I Quantum mechanics getting position operator from momentum

1. Sep 19, 2016

mertcan

hi, initially I want to put into words that I looked up the link (http://physics.stackexchange.com/qu...-the-momentum-representation-from-knowing-the), and I saw that $$\langle p|[\hat x,\hat p]|\psi \rangle = \langle p|\hat x\hat p|\psi \rangle - \langle p|\hat p\hat x|\psi \rangle = \langle p|\hat x\hat p|\psi \rangle - p\langle p|\hat x|\psi \rangle$$
But I can not understand how $$\langle p|\hat p,\hat x|\psi \rangle=p\langle p|\hat x|\psi \rangle$$ is possible.
why do we have $$p$$ and $$\hat p$$ in the former and just $$p$$ in the latter???? What is the logic and proof of this kind of transformation?????? AND why do we lose the $$\hat p$$ term????
I am looking forward to your valuable responses.....

2. Sep 19, 2016

Staff: Mentor

$|p\rangle$ is an eigenstate of the momentum operator:
$$\hat p | p \rangle = p | p \rangle$$
where $p$ is the value of the momentum in state $|p\rangle$.

3. Sep 19, 2016

mertcan

I know that fact $$p$$ is eigenvalue and $$p\rangle$$ is eigenvector but still can not use this situation in the equation I have given....A little bit more help....

4. Sep 19, 2016

vanhees71

What's the goal of these manipulations? For the term in question you use the self-adjointness of the momentum operator to get
$$\langle p|\hat{p} \hat{x} \psi \rangle=\langle \hat{p} p|\hat{x} \psi \rangle=p \langle p|\hat{x} \psi \rangle.$$

5. Sep 19, 2016

Demystifier

In the latter, we don't have just $p$. We have two $p$'s. One $p$ is in $\langle p|$ and the other $p$ is on the left from that. So we don't lose $\hat p$. We replace $\hat p$ with $p$ and put it on the left.

Perhaps you ask why can we put it on the left?