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Interpretation of quantum fields and their ground state energies.

  1. Jun 5, 2012 #1
    Quantum field theory predicts a value for the cosmological constant that is 123 orders of magnitude larger than the observed value (if one assumes the Standard Model to be correct up to the Planck scale of 10^19 GeV)! To theoretically predict the value of the cosmological constant, one must, I suppose, add up all the ground state energies of the quantum fields corresponding to the particles of the Standard Model. By introducing some cut-off to the integral over frequencies, one obtains a finite answer. I would like to understand where this (horrible) prediction exactly comes from. I am a graduate student in particle physics and have had just a little on QFT so far. I have a few questions about quantum fields and it would be great if anyone could help!

    I understand that within the QFT framework particles are interpreted as excited states of fields. I have seen that photons, for example, are the quanta of the electromagnetic field.
    (1) But does every particle of the Standard Model correspond (uniquely) to some field? So do we have, say, a down quark field for the down quark, a charm quark field for the charm quark, a tau neutrino field for the tau neutrino, a Z boson field for the Z boson and so on... ? Or do some particles, in a way, correspond to the same field?
    (3) Does the concept of a quantum field only makes sense for the elementary particles of the Standard Model? Or can one also speak of fields corresponding to composed particles, like mesons?
    (2) And is there just one quantum field of each type in the whole Universe? So is the vacuum filled with all different types of quantum fields that exist? Is the vacuum hence identical for all points in spacetime? Perhaps I do not understand very well what a quantum field is...

    I have seen that one can write the Hamiltonian of the electromagnetic field as an infinite sum of quantum harmonic oscillators. Then you immediately see that this field has some non-zero ground state energy which must contribute to the energy of the vacuum and in turn contribute to the cosmological constant.
    (4) Do all quantum fields corresponding to the particles of the Standard Model have a non-zero ground state energy? If yes, then I suppose this follows from the Heisenberg uncertainty principle. Can this contribution also be negative?
    (5) Can every quantum field be thought of as a collection of harmonic oscillators of all possible frequencies?

    I have quite a lot of questions... I still just miss the conceptual part of QFT. Any help would be greatly appreciated!

    Thanks in advance:)
     
  2. jcsd
  3. Jun 5, 2012 #2

    Bill_K

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    Sure, just assume that space is filled with one Planck mass per cubic Planck length. Perfectly reasonable! :wink: I think people like to repeat this ridiculous claim just to amuse their graduate students.
    They do if you write the Lagrangian wrong. :uhh: Write it as a normal ordered product and the ground state energy goes away.
     
  4. Jun 5, 2012 #3
    @TopQuark
    (1) Yes. it does
    (2) In principle, fields for composed particles are o.k. - otherwise there would be some "trick" to find out whether some particle is composed...
    (3) Yes, there is one quantum field for every particle if you want. Of course you could also say that an up- and down quark comprise a single field with two quantum numbers, that's a matter of taste.

    (4) Yes, the contribution for fermions is negative - that's one reason why people like supersymmetry: All the zero-point energies of the fermions cancel those of their bosonic partners.

    (5) In principle, yes - although things are a bit more involved when spin comes into the picture, but the principle is the same.

    Conceptually, I recommend the qftfield.info site by Bob Klauber, that should clear up a lot of things (did for me). You might also want to look at chapter 10 of Hatfields QFT of point particles and strings - he presents QFT in the Schrödinger picture using wave functionals, which at least to me is highly intuitive.

    If you can read German, you might also look at my Blog, where I struggle with these things at the moment:
    http://www.scienceblogs.de/hier-wohnen-drachen/2012/06/qft-fur-alle-wir-verstehen-nichts.php
    I'm still not sure if everything I understand is correct - so no guarantees here...

    @Bill K
    Normal ordering? Sure - first we say that commutators are all-important and then we just "normal order" things when we don't like the result. I've never heard of any justification for "normal ordering" beyond "it gets rid of the ZPE".
     
  5. Jun 6, 2012 #4
    @Sonderval
    Thank you very much! Things are more clear for me now.
    I will have a look at the qftfield.info site. My German though is not very good, but perhaps I can understand physics texts.
     
  6. Jun 6, 2012 #5

    Bill_K

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    Sonderval, I can't imagine what kind of justification you are expecting. The zero point energy is divergent and must be removed as part of the renormalization process. If you prefer, it can be done explicitly by introducing an ultraviolet cutoff, and in more complex situations this becomes necessary, but in the present case normal ordering accomplishes the same thing neatly and efficiently.
    TopQuark38, The correct location is http://quantumfieldtheory.info/ Warning: This reference is full of nonstandard personal opinions which in many cases are flagrantly incorrect.
     
