientation randomly. Due to
the presence of interactions small droplets or domains of correlated spins appear
as the temperature decreases. These droplets grow in size as T decreases closer
to Tc . At T = Tc , droplets or domains of correlated spins of all possible sizes
appear in the system. Lateral dimension of these droplets become of the order of
the wavelength of ordinary light. Upon shining light on the fluid at T = Tc , a
strong scattering is observed and the fluid appears as milky white. The
phenomenon is known as critical opalescence. Similarly in magnetic systems,
domains of correlated spins of all possible sizes appear in the system and a huge
neutron scattering cross section is observed at T = Tc . As T → Tc , there appears
droplet or domain of correlated spins of the order of system size. One may define
a length scale called correlation length which is the lateral dimension of the
droplets or domains of correlated spins. Therefore, the correlation length diverges
as T → Tc . One should note that the system does not correspond to a ordered state
at Tc and a completely ordered state is achieved only at T = 0 .
As the system approaches Tc , there are long wave-length fluctuations in density
in fluid or in the orientation of magnetic moments in the magnetic system. These
fluctuations occur at every scale. If ξ is the largest scale of fluctuation and a is
the lattice spacing, then the system appears to be self-similar on all length scales
x for a < x < ξ . At T = Tc , ξ is infinite and the system becomes truly scale
invariant. The correlation between the spins (or molecules) is measured in terms
of fluctuations of spins (or density) away from their mean values:
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G (si , s j ) = 〈( si − 〈 si 〉 )( s j − 〈 s j 〉 )〉 = r − ( d −2+η ) exp(−r/ξ )
where r is the distance between si and s j , ξ is the correlation length and η is
some exponent. At the criticality, ξ diverges to infinity and ( ⃗) decays as a
power law.
T < Tc
T ≈ Tc
T > Tc
Figure 2.7: Morphology of the system below, near and above the critical temperature Tc .
The black region represents the liquid (up spin) and the white space represents the gas (down
spin) in fluid systems (magnetic systems). A huge fluctuation in density (or in spin
orientation) appears as T → Tc .
Close to a critical point, the large spatial correlations which develop in the system
are associated with long temporal correlations as well. At the critical point, the
relaxation time and characteristic time scales diverge as determined by the
conservation laws. This is known as the critical slowing down. A relaxation
function φ (t ) may decay exponentially at long times as φ (t ) : e − t/τ , where τ is the
relaxation time. τ diverges at the critical point and the dynamic critical behavior
can be expressed in terms of the power law as τ ∝ ξ z , where z is called the
dynamic critical exponent.
2.6 Fluctuation and response functions
Apart from macroscopic thermodynamics quantities, statistical mechanics can
also provide information about microscopic quantities such as fluctuations and
correlation. Even if the system is in thermal equilibrium (constant T ) or
mechanical equilibrium (constant P ) or chemical equilibrium (constant µ ), the
energy E , magnetization M , number of particles N may vary indefinitely and
only the average values remain constant. It would be interesting to check that the
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thermodynamics response functions such as specific heat CV , isothermal
compressibility κ T or isothermal susceptibility χT are directly proportional to
the fluctuation in energy, density or magnetization respectively.
The fluctuation in energy is defined a
〈(ΔE ) 2 〉 = 〈( E − 〈 E 〉 ) 2 〉 = 〈 E 2 〉 − 〈 E 〉 2 .
By calculating 〈 E 2 〉 , it can be shown that
〈 (ΔE ) 2 〉 = −
∂〈 E 〉
= k BT 2CV
∂β
or
CV =
1
(〈 E 2 〉 − 〈 E 〉 2 ).
2
k BT
Thus the specific heat is nothing but fluctuation in energy.
The fluctuation in number of particles N is defined as
2
N k BT
∂〈 N 〉
〈(ΔN ) 〉 = 〈( N − 〈 N 〉) 〉 = 〈 N 〉 − 〈 N 〉 = k BT
=
κT
V
∂µ
2
2
2
2
where κ T is the isothermal compressibility. The isothermal compressibility is
) , one has
then proportional to density fluctuation. If
= ⁄(〈 〉
=
〈(
− 〈 〉) 〉
.
〈 〉
Similarly, the isothermal susceptibility is proportional to the fluctuation in
magnetization
k T
χT = B (〈 M 2 〉 − 〈 M 〉 2 ).
N
These are system-independent general results. Generally these fluctuations are
negligibly small at normal conditions. At room temperature, the rms energy
fluctuation for 1 kg of water is : 4.2 × 10 −8 J [T (k B CV )1/2 ] , whereas to change the
water temperature by 1 degree the energy needed is 1011 × CV . Since the heat
capacity grows linearly with the system size, the relative energy fluctuation goes
to zero at the thermodynamic limit.
The above relations show that the responses CV , κ T , χ T are linearly proportional
to the fluctuation in respective thermodynamic quantities - this is known as linear
response theorem.
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2.7 Correlation in terms of fluctuation and response
So far, the response functions are obtained as the thermal-average of
corresponding macroscopic variables from the knowledge of the probability
distribution of the microstates of the system. It can also be obtained in terms of
“microscopic” variables like spin or particle density at a point. A quantitative
way of doing it is through defining two point correlation functions, how the spins
or particle densities at different points are related. Below we will establish a
relationship between correlation, fluctuation and responses of the system for fluid
and magnetic systems.
Fluid: Density at any point ⃗ is given by the Dirac delta function ( ⃗) as
( ⃗) =
〈 〉
=
( − ).
A density-density correlation function is the correlation of the fluctuation of the
densities from its average values at ⃗ and ⃗′ and can be defined as
G( ⃗, ⃗′) = 〈( ( ⃗) − 〈 ( ⃗)〉) ( ( ⃗′) − 〈 ( ⃗′ )〉)〉 = 〈 ( ⃗) ( ⃗′)〉 − 〈 ( ⃗)〉〈 ( ⃗′)〉.
Since the system is spatially uniform (or translationally invariant) 〈 ( ⃗)〉 =
〈 ( ⃗′)〉 = , the average density of the system, the correaliton function can be
written as
G( ⃗, ⃗′) = 〈 ( ⃗) ( ⃗′)〉 −
.
As | ⃗ − ⃗′| → ∞, the probability of finding a particle at ⃗′ becomes independent
of what is happening at ⃗ ; i.e., the densities become uncorrelated. Hence,
G( ⃗, ⃗′) → 0 as | ⃗ − ⃗′| → ∞. However, at a short distance, the correlation
function G( ⃗) depends on as
/
G( ⃗) ~
where
is an exponent and
is called the correlation length.
On the other hand, the particle number fluctuation can be written as
〈(
− 〈 〉) 〉 = 〈
⃗( ( ⃗) − 〈 ( ⃗)〉)
= ∬ G( ⃗, ⃗′)
⃗
⃗’.
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⃗′( ( ⃗′) − 〈 ( ⃗′)〉) 〉
Or,
〈(
− 〈 〉) 〉 =
⃗′G( ⃗ − ⃗′) =
⃗
⃗′′G( ⃗′′)
where ⃗ ′′ = ⃗ − ⃗′.
The isothermal compressibility then can be expressed in terms of the
density-density correlation function as
=
〈(
− 〈 〉) 〉
=
〈 〉
〈 〉
⃗′′G( ⃗′′) .
Thus, the density fluctuation, isothermal compressibility and the density-density
correlation function are all interrelated quantities.
Magnets: The correlation between the spins on site i and j can be measured by
defining spin-spin correlation function
Γ ⃗ , ⃗ = 〈(
− 〈 〉)
−〈 〉 〉= 〈
〉 − 〈 〉〈 〉
where the spin si is at ⃗ and the spin sj is at ⃗ . If the spins are non-interacting,
〈 si s j 〉 = 〈 si 〉〈 s j 〉. For example, for a magnet at high temperature paramagnetic
phase, 〈 si 〉 = 0 and hence 〈 si s j 〉 = 0 . If the spins interact with each other, the
correlation function tells how correlated different parts of the system are. For an
interacting system, usually the interaction between two spins become
independent if they are infinitely separated and consequently, the correlation goes
to zero. Thus, as ⃗ − ⃗ → ∞, Γ ⃗ , ⃗ → 0. The correlations decay to zero
exponentially with the distance between the spins
Γ( ) ~
where
is an exponent and
(− / )
is the correlation length.
The spin-spin correlation function can be related to the fluctuation in
magnetization in the following way. The fluctuation in magnetization is given by
〈(
− 〈 〉) 〉 =
(
− 〈 〉)
−〈 〉 =
In continuum, the above equation can be expressed as
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Γ ⃗,⃗
〈(
− 〈 〉) 〉 =
⃗′ Γ( ⃗, ⃗′) =
⃗
⃗′′Γ( ⃗′′)
where ⃗ ′′ = ⃗ − ⃗′. The isothermal susceptibility now can be related to the
correaltion function as
=
〈(
− 〈 〉) 〉
=
⃗′′Γ( ⃗′′)
Thus, fluctuation in magnetization, isothermal susceptibility, spin-spin
correlation function all are inter-related phenomena.
2.8 Summary
As T → Tc , certain characteristic features appear in the system are called critical
phenomena. The characteristic features are: (i) the order parameter continuously
goes to zero, (ii) response functions diverge, (iii) fluctuations (in density or in
spin orientation) appear at all length scales, (iv) long range order appears in
density-density or spin-spin correlation, (v) correlation length diverges, etc.
However, it is already demonstrated in previous sections that increase in the
density or spin fluctuation, increase in the response function, correlation length,
appearance of long range order are all related phenomena. These singular
behaviour of thermodynamic quantities are essentially manifestation of
cooperative behaviour of molecular or spin-spin interaction among a large
number of molecules or spin angular momentum of atoms or ions. The
singularities associated with the thermodynamic quantities are described by
certain “critical exponents”. In the next chapter, we will define the critical
exponents for different thermodynamic quantities and will develop relations
among the critical exponents. The whole theory of critical phenomena will be
then developed in terms of critical exponents.
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Problems:
Problem 1: One mole of nitrous oxide becomes N 2 +O 2 at 25o C and one
atmospheric pressure. In this process the entropy increases by 76 Joule/K and
enthalpy decreases by 8.2× 10 4 Joule. Calculate the change in Gibb's free energy
and determine the stable phase under these conditions.
Problem 2: Show that (a ) for a fluid system, C P − CV = TVα P2 /κ T and
κ T − κ S = TVα P2 /C P where α P = (∂V/∂T ) P /V , the thermal expansion coefficient and
(b) for a magnetic system, C H − CM = Tα H2 /χ T and χ T − χ S = T α H2 /C H where
α H = (∂M/∂T ) H .
Problem 3: A system on N localized non-interacting paramagnetic ions of
spin-½ and magnetic moment µ in an external magnetic field H , is in thermal
equilibrium with a heat bath at temperature T. If the square deviation in
magnetization M is defined as 〈∆ 〉 = 〈 〉 − 〈 〉 , show that 〈∆ 〉 =
, where
is the isothermal susceptibility of the system.
−
Problem 4: For a ferromagnetic systems, described by Hamiltonian =
⃗ ∙ ⃗, show that G (T , H ) is a concave function of both T and H and F (T , M )
is a concave function of T but a convex function of M .
Problem 5: Using convexity property of free energy functions, demonstrate
geometrically discontinuous and continuous phase transitions for T < Tc and T > Tc
respectively for a magnetic system.
37
38
Chapter 3
Critical exponents and exponent
inequalities
39
3.1 Introduction
It was demonstrated in the previous chapter that different thermodynamic
quantities become singular as T → Tc . They exhibit either branch point singularity
or diverging singularity. The order parameter continuously goes to zero as T → Tc
and exhibits a branch point singularity since it becomes a double valued function.
The response functions and correlation length diverge and exhibit diverging
singularity. Long range order appears in density-density or spin-spin correlation
and it decays with power law. The singular behaviour of thermodynamic quantities
around the critical temperature can be described by power series. The leading
singularity of the power series in the limit T → Tc are characterized by certain
exponents called critical exponents. The power series for different thermodynamic
quantities and the associated critical exponents will be described below.
3.2 Critical exponents
The power series describing the thermodynamic quantities in the critical regime are
usually expressed in terms of the reduced temperature t = (T − Tc ) / Tc . In terms of
the reduced temperature, the power series of different thermodynamic quantities
and the associated critical exponents are given below.
The order parameter, density difference Δρ in fluid and spontaneous
magnetization M in ferromagnets, below Tc are given as
∆ = (− )
1 + (− )
+⋯
and
= (− )
1 + (− )
+⋯
respectively with
> 0, A and a are constants. Note that below , t is negative.
The exponent β describes the leading singularity of these quantities and is called
the critical exponent of the order parameter.
The specific heats at constant volume V or constant magnetic field H below and
above Tc are given as:
For
< ,
= (− )
>
and for
=
where
,
( )
and
[1 + ( − )
+ ⋯ ],
=
(− )
[1 + ( )
+ ⋯ ],
=
( )
[1 + ( − )
[1 + ( )
are the specific heat exponents below and above
40
+⋯ ]
+⋯ ]
respectively,
B and b are constants.
The isothermal compressibility κ T and isothermal susceptibility χT below and
above Tc are given as:
For
<
,
= (− )
and for
where
above
> ,
= ( )
[1 + ( − )
+ ⋯ ],
= (− )
[1 + ( )
+ ⋯ ],
= ( )
[1 + ( − )
[1 + ( )
+⋯ ]
+⋯ ]
and are the compressibility or susceptibility exponents below and
respectively, C and c are constants.
The critical isotherms at T = Tc are given by
−
=
−1
and
=
where D is a constant and δ is the critical isotherm exponent.
The divergence of correlation length ξ below and above Tc can be described as
(− )
( )
=
where
and
for
for
<
>
are the correlation length exponents below and above
.
The correlation functions for the fluid and magnetic systems at the critical point go
as
G (r ) =
1
r
d − 2 +η
and Γ (r ) =
1
r
d − 2 +η
at T = Tc
where d is the space dimension and η is an exponent.
It is now necessary to know how to extract the critical exponent describing the
leading singularity of a thermodynamic quantity when it is in the form of a power
series.
41
3.3 Extraction of critical exponents
Let us take a general function F (t ) as
( ) = | | 1+
+⋯
with
>0
and the function is singular at t = 0 . We are now interested in extracting the
exponent λ . Let us take the following limit
lim
→
ln| ( )|
ln
= lim
+
→ ln| |
ln| |
lim
→
ln| |
ln 1 +
+⋯
+ lim
=
→
ln| |
ln| |
Thus, the exponent λ is given by
λ = lim
t →0
ln | F (t ) |
ln | t |
(3.1)
Note that F (t ) is just not given by F (t ) = t λ .
For example, let us take a function F (t ) = At −2 /3 (t + b)2 /3 and find the exponent λ
describing the leading singularity of the function in the limit t → 0. As per
definition,
ln | F (t ) |
ln A 2
ln | t | 2
ln(t + b)
λ = lim
= lim
− lim
+ lim
t →0
t →0 ln | t |
ln | t |
3 t →0 ln | t | 3 t →0 ln | t |
2
2
2
= 0 − ×1 + × 0 = −
3
3
3
However, F (t ) = A | ln(t ) | + B and F (t ) = A − Bt 1/ 2 both have λ = 0 as per the
definition given in Eq.2.1. The first function has logarithmic singularity whereas
the second one has cusp like singularity. The above definition thus cannot
distinguish these two singularities. A modified definition can be adopted for the
critical exponent to distinguish such singularities.
If for a smallest integer j , the jth derivative of the function, F ( j ) (t ) =
as t → 0 , the exponent will be given by
λ = j + lim
t →0
ln | F ( j ) (t ) |
ln | t |
∂ jF
, diverges
∂t j
(3.2)
Using this definition, let us find the exponent λ that describes the logarithmic
singularity in F (t ) = A | ln(t ) | + B and the cusp like singularity in F (t ) = A − Bt1 / 2 .
42
1. F (t ) = A | ln(t ) | + B :
A
and lim F (1) (t ) → ∞
t →0
t
ln | F (1) (t ) |
ln A
ln t
Thus, λ = 1 + lim
= 1 + lim
− lim
= 1+ 0 −1 = 0 .
t →0
t
→
t
→
0
0
ln | t |
ln t
ln t
2. F (t ) = A − Bt1/ 2 :
1
F (1) (t ) = − Bt −1/ 2 and lim F (1) (t ) → ∞
t →0
2
(1)
ln | F (t ) |
ln( B / 2) 1
ln t
1 1
Thus, λ = 1 + lim
= 1 + lim
− lim
= 1+ 0 − = .
t →0
t →0
ln | t |
ln t
2 t →0 ln t
2 2
F (1) (t ) =
Therefore, the definition given in Eq.(3.2) can distinguish the logarithmic and cusp
like singularities. One then can calculate all the critical exponents associated with
the thermodynamic quantities defined in section 2, in the limit T → Tc following
either Eq. (3.1) or (3.2).
However, one may wonder why critical exponents are so important when they
contain less information than the complete function. Firstly, in experiments,
sufficiently close to
the behaviour of the leading term in the thermodynamic
quantities ( ) dominates and they can be described as ( )~ as → 0 .
Therefore it is easy to estimate the critical exponents but the full function may not
be. Secondly, there exist a large number of relations among the critical exponents
and they are not all independent. In fact, only two of them are independent (which
will be shown later). Thus, determining only two exponents one may obtain the
values of rest of the exponents.
System
T < Tc
α'
Fluids
0.10
CO 2
Xe
0.20
Magnets
Ni
- 0.30
EuS
- 0.15
CrBr 3
T = Tc
β
γ'
0.34
0.35
1.00
1.20
0.42
0.33
0.368
1.35
T > Tc
γ
ν'
δ
α
0.10
0.57
4.20
4.40
1.35
1.30
4.22
0.00
0.05
1.35
4.30
ν
1.215
Table 3.1: List of critical exponents for different fluid and magnetic systems. Data are taken
from Introduction to Phase transitions and Critical Phenomena, H. E. Stanley (Oxford University
Press, New York).
43
3.4 Values of critical exponents and their characteristics
A close look into the numerical values of critical exponents will be made here.
Critical exponents considered here are:
α ', β , γ ',ν ' for T < Tc , δ for T = Tc and
α , γ ,ν for T > Tc .
The numerical values of the critical exponents obtained experimentally are listed in
Table.3.1.
There are a number of observations can be made from the data given in the above
table. The first observation is that the values of the critical exponents for T < Tc and
those for T > Tc are almost the same. Thus the leading singularity of
thermodynamic quantities below and above Tc can be described by the same
critical exponents. The leading forms of different thermodynamic quantities are
listed below.
Fluid system
Order parameter:
Magnetic system
Density difference:
Spontaneous
magnetization:
Δρ : (Tc − T ) β
M : (Tc − T ) β
Critical isotherm ( T = Tc ):
Δρ : ( P − Pc )1/δ
M : H 1/δ
Response functions:
Compressibility:
Susceptibility:
κ T : | T − Tc |−γ
χT : | T − Tc |−γ
Specific heat:
Specific heat:
CV : | T − Tc |−α
CH : | T − Tc |−α
ξ : | T − Tc |−ν
ξ : | T − Tc |−ν
Correlation length:
Correlation function ( T = Tc ):
( ⃗)~
(
)
Γ( ⃗)~
(
)
The assumption that the critical exponent associated with a given thermodynamic
quantity is the same as T → Tc from above or below does not have any proper
justification at this stage but with the help of renormalization group it would be
proved later that they are same.
Secondly, it is remarkable to note that the transitions as different as ``liquid to gas''
and ``ferromagnet to paramagnet'' are described by almost same set of critical
exponents. The critical exponents are then somewhat universal. But the transition
44
temperature Tc is not universal and varies from material to material. Tc depends
on the details of interatomic interaction, lattice structure, etc. On the other hand,
critical exponents depend only on a few fundamental parameters such as spatial
dimensionality, spin dimensionality, symmetry of the ordered state, and presence
of symmetry breaking fields. Critical exponents do not depend on the lattice
structure or the type of interaction.
The third observation is that the exponents listed in Table 3.1 satisfy certain
relations among themselves. For example: the Rushbrooke inequality
α + 2 β + γ ≥ 2 , the Griffiths inequality γ ≥ β (δ − 1) , the Fisher inequality (2 − η )ν ≥ γ ,
the Josephson inequality dν ≥ 2 − α , etc. These exponent inequalities can be
derived from thermodynamic consideration. In the following we will be deriving
the Rushbrooke inequality α + 2 β + γ ≥ 2 , the easiest one.
As t → 0 from below the critical temperature, the scaling form of thermodynamic
quantities are given by M : ( −t ) β , CH : (−t ) −α and χT : ( −t ) −γ for H = 0 . From
∂M 
thermodynamic consideration, we know that χT (CH − CM ) = T α H2 , where α H = 

