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:

  1. 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).

  2. 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?

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    $\begingroup$ Answer to edit: I fear this impinges on the domain of foundations. But, let's take the naive realist stance that I'd like to be true. In it, every photon is in a pure state - the ontological state is pure. But, sometimes you don't know what that state is, and the best you can do is say that it's in state $|\psi(x)\rangle$ with probability $p(x)$; so we construct an epistemic state, the density matrix, that predicts the results of all experiments. So, to directly answer your questions, the photon is in the same ontological state for both of you, but in a different epistemic state. $\endgroup$ – Ronak M Soni Apr 22 '15 at 9:14

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.

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  • $\begingroup$ Ok that makes sense if one follows density matrix formalism. But I am still a bit confused. Since the the off diagonal matrix are zero, can I say that state of each photon is in classical superposition? Since the density matrices are same then the state are also same-both classical? but for my friend how knows the phase everytime he changes it, state of each photon is quantum. Whether state is in quantum or classical superposition should not be subjective, right? $\endgroup$ – sa101 Apr 21 '15 at 18:17
  • $\begingroup$ @sa101 The density matrix of a state is based on your knowledge of the system. If $\rho = \left|\psi\right\rangle\!\left\langle\psi\right|$ for some quantum state $\left|\psi\right\rangle$, then you have what is known as a pure state which represents the maximal amount of knowledge you could possibly know about a system. By contrast a so called "mixed state" represents a (non-unique) mixture of possibly pure states, and these probabilities represent your ignorance which is generally subjective. $\endgroup$ – Punk_Physicist Apr 21 '15 at 18:47
  • $\begingroup$ It all comes down to this: quantum theory is horrible at predicting the answers to single experiments; it can only give the average of the answers you'll get if you measure the same observable multiple times for the same state. Since in the random phase case you don't actually get the same state again, what you know about the state is that it is $|\psi (\phi)\rangle$ with probability density $P(\phi) = \frac{1}{2\pi}$, and so the answer you can predict with quantum theory is that of the state $\rho_{ensemble}$ above. $\endgroup$ – Ronak M Soni Apr 21 '15 at 18:57
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    $\begingroup$ I think the difference between a quantum and classical superposition is not always a good dichotomy. In particular, you broke the dichotomy with this nice thought experiment. The state is in a 'classical superposition' (= probabilistic mixture with no interference or other coherence effects) - with a 'superposition' element for every $\phi$ - of the 'quantum superpositions' $|\psi(\phi)\rangle$ - with a 'superposition' element for each channel - such that the final state looks like the 'classical superposition' over channels. $\endgroup$ – Ronak M Soni Apr 21 '15 at 19:04

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.

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  • $\begingroup$ So for $\theta=\pi$ /4 I can not distinguish two options? In that case, are the states that describe option 1 and option 2 the same? $\endgroup$ – sa101 Apr 21 '15 at 18:38
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    $\begingroup$ I think, as you've defined it, $P = 1$ (the identity operator). Also, you seem to be missing the fact that every photon in the second case is in a different random phase and therefore you can't do multiple measurements with the same $\theta$ - making the fact that the probabilities are different rather irrelevant. $\endgroup$ – Ronak M Soni Apr 21 '15 at 19:11
  • $\begingroup$ Yes, I did not realize that the question was asking for the scenario of a changing random phase. In my answer I was thinking of $\theta$ as the angle of the beam splitter and $\phi$ as the phase. The angle $\theta$ is to account for the possibility that a 50/50 beam splitter is not being used. If it is a 50/50 beam splitter then $\theta=\pi/4$ and you cannot distinguish between the scenarios. $\endgroup$ – Julien V Apr 21 '15 at 19:35

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