# What is the state of an entangled photon after its twin is absorbed?

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?

• Commented Dec 3, 2022 at 19:01
• @mmesser314 The above thread is about measurement and I am not sure the absorption in the black sheet which is inevitable does constitute a measurement. And don't know how it is interpretation dependent. Just one has to make an exp and check polarization of photon 2. Commented Dec 3, 2022 at 20:46
• There is a lot to read through. Check the second answer to the linked post. - How is a measurement performed? By interaction. ... Check the third answer - In summary, the Copenhagen interpretation says that if you measure one of an entangled pair, you instantly force the other into a state corresponding to the expected output, regardless of the speed of light. Commented Dec 4, 2022 at 2:06
• If a polarizing black sheet absorbs one photon, the other must have the perpendicular orientation, and its wave function must be consistent with this orientation. The wave function cannot be observed. The Copenhagen interpretation says the wave function collapsed to a new state, and ignores problems of speed of light delay in this collapse. The Evert interpretation says the wave function split in two, and you are on a branch where the wave function of the remaining photon has the perpendicular polarization. There are no observable differences between the two. Commented Dec 4, 2022 at 2:20
• Commented Dec 14, 2022 at 22:35

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.

• I am looking for experiment. Has not anybody made such experiment? Commented Dec 11, 2022 at 15:26
• I am asking what is the state of photon1 after 2 is gone. As far as I understand 1 is superposition v+h. But suppose a have a polarizing beam splitter v/h for ph2. The v part is absorbed so at this moment the ph2 is in h polarization only, right? The ph1/ph2 is not entangled and 1 must be in v only, as initial entangled state v1h2+v2h1 -> v1h2+O.h1. Commented Dec 11, 2022 at 17:51

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

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.

• Do you experiments of this kind (one photon of EPR pair absorbed)? What are their results? Commented Dec 11, 2022 at 16:44
• I have edited the answer to explain more fully. Commented Dec 11, 2022 at 19:40
• I was just wondering is it edited. Commented Dec 11, 2022 at 19:45
• I dont accept that measuring photon 1' polarization from an entangled pair does not influence the other one p2. That is not, true. They were both in h+v superpostion before measurement - entangled v1h2+v2h1. There was 50/50 chance that p1 get h in measurement M. After M p1 is h 50%. Then surely p2 is in v 100%. It was in superposition h+v and after M of p1 is in v. How is this not an imfluence?? Commented Dec 11, 2022 at 19:53
• Two entangled photons are ONE system not two separate systems. So if a part of the system changes they is no wonder the other part chamges too. It is like a absolite solid body in real realization. Commented Dec 11, 2022 at 19:58

What if the black sheet atoms absorb photons only with a certain polarisation?

They have used holes, rather than a black sheet, so this experiment may not be exactly what you want.

But,

I found the above activities very helpful!

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.

• No, the fact that the state changes does not mean that you can use it to communicate. Commented Dec 9, 2022 at 4:26
• Surely you must know that the state of two entangled particles and the mixed state of the remaining particle after one has been measured are two different states. Still that does not allow you to communicate. Commented Dec 10, 2022 at 2:15
• Do you experiments of this kind (one photon of EPR pair absorbed)? What are their results? Commented Dec 11, 2022 at 16:47
• @flippiefanus I am asking what is the state of photon1 after 2 is gone. As far as I understand 1 is superposition v+h. But suppose a have a polarizing beam splitter v/h for ph2. The v part is absorbed so at this moment the ph2 is in h polarization only, right? The ph1/ph2 is not entangled and 1 must be in v only, as initial entangled state v1h2+v2h1 -> v1h2+O.h1. Commented Dec 11, 2022 at 17:50
• It depends on what happens on the other side. I'll add some explanation. Commented Dec 12, 2022 at 8:19

Take the initial state of the pair and project onto the outcome of the measurement.

• I'm not sure, but I believe that the OP is asking whether or not the described setup does constitute a measurement, and why. Commented Dec 8, 2022 at 2:23
• I am asking what is the experimental result if one twin is absorbed? What is the twin is an sheet (conical) and inevitably is forced to absorb as v polarized? Is it possible it not to be absorbed when there is no way out and not possible to be absorbed as h polarized? Commented Dec 11, 2022 at 16:52
• I am asking what is the state of photon1 after 2 is gone. As far as I understand 1 is superposition v+h. But suppose a have a polarizing beam splitter v/h for ph2. The v part is absorbed so at this moment the ph2 is in h polarization only, right? The ph1/ph2 is not entangled and 1 must be in v only, as initial entangled state v1h2+v2h1 -> v1h2+O.h1. Commented Dec 11, 2022 at 17:51
• @Mercury: You haven't specified the initial state, so the final state can be anything. Specify the initial state, project to it, and you'll have your answer. Commented Dec 12, 2022 at 0:31

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.

• What does "absorbed in a superposition of horizontal and vertical polarization" even mean? Commented Dec 8, 2022 at 16:17
• if a photon (particle of light) is said to be "absorbed in a superposition of horizontal and vertical polarization," it means that the photon is in a state where it is simultaneously absorbed as both a horizontally polarized photon and a vertically polarized photon. Commented Dec 8, 2022 at 16:24
• But a superposition of H and V is e.g. |H>+|V>, which is a diagonally polarized photon, with a specific diagonal direction. Does it mean it is absorbed as a diagonally polarized photon? If yes, does this mean the absorption chances if I put a phase plate? Commented Dec 8, 2022 at 16:39
• when a photon in the superposition state |H>+|V> is absorbed, it can be absorbed as either a horizontally or vertically polarized photon, or some combination of the two, but it is not necessarily absorbed as a diagonally polarized photon. Commented Dec 8, 2022 at 16:50
• Then what does it mean to be "absorbed in a superposition of horizontal and vertical polarization", if it can be absorbed this way, that way, or another way? Commented Dec 8, 2022 at 16:54

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.

• Do you experiments of this kind (one photon of EPR pair absorbed)? What are their results? Commented Dec 11, 2022 at 16:45
• Almost every experiment that detects photons does this, since the emitted photon's state is theoretically entangled with its emitter, but you usually don't measure the emitter's state. Commented Dec 11, 2022 at 16:56
• I believe that about emitter in x and p. But not In polarization. I am asking what is the state of photon1 after 2 is gone. As far as I understand 1 is superposition v+h. But suppose a have a polarizing beam splitter v/h for ph2. The v part is absorbed so at this moment the ph2 is in h polarization only, right? The ph1/ph2 is not entangled and 1 must be in v only, as initial entangled state v1h2+v2h1 -> v1h2+O.h1. Commented Dec 11, 2022 at 17:46
• @Mercury No, 1 is in whatever polarization you detect it. If your detector distinguishes h and v, you're right, but what if your detector is rotated 45 degrees? Commented Dec 11, 2022 at 18:41