# In what sense is $\frac 1r-\frac d{dr}$ translation in polar coordinates?

This question seems to claim that the operation of $$R_n := \frac nr-\frac d{dr}$$ on smooth functions in polar coordinates is "translation to the right". Similarly $$L_n:=\frac nr+\frac d{dr}$$ is said to be translation to the left. Here $$n$$ is an arbitrary integer.

What does it mean? Why is it intuitively true?

The question goes on to conclude that since the Bessel function satisfies $$L_{n+1}R_n(J_n)=J_n$$, it means that the function $$J_n$$ is characterized as the function that "stays the same" when translating to the right then to the left. My end-goal is to understand why this follows. (So in particular what role do these $$n,n+1$$ play.)

EDIT.

I figured out small parts of it: The "left" and "right" shouldn't mean horizontal and vertical in Euclidean coordinates but rather in polar coordinates. The keyword in "radial momentum operator", which is what you get when you take the radial component of the momentum operator and symmetrize it. I guess the $$R_n$$ corresponds to translation in the $$\theta$$ direction.

However, this is still vague and I would like a more geometric viewpoint and the missing details. It would also be nice to translate it to a picture in the original Lie group $$E(2)$$ instead of the Lie algebra.

• I guess I don't quite follow: what if $f(r) = 1$? Then $L_n f = n / r$, which doesn't seem to me to be a translation of $f$ in any sense? Commented Jun 21, 2020 at 3:50
• @paulinho I don't know, but this should perhaps correspond to "moving farther away in the radial direction". Commented Jun 21, 2020 at 11:08