Describe an apparatus that would put two photons in a classically correlated state, such that both photons?

Question

have equal a priori probability to be in either polarization state (e.g. left- and right-hand circular polarization, or horizontal and vertical plane polarization) but the apparatus is designed so as to put them in have opposite polarization states. Thus measuring the polarization of one tells you that the other photon is in the opposite polarization state without having to make another measurement. (Note that this is not quantum weirdness!)

Answer 1

http://en.wikipedia.org/wiki/Spontaneous...

Answer 2

Either there is a contradiction in your question or I don't understand something in it correctly. Please fix me in that case and I'll try to improve my answer.

If a photon can be in either polarization state, you can't get this information by measuring it (you would need an unlimited source of identical photons to do so). Remember that performing any measurement on a qubit, which a polarization state of a photon represents, you can't get more than 1 bit of classical information.

Therefore, you can't build an apparatus to tell you it is horizontal, vertical, slanted anyhow, circular or elliptical. You only can build an apparatus that can tell horizontally polarized photons from vertical, for example. Other states get projected with some probabilities to one of these possibilities.

If you accept this and want, after the measurement, to have the other photon to be with certainty in the opposite (orthogonal) state, this is exactly what ones call entanglement and you the quantum weirdness and which you seem to reject.

In other words: no measurement can be done without spoiling the unknown state of the photon. Then somehow it must influence the other one as well to ensure the result you want. Without accepting superluminar communication, this is only possible to be done using quantum phenomena.

Edit: Pointing this out, I forgot to answer the main question. Mistress Bekki is right there, but the SPDC gives _entangled_ pairs of photons again (I think it is the primary source of such pairs for the according purposes).

Hope this helps!

Edit 2: This may be the answer... If one basis was preferred (let's say its H & V), you can take an entangled pair source *) and make a measurement in this basis as a part of the preparation. After this, pairs of photons polarized HV or VH come out at random and with equal probabilities.

There's a classical correlation, just with restricted usability: you can only measure the given photon in the same basis you chose above to get certainty in the other photon's state. There's neither entanglement nor any wavefunction collapses anymore, just "re-assuring" of the previous measurement's result which was forgotten.

* producing the Bell's Psi-minus state,

http://en.wikipedia.org/wiki/Bell_state

or http://en.wikipedia.org/wiki/Quantum_ent...

Edit 3: On the single photon polarization

I don't think the single photon measurement is overlooked by book authors. In fact, I have already told it above: it acts exactly like a qubit, which is nothing more than a two-level system. So that if you define polarization states |H> and |V>, all other states are their complex linear combinations with coefficients adding up to 1 in squared absolute value:

|ψ> = α |H> + β |V>, |α|^2 + |β|^2 = 1

If you draw a Bloch sphere of such system, there will be for example horizontal polarization on the "north pole" and vertical on the "south pole". The "equator" will be built by a transition from left circular through +45° slanted, right circular and -45° slanted back to left circular polarization. All other states will be somewhere inbetween according to which of |H> or |V> has bigger overlap with them.

For a complete measurement on such system, you must choose an orthonormal basis. Examples are horizontal / vertical, any other angle / perpendicular to it or left / right circular. The measurement collapses the wavefunction (or most probably kills the photon), giving you exactly 1 bit of information. This is surely not enough to tell what the state was prior to the measurement, and there's no way of exceeding this limit.

Let's have a look on your situation. If there's a truly random generator, you are exactly right when saying both of the photons will be in the mixed state 1/2*I. Unfortunately, any kind of classical correlation goes here away, mostly because of the last paragraph. If you measure anything without entanglement and without information about what the state could in fact be, you know absolutely nothing about the second photon **) and so it stays for you in 1/2*I. You know, the density matrix problems can sometimes become as puzzling as entanglement ;-)

**) To be precise, you know it isn't in the very basis state you have measured, this would be a contradiction. However, this excludes only a measure-0 subset of the possible states.

I'm sorry I haven't provided any references so far. I think you'd find some discussion on this topic in Harry Paul: Introduction To Quantum Optics.

P.S. I can't find the other question!

Edit 4: I'm sorry, I did not want to offend you... I can see you know what you are talking about very well. However, from the sole fact its state space is 2-dimensional, it follows we have no way of finding the complete state by a single measurement. The principles of QM say we can only use projective measurement (or some of its equivalent formulations) to get information about quantum state. Thus, we must construct some observable on the state space. Any observable on 2D space can have at most 2 different eigenvalues, so it is a yes-no experiment. Also the projection and thus destruction of the state is inevitable according to that principle. How could we use this to locate a point on a sphere?

Just now, I realized I can be wrong with the exact form of the density matrix of the second photon after measurement. I am sorry. However, I am absolutely sure it won't represent a pure state. If you measured the second photon as well, most possibly there would be some correlation of the results, but not 100% (which we would need to be able to say we know the other state without measurement).