# Pointwise convergence of fourier expansion

1. Jul 18, 2005

### hellfire

I am reading in Wald "General Relativity" page 394, that the expansion of a scalar field over an orthonormal basis with creation and annihilation operators does not converge pointwise. What does this mean for a quantum field and what are the consequences of this?

2. Jul 19, 2005

### George Jones

Staff Emeritus
I don't have Wald handy, so here's a guess. Localizing a particle to a point in 1-particle quantum theory results in a singular wavefunction (a delta function), and, similarly, Localizing a quantum field to a point in spacetime results in a singular field.

In non-relativistic quantum theory, consider a particle on a 1-dimensional ring of circumference $L$. This is like a particle in a 1-dimensional box, where the box has been bent around until the ends join. Because $x$ and $x + nL$ represent the same location on the ring for any integer $n$, the wavefunction satisfies the periodic boundary condition $\psi \left( x + L \right) = \psi \left( x \right)$ for all $x$.

A particle localized to the point $x = a$ should have a wavefunction that looks something like $\psi \left( x \right) = \delta \left( x - a \right)$. However, this doesn't satisfy the periodic boundary condition. A delta function is needed at $a + nL$ for each $n$, so

$$\psi \left( x \right) = \sum^{\infty}_{n = -\infty} \delta \left( x - \left(a + nL \right) \right).$$

For simplicity take $a = 0$, and try and write $\psi$ as the Fourier expansion

$$\psi \left( x \right) = \sum^{\infty}_{n = -\infty} c_{n} e^{i2\pi nx/L},$$

and solve formally for the Fourier coefficients $c_{n}$.

$$\begin{equation*} \begin{split} c_{n} &= \frac{1}{2\pi} \int_{-L/2}^{L/2} \psi \left( x \right) e^{-i2\pi nx/L} dx\\ &= \frac{1}{2\pi} \int_{-L/2}^{L/2} \delta \left( x \right) e^{-i2\pi nx/L} dx\\ &= \frac{1}{2\pi}, \end{split} \end{equation*}$$

since only one of the delta functions is within the range of integration. Thus, the wavefunction for a particle localized at $x = 0$ is

$$\psi \left( x \right) = \sum^{\infty}_{n = -\infty} \delta \left( x - nL \right) = \frac{1}{2\pi} \sum^{\infty}_{n = -\infty} e^{i2\pi nx/L}.$$

The Fourier expansion for $\psi$ doesn't converge pointwise because $\psi$ is a (singular) distribution, not a function. The expansion does converge weakly, i.e., in the sense of tempered distributions.

In quantum field theory, at point $x$ in spacetime, the field is represented by an "operator" $\Psi \left( x \right)$ that is really an operator-valued distribution with, for states $f$ and $g$, $<f|\Psi \left( x \right)|g>$ a distribution.

For my money, Chapter 9, Quantum Field Theory in Curved Spacetime, from Sean Carroll's new GR book is the best introduction in print to Hawking radiation. Carroll doesn't concern himself with mathematical niceties like operator-valued distributions but does give a nice physics-style intro to things like Bogolubov transformations. If you do want to see the math done carefully, stick with Wald or Wald.

Regards,
George

3. Jul 19, 2005

### hellfire

This is an excelent answer and very clear, thank you!

Last edited: Jul 19, 2005
4. Jul 19, 2005

### reilly

Fourier series and integrals converge in the mean. That is, Fourier expansions are based on least squares regression: Minimize the mean squared error ---
E(x)= |F(x)-f(x)|**2
over the entire approximation domain, ie minimize Integral { E(x) }, where f(x) is the function, and F(x) is the approximation. This scheme, due to Gauss, is often called RMS, root mean square approximation.

Point-wise convergence means that lim (F(x)-f(x))-> for all x (as something goes to zero or infinity. F could be Sum (1 to N) [F0 +F1*x + FN x**N) as N-> infinity.

The issue of wave functions for a localized particle is not the only manifestation of convergence issues. For example, there's the Gibbs Phenomena, in which the Fourier representation always oscillates around a point of discontinuity -- just look at a square waveform in an oscilloscope -- you get a very good fit to most of the length of the horizontal segments except at the endpoints, where the approximation gives an up-and-down curve. You get a good RMS fit at the corners, and a lousy pointwise fit at the corners. (Note that there's an approximation scheme based on minimising the absolute error, |F(x) - f(x)| = |E(x)|, as well as |E(x)|**M, all of which gets you to various forms of Hilbert Space, in which "E(x)" is considered a coordinate difference . The Euclidean norm (squared) of |E(x)|**2, is the squared norm of the Hilbert Space used in QM. So, in a sense, quantum theory is based on regression, just like a lot of economics.

Regards,
Reilly Atkinson

5. Jul 20, 2005

### George Jones

Staff Emeritus
This a very good point, which I decided to neglect when I answered hellfire's question. Hellfire asked specifically about quantum fields, and, using print.google.com, I saw that Wald p. 393 (access to p. 394 was restricted ) talks about operators, so I answered in terms Fourier expansions of quantum field operators, which don't even converge in norm.

I first ran into this when I learned in a high school electronic course that a square wave "equals"

sinx + (1/3)sin3x + (1/5) sin5x + ...

I decided to use a computer to graph the first n terms of the series for n = 1, 2, 3, ... . I was completely astonished and baffled by what I saw happening at the corners. It wasn't until university analysis classes that I really learned about different types of convergence, and it wasn't until even later that I read on my own about Gibb's phenomena.

Regards,
George