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I Why does the kinetic operator depend on a second derivative?

  1. Apr 9, 2016 #1
    The formula T = -(ħ/2m)∇2 implies that T is proportional to the second spatial derivative of a wavefunction. What is the origin of this dependence?

    In classical mechanics, T = p2/2m. Is it also the case in classical mechanics that p2/2m is proportional to a second spatial derivative? I have not been able to relate (d/dt)2 to d2/dx2 using classical mechanics.
    Thanks,
    Steven
     
  2. jcsd
  3. Apr 9, 2016 #2
    check the expression for T -i think it should be hbar^2 may be a typo.
    as we know the kinetic energy operator in coordinate representation is the above expression as differential operator.
    it can be derived from the operator form of momentum

    P(operator) = -ihbar. Del or in one dimension -ihbar. d/dx
    as T= P^2/2m the operator form will be -hbar^2 /2m . d^2/dx^2
     
  4. Apr 9, 2016 #3
    Thanks! Let me make sure I understand your explanation: squaring the p(operator) gives (-ihbar∇)^2 = -hbar^2*(∇∙∇), and ∇∙∇ = ∇^2 because the dot product of a vector with itself is equal to the magnitude of that vector squared. Therefore, T = p^2/2m = (-hbar^2/2m)∇^2. The kinetic energy of a classical wave would also be proportional to ∇^2, because the momentum of a classical wave is proportional to ∇.
     
  5. Apr 9, 2016 #4

    blue_leaf77

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    Yes.
    No, the macroscopic, nonrelativistic kinetic energy is always given by ##\frac{1}{2}mv^2##.
     
  6. Apr 9, 2016 #5

    bhobba

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    Strange but true, and one of the deepest discoveries of modern physics, the answer is symmetry. You will find the detail in chapter 3 of Ballemtine:
    https://www.amazon.com/Quantum-Mechanics-Modern-Development-Edition/dp/9814578584

    It involves advance math but is still worth taking a look at.

    Thanks
    Bill
     
  7. Apr 9, 2016 #6

    PeroK

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    You're getting close to 500 posts now. Time to learn some Latex!
     
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