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State space notion in QFT

  1. Dec 9, 2013 #1

    WannabeNewton

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    Hi guys! So this question has been bugging me a bit and I can't seem to find any textbook (at least, restricted to physics textbooks) that talks about it.

    In QM, the overarching formalism is clear. We have ##L^2(\mathbb{R})## and states are given by ##|\psi\rangle \in L^2(\mathbb{R})## such that ##\langle \psi |\psi\rangle = 1## (so really we consider the projective space of unit norm kets in ##L^2(\mathbb{R})## since ## |\psi\rangle \sim |\varphi \rangle## if they differ by a phase factor). Furthermore we have unbounded operators ##A: \text{dom}A\rightarrow L^2(\mathbb{R})## where ##\text{dom}A \subseteq L^2(\mathbb{R})## (if ##A## is bounded then of course ##\text{dom}A = L^2(\mathbb{R})## as usual; the continuity of ##A## in the operator norm topology will guarantee that we can always extend it as such). If ##\text{dom}A = \text{dom}A^{\dagger}## and ##A^{\dagger}|\psi \rangle = A |\psi \rangle## for all ##|\psi \rangle \in \text{dom} A## then as usual ##A## is self-adjoint (if ##A## is bounded then this just reduces to ##A = A^{\dagger}## of course) and these are the operators we care about in QM because they correspond to dynamical variables etc.

    But the coherence of this formalism isn't apparent to me in QFT. For example for scalar quantum fields we start with a classical scalar field ##\varphi## and its conjugate momentum ##\pi = \frac{\partial \mathcal{L}}{\partial \dot{\varphi}}##, along with the Poisson bracket relation ##\{\varphi(\mathbf{x},t), \pi(\mathbf{x'},t)\} = \delta(\mathbf{x}-\mathbf{x'})## and promote the classical fields to operator fields satisfying ##[\varphi(\mathbf{x},t), \pi(\mathbf{x'},t)] = i\hbar\delta(\mathbf{x}-\mathbf{x'})##. But as mathematical objects, how do the operator fields ##\varphi## and ##\pi## actually work i.e. what domain and codomain do they map between? Is the Fock space formalism even useful in this context (I mention Fock space because we are interested in multiparticle states for creation and destruction operators to act on)?

    Thanks in advance!
     
    Last edited: Dec 9, 2013
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  3. Dec 9, 2013 #2

    strangerep

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    In the standard rigorous treatments, quantum fields are operator-valued distributions.

    So, e.g., if we have a quantum field ##\psi(x)##, then for any Schwartz function ##f(x)##, we define
    $$
    \psi_f ~:=~ \int dx \; f(x) \psi(x) ~.
    $$
    Decomposing this into annihilation and creation operators, and acting with the latter on a vacuum state (a.k.a. "fiducial" state) we generate a Hilbert (Fock) space in which inner products involve integrals of Schwartz functions, hence converge ok. Thus we generate the Hilbert space by first constructing an algebra of suitable quantities, and act on a fiducial state to get a Hilbert space. Advanced QFT tries to do away with the Hilbert space (mostly), and just work with a ##C^*##-algebra of field observables.

    But there's lots of other issues. Finish Ballentine thoroughly first... :biggrin:
     
  4. Dec 9, 2013 #3

    atyy

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  5. Dec 10, 2013 #4

    dextercioby

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    The only resource on this matter which satisfies my taste is the book of 1975 by Bogolubov, Logunov and Todorov.
     
  6. Dec 10, 2013 #5

    George Jones

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    For free quantum field theory, one-particle state spaces are carrier spaces for unitary (consequently infinite-dimensional) irreducible representations of the Poincare (not Lorentz) group. These one-particle state spaces are used to construct multi-particle Fock spaces on which the field operators operate.
     
  7. Dec 11, 2013 #6

    WannabeNewton

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    Thanks for the replies guys! It took me quite a while to understand and absorb what was being said but I think I understand the basic framework now. Thanks for the references; Folland's QFT text also had an englightening discussion of said basic framework for anyone interested (section 5.2 in particular).

