That depends on the details of the interaction with the medium, that will lead to the emission of the photon (photons don't interact directly).

But it is possible to write mathematical models that are simpler, therefore can be studied quantitatively.

Dear Neil,

Great question. The short answer to your question is an unambiguous "YES".

The longer answer is genuinely interesting and essentially your question led to one of the most profound insights into the fundamentals of quantum physics, namely the No Cloning Theorem.

Back in 1982, the journal Foundations of Physics published a rather odd little article entitled "FLASH - A superluminal communicator based upon a new kind of quantum measurement" [N. Herbert, Found. Phys. 12, 1171 (1982)]. The paper proposed a scheme where there was a source of entangled photons in the control of Alice, who chooses to measure one half of the Bell pair in either the rectilinear or circular basis. The other photon is sent to Bob.

So far this is almost canonical E91 type quantum crypto (and as you can tell, older by 9 years). Where it differs is by the clear understanding that a single photon polarisation cannot be measured to arbitrary accuracy with no a priori information. (This is, of course, the source of the shared randomness in E91). So what Herbert proposed was to place a gain medium on the Bob side to amplify the single photon. Stimulated emission should give exact copies of the input photon, and so with an ensemble of identical photons, it is trivial to determine the polarisation of the incoming photon, and hence to determine the basis in which Alice's measurement encoded the photon. Because the Bell-type correlations are thought to be instantaneous (and the Gisin group has already put lower bounds on the speed of Bell correlations and they are orders of magnitude faster than c), then hey presto! Simple, superluminal communication!

Except it doesn't work, and it can't work. This is rather beautifully discussed in Asher Peres' paper "How the no-cloning theorem got its name" [Fortschritte der Physik 51, 458 (2003)]. As Peres points out (and indeed, he was one of the reviewers of the paper), the idea must be wrong because it violates special relativity: but the mechanism that protects relativity (and causality, and all of those things that we really want physics to have) is not obvious in the scheme.

The key to unlocking the secret of why the signally cannot work is in measurement. I think of the protocol this way: a photon comes into a gain medium, it must be measured to create two photons with the same properties. This measurement, although it is ultimately not destructive (and indeed is amplifying) still is a measurement of the photons' quantum numbers, including its polarisation. Hence the entanglement must be broken. This is at the heart of the no-cloning theorem, which states that an unknown quantum state cannot be cloned (ie copied).

So at it's simplest level, the entanglement is destroyed.

However, there are some caveats here. Firstly, there is remarkable work on quantum state amplification, where there are probabilistic amplifiers. Nothing that lets you get around superluminal propagation, but results that can be very useful for sensing (for example).

There is quantum erasure, where I think you might find your intuition of the third photon taking you. This is more commonly considered with the entanglement of a photon with a massive particle, and where appropriate measurement of the massive particle can be used to determine whether or not the entanglement is broken by the interaction with that massive particle.

And this is not unlike one of my favourite quantum protocols, again by Peres: Delayed Choice Entanglement Swapping [Peres, J. Mod. Opt. 47, 139 (2000)], and indeed I'm sure you could find a nice analogue of such work in a simpler three photon protocol. (perhaps already done - the field is large and memories finite ;) ).

Hope this helps

Andy Greentree

The answer depends on how you understand the physical meaning of the term "entanglement". This sense is not completely clear until now (see viXra:1712.0562).

Dear Vladimir,

thanks for that reference, i'll check out. - Certainly we must be clear about definitions.

Dear Andy,

thanks for your excellent answer - i kind of suspected the answer was yes. However, this brings another question. Say the entanglement between photon 1 and photon 2 is broken, because photon 1 stimulates a 3rd photon. Is then the original entangled information between photon 1 and photon 2, now distributed between all 3 photons?

So in a possible Bell test of this scenario, photon 1 and 3 would enter one half of an analyzer and photon 2 would enter the other half? So in spite of the fact stimulated emission took place, by collecting all three photons and processing, you still retain information about the entanglement?

many thanks,

Neil

To start with, great question. IMHO, the best one I ever seen here.

The answer, again IMHO, should be "no". Entanglement is about coherence, and coherence is preserved by stimulated emission. In classical ED, stimulated emission is simply the linear susceptibility. The same holds in macroscopic QED. Another point, stimulated emission is only possible in active media - unlike, e.g., refraction in passive media - and is therefore always accompanied by spontaneous emission. The latter is a killer of entanglement.

The bottom line is, stimulated emission preserves entanglement, but exists only under conditions where there are other processes reducing entanglement.

Dear All,

As i understood it classically stimulated emission photon was 'in-phase' with the stimulating emission, but the quantum picture of this is a bit more subtle, as phase of a single photon isn't defined, and if you could clone a photon, then quantum comms would not be secure.

Still perhaps in the presence of stimulated emission the entanglements just spread to three photons, or more, in the case of further stimulated emission processes?

In terms of the media, if the temperature of the medium is high compared to the frequency (kT>hf) then you're in the Rayleigh-Jean regime and the photon energy levels are well populated by thermal excitation, but still with empty states for transitions. For ambient temperature, this means frequencies below those of around infra red (lambda 10 microns). This means the emission from these media is 50% spontaneous and 50% stimulated.

I'm considering what might happen in a Bell test operating much lower in frequency, in the millimetre wave band, but with ambient temperature equipment. How and what effects might this stimulated and spontaneous emission might have in the Bell test?

many thanks for any suggestions.

Neil

I don't see why stimulated emission should necessarily destroy entanglement - it can 'clone' the incident photon state. However quantum optics tells us that stimulated emission by a single photon occurs at the same rate as spontaneous emission into the same mode, so the gain medium will produce indistinguishable spontaneously emitted photons into both the state of the incident photon and any orthogonal states, which will prevent any useful measurement of the cloned photon.

Well, some say fundamentally a photon cant be cloned, so i'm not sure. However, it does raise the question: if entanglement is destroyed by simulated emission, where does the information go to? Cant you just 'collect the pieces' (ie the photons) and extract the information from them somehow.

As for the 50% spontaneous / 50 % stimulated emission, perhaps a particular detection system might discriminate the spontaneous emission (as it is random in time) from the stimulated emission (as it has some coherence with the stimulating photon). I'm still trying to figure this...

Hi Neil,

It's not possible to tell them apart. Photons are just quantized excitations of a field. The rate at which an atom in its excited state in a field with n photons transfers its energy to the field scales as (n+1), which comes directly from the field creation operator equation

a+ |n> = sqrt(n+1) |n+1>

The terms 'stimulated emission' and 'spontaneous emission' are coined to refer to the effect of the 'n' photons as opposed to the '1', but you can't separate them as they're both part of the same process.

This document from Ben Simons lays out the theory in full

http://www.tcm.phy.cam.ac.uk/~bds10/aqp/handout_transitions.pdf

Hope this helps!

Jason

Hi Jason,

thank you very much for your good advice and the document from Ben Simons. Certainly Ben's work, together with that on his website, is an excellent overview of the subject. This will certainly be helpful in making some quantitative assessment of transition rates.

best wishes,

Neil

I suspect that stimulated emission in a degenerate four-wave mixing process would conserve entanglement. For instance, in a mirrorless parametric self-oscillator, as described by Yaris and Pepper in the 70’s. There, the first signal and idler photons produced from vacuum fluctuations are entangled and the successive buildup (equivalent to laser action) from distributed feedback would not alter this. One would have a large occupation number of entangled photons.

Thank you Thomas. There must surely be a mathematical proof of entanglement remaining after stimulated emission, but I'm not sure how this goes. Anyone know how this might look? many thanks N