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Momentum four vector for ground state.

  1. May 23, 2010 #1
    I am confused about the definition of ground state and its characteristics. For a field in spacetime do we mean the ground state to be the minimum of the potential or minimum of the total energy? Specifically what is the momentum four vector of the ground state ([tex] p^{\mu}=(...) [/tex]) for a massive and for a massless field? In my opinion, for massless case, [tex] p^{\mu}=(0,0,0,0) [/tex] and for massive case, [tex] p^{\mu}=(m,0,0,0) [/tex] . Is that correct? How does the potential [tex] V(x) [/tex] enter in the four momentum? In my expression for the four momentum, I have not included the potential in the four momentum.

    Finally what is the momentum four vector for a supersymmetric ground state?
    Last edited: May 23, 2010
  2. jcsd
  3. May 24, 2010 #2


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    I think you are confusing the ground state of the theory with the lowest energy state of a given particle. The ground state of a QFT should have zero momentum as this is the only state that does not violate Poincare symmetry; that fixes it to be [tex] p^{\mu}=(E,0,0,0) [/tex]. In the massless case (photons) you can reduce the photon energy towards zero by making its wavelength longer and longer. But that means you still have a one photon in your state.

    In the massive case you see that the true ground state of the theory with [tex] p^{\mu}=(0,0,0,0) [/tex] means that there is no electron at all. And this is the ground state of the QFT (not the lowest energy state of one electron, but of no electron). In addition, a state like [tex] p^{\mu}=(m,0,0,0) [/tex] would still carry spin, whereas [tex] p^{\mu}=(0,0,0,0) [/tex] does not. This applies to the photon case as well; a photon with infinite wavelength would still carry spin 1, that means you should consider a state w/o photon as the true ground state.
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