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What is the operational difference between a quantum field and projection operators?

  1. Mar 13, 2009 #1
    For free quantum fields, there are two types of observables indexed by wave-number, [tex]\tilde{\hat{\phi}}(k)[/tex], the Fourier transform of the local field, which can be written as [tex]a(-k)+a^\dagger(k)[/tex], and projection operators such as [tex]a(k)^\dagger\left|0\right>\left<0\right|a(k)[/tex], [tex]a(k_1)^\dagger a(k_2)^\dagger\left|0\right>\left<0\right|a(k_1)a(k_2)[/tex], etc. [strictly speaking, none of these are operators acting on the Fock space, they're operator-valued distributions, but there are well-established ways to finesse this distinction].

    In quantum optics, the latter construction is always used to model measurements, the local field is not used, as far as I know. Notably, the projection operator form of observables are not local if we smear them with test functions.

    Additionally, the projection form of observable gives a zero result in the vacuum state, which corresponds to the actual response of most experimental apparatuses to the vacuum state, whereas no local observable can give a zero result in the vacuum state.

    So, to rephrase the question, is there a well-established operational distinction between these two quite distinct types of self-adjoint operator? How does one go about measuring the field observable [tex]\tilde{\hat{\phi}}(k)[/tex] instead of the projection operator observable [tex]a(k)^\dagger\left|0\right>\left<0\right|a(k)[/tex]?
     
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  3. Mar 18, 2009 #2

    strangerep

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    Re: What is the operational difference between a quantum field and projection operato

    Hi Peter.

    Interesting question. (Sorry, I couldn't reply earlier - I've been out of action.)
    I suspect part of the answer lies in the mathematical details of this "finesse"
    since, naively, [itex]a(k)^\dagger\left|0\right>\left<0\right|a(k)[/itex] is not a projection
    operator (you're left with an inconvenient [itex]\delta(0)[/itex]).

    To define "measurement" of any observable, you've got to define linear functionals over
    the observable algebra (which could involve a Hilbert space representation, or a more
    general Banach-space approach over the operator algebra itself). In this sense, I'm not so
    sure they're quite as "distinct" as you say. It seems to me that one is an element of an abstract
    inf-dim algebra, while the other is some kind of representation of the operator algebra on
    a Hilbert space (obtained via a GNS construction by acting cyclically on a postulated
    vacuum vector). But the details of all this are tricky as you know. One must actually construct
    a rigged Hilbert space instead -- which is basically what the "smearing" tricks are all about.
    Without a rigged Hilbert space treatment, the meaning of "self-adjoint" is lost in this context.
     
  4. Mar 19, 2009 #3
    Re: What is the operational difference between a quantum field and projection operato

    Thanks strangerep, but, smearing with test functions, we have [itex][a_f^\dagger\left|0\right>\left<0\right|a_g, a_p^\dagger\left|0\right>\left<0\right|a_q]=(p,g)a_f^\dagger\left|0\right>\left<0\right|a_q-(f,q)a_p^\dagger\left|0\right>\left<0\right|a_g[/itex], where [itex](p,g)[/itex] is an inner product appropriate to the particular free quantum field. In other words, the [itex]\delta(0)[/itex] isn't inconvenient when we smear with test functions (which is why we do it). Test functions aren't pure delta functions, but they can be as close to delta functions as we like.
    The way I do things in my papers avoids rigged Hilbert spaces by constructing an abstract *-algebra and a state over the algebra, then using the GNS construction of a Hilbert space and of a representation of the *-algebra on a dense subset of the Hilbert space.

    That means no more than that we take a commutation algebra [itex][a_f,a_g^\dagger]=(g,f)[/itex], where [itex](f,g)[/itex] is a positive semi-definite inner product, and introduce a normalized vacuum vector [itex]\left|0\right>, \left<0\right.\left|0\right>=1[/itex], for which [itex]a_f\left|0\right>=0[/itex]. Then, we prove that [itex]\left<0\right|AA^\dagger\left|0\right>[/itex] is positive semi-definite, which makes [itex]\rho(A)=\left<0\right|A\left|0\right>[/itex] a state over the *-algebra. [itex](A,B)=\left<0\right|BA^\dagger\left|0\right>[/itex] is then an inner product on vectors of the form [itex]A^\dagger\left|0\right>[/itex] (provided we factor out the ideal of zero vectors), so we can close that vector space in that inner product.

