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Is the Hermitian operator $\hat{\mathcal{O}}=\hat{\phi}^{-}(x)\hat{\phi}^{+}(x)$, where $\hat{\phi}^{+}(x)$ is positive frequency part of the scalar field operator, a well defined observable in QFT?

Some background:

  1. In Phys. Rev. 130, 2529 (1963), Glauber argues that square-law detectors measure the average value of the product $\hat{E}^{-}(x)\hat{E}^{+}(x)$, i.e., $\langle\Psi|\hat{E}^{-}(x)\hat{E}^{+}(x)|\Psi\rangle$ where $\hat{E}$ is the electric field. This is the basis of quantum optics which is an extremely successful empirically tested theory. I'm viewing $\hat{\mathcal{O}}$ as a generalisation of this theory to the scalar field.

  2. Lubos Motl answers here: In what sense is a scalar field observable in QFT?:

    • "Every observable in the technical or mathematical sense (linear Hermitian operator on the Hilbert space) is, in principle, observable in the physical operational sense, too."

    Since $\hat{\mathcal{O}}$ a well-defined (modulo smearing by test functions etc) linear Hermitian operator, it would be according to this definition physically observable in principle.

  3. However, the commutator $[\hat{\mathcal{O}}(x),\hat{\mathcal{O}}(y)]$ is non-zero for x-y space like separated distances. One usually demands such a condition, for example on the observables in Wightman axioms, see for example the (micro)locality Axiom 4, http://www.maths.ed.ac.uk/~jthomas7/GeomQuant/Wightman-Axioms.pdf.

    I've tried to find out why such a condition is necessary. Ron Maimon gives the following nice physical reason, Is microcausality *necessary* for no-signaling?:

    • "when you have arbitrarily tiny external agents capable of measuring any bosonic field in an arbitrarily tiny region, then microcausality is obviously necessary for no signaling, since if you have two noncommuting operators A and B associated with two tiny spacelike separated regions, and two external agents wants to transmit information from A's region to B's, the agent can either measure A repeatedly or not, while another agent measures B a few times to see if A is being measured. The B measurements will have a probability of giving different answers, which will inform the B agent about the A measurement."

    But this assumes that the measurement of A is a projective measurement, and leaves the state of the field in an eigenstate of A (i.e., so that the B measurement will pull the state out of the A eigenstate and hence signal to the A agent superluminally). In the Glauber theory, one never knows the state of the field. The detector interacts with the field by absorbing particles. This doesn't leave the state of the field in an eigenstate of $\hat{\mathcal{O}}$. Nevertheless, the observable $\langle \hat{\mathcal{O}}\rangle$ is observed from the particle detection intensity.

One might try to rule out the Glauber theory by absurdity: design a gedanken experiment using the Glauber detector on two space like separated positions, and show that they can be used to signal faster than the speed of light. I've tried but not been able to establish such an experiment. I would accept any answer that achieves to do this.

Another possibility is that the micro-locality conditions in axiomatic quantum field theory too strong? Can generalised (non-projective) measurements be used to obtain expectation values of micro-local violating operators?

Any other comments on why my question is incorrectly posed or based on any misunderstandings would be appreciated.

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    $\begingroup$ The problem is that the model explicitly exhibits a dependence on a spatial coordinate, so it seems to describe a localized observable but, in view of its violation of commutation rules it cannot be considered as a localized observable. Therefore it is not physically obvious the meaning of that $x$. $\endgroup$
    – Valter Moretti
    Commented Feb 4, 2014 at 14:23
  • $\begingroup$ A naive question. Is this model of detector devoted to theoretically describe detectors of coherent states or just states containing a finite number of photons, or both? $\endgroup$
    – Valter Moretti
    Commented Feb 4, 2014 at 14:25
  • $\begingroup$ @V.Moretti, I was under the impression that the Glauber detector works for any initial field state, i.e., number states, coherent states or other. In your first comment, are you disputing the claim made by Glauber that the detector absorbs a particle at the position x? $\endgroup$
    – Jase Uknow
    Commented Feb 4, 2014 at 14:53
  • $\begingroup$ (I have re-written this comment due to some typos). Concerning the nature of the states, my idea was that, if you consider states $|\Psi>$ containing a defined and finite number of photons, $<\Psi|E^-(x)E^+(x)|\Psi>=<\Psi|:E^2:(x)|\Psi>/2$ so no problems with local variable arise (if I have understood your decomposition of $E$. I have assumed that $E^+$ is that containing $a_k$ so that $<0|E^-(x)E^+(x)|0>=0$). Regarding x, well, actually I am not disputing anything since this is the first time I encounter Glauber detector, I just try to understand better it $\endgroup$
    – Valter Moretti
    Commented Feb 4, 2014 at 17:17
  • $\begingroup$ This is an interesting point. Glauber understood that a detector which works on an absorption principle never measures the vacuum fluctuations. You are saying that an absorbing detector is somehow equivalent to a renormalized detector which absorbs and emits). $\endgroup$
    – Jase Uknow
    Commented Feb 4, 2014 at 17:24

