Quantum commutation relation $[e^{-x^{2}},e^{\alpha i p}] = ?$

Question

I have been trying for find a closed form solution, or at least something neat for the commutation relation $$[e^{-x^{2}},e^{\alpha i p}] = ?$$ (where $[x,p] = i\mathbb{I}$) but have had little luck. I have tried using BHC theorem but this does not get me very far. I think that there must be some simple relation that I am over looking.

Answer 1

You have set $\hbar=1$, so $p= -i\partial_x$ in the coordinate representation, so one of your operators is a bland Lagrange shift operator, and hence $$ [e^{\alpha i p}, f(x) ] = (f(x+\alpha)-f(x)) e^{\alpha i p} ~~~~\leadsto \\ [e^{-x^2}, e^{\alpha i p}]= - (e^{-(x+\alpha)^2}-e^{x^2}) e^{\alpha i p} . $$

(With a tip of the hat to @thedude 's comment! The linked WP article reminds you that $e^{i\alpha p} f(x) e^{-i\alpha p}= f(x+\alpha) $, in operator calculus language; when it acts on a constant, it reduces to just $e^{i\alpha p} f(x)= f(x+\alpha) $.)

Make sure to confirm for small α.