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Dirac Notation - Position and Momentum

  1. Sep 29, 2011 #1
    1. The problem statement, all variables and given/known data

    Show that [itex]\left\langle[/itex]x|p|x'[itex]\right\rangle[/itex] = [itex]\hbar[/itex]/i [itex]\partial[/itex]/[itex]\partial[/itex]x [itex]\delta[/itex](x-x')

    2. The attempt at a solution

    [itex]\left\langle[/itex]x|p|x'[itex]\right\rangle[/itex] = i[itex]\hbar[/itex] [itex]\delta[/itex](x-x')/(x-x') = i[itex]\hbar[/itex] [itex]\partial[/itex]/[itex]\partial[/itex]x' [itex]\delta[/itex](x-x') = [itex]\hbar[/itex]/i [itex]\partial[/itex]/[itex]\partial[/itex]x [itex]\delta[/itex](x-x')

    For the sake of formality I think I need an integral after my first equals sign which I think would be:

    [itex]\int[/itex][itex]\delta[/itex](x-x') p [itex]\delta[/itex](x-x') dx'= p[itex]\int[/itex][itex]\delta[/itex](x-x') p [itex]\delta[/itex](x-x') dx'

    However I'm not sure if a) it's needed, or b) if I set it up correctly. Any help would be appreciated!
     
  2. jcsd
  3. Sep 29, 2011 #2

    vela

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    Suggestion: If you're going to use LaTeX, use it for the entire expression instead of individual symbols. Fixed your post up:
    I actually don't follow what you're doing here at all. :wink: Try inserting a complete set:
    [tex]\langle x| \hat{p} |x' \rangle = \int \langle x | \hat{p} | p\rangle\langle p |x' \rangle \,dp[/tex]
     
  4. Sep 29, 2011 #3
    So, I was right... in that I was wrong! :)

    I'll go back and work with the complete set.
     
  5. Oct 3, 2011 #4
    I'm going to call out my own ignorance (probably related to having spent too much time with a QM book causing less clarity instead of more):

    [tex]\langle x| \hat{p} |x' \rangle = \int \langle x | \hat{p} | p\rangle\langle p |x' \rangle \,dp[/tex]

    I don't see how this is the complete set. (If that question made sense, I'm on a good start.)

    Shouldn't we have:

    [tex]\langle x| \hat{p} |x' \rangle = \int \langle x | \hat{p} | p\rangle dp \langle p |x' \rangle \[/tex]?
     
    Last edited: Oct 3, 2011
  6. Oct 3, 2011 #5
    Wait, does this work?

    [tex]\langle x| \hat{p} |x' \rangle = \int \langle x | \hat{p} | p\rangle\langle p |x' \rangle \,dp[/tex]

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

    [tex]\langle x| \hat{p} |x' \rangle = ( \hbar /i ) ( d/dx ) \delta(x-x')[/tex]
     
  7. Oct 3, 2011 #6

    vela

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    What are [itex]\langle p \vert x' \rangle[/itex] and [itex]\langle x \lvert \hat{p} \vert p \rangle[/itex] equal to?
     
  8. Oct 3, 2011 #7

    diazona

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    Just a side note:
    They're the same thing. [itex]\mathrm{d}p[/itex] is a pure number and thus commutes with everything, in particular [itex]\langle p|x'\rangle[/itex].
     
  9. Oct 4, 2011 #8
    The easier question to answer is what is [itex]\langle p \vert x' \rangle[/itex] equal to

    [itex]\langle p \vert x' \rangle = 1/ \sqrt{2 \pi \hbar} exp( (i/ \hbar) p x') [/itex]

    The more important question is then what does [itex]\langle x \lvert \hat{p} \vert p \rangle[/itex] equal, because I don't think I know anymore! Except to say that it has matrix elements (and I'm not sure what those are).
     
  10. Oct 4, 2011 #9

    vela

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    Surely you must know what [itex]\hat{p}\vert p\rangle[/itex] is equal to.
     
  11. Oct 4, 2011 #10
    Well my first gut reply is that

    [itex]\hat{p}\vert p\rangle = p\vert p\rangle[/itex]
     
  12. Oct 4, 2011 #11

    vela

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    Right, so [itex]\langle x | \hat{p} | p \rangle = \langle x | p | p \rangle = p\langle x | p \rangle[/itex].
     
  13. Oct 4, 2011 #12
    Ah right! Ok, so what's the mathematical reasoning that gets me from [itex]\langle x| \hat{p} |x' \rangle[/itex] to [itex]\int \langle x | \hat{p} | p\rangle\langle p |x' \rangle \,dp[/itex] ? I'm really trying to understand the basic underpinnings of what's going on.
     
  14. Oct 5, 2011 #13
    I'm considerably more comfortable with the original problem. Now I've got a follow up (thanks to yesterday's class). We have

    [itex]\langle x | \hat{p} | p \rangle = \langle x | p | p \rangle = p\langle x | p \rangle[/itex]

    If [itex]p = i / (\hbar) d/dx[/itex] and [itex]\langle x | p \rangle = 1 / ( \sqrt{2 \pi \hbar} ) exp[ (i / \hbar) p x) ][/itex] How do I go about determining matrix elements? Is it as simple as the product of p and [itex]\langle x | p \rangle[/itex] and then taking x = 0,1,2,...?
     
  15. Oct 5, 2011 #14

    vela

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    No. I don't understand what you're trying to do.
     
  16. Oct 5, 2011 #15
    I would like to determine matrix elements for [itex]\langle x | \hat{p} | p \rangle[/itex].
     
  17. Oct 5, 2011 #16

    vela

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    I think you're confusing the operator [itex]\hat{p}[/itex] with the eigenvalue [itex]p[/itex] of the momentum eigenstate [itex]\vert p \rangle[/itex].
     
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