Let's two photons are entangled in polarization after a laser beam passes through a Betha Barium Borate crystal. They take different paths and one of them (1) is absorbed in a black sheet. What is the state of the leftover photon (2)? Is it in superposition of polarization h/v or it must flip spontaneously in a certain polarization? What if the black sheet atoms absorb photons only with a certain polarization (say h)? Will the absorbed photon (1) take h polarization in the process of absorption and hence the second twin flip to v?
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At least under the many-worlds interpretation of QM, the leftover photon becomes entangled with the sheet. It enters into a macroscopic superposition in which its state becomes correlated with the state of the sheet that absorbed its twin.
Locally, nothing changes; the reduced density matrix for the leftover photon is the same just before and just after the absorption (assuming that process happens quickly).
But if the sheet consists of so many internal quantum degrees of freedom that the absorption process is effectively irreversible, and moreover in practice we observers can't conceivably perform any kind of controlled operation on the sheet's collective quantum degrees of freedom, then we say that the leftover photon has "decohered". In principle, there are still incredibly complicated correlations between its local observables and the state of the sheet - and very soon afterward, with anything else that the sheet interacts with. But in practice, these correlations are so complicated and nonlocal (e.g. many-point) as to be experimentally undetectable, and so the reduced density matrix for the leftover photon gives all the information that we can feasibly extract about that photon. Within our decohered "world" of experimentally accessible measurements, the photon is now essentially fully described by a classical probabilistic mixture - not a coherent superposition - of $h$ and $v$ states.
tparker
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Measuring one photon of an entangled pair doesn't affect the other member of the entangled pair. In the Heisenberg picture a system is described by quantum observables. The observables for a system only change as a result of interactions with that system. So the interaction with one photon doesn't change the observables of the other photon. These observables describe physical reality as being a more complex structure than the universe as described by classical physics that, in some approximations, resembles multiple non-interacting versions of the world as described by classical physics.
For each measurement there will be two versions of the measuring apparatus after the measurement. One of the versions of the measuring apparatus will record spin up, the other will record spin down. When a joint measurement is done on records of each result they then become correlated:
http://arxiv.org/abs/quant-ph/9906007
http://arxiv.org/abs/1109.6223.
Let's two photons are entangled in polarization after a laser beam passes through a Betha Barium Borate crystal. They take different paths and one of them (1) is absorbed in a black sheet. What is the state of the leftover photon (2)? Is it in superposition of polarization h/v or it must flip spontaneously in a certain polarization? What if the black sheet atoms absorb photons only with a certain polarization (say h)? Will the absorbed photon (1) take h polarization in the process of absorption and hence the second twin flip to v?
Photon 2 doesn't change as a result of an interaction between an absorber and photon 1. Photon 2's polarisation observables were unsharp before the absorption of photon 1 and remain unsharp after that. There are two versions of photon 2 before the absorption of photon 1 and there are two versions of photon 2 after the absorption.
If the black sheet absorbs photons with horizontal polarisation then any photon that passes through has vertical polarisation. If we place a detector after the sheet then any photon we measure has to have vertical polarisation. If we measure whether photon 2 has horizontal or vertical polarisation after photon 1 is absorbed and compare the results, then we will find that the polarisation of photon 2 matches that of photon 1. Each photon holds quantum information that can't be revealed by measurements on that photon alone, but only by comparisons of measurement results on the photons: locally inaccessible quantum information. This information is carried in decoherent (classical) channels and the correlation is only created when the information from one photon interacts with information from the other photon and that process takes place as a result of local interactions.
alanf
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The state of the remaining photon depends both on the nature of the initial entangled state as well as the nature of the absorption (or detection) of the other photon. To get an entangled photon with a BBO crystal, one needs to use type II phase matching, which would produce a Bell state for a single photon pair, given by $$ |\psi\rangle = \frac{1}{\sqrt{2}} |H\rangle_A |V\rangle_B + |V\rangle_A |H\rangle_B . $$
If the black screen absorbs the $B$-photon without any regard for the polarization, then the absorption process represents a trace over the $B$-degrees of freedom, leaving behind a mixed state of both polarization. That means that the remaining photon would be unpolarized, having equal probabilities for both states of polarization: $$ \rho_A = \text{tr}_B\{|\psi\rangle\langle\psi|\} = \frac{1}{2}(|H\rangle\langle H|+|V\rangle\langle V|)_A . $$ In effect, it represents a measurement with an observable given by the identity operator.
If the black screen absorbs one particular state of polarization, then it works like a detection, leading to a projective measurement. In that case the remaining photon would have the opposite polarization due to the way the photons were entangled by the preparation process. Say, for example, the screen measures the $V$ photon. Then it represents an observable given by $(|V\rangle\langle V|)_B$. The state of the remaining photon is then given by $$ \rho_A = N \text{tr}_B\{|\psi\rangle\langle\psi|(|V\rangle\langle V|)_B\} = (|H\rangle\langle H|)_A . $$ The $N$ is a normalization constant that we need because a projective measurement is not trace preserving.
flippiefanus
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What if the black sheet atoms absorb photons only with a certain polarisation?
https://apps.dtic.mil/sti/pdfs/AD1096363.pdf
They have used holes, rather than a black sheet, so this experiment may not be exactly what you want.
But,
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user353326
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When two photons are entangled in polarization, the state of each photon is dependent on the state of the other. In the scenario you describe, if one of the entangled photons (photon 1) is absorbed by a black sheet, the state of the remaining photon (photon 2) will change instantaneously. However, the exact state of photon 2 after the absorption of photon 1 will depend on the details of the situation.
If the black sheet absorbs photons of both horizontal and vertical polarization, then photon 1 will be absorbed in a superposition of horizontal and vertical polarization. In this case, photon 2 will also be in a superposition of horizontal and vertical polarization after photon 1 is absorbed.
If, on the other hand, the black sheet absorbs photons of only one polarization (say, horizontal), then photon 1 will be absorbed in that polarization. In this case, photon 2 will be in the opposite polarization (vertical) after photon 1 is absorbed.
Finnaly, the state of the remaining entangled photon (photon 2) will change instantaneously when the other photon (photon 1) is absorbed by the black sheet. The exact state of photon 2 will depend on the details of the situation, such as the polarization of the photons and the properties of the black sheet.
An example of the state of an entangled photon after its twin is absorbed is the phenomenon of "quantum teleportation." This is a process in which the state of one entangled photon is transferred to the other entangled photon, even when the photons are separated by large distances. In this case, the state of the photon that is absorbed will be "teleported" to the remaining photon, effectively making the two photons "exchange" their states. However, this process is probabilistic and depends on a number of factors, such as the efficiency of the teleportation process and the state of the entangled photons before the absorption occurs.
Pegasus
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What do the experiments say? With a quantum measurement, the measured state depends on what measurement is performed. If you assume that the photons individually have states before measurement, you get Bell's Inequality, and the experiments falsify this. It thus doesn't make sense to ask what the state of the photon is before measurement: all you can predict is the correlation between measurements.
John Doty
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