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Momentum operator

  1. Jan 30, 2012 #1

    dpa

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    why is complex number i involved in defining momentum operator p
    I mean Px=-ih......
    What has complex number to do with momentum.

    I do get however that i in other cases of quantum mechanics has to do with euler's formula that comes from harmonic nature of wave.
     
  2. jcsd
  3. Jan 30, 2012 #2

    bhobba

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    Its not a matter of definition - you can actually derive it from Galilaian invariance - you can find the derivation in Ballentine - Quantum Mechanics - A Modern Development. But aside from that the derivative part is a pure imaginary operator and real valued operators should - well be real - multiplying it by i makes it real.

    Thanks
    Bill
     
  4. Jan 30, 2012 #3

    Ken G

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    And if you think about it from the perspective of kinetic energy, you can say the same thing like this. Confining a particle in a small region requires giving the particle kinetic energy, and it also requires getting the particle's wave function to have a negative second derivative (so you can get the wavefunction to be large in the confined region and small outside of it). Since KE ~ P2, the momentum must inherit a meaning related to the square root of the kinetic energy. The square root of a second derivative is like d/dx, but a square root of a minus second derivative is like i d/dx.
     
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