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I have learned about the commutators, and read this:

$$[A, f(B)] = f'[A,B]+\frac{1}{2}f''([A,B]B+B[A,B])+\frac{1}{3!}f'''([A,B]B^2+B[A,B]+B^2[A,B])+...$$

then Simplified to

$$[A, f(B)] = [A,B](f'+f''B+\frac{1}{2}f'''B^2+...)=[A,B]\frac{df}{dB}$$

I do understand the first two equations, only don't understand is why the series $$(f' + f''B + \frac{1}{2}f''' B^2+...)$$ equals to $$\frac{df}{dB}$$

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    $\begingroup$ Is this from a reference? Which page? $\endgroup$
    – Qmechanic
    Nov 24, 2020 at 5:47
  • $\begingroup$ Is it assumed that $[[A,B],B]=0$? $\endgroup$
    – Qmechanic
    Nov 24, 2020 at 20:45

1 Answer 1

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This is simply an identity of Taylor expansion and has nothing to do with the fact that you have operators around, if $$f(x)=\sum_n \frac{f^{(n)}(0)}{n!}x^n$$ then $$\frac{df}{dx} = \sum_n \frac{f^{(n+1)}}{n!}x^n$$ which simplifies to the third equation.

Writing $\frac{df}{dB}$ is just a notation for $\frac{df}{dx}|_{x=B}$. The reason for that is that analytic functions of operators are defined by their Taylor series.

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  • $\begingroup$ I got it, thanks $\endgroup$
    – tinghaoliu
    Nov 24, 2020 at 16:24
  • $\begingroup$ They are not defined by the Taylor series. In fact for the default operators of quantum mechanics, $X$ and $P$, the Taylor series does not converge. $\endgroup$ Nov 24, 2020 at 22:34

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