Role of wavelength in the representations of the translation group

Question

In the representation theory of the translation group we have $U^p(x)=e^{ipx}$ where $p=2\pi/\lambda$. I know in quantum mechanics this ends up being momentum somehow. And we can also see that $p$ is like an angular speed on the complex plane. Is this basically the intuition behind these infinite representations? Kind of like, okay all the representations reach $x$ but at different speeds? Is there some other way to think of this intuitively?

Answer 1

I think you malformed your operator. In the position representation, the translation operator is $$ U^p(\Delta x) = e^{i \Delta x ~ \hat p/\hbar }= e^{\Delta x ~ \partial_x}, $$ which then amounts to Lagrange's shift operator, $$ U^p(\Delta x) f(x) = f(x+\Delta x). $$ Here, $\Delta x$ is a constant number not acted upon by $\partial_x$.

(I wouldn't go there, but if you really wished to consider $U^p( x) $ instead, you see this is $ e^{x\partial_x}= e^{\partial/\partial(\log x)}$, so that $$ U^p(x) f(x) = e^{\partial/\partial(\log x)} f(e^{\log x})= f(ex), $$ a mere scale transformation.)