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I Quantum mechanics getting position operator from momentum

  1. Sep 19, 2016 #1
    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. jcsd
  3. Sep 19, 2016 #2

    DrClaude

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    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##.
     
  4. Sep 19, 2016 #3
    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....
     
  5. Sep 19, 2016 #4

    vanhees71

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    Science Advisor
    2016 Award

    What's the goal of these manipulations? For the term in question you use the self-adjointness of the momentum operator to get
    $$\langle p|\hat{p} \hat{x} \psi \rangle=\langle \hat{p} p|\hat{x} \psi \rangle=p \langle p|\hat{x} \psi \rangle.$$
     
  6. Sep 19, 2016 #5

    Demystifier

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    Science Advisor

    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?
     
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