Arbitrary boundary conditions for spherical polar coordinates

In the lectures you learned that the Laplacian in 3D has the eigenfunctions $$ \Delta u_{nlm}(r,\theta, \phi) = \lambda_{nlm} u(r,\theta, \phi), $$ where $$ u_{nlm}(r,\theta,\phi) = R_{nl}(r) \tilde P_{lm}(\theta) \Phi_m(\phi), $$ where $\tilde P_{lm}(\theta) = C_{lm} P_{lm}(\cos \theta)$. As it is separable, it can be thought of a tensor product of three vectors living in separate spaces $1,2,3$, as $$ \ket{u_{nlm}} = \ket{R_{nl}}_1 \ket{\tilde P_{lm}}_2 \ket{\Phi_m}_3. $$ with $u_{nlm}(r, \theta, \phi) = \langle r \vert R_{nl} \rangle_1 \langle \theta \vert \tilde P_{lm} \rangle_2 \langle \phi \vert \Phi_m \rangle_3$. Since inner product on the joint space is defined as $$ \langle u \vert u’ \rangle := \int_0^\infty r^2 dr \int_0^\pi \sin\theta d \theta \int_0^{2\pi} d \phi ,u^*(r, \theta, \phi) u(r, \theta, \phi), $$ The individual factors can be normalised under the inner products $$ \langle R \vert R’ \rangle_1 := \int_0^\infty r^2 dr R^*(r) R’(r), $$ $$ \langle \tilde P \vert \tilde P’ \rangle_2 := \int_0^\pi \sin \theta d \theta \tilde{P}^*(\theta) \tilde{P}’(\theta), $$ and $$ \langle \Phi \vert \Phi’ \rangle_3 := \int_0^{2\pi} d\phi \Phi^*(r) \Phi’(r). $$ ${u_{nlm}}_{nlm}$ form a CONS (complete set of orthonomal basis) and therefore any solution can be written as $$ \psi(r, \theta, \phi) = \sum_{nlm} A_{nlm} R_{nl}(r) \tilde P_{lm}(\theta) \Phi_m(\phi), $$ or abstractly as $$ \ket{\psi} = \sum_{nlm} A_{nlm} \ket{R_{nl}}_1 \ket{\tilde P_{lm}}_2 \ket{\Phi_m}_3. $$ We can evaluate this abstract vector on a subset of variables, for example only for $r$, as: $$ \ket{\psi(r)} := \bra{r}_1\ket{\psi} = \sum_{nlm} A_{nlm} R_{nl}(r) \ket{\tilde P_{lm}}_2 \ket{\Phi_m}_3. $$ In the lectures, you had a simple boundary condition $u(R, \theta, \phi) = \cos(\theta)$. What if it is more complicated? Consider we have a boundary condition like $u(R, \theta, \phi) = f(\theta, \phi)$.

This can be written as an abstract vector in spaces $2, 3$ as $$ \ket{u(R)} = \sum_{nlm} B_{nlm} R_{nl}(R) R_{nl}(r) \ket{\tilde P_{lm}}_2 \ket{\Phi_m}_3. $$ The coefficient $B_{nlm}$ can be found by taking the inner product of $\ket{u(R)}$ with basis vectors for spaces $2, 3$, as $$ B_{nlm} = \frac{1}{R_{nl}(R)} \int_0^\pi \sin\theta d \theta \int_0^{2\pi} d \phi , \tilde P^*_{lm}(\theta) \Phi^*_m(\phi) f(\theta, \phi). $$ Then to obey the boundary conditions, we must have that $B_{nlm} = A_{nlm}$, such that $\psi(R, \theta, \phi) = u(R,\theta,\phi) = f(\theta, \phi)$.

For example, our boundary condition $u(R, \theta, \phi) = \cos(\theta)$ is proportional to $\tilde P_{10}(\theta) \Phi_m(\phi)$. Therefore $B_{nlm} = \delta_{l,1} \delta_{m,0}$.