# How to ascertain that the Rayleigh-Ritz variational method gives the exact value of the ground state energy?

So the Rayleigh-Ritz variational method can be used to calculate the ground state energy of a quantum system. If $$\phi(x)$$ is a suitable (square integrable) and normalised function of the coordinates of a system, then we can set $$E(\phi) = \langle\phi\rvert H\lvert\phi\rangle$$ We can obtain an upper bound on the ground state energy by minimising $$E(\phi)$$, the lower the value that we get, the better the approximation. Now my question, how can one know that this method, when possible, gives us the exact value of the energy if we do not know that value beforehand. Is there a way to check that we got an exact answer or just an approximation, without knowing the exact value beforehand?

• The Euler-Lagrange equation of the functional $\langle\phi|H|\phi\rangle/|\langle\phi|\phi\rangle$ is the eigenvalue equation. (Think of the eigenvalue as a Lagrange multiplier enforcing $\langle\phi|\phi\rangle=1$.) Since minima are critical points, you find that the minimizer is an eigenfunction. Since you're taking the minimum, this is the ground state. – Ryan Unger May 16 at 21:09
• @RyanUnger but the Euler Lagrange equation requires that the function has continuous second derivatives... Then it would only work if the eigenfunctions are in $C^2$, so that would not work for a discontinuous $V(x)$ – S V May 16 at 21:54
• Sure, you need some kind of elliptic theory if you want the EL equation to make sense classically. If you want technical details (which belong on math SE, not physics SE), check out Section XIII.1 in Reed--Simon (vol 4). – Ryan Unger May 16 at 23:47

I am not aware of any theorem that can tell you how close you are to the ground state energy. However,the energy fluctuations $$\Delta^2 E=\langle \psi| H^2 |\psi \rangle - \langle \psi| H |\psi \rangle^2$$ can be used as criteria, if the correct ground state is guessed then $$\Delta^2 E=0$$.
• This holds for any eigenstate, not only for the ground state, and for those you can check by simply applying $H$ to $\left| \psi \right>$ and seeing whether the result is $E \left| \psi \right>$ for some real number $E$. – Sebastian Riese May 16 at 21:13
• I agree that it works for all eigenstates, yet it gives an idea of how close you are to an eigenstate. I also thought about applying $H$ to $|\psi\rangle$, but when you do this unless you get the exact eigenstate the result is not $E|\psi\rangle$. – lakehal May 16 at 21:23