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What is the commutator of the exponential derivative operator and the exponential position operator? \begin{align} \left[\exp(\partial_x),\exp(x)\right] =\exp(\partial_x)\exp(x) -\exp(x)\exp(\partial_x)=~? \end{align}

I first wrote down the exponentials into there power series sum \begin{align} \left[\exp(\partial_x),\exp(x)\right] = \sum_{n=0}^{\infty} \dfrac{\partial_x^n}{n!} \sum_{m=0}^{\infty} \dfrac{\partial_x^m}{m!} - \sum_{n=0}^{\infty} \dfrac{x^n}{n!} \sum_{m=0}^{\infty} \dfrac{\partial_x^m}{m!}, \end{align} and then grouped like terms \begin{align} \left[\exp(\partial_x),\exp(x)\right] = \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \dfrac{ \left[ \partial_x^n x^m - x^n \partial_x^m \right] }{n!m!}, \end{align} and then evaluate the derivatives \begin{align} \left[\exp(\partial_x),\exp(x)\right] = \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \dfrac{ \left[ \partial_x^n (x^m) + x^m \partial_x^n - x^n \partial_x^m \right] }{n!m!}. \end{align} I am not sure if$$ \partial_x^n\left(x^m\right)~=~\frac{m!}{\left(m-n+1\right)!} \, x^{m-n} \,,$$and then I got stuck.

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  • $\begingroup$ Do you know how $[x,F(p)]$ behaves? Do you know how $[x^2, p]$ behaves? If the answer is yes to both, you should be able to do this question (though it certainly looks a bit lengthy in the math!); if not, then you may want to look into solving those two general cases first before taking a stab at this one. $\endgroup$ – Kyle Kanos Sep 14 '14 at 14:57
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    $\begingroup$ Have you heard of the Baker-Campbell-Hausdorff formula? $\endgroup$ – suresh Sep 14 '14 at 14:59
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Suresh is right, the Baker-Campbell-Hausdorff formula will save you. If $\left[A,B\right]$ commutes with $A$ and $B$ you have,

$$ \exp\left(A\right)\exp\left(B\right) = \exp\left(A+B+\frac{1}{2}\left[A,B\right]\right) \, .$$

Then the answer to your question is

$$ \exp\left(\partial_x\right)\exp\left(x\right) = 2 \exp\left(x+\partial_x\right) \sinh(1/2) \, .$$

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Besides the truncated BCH formula (which is also proven in this Phys.SE post and mentioned in Steven Mathey's answer), in practice one often wants to normal order the differential operators, i.e. putting all the $x$'s to the left of all the $\partial_x$. To this end the formula

$$\tag{1} e^{a\partial_x}f(x) ~=~ f(x+a)e^{a\partial_x}. $$

is useful. Hence the sought-for normal-ordered commutator becomes

$$\tag{2} [e^{a\partial_x}, e^{bx}] ~=~(e^{b(x+a)}-e^{bx})e^{a\partial_x}.$$

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  • $\begingroup$ It is this form of the BCH formula that I had in mind though it can be derived from Steven Mathey's answer as well. $\endgroup$ – suresh Sep 16 '14 at 0:13

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