How does the approximate energy relate with the trial function in variational method?

Question

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?

Answer 1

I will continue with your formula for the energy of a normalized state $|\phi\rangle$ (i.e. $|\phi\rangle=\sum_{n=0}^\infty c_n|\psi_n\rangle$ with $\sum_{n=0}^\infty |c_n|^2=1$): $$\begin{align} E &= \langle\phi|\hat{H}|\phi\rangle \\ &= \sum_{n=0}^{\infty}|c_n|^2E_n \\ &= |c_0|^2E_0 + \sum_{n=1}^{\infty} |c_n|^2E_n \\ &= \left(1 - \sum_{n=1}^{\infty}|c_n|^2\right)E_0 + \sum_{n=1}^{\infty}|c_n|^2E_n \\ &= E_0 + \sum_{n=1}^{\infty}|c_n|^2 (E_n-E_0) \end{align}$$ and hence $$E-E_0 = \sum_{n=1}^{\infty}|c_n|^2 (E_n-E_0)$$

Now the prerequisite was that $|\phi\rangle$ differs from $|\psi_0\rangle$ only by $O(\epsilon)$. Therefore for all $n\ne 0$ we have $c_n\approx O(\epsilon)$ and hence $|c_n|^2\approx O(\epsilon^2)$. So we finally have $$E-E_0\approx O(\epsilon^2)$$