    I was working through Srednicki's QFT text (specifically for course related reasons the coming spring semester) and the lack of any discussion on the foundational aspects of QFT had me running in circles :biggrin:
     
    Last edited: Dec 11, 2013
  8. Dec 11, 2013 #7

    atyy

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    You can't get any of that from a physics text like Srednicki. Apparently, in rigourous interacting QFT, the appropriate Hilbert space is not a Fock space. Also, there's Haag's theorem, which basically says the standard way of getting to Dyson's formula is wrong (apparently there's something called Fell's theorem which says it's ok anyway, or something like that). For many practical purposes, such as the standard model of particle physics, it's ok since it's just an effective theory, and doesn't have to make predictions for all energies. The most important conceptual tool from a practical viewpoint is Wilson's renormalization group, which I find easiest to understand in the context of classical statistical physics (like in the second volume of Kardar's stat mech text). Wilson's viewpoint is important, because it explains to physicists why renormalization is ok and not a bizarre process of subtracting infinities, eg. http://quantumfrontiers.com/2013/06/18/we-are-all-wilsonians-now/ . However, I've heard from several mathematicians that the renormalization group is still not on rigourous footing (apparently there's some problem with some functions not dying away fast enough). Similarly. physicists believe lattice QCD is quite useful for non-perturbative stuff, while the existence and mass gap in Yang-Mills are Clay problems.
     
    Last edited: Dec 11, 2013
  9. Dec 12, 2013 #8

    DarMM

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    Well, as you said, in Quantum Mechanics the Hilbert space is:
    ##L^{2}\left(\mathbb{R}^{nd}\right)##

    where ##d## is the number of spatial dimensions and ##n## is the number of particles.

    If the particles lived on another manifold ##Q##, instead of ##\mathbb{R}^{d}## then the Hilbert space would be ##L^{2}\left(Q^{n}\right)##, with ##n## again the number of particles.

    So we can see that the basic idea is that the Hilbert space should be the space of square integrable functions over the classical configuration space of the particles.
    However square-integrable with respect to what measure?
    ##L^{2}\left(Q^{n}, d\nu\right)##
    with ##d\nu## some measure.

    The Stone-von Neumann theorem guarantees that any choice of ##d\nu## results in the same physics, so in Quantum Mechanics you never bother listing the measure.

    Now the classical configuration space of a single scalar field is ##\mathcal{S}^{'}\left(\mathbb{R}^{d}\right)##. Hence the Hilbert space of a quantum field theory is:
    ##L^{2}\left(\mathcal{S}^{'}\left(\mathbb{R}^{d}\right)\right)##.

    However in this case the measure is important to list, as different measures result in different physics:
    ##L^{2}\left(\mathcal{S}^{'}\left(\mathbb{R}^{d}\right), d\nu\right)##

    This would seem pretty bad, as for every single scalar field theory you would have an infinite number of choices of Hilbert space, hence the Lagrangian/Hamiltonian would not be enough to know the physics, only Hamiltonian + Hilbert Space choice.

    The truth however is not so bad. In actuality any given Hamiltonian is only well-defined on a single Hilbert space (this is an extension of Haag's theorem). Haag's theorem is basically the result that the free-theory Hilbert space doesn't work for any interacting theory.

    Fock space enters because it turns out that the Hilbert space for the free-field has a basis where it takes the form of sums of Tensor products of Quantum Mechanical type Hilbert spaces. In other words:
    ##L^{2}\left(\mathcal{S}^{'}\left(\mathbb{R}^{d}\right), d\nu_{0}\right) = \oplus_{n=0}^{\infty}\otimes_{i = 1}^{n}\mathcal{H}^{i}##
    with ##d\nu_{0}## the measure for the free quantum field and ##\mathcal{H}## being a quantum mechanical type Hilbert space like: ##L^{2}\left(\mathbb{R}^{nd}\right)##.

