# How to interpret $\left[f(A), B\right] = \left[A, B \right]\frac{\partial f}{\partial A}$ - how to differentiate with respect to an operator?

From here and here I know the commutation relation for two operators are:

$$\left[f(A), B\right] = \left[A, B \right]\frac{\partial f}{\partial A}$$ if $$\left[A, \left[A, B \right]\right] =0$$ and $$f$$ is a reasonably behaved function.

What does $$\frac{\partial f}{\partial A}$$ mean? How can I differentiate with respect to an operator?

It works because functions of operators are usually defined by their Taylor series. If we plug that into the left side, the problem reduces to calculating $$[A^n, B]$$ (since the commutator is a linear operation). By repeatedly using $$[AB,C]=A[B,C]+[A,C]B$$ we find $$[A^n,B] = \sum_{k=0}^{n-1}A^k[A,B]A^{n-k-1}.$$ Now if $$[A,[A,B]]=0$$, as you write, we can switch all $$A$$ to one side and get $$[A^n,B] = \sum_{k=0}^{n-1}[A,B]A^{n-1} =[A,B]\,n A^{n-1}$$ which on the right is exactly the usual rule for differentiation with respect to $$A$$, if $$A$$ was a variable. So for the purpose of calculating the original commutator, you can just treat it as if it was a variable and derive with respect to it. Careful though whenever $$f$$ is a function of multiple non-commuting operators.
• So it would be accurate to write it like this: $$\left[f(\hat A), \hat B\right] = \left[\hat A, \hat B \right]\frac{\partial f}{\partial A}$$ yep? – zabop Feb 20 at 19:13
• If you write the argument of $f$ with a hat you need it in the denominator, otherwise it's inconsistent. – noah Feb 20 at 19:21