 ∂T  H
. Since the response functions ( χT , CH , CM ) are positive, CH must be given by
2
T  ∂M 
CH ≥


χT  ∂T  H
∂M 
β −1
As 
 : (−t ) , one has
 ∂T  H
(−t )−α ≥ (−t )γ + 2( β −1)
which implies −α ≤ γ + 2( β − 1) or α + 2 β + γ ≥ 2 .
The Rushbrooke inequality is thus due to positivity of the response function.
Similarly, the Griffiths inequality can be obtained from the convexity of free
energy, the Fisher inequality is due to the positivity of the correlation function for
any temperature and non-negative field, the Josephson inequality is obtained from
hyper-scaling.
However, from static scaling hypothesis one can show that these inequalities hold
as exact equalities. Thus, not all critical exponents are independent. It would be
shown later that only two of them are independent. It is therefore intriguing to
search for a theory which could explain why the critical exponents hang together.
45
Example: Consider an equation of state
H = aM (t + bM 2 );
for
a, b > 0
near the critical point t = 0 . Find the exponents β , γ , δ and verify the scaling
relation among them.
At t = 0 , H : M 3 , thus δ = 3 . For spontaneous magnetization, H = 0 and
M 2 : (−t ) or M : (−t )1/ 2 , thus β = 1/ 2 . For the susceptibility exponent γ , we define
1 H
=
= at + abM 2
χ M
Since, M 2 : (−t ) ,
1
= a (1 − b)t , and thus, γ = 1 . Therefore, γ = β (δ − 1) , the
χ
inequality holds as equality.
3.5 How to study critical phenomena?
One needs to develop a statistical mechanical theory for these systems to calculate
thermodynamic quantities taking temperature T as a parameter. Tuning T to Tc ,
the theory should be able to demonstrate the singular behaviour of all these
thermodynamic quantities. One then should be able to calculate the critical
exponents and verify the scaling relations among them. Finally, the universality
class should be classified.
However, the prescription provided in the statistical mechanics of non-interacting
systems for calculation of thermodynamic quantities from the free energy function
F = −k BT ln Z , Z being the canonical partition function, is no good. Consider the
following example:
A paramagnetic solid contains a large number N of non-interacting spin-1/2
particles, each of magnetic moment µ on fixed lattice sites. The solid is placed in
a uniform magnetic field H and is in thermal equilibrium at temperature T . Find
the magnetization M and susceptibility χ of the solid in the given field as a
function of temperature T .
To solve the problem, let us calculate the single particle partition function of the
system. Since these are spin-1/2 particles, each particle has two energy states in an
external field, spin magnetic moment aligned either along the applied field or
opposite to applied field. The energy values are given by
46
E = −µ ⋅ H = ± µ H .
The single partition function is then given by
Z1 = e− x + e x = 2 cosh x where x =
µH
.
k BT
Since the particles are localized and non-interacting, the total partition function Z
of the system is given by Z = Z1N = (2 cosh x) N . The Helmholtz free energy is then
F = − kBT ln Z = − Nk BT ln(2cosh x) .
∂F


