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# Perturbation theory and Borel summation

One of the most common way to solve a non-trivial quantum field theory is to use perturbation theory. Roughly speaking, if we are given a Hamiltonian of the form

$$H = H_0 + H_I$$

A sum of a free part and an interacting part, we then expand the interacting part in whatever exponential we are dealing with :

\begin{eqnarray} e^{-H} &=& e^{-H_0} e^{-H_I}\\ &=& e^{-H_0} \sum_{n = 0}^\infty \frac{(-H_I)^n}{n!} \end{eqnarray}

## The zero-dimensional case

The simplest case we can use is the zero-dimensional case, for which the configuration space is simply $\mathbb{R}$, the action is an ordinary function over $\mathbb{R}$, and the path integral a simple Lebesgue integral. The resulting partition function for the simplest interacting theory (in the Euclidian case) is

$$Z[\lambda] = \int_{\mathbb{R}} \exp(- x^2 - \lambda x^4)$$

Fortunately here, the integral in this case is exactly solvable. It is for instance solved in Gradshteyn and Ryzhik as the integral 3.323.3 : $$\int_{0}^\infty \exp(- \beta^2 x^4 - 2 \gamma^2 x^2) = 2^{-\frac{3}{2}} \frac{\gamma}{\beta} e^{\frac{\gamma^4}{2\beta^2}} K_{\frac{1}{4}} (\frac{\gamma^4}{2\beta^2})$$

With our own integral, this corresponds to $\lambda = \beta^2$, or $\beta= \sqrt{\lambda}$, and $\gamma = 2^{-\frac{1}{2}}$, as well as adding an extra factor of $2$ since we integrate on the entire real line :

$$Z[x] = \frac{1}{2\sqrt{\lambda}} e^{\frac{1}{8\lambda}} K_{\frac{1}{4}} (\frac{1}{8\lambda})$$

From that definition, you'll notice that this solution is only valid for $\lambda \geq 0$, and this is the crux of the issue here. Let's consider for a moment the case $\lambda < 0$. First, let's recast our function as

$$\exp\left[ x^2 (- 1 + |\lambda| x^2) \right]$$

This function has three extrema, $0$ and $\pm 2 / |\lambda|$.

...

What happens if we attempt the usual perturbative expansion on our partition function?

\begin{eqnarray} Z[\lambda] &=& \int_{\mathbb{R}} \exp(- x^2 - \lambda x^4)\\ &=& \int_{\mathbb{R}} \exp(- x^2) \exp(- \lambda x^4)\\ &=& \int_{\mathbb{R}} \exp(- x^2) \sum_{n = 0}^\infty \frac{(- \lambda x^4)^n}{n!}\\ &=& \int_{\mathbb{R}} \sum_{n = 0}^\infty (-1)^n\exp(- x^2) \frac{(\lambda x^4)^n}{n!}\\ \end{eqnarray}

So far, everything has been legal. The dangerous step comes up when we try to interchange the sum and integral sign here. By Fubini's theorem, we are allowed to exchange an integral and sum

$$\int_\Omega \sum_{n} f_n = \sum_{n} \int_\Omega f_n$$

if the norm of $f_n$ can be integrated thus. In other words,

\begin{eqnarray} \int_{\mathbb{R}} \sum_{n = 0}^\infty \exp(- x^2) \frac{(\lambda x^4)^n}{n!} < \infty \end{eqnarray}

...

Despite not being able to use Fubini's theorem, let's consider for now the series made from switching the integral and sum.

\begin{eqnarray} Z_N &=& \sum_{n = 0}^N \int_{\mathbb{R}} (-1)^n\exp(- x^2) \frac{(\lambda x^4)^n}{n!}\\ &=& \sum_{n = 0}^N \frac{\lambda^n}{n!} \int_{\mathbb{R}} (-1)^n\exp(- x^2) x^{n} &=& \sum_{n = 0}^N \frac{\lambda^n}{n!} F_n \end{eqnarray}

with

\begin{eqnarray} F_n &=& \int_{\mathbb{R}} (-1)^n\exp(- x^2) x^{n} \end{eqnarray}

As we do not yet know whether this sum converge, we will be leaving it as a partial sum for now. $F_n$ is simply the moment of a normal distribution,

## Borel summation

1. M. Flory et al., How I Learned to Stop Worrying and Love QFT

Last updated : 2021-01-25 15:30:13
Tags : physics , quantum-field-theory