1 Answer 1

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Microcausality holds for observables that obey microcausality; it doesn't hold for observables that don't obey microcausality. The previous sentence is a tautology but I had to write it in this way because it seems to me that the question implicitly tries to disagree with this tautology.

There is no general condition that "all observables" have to commute at spacelike separation. Instead, what (local or relativistic) quantum field theory demands is that "there exist" elementary observables that obey this condition, like the "elementary fields" or the "stress-energy tensor" field, and so on. This is needed for special relativity or Lorentz symmetry to hold.

The field operators $\hat \phi^\pm$ clearly don't obey this condition (and, consequently, their general functions don't obey this condition) because they may be expressed (it's their definition) as linear superpositions of the microcausality-obeying fields $\phi(x)$ at "almost all" spacetime points (although the points where both arguments are nearby dominate in some sense). But that doesn't mean that they are not observables. It only means that the apparatuses that may measure these observables should be thought of as "intrinsically extended", not pointlike, gadgets.

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  • $\begingroup$ I hope I'm not denying a tautology :) Perhaps I'm just missing the point of micro-causality: If A is micro-causal then it ensures that measurements A(x) are independent of A(y) for space like x-y. This strongly forbids faster-than-light signalling from such measurements. My question was, if an observable isn't micro-causal does the ability to observe it imply one can signal faster than light? The fact that the word "causal" appears in micro-causal suggests this is true, but how do I see that the Glauber detector does this? It's not a projective measurement, so Ron Maimon's ex. doesn't work. $\endgroup$
    – Jase Uknow
    Commented Feb 4, 2014 at 15:20
  • $\begingroup$ Also nice point about being able to express $\hat{\mathcal{O}}$ as superpositions of the microcausality-obeying fields. I realised the same thing but was confused about what it meant :) So your interpretation here is that the Glauber detector must describe a detection which is "intrinsically extended", do you have some understanding of how this reconciles with Glauber's original argument that the detector is absorbing particles at the position, x? It seems, superficially, to be a local operation. $\endgroup$
    – Jase Uknow
    Commented Feb 4, 2014 at 15:35
  • $\begingroup$ If an observable does not fulfil micro causality it just means that it is not localized: To measure it you should, in principle, consider the whole (spatial) universe! All conserved charged arising from the quantum version of Noether theorem are of this type. In practice, instead, you measure something like $\int_{R^3} J^0(t,x) f(x) d^3x$ where $f(x)=1$ on a large spatial region $\Omega$ and rapidly vanishes outside it. However this model does seem to encompass Glauber's idea of detector. $\endgroup$
    – Valter Moretti
    Commented Feb 4, 2014 at 16:52
  • $\begingroup$ @V.Moretti, perhaps then it is instructive to rephrase my question in terms of the measurement of charge. Let me assume the measurement corresponds to your definition above, take two \Omega$ regions spacelike separated, is it possible (even if exponentially improbable) to signal by measuring charge in this way? Is this how one proves that one must consider a measurement of charge as being over the whole space? i.e.,does one use a no-signaling argument to prove that the operator is non-local-- or is there another way to see that non-microcausality implies the measurement is non-local. $\endgroup$
    – Jase Uknow
    Commented Feb 4, 2014 at 18:17
  • $\begingroup$ Yes I think you can rephrase everything using those spatially smeared observables. (Actually a spacetime smearing is more useful for field observables). If the two regions $\Omega$ and $\Omega_1$ are spacelike separated, the associated integrated currents $J^0$ commute. The observables that may not commute with, say, $J^0$ integrated over $\Omega$, are those observables similarly localized in regions $\Omega'$ at different time such that $\Omega' \cap J(\Omega) \neq \emptyset$. $J(A)$ is the union of causal curves intersecting $A$. $\endgroup$
    – Valter Moretti
    Commented Feb 4, 2014 at 18:51

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