    In practical calculations, we can freely generate a Hilbert space from just a few test functions, say [itex]g_1,g_2[/itex] (we don't have to put every test function we might possibly use in there), obtaining everything of the form [tex](a_{g_1}^\dagger)^m (a_{g_2}^\dagger)^n\left|0\right>[/tex], closed in the inner product. That's a Hilbert space, and there's a natural action of [itex]a_{g_1}[/itex] and [itex]a_{g_1}^\dagger[/itex] on the dense set of vectors that we had before we did the closure.

    What I'm worrying at is that in quantum optics we use observables of the type [itex]a_f^\dagger\left|0\right>\left<0\right|a_f[/itex] almost exclusively (which is nonlocal, but who's counting). I don't see in quantum optics where we use the local observable [itex]\phi_f=a_{f^*}+a_f^\dagger[/itex]. These are so different as operators that there must be an operational difference. How do I measure [itex]\phi_f[/itex]? The key thing about [itex]\phi_f[/itex] is that, because it is local, it has a non-zero response to the vacuum, but, if we obtained a non-zero response to the vacuum we would ordinarily say, "Huh, what's wrong with my detector? It's detecting stuff when there's nothing there!"

    There's by now lots of published literature pointing out that the idea of particles is not supportable in QFT (even for free fields, though most of the blogosphere doesn't know it), so the fact that our detector gives a non-zero response to the field doesn't have to worry us much, but is there something in the literature that I've missed, that explains how to measure the field, instead of the particle content of the field? I think that last sentence gives a fairly good answer to the theoretical difference between the two types of measurements, but what's the operational difference?
     
  5. Mar 19, 2009 #4

    DarMM

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    Re: What is the operational difference between a quantum field and projection operato

    Hey Peter, just thought I would mention that for interacting field theories the appropriate representation of the canonical commutation relations does not allow for the [itex]a_f, a_f^\dagger[/itex] operators to have a state such that [itex]a_f\left|0\right>=0[/itex] for all [itex]f[/itex].
     
  6. Mar 19, 2009 #5
    Re: What is the operational difference between a quantum field and projection operato

    Fair enough. Quantum optics is a non-interacting field, but quantum optics is definitely not everything in QFT. But then, there is so far no rigorously constructed interacting quantum field in Minkowski space.

    The papers of mine that I was referring to either construct free quantum fields or construct Lie random fields, where interactions are possible for a rigorously constructed system that has a vacuum that is annihilated by all annihilation operators. Marcus kindly opened https://www.physicsforums.com/showthread.php?t=300853" on a paper I uploaded to the arXiv yesterday (which points to the arXiv paper and to a JMathPhys paper of just over a year ago). That, however, is a random field that explicitly models quantum fluctuations, which is quite closely related to a quantum field, but it is not a quantum field.

    Perhaps I will be hemmed into such a corner here that the question is no longer interesting? What's the operational difference between measurements of the field, [itex]\phi_f[/itex], in contrast to a measurement of particle content, [itex]a_f^\dagger\left|0\right>\left<0\right|a_f[/itex], in quantum optics?
     
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  7. Mar 20, 2009 #6

    strangerep

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    Re: What is the operational difference between a quantum field and projection operato

    Well, not from my point of view. The reason I haven't attempted a better answer is
    that I don't yet understand your question precisely enough...

    When you say "measurement of the field [itex]\phi_f[/itex]", I don't know what this means
    in a rigorous context. The word "measurement" usually means "extract a (set of) numbers",
    or some variation on that theme. I also don't know (in mathematically precise terms)
    what you meant earlier by "...field's response to the vacuum...". Did you mean a vev?

    I.e., before one can discuss "operational differences between measurements of
    [something modeled by a particular two mathematical entities]", both must be clearly
    and consistently defined. I don't think that's the case here since a sum of operators
    such as [itex]\phi_f[/itex] only makes sense if both ops have the same domain, and
    that in turn depends on the space they act within.

    Cheers.

    P.S: I'll take a look at your new paper this evening.
     