The magnetization M can be obtained as M = −   = N µ tanh x and the
 ∂H T
susceptibility χT =
Nµ2
1
2
. In Fig.3.1 and Fig.3.2, M / N µ and χT ( k BT N µ )
k BT cosh 2 x
are plotted against x = µ H k BT respectively.
(
Fig.3.1 Plot of M / N µ against x = µ H kBT .
2
Fig.3.2 Plot of χT × k BT N µ
)
against x .
One may note that for any finite temperature,
is zero if
is zero. Thus, there
is no spontaneous magnetization. Since there is no interaction among the spin
magnetic moments, due to temperature the magnetic moments are randomly
oriented and averaged out to zero. This is expected in a paramagnetic solid. The
zero field susceptibility at any finite temperature is also a finite quantity. The
response function is then not diverging at any finite temperature. Hence, there is no
47
phase transition in this problem. The reason is obvious. The origin of the second
order phase transition, the spin-spin interaction, is not taken care in this problem.
Therefore, one needs to develop a theory incorporating the spin-spin (or molecular)
interaction present in the system. However, as soon as the interaction is switched
on, the partition function cannot be given by the multiple of single particle partition
functions, i.e =
. One needs to integrate the full 6 dimensional phase
space in order to calculate the partition function. The difficulty is therefore two
fold, complexity in the interaction and largeness of the system. To handle the
situation one needs to develop suitable models incorporating tractable interaction.
Ultimately, one expects that the models should be solved exactly and through
understanding of critical phenomena would be possible.
In the next chapter we will be developing several such models and determine their
ground states.
48
Problems:
12
14
Problem.1 Determine the critical exponents λ for the functions (i) F (t ) = at + bt + ct ,
−2 3
23
2 −t
(ii) F (t ) = at (t + b) , (iii) F (t ) = at e , (iv) F (t ) = at ln | t | +b as t → 0 , where a, b, c
are constants. [Answers: (i) 1/4, (ii) -2/3, (iii) 2, (iv) 1]
49
50
Chapter 4
Models and Universality
51
4.1 Introduction
In this chapter some of the fundamental models developed for studying
interacting systems will be described. These models are spin-1/2 Ising model,
spin-1 Ising model, q-state Potts model, XY-model, Heisenberg model and nvector model. In these models of interacting systems, the details of all possible
complicated many body interactions are not taken into account. Rather,
interactions are included in a simplest possible way such that the models could
be solved exactly either analytically or numerically and the essential physics of
an interacting system can be understood. We will be using the magnetic
language and write down the Hamiltonian in terms of spin variables. However,
they will be applicable to non-magnetic systems also. The models will be
described on one, two or three dimensional regular lattices. The spin variables
will be assigned to the lattice sites of a given lattice.
4.2 Spin-1/2 Ising Model
A classical spin variable si , which takes +1 or − 1 values corresponding to the
states up or down, placed on each lattice site. Usually, the interaction among
the spins is limited (however, not restricted) to the nearest neighbor spins only.
The interaction energy or the exchange energy among two spins is given by J .
The Hamiltonian for such an interacting system is given by
H = − J ∑ si s j − H ∑ si
ij
i
where H is external magnetic field in units of energy and ij represents the
nearest neighbor interaction. The first term in the Hamiltonian is responsible
for the cooperative behavior. For = 0, the Hamiltonian corresponds to a
paramagnetic system.
Ground state configuration: First we set the external field
Hamiltonian is given by
= 0 and then the
H = − J ∑ si s j
ij
Consider two spins only along a one dimensional chain. Since the spins have
two states each, there are total 2 = 4 configurations possible.
52
There are two parallel configurations, both up spins and both down spins and
two anti-parallel configurations, one up and another down. The Hamiltonian for
the parallel and anti-parallel configurations are then given by
HP = − J and HA = J
The partition function and the free energy for N parallel and anti-parallel
configurations can be calculated as
Z P = ( e+2 J β ) N and FP = − NkBT ln Z = − NJ
Z A = ( e−2 J β ) and FA = − Nk BT ln Z = + NJ
N
Since the free energy corresponding to parallel configuration is lowest, the
ground state configuration of spin-1/2 Ising model will be either all spins up or
all spins down as shown below.
We now qualitatively discuss the possibility of phase transition in one and two
dimension using the spin-1/2 Ising Hamiltonian.
One dimensional Ising Model: Consider a chain of
spins all pointing up.
Now say one domain wall is introduced as shown below.
The change in interaction energy is ∆ = 2 . On the other hand, the domain
wall can be placed in
different ways (or places), the change in entropy is
given by ∆ =
ln =
ln . Therefore, the change in free energy is given
by
∆ =∆ − ∆ =2 −
53
ln
Since
is large, for ≠ 0, the second term in the free energy will dominate
which corresponds to the presence of domain wall. Since ∆ < 0, the
fluctuation in spin orientation will be cost free. No long range order in the spin
orientation will appear and thus there will be no spontaneous magnetization. On
the other hand, for = 0, the first term in the free energy will survive and the
ground state configuration is either all spins up or all spins down. A long range
order state is then possible only at = 0. Therefore, in one dimensional spin1/2 Ising model on phase transition will occur at any finite temperature except at
= 0.
Two dimensional Ising Model: Consider the following spin configuration on a
two dimensional square lattice. There are two up spin domains within a large up
spin domain. It can be checked that the total number of anti-parallel spins
adjacent to a domain is equal to the length of the domain wall in units of lattice
spacing. For the configuration given, the number of anti-parallel spins is 16 and
the length of the domain wall for both the domains is 16. Since each antiparallel spins costs 2 amount of energy, the internal energy change for a
domain wall of length would be ∆ = 2 .
ln where
is the number of all
The maximum entropy is given by =
possible domain configurations for the same number of anti-parallel spins. The
domain walls can be constructed by making a walk along the boundary. For
each step of walk there are three possibilities and thus = 3 . Therefore the
change in entropy would be ∆ =
ln 3.
54
The change in free energy is then given by
∆ = ∆ − ∆S = 2
−
ln 3
Now, there may exist a critical temperature
above which the second term in
the free energy will dominate. The second term is the entropy term and hence
there will be large number of domain walls. No long range order can be
established and no spontaneous magnetization can exists. On the other hand,
Below , the term coming from the interaction of the spins will dominate.
There will be less number of domain walls and long range order would be
possible. Also, spontaneous magnetization can exist. Therefore phase transition
is possible to explain with spin-1/2 Ising Hamiltonian at any finite temperature
in the space dimension two or above.
The Ising model can be solved exactly on one dimension and will be
demonstrated as example problem in one of the chapters. However, the phase
transition is only at = 0. Determination of exact partition function of 2 Ising
model even in absence of external field is a mathematically difficult task and it
is solved by Onsager in 1944. We will be describing some of the results
obtained in two dimensions in chapter 7 after discussing the transfer matrix
method. The 2 ising model in presence of external field and 3 Ising model
even in absence of external field remain unsolved. However, the properties of
these models are determined numerically.
The ising model is applied widely for many interacting two states systems. For
example it can be applied to study adsorption of hydrogen on the ( 110) plane of
iron. Each adsorption site has two states either occupied or vacant.
4.3 Application of spin-1/2 Ising model in other systems
Order-disorder transition of Beta brass: Beta-brass is a binary alloy consists
of equal number of Cu and Zn atoms. Each sub-lattice has simple cubic
structure and they inter-penetrate into each other and form a body centre cubic
structure. At room temperature, neither of the sub-lattices contains atoms of
other type and it is called a ordered state. However, above a critical temperature
= 733 both the sub-lattices are occupied by both the atoms and it is called
a disordered state. The order parameter of the transition can be defined as the
55
difference between the concentration of Cu and Zn atoms on a chosen sublattice. For the Cu sub-lattice, it can be defined as
Δc =
cCu − cZn
cCu
At room temperature, ∆ is equal to one whereas at above
it is going to be
from a order to a
zero. Thus there is a continuous phase transition at =
disordered state. In order to study such a phase transition, we introduce a two
states variable = ±1, = 1 if a site is occupied with Cu and = −1 if a
site is occupied with Zn. The phase transition can be studied constructing a
Hamiltonian including all the interactions among Cu and Zn atoms in the Beta
brass. There could be three different types of interaction in this system, Cu-Cu,
Zn-Zn and Cu-Zn (or Zn-Cu) and the corresponding interaction energies can be
,
, and
.
taken as
The model can also be applied to study the liquid-gas transition also. One of the
spin states can represent the liquid state and the other can represent the gas state.
It can be seen later that the critical exponents of 3 Ising model are the same as
that of liquid-gas transition.
4.4 Spin-1 Ising model
Spin variable si takes 0 and ±1 values. The general Hamiltonian can be written
as
H = − Jαβ ∑ siα s βj
with
α , β = 0,1, 2
ij
Taking all possible terms, the Hamiltonian takes the form
H = − J ∑ si s j − K ∑ si2 s 2j − D ∑ si2 − L ∑ ( si2 s j + si s 2j ) − H ∑ si
ij
ij
i
ij
i
with no cubic term because
= . Spin-1 Ising model exhibits a complex
critical behavior because of its enlarged parameter space.
56
Ground state configuration: Consider a 1 chain of
spin-1 Ising spins,
= 0, ±1. For simplicity put = 0 and = 0. The Hamiltonian is then given
by
H = − J ∑ si s j − K ∑ si2 s 2j − D∑ si2
ij
ij
i
The ground state is expected to be one in which all spins are at the same state,
i.e., ( 0,0) ; ( 1,1) ; (−1, −1). Let us calculate the energy for these configurations.
= 0,
= −2
−2
,
−
= −2
−2
−
The states ( 1,1) & (−1, −1) are then degenerate.
Spin-1 Ising model is applied to analyze superfluid transition in 3He and 4He
mixture, condensation and solidification of fluid.
4.5 q-state Potts model
In the Potts model, a q-state spin variable = 1,2,3, ⋯ , is placed at each
lattice site. The Hamiltonian for the spin interaction is given by
H = − J ∑ δσ iσ j
ij
where is Kronecker delta. So the energy of two neighboring spins is – if they
are in the same state and zero otherwise. The ground state is q-fold degenerate.
In two dimensions, the Potts model describes a continuous phase transition to a
paramagnetic phase for ≤ 4 whereas the transition is first order for > 4.
It can be shown that the q=2 Potts model is equivalent to spin-1/2 Ising model
by replacing the Kronecker delta function in the Potts Hamiltonian in terms of
spin-1/2 Ising variable as
1
1
δσ iσ j = (1 + si )(1 + s j ) + (1 − si )(1 − s j )
4
4
1 1
= + si s j
2 2
where σ i , σ j = 1, 2 and si , s j = ±1 . The Potts Hamiltonian then can be written as
57
H = − J ∑ δσ iσ j = −
ij
J
1
si s j − zJ
∑
2 ij
2
where is the coordination number of the lattice. Therefore, the q=2 Potts
model is essentially an Ising model with interaction strength ′ = / 2 with a
shift in ground state energy by / 2.
However, the q=3 Potts model is not equivalent to spin-1 Ising model. The Potts
model has a three-fold degenerate ground state whereas spin-1 Ising model with
the Hamiltonian
H = − J ∑ si s j − K ∑ si2 s 2j − D ∑ si2
ij
ij
i
there are two ground states and one doubly degenerate. However, in one
dimension if 2( + ) + = 0 for spin-1 Ising model, the two models become
identical.
Absorption of krypton on the basal plane of graphite can be studied with q=3
Potts model.
4.6 XY model
The Ising model has very restricted application to magnetic systems. The Ising
spins have only two states, either parallel to the applied field or anti-parallel to
it. No other orientation than up or down is possible and thus the spin orientation
are highly anisotropic in spin space. MnF2 is a magnetic system where such
model could be applied. However, there are many physical systems where spin
orientation away from the quantization axis occur. One of the modified models
is XY model where each spin is a two dimensional unit vector ⃗. The interaction
Hamiltonian is given by
ℋ=−
+
−
⃗∙ ⃗
〈 〉
where , are the labels of Cartesian axes in spin space. It has a conventional
phase transition at a finite temperature for > 2 whereas for = 2, there is a
transition at finite temperatures to an unusual ordered phase with quasi long
range order known as the Kosterlitz–Thouless transition.
58
4.7 Heisenberg model
In the case of classical Heisenberg model, the spin variables are the isotropically
interacting three dimensional unit vectors. The Hamiltonian is given by
ℋ=−
⃗ ∙⃗ −
⃗∙⃗
〈 〉
where ⃗ is the external field. The quantum mechanical Heisenberg model can
be written as
H = − J ∑ (σ ixσ xj +σ iyσ jy + σ izσ jz ) − H ∑ σ iz
ij
i
where σjs are quantum operators, the Pauli spin matrices.
The classical Heisenberg model can be considered as → ∞ limit of the
quantum Heisenberg model. In the classical limit, the Heisenberg spin can take
on an entire continuum orientation instead of a finite number (2 + 1) of
discrete orientations. The magnitude of the spin
( + 1) has to be
normalized 1. Though the classical approximation is unrealistic at low
temperatures, it is extremely realistic near the critical temperature . The
critical exponents are found to be independent or very weakly dependent on the
spin quantum number and thus the spin dependence can be neglected.
The quantum models can be mapped on to classical model models in one higher
dimension. There are exact results for 1 quantum systems whereas classical
models are solved exactly in two dimensions. Moreover, the Heisenberg model
exhibits continuous phase transition at a finite temperature for > 2 whereas
the same occurs in Ising model for > 1.
The Heisenberg model provides a reasonable description of the properties some
magnetic materials, such as EuS, and able to describe ferromagnetism.
4.8 Models and universality
The values of the critical exponents are estimated studying the above models at
their respective critical points using several analytical and numerical methods.
Some of the techniques will be described in later chapters. A brief description of
the methods and results obtained is described here. Mean field theories are often
applied in which the molecular interaction (or spin-spin interaction) in the
59
Hamiltonian is replaced by an average (or mean) field. The ``interacting system''
is then approximated as a ``non-interacting system'' in a self-consistent external
field. The approximation enables one to obtain the partition function and hence
the free energy exactly. The mean field exponents however do not agree with
those of the real magnets or fluids (see table 1). This is because of the fact that
fluctuations are ignored in mean field approximation. The theory is expected to
be consistent and the exponents would be correct at space dimensionality d ≥ dc
where d c is called the upper critical dimension. It depends on the model, for
Ising model and φ 4 Landau-Ginzburg model d c = 4 . Moreover, mean field
wrongly predicts phase transition in d = 1 at finite temperature.
Since evaluation of partition function for an interacting system is difficult, the
exact solution of these models near the critical point is rarely available.
Moreover, the partition function as well as the free energy function becomes
singular at T = Tc . As it is already mentioned, the spin- 1/2 Ising model is exactly
solved in d = 1 and in d = 2 with zero fields. In d = 3 , exact solution of Ising
model is not obtained. On the other hand, exact solution of quantum models
exits only in d = 1 . Transfer matrix technique is often useful to obtain exact
solutions. In this technique, the partition function is obtained as a matrix and the
free energy is obtained as a function of the largest eigenvalue.
Renormalization group (RG) technique provides a direct method of calculating
the properties of a system arbitrarily close to the critical point. RG is able to
explain the universality, explains why the free energy function is a generalized
homogeneous function and estimates the values of the critical exponents. The
essence of RG is to break a large system down to a sequence of smaller systems.
Instead of keeping track of all the spins (or particles) in a region of size ξ , the
correlation length, the long range properties are obtained from short range
fluctuations in a recursive manner. Each time after eliminating the short range
fluctuations, the system is restored to its original scale. Most often the recursion
relations of the system parameters are found to be complicated and estimation
of critical exponents becomes difficult.
Apart from these analytical techniques, two numerical techniques, series
expansion and Monte Carlo, are also used rigorously. In series expansion, the
partition function is obtained exactly upto certain terms. Determining the radius
of convergence and the singularity, the values of the critical exponents are
determined. In Monte Carlo technique, statistical averages of thermodynamic
quantities are obtained by generating a large number of equilibrium
configurations with right Boltzmann weight following Metropolis algorithm.
60
In Table 4.1, we summarize the values of the critical exponents obtain in
different models applying some of the techniques mentioned above.
Models
2 − d Ising
3 − d Ising
2 − d Potts,
q=3
3 − d X-Y
3− d
Heisenberg
Mean field
Symmetry
of α
order parameter
2 -component
0
(log)
scalar
2 -component
0.10
scalar
1/3
q -component
scalar
2 − d vector
0.01
3 − d vector
− 0.12
0
(dis)
β
γ
δ
ν
η
1/8
7/4
15
1
1/4
0.33
1.24
4.8
0.63
0.04
1/9
13/9
14
5/6
4/15
0.34
0.36
1.30
1.39
4.8
4.8
0.66
0.71
0.04
0.04
1/2
1
3
1/2
0
Table 4.1: List of critical exponents obtained from different analytical models. Data have
been taken from reference [7].
It can be seen that the values of the critical exponents depend on the space
dimension as well as on the symmetry of the order parameter. A class of
systems found to have the same values of the critical exponents if the number of
components of order parameter and dimension d of the embedding space are
the same. Hence, these critical exponents define a universality class. In other
words, those systems which have the same set of critical exponents are said to
belong to the same universality class. For example, nearest neighbor Ising
ferromagnets on square and triangular lattices have identical critical exponents
and belong to the same universality class. Critical exponents of d = 3 Ising
model are similar to those of fluid systems because of the same symmetry of the
order parameter.
61
Problems:
Problem. 1: If the first neighbor interaction energy is
and the second
neighbor interaction is , find the ground state configuration of spin-1/2 Ising
model with first and second neighbor interaction on 2 square lattice.
Problem. 2: In beta brass, the interactions between different atoms can be taken
as
,
and
for copper-copper, zinc-zinc and copper-zinc (or zinccopper) interactions respectively. If = +1 represents occupation of site by
Cu and = −1 represents occupation of site by Zn, the Hamiltonian can be
written as
H = − J ∑ si s j − H ∑ si + C.
ij
=− (
Show that
= (
+
+
+2
i
−2
),
=− (
−
) and
). If there are equal n umber of Cu and Zn atoms,
show that the above Hamiltonian reduces to zero field spin-1/2 Ising
Hamiltonian with a constant shift in ground state energy by .
Problem. 3: Find the ground state energies of spin-1 Ising model described by
the Hamiltonian
H = − J ∑ si s j − K ∑ si2 s 2j − D ∑ si2 ,
ij
ij
i
si = 0, ±1
on 2 square lattice. Find the condition for which they will be equal to the
ground state energies of = 3 Potts model.
62
Chapter 5
Mean field theory
63
5.1 Introduction
It is now time to develop a quantitative theory of phase transition which will be
able to determine the critical temperature and provide the estimates of the critical
exponents. Knowing the values of the critical exponents one then can verify the
scaling relations among them. However, before trying the exact solutions of the
models discussed in the previous chapter, one could try some approximate
analytic method or some phenomenological theory. The simplest one is the mean
field theory.
Mean field approximation for the interaction can be made in two ways. In the first
approach or classical approach, an interacting system is approximated by a
non-interacting system in a self-consistent external field. In the second approach,
an approximate free energy is expressed in terms of an unknown parameter and
the free energy is then minimized with respect to that parameter.
The method is applicable to interacting many particle systems such as fluids,
magnets, binary alloys, superfluid, superconductors, etc. In the following we will
be applying the mean field method to the magnetic system as well as to the fluid
system. First we will discuss the mean field theory for fluid system and then for
magnetic system.
5.2 Mean field theory for fluids
In this section, mean field equation of state for fluids will be derived. The critical
point for the mean field equation will be identified and the mean field values of
the critical exponents will be estimated. It would be interesting to notice that the
mean field equation of state will be nothing but the van der Waals equation of
state for non-ideal gases.
The intermolecular interaction in a fluid system is often represent by
Lennard-Jones potential
  r0 12  r0  6 
E ( r ) = 4ε    −   
 r  
  r 
where is the interatomic separation, is the depth of the potential,
is a
finite distance at which the potential is zero. The potential is minimum at
64
=2 /
and the value of the potential at
is – . As per mean field
approximation, an atom moves in an average field due to all other atoms or
molecules. We will be following the mean field theory developed by Reif. In
Reif’s mean field theory, it is assumed that the particles move independently in an
constant effective potential of the form
E (r ) =
r<r
{ ε∞ otherwise
0
The Hamiltonian of the system is then given by ℋ =
energy is given by
=−
ln
where
+ ( ) and the free
is the canonical partition function.
Let us consider the fluid system is consisting of
particles (atoms or molecules)
and enclosed in a volume . The system is in thermal equilibrium at temperature
. The canonical partition function of the system is given by
N
+∞

