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How does the quantum vacuum appear to different inertial observers?

  1. Jul 14, 2013 #1
    According to QM the ground state of the quantized empty space is not considered to be empty but as a fluctuating sea of virtual particles creating and annihilating continuously. In the case of virtual particles with mass I have a problem with this. According to which reference inertial frame are these supposed to appear? If two inertial observers look at the same region of space (say one in a laboratory at rest with earth, the other one traveling throughout space at nearly the speed of light, but both examining the same vacuum), would they agree? According to QFT do they see the same Lorentz transformed particles appear, or something completely different?

    Please note that I'm not asking about the Unruh effect, I'm considering only inertial reference frames.
     
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  3. Jul 14, 2013 #2

    Bill_K

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    No, that's just what we tell people who don't have enough physics background to understand the quantum mechanical truth. The vacuum state is not at all "rapidly fluctuating", in fact it's time independent. It's a time-independent superposition of states containing various numbers of particles. Saying it fluctuates is like saying Schrodinger's cat rapidly fluctuates between being alive and dead.

    The vacuum state is Lorentz invariant. It appears the same in all reference frames.
     
  4. Jul 14, 2013 #3
    Do you mean a superposition of non-interacting particle states? Because if you mean the particle eigenstates of the full-interacting Hamiltonian, then it would no longer be the vacuum if it contained those states.
     
  5. Jul 14, 2013 #4

    Bill_K

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    No of course not, not real particle states. Virtual particle pairs. The same virtual particle pairs that are said to "rapidly pop in and out of existence". Only they don't pop.
     
  6. Jul 14, 2013 #5
    What do virtual particle states look like? The eigenstates of the full Hamiltonian form a basis. The virtual particle states would have to be a linear combo of the fully interacting eigenstates?

    As an example, the vacuum for a free-field theory should look like the product of the zero-point energy wavefunctions for a collection of uncoupled harmonic oscillators, exp[-x^2 ωk] where ωk is the energy of a mode.

    The interacting vacuum I have no idea what it is, but if it's a linear combo of virtual particle pairs, then what are the virtual particle pair states and what do they look like?

    The way that I see in textbooks is usually they express the interacting vacuum |Ω> as a linear combination of the noninteracting eigenstates |n> of H0, |Ω>=|n><n|Ω>. Then there is a trick that can be done by applying eiHt to both sides, give time t a bit of an imaginary part, and then because the ground state decays slower in that limit, you can express
    |Ω> in terms of |0>.

    So I've seen |Ω> written as a linear combination of on-shell, non-interacting states |n>, but not of virtual particle states.
     
  7. Jul 14, 2013 #6

    Bill_K

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    I did not say virtual particle states. I think you're not even being serious with me, geoduck.

    As I'm sure you're well aware, virtual particles arise in perturbation theory. The transition from an in state to an out state is expressed as a sum over Feynman diagrams. Each diagram contains internal lines, which stand for virtual particles. Are we clear so far?

    The vacuum-to-vacuum amplitude consists of diagrams containing virtual pairs of consisting of particles and their antiparticles. The total amplitude is a sum over all such diagrams.
     
  8. Jul 14, 2013 #7
    I must have gotten confused. I thought you meant the virtual particles are states.

    It's true that the virtual particles don't have to be on-shell. But that I believe is due to a weird trick where a 3-dimensional momentum integral is made into a 4-d one and the poles bring things back on shell. The propagator can be derived by inserting on-shell states between the θ(x) and θ(y) fields in the expression: <Ω|θ(x)θ(y)|Ω>. The completeness relation requires that the states are over all total 3-momenta. But a trick converts this to 4-momenta which goes off-shell.

    However, the insertion of complete states between the fields in <Ω|θ(x)θ(y)|Ω> doesn't really involve the vacuum |Ω>, at least I don't think. Well maybe to calculate expressions like <Ω|θ(x)|k>, there needs to be connection between <Ω| and |k>. It's actually hard for me to see why virtual particles are properties of the vacuum. It seems like virtual particles are more the result of inserting |k><k| between two fields then anything to do with the vacuum |Ω>.
     
  9. Jul 16, 2013 #8

    tom.stoer

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    Virtual particles are artifacts of perturbation theory.

