# Momentum and fourier series

1. Oct 23, 2009

### RedX

Both (i d/dx) and (-i d/dx ) are Hermitian. For some reason (-i d/dx ) is chosen to be the momentum operator, and the consequences are that [x,p]=ih (and not -ih), and that $$e^{ipx}$$ is an eigenvalue of momentum p (and not -p).

Is there any fundamental reason why [x,p] can't be -ih, and $$e^{ipx}$$ can't have an eigenvalue -p, so that the momentum operator can be (i d/dx) ?

When dealing with Fourier series, $$f(x)=\int d^3p \mbox{ } f(p) e^{-ipx}$$ would be incorrect, right? It would have to be $$f(x)=\int d^3p \mbox{ } f(p) e^{ipx}$$ if you choose (-i d/dx )?

Do most math books use $$f(x)=\int d^3p \mbox{ } f(p) e^{-ipx}$$ or $$f(x)=\int d^3p \mbox{ } f(p) e^{ipx}$$ for their definition of a Fourier series? Which convention do you use for a Fourier series?

2. Oct 24, 2009

### Ben Niehoff

There is no fundamental reason. In fact, all quantum mechanics (indeed, all mathematics) remains valid under the substitution, everywhere, of i -> -i. The reason is that there is some ambiguity in defining i in the first place. There are two square roots of -1, after all.

The convention I use for Fourier integrals is

$$f(x) = \int \frac{d^3p}{(2\pi)^3} \; f(p) e^{ipx}$$