    This allows you to interpret free quantum field states as being like states of multiple quantum mechanical particles.
     
  10. Dec 12, 2013 #9

    Avodyne

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    One way to think about QFT is to put it on a lattice: then it becomes QM, and all these extra issues go away. The problem is then to demonstrate Lorentz invariance at long distances.

    See also section 10.5, "How to stop worrying about Haag's theorem", in The Conceptual Framework of Quantum Field Theory by Anthony Duncan (Oxford, 2012).
     
    Last edited: Dec 12, 2013
  11. Dec 12, 2013 #10

    atyy

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    My favourite way is also to put it on a lattice. However, can the lattice support chiral gauge theories? As recently as 2009, Kaplan http://arxiv.org/abs/0912.2560 wrote "there is currently no practical way to regulate general nonabelian chiral gauge theories on the lattice." A similar sentiment was expressed by Poppitz and Shang http://arxiv.org/abs/1003.5896 "we do not yet have a method of approximating an arbitrary chiral gauge theory by latticizing and then simulating it on a computer even in principle."
     
    Last edited: Dec 13, 2013
  12. Dec 13, 2013 #11

    Avodyne

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    Yes, chiral gauge theories are problematic for the lattice, so we don't yet have a rigorous definition of them. My point was that lattice regularization is the only known method for defining any interacting QFT in 3+1 dimensions.
     
  13. Dec 13, 2013 #12

    DarMM

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    True, although bear in mind that the lattice doesn't fully define a QFT in 3+1 dimensions as there is no proof that the ##a\rightarrow0## limit exists for any field theory, where ##a## is the lattice spacing.
     
  14. Dec 13, 2013 #13

    Avodyne

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    There is no proof, and in fact it's widely believed that an interacting theory must be asymptotically free to exist in the [itex]a\to 0[/itex] limit. Non-asymptotically free theories (like QED, [itex]\varphi^4[/itex], and the Standard Model) are widely believed not to exist without an ultraviolet cutoff.

    Which is why I think it's a waste of time to try to do QFT rigorously with C* algebras or any other method that doesn't select asymptotically free theories as special at the beginning.
     
  15. Dec 14, 2013 #14

    DarMM

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    I'm not so sure about the Standard Model, don't the calculations of D.J.E. Callaway and others suggest that ##\phi^4## and a ##U(1)## gauge theory in the presence of a ##SU(N)## gauge field are asymptotically free.

    How does this follow, if you don't mind me asking. Any formalism should work in all dimensions and asymptotic freedom isn't special in lower dimensions.
     
  16. Dec 15, 2013 #15

    Avodyne

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    This is possibile in principle, but is not true for the Standard Model (where the values of all couplings are now known).

    As I understand it, the folks doing C* algebras and related stuff are trying to rigorously construct interacting QFTs. But if you try to rigorously construct a QFT that doesn't actually exist, you should fail!! So it seems to me that a sign that you are on the right track is that your methods are able to distinguish between theories that exist and theories that don't exist (assuming that the conventional wisdom on these theories is correct).

    If the conventional wisdom is not correct, and a theory like ##\phi^4## in 3+1D actually exists, then your methods should give some hint as to how it manages to escape from no-go theorems like Rosten's, arXiv:0808.0082. But AFAIK the C* folks have nothing to say about key issues like this. So I just don't believe that the whole approach is worthwhile (though of course this is just a personal opinion that's worth what you paid for it!).
     
    Last edited: Dec 16, 2013
  17. Dec 18, 2013 #16

    DarMM

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    I thought that the Standard Model's parameters were beneath the values imposed by the asymptotic freedom constraint. If they aren't, do you have a good reference for this? Not that I don't believe you, but obviously I need to do some updating on my knowledge!

    Yes, indeed.

    The methods do show that. If the theory doesn't exist construction fails. You can prove that the continuum limit doesn't exist, or breaks one of the Wightman axioms.
    The real problem is how difficult it is to control the continuum limit and gain any information about the theory which exists in that limit.

