# Taylor expanding a function of an operator?

I am trying to understand the following description in my quantum mechanics textbook:

Let $$F(\hat{A})$$ be a function of an operator $$\hat{A}$$. If $$\hat{A}$$ is a linear operator, we can Taylor expand $$F(\hat{A})$$ in a power series of $$\hat{A}$$: $$F(\hat{A})=\sum_{n=0}^{\infty}a_{n}\hat{A}{}^{n},$$

I'm a little confused about the Taylor expansion of the operator $$A$$. Should the Taylor expansion involve the derivatives of $$A$$? Considering this is an operator, I think that taking derivatives should be valid here, but it seems like this is implying the derivatives are the same as the operator itself, which only seems to be the case for a specific subset of operators.

• Note that there is no guarantee that an arbitrary function of $\hat{A}$ can actually be expanded in a Maclaurin series around this way. It doesn't work for $\hat{A}{}^{-1}$, for example, even if $\hat{A}$ is invertible.
– Buzz
Aug 23, 2023 at 20:00

You are Taylor expanding the function $$F(\hat{A})$$, not the operator $$\hat{A}$$, so the coefficients contain derivatives of $$F$$, not of $$\hat{A}$$. Explicitly, $$$$F(\hat{A}) = \sum_{n=0}^\infty \frac{F^{(n)}(0)}{n!} {\hat{A}}^n,$$$$ where $$F^{(n)}$$ is the n-th derivative of the function $$F$$.
Derivatives of an operator can appear if $$\hat{A}$$ depends on some continuous parameter, like time, $$\hat{A}(t)$$. Then you can Taylor expand the operator in powers of $$t$$ and each coefficient is an operator. This is not what the textbook is discussing.