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Decoherence  How not to break entanglement? 
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#1
Feb1806, 03:43 AM

P: 37

Assume one photon of a momentum entangled pair of photons is absorbed by an isolated atom. Is there now an entanglement between the other photon and this atom? If so, in what way is the atom entangled with the photon?
If the atom after a while emits a photon, under what conditions would the entanglement survive and get transferred to the emitted photon (i.e. that the photon would remain coherent)? I think this is the same question as to why would a photon travelling through glass or through air remain coherent (if it is absorbed and reemitted). Now assume an entangled photon (of a pair of momemtum entangled photons) gets absorbed by an isolated ensemble of atoms bound together with many degrees of freedom. Does the entanglement survive? Does the entanglement get dissipated throughout the ensemble, but survives with intact total amplitude at the photon? Does it get attenuated? Or does the entanglement get completely broken/destroyed? Thanks to all. 


#2
Feb1806, 06:36 AM

P: 87

Here's an article that you might find useful: http://www.newscientist.com/article.ns?id=dn2564
Here's an excerpt: "Altewischer found the photons created waves of electrons on the gold surface called plasmons that passed through the holes and reemitted the photons on the other side. Measurements showed that the emitted photons were still entangled." 


#3
Feb1806, 07:03 AM

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These are very interesting thoughts! First of all, it is probably necessary to make a distinction between two kinds of interpretations: MWI/decoherence interpretations and collapse interpretations. The reason is that in MWI interpretations, entanglement grows, and irreversible entanglement is at the origin of the lack of interference  while in collapse interpretations, the collapse kills entanglement, and is  as such  responsible for the lack of interference.
The point is that "entanglement" is not, as such, an observable phenomenon; quantum interference is. As I said already in a few posts, if we have two states of a system, a> and b>, then an observable that looks for interference between both is an observable that has as eigenstates a>+b> (eigenvalue 1) and a>b> (eigenvalue 0). In the typical 2slit experiment, the states a> and b> correspond to "through slit 1" and "through slit 2" respectively, and the observable is the intensity observed at, say, a peak of the interference pattern. When the probability for 1 is 100% and the probability for 0 is 0%, then we have an interference pattern ; when it is 0% for 1 and 100% for 0, we have the complementary interference pattern ; and when it is 50% for 1 and 50% for 0, we have NO interference. Now, I think that what you call "entanglement preserved" is a special kind of interference which shows as correlations between measurements on two systems. Indeed, if we have an entangled system: a>c> + b>d>, it is easy to show that the interference between a> and b> as such, disappears, but that the interference between a>c> and b>d> DOES show up. A typical way of seeing this, is to test for a correlation between (a>+b>) and (c>+d>). Indeed, a>c> + b>d> = 1/2{ (a>+b>)(c>+d>) + (a>  b>)(c>d>)} This is what's typically happening in "quantum erasure" experiments, where we find coincidences between, say c>+d> events and an interference pattern on the a/b side. This correlation (between an interference pattern on the c/d side, and an interference pattern on the a/b side) is usually what is seen as proof of "entanglement" between the a/b and the c/d system. Now, if this a/b  c/d system interacts with a third system, we will end up with: a>c>e> + b>d>f>, and now, there will be NO correlation anymore between the c>+d> observable, and the a>+b> observable (in other words, triggering on the c>+d> state will not give rise anymore to the a>+b> interference pattern. This is simply because the e> and the f> are not an identical state anymore, which can factor out (as was the state of the e/f system before it interacted with the a/b and c/d system). In the MWI/decoherence view, we simply say the above: the abcd system got entangled with another system (the ef system) and as such, destroyed the interference (or coherence) between the ab and cd system. So in this view it is the EXTRA ENTANGLEMENT which destroyed the coherence between the ab and cd system (which served as a proof of their entanglement between them). This is the same mechanism that was responsible for the disappearence of the IP for the ab system alone. When the ab system got entangled with the cd system, the interference pattern of the a>+b> observable disappeared, only to be seen as a CORRELATION between the IP of the ab and the cd system. In the same way, the entanglement between the (abcd) system and the ef system made this interference disappear, and appear as a correlation between an IP of the ab,cd and ef systems. But as we're not going to do an interference measurement on the ef system, this is just saying that the evidence for entanglement between ab and cd has been destroyed (decohered). In the projection view, one can consider that the interaction with the e/f system considered a measurement in the {a>c> ; b>d>} basis, and so we now have, with 50% probability, the PRODUCT state a>c> OR the product state b>d> (and hence entanglement is GONE). Of course this view gets in troubles if we are going to test for an interference between the ef system and the ab/cd system. Now, to answer your questions: the two photons are given by the a>c>+b>d> state. If the cd photon interacts with the atom, we will have: a>e> + b>f>, where the e/f states are atom states (the cd photon has disappeared). So yes, the entanglement is now transmitted to the photon/atom system. And now, how are we going to emit a photon from these two different atom states, so that we get back our original pair of entangled photons ? The answer is simply: the FINAL STATE of the atom, in the two branches, must be identical. The e> state must go to a g>k> state, and the f> state must go to a g>l> state. So the atom must return to the same state g> (and the emitted photons to k> and l>). If not, the atom will STAY entangled with the two terms, and as such play the role of the decoherence inducing "environment". Because if the final state is the same (for the atom), we will now have: a>g>k> + b>g>l> g> factors out: g> (a>k> +b>l>) and we now have an entangled photon pair given by a/b and k/l, with the atom "outside" of the entanglement. It is clear that there is no "projection" version of this thing happening... cheers, patrick. 


