Category: Physics

Anharmonic oscillator II

Remember that the perturbative series for the anharmonic oscillator is

Anharmonic oscillator I

$\displaystyle E_0(\epsilon) = \frac{1}{2} + \frac{3}{4}\epsilon - \frac{21}{8}\epsilon^2 + \frac{333}{16}\epsilon^3 + \mathcal{O}(\epsilon^4)$

We used Padé approximants to compute the ground state energy $E_0(\epsilon)$ . Now we aim to calculate $E_0(\epsilon)$ using Borel summation. The (formal) Borel sum is given by

$\displaystyle \int_{0}^{\infty} e^{-t} \sum_{n=0}^\infty \frac{a_n}{n!} (\epsilon t)^n \, dt.$

Using the first three coefficients of the perturbative series, the truncated Borel transform is approximated by

$\displaystyle \mathcal{B}E_0(t) \approx \frac{1}{2} + \frac{3}{4} t - \frac{21}{16} t^2,$

so the truncated Borel sum writes

$\displaystyle E_0(\epsilon) \approx \int_0^\infty e^{-t} \left(\frac{1}{2} + \frac{3}{4} \epsilon t - \frac{21}{16} \epsilon^2 t^2 \right) dt = \frac{1}{2} + \frac{3}{4} \epsilon - \frac{21}{8} \epsilon^2.$

since

$\displaystyle \int_0^\infty \frac{1}{2} e^{-t} \, dt = \frac{1}{2}$

$\displaystyle \int_0^\infty e^{-t} \cdot \frac{3}{4} \epsilon \, t \, dt = \frac{3}{4} \epsilon$

$\displaystyle \int_0^\infty e^{-t} \cdot \left( -\frac{21}{16} \epsilon^2 t^2 \right) dt = -\frac{21}{8} \epsilon^2$

This doesn’t really seem like progress, since the first terms of the Borel sum are identical to those of the perturbative expansion. The perturbative expansion $E_0(\epsilon) = \sum_{n=0}^\infty a_n \epsilon^n$ above has coefficients that grow factorially. Bender and Wu showed that for large $n$ ,

$\displaystyle a_n \sim -r\, (-1)^n \left(\frac{3}{2}\right)^n \Gamma\!\left(n+\tfrac12\right), \qquad r>0.$

Since

$\displaystyle \Gamma\!\left(n+\tfrac12\right) \sim n!\, n^{-1/2},$

$formule avec n! en rouge et gras$

thus the series diverges for all $\epsilon \neq 0$ .

Borel summation improves convergence by dividing out this factorial growth.

In summary, even with just a few terms, Borel summation correctly recovers the perturbative results for the anharmonic oscillator then It turns the divergent series into a well-defined and useful result.

December 11, 2025
Anharmonic oscillator I

We would like to illustrate the use of Padé approximants in the context of the anharmonic oscillator. A harmonic oscillator is an oscillating system that experiences a restoring force. Anharmonicity is the deviation of a system from being a harmonic oscillator. The anharmonic oscillator is described by the following differential equation:

$\displaystyle \left(-\frac{d^2}{dx^2} + x^2 + x^4\right) \phi(x) = E_n \phi(x)$

This differential equation is very hard to solve exactly. In order to obtain an approximate solution, we can use perturbation theory. By inserting a small dimensionless parameter $\epsilon$ , the equation becomes:

$\displaystyle \left(-\frac{d^2}{dx^2} + x^2 + \epsilon x^4\right) \phi(x) = E_n(\epsilon) \phi(x)$

If we are interested in the energy levels, the basic idea is to write $E_n(\epsilon)$ as a geometric series. For the ground state $E_0$ we write:

$\displaystyle E_0(\epsilon) = \sum_{n=0}^{\infty} E_{0,n} \epsilon^n$

The solution is a divergent perturbation series (C.M. Bender and T.T. Wu, Phys. Rev. 184, 1969):

$\displaystyle E_0(\epsilon) = \frac{1}{2} + \frac{3}{4}\epsilon - \frac{21}{8}\epsilon^2 + \frac{333}{16}\epsilon^3 + \mathcal{O}(\epsilon^4)$

Let’s calculate the $P(1,1)$ approximant of $E_0$ using the techniques presented in post Computing Padé approximants:

$\displaystyle P(1,1) = \frac{A_0 + A_1 \epsilon}{1 + B_1 \epsilon} = C_0 + C_1 \epsilon + C_2 \epsilon^2$

Solving the linear systems for computing Padé approximants sequentially leads to:

$\displaystyle C_1 B_1 = - C_2 \implies B_1 = -\frac{C_2}{C_1}$

$\displaystyle \frac{3}{4} B_1 = \frac{21}{8} \implies B_1 = \frac{84}{24} = \frac{7}{2}$

$\displaystyle \begin{pmatrix} C_0 & 0 \\ C_1 & C_0 \\ \end{pmatrix} \begin{pmatrix} 1 \\ B_1 \end{pmatrix} = \begin{pmatrix} A_0 \\ A_1 \end{pmatrix}$

$\displaystyle \begin{pmatrix} \frac{1}{2} & 0 \\ \frac{3}{4} & \frac{1}{2} \\ \end{pmatrix} \begin{pmatrix} 1 \\ \frac{7}{2} \end{pmatrix} = \begin{pmatrix} \frac{1}{2} \\ \frac{3}{4} + \frac{7}{4} \end{pmatrix}$

$\displaystyle P(1,1)_{E_0} = \frac{\frac{1}{2} + \left(\frac{3}{4} + \frac{7}{4}\right)\epsilon}{1 + \frac{7}{2}\epsilon}$

Setting $\epsilon = 1$ (to recover the initial differential equation) we have:

$\displaystyle P(1,1)_{E_0} = \frac{\frac{1}{2} + \frac{3}{4} + \frac{7}{4}}{1 + \frac{7}{2}} = 0.6666$

The $P(4,4)_{E_0} = 1.3838$ and the exact solution is 1.3924.

The relative error (definition here) of the approximant $P(4,4)_{E_0}$ is:

$\displaystyle \text{Relative error} = \frac{1.3838 - 1.3924}{1.3924} = -0.0062$

It’s therefore interesting to note that in order to solve a very difficult differential equation, we can use perturbation theory. In the case of the anharmonic oscillator, the perturbative series is divergent. Using Padé approximants, we can make use of a divergent perturbative series and approximate the exact answer arbitrarily.

September 18, 2025