Editing Ferroelectricity (section)

==Theory==
An introduction to Landau theory can be found here.<ref name="P. Chandra">{{Cite arXiv|author1=P. Chandra |author2=P.B. Littlewood |title=A Landau Primer for Ferroelectrics|eprint=cond-mat/0609347 |year=2006 }}</ref>
Based on [[Ginzburg–Landau theory]], the free energy of a ferroelectric material, in the absence of an electric field and applied stress may be written as a [[Taylor series|Taylor expansion]] in terms of the order parameter, {{mvar|P}}.  If a sixth order expansion is used (i.e. 8th order and higher terms truncated), the free energy is given by:

<math display=block>\begin{align}
  \Delta E =&\quad\, \tfrac{1}{2}\alpha_0 (T - T_0) (P_x^2 + P_y^2 + P_z^2) \\[4pt]
  &+ \tfrac{1}{4} \alpha_{11} (P_x^4 + P_y^4 + P_z^4) \\[4pt]
  &+ \tfrac{1}{2} \alpha_{12} (P_x^2 P_y^2 + P_y^2 P_z^2 + P_z^2P_x^2) \\[4pt]
  &+ \tfrac{1}{6} \alpha_{111} (P_x^6 + P_y^6 + P_z^6) \\[4pt]
  &+ \tfrac{1}{2} \alpha_{112} \bigl[ P_x^4(P_y^2 + P_z^2)
 + P_y^4(P_x^2 + P_z^2) + P_z^4(P_x^2 + P_y^2) \bigr] \\[4pt]
  &+ \tfrac{1}{2} \alpha_{123} P_x^2P_y^2P_z^2
\end{align}</math>

where {{math|''P<sub>x</sub>'', ''P<sub>y</sub>'', ''P<sub>z</sub>''}} are the components of the polarization vector in the {{math|''x'', ''y'', ''z''}} directions respectively, and the coefficients, {{math|''&alpha;{{sub|i}}'', ''&alpha;{{sub|ij}}'', ''&alpha;{{sub|ijk}}''}} must be consistent with the crystal symmetry.  To investigate domain formation and other phenomena in ferroelectrics, these equations are often used in the context of a [[Phase field models|phase field model]].  Typically, this involves adding a gradient term, an electrostatic term and an elastic term to the free energy.  The equations are then discretized onto a grid using the [[finite difference method]] or [[finite element method]] and solved subject to the constraints of [[Gauss's law]] and [[Linear elasticity]].

In all known ferroelectrics, {{math|''&alpha;''{{sub|0}} > 0}} and {{math|''&alpha;''{{sub|111}} > 0}}.  These coefficients may be obtained experimentally or from ab-initio simulations.  For ferroelectrics with a first order phase transition, {{math|''&alpha;''{{sub|11}} < 0}},  whereas {{math|''&alpha;''{{sub|11}} > 0}} for a second order phase transition.

The ''spontaneous polarization'', {{mvar|P<sub>s</sub>}} of a ferroelectric for a cubic to tetragonal phase transition may be obtained by considering the 1D expression of the free energy which is:

<math display=block>
\Delta E = \tfrac{1}{2}\alpha_0 (T-T_0)P_x^2 + \tfrac{1}{4}\alpha_{11}P_x^4 + \tfrac{1}{6}\alpha_{111}P_x^6
</math>

This free energy has the shape of a double well potential with two free energy minima at {{math|1=''P{{sub|x}}'' = ''P{{sub|s}}''}}, the spontaneous polarization.  We find the derivative of the free energy, and set it equal to zero in order to solve for {{mvar|P<sub>s</sub>}}:

<math display=block>\begin{align}
  \frac{\partial \Delta E}{\partial P_x} &= \alpha_0(T-T_0)P_x + \alpha_{11}P_x^3 + \alpha_{111}P_x^5 \\[4pt]
  \implies 0 = \frac{\partial \Delta E}{\partial P_x} &= P_s \bigl[ \alpha_0(T-T_0) + \alpha_{11}P_s^2 + \alpha_{111}P_s^4 \bigr]
\end{align}</math>

Since the {{math|1=''P<sub>s</sub>'' = 0}} solution of this equation rather corresponds to a free energy ''maxima'' in the ferroelectric phase, the desired solutions for {{mvar|P<sub>s</sub>}} correspond to setting the remaining factor to zero:
<math display=block>
\alpha_0(T-T_0) + \alpha_{11}P_s^2 + \alpha_{111}P_s^4 = 0
</math>

whose solution is:

<math display=block>P_s^2 = \frac{1}{2\alpha_{111}} \left[-\alpha_{11} \pm \sqrt{\alpha_{11}^2 + 4\alpha_0\alpha_{111} (T_0-T)} \;\right]</math>

and eliminating solutions which take the square root of a negative number (for either the first or second order phase transitions) gives:

<math display=block>P_s = \pm \sqrt{\frac{1}{2\alpha_{111}} \left[-\alpha_{11}+\sqrt{\alpha_{11}^2 + 4\alpha_0\alpha_{111} (T_0-T)} \;\right]}</math>

If <math>\alpha_{11}=0</math>, the solution for the spontaneous polarization reduces to:

<math display=block>P_s = \pm\sqrt[4]{\frac{\alpha_0 (T_0-T)}{\alpha_{111}}}</math>

The hysteresis loop ({{mvar|P<sub>x</sub>}} versus {{mvar|E<sub>x</sub>}}) may be obtained from the free energy expansion by including the term {{mvar|&minus;E<sub>x</sub>P<sub>x</sub>}} corresponding to the energy due to an external electric field {{mvar|E<sub>x</sub>}} interacting with the polarization {{mvar|P<sub>x</sub>}}, as follows:

<math display=block>
\Delta E = \tfrac{1}{2} \alpha_0(T-T_0)P_x^2 + \tfrac{1}{4} \alpha_{11}P_x^4 + \tfrac{1}{6} \alpha_{111}P_x^6 - E_x P_x
</math>

We find the stable polarization values of {{mvar|P<sub>x</sub>}} under the influence of the external field, now denoted as {{mvar|P<sub>e</sub>}}, again by setting the derivative of the energy with respect to {{mvar|P<sub>x</sub>}} to zero:
<math display=block>\begin{align}
  \frac{\partial \Delta E}{\partial P_x} &= \alpha_0(T-T_0)P_x + \alpha_{11}P_x^3 + \alpha_{111}P_x^5 - E_x = 0 \\[4pt]
  E_x &= \alpha_0(T-T_0)P_e + \alpha_{11}P_e^3 + \alpha_{111}P_e^5
\end{align}</math>

Plotting {{mvar|E<sub>x</sub>}} (on the X axis) as a function of {{mvar|P<sub>e</sub>}} (but on the Y axis) gives an S-shaped curve which is multi-valued in  {{mvar|P<sub>e</sub>}} for some values of {{mvar|E<sub>x</sub>}}.  The central part of the 'S' corresponds to a free energy [[Second derivative test|local maximum]] (since <math>\tfrac{\partial^2 \Delta E}{\partial P_x^2}<0</math> ).  Elimination of this region, and connection of the top and bottom portions of the 'S' curve by vertical lines at the discontinuities gives the hysteresis loop of internal polarization due to an external electric field.