# Translation on $x$-operator

I need to show that given $$[x, p_x]=i\hbar$$ 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!

• If $a$ is a small parameter you can discard terms $\mathcal O(a^2)$ ($a^2$ and higher terms in the expansion). Apr 25, 2021 at 21:47
• Are you familiar with Lagrange's shift operator? Apr 26, 2021 at 0:35
• Not at all, I'll look it up Apr 26, 2021 at 0:38
• Please note that $f(x)$ is a number, so $e^{i a p_x / \hbar} f(x) e^{-iap_x / \hbar}$ is nonsense. It's much better to think as $f$ as the thing you're operating on and write $(e^{i a p_x / \hbar} f e^{-i a p_x / \hbar})(x) = f(x + a)$. Apr 26, 2021 at 0:41
• @DanielSank In the present context, that doesn't seem to be what OP means. It seems more likely that $f(x)$ is an operator-valued function of the position operator. This calculation makes sense in that context so long as $f$ admits a Taylor expansion. For the OP, you should look up the Baker-Campbell-Hausdorff formula. Apr 26, 2021 at 3:27

There's a few different ways to tackle this. One is to expand both $$e^{iap_x/\hbar}$$ and $$f(x)$$ as power series and multiply them. This will give you terms like $$p_x^nx^m$$. Using the canonical commutation relation $$[x,p_x]=i\hbar$$, you can derive a general expression for $$[x^m,p_x^n]$$ and use this to commute the $$x$$ and $$p_x$$ terms. Then collect like terms and show that

$$e^{iap_x/\hbar}f(x) = f(x+a)e^{iap_x/\hbar}$$

It might be instructive to work backwards and expand $$f(x+a)e^{iap_x/\hbar}$$ in terms of power series, so you know what you are working towards.

• Could you go into detail about the general expression for $[x^m, p^n]$? I'm having trouble working it out Apr 25, 2021 at 23:36
• @GustavoSchranckHabermann Hint: $[AB,C]=A[B,C]+[A,C]B$. So $[A^2,B]=A[A,B]+[A,B]A$ and $[x^2,p_x]=2i\hbar x$. You can do something similar for $[x^3,p_x]$ and such until you see the pattern.
– Chris
Apr 25, 2021 at 23:59

The trick is to use the Taylor expansion and the fact that $$e^Aa^{-A}=1$$: $$e^Af(B)e^{-A} = e^A\sum_{n=0}^{+\infty}\frac{f^{(n)}(0)B^n}{n!}e^{-A}= \sum_{n=0}^{+\infty}\frac{f^{(n)}(0)\left(e^ABe^{-A}\right)^n}{n!}= f\left(e^ABe^{-A}\right)$$

• That was really helpful, using this property and the Prahar Mitra's sugestion below i managed to solve pre problem, thanks for you answer Apr 27, 2021 at 16:11

First show that $$e^{iap_x/\hbar} x e^{-iap_x/\hbar} = x+a$$ Then, show it for a general function by Taylor expanding $$f(x)$$ around $$x=0$$ and then using the property above.

• Unfortunetly i can't accept two answers to the post but your suggestion combined with the previous answer by Vadim solved my problem, Thanks for your suggestions Apr 27, 2021 at 16:10