 p2
 
1  3
1  − β p2 / 2 m 
3
+ E (r)   = 3N  ∫ e
Z = 3 N  ∫ d p ∫ d r exp  − β 
dp 
h 
h  −∞
 2m
  


 2 mπ k BT 
=

2
 h

 2 mπ k BT 
=

2
 h

3N / 2
3N
 e− β E ( r ) d 3r 
∫

N
N
3N / 2
 r0 − β E ( r ) 3 ∞ − β E ( r ) 3 
 2 mπ k BT 
−∞
− βε N


e
d
r
e
d
r
V
e
(
V
V
)
e
+
=
+
−
∫

ex
ex


2
∫r

h


 0

0

3N / 2
 (V − Vex ) e − βε 
N
where
represents the excluded volume by the hard core of the molecules. The
pressure in the fluid is given by
∂
 ∂F 
 ∂
P = −
= k BT 
( ln Z ) = NkBT {ln(V − Vex ) − βε } .

∂V
 ∂V T
 ∂V
T
is expected to be proportional to the number of
The excluded volume
particles (atoms or molecules) . Similarly, the constant effective energy ̅
should be proportional to the density ⁄ of the particles in the system.
and
̅ are then given by
b
Vex =
N = bn
NA
and
a N
ε =− 2
NA  V
65
an 2

=−
NV

where ⁄
is the
and ⁄
are the respective proportionality constants,
= ⁄ . On substitution of
and ̅ in the
Avogadro number and
expression for and taking the volume derivative, one obtains
 1
β an 2  nRT
an 2
P = Nk BT 
,
−
=
−
2 
2
V
bn
NV
V
bn
V
−
−


which is precisely the van der Waals equation of state.
5.2.1 Determination of critical point and the critical exponents
Critical point: The critical point can be identified considering the fact that both
and
=
,
=
vanishes for the
,
=
−
isotherm at the critical point,
.
Consider the mean field (or the van der Waals) equation of state for one mole
( = 1) fluid
P=
RT
a
− 2.
V −b V
To find ( , , ), one has to solve the following three equations:

RTc
RT
2a 
2a
 ∂P 
=
−
+
=−
+ 3 = 0,

 ∂V 
2
2
3 

Tc ,Vc  (V − b ) V 
(Vc − b ) Vc
Tc ,Vc
 2 RT
 ∂2P 
6a 
=
−
=


3
2 
4 
 ∂V Tc ,Vc  (V − b ) V 
Tc ,Vc
and Pc =
2 RTc
(Vc − b )
3
−
6a
= 0,
Vc4
RTc
a
− 2
Vc − b Vc
Solving the above three equations, one finds the critical point
Tc =
8a
a
, Pc =
and Vc = 3b.
27bR
27b 2
66
Critical exponents: In order to obtain the critical exponents, we will write the
equation of state in terms of reduced variables ( , , ) :
=
=
=
−
−
−
=
− 1,
where
=
=
− 1,
where
=
=
− 1,
where
=
Multiplying both sides of the van der Waals equation by 27 /
+
3
one obtains
3 −1 =8
and in terms of reduced variables
(1 + p ) + 3 (1 + v ) −2  3 (1 + v ) − 1 = 8(1 + t ),



Multiplying both sides by ( 1 + ) one has
3 
 7
2 p  1 + v + 4v 2 + v 3  = −3v 3 + 8t (1 + 2v + v 2 )
2 
 2
Exponent : To obtain critical isotherm exponent
The equation of state at = 0 becomes
one needs to set
= 0.
−1
3  7
3 
3  7

p = − v 3 1 + v + 4v 2 + v 3  = − v 3  1 − v + L 
2  2
2 
2  2

Since the leading term on the right is cubic and
~
, then
= 3.
diverges as
Exponent
: As → 0 , the isothermal compressibility
~| | along the critical isochore = 0. The isothermal compressibility is
given by
κT = −
Vc  ∂v 
1  ∂V 

 =−
 
V  ∂P T
VPc  ∂p T
67
∂p
 
−
+
Since   = −
, along the critical isochore
2
2
 ∂v T
[3(1 + v) − 1] [3(1 + v) − 1] (1 + v)3
24
24t
6
 ∂p 
  = −6t and the isothermal compressibility is give by
 ∂v T
κT =
Therefore
1 −1
t
6 Pc
= 1.
Exponent : The order parameter exponent
equation of co-existence curve at = 0. Putting
one has
can be obtained from the
= 0 in the equation of state,
1 + 3 (1 + v )−2  3 (1 + v ) − 1 = 8(1 + t ),



or (1 + 3(1 − 2v ) ) (2 + v ) = 8(1 + t ),
or 2(2 − 3v)(2 + 3v) = 8(1 + t ),
8
or v 2 = − t ,
9
or v ≈ (−t )1/ 2
Thus,
= 1/ 2.
Exponent
given by
: The specific heat C has a finite discontinuity at
CV = 6 Nk B
=
Since
~| |
, then
for T < Tc
3
Nk B for T > Tc
2
= 0.
68
=
and it is
The values of the exponents obtained for a fluid system are listed below.
α
0
β
1/2
γ
1
δ
3
The values of most of the exponents are found different from those obtained
experimentally (listed in chapter 3) for fluids near the critical point. It should be
emphasized here that in the mean field approximation it is assumed each particle
moves in a mean field due to all other particles. On the other hand, it is generally
believed that the inter particle interactions are of extremely long range. Thus,
mean field approximation is unrealistic for real fluids.
It can be checked that the critical exponents satisfy the scaling relations:
α + 2 β + γ = 2, γ = β (δ − 1), etc. as exact equality and not as inequalities obtained
by thermodynamic considerations.
5.3 Mean field theory for magnetic systems
We will now consider the interacting Hamiltonian of spin-1/2 Ising Model. Note
that the non-interacting Ising spins = ±1 in an external field
described by
the Hamiltonian ℋ = − ∑
does not show any phase transition.
Spontaneous magnetization goes to zero even at
= 0 . There are two
approaches: the classical approach by Weiss and the minimization of free energy
by Bragg-Williams. We will develop the mean field equation of state applying
both the approaches.
5.3.1 Classical approach:
The spin-1/2 Ising Hamiltonian in presence of an external magnetic field
given by
H = − J ∑si s j − h∑si ,
〈 ij 〉
is
si = ±1
i
where ℎ =
and
is the magnetic moment of each spin. In the classical
mean field approximation invented by P. E. Weiss, the spin-spin interaction is
replaced by an average magnetic field in which the spin is sitting and the field
should be proportional to magnetization. The self-consistent field is then given by
z
Bm = J ∑ si = Jzm
i =1
where
is the coordination number (the number of nearest neighbors) of the
69
lattice and
is the magnetization per spin. Hence, the mean field Hamiltonian
for single spin can be written as
z
H i = − si ( J ∑ s j + h) = −( Jzm + h) si
j =1
Note that all details of the lattice structure is now lost. Only , the coordination
number, can not describe a lattice. For example, both for 2 triangular lattice
and 3 simple cubic lattice the coordination number is = 6. If the system is
in thermal equilibrium with a heat bath at temperature , the single particle
canonical partition function is given by
Z1 = eβ ( Jzm+h ) + e− β ( Jzm+h ) = 2cosh β ( Jzm + h)
where
= 1/
. Hence, the free energy density is
f = −k BT ln Z1 = −k BT [ln 2 + ln{cosh β ( Jzm + h)}] .
The per spin magnetization m then can be calculated as
m=−
∂f
Jzm + h
= tanh
kBT
∂h
This is the self-consistent mean field equation of state.
5.3.2 Minimization of free energy:
Consider a system described by the Hamiltonian ℋ . The corresponding free
ℋ
energy of the system is given by = −
where = ∑{ ]
is the
canonical partition function of the system.
is the true free energy of the
system. Let us take a trial Hamiltonian ℋ whose corresponding free energy is
ℋ
=−
where
= ∑{ ]
. Now say,
H = H ′ +H 0
and the partition function is given by
Z = ∑ e − βH = ∑ e− βH ′e − βH 0
{si }
{si }
70
Therefore one has,
Z
=
Z0
∑e
{si }
− βH
∑e
′e − β H 0
= e− β H ′
− βH 0
0
{si }
where 〈⋯ 〉 denotes an average taken in the ensemble defined by ℋ . Due to
convexity property,
e− βH ′
and it follows that
0
≥e
−β H ′
0
Z
−β H ′ 0
≥e
or ln Z − ln Z0 ≥ −β H ′
Z0
0
then one has
F ≤ F0 + H − H 0
0
This is called Bogoliubov inequality. If the trial Hamiltonian is given by
H 0 = −λ ∑ si
i
where the unknown parameter is the effective field, the mean field free energy
can be obtained minimizing
+ 〈ℋ − ℋ 〉 by varying . The single particle
partition function
and free energy
corresponding to ℋ is given by
z0 = e+ βλ + e− βλ = 2cosh βλ, F0 = − NkBT ln z0 = − NkBT [ln 2 + ln cosh βλ ].
and then
=−
=
ℎ . The second term on the right hand side of
Bogoliubov inequality can be calculated as
H −H0
0

