Why do radio signals have an absolute phase angle, but visible light photons don't? In quantum mechanics, we are usually told that the absolute phase angle of a wave function has no physical meaning, and that it's only the difference in angle that matters. That is, if we take a wave function ψ() and add 90° to the entire thing, it still represents the exact same physical situation. But I'm having a hard time nailing down exactly how that blends into the classical notion of a phase angle in low-frequency radio waves.
In radio waves, the phase angle of a radio wave is pretty obvious. I can just set up an antenna to receive a signal and plot a graph of it against time. There's a phase where the EM field is changing rapidly (i.e. in the middle of a sine wave), and there's a phase where it's changing slowly (i.e. at the extremes of a sine wave). This notion of a phase angle is also perfectly good at explaining any interference patterns involving radio waves. Importantly, the absolute angle is physically measurable. 0° is distinguishable from 90°.
Compare that to quantized visible light, when we do interference experiments with a single-photon light source. We talk about an abstract wave function, and we use its complex phase angle to determine the probability of when and where we will detect a photon. Why is it that in this case, we accept that the absolute phase angle has no physical meaning? Maybe there is some experiment we haven't yet thought of, which will distinguish the rising and falling edges of a photon?
I am just having trouble connecting the dots between the two notions of a phase angle. As we move between frequency domains, when do we exactly lose this phase information?
 A: The coherent state (aka: The Glauber state), is an eigenstate of the annihilation operator:
$$ \hat\alpha|\alpha\rangle = \alpha|\alpha\rangle $$
where $\alpha$ is a complex number:
$$ \alpha = |\alpha|e^{i\theta}$$
where $\theta$ is the phase. In the photon number (Fock) basis, it is:
$$ \alpha = e^{-\frac{|\alpha|^2}2}\sum_{n=0}^{\infty}{\frac{\alpha^n}{\sqrt n}}|n\rangle$$
From there you can compute phase noise and amplitude noise and photon number noise and all that.
The amplitude evolves as shown in the figure (see:https://en.wikipedia.org/wiki/Coherent_state)

as the amplitude increases, the relative size of the noise decreases and approaches a classical sine wave.
A: The phase of an electromagnetic wave is based measurable quantities, but the phase of a quantum wave function is not.  So you can measure or predict the phase of a light wave, but not a quantum wave function.
Consider an electromagnetic wave.  Suppose the wave source is an antenna with oscillating current flowing in it.  Then the wave phase at any place and time is determined by the distance from the source and the phase of the current.  Those are measurable quantities, and so the wave phase is predictable and physically meaningful.
However, the phase of a quantum wave function is not based on any measurable quantity that we know of.  Moreover, the wave function itself only has meaning when used to compute a probability density (e.g. the chance of a particle being within a region of space).  The probability density itself has physical meaning.  But in computing the probability density, the phase information is lost.  It has no physical meaning in the sense that it's not measurable or predictable.
