Suppose we have a tensor product Hilbert space $\mathcal{H} = \mathcal{H}_1 \otimes \mathcal{H}_2$ and we have a Hamiltonian defined thereon which is given by $H = H_{1e} \otimes I+ H_1 \otimes H_2$. Suppose also we know an eigenbasis $|E_{2i} \rangle$ of $H_2$. What can we say then about the overall eigenbasis of $H$? Can we "try" eigenstates of the form $|\psi \rangle |E_{2i} \rangle$, and so find that $$H|\psi \rangle |E_{2i} \rangle = (H_{1e}+ E_{2i}H_1\otimes I)|\psi \rangle |E_{2i} \rangle$$ and then solve this eigenvalue problem in $\mathcal{H}_1$ by considering the eigenstates $|\psi \rangle$ of $H_{1e}+ E_{2i}H_1$ in $\mathcal{H_1}$ (we see immediately that $|\psi \rangle |E_{2i} \rangle$ will be an $H$ eigenstate in this case)? My chief concern, and the gist of this question, is how do I know I am not missing any possible $H$ eigenstates with this "trial" method?
For example/context, in the case of a fully separable Hamiltonian, say $H = H_{1e} \otimes I+ I \otimes H_{2e}$, one can verify by explicit calculation that the product basis $|E_{1j} \rangle|E_{2i} \rangle$ are eigenstates. Then (and this is the key step which I don't know above), one can use the particular definition of a tensor product space which says that it is the vector space spanned by any product basis (a basis formed by products of bases of each $\mathcal{H}_i$). Thus, since the $|E_{1j} \rangle$ and $|E_{2i} \rangle$ are eigenkets of the self-adjoint operators $H_{1e}$ and $H_{2e}$, respectively, we have a complete eigenbasis of $H$.
In case this question is not clear, it is motivated by Problem 7.10 in Ballentine's quantum text, which reads:
Two spin 1/2 particles interact through the spin-dependent potential $V(r)$ = $V_1(r)+V_2(r)σ(1)·σ(2)$. Show that the equation determining the bound states can be split into two equations, one having the effective potential $V_1(r)+V_2(r)$ and the other having the effective potential $V_1(r)-3V_2(r)$.
Of course, the idea here is that we want to use the eigenkets of $σ(1)·σ(2)$ which were obtained in an earlier problem, but my concern is around using trial eigenstates of the form $|{\psi_0}\rangle|{1/2,1/2,0,0}\rangle$ and $|{\psi_1}\rangle|{1/2,1/2,1,0/\pm 1}\rangle$ ($|j_1,j_2,J,M\rangle$ is my notation for total angular momentum eigenstates).
P.S. I would greatly appreciate it if you can comment on why Ballentine mentions bound specifically ("that the equation determining the bound states can be split into two equations" -- i.e. why bound in particular) when this seems to be general, whether we are looking for bound or unbound states of $H$.