    Well to construct a field theory in a C*-algebraic sense, you have to attach a C*-algebra to every point in space and then prove there exists a good set of states on that algebra. This can be done easily enough. The real problem with the algebraic approach is identifying an individual theory. So let's say you have a C*-algebra on every open region of spacetime. To say a theory is ##\phi^4##, you have to demonstrate that that algebra can produce from operator valued distirbutions (fields) exponentiated, i.e. for every region ##\mathcal{O}## the operators are all formed from combinations of operators like:
    ##e^{\int{\phi f dx}}##
    where ##f## is then supported in ##\mathcal{O}##. Then showing that the operator generating the operator algebra (##\phi##) evolves according to something involving ##\phi^4##, which should be the point-like limit of the automorphism that evolves the algebra.
    It's simply that this is very hard to do.

    Proving that ##\phi^4## doesn't exist would mean showing that any scalar operator algebra cannot be generated from a field obeying something like the ##\phi^4## hamiltonian. Which again is very hard.

    However, I don't think the C*-algebra approach is particularly weak in this regard, there is no method of constructing field theories in 3+1 dimensions which has advanced to the point where it can say anything like this. Otherwise we would have rigorous control over quantum field theories.
     
  18. Jul 31, 2014 #17

    A. Neumaier

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    Please give a reference for the first statement. I thought that Haag's theorem says that there is no single Hilbert space in which the dynamics works!?
     
  19. Sep 27, 2014 #18

    DarMM

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    I only saw this now.

    Unless I'm mistaken I think any of the papers by Glimm, Jaffe, Osterwalder, Schrader, Frohlich, e.t.c. which construct the interacting Hilbert space would disprove this as the dynamics are well-defined on such a space. If there was no Hilbert space on which the interacting dynamics were defined there would be no interacting theory right?
     
  20. Sep 28, 2014 #19

    aleazk

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    Hi, a free Klein-Gordon classical field is a linear system. These systems are quantized by taking certain subspace of positive frequency solutions of the equations of motion. Using the symplectic form, an inner product can be defined and the completion of this gives you a Hilbert space. Then you take the Fock space from this, and using the annihilation/creation operators, you define the elements of the Weyl algebra of the system. This can be done also for a finite collection of harmonic oscillators. The main difference is that for the finite dimensional phase space of the harmonic oscillators, the Stone-von Neumann theorem holds, while this is not true for the field system. So, in the finite dimensional case, the above construction is the same as the usual tensor product of the Hilbert space of one oscillator. In the field case, different definitions of 'positive frequency solutions' give rise to different and inequivalent constructions. Check Wald's 'Quantum field theory in curved spacetime'. Also, the infinite sums that define the 'field operators' don't converge, so they have to be reinterpreted as operator-valued distributions rather than operator-valued functions.

    If you are quantizing the field in flat spacetime, you have a natural notion of positive frequency solutions given by the time coordinate of some inertial observer. The Hilbert space in this case is isomorphic to L^2(Xm, du), where Xm is a positive mass orbit of the natural action of the Lorentz group in momentum space and du the induced invariant measure. But this space is precisely the representation space of the zero spin representation of the Poincaré group (check Varadarajan's "Geometry of quantum theory" for all this). In fact, you can check that the generators of the spacetime translations correspond to the quantized energy-momentum of the field. In this sense, if you define a quantum elementary relativistic particle as an irreducible representation of the Poincaré group, the above comments show that the Fock space for these particles can be naturally interpreted as the basic Fock space you use for constructing a particular quantization of the Klein-Gordon field (in the sense discussed in the first paragraph). And this is the one that Folland studies in section 5.2 of his book.
     
  21. Sep 30, 2014 #20

    A. Neumaier

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    Yes. But they construct theories only in lower dimensions, and I thought Haag's theorem was for dimension 1+3. In any case, Haag's theorem must be dimension dependent, as it clearly fails in 1+0 dimensions.

    So what is the precise mathematical claim in Haag's theorem? (With reference, please.)
     
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