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I stumbled upon an equation that could be an identity, but I'm not sure. $$ e^{a(\partial_x + f'(x))} e^{-a \partial_x} = e^{f(x+a) - f(x)} $$

The operator exponential can be understood as a power series $$ e^{a\partial_x} g(x) = (1 + a\partial_x + \frac{a^2}{2!} \partial_x^2 + \ldots) g(x) = g(x+a) $$

Let $X = a(\partial_x + f'(x))$ and $Y = -a \partial_x$. $$ [X,Y] = -a^2 [f'(x), \partial_x] = a^2 f''(x) \\ $$ $$ [X,[X,Y]] = -[Y,[X,Y]] = a^3 f^{(3)}(x) $$

I checked the first few terms of the BCH formula. \begin{align*} &X + Y + \frac{1}{2}[X,Y] + \frac{1}{12} [X,[X,Y]] - \frac{1}{12} [Y,[X,Y]] + \ldots \\ & \stackrel{\text{?}}{=} a f'(x) + \frac{a^2}{2!} f''(x) + \frac{a^3}{3!} f^{(3)}(x) + \ldots \\ &= f(x+a) - f(x) \end{align*}

Bio
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1 Answers1

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I don't think it's a good idea to use BCH formula. Here is my proof.

Set $I(x)=e^{f(x)}$ we get $\partial_x+f'=I^{-1}\circ\partial_x\circ I$, since $(Ig)'=(f'g+g')I$.

Therefore $\exp(a(\partial_x+f'))=\exp(I^{-1}\circ a\partial_x\circ I)=I^{-1}\circ\exp(a\partial_x)\circ I$, which means $$e^{a(\partial_x+f')}e^{-a\partial_x}g(x)=I^{-1}\circ e^{a\partial_x}(I(x)g(x-a))=I(x)^{-1}I(x+a)g(x)=e^{-f(x)+f(x+a)}g(x).$$

Liding Yao
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