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 $0$?
Best Answer
In my opinion, manipulations involving $\hat p$ and position bras and kets are most easily done by considering the action of $\hat p$ on the position bras, which is simply $$ \boxed{ \vphantom{\begin{array}{}make\\the box\\taller\end{array}} \quad\,\,\, \langle x|\hat p=-i\hbar\frac{\text d}{\text dx}\langle x|. \quad\,\,\,} \tag 1 $$
You can get this easily by seeing that for any state $|\psi\rangle$ with position-representation wavefunction $\psi(x)=\langle x|\psi\rangle$, the action of the momentum operator on the state gives a derivative on the wavefunction. That is, $$\langle x|\hat p|\psi\rangle =-i\hbar\frac{\text d}{\text dx}\langle x|\psi\rangle.$$ Since this equation holds for all states $|\psi\rangle\in\mathcal H$, you can "cancel $|\psi\rangle$ out". (More technically, since the action of the bras $\langle x|\hat p$ and $-i\hbar\frac{\text d}{\text dx}\langle x|$ is the same for all vectors, they must be equal as linear functionals.)