# 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. May 16, 2019 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 $V(x)$
– user137661
May 16, 2019 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). May 16, 2019 at 23:47
• @ryan 's answer Is correct provided $E$ is an eigenvalue. In general you should take the infimum. The latter exists as long as $H$ is bounded from below, which is generally assumed.
– lcv
Oct 31, 2019 at 1:00

There is no way of knowing how good or bad is the approximate eigenvalue that comes out of the Rayleigh-Ritz method. This is probably the biggest weakness of the method. Nevertheless, one can make the following qualitative observations.

Even if the ansatz is bad, the energy can be quite good. For instance, imagine a trial ground state of the form \begin{align} \vert \tilde 0\rangle = \vert 0\rangle + \alpha \vert 1\rangle, \end{align} i.e. an ansatz ground state $$\vert\tilde 0\rangle$$ where the exact ground state $$\vert 0\rangle$$ is contaminated by some fraction $$\alpha$$ of the first excited state. Then it follows immediately that \begin{align} E=\frac{\langle \tilde 0\vert H\vert\tilde 0\rangle}{\langle \tilde 0\vert\tilde 0\rangle}= \frac{E_0+\vert\alpha\vert^2 E_1}{1+\vert\alpha\vert^2} \approx E_0 \left(1+\vert\alpha\vert^2 \frac{(E_1 -E_0)}{E_0}\right) \end{align} so the error is “only” of a few percent if $$\vert\alpha\vert\approx 1/10$$ and $$\frac{(E_1 -E_0)}{E_0}$$ is not too large.

The more quantitative way of increasing confidence in $$E$$ is to increase the number of variational parameters. Thus, as a first step in estimating the ground state energy of a harmonic oscillator, we might consider the trial state \begin{align} \psi_1(x;\alpha)=\left\{\begin{array}{cc} (x-\alpha)^2(x+\alpha)^2&\hbox{if } \vert x\vert\le \alpha \\ 0&\hbox{otherwise}\, , \end{array}\right. \end{align} with $$\alpha$$ a variational parameter. Following minimization w/r to $$\alpha$$, the resulting energy is $$E=\frac{1}{2}\hbar\omega \sqrt{\frac{12}{11}}\approx 1.04 E_0$$ and is quite close to the exact value $$E_0$$ (even if $$\alpha=(33)^{1/4}\sqrt{\hbar/m\omega}$$ is not very close to the turning points of the function, showing that the function extends significantly beyond the classical turning points.)

Now, introduce a second variational parameter $$\beta$$ with an “improved” guess \begin{align} \psi_2(x;\alpha, \beta)=\left\{\begin{array}{cc} (x-\alpha)^2(x+\alpha)^2(1+\beta x^2)&\hbox{if } \vert x\vert\le \alpha \\ 0&\hbox{otherwise}\, . \end{array}\right. \end{align} which is still an even function of $$x$$ with no nodes. This improved guess leads to complicated simultaneous equations arising from $$\frac{\partial E(\alpha,\beta)}{\partial\beta}=0\, ,\qquad \frac{\partial E(\alpha,\beta)}{\partial\alpha}=0$$ but the final resulting value of the energy is $$E=\frac{1}{2}\hbar\omega \sqrt{\frac{26}{25}}\approx 1.02E_0$$.

Likewise, using $$(x^2-\alpha^2)^3$$ produces very little change in energy from the calculation using $$\psi_1(x;\alpha)$$, suggesting the original ansatz of $$\psi_1(x;\alpha)$$ produced a value of energy reasonably close to the exact value.

Of course this is no guarantee of anything, and it might well be that the family of increasingly complicated tests functions that one uses to see some “convergence” of the estimated $$E$$ are inherently inferior to others. Laughlin won a Nobel Prize by producing for the fractional quantum Hall effect a better trial wave function than the competition.

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 $E$. May 16, 2019 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 $E|\psi\rangle$. May 16, 2019 at 21:23