Ex 5958: Mean field approximation for the Model of ferroelectricity
Submitted by: yair massury
The problem:
Model of ferroelectricity: Consider electric dipoles p on sites of a simple cubic lattice which point
along one of the crystal axes, ±h100i. The interaction between dipoles is
U=
p1 · p2 − 3(p1 · r)(p2 · r)/r2
4π0 r3
where r is the distance between the dipoles, r = |r| and 0 is the dielectric constant.
(a) Assume nearest neighbour interactions and find the ground state configuration. Consider
either ferroelectric (parallel dipoles) or anti-ferroelectric alignment (anti-parallel) between
neighbours in various directions.
(b) Develop a mean field theory for the ordering in (a) for the average polarization P at a given
site at temperature T: Write a mean field equation for P (T ) and find the critical temperature
Tc .
∂P
(c) Find the susceptibility χ = ∂E
at T > Tc for an electric field E||h100i, using the mean
E=0
field theory.
The solution:
(a)
U=
2
pi · pj − 3(pi · rij )(pj · rij )/rij
4π0 r3
(1)
For rij ||pi , pj which mean’s that rij ||x̂ we get:
Uij =
−2pi · pj
4π0 r3
(2)
The lowest enrgy is when pi = pj so that Uij = −2U0 where U0 =
For
rij ⊥pi , pj which mean’s that
Uij =
p2
4π0 r3
rij ⊥x̂ we get:
+pi · pj
4π0 r3
(3)
So pi = −pj (in opposite direction),which mean’s that Uij = −U0
1
The ground state:
FIG.1 We have a chain in x̂ direction as each chain four
different neighboring chains with opposite polarity.
(b) Lets define two sublattice:
A=positive moment +, which means that y+z=even.
B=negative moment -, which means that y+z=odd.
we get the Hamiltonian:
rij ||x̂
H = −2Uo
rij ||ŷ
rij ||ẑ
X pi · pj
X pi · pj
X pi · pj
X
+ Uo
+ Uo
− Eelctric
pi
2
2
2
p
p
p
i∈A,j∈A
i∈A,j∈B
i∈A,j∈B
i∈A,i∈B
rij ||x̂
rij ||ŷ,ẑ
X
X
X pi · pj
X pi · pj
+
2U
−
E
p
−
E
pi
o
i
el
el
p2
p2
= −2Uo
i∈B,j∈B
i∈B,j∈A
i∈A
(4)
i∈B
we make a mean field approximation (not curfully) to get:
HA = −2Uo PA
X pi
X pi
X
+
2U
P
−
E
pi
o
B
el
p2
p2
(5)
X pi
X pi
X
pi
+ 2Uo PA
− Eel
2
2
p
p
(6)
i∈A
HB = −2Uo PB
i∈A
i∈B
i∈A
i∈B
i∈B
H = HA + HB to find the enrgy for one site .we took pA = −pB (in opossite direction)
1
P2
PB (PA )
P2
PA (PB )
E=
− Eel PA − 2U0 2 + 2U0
− Eel PB
−2U0 2 + 2U0
2
P
P2
P
P2
⇒ E = −4Uo
(7)
In zero filed we look at average filed Pav
MF
HE=0
= −4U0 Pav
Pav =
+pe
e
β4U0 Ppav
β4U0 Ppav
pi
p2
(8)
− pe
−β4U0 Ppav
−β4U0 Ppav
+e
Pav
= p tanh β4U0
p
(9)
To find Tc we take the first order in tanh β4U0 Ppav
Pav
Pav
= T anh(β4U0
)
p
p
→
Pav
Pav
= β4U0
p
p
2
⇒ Tc = 4U0
(10)
To see the behaviour near the critical temperature:
( Ppav )3
Pav
Tc
Pav 2
−1 Pav ∼ Pav
β4U0
= T anh (
)=
+
→ 3( − 1) = (
)
p
p
p
3
T
p
∂P
∂E E=0
(c) To Find the susceptibility χ =
tanh−1 (x)
(11)
at T > Tc . we take first order in Taylor Series for
PA
PB
PA = p tanh β2U0
− β2U0
+ βEp
p
p
PA
PB
− β2U0
+ βEp
PB = p tanh β2U0
p
p
PA
PA
= β2U0
tanh
p
p
PB
PB
tanh−1
= β2U0
p
p
PA
PA
= β2U0
→
p
p
PB
PB
→
= β2U0
p
p
−1
P =
N
(PA + PB ) ⇒
2
χ=
∂P
|E=0 = N βp2
∂E
PB
p
PA
− β2U0
p
PB
− β2U0
p
PA
− β2U0
p
− β2U0
(12)
+ βEp
+ βEp
+ βEp
+ βEp
P = N βEp2
(13)
(14)
(15)
3