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Homework Help: More operator questions

  1. Oct 31, 2007 #1
    Consider
    [tex]\langle x'|\hat{p} \hat{x} | x \rangle[/tex].
    Are these steps correct?
    1. [tex]\hat{x}[/tex] operates on the x eigenstate to get
    [tex]\langle x'|\hat{p} x | x \rangle[/tex].
    2. x is a c-number so can be pulled out to get
    [tex]x \langle x'|\hat{p} | x \rangle[/tex].
    3. [tex]x \langle x'|\hat{p} | x \rangle = x \frac{\partial}{\partial x'}\delta (x'-x)[/tex]

    The part I'm most wary on is 2 where I pulled x out. It seems like the p operator is acting on it so I shouln't be able to do that.

    Alternatively, could I do the following:
    1. have p operate on [tex]\hat{x}[/tex] first to get
    [tex] \frac{\partial}{\partial x'}x' \delta (x'-x)[/tex]
    2. Which becomes
    [tex] \delta (x'-x) + x' \delta'(x'-x)[/tex]

    Is one of these two routes incorrect?
     
  2. jcsd
  3. Oct 31, 2007 #2
    Yes, you can do that. Operators do not act on c-numbers.

    Eugene.
     
  4. Oct 31, 2007 #3

    Avodyne

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    Science Advisor

    Both are correct (except you dropped a factor of -i*hbar in line 3, but you also dropped it in the 2nd part). You can check that your two final expressions are equal by integrating them against a test funtion of either x or x'.
     
  5. Nov 1, 2007 #4
    Won't you get the same thing for
    [tex]\langle x'|\hat{x} \hat{p} | x \rangle[/tex]
    though? That would imply the two operators commute.
     
    Last edited: Nov 1, 2007
  6. Nov 1, 2007 #5

    Hurkyl

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    It's not a problem because [itex]\hat{p}[/itex] is not a differential operator: [itex]\hat{p}[/itex] is a linear operator acting on states. Because it's linear, it commutes with all complex numbers.


    [itex]\hat{p}[/itex] only turns into a differential operator when everything's written out in the position representation -- in that case, you would have

    [tex]
    \langle x' | \hat{p} \hat{x} | x \rangle
    =
    \int_{-\infty}^{+\infty} \delta(x' - \xi) \left(-i \hbar \frac{\partial}{\partial \xi} \right)
    \xi \delta(x - \xi) \, d\xi[/tex]

    and now it's differently obvious, since x clearly commutes with [itex]\partial / \partial \xi[/itex].
     
    Last edited: Nov 1, 2007
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