  8. Mar 20, 2009 #7
    Re: What is the operational difference between a quantum field and projection operato

    I intend to distinguish representation of results of a measurement in a relatively theoretical way, as expected values of an operator in a state, such as, in the vacuum state, VEVs (which can be summarized in terms of probability measures over a probability space that is determined by the eigenvalues of the operator), and representation of results of measurement at a relatively phenomenological level, which lists specific events in a detector possibly placed near a device that prepares the state (and statistics of the list of events that we compare for statistical significance with the probability densities, or perhaps that we use for parameter estimation). I also intend to distinguish between a theoretical description of a measurement as one operator rather than as another, and a phenomenological (or operational) description of a measurement as a detailed list of pieces of metal, plastic, electronic circuits and the precise relationships of position and electrical connections between those components.

    With this kind of distinction in mind, a measurement of [itex]\phi_f[/itex] in a pure state [itex]\left|\psi\right>[/itex] at the theoretical level is modeled by expectation values [itex]\left<\psi\right|\phi_f\left|\psi\right>[/itex], [itex]\left<\psi\right|\phi_f^2\left|\psi\right>[/itex], etc. We can generate the characteristic function of such measurements in the vacuum state as [itex]\left<0\right|e^{i\lambda\phi_f}\left|0\right>[/itex], obtaining for the free field [itex]e^{-\lambda^2(f,f)/2}[/itex], so it's a Gaussian probability measure, with non-zero variance for choices of [itex]f[/itex] that have on-mass-shell components (because the inner product [itex](f,f)[/itex] projects to on-mass-shell components).

    For measurement of [itex]a_f^\dagger\left|0\right>\left<0\right|a_f[/itex], we obtain zero in the vacuum state, no matter what we choose for [itex]f[/itex], which corresponds pretty well with how we tune our measurement devices, and certainly with the practice in quantum optics. We all know there's a dark rate associated with any detector, which we try to minimize, but we can't eliminate the dark rate completely, because that would need a device to be in a totally stable thermodynamic state, so there'd be no detector events ever. How much of the signal from a detector is labeled "the dark rate" depends on how the signal changes when we put a new piece of apparatus near the measurement apparatus, either a thick lead plate, say, to reduce the background, or a gamma ray emitting source, to see what the response of the measurement device changes to.

    The irreducible "dark rate" is effectively a measurement of the local field in the vacuum state, assuming we really have eliminated even the neutrino background (which we can block by a different amounts by placing the detector at different depths inside the earth), but I guess each measurement device implements a different measurement of the vacuum. How can I be sure that I'm measuring [itex]\phi_f[/itex], not [itex]\phi_g[/itex], when I measure the dark rate of a given measurement device? In QM, we're used to dimensions of the apparatus determining what measurement we think we're making (such as the dimensions of a grating and the relative placing of a detector determining what wavelengths are measured by an apparatus), but what details of the geometry and internal structure of a measurement device determines what measurement we are making of the local field we have implemented when we measure its dark-rate?
    I take this to refer to my "The key thing about [itex]\phi_f[/itex] is that, because it is local, it has a non-zero response to the vacuum". I mean by this that the vacuum expectation value of [itex]\phi_f[/itex] is of course zero, but the variance of such measurements is non-zero, [itex]\left<0\right|\phi_f^2\left|0\right>=(f,f)[/itex].
    The mathematics may be OK, but clear and consistent definition would be a first in Physics! The domains of both [itex]\phi_f[/itex] and [itex]a_f^\dagger\left|0\right>\left<0\right|a_f[/itex] include all states in the vacuum sector that are finitely generated as [itex](a_{g_1}^\dagger)^m(a_{g_2}^\dagger)^n\left|0\right>[/itex], etc., which is a dense subspace of the Hilbert space, but I suppose their domains in the Hilbert space that is constructed as a closure in the norm are slightly different, larger than the finitely generated space in different ways. Still, I think the common dense subspace gives us enough to have a discussion?
    It's far from perfect, so there are definitely things you can say about it, even if only to repeat Haelfix's skepticism. I hope you find it interesting enough or curious enough to make a comment.
     
  9. Mar 20, 2009 #8

    strangerep

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    Re: What is the operational difference between a quantum field and projection operato

    [itex]e^{i\lambda\phi_f}\left|0\right>[/itex] looks to me like the well-known
    Glauber coherent states (the infinite-mode case, that is). But since you talk a fair bit
    about quantum optics, I imagine you've already studied the "bible" by
    Mandel & Wolf ("Optical coherence and quantum optics") ?

