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Square of a momentum operator

  1. Aug 25, 2012 #1
    know I'm missing something obvious.

    for a momentum operator p = -iħ d/dx

    if I square the -iħ part I get (+1)ħ2

    but I believe the correct value (as in the kinetic energy of the Hamiltonian) is

    -ħ/2m d2/dx2.

    how is the value of the term -ħ/2m where the square of -i = +1?

    Thanks!
     
  2. jcsd
  3. Aug 25, 2012 #2

    jedishrfu

    Staff: Mentor

    i * i = -1 and so -i * -i = -1 * i * -1 * i = 1 * i * i = i * i = -1
     
  4. Aug 25, 2012 #3
    Kinetic energy is given by
    K = (1/2)*m*v^2 = (p^2)/(2m).
    (-ih)^2 is actually equal to -h^2, because (-i)*(-i) = -1.
    Hence, K = (p^2)/2m = (-h^2/2m)*(d^2/dx^2)
     
  5. Aug 26, 2012 #4
    this suggests, i x i = -1 and -i x -i = -1 x i x -1 x i = 1 x -1 = -1

    right?
     
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