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Does every quantum field have a non-zero groundstate?

  1. Apr 12, 2012 #1
    Does every quantum field have a non-zero groundstate? I understand that the energy ground state is non-zero, due to the uncertainty principle virtual particles pop in and out of the vaccum. But is this true for any quantum field (higgs field, for example)?

    Thanks, Mark
     
  2. jcsd
  3. Apr 12, 2012 #2

    Jano L.

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    Probably you refer to the fact that the lowest possible energy of harmonic oscillator, defined in a standard way as the first eigenvalue of the operator

    [tex]
    H=\frac{p^2}{2m}+\frac{1}{2}m\omega^2 x^2
    [/tex]

    is [itex]\hbar \omega /2[/itex], which is positive.

    However, it is important to realize that energy can be defined up to arbitrary constant, so the value of the energy alone does not tell anything important.

    The really important thing is that the position and momentum have nonzero statistical scatters, which imply particle is still moving when the wave function is that adjoined to the lowest eigenvalue of energy.

    I do not know much about what happens in quantum theory of field, but I have heard there is not even ground-state solution known (vacuum).
     
  4. Apr 16, 2012 #3
    Thanks very much Jano, that help. -Mark
     
  5. Apr 16, 2012 #4
    Yes, that is true for any quantum field. When you measure very accurately the vacuum for a given field, you will measure non-zero magnitudes for this field. But then also, a split moment later the magnitude and/ or the direction of the field will be completely different.

    How large the magnitudes and the fluctuations are, depends on how accurate you measure. When you measure at very short distances and very short time intervalls, the magnitudes and fluctuation will be higher.

    For the most fields the fluctuations cancel each other out and give the field a vanishing vacuum expectation value (VEV). The values vary from one point to another and one time to the next, resulting in a zero VEV.

    For one field that this is assumed not true and a non-zero VEV is postulated, and that is the Higgs field.

    Also very important, note the difference between the magnitude of a field and the energy or energy density of a field. The energy density depends only on the field strength, not on its direction. So while the field fluctuates back and forth its energy density does not average out to zero. Quite to contrary, if you zoom in into smaller space and time intervals, the energy becomes infinte! Normally, physicist stop at the Planck scale, saying at that point some new and yet unknown physics must come into play. But even though this prevents the energy of the vacuum from becoming infinite, when we include all the vacuum fluctuation (of all kind of fields that exist!) down to planck scale, we still get a crazy huge number for the energy density of the vacuum! 10^88 tons per cubic centimeter!

    Of course, this not what we observe. And that what makes it one of the biggest puzzle in physics.
     
    Last edited: Apr 16, 2012
  6. Apr 17, 2012 #5

    Demystifier

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    Bosonic fields have positive groundstate energy, while fermionic fields have negative groundstate energy. Supersymmetric theories have equal number of bosonic and fermionic fields, so that their groundstate energies cancel up.
     
  7. Apr 17, 2012 #6
    Thanks Lapidus, that helps me think a little more on what spontaneous symmetry breaking means.

    Thanks Demystifier - can I build on your answer? I think it must be related to this statement "The wavefunction is said to be symmetric (no sign change) under boson interchange and antisymmetric (sign changes) under fermion interchange. This feature of the wavefunction is known as the Pauli principle."

    Can you tie the energy groundstates for fermions vs. bosons to this statement?

    Thanks, Mark
     
  8. Apr 17, 2012 #7

    A. Neumaier

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    The vacuum _is_ the ground state of every quantum field theory. It is completely inert; nothing can pop in and out of the vacuum = ground state. (One needs energy provided by fields to create particles. The presence of a field means that one is no longer in the ground state.)
     
  9. Apr 18, 2012 #8

    Demystifier

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    Yes, these two things are definitely related, but a detailed explanation would require more formalism (which can be found in many textbooks on quantum field theory.)
     
  10. Apr 18, 2012 #9
    Are you saying there is no such thing as vacuum energy? That the vacuum is completely empty, devoid of anything? When vacuum is completely inert, what does that imply for the cosmological constant problem?
     
  11. Apr 19, 2012 #10

    Demystifier

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    No, he is (correctly) saying that vacuum is inert, that no changes happen in vacuum. Being inert is not the same as having no energy.
     
  12. Apr 19, 2012 #11
    How do you know? When I look/ measure for fields in the vacuum, the vacuum is a buzzling sea of fluctuations.
     
  13. Apr 19, 2012 #12
    Also, how can something have energy but does not change in quantum physics?

    The particle in the groundstate of the harmonic oscillator does not sit still like in classical physics.
     
  14. Apr 19, 2012 #13

    martinbn

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    A pedantic remark, you don't mean non-zero as the ground state is always a non-zero vector.
     
  15. Apr 19, 2012 #14

    Demystifier

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    I calculate it from the theory.

    I guess you calculate it from the theory as well, since I think there is no such experiment performed in practice. But even the theory says that, to measure something, you need a measuring apparatus. And clearly, the presence of the measuring apparatus implies that the total state is not the vacuum.
     
  16. Apr 19, 2012 #15

    Demystifier

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    The wave function of the ground-state (or any energy eigenstate) has a trivial time dependence proportional to exp(-iEt), so the probability density of any observable is time-independent. In other words, nothing observable changes.

    To have any observable change at all in quantum physics, the system must be in a state which is not an energy-eigenstate.
     
  17. Apr 19, 2012 #16

    A. Neumaier

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    Yes. Everything you can compute about the vacuum is constant.
    Nothing, as currently there is no accepted quantum field theory featuring the cosmological constant.
     
  18. Apr 19, 2012 #17

    A. Neumaier

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    How do you look at or measure fields in the vacuum? If you can, you don't have a vacuum.

    However, nobody ever has measured or seen a buzzling sea of fluctuations. This is just fashionalbe visualization imagery for laymen.
     
  19. Apr 19, 2012 #18

    A. Neumaier

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    It only changes an immaterial phase without any influence on measurements.
     
  20. Apr 19, 2012 #19

    Demystifier

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    How about the correlation function <0| phi(x) phi(y) |0> ?
     
  21. Apr 19, 2012 #20

    A. Neumaier

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    It is translation invariant; changing x and y by the same 4-vector doesn't change the correlation function. Thus what is observable about it is time-independent. (Relative times have no simple meaning in quantum field theory.)
     
    Last edited: Apr 19, 2012
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