can one distinguish between superposition with randomized phase and classical probability?(an experiment) I hope that the following experiment will help me understand the topic better.
Let's say my friend is sending me photons via two channels and he is doing it in one of the following two ways:


*

*he is sending one photon at the time in one of the two channels. Sometimes through first, sometimes through second channel(i dont know which one unless i measure them).

*he is using a beam splitter and sending photons in superposition of through two channels, but he also adds some unknown phase to one of the channels each time he send a photon.
So I am receiving photons and I can do whatever I want with them. How could I determine whether he is doing option 1. or option 2. Since he is adding some different phase each time I guess that by recombine two paths I could not conclude much
EDIT: After reading the answers by Ronak and Julien I figured I have one more question that is puzzling me. In case there is no experiment that can distinguish two options, can I say that the states describing them are the same? Are they also the same for my friend how knows how much phase is changed each time?
 A: In fact you can't, and the formalism of quantum theory supports that.
Suppose, in the first case he's sending it in channel 1 with probability $p$. Then, the state of the photon you get is
$$\rho_1 = p |1\rangle \langle1| + (1-p) |2\rangle \langle 2|.$$
Suppose now he's doing the second thing and he's sending it through channel 1 with amplitude $\sqrt{p}$. Then, the photon you get is in the pure state
$$|\psi\rangle = \sqrt{p} |1\rangle + e^{i \phi} \sqrt{1-p} |2\rangle,$$
where $\phi$ is the random phase. Consider the density matrix
$$\rho'(\phi) = p |1\rangle\langle 1| + \sqrt{p(1-p)} e^{-i\phi} |1\rangle\langle 2|+ \sqrt{p(1-p)} e^{i\phi} |2\rangle\langle 1| + (1-p) |2\rangle\langle 2|.$$
Now, you get $n$ photons, each at random phases. As $n \to \infty$, the density matrix of the ensemble is
$$\rho_{ensemble} = \int_0^{2\pi} \frac{d\phi}{2\pi} \rho'(\phi) = \rho_1.$$
So, since the density matrix predicts the results of all series of measurements and unitary operations, these two cases aren't distinguishable.
A: You can try to distinguish between the two cases by measuring whether the photon came from the first channel or the second channel.
In the first scenario if he is sending the photons in one at a time and assuming he is randoly choosing which channel then you should see a photon in either channel 50% of the time.
In the second scenario there is a beam splitter which means that the photon is now in an entangled state of being in either channel 1 or 2.
$$ \left|\psi\right> = c_1\left|1\right> +c_2 \left|2\right> $$
Lets assume that the beam splitter is at some angle $\theta$ and that there is a phase difference $\phi$ added to beam 1.
$$ \Rightarrow \left|\psi\right> = \cos\theta e^{i\phi}\left|1\right> +\sin\theta \left|2\right> $$
Taking a measurement on your end is then equivalent to using the projection operator $P =  \sum_i \left|i\right>\left< i\right|$ and we can find the expectation value of that measuremnt.
$$ \left<\psi\right| P\left|\psi\right> = \cos^2\theta \left<1|1\right> + \sin^2\theta \left<2|2\right> $$
The phase factor goes away which makes sense since it should not change the expectation value of an Hermitian operator.
You will only be able to detect a difference in scenario 1 and 2 if $\theta \neq \pi/4$. You can get different results if you polarize your beams before passing them through either channel.
