Linearity of Schrödinger equation and perturbation theory

So, I was studying quantum mechanics and reached the point where perturbation theory is discussed. It is my first time in this topic, and something called my attention: it was said that we need perturbation theory when there is this hamiltonian that we cannot solve and them we can write:

$$H = H_0 + \lambda V$$

where $$H_0$$ is a hamiltonian that we can solve - or better put, is a hamiltonian that we KNOW how to solve -, and $$V$$ is another one that we can solve or not, that we do know how to solve or we do not, the question is that their sum, $$H$$, is the one that we cannot solve.

Ok, $$H$$ may be written in something way that is really hard to solve, or impossible, so we perturb the system and solve an easier problem. My question is: if we know how to solve $$H_0$$ and $$V$$, why we cannot solve $$H$$ = $$H_0 + \lambda V$$? The Schrödinger equation is linear.

What am I missing?

• Are you asserting that if we know the eigenfunctions of $H_0$ and $V$, then we can simply add them together to get the eigenfunctions of $H_0 + \lambda V$? – J. Murray Aug 29 at 20:51

When people say "The Schrödinger equation is linear." they mean that for a particular $$\hat{H}$$, given two (or more solutions) $$\Psi_i$$ to $$\hat{H}\Psi = i \hbar \frac{\mathrm{d}}{\mathrm{d}t} \Psi \,$$ then $$\Psi_\text{combination} = \sum_i c_i \Psi_i$$ with $$c_i \in \mathbb{C}$$ for all $$i$$ is also a solution.
They don't mean that given two different Hamiltonians $$\hat{H}_i$$ whose solutions are know you can trivially find solutions for $$\hat{H}_\text{combination} = \sum_i a_i \hat{H}_i$$.