Recognitions:

 Quote by Loro Dirac actually says in the book that: $\hat{\partial_q} |\psi> = 0$ Where $|\psi>$ is the standard ket.
Where does Dirac say that? If you're looking at his eq(11) on p89, what he
actually writes is
$$\frac{d}{dq} \psi \rangle ~=~ \frac{d\psi}{dq} \,\rangle$$
Note that absence of the ##|##''. Dirac is using his own notation, but the
versions of that notation presented by more modern authors are different.

I checked a few other things in his book, including other parts of his notation.
IMHO, once one understands Dirac's notation properly, nothing is wrong.

@Tom: exactly what part of Dirac's explanation do you think is wrong??

 Where does Dirac say that?
I'm looking at eqn (12) on page 89, or (23) on page 91. I didn't really get why that is. Maybe it's as tom.stoer described?

And my main problem is actually with (46) on page 94 - that there isn't a minus there, as it would follow from sloppy calculations. This is mind-boggling to me!
 Coming back to eqn 46 - I've just understood it. I thought there was a contradiction. Thanks everyone for giving me an impulse for thinking about all this more carefully. I would still love to hear replies about $\hat{\partial_q} |\psi>$ though. It's very interesting, and thought-provoking.

Recognitions: