0
$\begingroup$

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!

$\endgroup$
12
  • $\begingroup$ If $a$ is a small parameter you can discard terms $\mathcal O(a^2)$ ($a^2$ and higher terms in the expansion). $\endgroup$
    – Charlie
    Apr 25, 2021 at 21:47
  • $\begingroup$ Are you familiar with Lagrange's shift operator? $\endgroup$ Apr 26, 2021 at 0:35
  • $\begingroup$ Not at all, I'll look it up $\endgroup$ Apr 26, 2021 at 0:38
  • $\begingroup$ 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 $(e^{i a p_x / \hbar} f e^{-i a p_x / \hbar})(x) = f(x + a)$. $\endgroup$
    – DanielSank
    Apr 26, 2021 at 0:41
  • 1
    $\begingroup$ @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. $\endgroup$ Apr 26, 2021 at 3:27

3 Answers 3

2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Could you go into detail about the general expression for $[x^m, p^n]$? I'm having trouble working it out $\endgroup$ Apr 25, 2021 at 23:36
  • $\begingroup$ @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. $\endgroup$
    – Chris
    Apr 25, 2021 at 23:59
0
$\begingroup$

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

$\endgroup$
1
  • 1
    $\begingroup$ That was really helpful, using this property and the Prahar Mitra's sugestion below i managed to solve pre problem, thanks for you answer $\endgroup$ Apr 27, 2021 at 16:11
0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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 $\endgroup$ Apr 27, 2021 at 16:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.