 βλ ∑i si
J
s
s
(
λ
h
)
s
−
+
−

∑s  ∑ij i j
∑i i e
{} 

=
βλ ∑ si
∑e i
{ s}
= − J ∑ si
ij
0
sj
0
+ (λ − h)∑ si
i
0
where 〈 〉 = 〈 〉 = . Thus, for system of
coordination number , 〈ℋ − ℋ 〉 is given by
71
spins on a lattice of
H −H 0
0
1
= − NJzm 2 + (λ − h) Nm
2
where the factor of 1/2 is to take care of double counting of pairs over the whole
lattice. Then by Bogoliubov inequality one has
1
F ≤ F0 − NJzm 2 + (λ − h) Nm
2
Taking derivative of the right hand side with respect to
could determine as,
and setting it to zero we
∂ (RHS) ∂F0
∂m
∂m
=
− NJzm
+ (λ − h ) N
+ Nm
∂λ
∂λ
∂λ
∂λ
∂m
∂F0
,
= − N ( Jzm + h − λ )
= − Nm.
∂λ
∂λ
The derivative to become zero,
+ ℎ − should be zero. Therefore,
=
+ ℎ and hence the magnetization is given by
m = tanh βλ = tanh β ( Jzm + h)
the same mean field equation as obtained by Weiss. Note that, tanh is due to Ising
spins (two states problem) and nothing to do with mean-field approximation.
If ℎ = 0,
=
is the mean field. The mean field free energy is then given by
1
F = − NkBT ln(2cosh β Jzm) + NJzm2 .
2
5.3.3 Solution of mean field equation of state:
= tanh (
+ ℎ) can be solved
The self-consistent MF equation
graphically as shown in Fig.5.1. Since we are interested in zero field
magnetization, let us put ℎ = 0. The MF equation then can be recast as =
. The solutions are the intersections of these two curves.
and = tanh
There are two regimes, >
and < .
= 0. It corresponds to MF free
For > , there is only one solution at
energy = 0. For < , the curves intersect at three different values, 0,± .
72
It can be checked that ( 0) > (± ). Thus, ± are the acceptable solutions
and they correspond to ferromagnetism. Therefore, MF theory, approximating an
interacting system as a non-interacting system in a self-consistent field,
demonstrates ferromagnetism.
Figure 5.1 Plot of
At
=
= tanh
and
=
. Intersection points provide the solution.
, the slopes of the two curves matches at
sech
=0. Thus, one has
= 1.
The critical temperature is then given by Tc = Jz/k B . The positive solutions are
plotted against /
in Fig.5.2. it can be seen that
continuously goes to zero
at = .
depends only on the coordination number
of the lattice.
Note that,
However, alone can not define a lattice uniquely. For example, triangular
lattice in 2 dimensions and simple cubic lattice in 3 dimensions both have
coordination number = 6. According to MF theory, these two lattices would
have should have the same critical temperature which is contradiction to
experimental observation. For a two dimensional square lattice, the coordination
number is = 4. Hence the critical temperature as per mean field theory would
⁄ = 4 whereas the exact value of
be
as obtained from the Onsager’s
73
⁄ = 2.269185 . Moreover, MF wrongly predicts phase
solution is
transition in one dimensional Ising model at a finite temperature
=2 / ,
since = 2 for a one dimensional lattice. It is known from the exact solution
that for one dimensional Ising chain there is no phase transition except at = 0.
Figure 5.2 Plot of spontaneous magnetization
at = .
against
/
.
continuously goes to zero
5.3.4 Determination of mean field critical exponents:
It is customary to use reduced temperature in analyzing the physical quantity at
the criticality. It is defined as
t=
T − Tc
Tc
or T = Tc (1 + t ) =
Jz
Jz
1
(1 + t ) or β Jz =
=
,
kB
kBT (1 + t )
and then the mean field equation of state in terms of reduced temperature
be written as
m
m = tanh β ( Jzm + h) = tanh(
+ β h).
1+ t
= −1 corresponds to
= 0 and
= 0 correspond to
74
.
can
Magnetization: Since we are interested in spontaneous magnetization, we set
ℎ = 0. The MF equation is then given by
m
m = tanh(
)
1+ t
As
→ 0,
m=
is small and tanh can be expanded as
m
m3
m5
m3
−
+
=
−
−
+ O{mt 2 , m3t , m5}
O
{
}
m
(1
t
)
3
5
1 + t 3(1 + t )
(1 + t )
3
m2
−t =
+ O{t 2 , m2t , m4 }
3
1
m ≈ (−t )1 2 with β =
2
or more and as
All correction terms are of the order of
negligible.
→ 0, they become
Isothermal susceptibility: Since the susceptibility is the change in
magnetization due to change in external magnetic field, we need to keep ℎ in the
mean field equation. Hence, we take the mean field equation of state as
m
+ β h).
m = tanh β ( Jzm + h) = tanh(
1+ t
For small ℎ and
→ 0,
m
h
+
+ O(m3 , m3 h, mh 2 , m5 )
1 + t k BT
1 
h
h (1 + t )

or m  1 −
, or m ≈
≈
k BT t
 1 + t  k BT
m≈
As Tc (1 + t ) = T , m ≈
Note that χ =
h 1
.
k BTc t
Since χ =
1 1 −γ
∂m
: t
, χ=
k BTc t
∂h
with γ = 1.
C
is the Curie-Weiss law for ferromagnetism. The exponent
T − Tc
γ = 1 is a consequence of m ≈ h for small h , i.e; a linear response of free
spins. Experimental values of the exponent γ are obtained as 1.22 for CrBr3 and
1.35 for Ni.
75
Critical isotherm: Let us take m = tanh β ( Jzm + h) . Since kBTc = Jz , at
= ,
h
h m3
m = tanh(m + ) = m + − + O(m2 h, mh2 + m5 )
Jz
Jz 3
Hence, ℎ~
related.
with
= 3. Note that here at
=
,
and ℎ are not linearly
Specific Heat: In order to calculate the specific heat about the critical point, one
needs expanding the mean field free energy
f = −kBT ln[2(cosh β Jzm)]
in terms of
up to order of
, knowing cosh = 1 + + + ⋯ . The
!
!
expansion can be obtained in terms of the reduced temperature , using
=0
/
for >
and
= (−3 )
for < . Differentiating twice with respect
to the temperature, it can be shown that
3
NkB for T < Tc
2
C =0
for T > Tc
C=
=
Thus, there is a discontinuity in at
must be equal to zero in mean field.
. Since
~| |
, the exponent
5.3.5 Critical exponents of correlation length and correlation function:
Correlation length: In order to obtain the correlation length exponent, let us start
with a periodic field ℎ = ℎ exp( ⃗ . ⃗). The response is also expected to be
sinusoidal,
=
exp( ⃗ . ⃗) . The magnetization at any point ⃗ can be
written as
( ⃗) =
( ⃗) +
⃗
( ⃗ − ⃗ ) ℎ( ⃗ )
where
is the magnetization due to applied field and second term represents
due to interaction of spins at ⃗′ and ⃗ , where ( ⃗ − ⃗ ′ ) is a non-local
susceptibility. As per mean field theory, the internal field is expected to be
∑
( ⃗ ′ ) where ( ⃗ ′ ) =
exp( ⃗. ⃗ ′ ) . The magnetization ( ⃗) at ⃗ is
76
then given by
( ⃗) =
ℎ( ⃗) +
⃗
(⃗ )
where
is the zero field susceptibility that diverges. The Fourier transform of
the magnetization is given by
=
⃗ ℎ( ⃗)
=
=
⃗∙ ⃗
( ⃗)
⃗
⃗∙ ⃗
+
⃗
⃗ ∙( ⃗ ⃗ )
⃗
ℎ +
where is a constant. The Fourier component of the susceptibility then can be
obtained as
χk =
Since
~1/ where
susceptibility as
∂mk
χ0
=
∂hk 1 − aJ χ0 k 2
is the reduced temperature, one could rewrite the
χk =
1
: (ξ −2 + k 2 )−1
t − aJk 2
where is the correlation length, the length scale that determines the decay of
as well as Γ( ) the two point correlation function. Since ~
, then
= 1/ 2.
Correlation function: The correlation between the spins on site i and j can be
measured by defining spin-spin correlation function
Γ ⃗ , ⃗ = 〈(
− 〈 〉)
−〈 〉 〉= 〈
〉 − 〈 〉〈 〉
where the spin si is at ⃗ and the spin sj is at ⃗ . The correlation decays to zero
exponentially with the distance between the spins
Γ( ) ~
(− / )
where = − 2 + and is the correlation length. Since at
the correlation function is expected to be
77
=
,
→ ∞,
Γ( r ) =
1
r d − 2+η
.
Taking the Fourier transform, one has
rr
1
1
Γ( k ) = ∫ d 3 r d −2 +η e−ik .r = ∫ r 2 dr sin θ dθ dφ 1+η e−ikr cosθ , taking d = 3
r
r
∞
π
dr
= 2π ∫ −1+η ∫ sin θ e−ikr cosθ dθ
r
r =0
θ =0
Taking
∞
= 2π
dr  2sin kr 
−

