# How far off can a variational test function be if it results in very precise estimate of an eigenvalue?

I'm reading about variational principle in quantum mechanics, which basically states that smallest (algebraically) eigenvalue $E_0$ of an Hermitian operator $H$ can be calculated as the minimum of the expression

$$\tilde E_0=\frac{\left\langle\psi\right|H\left|\psi\right\rangle}{\langle\psi\mid\psi\rangle}$$

over the space of test functions $\psi$ in the domain of $H$. But there is a caveat, as said on this page:

There is no a priori guarantee whatsoever that the trial state ${|\psi(\mathbf{x}_*)\rangle}$ corresponding to the minima of the energy expectation value ${E(\mathbf{x})}$ at ${\mathbf{x}_*}$ has any resemblance to the ground state.

I wonder, how far off can such a trial state be, if it gives some high precision estimate of the eigenvalue? What is an example of a really "bad" case where the test function is very different from the true eigenfunction, but still estimates the eigenvalue very precisely (i.e. eigenvalue is correct up to $N$ decimal places while test function's precision as estimate of eigenfunction is much worse)?

• For example, if $H$ has a second eigenvalue that is very close to the smallest, we might get into trouble. – Ben Grossmann Mar 1 '17 at 12:58
• When the boys in quantum chemistry calculate the ground states of molecules etc, they do expand their trial wavefunctions in the incomplete, however large basis of Gaussians centered where appropriate, compute the energy and then variate, results for the energy are good, but there is no chance that this bunch of Gaussians is even nearly close to the true eigen function. – Kiryl Pesotski Mar 1 '17 at 13:10