Question

Show that
[ x,e iap ] = -ae iap
h h

Answer 1

In order to evaluate commutator $[x, e^{i a p}]$, we need to expand the exponent into Taylor series, i.e. $e^{i a p} = 1 + i a p + \frac{(i a p)^2}{2!} + \frac{(i a p)^3}{3!} + ...$.

Since the commutator is linear in both arguments, we need to evaluate commutators like $[x, p^n]$. Using Leibniz rule for commutators and canonical commutation relation $[x, p] = i \hbar$:

$[x, p^n] = (i \hbar) p^{n-1} + p (i \hbar) p^{n-2} + ... + p^{n-1} (i \hbar) = i \hbar n p^{n-1}$.

Hence,

$[x, e^{i a p}] = [x, 1 + i a p + \frac{(i a p)^2}{2!} + \frac{(i a p)^3}{3!} + ...] =$

$= (i \hbar)[(i a)^1 + \frac{(i a)^2}{2!} \cdot 2 \cdot p^1 + \frac{(i a)^3}{3!} \cdot 3 \cdot p^2 +... = (i \hbar) (i a) \sum_{n=0}^{\infty} \frac{(i a p)^n}{n!} = - a \hbar e^{i a p}$.