### Solution

A1  ?? Write down the conditions for a local energy minimum, if energy is given by equation $(1)$. For which values of $T$ the spontaneous polarization would be possible?

Conditions of local minimum for a function $f(x)$ are:$\begin{cases}f'(x)=0,\\f''(x) > 0.\end{cases}$In our case:$\begin{cases}\dfrac{\partial E}{\partial P}=\alpha(T-T_0)P+\beta P^3+\gamma P^5,\\\dfrac{\partial^2 E}{\partial P^2}=\alpha(T-T_0)+3\beta P^2+5\gamma P^4,\end{cases}$so the conditions of local minimum transform to the form:

Answer: $\begin{cases}P\cdot\left[\alpha(T-T_0)+\beta P^2+\gamma P^4\right]=0\\\alpha(T-T_0)+3\beta P^2+5\gamma P^4 > 0\end{cases}$

For segnetoelectric material to have spontaneous polarization there must be $P\ne0$. This is only possible for $T < T_0$. In fact:

• If $T\ge T_0$, then the potential only has one minimum at $P=0$. The first equation has only one soultion which doesn't fit into the second equation.
• If $T < T_0$, then the potential has two local minima. The first equation has a total of three solutions, one of which ($P=0$) doesn't fit into the second equation and in fact corresponds to a local maximum.

Answer: Spontaneous polarization is possible for $T < T_0$

A2  ?? Derive an equation for equilibrium values of spontaneous polarization, $P_0$.

We are solving the equation:$-\alpha(T_0-T)+\beta P^2+\gamma P^4=0.$This is a quadratic equation for $P^2$ and its soultions are:$P_0^2=\frac{1}{2\gamma}\left[-\beta\pm\sqrt{\beta^2+4\alpha\gamma(T_0-T)}\right].$As $P^2 > 0$, we are left with only $+$. So:

Answer: $P_0=\pm\sqrt{\frac{1}{2\gamma}\left[\sqrt{\beta^2+4\alpha\gamma(T_0-T)}-\beta\right]}$

A3  ?? Expression from the previous task simplifies significantly in the limit $T\to T_0$. Find the values of $P_0$ in this limit. Express your answer in terms of $\alpha$, $\beta$ and $\Delta T=T_0-T$.

In the limit $T\to T_0$ the expression under the square root can be expanded into Taylor series as:\begin{aligned}P_0&=\pm\sqrt{\frac{1}{2\gamma}\left[\beta\sqrt{1+\frac{4\alpha\gamma(T_0-T)}{\beta^2}}-\beta\right]}=\\&=\pm\sqrt{\frac{\beta}{2}\left[1+\frac{2\alpha\gamma(T_0-T)}{\beta^2}+\ldots-1\right]}=\\&=\pm\sqrt{\frac{\alpha\Delta T}{\beta}}.\end{aligned}

Answer: $P_0=\pm\sqrt{\frac{\alpha\Delta T}{\beta}}$

B1  ?? Derive an expression for the electric susceptibility of a spontaneously polarized ferroelectric.

In the limit we are using $\gamma\ll \beta^2/\alpha$, so one can conclude that on scales of $P\sim\sqrt{\alpha/\beta}$ the last term in Taylor series $(1)$ can be neglected. Then the energy in a presence of electric field is:$W=-EP-\frac{1}{2}\alpha(T_0-T)P^2+\frac{1}{4}\beta P^4\implies\\\implies \frac{\partial W}{\partial P}=-E-\alpha\Delta T\cdot P+\beta P^3.$To find $\chi$ we substitute $P=\sqrt{\dfrac{\alpha\Delta T}{\beta}}+\chi E$ into the equation $\dfrac{\partial W}{\partial P}=0$ and expand ot to the first order of magnitude by $E$:\begin{aligned}0&=-E-\alpha\Delta T\left(\sqrt{\dfrac{\alpha\Delta T}{\beta}}+\chi E\right)+\beta \left(\sqrt{\dfrac{\alpha\Delta T}{\beta}}+\chi E\right)^3=\\&=-E-\alpha\Delta T\left(\sqrt{\dfrac{\alpha\Delta T}{\beta}}+\chi E\right)+ {\alpha\Delta T}\left(\sqrt{\dfrac{\alpha\Delta T}{\beta}}+3\chi E\right)=\\&=-E+2\alpha \Delta T\chi E\implies \chi=\frac{1}{2\alpha\Delta T}.\end{aligned}

Answer: $\chi=\frac{1}{2\alpha\Delta T}$