  7. Jun 7, 2012 #6
    @BillK
    I think it is more honest conceptually to actually write the removal of the ZPE as a renormalisation procedure - this at least acknowledges that there is some kind of problem. Just saying "We do normal ordering" (when in all other cases we insist on proper commutation relations and say "never ever exchange operators without using commutators") seems very confusing to me, especially to beginners.

    You are right that the qftinfo-site has many non-standard opinions, but the author is - as far as I can tell - honest enough to always state when what he writes is not the standard. I like the site because he explains many things others don't even mention (wave packets, general state of a field etc.). And he gives lots of pedagogical hints and helps - I, at least, learned and understood a lot from there that I did not get in other books.

    If there is something that is flagrantly incorrect, as you say, you should perhaps notify the author as he is currently creating a book from his site.
     
  8. Jun 7, 2012 #7

    vanhees71

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    Normal ordering is perfectly fine as long as one defines it properly. First of all it is important to note that normal ordering is always with respect to a basis and particularly the vacuum. You have to define the Fock-space basis first by introducing a single-particle basis and a vacuum state an the corresponding creation and annihilation operators. This is unambiguously only possible for free particles in a background free situation. Even if you invoke external classical background fields only (e.g., a general-relativistic curved space-time) already a lot of complications arise even in the non-interacting case, and it is not even so clear anymore what physically meaningful particle states are. You may have a look at the textbook by Calzetta and Hu on nonequilibrium QFT.

    In usual quantum-field theory without a background field and for free particles normal ordering is however perfectly fine. When it comes to perturbation theory, you still can use normal ordering for the interaction part of the Hamiltonian and avoid some tadpole diagrams from the very beginning. You can of course also forget normal ordering from the very beginning and just renormalize the action (and with that all proper vertex functions) by the most general counter terms allowed by the symmetries underlying your model. Both procedures are equivalent. The version without normal ordering is sometimes more convenient since you don't introduce trouble with gauge invariance via normal ordering, but I'm not sure whether this is an issue at all at the end. In the path-integral formalism, usually one doesn't use any kind of normal ordering anyway.

    In the operator approach, usually I perfer to use normal ordering since then at least in the perturbative sense you can make sense to otherwise ill-defined operator products of local fields at the same space-time point. These are not well defined from the very beginning since the canonical equal-time commutation relations (or in the Schroedinger picture of the time-independent field operators) involve distributions like the Dirac [itex]\delta[/itex] distributions and its derivatives. So in a way you must give sense to the operator products you also write down in the very beginning into your Lagrangian and the Hamiltonian derived from it.

    Also normal ordering is a nice systematic tool to derive the Feynman rules from Wick's theorem, which is general if understood in terms of a operator identity but applies in its usual form only with your Statistical Operator, describing the system's state, is the exponential of a one-body operator or a pure (ground) state, because in the general case expectation values of normal-ordered operator products do not necessarily vanish.

    Here are my notes on QFT:

    http://fias.uni-frankfurt.de/~hees/publ/lect.pdf
     
  9. Jun 7, 2012 #8
    @vanhees
    I agree, sorry if it sounded as if I was generally opposed to normal ordering - I just think it is a confusing way of getting rid of ZPE, especially for newcomers, for the reasons mentioned.
    Bill's statement - that Lagrangians without NO are wrong - was a bit too simplistic for my taste, that's all.
     
  10. Jun 7, 2012 #9

    Bill_K

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    Sonderval, I believe where we diverge is the interpretation. In your comments (and also in the qftfield.info site) there's the suggestion that normal ordering is a transformation during which the commutation relations have been temporarily suspended. (!) Don't agree.

    Any quantity like the Hamiltonian which involves a product of field operators at the same spacetime point will be divergent. Part of the process in defining a quantum field theory is to give meaningful definition to such expressions. Normal ordering is one way to do that. We say, for example, that the Hamiltonian is not ½(a*a + aa*) (and never was), it's really a*a. Normal ordering is not a transformation of anything, it's an ab initio definition! And in that sense the second expression is 'right' and the first expression is 'wrong'. The only connection is that one motivates the other.
     
  11. Jun 7, 2012 #10
    @Bill
    Thanks for that, I never understood it that way. I'll need to think about this, but it sure gives more meaning to normal ordering which always seemed arbitrary to me.
    So, apologies for the harsh words above.
     
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