    In this context the following is important:
    1) Virtual particles are not Hilbert space states but propgatators, i.e. special matrix elements. So they are not subject to "forming a basis", "having properties in terms of quantum numbers", "being eigenstates of observables", ...
    2) In addition these virtual particles cannot be physical observables b/c virtual particles can even be gauge dependent. You can formulate QCD in physical gauges using two transversal gluon polarizations, or you can use unphysical gauges with four gluon polarizations plus ghosts (cancelling unphysical polarizations). So two observers may not even agree whether they observe a ghost or not - dependent on their gauge choice. But this is nonsense, the observation cannot depend on the gauge fixing condition I am writing on a sheet of paper.
     
  10. Jul 17, 2013 #9
    When you say special matrix elements, do you mean just one component of a matrix, the vacuum-vacuum element (vacuum expectation value)?

    It's not just in popular books, but I believe I've seen textbooks that say that the vacuum has particles and antiparticles in it, and this does things like shield the electric charge in QED (or strengthen it in QCD). I'm wondering are these particles and antiparticles in the vacuum virtual particles?

    I was thinking maybe what they mean is that <0|θ(x)|k><k|θ(y)|0>, where the sum is over Hilbert space states |k> (including multiparticle ones), is nonzero in any interacting theory. So out of the vacuum the field θ(x) can connect to a multiparticle state |k>. But would |k> have to be a particle/antiparticle state to conserve charge? When you sum over all states |k> including multiparticle ones, this does become the propagator.

    Also, I find it odd that sometimes textbooks talk about creation and annihlation operators for an interacting theory. I thought creation and annihlation operators are a result of the harmonic oscillator structure of the Hamiltonian, so making the Hamiltonian not harmonic by adding interactions disallows the creation/annihilation construction.

    It's all rather confusing to me at this point.
     
  11. Jul 18, 2013 #10

    tom.stoer

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    The states |k> and the propagators are two different entities. In Feynman diagrams the propagators (= internal lines) a called virtual particles, but as you see from your construction they are related with the states |k>, but they are not identical.

    Looking at the algebraic structure of the propagators you see that they do not carry quantum numbers or eigenvalues like momentum, spin, charge etc. They only mediated between states having these properties. And they are gauge-dependent presenting any direct physical interpretation!

    Strictly speaking for the states |k> you use in the sum, you need not use any special selection rules or conservation laws. The theory itself guarantuees that matrix elements violating them become exactly zero (this is like in QM).

    Regarding creation/annihilation operators: first of all they are nothing else but the Fourier modes of your original field operators, so introducing them and rewriting H and other observables in terms of them is just an algebraic identity. The interaction terms become non-linear or even non-local expressions in terms of these operators, but it's an identity. The problem is that due to Haag's theorem the free-particle Fock space and the Hilbert space of the interacting theory are different, therefore the free-particle operators are not defined in the interacting case (but this is only one of many subtleties in QFT and can be avoided, e.g. by quantizing in a finite volume and using the infinity volume limit on the level of matrix elements)
     
    Last edited: Jul 18, 2013
  12. Aug 1, 2013 #11

    DarMM

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    If there is a cut-off in the interacting terms (usually both an ultraviolet and infrared cutoff are needed), then the interacting field and the free field can be well-defined in the original Hilbert space.

    Of course since the interacting and free cases have different Hamiltonians, they will have a different spectrum, different eigenstates. Of course, like any two Hamiltonians, you can expand the states of one in terms of the other and hence the interacting vacuum can be written as a superposition of
    free n-particle states.

    However this no more means the interacting vacuum is "full of particles" than the Helium ground state is "full of two hydrogen energy levels", just because you can expand the Helium ground state in
    terms of the energy levels of two hydrogen atoms.

    Of course all of this is only with cutoffs. Remove the cutoffs and the two theories live in different Hilbert spaces anyway and so the expansion is impossible.
     
  13. Aug 1, 2013 #12
    Aren't all Hilbert spaces the same, namely the space of all possible field values at each point in spacetime? In QM for example isn't the position space a suitable basis for different Hamiltonians?
     
