An interesting question, particularly in regards the photon, as you have a choice of what kind of phase to think about.  There is the phase of an electromagnetic (EM) wave of which the photon is an energy quantum, and there is the phase of a Schrodinger (or de Broglie) wave that is a complex quantity related to the location of the photon.

It is in the context of de Broglie-Schrodinger waves that the uncertainty principle is valid.  You may have a laser, emitting coherent photons, and you may reduce the intensity of the beam to where only one photon at a time is present, and you may know both the frequency and phase in the EM sense as accurately as you please.  But you still have no idea where individual photons are, or when they will be detected.

The EM frequency of a photon gives its energy level E.  The uncertainty principle holds that we cannot then determine precisely the time of detection of a photon of precisely known frequency but makes no prohibition about phase:  ΔEΔt ≤ ћ/2

On the other hand, the EM phase of a photon (or field) gives precisely its momentum.  For this the uncertainty principle is ΔpΔx ≤ ћ/2 and we find that if we precisely know phase, we have no idea where the photon is.  But there are no restrictions on frequency.

So, frequency and phase are NOT a quantum conjugate pair.  It is possible that a photon may have ... or not have ... a precisely defined frequency or phase, but measuring one does not disturb the other.

Note that if we construct a measurement to determine the phase of a photon, we will always be able to measure it.  By doing so, we scramble its position, so that it does not have simultaneously precise position and phase (momentum).  However, we may turn around and then attempt to measure the position and we will get an answer to that too, but the phase will be uncertain.  Uncertain, or unknown, does not mean that it does not have a phase.

The experiment I just described is a bit difficult with photons, because you have to absorb at least part of the energy to detect it.  By scattering, you may only absorb a small part, but nonetheless the frequency is a bit changed and by definition it is incoherent with its previous state.  A gamma ray may knock loose a few electrons and continue on as a lower frequency gamma ray, for example, but it's position has been determined.

So the momentum-position experiment cannot be repeated indefinitely without gradually changing the photon into a lower and lower frequency photon.  Usually the conjugate property scrambling is illustrated with spin or polarization, which may be modified several times before detecting and thus destroying the photon.

Hossein, please recall how the concept of a photon came to life. 1.You have Maxwell's equations, 2. you look for wavy solutions (all classically),3. you get a classical Lagrangian for EM fields and the classical action accordingly. You quantize this classical action. Frequency in this case is an energy (Planck constant is unity in selected system of units). The uncertainty relation is between energy and time. Your n is related to a NUMBER of quanta while your (delta fee) is related to a specific phase for a specific photon. This is like comparing apples with oranges...Sorry...

Dear prof. Kholodenko, I would thank for your frank answer, but please pay attention to attached file.

Thanks

Hossein, formally you are correct.But again, when it comes to delta t you are referring to a specific photon while delta n relates to all of them. Look, one has always start with classical mechanics. And when it comes to Heisenberg-type inequalities, these are always between the conjugate variables(in the sense of Poisson's brackets), e.g.  coordinate and momentum. So, figure out how n is related to the phase fee. As far as I am concerned, not related at all!

And what about magnitude? Has a photon magnitude?

Dear Hossein,

Per Richard Feynman, "I think I can safely say that nobody understands quantum mechanics."     Per Steven Weinberg, "There is now in my opinion no entirely satisfactory interpretation of quantum mechanics."

These expressions of uncertainty still exist today because QM was empirically derived based on the observations and measurements of the behavior of particles and forces - NOT on the underlying mechanisms of the particles and forces that give such observations and measurements. For example, we need to know what particles and the four forces look like and how they work before we can take the uncertainty out of QM.

Hence, you shouldn't expect to get an entirely satisfying answer to your question and questions like these that probe the depths of QM. Once we understand how particles and forces work, then we will be able to conceptually explain how and why the observations and measurements of them occur as they do, without having to invoke QM. Due to the randomness of particles, a probabilistic mathematical platform will still be required in some cases, but we will understand why we get the measurements and observations that we do. Please let me know if you have any comments

Regards,

Dan Correnti

In De Broglie formalism, massive particles have a phase velocity given by V = c^2/v, where v is the particle velocity and c the velocity of light. For photon, v = c. Therefore phase velocity of photon is also c. So if one accepts that the photon has a phase velocity, then how can you say that it does not have a phase?

Dear Hossein,

   As a first comment, saying that a photon has an uncertain phase does not mean that it has no phase, or that phase is meaningless for a photon.  

   In case it helps, please see the attached file.

Regards.  

Article On the Wave-Particle Duality: A Comment

  Being wavelike falls short of being a classical wave. A single quantum particle (photon included) is neither exactly classical particle, nor exactly classical wave. To form the latter, you need a very large number of photons, and to make it monochromatic and plane, all of them must be in one state (pure ensemble). An attribute of classical monochromatic plane wave is definite phase, which is reflected in

                         E(x, t) = E0 cos (kx-omega*t+ϕ)          (1).

But for a single particle, we must use instead de Broglie's complex expression 

                         E(x, t) = E0 exp(i(kx-omega*t+q)),      (2)

and its probabilistic interpretation makes it ambivalent to multiplication by exp(iq') with an arbitrary phase q' (invariance under gauge transformation). As the result, the total phase ϕ=(q+q') automatically becomes indeterminate. This is consistent with ΔnΔϕ≥1 since Δn=0 for a single particle. But for a pure ensemble with n>>1 we can have a non-zero Δn and even make it arbitrarily large in the classical limit. This can bring phase indeterminacy Δϕ down to zero and allow us to get back to the classical wave (1). Neither of these operations affects frequency, so a huge ensemble of photons all in one state makes a classical wave, but a single photon with the same frequency becomes phase-indeterminate.  

Sorry, I used the term "gauge transformation" instead of "global phase transformation" in my answer (under Eq. (2)). Both transformations are related to each other, but the gauge transformation acts on 4-vector of the EM field, whereas the phase transformation acts on a wave function. I hope that my typo will not cause any confusion, since the true meaning must be evident from the context (it was all about phase indeterminacy in the wave function of a single photon).  

In Standard Model, elementary particle such as photon are considered to be point-like and structureless. But recently experiment shows photons have been made to both change colour and shape-shift.

Time-resolved Scattering of a Single Photon by a Single Atom

https://arxiv.org/pdf/1604.08020v1.pdf

How the concept of unstructured photon is able to describe the different shapes and length of photon? In addition to shapes of photons, how two opposites’ charged particles such as electron and positron absorb and emit neutral and unstructured photons?

However, in CPH theory photons are combination of positive and negative virtual photons. Photon is a very weak electric dipole that is consistent with the experience and these articles are asserted. In addition, this property of photon (very weak electric dipole) can describe the absorption and emission energy by charged particles. In additon, a photon is made up of sub quantum energies and frequency of photon is depended to the number of sub quantum energies and interaction between them in structure of photon.

https://www.researchgate.net/publication/303988070_Generalization_of_the_Dirac%27s_Equation_and_Sea?ev=prf_pub

Article Generalization of the Dirac's Equation and Sea