= Solution
For $V(-x)=V(x)$, the <parity operator> $P\psi(x)=\psi(-x)$ commutes with the <Hamiltonian operator>. Also $P^2=I$, so its <eigenvalues> are $\pm1$ and the <energy eigenspaces> admit a <basis> of definite <parity>. More strongly, <nondegeneracy of one-dimensional bound states> says that each <bound state> <energy> here is nondegenerate: the <Wronskian> of two solutions at the same <energy> is constant by the <Schrödinger equation>, and decay at infinity makes it zero, forcing linear dependence. Thus each <bound state> itself is even or odd.
An <even function> obeys the <Neumann boundary condition> $\psi'(0)=0$; an <odd function> obeys the <Dirichlet boundary condition> $\psi(0)=0$. Write
$$
K_0=\frac1\hbar\int_0^{x_1}\chi(x)\,dx,\qquad \phi(E)=2K_0=\frac1\hbar\int_{-x_1(E)}^{x_1(E)}\sqrt{2m[V(x)-E]}\,dx.
$$
Smooth reflection symmetry gives $\chi'(0)=0$, while $K'(0)=-\chi(0)/\hbar$. Hence, for the central <WKB approximation> from part (c),
$$
\psi_R(0)\simeq\frac{F}{\sqrt{\chi(0)}}\bigl[2\cos\theta\,e^{K_0}+\sin\theta\,e^{-K_0}\bigr],
$$
$$
\psi_R'(0)\simeq\frac{F\sqrt{\chi(0)}}\hbar\bigl[-2\cos\theta\,e^{K_0}+\sin\theta\,e^{-K_0}\bigr].
$$
The two <boundary conditions> consequently give the <WKB quantization condition for a symmetric double well>:
$$
\boxed{\tan\theta(E)=C_\pm e^{\phi(E)},\qquad C_+=2\ \text{for even parity},\qquad C_-=-2\ \text{for odd parity}.}
$$
The full <wavefunction> is $\psi_+(x)=\psi_R(|x|)$ in the even case and $\psi_-(x)=\operatorname{sgn}(x)\psi_R(|x|)$ in the odd case, with the value at zero fixed continuously. The relevant <boundary condition> makes these extensions differentiable at zero. Both <action integrals> $\theta,\phi$ depend on $E$ through the integrands and the <classical turning points>.
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