# What will happen to an entangled pair of photons if one of them is absorbed and remitted by an atom?

Imagine if I have an entangled photon pair in linear polarisation, let them be named photon A and photon B, and if photon A is absorbed by an atom and then re-emitted what would be the state of photon B?

I can think of the following options:

a) The photon A and photon B will still be entangled.

b) The entanglement between photon A and photon B will be broken but photon B will be in still superposition of all eigen states of linear polarisation.

c) The photon B will not be in superposition but any of the random state of linear polarisation but we don't know what state it is.

• It's entangled with both the state of its partner and the atom it interacted with. Dec 17, 2022 at 12:53
• Is linear polarisation of an atom a conserved quantity ?? And the emitted photon and the absorbed photon by the atom don't have the same polarisation right ? Dec 17, 2022 at 13:00
• Merely tracking polarization is no longer sufficient to track the entanglement: you must consider more components of the combined wavefunction. Dec 17, 2022 at 13:09
• @MeruguArunKumar Angular momentum should be conserved in the absorption, so circular polarization should be conserved. So this should be a better basis for what you think of. Linear polarization is, of course, just a sum of two circular components, but if the phase is unknown (because it depends on the emission time and you don't have good enough control/knowledge of that) this will scramble the information. Dec 17, 2022 at 14:27
• Similar question answered physics.stackexchange.com/questions/739483/… Dec 17, 2022 at 17:27

What will happen to one of the entangled particle if its entangled partner is been absorbed and remitted by an atom?

It depends on what is meant here by absorbed and remitted, since these terms are used in somewhat different meaning, depending on the context (see, e.g., the discussion here.)

E.g., if we can neglect the atom interaction with the rest of the universe (e.g., we have an atom flying in vacuum in a cavity, tuned to be poorly coupled to the modes other than the two photons), then the atom will become entangled with the photons - what we will get is a bit more complex than two entangled photons, but it is still an entangled state. The Hamiltonian describing this situation is something like: $$H = \Delta\sigma_z + \omega_aa^\dagger a + \omega_bb^\dagger b + \lambda_a\sigma_x(a^\dagger+a)+ \lambda_b\sigma_x(b^\dagger+b)$$

On the other hand, the atom might be coupled to a photon bath, with the radiative lifetime much shorter than the observation time. In this case the absorption would be effectively a measurement/detection process, i.e., the entanglement would be surely destroyed. The possible model Hamiltonian is $$H = \Delta\sigma_z + \omega_aa^\dagger a + \omega_bb^\dagger b + \lambda_a\sigma_x(a^\dagger+a)+ \lambda_b\sigma_x(b^\dagger+b) +\\ \sum_k\omega_k c_k^\dagger c_k + \sum_k\lambda_k \sigma_x(c_k^\dagger+c_k)$$

It depends on which degrees of freedom are entangled, and how the absorption and emission process works exactly. It is certainly possible to design this in such a way that case a) happens -- A and the re-emitted B are still entangled, in exactly the same way as A and the original B photon were.

For instance, https://www.nature.com/articles/nature11023 reports experiments where the state of a photon is mapped onto an atom, and conversely, the state of the atom is mapped back onto an emitted photon -- such a setup would do precisely what I describe above: The state of the absorbed photon (which includes any kind of entanglement of B, due to the linearity of quantum theory) would be stored in the atom, and subsequently be mapped back onto the re-emitted photon.

On the other hand, c) can also happen -- in fact, it is always an accurate description, regardless of the process, if you only look at photon B. Then, however, the information carried by the entanglement should be gone somewhere, meaning that the photon is still entangled with something, just not the new photon A.

b), on the other hand, never happens. In fact, I don't know what a "superposition of all eigen states of linear polarisation" would precisely be.

Don't be confused by entanglement .... in downconversion we create 2 photons ...we know they have polarization relative to each other ..... BUT the photons are free to do whatever they want and they DO NOT effect each other after the initial creation. When you think of entanglement just think OK if we know A is vertical then we know B must be horizontal for example. There are quite a few papers that are trying to imply that the particles are entangled forever and by measuring one of them we force it to a certain polarization and therefore the other must get forced as well .... even though they could be miles and miles apart .... this science is proven wrong.

When a photon in general is absorbed in any molecule/atom and if there is an emission .... the emission process is most often random in polarization/direction ... example fluorescence. If we do use a process where polarization is known .... then yes we can measure polarization.

To elaborate on Roger Vadim's answer: The bottom-line here is that the system is described by some Hamiltonian the determines the evolution in time of the system's initial state. The time evolution will conserve only the information that is described by the symmetries of the Hamiltonian. Any other information will be dispersed into the correlations between the system's various degrees of freedom.

Considering the case in questions, the emitted photon will be emitted into one of the allowed states as dictated by the conservation of angular momentum + energy. As Roger Vadim mentioned, the Hamiltonian may be complicated and pertaining to many degrees of freedom besides the atom and the photons of the initial states. Since there are many states into which the system may evolve, the entanglement will be greatly reduced.