I have been going through some problems in Sakurai's Modern QM and at one point have to calculate $\langle \alpha|\hat{p}|\alpha\rangle$ where all we know about the state $|\alpha\rangle$ is that $$\langle x|\alpha\rangle=f(x)$$ for some known function $f$. ($|\alpha\rangle$ is a Gaussian wave packet.) Sakurai says that this is given by:

$$\langle p\rangle = \int\limits_{-\infty}^{+\infty}\langle\alpha|x\rangle\left(-i\hbar\frac{\partial}{\partial x}\right)\langle x|\alpha\rangle dx.$$

I am wondering how we get to this expression. I know that we can express

$$|\alpha\rangle =\int dx|x\rangle\langle x|\alpha\rangle$$

and

$$\langle\alpha|=\int dx\langle\alpha|x\rangle\langle x|,$$

so my thinking is that we have:

$$\langle\alpha|\hat{p}|\alpha\rangle =\iint dx dx'\langle\alpha|x\rangle\langle x|\hat{p}|x'\rangle \langle x'|\alpha\rangle, $$

and if we can 'commute' $|x\rangle$ and $\hat{p}$ this would become: $$\iint dxdx'\langle\alpha |x\rangle\hat{p}\langle x|x'\rangle \langle x'|\alpha\rangle,$$ which is the desired result as $$\langle x|x'\rangle=\delta(x-x').$$ Is this approach valid?

I think my question boils down to: Does the operator $\hat{p}$ act on the basis kets $|x\rangle$ or on their coefficients? In the latter case, if we had some state $|\psi\rangle = |x_0\rangle$ for some position $x_0$, then would we say that for this state $$\langle p\rangle =\langle x_0|\left(-i\hbar\frac{\partial}{\partial x}\right)|x_0\rangle = 0$$ as the single coefficient is $1$ and the derivative of $1$ is $0$?