Some understandings of ferroelectricity

Some conceptional paradox with ferroelectricity (FE).

Figure 1

1) As shown in Fig. 1, there is a one-dimensional system that consists equally N anions with charge -q and N cations with charge +q, separated by distance a. This system definitely has spontaneous permanent dipole moment p=Nqa, but why it is not FE?

Note that normally FE is defined for bulk (infinite) system, in which case not so many systems have FE. However, for a finite non-metallic system, spontaneous dipole moments are not rare at all. Figure 1 is actually a good example. Then the paradox is that the real systems are finite anyway, how come they are not normally FE?

The answer has to do with the imperfectness of the real system. One can see in Fig. 1, the direction of the dipole moment depends on the type of the atoms on the surface, which is random for a real system. For example, we don't normally have a NaCl sample that has a surface only with Na atoms and the opposite surface with only Cl atoms. This randomness will eventually cancel out these "spontaneous dipole moments", making the total dipole moment of the sample negligibly zero.

Figure 2

2) The system shown in Fig. 2 is trully FE.

For this system in panel A, the dipole moment is p=Nqa. But if the surface atoms change to panel B, the dipole moment is p=-Nqb. Therefore the randomness of the surface can not cancel the dipole moment, which on average is p=Nq(a-b). This is why the system is FE.

3) To summarize the cases we discussed above, to determine if a system is FE. We have to test if the system (infinite) is central-symmetric. In other words, whether you can find an unit cell that has an inversion symmetry. For the system in Fig. 1, it is central symmetric. But for system in Fig. 2, it is not. That's why it is FE.

Low frequency dielectric constant

Starting from Lorentz model:

md²x/dt²+mΓdx/dt+mω₀²x=qE

which is the equation of motion for a bond electric charge q with mass m in an oscillating field E, where ω₀ is the frequency of the Harmonic potential of the charge and Γ is the dampling parameter.

Take the oscillating form of x and E:

x=x₀e^-iωt and
E=E₀e^-iωt

one get the solution:

x=q/m/(ω₀²-ω²-iΓω).

The the dielectric constant will be (notice that P=xq):

ε=1+P/(Eε₀)=1+q²/mε₀/(ω₀²-ω²-iΓω)

For a low frequency measurement ω₀>>ω,which is valid for normal dielectric measurement because the lowest ω₀ is normally at the 10¹² Hz range.

In this case, we can ignore the high order term ω²,

ε=1+q²/mε₀/(ω₀²-iΓω).

Imaginary part:

The imaginary part is

ε₂=q²Γω/mε₀/(ω₀⁴+Γ²ω²).

Experimentally, a maximum is always observed for the temperature dependence of ε₂, here we can calculate the maximum by solving

dε₂/dT=0.

Note that the only temperature dependent parameter is Γ.

Then one gets:

Γ=ω₀²/ω at dε₂/dT=0.

Now let's look at the parameter Γ. It is a dampling parameter, which should be proportional to relaxation time τ, which satisfies the Arhenious relation:

τ=τ₀e^E_a/(k_BT)

where Ea is activation energy (or energy barrier to overcome the).

We can even define:

τ=Γ/ω₀².

Then

τ₀e^E_a/(k_BT)=1/ω should give us the temprature where the maximum ε₂ is, which is

E_a/(k_BT_max)=-log(τ₀ω).

Experimentally, fitting the the relation between T_max and ω, one gets the activation energy E_a, which is what most paper shows.

Real part:

The real part dielectric constant is

ε₁=q²ω₀²/mε₀/(ω₀⁴+Γ²ω²).

One can see that it increases monotonic with temperature simply because of the temperature dependence of Γ.

Derivation of optical conductivity σ:

Noticing that E=D-P= ε₀(εE-P), one gets

P=ε₀(ε-1)E (1)

The electric current density (considering E=E₀e^-iωt):

J=dP/dt=-iωε₀(ε-1)E (2)

Compare (2) with the Ohm's law:

J=σE (3),

one gets:

σ=-iωε₀(ε-1)=ωε₀ε₂-iωε₀(ε₁-1), therefore

σ₁=ωε₀ε₂ and
σ₂=-ωε₀(ε₁-1)

Derivation of absorption coefficient α.

Starting from Maxwell equations:
∇D=ρ
∇B=0
∇×E=∂B/∂t
∇×H=∂D/∂t (1)

To get a solution of EM wave, one need to use

B=μμ₀H and D=εε₀E and plug into the last two equations of (1).

Then we have:

∇²E+μμ₀εε₀∂²E/∂t²=0, the solution is

E=E₀e^-i(ωt+qr) (2)

where ω is the frequency and q is the wave vector and

q=ω√(μμ₀εε₀).

Note that normally μ=1 for non-magnetic material (or even for magnetic material at high frequency) and μ₀ε₀=1/c².

Therefore:

q=ω/c√ε (3).

Plug (3) into (2) and using the definition √ε=n+ik for imaginary dielectric constant (whee n is the refractive index and k is the extinction coefficient), one gets:

E=E₀e^{-iω(t+√εr/c)}=E₀e^{-iω[t-(n+ik)r/c]}=E₀e^{-iω[t-(n+ik)r/c]}e^-ωkr/c (4).

One can see that the electric field decays if there is an imaginary part of the dielectric constant.