#4
Feb1806, 05:40 PM

P: 37

Decoherence  How not to break entanglement?
Thanks Vanesch and Bruce2g. I understand the article shows that "waves of electrons" (plasmons) remain entangled with the photon, and can travel a distance and remain coherent, and then emit an entangled photon  and cease to be entangled themselves. I wonder if this is the same effect of a photon simply getting reflected in a silvered mirror and remaining coherent.
In which case, what happens to the recoil? The reflection on the silvered mirror generates a measurable recoil and backforce. Shouldn't this destroy the entanglement? In other words, consider this doubleslit experiment: Assume we have a beam splitter and a full mirror on one of the two paths after the BS, and we shoot a single (unentangled) photon at the BS and observe the interference pattern (one path is reflected in the mirror). Could we not measure the recoil on the mirror and obtain whichpath information and interference on the same photon? Does it have to do with the "amount of uncertainty" about the mirror? Like, if you are unsure about the miror's initial condition, then you will not be able to measure the recoil, and so you will not obtain whichpath information? Like if the mirror is itself in a superposed state? But if you were "certain" about the mirror's initial condition, then you would be able to measure the recoil, and the interference pattern would disappear? And the only way to be "certain" about the mirror's initial condition, is to be entangled with the mirror? And the only way to be entangled with the mirror is to isolate it? And once the mirror is isolated, the incoming photon will then entangle with it? And once the photon entangles with the mirror, then the interference pattern will be lost? Does this rambling make sense? What is the relationship between "certainty", "isolation", and "degrees of freedom"? 


#5
Feb1906, 02:00 AM

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The way to avoid that, is to make sure that, whether or not, the photon went on the path with the mirror or not, the final state of the mirror is the same (so that you cannot deduce from the state of the mirror whether the photon went by or not). In the decoherence view, this state of the mirror then factors out, in the collapse view, the mirror didn't measure the path of the photon (because its final classical state doesn't contain different information concerning the path of the photon). In practice, you do this by screwing the mirror tightly to the optical bench, so that all potential momentum transfer will be compensated for by whereever the photon will be hitting (and as thus, induce an opposite momentum). For an optical photon and a heavy mirror, the recoil will change the state of the mirror so little, that the quantum state after recoil cannot really be considered orthogonal (but rather almost identical). As such, it factors out "almost completely" and the small remnant will induce a very tiny decrease in the IP. This can be classically understood as some phase noise on the beam going over the mirror, if the mirror is moving a bit. 


