I know that operators in quantum mechanics act on states (kets, assumed to be normalized), and return another state. So, an operator such as $\hat{p}$ when acted on a ket $|\psi\rangle$ will give me another ket $|\phi\rangle$. Thus, $$\hat{p}|\psi\rangle = |\phi\rangle.\tag{1}\label1$$ Also, $\langle x|\psi\rangle =\psi(x)$ if $x$ is the position. Now, since $\psi(x)$ is a complex number, what I don't understand is how can the operator $\hat{p}$ cat on $\psi(x)$? How does the expression $\hat{p}\psi(x)$ make sense if $\psi(x)$ does not lie in the domain of the operator/function $\hat{p}$? And how does $\hat{p}\psi(x) =\langle x|\hat{p}|\psi\rangle$?
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Qmechanic
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Apoorv Potnis
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1$\hat p\psi(x):= \langle x\vert \hat p\vert \psi\rangle$ is a definition. – ZeroTheHero Jul 31 '20 at 16:26
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5You might have it easier if you see $\hat{p}\psi(x)$ as $(\hat{p}\psi)(x)$. – ɪdɪət strəʊlə Jul 31 '20 at 17:01