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Imposing those two boundary conditions on the ansatz gives, with $\kappa =(\epsilon -{\epsilon }_0)/(\epsilon +2{\epsilon }_0)$,Thus the internal field is uniform and the external correction has dipolar angular form.

For a linear dielectric, $D=\epsilon E$ (equivalently $D={\epsilon }_{0}E+P$). With no free charge, Gauss law gives $\nabla \cdot D=0$. Across an interface without free surface charge, $D_n$ is continuous; electrostatic Faraday law makes $E_t$ continuous.

Writing $x=m^2$, nonzero equilibria satisfy $a(T-T_c)+bx+c{x}^2=0$. For $b<0$, coexistence occurs at $x=-3b/(4c)$ and $T-T_c=3b^2/(16ac)$, so magnetization jumps and the transition is first order.

For $b=0$, $m=0$ above $T_c$ and $m^4=a(T_c-T)/c$ below: this is the tricritical continuous case. Below $T_c$,so the extra heat capacity is $a^{3/2}T/[4\sqrt{c(T_c-T)}]$ and diverges on approach from below; above it is absent.

Minimization gives $m=0$ for $T>T_c$ and $m=\pm \sqrt{a(T_c-T)/b}$ below. ThusUsing $S=-F_T$ and $C=-T{F}_{TT}$, entropy is continuous but the low-temperature heat capacity has the additional $T{a}^2/(2b)$. The jump is $T_c{a}^2/(2b)$, so the transition is second order.

At fixed temperature, volume, and particle number, spontaneous changes satisfy $\Delta F\le 0$ because $F=E-TS$ accounts for heat exchanged with the reservoir. Equilibrium therefore minimizes the Helmholtz free energy.

A first-order phase transition has a discontinuous first derivative of the equilibrium free energy, such as entropy or volume, and latent heat. A second-order transition has continuous first derivatives but a discontinuous or divergent second derivative.

Filled bands separated by gaps explain the distinction between electrical insulators and conductors: a completely filled band cannot change its net crystal momentum under a weak field, whereas a partially filled band supports current.

Nondegenerate perturbation theory fails where free energies related by a reciprocal vector coincide, at $k=l\pi /a$. Diagonalizing the two-state block places the gap centre at ${\hslash }^2(l\pi /a{)}^2/(2m)+\lambda U_0$ and gives width $2\lambda |U_l|+O({\lambda }^2)$.

$E_0(k)={\hslash }^2{k}^2/(2m)$. Periodic boundary conditions give $k=2\pi n/L$. Orthogonality of plane waves gives $⟨k|U|k\prime ⟩=U_l$ when $k\prime -k=2\pi l/a$, and zero otherwise.

Move to a rotating frame. With detuning $\Delta ={\omega }_1-{\omega }_0+\omega$ (changing sign with the phase convention), the constant two-level Hamiltonian has generalized Rabi frequency $\Omega =\sqrt{{\Delta }^2+4v^2}$. Starting in state zero,The transition envelope is maximized on resonance, $\Delta =0$, namely $\omega ={\omega }_0-{\omega }_1$ for the matrix phases printed in the question.

First-order time-dependent perturbation theory givesThe lowest nontrivial transition probability is $|c_1(t){|}^2$, with errors of higher order in $V$.

The covering graph has at least one primitive cycle of lengths two and four and two inequivalent primitive cycles of length five. Thus the guaranteed minimum numbers of distinct orbits are respectively $1,1,2$.

Order a period-three orbit as $a<b<c$. Its images produce two subintervals $I_0,I_1$ with covering relations $I_0\to I_0,I_1$ and $I_1\to I_0$. Any closed walk in this directed graph gives, by the interval-covering lemma and the intermediate value theorem, a fixed point of the corresponding iterate. Primitive closed walks of every length exist, giving an orbit of every positive period; this is the period-three case of Sharkovsky theorem.

Since $y=u/x$, two real independent approximations are

The Liouville-Green approximation givesHereThe control ratio $|q\prime |q^{-3/2}=O(x^{-1/2})$ tends to zero, verifying consistency.

The standard removal of the first derivative, $y=u/x$, gives $u\prime \prime +q(x)u=0$ with $q(x)=x^{-1}-x^{-2}$.