Commutator and Taylor series in quantum mechanics

Question

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 $$frac{df}{dB}$$

Yarden Sheffer · Answer

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.

Commutator and Taylor series in quantum mechanics

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