PeterDonis said:
You're not the only one posting in this thread. You said you were using standard terminology. Presumably
@vanhees71 is too. Here is what he said:
But of course it isn't exactly a "single-photon Fock state", as you pointed out in response to him.
So maybe my confusion is about what "standard terminology" is. Is it "standard terminology" to add a "finite spectral width" qualifier when describing states, but not a "finite number width" qualifier? That seems to be what
@vanhees71 was doing.
I don't think that the Reeh-Schlieder or Haag's theorem are very important concerning relativistic QFT as a physical theory describing phenomenology. Of course, these mathematical theorems make it clear that in fact relativistic QFT of interacting particles is not mathematically strictly defined, and what's used with great success is renormalized perturbative QFT, where these theorems are irrelevant. They are of course highly relevant in the investigation of the mathematical properties of QFTs in the sense of algebraic or axiomatic approaches.
It's also clear that strictly speaking the "photon number" is not an observable to begin with, because it's not a gauge-covariant quantity. What's observable are detector clicks due to the presence of the electromagnetic field. What can be prepared are photon states, which are as close to single-photon states as you want to get them.
It's as in non-relativistic quantum mechanics with momentum (and energy for free particles) eigenstates: They are no true states, and you cannot strictly speaking prepare any particle in a momentum eigenstate, because the plane wave ##\exp(\mathrm{i} \vec{x} \cdot \vec{p})## is not square integrable and thus no representant of a Hilbert-space vector in position representation. What you can, however, prepare are "wave packets" with an arbitrarily sharp momentum.
The same is true for "single-photon" states. Of course the plane-wave momentum eigenstate (in quantum-optics lingo a "single-mode one-photon Fock state" is not a true state too, but you can build normlizable one-photon Fock states in the form
$$|\Psi \rangle=\int \mathrm{d}^3 p \frac{1}{\sqrt{(2 \pi)^3 2 \omega_{\vec{p}}}} A(\vec{p}) \hat{a}^{\dagger}(\vec{p},\lambda) |\Omega \rangle,$$
where ##A(\vec{p})## is square integrable. The creation-annihilation operators obey the (non-covariant) commutation relations ##[\hat{a}(\vec{p},\lambda),\hat{a}^{\dagger}(\vec{p}',\lambda')=\delta^{(3)}(\vec{p}-\vec{p}') \delta_{\lambda \lambda'}##; ##\lambda \in \{1,-1\}## denotes the helicity eigenstates . All this is in the Coulomb gauge for free fields which leads to transverse four-potential fields, ##A^0=0##, ##\vec{\nabla} \cdot \vec{A}=## (which completely fixes the gauge at the cost of lost manifest Poincare invariance).
In this convention the total number operator for free photons is
$$\hat{N}=\sum_{\lambda} \int_{\mathrm{R}^3} \mathrm{d}^3 k \hat{a}^{\dagger}(\vec{k},\lambda') \hat{a}(\vec{p},\lambda').$$
Now we have
$$\hat{N} |\Psi \rangle = \int_{\mathbb{R}^3} \mathrm{d}^3 p \frac{1}{\sqrt{(2 \pi)^3 2 \omega_{\vec{p}}}} A(\vec{p}) \int_{\mathbb{R}^3} \mathrm{d}^3 k \sum_{\lambda'} \hat{a}^{\dagger}(\vec{k},\lambda') \hat{a}(\vec{k},\lambda') \hat{a}^{\dagger} (\vec{p},\lambda) |\Omega \rangle.$$
Since ##\hat{\vec{k},\lambda'}|\Omega \rangle=0## we can as well write
$$\hat{N} |\Psi \rangle= \int \mathrm{d}^3 p \frac{1}{\sqrt{(2 \pi)^3 2 \omega_{\vec{p}}}} A(\vec{p}) \int_{\mathbb{R}^3} \mathrm{d}^3 k \sum_{\lambda'} \hat{a}^{\dagger}(\vec{k},\lambda') [\hat{a}(\vec{k},\lambda'), \hat{a}^{\dagger} (\vec{p},\lambda)] |\Omega \rangle.$$
and use the commutator relation
$$[\hat{a}(\vec{k},\lambda'), \hat{a}^{\dagger} (\vec{p},\lambda)]=\delta^{(3)}(\vec{p}-\vec{k}) \delta_{\lambda \lambda'}$$
to get
$$\hat{N} |\Psi \rangle = |\Psi \rangle,$$
i.e., you have a single-photon state, which is normalized to 1, ##\langle \psi|\psi \rangle##.
Of course there's no contradiction to the Reeh-Schlieder theorem since this is not a "localized state" in any sense. Photons can't be localized in any strict sense since they don't even admit the definition of a position operator.
Of course you can't measure the photon number ##N## but only true observables. AFAIK the first preparation of single-photon states have been realized by Clauser in 1973. Of course there a true observable was measured (what else can you meausure), i.e., the registration of the single photons by photo multipliers and clearly the expected non-classicality of the photo effect for such single-photon states have been demonstrated:
https://doi.org/10.1103/PhysRevD.9.853
https://escholarship.org/content/qt3wm0v847/qt3wm0v847.pdf?t=p0lmks (for a freely downloadable scan of the preprint)