−1+η 
r
kr 

r =0
=
, one may show that
∫
Γ( k ) = −4π k
−2 +η
=
,
= 0,
=
dx  sin x 
−2 +η
: k
−1+η 
 x 
x =0 x
∫
=[
On the other hand, we know that
at
∞
Γ( )~
+
. Thus,
]
and
=
Γ( ). Since
= 0.
5.3.6 Scaling relations and upper critical dimension:
The values of the mean field critical exponents for magnetic systems are listed
below:
α
0
β
1/2
γ
1
δ
3
ν
1/2
η
0
It could be checked that the values of the critical exponents obtained in mean field
theory are different from those obtained experimentally (given in chapter 2) for
magnetic systems. This disagreement may be due to the mean field
approximation itself. Note that in the mean field approximation, interacting spins
are assumed to be non-interacting and every spin feels the same field due to all
other spins. On the other hand, magnetism is the outcome of long rang
cooperative behavior. The mean field approximation seems to be a drastic and
unrealistic approximation for magnetic systems.
It is now important to verify the thermodynamic scaling relations among the
78
critical exponents. The scaling relations are:
1
α + 2 β + γ = 0 + 2 × + 1 = 2,
2
1
(3 − 1) = 1,
2
1
γ = ν (2 − η ) = (2 − 0) = 1
2
γ = β (δ − 1) =
satisfied as exact equality and not as inequalities obtained by thermodynamic
considerations. As they are satisfying the scaling relations, they are not all
independent. Only two of them are independent. Moreover, the values of the
critical exponents obtained for any dimension.
One may notice that the values of the critical exponents obtained for the fluid
system are exactly same as that of the critical exponents obtained for the
magnetic system. Two widely different systems have the critical exponents. The
critical behavior is thus universal.
One important information can be obtained assuming the scaling relation
2− =
, where
is the space dimension. Since = 0 and ν = 1/2 , the
space dimension must be = 4. This means that mean field critical exponents
are exact at an upper critical dimension d c = 4 .
5.4 Bethe approximation
In the MF calculation, the interactions among the spins are neglected upto nearest
neighbor level. It could be the reason why the transition temperatures as well as
the values of the critical exponents are away from the exact values. It would be
interesting to see whether the value of the transition temperature can be made
closer to the exact values by considering a cluster of interacting spins in the
effective field due to the other spins. One such important approximation is the
Bethe approximation. In this approximation, a central spin should interact with
all its nearest neighbour spins and such an interacting spin cluster would be
subjected to an effective field due to next nearest neighbour spins to the cluster.
Below, we will be discussing the Bethe approximation in both one and two
dimensions.
79
5.4.1 Bethe approximation for one dimensional Ising model:
Consider a three spin cluster in an Ising chain, the central spin
and
as shown in the following figure.
neighbouring spins
s0
s1
and two
s2
Figure 5.3 A linear chain of Ising spins.
The Hamiltonian is given by
H = − J ( s1s0 + s2 s0 ) − ( Jm + h)( s1 + s2 ) − hs0 ,
si = ±1
where Jm is the effective field and h is the external field in units of energy. If the
system is in thermodynamic equilibrium at temperature T, the canonical partition
function is given by
Z=
∑
s0 , s1 , s2 =±1
exp {β J ( s1 s0 + s2 s0 ) + β ( Jm + h)( s1 + s2 ) + β hs0 }
The partition function can be written as a sum of two terms, one corresponding to
s0=+1 and s0=-1 as
=
+
Note that there are 2 = 8 states out of which 4 corresponds to
and 4
corresponds to
. Different state configurations are given in the table below.
,
( : ↑)
( : ↓)
+
Both up
( ↑↑↑)
( ↑↓↑)
2
1 up, 1 down
2 ( ↓↑↑)
2 ( ↓↓↑)
0
80
Both down
( ↓↑↓)
( ↓↓↓)
-2
Therefore,
(
=
)
(
+2+
)
and
(
=
Since
+2=(
+
+2+
+ ℎ )] +
The magnetization corresponding to the spin
〉=
〈
(
)
) , the partition function can be written as
+
[2 cosh ( +
=
)
[2 cosh ( −
− ℎ )]
is then given by
1
{}
=
1
[2 cosh ( +
+ ℎ)] −
[2 cosh ( −
whereas the magnetization corresponding to the spin
〉=
〈
− ℎ)]
= 1,2 would be
,
1
{}
=
=
1
1
(
)
−
[4 sinh ( +
(
)
+ ℎ ) cosh ( +
[4 sinh ( −
−
(
+
)
−
(
)
+ ℎ )]
− ℎ ) cosh ( −
− ℎ )]}
For simplicity cone could set ℎ = 0. Since the system is translationally invariant,
one should have 〈 〉 = 〈 〉. This yields
[2 cosh ( +
= 4 sinh ( +
or,
)] − [2 cosh ( −
) cosh ( +
cosh ( +
cosh ( −
Taking ℎ ′ =
)]
) − 4 sinh ( −
) cosh ( −
=
)
cosh ( +
) cosh ( −
) − sinh ( −
) − sinh ( +
, one has
81
)
=
)
)
cosh ( + ℎ′)
=
cosh ( − ℎ′)
( ℎ′) =
At
=
, one should have
ℎ′
′
=
2
′
′
′
= 2 . On the other hand,
sinh
cosh
cosh
=2
tanh
Hence the condition of criticality is given by
tanh
= 1,
or
=0
Therefore according to Bethe approximation there is no phase transition at any
finite temperature except at = 0 for one dimensional Ising chain as it would be
obtained by exact calculation in the next chapter.
5.4.2 Bethe approximation for Ising model on 2-dimensional square lattice:
Consider interacting five spin cluster as shown in the figure below in which
interacts with , ,
and
. Each , = 1,2,3,4 is subject to the effective
field of three nearest neighbor spins, 3 . The system is under an external
magnetic field and at a temperature .
s2
s1
s0
s3
s4
Figure 5.4 Ising spins on a square lattice. The central spin is interacting with four nearest
neighbours.
82
The Hamiltonian of the system is given by
(
ℋ=−
+
+
−ℎ(
)−3
+
+
+
+
(
+
+
)
+
+
)
where ℎ is the external magnetic field in units of energy. The canonical partition
function is given by
exp {
=
(
+
+
) +3
+
(
+
+
+
)
±
+ ℎ(
+
+
+
)}
+
The partition function can be written as a sum of two terms, one corresponding to
s0=+1 and s0=-1 as
=
+
Note that there are 2 = 32 states out of which 16 corresponds to
and 16
corresponds to
. Different state configurations are given in the table below.
,
, ,
( : ↑)
( : ↓)
=
All up
↑
↑
↑
↑
↑
↑
↓
↑
4
↑
↑
1 down, 3
up
4
↑
↓ ↑ ↑
↑
4
↑
↓ ↓ ↑
↑
2 down, 2
up
6
↓
↓ ↑ ↑
↑
6
↓
↓ ↓ ↑
↑
3 down, 1
up
4
↓
↓ ↑ ↓
↑
4
↓
↓ ↓ ↓
↑
All down
2
0
-2
-4
83
↓
↓
↑
↓
↓
↓
↓
↓
↓
↓
So the partition function is given by
(
=
)
+
+
(
+
(
)
(
)
(
+4
)
〉=
)
(
+6+4
)
)]
+
[2 cosh ( + 3
=(
) , the partition function
+
+ ℎ )] +
[2 cosh ( − 3
The magnetization corresponding to the spin
〈
(
+6+4
)]
Since
+4
+6+4
then can be written as
=
(
+4
− ℎ )]
is then given by
1
{}
=
1
[2 cosh ( + 3
+ ℎ )] −
[2 cosh ( − 3
whereas the magnetization corresponding to the spin
be
〉=
〈
,
− ℎ )]
= 1,2,3 or 4 would
1
{}
=
1
2
sinh ( + 3
−2
+ ℎ)[2 cosh ( + 3
sinh ( − 3
+ ℎ)]
− ℎ )[2 cosh ( − 3
− ℎ )] }
For simplicity cone could set ℎ = 0. Since the system is translationally invariant,
one should have 〈 〉 = 〈 〉. This yields
[2 cosh ( + 3 ) ] − [2 cosh ( − 3
= 16 sinh ( + 3 ) cosh
− 16 sinh ( − 3 ) cosh
or,
cosh
cosh
Taking ℎ ′ = 3
( +3
( −3
) cosh ( − 3
=
)
cosh ( + 3
, one has
84
)]
( +3
( −3
) − sinh ( − 3
) − sinh ( + 3
)
)
)
=
)
( ℎ′) =
At
=
, one should have
ℎ′
′
=
6
( + ℎ′)
=
( − ℎ′)
cosh
cosh
′
′
′
= 2 . On the other hand,
sinh
cosh
cosh
=6
tanh
Hence the condition for criticality is given by
coth
= 3,
or
≈ 2.885419
This is close to the exact value of
on the square lattice for which
2.269185. The values of the critical exponents however do not improve.
⁄ =
Other interacting spin clusters may also be considered as given in Problem 1 but it
is not remarkable as in the case of Bethe
is found that the change in
approximation.
5.5 Landau theory of phase transition
In the previous sections, mean field (MF) theory was discussed considering
different models. It was observed that MF not only can reproduce the
experimentally observed behavior at the critical point but also the critical
behavior is found independent of the detailed features of the models. It is
therefore natural to have a simple theory without involving detailed interactions
in it but should be good enough to describe critical phenomena. Since
thermodynamic potential can provide information about thermodynamic
quantities, Landau developed a simple theory for critical phenomena guessing the
form of the thermodynamic potential.
5.5.1 Landau potential:
The Landau potential ( , ℎ, ) should be such that its minima with respect to
should describe the thermodynamic properties of the system at = . Since
the free energy usually obtained as a function of the intrinsic variables such as
and ℎ and one may be tempted to consider the Landau potential as ( , ℎ ) =
( , ℎ). Since the free energy contains most of the information about the critical
singularity, the free energy can be represented as a Taylor series expansion
85
around the critical point (
,ℎ = 0) . However, at the critical point, the
=−
magnetization
is a multivalued function for
=−
isothermal susceptibility
<
tends to infinity as
. The
.
→
Therefore, such an expansion is going to be highly singular because the
derivatives appearing in the Taylor series expansion are singular. Another form
of the Landau potential ( , ) can be obtained through a Legendre
transformation
( , ) = ( ,ℎ ) − ℎ .
Now the first derivative
= −ℎ and the second derivative
either finite or zero as
. Thus an expansion in
→
=−
are
is possible.
The structure of Landau potential ( , ) :
(i)
Should be analytic,
(ii) Should have the same symmetry of the Hamiltonian. That is ( , ) =
( ,− ) and hence an even function.
(iii) For ≥ , = 0. For < ,
=±
and two phases co-exists.
-----------------------------------------------------------------------------------------------Note: Taylor’s series expansion of a function of two variables
Consider a function ( , ) having singularity at a point ( , ). The function
can be expanded around ( , ) in Taylor’s series as
( , )= ( , )+( − )
+
+( − )
1
( − )
2!
+( − )
+ 2 ( − )( − )
,
+⋯=
( − )
( − ) and s are different derivatives of the function
where
=∑
( , ) with respect to and .
------------------------------------------------------------------------------------------------
86
Following the above expansion, the Landau free energy ( , ) can be written
as
( , )= ( )+ ( )
+ ( )
+⋯
Since, ( , ) is an even function of , no odd terms appear in the expansion.
The coefficients are function of the reduced temperature = ( − ) ⁄ only.
The Helmholtz free energy is a convex function of magnetization . Therefore,
( , ) must be convex of
and thus the coefficients must be positive. The
coefficients can be expanded in terms of as:
( )=
( )=
( )=
+ ⋯,
+
+ ⋯,
+⋯
⁄
= 1⁄ = 0 as
Since
free energy can be taken as
→ 0, one has
= 0. Therefore, the Landau
A(t , m ) = a0 + a2 tm 2 + a4 m 4
Since higher order terms cannot alter the critical behavior of the system, one may
terminate the series at the fourth order in .
5.5.2 Critical behavior with Landau potential:
At equilibrium, the Landau potential A(t , m ) = a0 + a2tm 2 + a4 m 4 must have a
⁄
= 0 . Hence,
minima with respect to , therefore
2a2tm + 3a4 m 2 = m (2a2t + 3a4 m ) = 0
or,
m = 0 and m = ± −
2 a2 t
3a4
Therefore, for t ≥ 0, m = 0 and for t < 0, m = 0, ±
2 a2 t
3a4
In order to determine which solution corresponds to maxima and which solution
corresponds to minima of the potential, we consider the second derivative of the
potential.
87
∂2 A
= 2 a2 t + 6 a4 m
∂m 2
∂2 A
= 2 a2t = −2a2 t
for, t < 0 and m = 0,
∂m 2
Since it is negative m = 0 corresponds to a maxima.
2 a2 t
2a2t ∂ 2 A
=
−
+
= 2 a2 t
For t < 0 and m = ± −
,
2
6
a
t
a
2
4
3 a4 ∂ m 2
3a4
Since it is positive m = ± −
Plot of the Landau potential
can be seen that
Figure 5.5 Plot of
taken as
= 0,
2a2t
corresponds to minima.
3a 4
( , ) against
for different
is given below. It
( , ) against
for different values of . The values of the constants are
= 4 and
= 1⁄2 .
the potential has a single minimum at
= 0 for ≥ 0 and it has two minima at
± for < 0 . As decreases, the value of | | increases. For = 0 or
= −1 , the value of spontaneous magnetization
reaches its maximum
possible value. The values of ( ) obtained from the positions of minima of
( , ) are plotted against in Fig.6.2 below. It can be seen that ( ) is
but is a double valued function below .
zero for ≥
88
Figure 5.6 Plot of
( ) versus
.
It is surprising that singular behavior is obtained from a regular expansion. This is