By the definition of α:

E²=E₀²e^-αr, we get

α=2ωk/c.

Parity and spin forbidden optical excitations

In optical excitation, we are normally looking at the transition matrix element:

M_ij^E=<φ_i|exE|φ_f>, where φ_i and φ_j are the wave functions of the initial and final states, E is the electric field.

Following observations are important concerning the M_ij^E.

• M_ij^E vanishes for a lattice structure with inversion symmetry. In this case the excitation is called parity forbidden excitation.
• The operator exE does not involve spin degree of freedom. So in order to have non-zero, the spin of φ_i and φ_j has to be the same. If the initial and final states have different spin, the excitation is called spin forbidden excitation.
  

How come we do see the "forbidden" excitation? Here are what happens in real material to circumvent the forbiddenness.

	Parity forbideness	Spin forbideness
Single site	Existence of phonon reduces the symmetry magnetic dipole excitation MijM=<φi|μB|φf>	spin-orbit coupling
Collective		Exciton + magnon
		Exciton + Exciton

Why magnetic dipole transition intens...

Why magnetic dipole transition intensity is so low in absorption spectra compared to the electric dipole transition.

In optical process, both electric and magnetic dipole moment respond to the EM wave, causing additional energy

H'=H_E+H_M,

where H_E=pE and H_M=μB.

One can estimate the order of the magnitude difference by taking the ratio:

R=(pE/μB)².

In EM wave, one has E/B=C/n, where C is the speed of light, and n is the refractive index.

For charge transfer optical process, it is reasonable to assume that p~eA=1.6e-19×1e-10 =1.6e-29Cm.

For magnetic dipole moment μ, Bohr magneton is always a good estimation for the order of magnitude.

Therefore, the ratio between the typical oscillator strength of electric charge transfer and magnetic dipole excitation is:

R~(1.6 e-29*3e8/9.24e-24/n)²=(5.2×10²/n)² >> 1
.

This is the reason that the magnetic dipole transition is normally not intense.

Transition temperature, magnon frequency ω and exchange interaction J

Ferromagnetic material

Material	Curie Temperature Tc (K)	Exchange interaction J (meV)	Magnon frequency ħω	Magnetization (M) and susceptibility (χ)
	Mean field theory: 2/3*Z[S(S+1)]J		ħωk=2ZJS(1-γk) γk=∑nncos(kδ)/Z Z is the number of nearest neighbor, δ is the nearest neighbor vector. Maximum: (k=π/a) ħωmax=4ZJS For 1 D system, maximum value will be ħωmax=8JS	Bloch theorem ΔM/M0 ∝ T3/2 Mean field theory: χ=Nμ2/[3kB(T-Tc)]
EuO	69.1 (K)		0.5 meV (k=0.2 A) 3.2 meV (k=0.6 A) 4.6 meV (k=0.8 A) 5.5 meV (k=1.0 A)

Materials	Structure	TM (K)	TN (K)	Weak FM	Spin orientation	Exchange interaction J	Magnon frequency ħω	Magnetization (M) and susceptibility (χ)
			The relation between TN and J is missing, I could not find it anywhere.				ħωk=2ZJS√{1-γk2} Taking into account of anisotropy and external field ħωk = 2ZJS√{(1+gμBBA/(2ZSJ))-γk2} ± gμBB At zero external field, we can get finite ħωk even for k=0  because of anisotropy. ħωmin = 2ZJS√{gμBBA/(2ZSJ)(1+gμBBA/(4ZSJ))} If we apply an external field, we can get spin flop when the mode of one branch of the spin wave goes to zero (instability) Bf=√{4ZJSBA/(gμB)} In the case that anisotropy is much smaller than J, in other words: gμBBA>>2ZSJ, we get ħωmax=2ZJS,   k=π/(2a) Then, ħωmin = √{gμBBA*(2ZSJ)	Mean field theory: χ=C/(T+TN)
MnF2			67.4 K				k‖z ħωmax=54.8 cm-1 k‖x ħωmax=50.4 cm-1 because of anisotropy ħωmin=8.715 cm-1 e.g. μBBA = 0.737 cm-1 then we can calculate (using the first order): ħωmin=√{gμBBA*ħωmax}=8.99 cm-1
BiFeO3	Rhombohedra (distorted perovskite)		640 K	supposably weakly ferromagnetic below TN			ħωmax=445 (assuming kBTN=ħωmax) ħωmin= 18.2 cm-1 We can estimate the anisotropy energy: μBBA = (ħωmin)2/(għωmax) = 0.372 cm-1, corresponding to BA=0.8 T, which does not look right. In any case, there is no real data on ħωmax., if we can succesfully get the spin flop, then we probably can get ħωmax.
Fe2O3 Hematite	Rhombohedra (corundum)	260 K	950 K	weakly ferromagnetic between TN and TM	‖[111]c below TM ⊥[111]c between TM and TN
Cr2O3	Rhombohedra (corundum)		312 K			94 K	ħωmax=600 K
RbMnF3	cubic perovskite		82 K			J1=3.4 K	ħωmax=100 K (Windsor CG and Stevenson RWH, Proc.Phys. Soc. 87, 501) ħωmin ~ 0
KMnF3	cubic perovskite		88K