Past exam of the further quantum theory course of the University of Oxford / 2023 / 3 / d / 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 $\pm 1$ 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 ${\psi }^{\prime }(0)=0$; an odd function obeys the Dirichlet boundary condition $\psi (0)=0$. WriteSmooth reflection symmetry gives ${\chi }^{\prime }(0)=0$, while $K^{\prime }(0)=-\chi (0)/\hslash$. Hence, for the central WKB approximation from part (c),The two boundary conditions consequently give the WKB quantization condition for a symmetric double well:The full wavefunction is ${\psi }_{+}(x)={\psi }_R(|x|)$ in the even case and ${\psi }_{-}(x)=\mathrm{sgn}(x){\psi }_R(|x|)$ in the odd case, with the value at zero fixed continuously. The relevant boundary condition makes these extensions differentiable at zero. Both action integrals $\theta ,\varphi$ depend on $E$ through the integrands and the classical turning points.