A.7 Accuracy of the variational method

This note has a closer look at the accuracy of the variational method.

Any approximate ground state solution $\psi$ may always be written as a sum of the eigenfunctions $\psi_1,\psi_2,\ldots$:

$$\psi = c_1 \psi_1 + \varepsilon_2 \psi_2 + \varepsilon_3 \psi_3 + \ldots$$

where, if the approximation is any good at all, the coefficient $c_1$ of the ground state $\psi_1$ is close to one, while $\varepsilon_2, \varepsilon_3, \ldots$ are small.

The condition that $\psi$ is normalized, $\langle\psi\vert\psi\rangle=1$, works out to be

$$1 = \langle c_1\psi_1 +\varepsilon_2\psi_2+\ldots \vert ... ...rangle = c_1^2 + \varepsilon_2^2 + \varepsilon_3^2 + \ldots$$

since the eigenfunctions $\psi_1,\psi_2,\ldots$ are orthonormal.

Similarly, the expectation energy $\langle{E}\rangle=\langle\psi\vert H\psi\rangle$ works out to be

$$\langle E\rangle = \langle c_1\psi_1 +\varepsilon_2\psi_2+... ...^2 E_1 + \varepsilon_2^2 E_2 + \varepsilon_3^2 E_3 + \ldots$$

Eliminating $c_1^2$ using the normalization condition above gives

$$\langle E\rangle = E_1 + \varepsilon_2^2 (E_2-E_1) + \varepsilon_3^2 (E_3-E_1) + \ldots$$

One of the things this expression shows is that any approximate wave function (not just eigenfunctions) has more expectation energy than the ground state $E_1$. All other terms in the sum above are positive since $E_1$ is the lowest energy value.

The expression above also shows that while the deviations of the wave function from the exact ground state $\psi_1$ are proportional to the coefficients $\varepsilon_2,\varepsilon_3,\ldots$, the errors in energy are proportional to the squares of those coefficients. And the square of any reasonably small quantity is much smaller than the quantity itself. So the approximate ground state energy is much more accurate than would be expected from the wave function errors.

Still, if an approximate system is close to the ground state energy, then the wave function must be close to the ground state wave function. More precisely, if the error in energy is a small number, call it $\varepsilon^2$, then the amount $\varepsilon_2$ of eigenfunction $\psi_2$ “polluting” approximate ground state $\psi$ must be no more than $\varepsilon/\sqrt{E_2-E_1}$. And that is in the worst case scenario that all the error in the expectation value of energy is due to the second eigenfunction.

As a measure of the average combined error in wave function, you can use the magnitude or norm of the combined pollution:

$$\vert\vert\varepsilon_2 \psi_2 + \varepsilon_3 \psi_3 + \l... ...\vert\vert = \sqrt{\varepsilon_2^2+\varepsilon_3^2 +\ldots}$$

That error is no more than $\varepsilon/\sqrt{E_2-E_1}$. To verify it, note that

$$\varepsilon_2^2 (E_2-E_1) + \varepsilon_3^2 (E_2-E_1) + \l... ...-E_1) + \varepsilon_3^2 (E_3-E_1) + \ldots = \varepsilon^2.$$

(Of course, if the ground state wave function would be degenerate, $E_2$ would be $E_1$. But in that case you do not care about the error in $\psi_2$, since then $\psi_1$ and $\psi_2$ are equally good ground states, and $E_2-E_1$ becomes $E_3-E_1$.)

The bottom line is that the lower you can get your expectation energy, the closer you will get to the true ground state energy, and the small error in energy will reflect in a small error in wave function.