How does the derivative of an operator valued function act on vectors?

Question

Let $A(x)$ be an (linear) operator valued function a real parameter $x$.
Is it true that: $$\left[\frac{d}{dx}A(x)\right]\left|\psi\right> = \frac{d}{dx}\left[A(x)\left|\psi\right>\right],$$ where $\left|\psi\right>$ is a constant (independent of $x$) vector (belonging to the space on which $A(x)$ acts)?

Does a similar relationship hold for an integral of $A(x)$ w.r.t. $x$?

Answer 1

Yes. Recall the following facts:

1. The definition of the derivative of an operator which depends on some parameter is given by $$\frac{d}{dx}A(x) = \lim_{\epsilon\rightarrow 0}\frac{1}{\epsilon}[A(x+\epsilon)-A(x)]$$
2. For all $|\psi\rangle$ in the vector space, the limit of a family of operators $A(x)$ is defined by $$\left[\lim_{x\rightarrow x_0} A(x)\right]|\psi\rangle = \lim_{x\rightarrow x_0} [A(x)|\psi\rangle]$$
3. The sum/difference of two operators is defined by $$(A+B) |\psi\rangle = A|\psi\rangle + B|\psi\rangle$$
4. The definition of the derivative of a vector which depends on some parameter is $$ \frac{d}{dx}|\psi(x)\rangle = \lim_{\epsilon\rightarrow 0} \frac{1}{\epsilon}\big[|\psi(x+\epsilon)\rangle - |\psi(x)\rangle\big]$$

and so $$\left[\frac{d}{dx}A(x)\right]|\psi\rangle = \left[\lim_{\epsilon\rightarrow 0} \frac{1}{\epsilon} [A(x+\epsilon)-A(x)\right] |\psi\rangle = \lim_{\epsilon\rightarrow 0}\frac{1}{\epsilon}[A(x+\epsilon)|\psi\rangle - A(x)|\psi\rangle]$$ $$ = \frac{d}{dx}\left[A(x)|\psi\rangle\right]$$

You can perform a similar series of steps to construct an integral as well.