    There's also a smaller, more recent monograph by Ulf Leonhardt,
    titled "Measuring the quantum state of light" which delves deeper
    into the links between theory and experiment which sound relevant
    to what you're asking. If you haven't seen it, there might be some useful
    partial-answer nuggets therein.

    Something similar to this comes under the heading of "squeezed states" and/or
    multi-photon coherent states. (M&W talk about the case of 2-photon
    coherent states.) The way you construct nonstandard 2-particle states
    in your paper (such that they're orthogonal to your 1-particle states)
    also reminds me of 2-photon coherent states.

    Commenting is easy. Making _helpful_ comments is rather more difficult.
    The paper is hard to absorb beyond a superficial level. I got lost
    almost immediately amongst all those [itex]\xi[/itex] functions, and I'd
    need time to work through it pen-in-hand (time which I don't have right
    now, unfortunately).

    So far, it looks like you're doing a quantum deformation of the
    classical inf-dim commutative algebra of test functions. (The passage
    from [itex]f[/itex] to the noncommutative algebra of [itex]\phi_f[/itex], etc, looks like
    a type of quantum deformation.)

    Then you investigate linear functionals over this noncommutative
    algebra (the main subject of your latest paper being to show
    positive-definiteness). However, it's not clear whether you've also
    explored the cohomological aspects of this construction (which might
    get in the way when you try to Lie-integrate the algebra).
     
  10. Mar 21, 2009 #9
    Re: What is the operational difference between a quantum field and projection operato

    I've looked at Mandel & Wolf quite a few times, but I'd have to say that I haven't plowed the field. Quantum optics is a very practical field, which throws up some real gems, but in some sense it has quite a few distractions. Of course the only way to see what is a distraction and what is fundamental is to plow the field.
    Ugggh, reading! But it sounds like reading I have to do, together with another look through M&W. There's too much plowing of other fields to do.
    I'm not asking you to engage with this, but I think part of this is not right. The operators [itex]\phi_f[/itex] form a commutative algebra C, say, but they are constructed as a subalgebra of a non-commutative algebra A, say, of creation and annihilation operators. We can construct a vacuum state over A, which gives us a nontrivial state over C.
    I think I see it slightly differently, if I attempt to make a higher-level mathematical view of what I'm doing (which doesn't come naturally to me at this point). The map from [itex]f[/itex] to [itex]\phi_f[/itex] preserves the additive structure, but multiplication of test functions is not preserved; multiplication of the operators is essentially a symmetric tensor product. Alternatively, I think we can regard the map as (possibly a functor between categories) from the inf-dim commutative algebra of addition and symmetric tensor products into C, which is rather different. From a classical point of view, this is fairly natural to do because we are interested in generating probability spaces, which operators acting on a Hilbert space are very good at, so we want to map some index space into an algebra of operators.
    I hope you can elaborate on what you mean by Lie-integration. Despite the name of Lie random field, I'm not thinking of this as truly a Lie algebra. We use the operators to generate characteristic functions of probability densities, [itex]\left<\psi\right|e^{i\lambda\phi_f}\left|\psi\right>[/itex], for example, or more fundamentally we just generate expectation values such as [itex]\left<\psi\right|\phi_f^m\left|\psi\right>[/itex], then we're done, at least for empirical content. I'm not sure I can see where an obstruction to this construction might come from. Perhaps you're referring to there being an obstruction to extending this manifestly Poincaré covariant construction to a representation of the Poincaré group, which might make modeling impossible because we would not be able rigorously to construct approximations to experimental data?
     
  11. Mar 22, 2009 #10

    strangerep

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    Re: What is the operational difference between a quantum field and projection operato

    Welcome to research. :-)

    Oops, yes. As written, it's wrong and I should have said "...passage from
    [itex]f[/itex] to [...] [itex]a_{f^*}, a^\dagger_f[/itex]. I was mentally
    accompanying your [itex]\phi_f[/itex] by its orthogonal partner
    [itex]\pi_f := a_{f^*} - a^\dagger_f[/itex]. Although... I now don't recall
    you mentioning the latter, but without it I don't think you have a dynamics.