  14. Aug 1, 2013 #13

    DarMM

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    Just so that I understand how to answer, what do you mean by the Hilbert space being the space of all possible field values? Also, what do you mean by position space being a suitable basis?
     
  15. Aug 1, 2013 #14

    tom.stoer

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    I guess DarMM refers to Haag's theorem

    http://en.wikipedia.org/wiki/Haag's_theorem

    which says that in QFT different representations of operators (H, ...) are not unitarily equivalent and produce different physical results, even if the Hilbert spaces, on which the representations are implemented, are isomorphic (all separable Hilbert spaces are isomorphic).
     
    Last edited: Aug 1, 2013
  16. Aug 1, 2013 #15
    The wave functional forms a complete basis. The Fock basis is another basis. Both completely describe the system?

    In qm you can have many different hamiltonians, but the position basis is always the same, and the energy eigenstates for all the hamiltonians should be equivalent to using the position basis, so aren't these systems in the same vector space?
     
  17. Aug 1, 2013 #16

    tom.stoer

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    In QM you have not "always the same vector space" but you have different isomorphic Hilbert spaces (position space, momentum space, ...) on which you can represent all operators x, p, L, H, ... You can construct unitary mappings between these Hilbert spaces which guarantuee that physical results (the spectrum of observables like H) do not depend on the choice of the Hilbert space.

    In QFT Haag's theorem says that there is no such unitary mapping in general, so different choices i.e. different Hilbert spaces will result in different physics.
     
  18. Aug 1, 2013 #17
    I thought the position and momentum states are vectors in the same vector space, not in different vector spaces isomorphic to each other. Isn't this why you can write momentum states as a superposition of position states, <x|p>=e^(ipx)?

    I thought Haags theorem concerned renaming of fields, or field redefinitions. So for example representing a complex scalar field as two real fields will yield the same physical results, i.e., the redefinition [itex]\phi=\frac{1}{\sqrt{2}}(\phi_1+i\phi_2) [/itex] and it's conjugate give the same results.

    What I always wondered was what about this Lagrangian: [itex]\mathcal L=\partial_\mu \phi^\dagger \partial^\mu\phi-m^2\phi^\dagger\phi[/itex]. Can you perform the field redefinition:
    [itex]\phi(x)=e^{-imt}\xi(x) [/itex] and it's conjugate, which would get rid of the mass2 term? The new Lagrangian (which is written in terms of the [itex]\xi(x) [/itex] fields) you get when you do that will look really different, but can you always redefine the field in this way and get a new Lagrangian which will give you the same results as the old Lagrangian? For classical field theory, the Euler-Lagrange equations have this type of invariance, but for QFT?
     
  19. Aug 1, 2013 #18

    Avodyne

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    Yes, you can do this in QFT. Results that are not sensitive to how the field is defined (such as the S matrix) will be the same.

    If you do a nonlinear transformation, though, like ##\phi(x)=\xi(x)+c\xi(x)^3## for some constant ##c##, then you must also do an appropriate transformation on the functional measure ##\cal D\phi##; this will add new effective terms in the lagrangian that must be included when computing loop diagrams.
     
  20. Aug 3, 2013 #19

    DarMM

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    In the case of a free field theory, the wave functionals form a complete basis (or rather a countable set of them do). This basis can be unitarily related to the Fock basis. However for interacting theories the Wave Functional basis can not be unitarily related to the Fock basis.
    Even worse, in the Fock basis the interacting Hamiltonian cannot be defined, so only the appropriate Wave Functional basis can be used.

    Of course I speaking in a mathematically rigorous sense. In normal perturbation theory, you put in a cutoff, which allows you to use the Fock basis.
     
  21. Aug 3, 2013 #20

    tom.stoer

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    Assume we start with a QFT defined on the compact 3-torus; or we start with lattice gauge theory; afaik these a loop-holes of Haag's theorem, so one can use Fock-operators to write down (and regularize) the full Hamltonian consistently; of course the theory is by no means trivial, even the ground state |Ω> defined as the state which minimizes <H> is complex and does in no way correspond to the Fock state |0>

    Where exactly does the theory break down when sending the torus to infinity, or when looking at the continuum limit of the lattice?
     
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