How could this wonderful relation $[A,f(B)]=[A,B]f^\prime(B)$ of commutator in quantum mechanics be true?

Question

I am reading Solid State Physics Essential Concepts by David W. Snoke when I encounter this relation $$[A,f(B)]=[A,B]f^\prime(B),\tag{4.4.15}$$ which is labled (4.4.15) in page 226. Snoke didn't give any proof of it, nor did he give any restrictions of this formula.

I tried to prove it at the first time but failed. Then I realized it is too good to be true. But I do know in some cases it holds. For example, suppose we have the simple operator $\hat{v}=\frac{\hat{p}}{m}$, and we try to calculate $e^{-ik\hat{r}}\hat{v}e^{ik\hat{r}}$. If we use this formula, the deduction is straightforward and the result is right.

But when we try to calculate the result of $e^{-ik\hat{r}}\hat{H}e^{ik\hat{r}}$, where $\hat{H}=\frac{\hat{p}^2}{2m}+V(\hat{r})$, this formula leads us to wrong answer.

So maybe this formula holds in some cases? Do you have any idea about proving it under some constrictions?

Answer 1

Let me expand on my comment. Since $[X,\,YZ]=[X,\,Y]Z+Y[X,\,Z]$,$$[A,\,f(B)g(B)]=[A,\,f]g+f[A,\,g],$$matching the desired Leibniz rule. The case $Y=Z=B$ shows $f=B^2$ works iff $[A,\,B]$ commutes with $B$ (proof is an exercise). If that is true, the aforementioned Leibniz rule covers all monomial $f$ by induction on degree, then all polynomial $f$ by the commutator's bilinearity. For Taylor-function choices of $f$, we have to worry about convergence. See if you can work out which conditions are needed for this, for example, to make sense:$$[A,\,e^{tB}]=\left[A,\,\sum_{n\ge0}\tfrac{t^n}{n!}B^n\right]=\sum_n\tfrac{t^n}{n!}[A,\,B^n]=\sum_n\tfrac{t^n}{n!}nB^{n-1}=te^{tB}.$$Snoke asks the reader to use these ideas to prove an equation that reduces to$$e^{-za_k^\dagger}a_ke^{-z^\ast a_k}=e^{-z^\ast a_k}e^{-|z|^2}(a_k+z)e^{-za_k^\dagger}.$$Take $A:=a_k,\,B:=a_k^\dagger\implies[A,\,B]=1$, then give it a try.