1
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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 $g = f'$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.