Small doubt on derivatives acting on kets/bras I have a quick, silly question. If $\psi(x):=\langle x|\psi\rangle$, does the bra $\langle x|$ 'go through' the $\partial_x$ operator, as in $$\langle x|\partial_x|\psi\rangle=\partial_x\psi(x) \quad ? $$ If not, how can I deal with an object of the form $\partial_x|x\rangle  ?$
 A: Sort of, yes. The correct expression is
$$
\langle x| \hat p|\psi \rangle= \frac{\hbar} {i} \partial_x \langle x  |\psi \rangle,
$$
which lots of people use the mnemonic $\hat p_x =\frac{\hbar} {i} \partial_x$ for, knowing what we all mean: the momentum operator in the x-representation.
My best mnemonic is the rigorous expression
$$
\hat p= \int\!\!dx ~ |x\rangle (\hbar/i)\partial_x \langle x|=  \int\!\!dp ~ |p\rangle  p\langle p|, 
$$
from which you may derive correct expressions.
A: Just in case you find it useful, I'll give you an alternative way to do it. You have to use a bunch of book-keeping relations. Namely:

*

*Completeness relation (for $p$ will suffice)
$$
I=\int dp\left|p\right\rangle \left\langle p\right|
$$
($I$ being the identity, AKA "doing nothing.")


*Wave functions in the $x$ and $p$ representations
$$
\psi\left(x\right)=\left\langle \left.x\right|\psi\right\rangle 
$$
$$
\hat{\psi}\left(p\right)=\left\langle \left.p\right|\psi\right\rangle 
$$


*Relation between wave functions in the $x$ and $p$ representations
$$
\psi\left(x\right)=\frac{1}{\sqrt{2\pi\hbar}}\int dp\:e^{ipx/\hbar}\hat{\psi}\left(p\right) 
$$
(Fourier transform) where the $\sqrt{2\pi\hbar}$ is just for convenience.
Now, your question requires to clarify first: What does the action of the momentum operator $P$ on an abstract state $\left|\psi\right\rangle $ look like in the $x$-representation? That must be,
$$
\langle x|P|\psi\rangle=\int dp\langle x|P|p\rangle\left\langle \left.p\right|\psi\right\rangle =\int dp\:p\left\langle \left.x\right|p\right\rangle \left\langle \left.p\right|\psi\right\rangle =
$$
$$
=\frac{1}{\sqrt{2\pi\hbar}}\int dp\:pe^{ipx/\hbar}\left\langle \left.p\right|\psi\right\rangle =\frac{1}{\sqrt{2\pi\hbar}}\int dp\:\frac{\hbar}{i}\partial_{x}e^{ipx/\hbar}\left\langle \left.p\right|\psi\right\rangle =
$$
$$
=\frac{\hbar}{i}\partial_{x}\left(\frac{1}{\sqrt{2\pi\hbar}}\int dp\:e^{ipx/\hbar}\left\langle \left.p\right|\psi\right\rangle \right)=\frac{\hbar}{i}\partial_{x}\psi\left(x\right)
$$
The expression you use $\langle x|\partial_x|\psi\rangle$ is somewhat cavalier. You should not assume any $x$-dependent form for $\left|\psi\right\rangle $, as abstract states have no such dependence. Their wave functions do.
A: Note that $\langle x|\psi\rangle$ is the projection of $|\psi\rangle$ in the basis $|x\rangle$. Your $\partial_x$ is the form of a certain operator, say $\hat p$, in this basis. In bra-ket notation:
$\langle x|\hat p|\psi\rangle=\partial_x \psi(x)$
