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QED vacuum and Lorentz invariance

  1. Mar 17, 2015 #1
    The measured energy density of the vacuum has a disturbing discrepance with the one theorized by imposig Poincare invariance in QFT, usually referred to as the "vacuum catastrophe".
    On the other hand the Heisenberg indeterminacy principle leads to a nonzero vacuum expectation value for the Hamiltonian of the quantum fields values. This seems independent of perturbative or non-perturbative treatments unless the HUP is considered itself a mathematical artifact.
    What would be the arguments countering the above apparent obstructions to Lorentz invariance of the QED vacuum?
     
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  3. Mar 18, 2015 #2
    Admitedly this is a difficult question, mathematically it comes down to the theorem by Haag that says that in the interacting relativistic quantum field theory the fields polarize the vacuum, and therefore its vacuum state lies inside a renormalized Hilbert space that is not unitarily equivalent to the Hilbert space of the free field. This in practice means that whatever the existing interacting theory is that is giving the outstanding results about the magnetic moment of the electron, the lamb shift etc, it is mathematically not a Lorentz covariant interacting QFT.
    This awkward result is usually ignored in QFT textbooks and QFT literature in general but hasn't been mathematically contested since Haag published it in 1955. But there are practical consequences, like obstructions to certain Wightman axioms, or to the assumption of a Poincare invariant vacuum needed for a rigorous proof of the spin-statistics theorem...
     
  4. Mar 19, 2015 #3

    strangerep

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    Depends what you mean by "Lorentz covariant". The operators in an interacting QFT do satisfy the Poincare commutation relations (it's called an "interacting representation"). The vacuum state in the interacting theory is defined to be Poincare invariant under the interacting representation. The tricky bit is that it cannot (in general) be the same vacuum as in the free theory (which is constructed using a different representation of the Poincare group).
     
  5. Mar 20, 2015 #4
    Yes, and this means that since their representations are not unitarily equivalent one of them must not even be unitary, and that is quite a disaster. My point about Lorentz symmetry being that in the nonrelativistic case one can change from the interaction picture to any other picture without that problem so it is obvious that it is the introduction of this symmetry that results in inequivalent vacuums.
     
  6. Mar 20, 2015 #5

    strangerep

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    Sorry, that's just nonsense.

    Have you studied Bogoliubov transformations? If so, from which textbook(s)?

    Well, it's true that passing from Galilean symmetry to Lorentz symmetry gives rise to inequivalent vacua, but not for the reasons you suggest.

    What's happening is that any unitary irreducible representation ("unirrep") of the Lorentz group is necessarily infinite-dimensional. In such cases, the usual Stone von-Neumann uniqueness theorem (concerning unitary equivalence of the unirreps of the canonical commutation relations) does not hold (unlike the situation in nonrel QM).

    It is the failure of the S-vN thm in the inf-dim case which gives rise to the ambiguity of representation of the canonical commutation relations in QFT.
     
  7. Mar 20, 2015 #6
    I concede that particular sentence is not correct as written. Can you give your take on the physical meaning of inequivalent unitarily representations?

    I didn't suggest any reason. What you explain IS the reason, and I'm not sure what you refer to by "ambiguity".
     
  8. Mar 20, 2015 #7
    I would also be interested in how are the renormalized (truncated) interacting QFTs like perturbative QED unitary.
     
  9. Mar 20, 2015 #8
    I think it's interesting that Dirac seemed to play both sides of the field on this. On one hand, he framed his most well known theoretical ideas and procedures assuming Lorentz covariance as a first principle. But his development of the idea of magnet monopoles as a basis for quantization (see Jackson's Classical Electrodynamics) and later efforts with topological solitons by others such as Michael Atiyah, Gerard 't Hooft, and Alexander Polyakov put the underlying representations in SU().

    Magnetic monopoles seen in the lab
    http://physicsworld.com/cws/article/news/2014/jan/30/magnetic-monopoles-seen-in-the-lab
     
    Last edited: Mar 20, 2015
  10. Mar 20, 2015 #9

    strangerep

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    Physical meaning? Well, I've been puzzling about that for many years.

    In condensed matter theory, they're related to quasi-particles and a ground state which is distinct from the free vacuum.

    In general QFT, there are hints that they have a role in formulating interacting theories. Typically, one can construct a manifold of IURs, -- essentially a set of disjoint Fock spaces indexed by a continuous parameter. But various people have been working with that idea for decades -- yet never achieving a harvest of edible fruit, istm.

    Dirac noticed it a long time ago -- see his "Dead Wood" paper and Yeshiva QFT lecture notes -- the presence of an interaction term in the Hamiltonian is so "violent" (his term) that one is ejected from the free Fock space in an infinitesimal time.

    Kibble constructed a particular manifold of these UIRs (i.e., disjoint Fock spaces) to show how IR divergences in QED could be handled more satisfactorily (giving a finite S-matrix, not merely finite cross-sections).

    The existence proof of at least one (low-dimensional) QFT involves (iirc) constructing a sequence of UIRs, and showing that the process converges.

    My suspicion is that "physical meaning" of UIRs is not quite the point. Rather, they reveal that we still haven't found quite the right mathematical framework to construct physically useful functionals (or generalizations thereof) over infinite dimensional algebras. YMMV.
     
