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Why forces are modeled as virtual particles in quantum mechanics?

  1. Apr 13, 2008 #1
    Why forces are modeled as virtual particles in quantum mechanics and how virtual particles can transmit force between two real particles?
     
  2. jcsd
  3. Apr 13, 2008 #2
    Amplitudes in quantum field theory / quantum mechanics are usually calculated via path integrals, which is a sum over possible "paths" a system might take through configuration space. In ordinary quantum mechanics, the path integral is a sum over trajectories the particle can have, hence the name, but these trajectories need not look physical, i.e, they need not conserve energy/momentum/whatever and they look more like random walks. But this is ok because they are intermediate states, and aren't actually observed.

    So in quantum field theory, the path integral can similarly be written as a sum over intermediate states, and this is where virtual particles come in.

    While the above answer is more insightful, I'll whet your appetite with something less rigorous. Lets say you have a Hamiltonian [tex]H_0[/tex] of which free particles are eigenstates, and you turn on an interaction of the form [tex]H_i = q \bar\psi \gamma^\mu A_\mu \psi[/tex], which is an electromagnetic coupling between say, an electron and a photon. Now, originally your time evolution operator was [tex]U(t, 0) = e^{-i H t}[/tex], but now there is an interaction term, so we get [tex]U(t,0) \approx T \exp\left[ - i \int\!dt\, ( H_0 + q \bar \psi \gamma^\mu A_\mu \psi ) \right] [/tex] (I write it in this way because operators at different times no longer commute, so you have to put time ordering [tex]T[/tex] in.)

    For arguments sake, then, lets compute the transition amplitude for a state of two electrons to a state with two electrons. You'll essentially want to calculate [tex]\left\langle e,e | U(t, 0) | e,e \right\rangle[/tex]. Now when you expand that exponential, you'll get terms with one or more interaction terms in it, and the terms in your expansion can be represented as Feynman diagrams. To finish the calculation, you "contract" each term which means that you evaluate the expectation value perturbatively, and pairwise contractions of operators give you "propagators" which are represented by internal particles propagating, and "vertices" which are represented by interactions between particles.

    I've waved my hands a bit here, and what I've written down isn't entirely correct (Hell, I don't even think my interaction term is correct... where did covariance go? heh heh), but I hope you'll get the idea. I normally deal with path integrals and not in the "interaction picture" as this is called, so things are done more cleanly and more efficiently.
     
    Last edited: Apr 13, 2008
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