next up previous
Next: Convergence of the Fourier Up: No Title Previous: The Bessel function and

Borel Summation

Sometimes it turns out that divergent sums have a meaning in physics. For example, if one adds a correction to a potential in the Schrödinger equation, then it may be possible to compute a few terms in a perturbative expansion (i.e. an expansion where the correction to the potential is treated as a ``small'' perturbation) near the original potential. It may also be possible to estimate the behavior of the n'th order in the perturbative series, and the resulting series then often turns out to be divergent. However, there does exist a perfectly reasonable exact answer to this problem, provided the additional potential is sufficiently well behaved.

In some cases the exact answer can be obtained from the divergent series by a Borel summation of this series. The Borel sum can be defined the following way: Consider the series

\begin{displaymath}\Sigma=\sum_n a_n,
\end{displaymath} (63)

which we assume to be divergent. Making use of the integral representation of n! we have

\begin{displaymath}\Sigma=\sum_n n!\frac{a_n}{n!}=\sum_n\frac{a_n}{n!}\int_0^\infty dt
e^{-t} t^n .
\end{displaymath} (64)

This step is perfectly valid. The Borel summation consists in interchanging the sum and the integral. This procedure is not valid for the divergent sum from a rigorous point of view, but leads to the following definition of the Borel sum of $\Sigma$,

\begin{displaymath}\Sigma_{\rm Borel}=\int_0^\infty dt e^{-t}\sum_n\frac{a_n}{n!}t^n.
\end{displaymath} (65)

If the series $\sum a_nt^n/n!$ converges, the Borel sum is thus well defined. If this is not the case, the method can be repeated.

As an example, let us consider the sum

\begin{displaymath}\sum_0^\infty n!x^n,
\end{displaymath} (66)

which is obviously divergent for all $x\neq 0$. In the case of the Schrödinger equation one could e.g. imagine that x is a parameter which measures the strength of the additional piece in the potential, so the series above would then be the result of a perturbative expansion.

Proceeding as before, the Borel sum is given by

\begin{displaymath}\sum_0^\infty n!x^n=\int_0^\infty dt e^{-t}\sum_{n=0}^\infty (xt)^n
=\int _0^\infty dt\frac{e^{-t}}{1-xt},
\end{displaymath} (67)

which is finite for x<0. If we are lucky, this would represent the exact answer. In some cases it is possible to show that the exact answer to some problem (e.g. in quantum mechanics), when expanded perturbatively, is Borel summable.

next up previous
Next: Convergence of the Fourier Up: No Title Previous: The Bessel function and
Mette Lund