DrChinese said:
1. Vanhees71, post #15 of original thread: "Forget about paths here! It's really crucial to understand in this context that there are no paths and that photons are indistinguishable bosons!" Ah, I don't think I misunderstood anything - and certain not on purpose.
DrChinese said:
Vanhees71, post #106 above: "The indistinguishability of the two "paths" is the key feature of the entanglement of the two photons we discuss here." No significant issue with the science of that post, a couple of points to tidy up.
This is indeed crucial: The "paths" are indeed indistinguishable here, i.e., there is a coherent superposition of two possibilities, which you always insist to call "paths": "path 1": the photon with momentum ##\vec{k}_1## has polarization H and the photon with momentum ##\vec{k}_2## has polarization V and "path 2" vice versa. This clumsy verbal description is made clear by the math,
$$|\Psi \rangle=\frac{1}{\sqrt{2}} [\hat{a}^{\dagger}(\vec{k}_1,H) \hat{a}^{\dagger}(\vec{k}_2,V) - \hat{a}^{\dagger}(\vec{k}_1,V) \hat{a}^{\dagger}(\vec{k}_2,V)] |\Omega \rangle.$$
DrChinese said:
a. When discussing entangled photon pairs, and a diagram specifies a PDC source for the sake of discussion: the mechanics of PDC are not relevant. And yes, it is a factual (but completely technical) statement that basic entangled PDC requires indistinguishability of the 2 photons' sources. Specifically, in your diagram, you must not be able to determine if one particular photon came from the V cone or from the H cone (and vice versa). You could choose to pick up the photon pair such that one is known to be from the V cone - such pairs will not be polarization entangled.
No it's not simply a technical statement, it's what's defining entanglement in the first place! Of course, I could choose to pick up another photon pair as you describe, but now you yourself admit: "such pairs will not be polarization entangled." As soon as what you call "paths" are distinguishable, you don't have (complete) entanglement anymore and you prepare a different state, like the unentangled product state
$$|\Psi' \rangle=\hat{a}^{\dagger}(\vec{k}_1,H) \hat{a}^{\dagger}(\vec{k}_2 V) |\Omega \rangle.$$
I used PDC as an example, because it's pretty easy to understand, why one has entangled photons for this specific choice of the momenta, i.e., on the interaction points of the cones. The same basic "mechanism" is quite common to the production of entangled states though (e.g., the original cascade in Aspect's experiment).
DrChinese said:
b. Again, in general, polarization entangled photon pairs have no path indistinguishability requirement. They don't need to be path entangled, they don't need to be momentum entangled, or indistinguishable on either of those bases.
You contradict yourself, because two lines before you said it yourself, if you choose a situation, where you know on which cone the photon with momentum ##\vec{k}_1## is located (i.e., which polarization it has), you don't get an entangled state.
I still don't understand, what you mean by "path entangled". The photons are either in an entangled state or not!
DrChinese said:
Hopefully, the science of these 2 points is not in question.2. I agree with this, no issue. Hopefully we can leave it at this.