Is there a good reference for the distinction between projection operators in QFT, with an eigenvalue spectrum of $\{1,0\}$, representing yes/no measurements, the prototype of which is the Vacuum Projection Operator $\left|0\right>\left<0\right|$, which allows the elementary construction of a panoply of projection operators such as $$\frac{\hat\phi_f\left|0\right>\left<0\right|\hat\phi_f^\dagger}{\left<0\right|\hat\phi_f^\dagger\hat\phi_f\left|0\right>},\qquad \frac{\hat\phi_f\hat\phi_g\left|0\right>\left<0\right|\hat\phi_g^\dagger\hat\phi_f^\dagger}{\left<0\right|\hat\phi_g^\dagger\hat\phi_f^\dagger\hat\phi_f\hat\phi_g\left|0\right>},$$ or of higher degree; in contrast to (smeared) field operators such as $\hat\phi_f$, which have a continuous spectrum of eigenvalues? I see this as effectively the distinction between, respectively, the S-matrix and the Wightman field as observables.

I'm particularly interested in anything that considers the operational difference between these different classes of QFT observables in detail. It's obvious that the projection operators are nonlocal, insofar as they clearly don't satisfy microcausality, in contrast to the requirement of microcausality for the field operators. It also seems that the field operators cannot be used on their own to construct models for the detection of a particle, which is a yes/no event, without introducing the vacuum projection operator (but is there a way of constructing projection operators without introducing the vacuum projection operator? EDIT: yes, obviously enough, "is the observed value in the range $[a..b]$" is a yes/no observable, etc., etc., ....)

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    $\begingroup$ These projections are rank 1 this is possible because they are non-local so they are in a type I factor. While local projections are generally in a type III_1 factor and in this all projections are equivalent. $\endgroup$ – Marcel Mar 3 '12 at 18:32
  • $\begingroup$ Quite dense, but rather helpful, thanks! $\endgroup$ – Peter Morgan Mar 4 '12 at 1:41
  • $\begingroup$ @Marcel So the vacuum state over the Wightman * -algebra of unbounded operators allows the GNS-construction of a Hilbert space $H$, and so of a W* -algebra of bounded observables $\mathcal{B}(H)$. That algebra is a Type III${}_1$ factor. Adding the (bounded) vacuum projection operator to the algebra of observables, which is nonlocal but natural insofar as the GNS-construction provides a vacuum vector in $H$ (and which is implicit whenever we use a transition probability, which is to say for very many empirically successful uses of QFT), converts that Type III${}_1$ factor into a Type I factor? $\endgroup$ – Peter Morgan Mar 4 '12 at 17:29
  • $\begingroup$ The W*-algebra generated by a local algebra and the projection on the vacuum vector should be all $\cal B(H)$ because by Reeh-Schlieder you become all finite rank projections. $\endgroup$ – Marcel Mar 4 '12 at 22:00

It is difficult to answer your question as it has no clear focus.

Projection operators play hardly any role in QFT, as the measurement problem is extremely rarely discussed in this context. Your projection operators are even more special as they are all rank 1. To construct projection operators, take any observable (e.g., a smeared Hermitian field), and integrate its spectral projection measure over an interval. Apart from that, the interpretation of projection operators is identical to the case of quantum mechanics.

The S-matrix (though the main observable object in QFT) is not an observable in the sense this term is used in QM, as it is unitary rather than Hermitian. Moreover, measuring a scattering cross section derived from the S-matrix has nothing to do with the kind of measurement discussed in the foundations of QM.

The same holds for other measurable consequences of QFT such as the Lamb shift, field expectations (which lead to hydrodynamical equations), or field correlations (which lead to kinetic equations).

Thus QFT has very little use for discussions about measuring ''observables'' in the textbook sense.

  • $\begingroup$ I rarely achieve "focus", sadly. Transition probabilities use the vacuum projection operator: eg. $\hat A=\hat\psi_f^\dagger|0\left>\right<0|\hat\psi_f$ measured in the state $\omega(\hat A)=\langle 0|\hat\psi_g\hat A\hat\psi_g^\dagger|0\rangle$ (assuming appropriate normalizations) is the ubiquitous transition probability $|\langle 0|\hat\psi_g\hat\psi_f^\dagger|0\rangle|^2$. Insert S-matrix if desired. We can construct projection operators of arbitrary finite rank using mixtures of such observables, or of infinite rank if we admit completion in the operator norm. Hopefully that helps? $\endgroup$ – Peter Morgan Mar 12 '12 at 18:52
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    $\begingroup$ I see. So you don't treat the S-matrix as observable, as you had written, but compose it of may different observables, one for each scattering state. Then maybe this is the difference you asked for - a smeared field operator is a single observable (though one usually observes only its ensemble mean or its correlations, not a realization as in case of a projection operator). --- "the operational difference between these different classes of QFT observables in detail" - canyou focus this, please? what kind of answer do you expect? $\endgroup$ – Arnold Neumaier Mar 12 '12 at 19:05
  • $\begingroup$ Your comment → +1. Thanks. A smeared field observable is always presentable as a sum of other smeared field observables. The operator-valued distribution $\hat\psi(x)$ is not, but it's also improper. With apologies, I'm not yet sure how to make the question precise. The S-matrix is unitary, but it's nonetheless more-or-less the principal observable in QFT, insofar as the S-matrix describes the "evolution" of transition probabilities. Insofar as I have expectations for the kind of answer, they are prejudices. I think you might be right that your comment contains an answer of sorts. $\endgroup$ – Peter Morgan Mar 12 '12 at 21:10
  • $\begingroup$ I have a somewhat deviant non-mainstream view since I noticed that the textbook definition of quantum measurements and observables covers only a tiny fraction of actual measurements and hence is quite misleading. It took me a while to find my own constructive version. See Chapter 10 (in particular Sections 10.4/5) of my book lanl.arxiv.org/abs/0810.1019 $\endgroup$ – Arnold Neumaier Mar 12 '12 at 21:20

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