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Attractive forces

  1. Aug 31, 2009 #1

    The unperturbed energy of the ground state a quantum system is denoted E0. Now the perturbed energy of the ground state of the same quantum system is denoted E'.

    If we have that delta V = E' - E0<0, then my book says that the perturbation is an attractive force.

    Why is that?
  2. jcsd
  3. Aug 31, 2009 #2
    It depends on how exactly "force" is defined. Suppose the strength of the perturbation is controlled by a physical degree of freedom. E.g. consider two atoms at some distance d, and take the perturbation to be the electrostatic interaction between the two atoms. If d is infinite you have the unperturbed Hamiltonian, and the smaller you make d, the larger the perturbation becomes. If the perturbation to the energy is negative, then that means that the total energy of the atoms held at rest would become less if you were to move them closer.

    If the atoms can move freely, then due to conservation of energy their kinetic eergy would have to increase. So, the "total energy at rest" gives you the effective potential energy and minus the gradient of that is the force.
  4. Aug 31, 2009 #3
    It arises when looking at two neutral, polarizable atoms. Here we show that the perturbation (the Coulomb interaction between the atoms) causes an attractive potential proportional to the inverse sixth power of their separation.

    This is the van der Waals interaction.

    What I cannot understand is, why delta V = E - E0 (see first post) implies an attractive force.
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