quantum mechanics - Matrix elements of momentum operator in position representation

I have two related questions on the representation of the momentum operator in the position basis.

The action of the momentum operator on a wave function is to derive it:

$$\hat{p} \psi(x)=-i\hbar\frac{\partial\psi(x)}{\partial x}$$

(1) Is it ok to conclude from this that:

$$\langle x | \hat{p} | x' \rangle = -i \hbar \frac{\partial \delta(x-x')}{\partial x}?$$

And what does this expression mean?

(2) Using the equations:

$$ \frac{\langle x | \hat{x}\hat{p} | x' \rangle}{x} = \frac{\langle x | \hat{p}\hat{x} | x' \rangle}{x'} = \langle x | \hat{p} | x' \rangle $$

and

$$\langle x | [\hat{x},\hat{p}]|x'\rangle=i\hbar \delta(x-x')$$

one can deduce that

$$\langle x | \hat{p} | x' \rangle = i \hbar \frac{\delta(x-x')}{x-x'}$$

Is this equation ok? Does it follow that

$$\frac{\partial \delta(x-x')}{\partial x} = - \frac{\delta(x-x')}{x-x'}?$$

Answer

1) Notice that by inserting a complete set of position states we can write $$ \hat p \psi(x) = \langle x|\hat p|\psi\rangle = \int dx'\langle x|\hat p|x'\rangle\langle x'|\psi\rangle =\int dx'\langle x|\hat p|x'\rangle \psi(x') $$ so if we set $$ \langle x|\hat p|x'\rangle = -i\hbar \frac{\partial}{\partial x}\delta(x-x') =i\hbar \frac{\partial}{\partial x'}\delta(x-x') $$ then we can use integration by parts to obtain $$ \hat p \psi(x) =i\hbar \int dx'\frac{\partial}{\partial x'}\delta(x-x') \psi(x') = -i\hbar \int dx'\delta(x-x') \frac{d \psi}{dx'}(x') = -i\hbar \frac{d\psi}{dx}(x) $$ So your expression is correct. The derivative of a delta function is essentially defined by the integration by parts manipulation that I just performed; in fact derivatives of distributions in general are defined in an analogous way. See this lecture for example.

Hope that helps; let me know of any typos!

Cheers!