Translation on X operator

Question

I need to show that given $[x, p_x]=i\hbar$ then the following is true: $$ e^{iap_x/\hbar}f(x)e^{-iap_x/\hbar}=f(x+a) $$ for a general function $f(x)$. I've tried using Taylor Series for both exponentials but it only seem to get more complicated and I'm not sure if thats the right approach. Any help is much appreciated!

Answer 1

I've managed yo solve the problem using the following: First i showed that $$ e^{iap_x/\hbar}xe^{-iap_x/\hbar}=f(x+a) $$ To do that i used the following properties: Given that $[x,p_x]=i\hbar$ and $[x[x,p_x]]=0$ then $$ [f(p_x),x]=f'(p_x)[p_x,x]=-i\hbar f'(x) $$ So for the exponentials we have $$ [e^{iap_x/\hbar},x]=-i\hbar\frac{ia}{\hbar} e^{iap_x/\hbar}=e^{iap_x/\hbar}x-xe^{iap_x/\hbar}\\ \Rightarrow e^{iap_x/\hbar}x=(x+a)e^{iap_x/\hbar} $$ Substituting back in the first equation: $$ e^{iap_x/\hbar}xe^{-iap_x/\hbar}=(x+a)e^{iap_x/\hbar}e^{-iap_x/\hbar}=(x+a) $$ Now i can use the propertie $f(e^{-\lambda A}Be^{\lambda A})=e^{-\lambda A}f(B)e^{\lambda A}$ valid since $[x[x,p_x]]=0$. Since $\hat{X} = x$ we have $$ e^{iap_x/\hbar}f(\hat{X})e^{-iap_x/\hbar}=e^{iap_x/\hbar}f(x)e^{-iap_x/\hbar}\\ =f(e^{iap_x/\hbar}xe^{-iap_x/\hbar}) =f(x+a) $$