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Negative kinetic energy?

  1. Jun 5, 2007 #1
    For a particle trapped in a potential that is greater than its total energy, the physical intepretation is that it has a kinetic energy<0. Is that reasonable? Since its potential is goverened by the potential of the well it is in and is fixed. The only value that can change is kinetic energy. OR in that situation should be abolish the term kinetic energy and define something else?
  2. jcsd
  3. Jun 5, 2007 #2


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    Kinetic Energy is never negative. If a Particle is bound in some kind of potential its TOTAL ENERGY is negative (this is characteristic of a bound system, classical or quantum), but its KINETIC ENERGY is still positive.
  4. Jun 6, 2007 #3

    Kinetic energy can be negative in GR.
  5. Jun 6, 2007 #4
    In what context, careful?
  6. Jun 6, 2007 #5
    Are you sure about that? I mean, the wavevector of a particle inside a barrier is complex so that we get exponentially decaying wavefunctions instead of oscillating ones, right? This means that the kinetic energy [tex]\hbar^2 k^2 / 2 m < 0[/tex] since [tex]k[/tex] is complex.
  7. Jun 6, 2007 #6


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    Kinetic energy is defined by the positive hermitian operator
    [tex]\hat{p}^2 / 2 m[/tex]
    The operator [tex]\hat{p}[/tex] is hermitian so its eigenvalues cannot be imaginary, but only real.
    A wave function [tex]e^{kx}[/tex], which is formally an eigenstate of the momentum operator with an imaginary eigenvalue, is not square integrable even as a functional (a plane wave is not square integrable either, but it is square integrable at least as a functional), so it is not a physical state. In a barrier, the wave function has the form [tex]e^{kx}[/tex] only on a small portion of space, not everywhere, so such a wave function is not really an eigenstate of the momentum operator. The momentum is a property of the whole wave function, not of a wave function on a small portion of space.

    Although the local momentum does not make sense in the usual formulation of QM, it does make sense in the Bohmian formulation. In this case, the local momentum associated with a wave function proportional to [tex]e^{kx}[/tex] is - zero. A local particle does not move and does not have a kinetic energy in this region.
    Last edited: Jun 6, 2007
  8. Jun 6, 2007 #7
    Cesiumfrog, write out the Hamiltonian constraint, the terms quadratic in the momentum are indefinite : 1/2 (P_c^c)^2 - P_ab P^ab. The ``potential'' term is the densitized Ricci curvature of the spacelike metric and can be positive and negative as well.
    Last edited: Jun 6, 2007
  9. Jun 9, 2007 #8
    I'm reminded of quantum tunneling. When, say, an electron is tunneling through a potential barrier then during the time the particle is inside the barrier the particle's kinetic energy is zero. This, of course, assumes that the potential energy of the barrier is zero at a certain point so we can meaningfully say that the particle is inside the potential and the potential is zero otherwise (or something of that manner.

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