# What am I REALLY doing when I take the Fourier transform of the momentum operator

I was playing around with some equations and found a surprising relationship when I took the fourier transform of the momentum operator

Define $\hat P = \frac{\hbar}{i} \partial_x$, then $F(\hat P) = k\hbar$

That's weird, I'm applying an operator on another operator to get some number that happens to be the de Broglie expression for momentum

Let's do the same for Energy, then $F(E) = F(i\hbar\partial_t) = w\hbar$

Doubly weird, de Broglie expression for energy.

So two questions I have in mind:

1. Are $k\hbar$ and $w\hbar$ operators themselves?
2. The definition of an operator is a map that maps one vector space to another, so what I am really doing in this case? Which vector space is getting mapped to where?
• Fourier transformation is not meant to apply to operators, but to functions. How did you arrive at $F(\hat P) = k\hbar$? – Ján Lalinský Feb 7 '15 at 10:24
• @JánLalinský you can Fourier-transform an operator like this: $\hat P'=\hat F\hat P\hat F^\dagger$, where $F^\dagger=F^{-1}$ is the Fourier transform unitary operator. – Ruslan Feb 7 '15 at 12:00
• Yes, that may be what Math Newb did. – Ján Lalinský Feb 7 '15 at 12:11
• It's well known that differentiation (integration) in one space is multiplication (division) in the Fourier dual space. In $k$ space, multiplication by $k$ is an operation just as, in $x$ space, multiplication by $x$ is. Indeed, we have $x \rightarrow -i\frac{\partial }{\partial k}$ – Alfred Centauri Feb 7 '15 at 15:06
• $k\hbar$ is an operator. More precisely, it is $k \hbar {\bf 1}$ where ${\bf 1}$ is the identity operator. – Prahar Feb 7 '15 at 16:02

($\hbar$ omitted in the following.)

That is not weird, it is one of the crucial properties of the Fourier transform $F(\bar{})$ that

$$F(\partial_x f) = \mathrm{i}p F(f)$$

i.e. differentiation by one variable becomes multiplication with the Fourier conjugate variable and vice versa. Because of this, Fourier transformation is a powerful tool to solve differential equations.

For your question as such, you do not apply the Fourier transform to operators such as you have written there. The fourier transform applies to functions of position $f(x)$ and turns them into functions of momentum $F(f)(p)$, but you can easily see that

$$F(\hat{P}\psi(x))(p) = pF(\psi)(p)$$

so, in a certain sense, the momentum operator is indeed simply turned into multiplication with the momentum. This is not surprising, since the momentum basis is exactly the basis in which the momentum operator is diagonal. And multiplication with a variable is a perfectly fine thing for an operator to be - there's nothing fishy about the operator $\hat{P} = p \cdot$.