  11. Mar 20, 2015 #10

    strangerep

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  12. Mar 20, 2015 #11

    bhobba

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    I went to read it with great excitement thinking this was a big breakthrough.

    Unfortunately sloppy reporting is all it is :frown::frown::frown::frown::frown::frown::frown::frown::frown:.

    Thanks
    Bill
     
  13. Mar 21, 2015 #12
    Completely agree :) I grabbed that from a google search and thought that it might be roughly the latest news on the subject. But on reading the article, there was no substantiation to the claims. But that doesn't dismiss the potential relevance of the research that Atiyah and many others are engaged in...
     
  14. Mar 23, 2015 #13

    vanhees71

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    Well, in a sense this are magnetic monopoles, but not the elementary ones, Dirac was after, but quasi-particle excitations in condensed matter (socalled spin ice) or using a Bose condensate to create corresponding magnetic fields. The article is, however, pretty misleading. Usually, the only chance to understand, what's really going on is to read the publication the popular-science writer pretends to simplify. The problem is that not all popular science writers are a Weinberg or a Lederman and thus are unaware of Einstein's warning, to simplify things as much as possible but not more simple that that.
     
  15. May 31, 2015 #14
    I just found this old thread and on rereading this post I guess I should have obviously replied with the Haag's theorem itself, wich asserts that the "interacting representation" doesn't exist except when there is no interaction. So in the presence of interaction(the relevant physical case by the way) there is no Poincare invariant vacuum state that can be assumed either, wich was my point in #2.
     
  16. May 31, 2015 #15

    strangerep

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    Well, it's a bit trickier than that. Haag's thm just means that you can't "shoehorn" the interacting fields sensibly into the Fock space constructed from the free fields.

    I was using the term "interacting representation" more generally, i.e., a representation of the Poincare group operators constructed from the interacting fields. Weinberg vol 1 talks about to construct such operators. One still assumes a Poincare-invariant vacuum state, but it does not lie in the free Fock space (i.e., it's not the same as the vacuum state of the free theory).
     
  17. Jun 1, 2015 #16
    Yes, we agree on this. What I'm saying is that there's more to the theorem than you seem to imply. The interacting fields representation is fine asymptotically,with the Poincare invariant vacuum, so as long as you don't actually apply it to an interaction there is no problem. But in practice when you want to calculate something and get some concrete result for instance for the S-matrix you need to take the limit to either plus or minus infinity and the operator, say U(-∞,t) is no longer unitary and the vacuum can no longer be assumed Poincare invariant.
    Of course in practice there is the Wilsonian view that explains all awy by arguing that the local physics observed is insensitive to whatever happens at high energies and long distances so we can act as if there is no Poincare symmetry or unitarity problem, but at a purely formal level the issues remain.
     
  18. Jun 1, 2015 #17

    strangerep

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    Huh? In QFT, one typically assumes that, at asymptotic times, we can use the free representation. One also assumes "asymptotic completeness", meaning an assumption that the physical fields at all times can be expressed sensibly in terms of the free fields. But that turns out to be a problematic assumption in general.

    You say "the vacuum", but... which vacuum? The vacuum of the free theory does not coincide with the vacuum of the interacting theory in general. Yet, the free vacuum is Poincare invariant in the free theory (since the Poincare operators are expressed in terms of the free theory operators), and the interacting vacuum is also Poincare invariant in the interacting theory (since then the Poincare operators are now expressed in terms of the interacting fields). This is an example of how there can be inequivalent representations of a symmetry algebra.

    Not sure there's much more I can usefully say about this.
     
  19. Jun 2, 2015 #18

    vanhees71

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    You have to do the "adiabatic switching" right in some way. In the path-integral formulation, e.g., it implies to add a little imaginary part, appropriate for the Green's function you like to calculate. It's very important to do the adiabatic switching right and it is very difficult, if not impossible, to make sense of a particle interpretation for interacting fields. For a drastic example, see

    F. Michler, H. van Hees, D. D. Dietrich, C. Greiner, Asymptotic description of finite lifetime effects on the photon emission from a quark-gluon plasma
    Phys. Rev. D 89, 116018 (2014)
    http://arXiv.org/abs/1310.5019
     
  20. Jun 2, 2015 #19
    I meant interacting field theories like the SM. You can use a free representation because there is no interaction representation in the presence of interactions(i.e.Haag's theorem), that is what makes the completeness assumption you mention "problematic" as you call it in the presence of divergence as explained below.
    I was referring to the interacting field theory vacuum, let's use as example of interacting field theory perturbative QED to avoid confusion. Its vacuum is Poincare invariant in some abstract sense if you want in the asymptotic expansion, but it is useless in practice since it is divergent and no valid result can be obtained from it. You need to renormalize and take the limits I commented above to get sensible(and in fact very good approximations to experiment) results, and in doing so you break Poincare invariance. In other words the QED vacuum is not realizable in nature.
    It's no use to to blame all this on the perturbative approach as long as there is no at least as good an approximations with non-perturbative models.
     
    Last edited: Jun 2, 2015
  21. Jun 2, 2015 #20
    I know nothing about "adiabatic switching", can you elaborate on it?
    You mean that no true vacuum makes hard to talk about particles?
     
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