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I am trying to calculate the following thing $$e^{A \cdot \beta \frac{\partial}{\partial \beta}}e^{i\eta \beta},$$ where $A,\eta$ are constants. Any idea how to write $e^{A \beta \partial/\partial\beta}$ in series? In general $$e^{A \cdot \beta \frac{\partial}{\partial \beta}} = \sum\limits_{n=0}^{\infty} \frac{A^n}{n!}\left(\beta \frac{\partial}{\partial \beta} \right)^n$$ and for $n=2$ I get $$\left(\beta \frac{\partial}{\partial \beta} \right)^2 = \beta \frac{\partial}{\partial \beta} \beta \frac{\partial}{\partial \beta} = \beta \frac{\partial}{\partial \beta} + \beta^2 \frac{\partial^2}{\partial\beta^2}$$

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It is well known that the stirling numbers of the second kind $S(n,i)$ occur in the formula \begin{eqnarray*} \left( \beta \frac{ \partial}{\partial \beta} \right)^n =\sum_{i=j}^{n} S(n,j) \beta^{j} \frac{ \partial^j}{\partial \beta^j}. \end{eqnarray*} So your quantity becomes \begin{eqnarray*} e^{A \cdot \beta \frac{\partial}{\partial \beta}} e^{i\eta \beta} &=& \sum\limits_{n=0}^{\infty} \frac{A^n}{n!}\left(\beta \frac{\partial}{\partial \beta} \right)^n e^{i\eta \beta} &=& \sum\limits_{n=0}^{\infty} \frac{A^n}{n!} \sum_{j=1}^{n} S(n,j) \beta^{j} \eta^j i^j e^{i\eta \beta} =\color{red}{\operatorname{exp}( i\beta \eta(e^A))}. \end{eqnarray*} See here for more info on the Stirling numbers of the second kind https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind

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  • $\begingroup$ Also some properties of the Stirling polynomials here $\endgroup$
    – WoofDoggy
    Commented Nov 7, 2017 at 22:24
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For any operator $T$, $e^{AT}$ formally satisfies the differential equation $$ \dfrac{\partial}{\partial A} e^{AT} = T e^{AT}$$ If $$v(A,\beta) = \exp\left(A \beta \dfrac{\partial}{\partial \beta}\right) e^{i\eta\beta}$$ this says

$$ \dfrac{\partial}{\partial A} v(A,\beta) = \beta \dfrac{\partial}{\partial \beta} v(A,\beta) $$

We have boundary condition $v(0,\beta) = e^{i\eta \beta}$.

The solution, obtained by the method of characteristics, is

$$ v(A,\beta) = \exp(i \eta \beta e^{A})$$

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