because the value of magnetization which minimized free energy is itself a
singular function of the expansion coefficients that depend on the field as well as
temperature. Since
continuously goes to zero at = , this describes a
second order phase transition. Therefore, one should look for critical exponents
describing with singularities associated with different thermodynamic quantities.
5.5.3 Determination of critical exponents:
The Landau potential considered here is given by
A(t , m) = a0 + a2 tm 2 + a4 m 4 .
Taking different derivatives of the above potential different thermodynamic
quantities will be determined and their critical behavior will be extracted in the
→ 0 limit.
Magnetization: The magnetization can be obtained by setting
∂A
= 2a2tm + 4a4 m3 = 0
∂m
For t < 0, m =
a2
(−t )1 2 ,
2a4
89
1
hence β = .
2
⁄
= 0.
Specific heat: Since for
< 0,
=−
⁄2
,
a2
a22 2
a22 2
A(t , m) = a0 + a2t (−
t ) + a4 ( 2 t ) = a0 −
t .
2a4
4 a4
4 a4
⁄
would be a constant. On the other hand, for > 0 ,
Hence,
=
= 0. Therefore =
and hence
= 0. Thus, there is a discontinuity in
at = . Since ~| | , the exponent must be equal to zero.
Isothermal susceptibility: Since the susceptibility is the change in
magnetization due to change in external magnetic field, we need to keep ℎ
dependence in the Landau potential. Hence, the Landau potential takes the form
A(t , m) = a0 + a2tm 2 + a4 m 4 − hm.
From the above potential the inverse susceptibility can be obtained as
1 ∂2 A
− =
= 2a2t + 12a4 m2 .
2
χ ∂m
Since for
< 0,
=−
⁄2
,
χ −1 = −(2a2 − 12a4 ×
with
= −4
and
′
a2
)t = C− (−t )
2a4
= 1. On the other hand, for
> 0,
= 0. Hence,
χ −1 = −2a2t = C+ t
with
= −2 and = 1. Therefore the susceptibility exponent
⁄ = 1 ⁄2 .
the pre-factor ratio
= 1 with
Critical isotherm: Since critical isotherm describes the dependence of
at = 0, one should consider the field dependent Landau potential.
A(t , m) = a0 + a2tm 2 + a4 m 4 − hm.
90
on ℎ
Since
For
⁄
= 0,
= 0 then ℎ~
∂A
= 2a2tm + 4a4 m3 − h = 0.
∂m
with
= 3.
Thus the values of the critical exponents obtained from the Landau theory,
= 0, = , = 1, = 3 are all mean field values and unable to represent
experimental situations.
Two important points to note: First, in the Taylor series expansion it is
assumed that the derivatives of the function with respect to its arguments exits
and are finite. The expansion of Landau free energy will not be valid for a system
whose specific heat diverges to infinity as → . Second, the Landau potential
has an unstable region with a negative susceptibility for
<
. In a
thermodynamic potential existence of such an unstable region is not possible. In
that sense, Landau potential is not truly a thermodynamic potential.
91
Problems:
Problem 1: Consider a spin-1/2 Ising system on a two dimensional square lattice
in presence of an external magnetic field H and nearest neighbor interaction J
(>0). (a) Consider a pair of nearest neighbor spins as shown in Fig.I below and (b)
consider a four spin cluster as shown in Fig.II below in the effective field. Obtain
the self-consistent equation for the magnetization associated with one of the spins
in these cases and compare with the mean field as
in both the cases. Calculate
well as with the exact results.
s1
s2
Fig.I
s1
s2
s4
s3
Fig.II
[Hint (a): Each spin is subjected to the mean field of three other spins. The
Hamiltonian is
− (3
ℋ=−
+ ℎ )(
)
+
where ℎ is the external magnetic field in units of energy. The partition function
is given by
+ (3
exp {
=
+ ℎ)(
+
)}
±
Since there are four states ( ↑↑) , ( ↑↓ ) , ( ↓↑) and ( ↓↓) , the partition function can
be obtained as
=
(
)
=2
+ 2
+
cosh 2 ( 3
The magnetization due to any of the spins
92
(
+ ℎ) + 2
is given by
)
〉 = ( , ℎ) =
〈
1
(
)(
)
{}
sinh 2 ( 3
+ ℎ)
=
cosh 2 ( 3 + ℎ ) +
Set ℎ = 0, and
|
⁄
= 1. This leads to
6
1+
=1
⁄ ≈ 3.78.
Solving this equation graphically, one finds
Hint (b): Each spin is subjected to the mean field of two other spins. The
Hamiltonian is
ℋ=− (
+
+
) − (2
+
+ ℎ)(
+
+
+
)
where ℎ is the external magnetic field in units of energy. The partition function
is given by
exp{ (
=
+
+
+ ℎ )(
+
)
+
±
+ (2
+
+
)}
↑ ↑
↑ ↓
↓ ↑
↑
, 4
, 4
, 4
↑ ↑
↑ ↑
↑ ↑
↓
↓
. The partition function then can be obtained as
↓
↑
,
↓
Note that there are sixteen states:
2
↑
↓
=
↓
↓
,
↑
↓
(
)
+2
=2
(
+
)
cosh 4 ( 2
+4
〉 = ( , ℎ) =
=
Set ℎ = 0, and
1
⁄
1
)
+
+ ℎ ) + 8 cosh 2 ( 2
The magnetization due to any of the spins
〈
(
(
)
+ ℎ) + 2
(
)
sinh 4 ( 2
|
(
+ ℎ ) + 4 sinh 2 ( 2
= 1. This leads to
93
+4
is given by
)(
)
{}
2
+4
+ ℎ)
4
cosh 4
+1
=1
+3
⁄ ≈ 3.5.]
Solving this equation one finds
Problem 2: The classical Heisenberg model used for ferromagnetism is
described by
=−
⃗ ∙⃗ −
ℎ⃗ ∙ ⃗
〈 〉
where ⃗ is a 3 -dimensional unit vector in a d -dimensional lattice. Consider
J > 0 and let the external field ℎ⃗ = ⃗ be uniform and along z-direction. Using
, show that the Hamiltonian can be obtained
mean field approximation, 〈 ⃗ 〉 =
as
ℋ = −(
+ ℎ)
where c is the coordination number of a d -dimensional lattice. Assuming s z
can take on an entire continuum of orientations, obtain the per spin magnetization
m as
m = coth (
cJm + h
1
)−
.
k BT
(cJm + h)/k BT
Find the critical temperature Tc = cJ/(3k B ) and the values of the critical exponents
β , γ and δ . Check that the values of β , γ and δ are the same as those obtained
in Ising mean field.
[Hint: The partition function is
=
(
)
sin
=
+ ℎ)
4 sinh (
(
+ ℎ)
Then the magnetization is obtained as
=
∂ln
( ℎ)
The exponents can be obtained by expanding the equation of state as it is done for
the MF theory.]
94
Problem 3: Consider a Landau free energy of the form
A(t , m) = a(t ) +
b(t ) 2 c(t ) 4 d (t ) 6
m +
m +
m
2
4
6
where = ( − )/
is the reduced temperature and
is the magnetization.
If ( ) > 0 and ( ) < 0, show that this free energy represents a first order
transition at a critical point = . Derive an expression for the latent heat of the
transition.
= −3 ⁄4 by equation the free
Hints: Find the jump in magnetization
(
)
energy
,
= ( , 0) at = . Calculate the entropy = − ⁄
and estimate the latent heat = [ ( ) − ( ) ].
95
96
Chapter 6
Tranfer matrix method
97
6.1 Introduction
In the previous two chapters we have developed approximate analytical theories
for continuous phase transition. The interacting system was approximated as a
non-interacting system in an external mean field arising from the particle-particle
(spin-spin) interaction. The limitation of mean field type theories is that usually
they do not represent the experimental situations. The aim of this chapter is to
develop an exact analytical theory for the continuous phase transition. One
dimensional problems will be exactly solved. However, a brief discussion of
spin-1/2 Ising model in two dimensions will be given.
6.2 The methodology
In this approach, the partition function of the interacting system will be written in
terms of trace of a matrix called the transfer matrix. The canonical partition
function can be represented as
Z = tr Τ qN× q
where the trace is over all possible states,
is usually the system size and the
order of the matrix depends on the range of interaction as well as on the number
of states of each spin variable. Since the trace of a matrix is the sum its
eigenvalues,
q
tr Τ q×q = ∑ λi .
i =1
In order to estimate the eigenvalues of T, one needs to diagonalize the matrix. A
matrix is diagonalizable if there exists an invertible matrix S such that
Τ = SDS −1
where the columns of S are the eigenvectors of T and the diagonal elements of D
are the eigenvalues of T. The partition function then can be obtained as
Z = tr Τ
N
q× q
q
= tr( SDS SDS L SDS ) = tr(SD S ) = tr(D ) = ∑ λiN .
−1
−1
−1
N
−1
N
i =1
Since the partition function is obtained in terms of the eigenvalues of the transfer
matrix, the thermodynamic properties will be then described by the eigenvalues
98
and the eigenvectors of the transfer matrix. It can be shown that the free energy
density depends only on the largest eigenvalue of the transfer matrix. Say the
eigenvalues of T are λ={λ1 , λ2 ,λ3 , .... λq } which are arranged in descending
order and λ1 is the largest eigenvalue. The free energy density is then given by
= lim
→∞
(−k T lnZ)
= −k T lim
ln {λ + λ + ⋯ + λ }
→∞
= −k T lnλ − k T lim
≅ −k T lnλ .
ln 1 +
→∞
λ
λ
+ ⋯+
λ
λ
It is important to note that the free energy density depends only on the largest
eigenvalue. Hence, the thermodynamic quantities which are different derivatives
of the free energy density will also depend on the largest eigenvalue. In order to
have a physically sensible free energy, the largest eigenvalue must be positive
and non-degenerate. It can also be shown that the correlation length is given
by the two largest eigenvalues as
1
λ
= ln( 1 ).
ξ
λ2
The above prescription will now be applied to one and two dimensional spin
models to obtain exact results.
This method is applicable whenever the partition function can be expressed as a
product of matrices. As the model gets complicated, the transfer matrix also gets
complicated. The usefulness of the method depends whether the matrix is
diagonalizable or not, either analytically or numerically.
6.3 One dimensional spin-1/2 Ising model
The Hamiltonian for the 1d spin-1/2 Ising model with nearest neighbour
interaction in presence of an external field is given by
ℋ = −J
SS −h
S
where 〈 〉 represents nearest neighbor interaction and the field ℎ is units of
energy, and
= ±1. Consider a linear chain of
spins with periodic boundary
99
condition as shown in the figure below. Under the periodic boundary condition,
≡ . However, the choice of periodic boundary condition becomes
irrelevant in the limit → ∞.
The canonical partition function is given by
Z = ∑{ } eβ (
= ∑ { } eβ
= ∑{ }
.
where
given by
,
)
⋯
β
(
)
.eβ
(
β
. eβ
⋯
(
)
)
……eβ
β
(
)
…
β
= eβ
(
β(
= e
e
Since the matrix is identical for (
is then given by
Z=
= Tr
)
= ±1 the transfer matrix is
. Since
)
β
e
eβ(
), (
=
β
)
), and so on, the partition function
λ =λ +λ
{}
where λ and λ are the eigenvalues of the 2 × 2 transfer matrix.
Note that the transfer matrix builds up the lattice step by step. The rth power of T
adds spin
and trace upto
. This step then puts a bond between rth and
th
(r-1) spins.
6.3.1 Eigenvalues and eigenvectors of T:
In order to obtain the eigenvalues, one needs to solve the determinant equation
| −λ | = 0
100
or,
eβ(
)
β
e
−λ
eβ(
β
e
)
−λ
=0
or, λ − λeβ 2 cosh(βh) + 2 sinh( 2βJ ) = 0
Hence, λ
So
,
= eβ cosh(βh) ±
e
β
β
sinh ( βh) + e
is the largest eigenvalue and hence the free energy density is given by
=−
Check that as
expected.
.
= eβ cosh( βh) +
ln
→ 0, = −
ln 2 and as
e
β
sinh ( βh) + e
→ ∞, = −( + ℎ) as it is
The eigenvectors can also be obtained. Consider that | > and |
orthogonal eigenvector of the transfer matrix. So <u1|u2>=0 and
<u1|u1>=<u2|u2>=1.
Let
|u > = (α ,α )
One has,
Using
,
|u > = (α , −α )
|u > = λ |u > and
and hence, α + α = 1
α
α
=0
−λ
−λ
T
α
=
α
T
Since,
> are two
|u > = λ |u >
|u > = λ |u >, one has
T −λ
T
T
T −λ
As a solution one may obtain:
β
α +α =1
One has
1
α± = (1 ±
2
eβ sinh( βh)
e
β
sinh ( βh) + e
β
)
Since the eigenvalues and the eigenvectors of the transfer matrix is known, all the
thermodynamic properties can be derived now.
101
6.3.2 Magnetization:
The magnetization
can be calculated either from the expectation of the spin
matrix operator or from the free energy. Calculation of
in both ways is given
below.
The magnetization
is given by
m = 〈S〉 = ⟨u | S|u ⟩
where S = ∑{ } |s >
< s| =
1
α )
0
= (α
0
−1
1
0
α
α
0
, then
−1
eβ sinh( βh)
= α −α =
e
β
sinh ( βh) + e
e
β
β
In terms of free energy,
= −k T lnλ = −k T ln eβ cosh(βh) +
sinh ( βh) + e
β
The magnetization can be obtained as
=−
∂
∂
ln eβ cosh(βh) +
∂h
eβ sinh( βh)
= −k T
ℎ
=
e
β
sinh ( βh) + e
e
β
sinh ( βh) + e
β
β
The same expression for the magnetization is obtained. The variation of
ℎ for different is shown in Fig.6.2.
with
= 0 for any finite
Now there are a few points to notice. If ℎ = 0 ,
temperature T. Therefore there is no spontaneous magnetization at any finite
temperature and hence no phase transition at any finite temperature. If = 0
and ℎ is small then
= ±1 . On the other hand, if → ∞ or J=0 then
