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Quick question: Momentum operator in QM

  1. Feb 1, 2013 #1
    1. The problem statement, all variables and given/known data

    There are two ways to write the momentum operator, p = (-i hbar d/dx) and p = (hbar / i)d/dx. How do you go from one to the other?

    2. Relevant equations

    The two I gave above.



    3. The attempt at a solution

    I tried to see if -ih = h/i by squaring both sides, but one came out positive and the other negative. Thanks for the help!
     
  2. jcsd
  3. Feb 1, 2013 #2
    -i^2 = 1, not -1. This makes (-i hbar)^2 not equivalent to (hbar / i)^2
     
    Last edited: Feb 1, 2013
  4. Feb 2, 2013 #3
    Since when did [itex](-i)^2 = 1[/itex]?
     
  5. Feb 2, 2013 #4

    Intrastellar

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    Gold Member

    [itex](-i)^2 = (-1)^2(i)^2[/itex]
     
  6. Feb 2, 2013 #5
    [itex] i^{-1} =- i [/itex]
    Why?
    [tex]
    i^{-1} = e^{-ln(i)}=e^{-ln(e^{i\pi /2})}=e^{-i\pi /2}=-i
    [/tex]
    since [itex]e^{\pm i\pi /2}= cos(\pm\pi /2) + isin(\pm\pi /2)=\pm i [/itex]
     
  7. Feb 2, 2013 #6

    CompuChip

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    Science Advisor
    Homework Helper

    Or you just multiply 1/i by i/i to get i/i^2 = i/(-1) = -i.
     
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