I tried to do the usual procedure and expand the commutator, but couldn't proceed after I Taylor-expanded $f(\hat x)$.

$$\Big[f(\hat x),\frac{d}{dx}f(\hat x)\Big]=$$

$$f(\hat x)f'(\hat x)-f'(\hat x)f(\hat x)=$$

$$\sum_0^{\infty}\frac{f^{n}(\hat x_0)}{n!}(\hat x-\hat x_0)^n\sum_0^{\infty}\frac{f^{n+1}(\hat x_0)}{n!}(\hat x-\hat x_0)^n-\sum_0^{\infty}\frac{f^{n+1}(\hat x_0)}{n!}(\hat x-\hat x_0)^n\sum_0^{\infty}\frac{f^{n}(\hat x_0)}{n!}(\hat x-\hat x_0)^n$$

And from here I don't know what to do. Maybe it is not possible to know that commutator without more knowledge of $f$.

Could you direct me into the right direction?

I want this commutator to be 0 because it would make my life much easier in a homework problem, that is how I encountered this question.


A function $f : \mathbb{R}\to\mathbb{R}$ can be applied to any self-adjoint operator $A$ by acting on its eigenvalues. This is the definition of what it means to apply the function to an operator.

That is, if $A$ has eigenvalues $\lambda_i$ with an eigenbasis $\lvert \lambda_i\rangle$, then $f(A)\lvert \lambda_i\rangle = f(\lambda_i)\lvert \lambda_i\rangle$. In particular, $f(A)$ is diagonal in the basis where $A$ is diagonal.

Therefore, for any two functions $f,g$, the operators $f(A),g(A)$ are simultaneously diagonal in the eigenbasis of $A$. Two operators that are simultaneously diagonal commute, so $f(A)$ and $g(A)$ always commute regardless of what their actual functional form is, so in particular when $g = f'$.

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