= tanh( βℎ), corresponds to a paramagnetic phase. Therefore, there is a phase
transition only at = 0 in the case of 1 Ising model. Such a conclusion was
qualitatively obtained from free energy consideration in Chapter-4.
102
=
>0
( , )
Figure 6.2 Plot of magnetization versus magnetic field for different temperature T.
6.3.3 Isothermal susceptibility:
The isothermal susceptibility can be obtained as
χ =
=
=
∂
∂h
βe cosh βh e
β
sinh ( βh) + e
βe cosh βh
e
β
sinh ( βh ) + e
For h=0,
=
β
β
− e sinh βh cosh βh e
β
e sinh (βh) + e β
e sinh βh cosh βh
−
β
(e sinh (βh) + e β ) ⁄
β
sinh ( βh) + e
β
.
As → 0,
→ ∞. Thus the criticality is at = 0 only. However, note that the
singularity is not of the power law type instead it is exponential.
6.3.4 Specific heat:
= −T
The specific heat can be obtained as
where
with
λ = eβ cosh(βh) +
e
103
β
sinh ( βh) + e
β
.
=−
T lnλ
Therefore,
is given by
=T
Since
sech
|
T
(
T lnλ ) = 2
=−
tanh
T
∂ ln λ
+
∂T
and
T
∂ ln λ
∂T
|
=
tanh
+
one has
( ℎ = 0) =
sech
→ 0 as expected.
This is a smooth function of temperature . As → 0,
There is no critical point at a finite temperature which corresponds to singular
.
behavior of
6.3.5 Correlation function:
Let us consider the spin-spin correlation in the Ising chain. Since the effect of
external field is not important, we set ℎ = 0. The Hamiltonian and the canonical
partition function are given by
H = −J
S S
,
eβ
Ζ=
and
{ }
The correlation function for a pair of spins
is given by
( ) = 〈S S
and
separated by a distance
〉 − 〈S 〉〈S
〉
Since the calculation is for a finite temperature > 0 and the criticality of the
Ising chain is at
= 0 , in absence of external field there will be no
〉 = 0. Hence,
magnetization, i.e., 〈 〉 = 〈
( ) = 〈S S
Since for Ising spins always
( ) = 〈S S
〉
= 1, the correlation function can be written as
S
S
⋯S
104
S
S
S
〉
For nearest neighbor interaction only, the above function can be written as
( ) = 〈S S
〉〈S
S
〉 ⋯ 〈S
eβ
= eβ
〉
〉 among two nearest neighbours, we
In order to calculate the correlation 〈
calculate the partition function
=
S
+e
β
= 2 cosh (βJ) = λ (h = 0)
±
for S
= +1 or − 1. Hence,
< SS
> =
eβ
∑{ } S S
=
1 ∂(ln
β
∂J
)
= tanh (βJ)
Therefore, the correlation function is given by
( ) = tanh ( βJ ) = coth ( βJ )
For
> 0, the correlation function for the Ising chain is expected to be
( )=e
ξ
where ξ is correlation length. Hence,
or,
On the other hand,
ξ
= coth ( βJ )
r
− = −r ln coth (βJ) or, ξ
ξ
e
λ
λ
eβ + e
= β
e −e
Therefore,
ξ
= ln
= ln coth ( βJ)
β
β
= coth (βJ)
λ
λ
where λ > λ . Though this relation is proved for a linear Ising chain this is a
general result. Now we take different limits of temperature and verify the
expected values of the correlation length.
As βJ → ∞ ( T → 0) ;
coth (βJ) → 1, therefore, ξ
105
= 0 or, ξ = ∞
As βJ → 0 ( T → ∞) ;
coth (βJ) → ∞, therefore, ξ
= ∞ or, ξ = 0
Hence, the correlation length assumes the right limiting values.
Now we will be considering a few examples in one dimension for which exact
partition function can be obtained. The free energy and the correlation length
expressions will be obtained and their values in different limiting conditions will
be verified.
Example 1: Consider one dimensional spin-1 Ising model for
spins with
nearest neighbour interaction in absence of external magnetic field. The
Hamiltonian is given by
ℋ = −J
SS
where
= +1,0, −1 and
= . We will construct the transfer matrix and
obtain the free energy density and the correlation length. Their values will be
checked in the → 0 and → ∞ limits.
The canonical partition function is given by
eβ
=
{}
∑
eβ
=
)
.eβ
…eβ
= Tr
{}
where
= eβ α β . Since
and
have three states each, the transfer
matrix would be a ( 3 × 3) matrix as given by
eβ
=
1
e β
1
1
1
e
β
1
eβ
The transfer matrix will have three eigenvalues, say,
function is then given by
Z= λ + λ +λ
,
The eigenvalues of the transfer matrix can be obtained as,
| − λI | = 0
106
&
. The partition
eβ − λ
1
e β
or,
or,
e −e
1
1−λ
1
+λ λ − e +e
e
β
=0
1
e −λ
β
+ 1 λ+ e + e
−2
=0
Solving the above equation one has,
λ = 2 sinh(βJ)
λ = [1 + 2 cosh(βJ) +
( 2 cosh( βJ ) − 1 ) + 8 ]
λ = [1 + 2 cosh( βJ ) −
( 2 cosh( βJ ) − 1 ) + 8 ]
It can be checked by plotting the eigenvalues against
that
λ >λ >λ
Therefore the free energy is given by
=−
T ln λ = −
T ln 1 + 2 cosh(βJ) +
( 2 cosh( βJ ) − 1) + 8 ⁄2
As T → 0, λ ≈ e hence = −J as expected. On the other hand, if
= −k T ln (3), the entropy term.
→ ∞,
The correlation length can be obtained as
= ln
= ln 1 + 2cosh
− ln sinh
As T → 0, ξ → 0 hence
→ ∞, hence
→ 0.
+
4cosh
− 4cosh
+9
− 2 ln 2
→ ∞ as expected. On the other hand, as
→ ∞,
Example 2: Consider the one dimensional q-states Potts model which is
described by the Hamiltonian
ℋ = −J
δ
107
where σ = 1,2,3 ⋯ ,q and is the Kroneker delta. Due to periodic boundary
condition
=
. Construct the transfer matrix and find that largest
+ − 1 and the remaining eigenvalues are all degenerate and
eigenvalue is
takes the value
− 1. Obtain the free energy density and correlation length
for this model and check limit T → 0 and T →∞.
The canonical partition function is given by
eβ
Ζ=
∑
{σ }
where
eβ δσ
=
σ
.eβ δσ
σ
…eβ δσ
σ
{σ }
αβ
=
.
…
{σ }
= eβ δαβ = eβ − 1 δαβ + 1.
The transfer matrix in this case is going to be a ( × ) matrix and it is given by
=
eβ
1
⋮
1
1 … 1
eβ … 1
⋮
⋱ ⋮
1 ⋯ eβ
The partition function is then given by
= Tr
=
|u > = λ|u > where the
The eigenvalues can be obtained by solving
, ,⋯ ,
. Hence the system of linear
eigenvector is given by | 〉 =
equations are given by
eβ
1
⋮
1
Or,
1 … 1
eβ … 1
⋮
⋱ ⋮
1 ⋯ eβ
+
=
,
u
u
⋮
u
u
u
=λ
⋮
u
+
=
,
⋯
+
Thus one may write
108
=
,
+
=
−
Or
+1
−
=
The right hand side of the above equation is independent of the index . So if
+ 1 ≠ 0, then all
are the same. Let us set them equal to 1 and say,
−
the corresponding eigenvalue of this eigenvector is . Then,
=
+
−1
Thus one eigenvalue corresponding to an eigenvector is obtained. The only
other possibility is
−
+ 1 = 0 which corresponds to ∑
= 0. Thus
the second eigenvalue is
=
−1
corresponding to an eigenvetor whose element sum is zero. Since the transfer
matrix has eigenvectors, all ( − 1) eigenvectors then should have the same
eigenvalue .
Since
>
, the free energy density is given by
=−
As T → 0,
term.
T ln λ = −
≈ −J as expected and as
T ln
+
→ ∞,
≈−
−1
T ln ( ), the entropy
The correlation length can be obtained as
= ln
As T → 0, ξ → 0 hence
→ ∞, hence
→ 0.
= ln
+
−1
−1
→ ∞ as expected. On the other hand, as
109
→ ∞,
6.4 Ising model in two dimensions:
Now we will consider Ising model in two dimensions. Consider a ( × )
square lattice consisting of
rows and
columns with an Ising spin on each
lattice site. There is only nearest neighbor interactions among the spins and a
periodic boundary condition is assumed in both the directions. In absence of
external magnetic field, the Hamiltonian of the system is given by
ℋ=−
−
where denotes the rows and represents spins along a row. The canonical
partition function is then given by
ℋ
=
= Tr(
)=
{ }
where the transfer matrix is of ( 2 × 2 ) order. However, diagonalization of
such a matrix is found to an extremely difficult task. The first exact result of the
problem was obtained by Krammers and Wannier in 1941who located the .
Onsagar in 1944 derived an explicit expression for the free energy density by
determining the largest eigenvalue. The free energy density is given by
( ) = − ln( 2 cosh 2
where
= 2 sinh 2
( )=
⁄cosh 2
)−
1
2
ln
1+
1−
2
sin
. The internal energy per spin is then given by
( )]
[
= −2 tanh 2
+
sin
2
1−
The integral,
110
sin
1+
1−
sin
sin
=
1−
sin
1+
sin
=
= −
+
1−
sin
1− 1−
1−
1
sin
sin
[1 − (1 −
1−
sin
sin
)]
The internal energy is then given by
( ) = −2 tanh 2
1
2
+
−1 +
1
1−
sin
The derivative,
′
=
= 2 tanh 2
1
−1,
= 2 [ coth 2
] and
− 2 tanh 2
+ ′ =1
Then the internal energy is obtained as
( ) = − coth 2
1+
2
= − coth 2
− 1) ( )
( 2 tanh 2
1+
2 ′
( )
where
⁄
( )=
1−
sin
is the complete elliptic integral of the first kind.
The specific heat ( ) can be obtained by taking one more derivative of ( )
with respect to temperature . In order to obtain ( ), one needs to use the
following relations:
′
= 8 tanh 2
(1 − tanh 2
),
and
111
( )
=
1
( )
− ( )
(1 − )
where
⁄
( )=∫
1−
sin
is the complete elliptic integral of the
second kind. The specific heat ( ) then can readily be obtained as
1
( )=
2
(
)
coth 2
2 ( ) − 2 ( ) − ( 1 − ′)
2
+
′
( )
The elliptic integral ( ) has a singularity at = 1 or ′ = 0 . Thus all
thermodynamic quantities should have singularity at =
defined by
=1
or ′ = 0. Thus, the critical temperature can be obtained as
2 tanh
2
− 1 = 0,
or
= 0.4406868,
or
= 2.269185
Other conditions of criticality can also be defined as
)
= 1,
)
2 sinh(2 ⁄
cosh (2 ⁄
or sinh
or cosh
2
2
= 2 tanh
2
=
2
√2
= √2,
=1
In the neighborhood of the singularity, the elliptic integrals can be approximated
as
( ) ≈ ln
Thus near
4
and
| ′|
( ) ≈1
, the specific heat is given by
1
( )≈
8
ln
4
− 1+
| ′|
4
At the same time ′ can be approximated as
′
≈ ′ +
′
( −
) = 2√2
Therefore the singularity in the specific heat as
112
1−
→
is given by
1
( ) ≈−
8
ln 1 −
+ constant
Thus the specific heat diverges logarithmically to infinity as → . The
specific heat singularity then differs from mean field prediction of a
discontinuous jump. The critical exponent describing the specific heat singularity
is then given by = 0(log).
The temperature dependence of the spontaneous magnetization is given by
( )=
1 − sinh
⁄
2
0
As
As
→ 0, ( ) ≈ 1 − 2exp ( −8 ⁄
→
,
for
≤
for
≥
) represents a slow variation with
.
⁄
( ) ≈ 8√2
Thus the order parameter exponent
the mean field value.
1−
is 1⁄8 which is very different from 1⁄2 ,
The zero field isothermal susceptibility is is found asymptotically as
~| |
⁄
where = ( − ) ⁄ . The susceptibility exponent is then given by
which is again very different from 1, the mean field value.
= 7 ⁄4
The correlation length and correlation function exponents are also estimated as
= 1 and = 1⁄4 respectively. Again the exponent values are very different
from their classical values 1/ 2 and zero respectively.
Therefore the exact values of the critical exponents of two dimensional Ising
model is obtained and found very different from those obtained in the men field
theory. A comparison of the values of the critical exponents obtained in these
methods is given in the table below.
113
Mean field
Exact
0 (dis)
0 (log)
1/2
1/8
1
7/4
1/2
1
0
1/4
= 2−
The critical exponents satisfy the scaling relations + 2 + = 2,
etc. as exact equality. Since the model is not yet solved in presence of external
magnetic field, a direct measurement the critical isotherm exponent is not
possible. However, assuming the scaling relation = ( − 1) holds as exact
equality one has = 15 which is again very different from the mean field value
= 3.
The Ising model is not yet solved even in absence of external field in three
dimensions.
114
Problems:
Problem 1. Calculate the canonical partition function
by transfer matrix
method for linear Ising chain of
spins with nearest neighbor interaction
and next nearest neighbor interaction
as shown in the figure below and in
absence of external field.
Find the free energy density and the correlation length.
[Hint: The Hamiltonian of the system is given by
ℋ = −J
Take a
S S .S
as
SS
−J
SS
new variable σ = S S = ±1 .Then check that σ σ
=
S = S S . Rewrite the Hamiltonian in terms of the new variable
N
N
σi − J2
ℋ = −J1
i=1
σi σi+1
i=1
Rest of the calculation is that of the spin
problem in Lecture-I.]
Ising model given as a demonstration
Problem 2. The nearest neighbour interaction along the chains of a double Ising
chain is J and that among the spins in the two chains is J as shown in the figure
below. Both the chains contain
spins.
Show that the transfer matrix of the system is given by
115
1 1
1
1
where A = e
k = βJ .
, B=e
1 1
1
1
,C= e
, D=e
and k = βJ ,
[ Hint: The Hamiltonian is given by
ℋ=J
(S S
− S′ S′
)−J
S S′
And the partition function is
Z=
{}
1
{exp [βJ S S + S′ S′ + βJ (S S′ + S S′ )]. exp [βJ S S + S′ S′
2
+
1
βJ (S S′ + S S′ )] ……….
2
If k = βJ , k = βJ , the transfer matrix is given by
⟨
′| |
′
⟩ = exp k ( S S
+ S′ S′
]
116
)+
k
( S S′ + S
2
S′
)