#6
Feb2206, 01:03 AM

P: 87

You might want to look at Roger Penrose and friends, at http://www.qubit.org/research/Photon.../collapse.html, who are hoping to put a tiny (10^13 kg) mirror in a state of macroscopic superposition(!).
I haven't read the whole paper yet, but I think it touches on the issues you've brought up. Here's an interesting excerpt: "In the absence of decoherence, after a full period, the system is in the state 1 p2 (j0iAj1iB + ei22j1iAj0iB)ji, such that the mirror is again disentangled from the photon. Full interference can be observed if the photon is detected at this time. If the environment of the mirror "remembers" that the mirror has moved, then, even after a full period, the photon will still be entangled with the mirror's environment, and thus the interference for the photon will be reduced. Therefore the setup can be used to measure the decoherence of the mirror.t:" Let me know if this sounds relevant or not. I like Penrose a lot because I think that deep down he wants to demonstrate a macroscopic quantum phenomenon. And I would certainly love to see him do it! Bruce 


#7
Feb2306, 06:19 PM

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P: 1,082

First of all, an isolated atom cannot absorb a photon without some other change  simple absorbtion does not conserve energy and momentum. That pretty much ends any discussion of entanglement for that situation. So, what's the simplest reaction possible under your circumstances? Either the atom emits one or more photons from the atom's "acceleration", or, gets involved with the bulk matter through plasmons, probably phonons in some circumstances  as is the case with slow neutron scattering in crystals. The two situations are very different: entanglement will go away quickly when bulk matter modes are at issue  the physics goes something like this; the interactions involved here generally take the form of a Poisson process. So energy and angular momentum from an absorbed photon will be distributed within the ensemble in basically a random fashion  goodbye entanglement. The corrected 'isolated atom" situation can be thoroughly examined and analysed along the lines of CohenTannoudji et al in AtomPhoton Interactions. Basically, the isolated atom, with correct recoil dynamics, situation is a scattering problem. Roughly, the first 290 pages of this book provide what you need to know to do a thorough investigation of your entanglement problem. (I think the answer is pretty clear, the entanglement will be lost or greatly diminished  the entanglement signal to noise ratio grows with the opportunities to share with all what was once shared only with the other photon. The combinatorials say it all.) (It's possible, I suppose, that some crystals might allow backwards lasing. If there is appropriate matter induced internal coheherence in the reponse to a photon, it's possible that entaglement prospers. Above my pay grade.) Regards, Reilly Atkinson/ 


#8
Feb2306, 10:09 PM

P: 37

B2) Now if indeed photons are absorbed, would this be the same effect that would retain entanglement? That is if one photon of an entangled pair is absorbed by an air particle, then during the time that the photon is fully absorbed, would it be that this particle is entangled with the other photon? In photonic entanglement experiments using down converters, I do not believe the experimenters bother to create a vacuum, so I believe that air absorbtion reemission happens frequently. It seems to me that we are just rearranging the deck of cards. If our ensemble is a lattice of 4 atoms, then we would have diminished entanglement per atom, but the ensemble as a whole still retains FULL entanglement. So now increase the size of the ensemble. Is FULL entanglement retained when we integrate that over the ensemble  and if not, what is the effect responsible for erasing the entanglement correlation? In a classical information metaphor, it is one thing to lets say do a lossless compression on information and render it intelligible, but in principle retain the information intact. It is another thing to do a lossful operation, introducing noise, and delete data. I think you are saying that "in principle" entanglement correlation is lost because information is simply the energy and momentum, and that comingles with other such phenomena and hence loses its coherency. But I am not sure, and I don't know how to say it right, but it appears to me that entanglement correlation is a phenomena more than energy and momentum. 


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