orisomech said:
Sorry, I'm kind of lost you. From what I learned in QED photons are described in momentum space, so a photon here or in the other side of the universe are equivalent if they have the same momentum, which is obviously wrong.
How is QED describe the photon different than I described?
Well, the usual description of relativistic QT is quantum field theory. A photon wave function makes not too much sense, at least not when it comes to interacting photons, and all we know about photons is through interactions with matter.
As massless quanta photons are utmost relativistic, and there's nothing that can hinder them from being produced and annihilated whatsoever, i.e., since they are massless, there's no energy gap, i.e., the slightest acceleration of a charged particle produces soft photons. That's why there's no limit, in which a single-particle wave description for photons interacting with charged matter could be valid.
So the right description is a theory, where photons can be created and annihilated through interactions with charged particles/matter enabling us to describe, how they are observed, and the most convenient (in fact also the only way making physical sense) is relativistic quantum field theory.
You start with quantizing the electromagnetic field coupled to charged particles. Full QED uses some (qantized) relativistic field for the charged particles too. The first and most simple physically relevant case is a system of fields, describing electrons and positrons interacting with the em. field (photons). You can as well also look at the approximation, where the matter particles are described by non-relativistic QT (in both 1st and 2nd quantized form).
You'll soon realize that there is some problem due to the gauge invariance of electromagnetism when trying to canonically quantize the em. field, described by the four-potential. The most simple way is to fix the gauge using the Coulomb-gauge condition ##\vec{\nabla} \cdot \vec{A}=0## and realizing that the temporal component ##A^0## is not an independent field but given by the Poisson equation,
$$\Delta A^0 = -\rho,$$
with the solution
$$A^0(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \frac{\rho(t,\vec{x}')}{|\vec{x}-\vec{x}'|}.$$
Then you can just quantize the field components ##\vec{A}## in the canonical way, taking into account the gauge constraint.
You end up with a Hamiltonian consisting of a free-field part, a matter part (including the static Coulomb interaction), and an interaction part, with which you can start to do perturbation theory in the interaction picture (carefully avoiding trouble related to Haag's theorem by first working in a finite volume with periodic boundary conditions for the fields).
The free em. field has an interpretation in terms of photons. It has two polarization-degrees of freedom (as can be immediately seen from the Coulomb-gauge condition), and you can expand the quantized em. field in terms of annihilation (associated with the positive-frequency modes) and creation (associated with the negative-frequency modes) operators, annihilating and creating a single photon in a given momentum-polarization state. The Hilbert space is realized as a socalled Fock space. The demand of microcausality, i.e., that the local observable operators should commute at space-like separation of their arguments, and of the existence of a ground state of lowest energy forces you to quantize the em. field as bosons (which is of course in accordance with observations). This holds true for any field with an integer spin, but photons are special, because they are massless; they have not three spin-degrees of freedom as expected for a massive particle with spin 1 but only two polarization states, which can be chosen as the helicity eigenstates; helicity is the projection of total (sic!) angular momentum of the quantum to the direction of its momentum, and there are only two values left, namely ##\lambda=\pm 1##, referring to the left- and right-circular polarized em. plane wave modes.
Then you can also quantize the (massive) charged particle part too. In this case it also contains the Coulomb interactions, giving rise to bound states. If you have a proton and an electron state, they form hydrogen atoms, which are described in the non-relativistic limit and neglecting spin as you know it from the QM 1 lecture, but there may also be scattering of an electron and a proton.
Then the interaction of such a system of charged matter particles with the em. radiation field leads to the description of all kinds of radiative processes, including spontaneous and stimulated emission as well as absorption of photons on a hydrogen atom leading to transitions from one to another bound state. In lowest order that's all a nice exercise in time-dependent perturbation theory.
Then there are also higher-order corrections, which are very tricky, because they give rise of infinite integrals. It took the physicists quite a while to resolve this issue by perturbative pertubation theory. In fact to get the issue resolved, the famous measurement of the Lamb shift, as a radiative correction of this kind, in the hydrogen spectrum triggered the theorists to finally resolve this issue with the divergences and to invent renormalization theory (among them Bethe, who got the issue solved first within the non-relativistic theory for hydrogen, Feynman, Schwinger, and Tomonaga, for which work the latter 3 got the Nobel prize in physics; Bethe got his Nobel a bit later for the explanation how stars shine).
In all this you nowhere need the idea of a position for a photon. The theory, however, of course also describes the detection of photons. One way is to use the photoelectric effect. In the most simple form the photon scatters on the condactance electron of a metal, hitting it out of the metal (i.e., a transition from a bound state of the electron in the metal to a scattering state of this electron). This "photoelectrons" can then be measured. Seen as a measurement of the "position" of an electron, it's just defined by the location of the detector, which as a massive object of course has a position observable.
This is quite analogous to the observation of visible light: What you precept is just the interaction of a classical electromagnetic wave with the retina in your eye, producing an electric pulse, which is later processed in your brain. What you call "intensity of the light" physically is given by the energy density of the electric field, and this can be easily defined by the field operators of this field.
Also what I've just called "a classical em. field" is quantum-field-theoretically described by a socalled coherent state, which is not a state of definite photon number like the Fock states used to define the Hilbert space for the free em. field but a superposition of such states with any number of photons, defining a state of well-defined phase rather than of well-defined photon number.
