While studying the variational principle in Quantum Mechanics, I came across the following problem: I want to relate the difference between the exact ground-state wave function $\psi_0$ and an approximation $\phi$ with the difference between the exact energy $E_0$ and the approximate energy $E$. For instance, if I know that $\phi$ and $\psi_0$ differ by at most, say, $O(\epsilon)$ over all space, can I say that $E-E_0 \approx O(\epsilon)$ (or $O(\epsilon^2)$, $O(\epsilon^3)$, $\ldots$)?
It is intuitive to think that the closer $\phi$ and $\psi_0$ are, the better the energy approximation gets, but I am not able to compare them quantitatively.
Without loss of generality, suppose that both $\phi$ and $\psi_0$ are normalized. By writing $\phi$ as a linear combination of $\psi_n$, we get $$ \phi=\sum_{n=0}^{\infty} c_n\psi_n $$ such that $\sum_{n=0}^{\infty} |c_n|^2=1$.
Therefore $$ \hat H |\phi\rangle=\sum_{n=0}^{\infty} c_n\hat H|\psi_n\rangle = \sum_{n=0}^{\infty} c_nE_n|\psi_n\rangle $$
Hence $E= \langle{\phi|\hat H| \phi}\rangle=\sum_{n=0}^{\infty} |c_n|^2E_n$.
From here, it is easy to prove that $E \geq E_0$, but how can I compare them?