    There's a Wiki page for it but it's in German last time I checked. Google
    can do an automatic translation (which is not very good, but allowed me
    to understand the idea). For a typical Lie group (considered as a
    manifold), we can take an infinitesimal neighbourhood of the identity
    (tangent space) which is basically just the corresponding Lie algebra
    in less high-falutin' language. Lie integration is the reverse
    procedure: start with a Lie algebra and attempt to construct the full
    manifold from it. AFAIU, this is not always possible for arbitrary Lie
    algebras, but I'm in the early stages of learning about this stuff
    myself, so I can't really explain much more about it. There was an item
    at the n-Category Cafe a while back touting the slogan that
    "Quantization is Lie Integration". It's an interesting perspective, but
    I still don't understand it very well. Maybe you shouldn't worry about
    it too much right now. There's plenty of more important items to think
    about, such as interactions.

    Your [itex]a_{f^*}, a^\dagger_f[/itex] together with Jacobi identity
    and your structure equation:
    [tex]
    [a_f , a^\dagger_g] ~=~ (f,g) + a_{\xi(g,f)} + a^\dagger_{\xi(g,f)}
    [/tex]
    certainly looks like an inf-dim Lie *-algebra to me.
     
  12. Mar 22, 2009 #11
    Re: What is the operational difference between a quantum field and projection operato

    I presume you're not working in this formalism, but if you ever do you should note that [itex]a_{f^*} - a^\dagger_f[/itex] is a Lorentz invariant linear functional of [itex]f[/itex], so it's not the canonical momentum, which depends on a choice of a space-like hypersurface. The canonical momentum can be constructed from the textbook formulas for it in terms of [itex]a(k)[/itex] and [itex]a^\dagger(k)[/itex]. Also, [itex]i(a_{f^*}-a^\dagger_f)[/itex] is not orthogonal to [itex]a_{f^*}+a^\dagger_f[/itex], though of course you were speaking loosely.

    What I'm doing can be understood to deprecate the role of dynamics. Indeed, it can be understood as a block-world formalism at the relatively theoretical level of probability measures (but without having anything to say about the nature of the world at the more phenomenological level of statistical descriptions of large numbers of recorded events). I think it's a good thing to eliminate objects that are as singular as the Hamiltonian and the Lagrangian (they're singular even for free fields, but of course for interacting fields they are truly terrible for anyone who's remotely constructivist), but of course dynamical thinking is rather entrenched. A random field formalism doesn't need a dynamics because it algebraically generates correlations and joint probability measures over observables that are at time-like separation.
    Thanks, strangerep. I guess that's what I thought you might mean. The construction of a Hilbert space is the main thing, and the GNS construction is so extraordinarily simple that to my knowledge there aren't any obstructions. It will be good to keep track of this, because obstructions to Lie integration could be useful for classification of sectors of states over the random field.

    I don't want to push this too far, because I'm sure some things are common to Lie algebras, but I think the central scalar component makes it not strictly a Lie algebra. I think the scalar component is essential to the behavior, certainly to the generation of VEVs. In the first instance, it's only the ability to construct probability measures over observables that's interesting to the Physics, but there is also the classification of states over the algebra to consider, and I'm sure other issues will come up. I'm particularly interested in questions of what non-thermal superselection sectors there might be.
     
  13. Mar 23, 2009 #12

    strangerep

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    Re: What is the operational difference between a quantum field and projection operato

    Yes, I was speaking loosely. (Part of me already knew I'd get into trouble for saying
    it that way.) As to the rest, I'll re-use the excuse that your papers are difficult to absorb
    and apparently I absorbed even less than I thought. I revisited your previous "Lie fields
    revisited" paper again (briefly) so I now think I understand your:

    OK, I get it (I think). Dynamical behaviour is (sort-of) already embedded into the Lie
    algebra.


    Some bona fide Lie algebras have a central element (Heisenberg being the
    prototypical example). All that's needed for a Lie algebra is the vector space
    structure of the (associative) algebra, an anticommuting Lie product,
    and Jacobi identity.
     
  14. Mar 23, 2009 #13
    Re: What is the operational difference between a quantum field and projection operato

    Thanks for all your replies, Strangerep. It's been very helpful.
    I think Yes, if we add active translation, as in my JMP paper.
    OK. I take the scalar component to be a generator of the algebra. Will think about this. My first thought is that the central scalar component makes it not a semi-simple Lie algebra. As well as reading Ulf Leonhardt some. Thanks again.
     
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