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Vacuum fluctuations

  1. Jun 26, 2011 #1
    Do fields have quantum fluctuations and are non-zero in the vacuum?

    Non-zero in the same way that for a harmonic oscillator the wave function is non-zero for values of the position of the harmonic oscillator?

    Or are they considered as 'mathematical fiction' just as 'virtual' particles?

    thanks
     
  2. jcsd
  3. Jun 26, 2011 #2
    As far as I know the term "vacuum fluctuation" is assigned just to virtual particles spontaneously coming to "existence".
     
  4. Jun 26, 2011 #3

    alxm

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    Let's take a case from standard non-relativistic QM:

    Consider two noble gas atoms in vacuum, separated by a significant distance. They interact via the Coulomb potential (so it's instantaneous, no fields or mediation going on here). So the total wave function of the system can't be the sum of the wave functions of the two atoms, since there's an interaction going on. It's basically an entangled state, the state of atom 1 is not independent of the state of atom 2.

    This gives rise to the London dispersion force - the total energy of the true system is lower than the system of two non-mutually-interacting atoms. You can describe this as being due to the quantum uncertainty or 'fluctuations' in electron density since the contribution is by definition the contribution from the instantaneous, position-dependent interaction between the electrons. But it's the ground state and so it's time-independent, so there are no actual measurable temporal 'fluctuations' going on.

    So how do you calculate that interaction? Again, assume you know the wave functions of the isolated individual atoms. So you can use that as a basis and use the Ritz variational method or perturbation theory. As I already said, the ground states of the two isolated atoms is not the true ground state of the interacting pair of atoms. So if I describe my interacting wave function in that basis, I'll have contributions from the "virtual" excited states of those non-interacting atomic wave functions. That does not mean that those atoms are actually in an excited state, because those states are not eigenstates of the true Hamiltonian. These "excitations" only exist because of the choice of basis. The electrons aren't in excited states - which should be obvious: If you have two atoms in their electronic ground states and bring them into proximity, lowering their mutual ground state energy, how could that lead to a real excitation?

    Now realize that the electromagnetic field is itself quantized and subject to the same kind of quantum behavior. If you take that into account in the above situation, you have an additional contribution to the dispersion force called the Casimir effect. But if you like, you can also simply consider a single atom and its interaction with the field, rather than another atom. Again you have contributions to its behavior because of that interaction. And again, you can describe those contributions in terms of 'virtual' excited states, which in the case of the field is what you call 'virtual particles'. And the reason is the same, you're starting from a non-interacting description.
     
  5. Jun 28, 2011 #4
    Thanks for the anwers, especially alxm, nice one!
     
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