In the two constant regions define
$$
\omega_\pm=\sqrt{k_z^2+a_\pm^2|\mathbf k|^2}.
$$
The normalized <positive-frequency solution>[positive-frequency modes] are
$$
\boxed{u_\pm(\eta)=\frac{e^{-i\omega_\pm\eta}}{\sqrt{2\omega_\pm}}},
$$
up to the common spatial Fourier normalization. They satisfy the unit Wronskian condition
$$
i(u_\pm^*u_\pm'-u_\pm u_\pm^{*\prime})=1,
$$
which is the mode form of the <Klein-Gordon inner product>. Because the mode equation contains no delta function at $\eta=0$, both $\Phi$ and $\Phi'$ are continuous there.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 311 / 4 / b / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-311/4/b/ii/solution