Past exam of the condensed matter physics major option course of the University of Oxford / 2016 / 2

Under spatial inversion the electric polarization, a polar vector, changes from $\mathbf{P}$ to $-\mathbf{P}$. Neumann's principle requires an equilibrium property of a phase retaining inversion symmetry to be unchanged. Thus $\mathbf{P}=-\mathbf{P}$, forcing $\mathbf{P}=0$: a centrosymmetric phase cannot itself be ferroelectric. A transition can, of course, remove that symmetry.

For an inversion-symmetric parent phase the Landau theory free energy satisfies $F(\mathbf{P})=F(-\mathbf{P})$. Every polynomial term of odd total degree changes sign and must have zero coefficient. Restricting to one fixed direction gives an even scalar expansion in $P$.

For the full prism point group, a threefold rotation eliminates any invariant basal component of a polar vector, while the horizontal mirror eliminates its axial component. Therefore the phase preserving all of $D_{3h}$ is nonpolar, despite lacking inversion. A polar phase would have to break some of these symmetries.

The indicated six vertex directions comprise three basal directions and their horizontal-mirror partners. Crucially, this orbit does not contain the negatives of those vectors. Choose $z$ along the prism and $x$ toward one basal vertex. The cubic polynomialis invariant under the prism's rotations and mirrors. Along a vertex direction $\mathbf{P}=P\hat{\mathbf{n}}$ with a nonzero basal component, it is proportional to $P^3$, so odd scalar terms are symmetry-allowed. Along the prism axis the horizontal mirror instead sends $P_z\mapsto -P_z$, forcing an even free energy. Merely omitting inversion is insufficient to decide either polarity or the allowed invariants.

Differentiating the Landau theory free energy givesIts stationary values arewhen the discriminant is nonnegative. At $P_0$, the curvature is $a(T-T_c)$: a local minimum above $T_c$ and a local maximum below it. Exactly at $T_c$, $F^{\prime \prime \prime }(0)=-2b\ne 0$, so the origin is a stationary inflection point, not a stable minimum.

At a nonzero stationary point, eliminating $T-T_c$ yields $F^{\prime \prime }=P(2cP-b)$. Between $T_c$ and the upper spinodal, $P_{+}$ is a minimum and $P_{-}$ a maximum. Below $T_c$, $P_{-}<0$ becomes another local minimum, while $P_{+}>0$ is the lower minimum. Thus loss of local stability at $T_c$ is distinct from the equilibrium transition temperature.

The two nonzero stationary points merge when their discriminant vanishes, giving the upper spinodalAt a nonzero stationary point, $a(T-T_c)=bP-c{P}^2$, so its free energy reduces to $F=P^3(b/6-cP/4)$. Equality with $F(0)=0$ therefore requires $P=2b/(3c)$. Substitution givesBoth have positive curvature $2b^2/(9c)$ at coexistence. The intervening maximum is at $P=b/(3c)$. For $T_2<T<T_1$ the positive-polarization minimum is metastable; for $T_c<T<T_2$ the zero-polarization minimum is metastable.

In equilibrium without hysteresis, the system selects the lower free-energy minimum. At $T_2$ the order parameter jumps between $0$ and $2b/(3c)$, which identifies a first-order phase transition. Differentiating the minimized free energy does not add a $dP/dT$ contribution because $F^{\prime }=0$. Thus the polarization-dependent entropy is $S=-a{P}^2/2$, givingThe heat absorbed on heating from the ordered to the disordered phase isThe same amount is released on cooling. If $\Delta S$ is defined in the opposite direction, the signed transition heat is correspondingly negative.