# No-cloning and stimulated emission

Hi,
I'm trying to find out why stimulated emission (creation of an identical photon) doesn't violate the no-cloning theorem. There are lots of different opinions on this on the internet, e.g. "the second photon is not exactly identical, the energy is slightly different" or "since stimulated emission only works for a certain frequency, it doesn't copy general states". However, those explanations don't say anything about the polarization. If it were possible to copy a general polarization (even with a small error), one could build a superluminal communication device.

The most convincing explanation I've found is that spontaneous emission disturbs the process just enough such that copies are "imperfect enough" so they cannot be used for superluminal communication.
Can anyone confirm this?

Strilanc
I don't know anything about stimulated emission, but one possible reason is that the copy is entangled instead of independent. That is to say, the atom acts like a controlled-not gate with the first photon being the control and the second photon being the effect.

There is no gate that can map the state ##\left( \alpha \left| 0 \right\rangle + \beta \left| 1 \right\rangle \right) \left|0\right\rangle## to the state ##\left( \alpha \left| 0 \right\rangle + \beta \left| 1 \right\rangle \right)^{\otimes 2}## for all ##\alpha## and ##\beta##, but it's trivial to instead map to ##\alpha \left| 00 \right\rangle + \beta \left| 11 \right\rangle##.

bhobba
Mentor
The no cloning theorem states that it is not possible to build a machine that clones a previously unknown and arbitrary state. That's not what stimulated emission does - it does not clone an arbitrary but unknown state. Only a certain range of frequencies can be used to stimulate emission so the state being produced is not arbitrary. Another issue is spontaneous emission which means there is a probability of two (or more) photons being produced so its not an exact clone anyway.

http://arxiv.org/abs/quant-ph/0205149

Thanks
Bill

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Demystifier
Well you can encode an arbitrary two-dimensional state into the polarization of a photon, so the frequency isn't an issue.

So I guess it comes all down to spontaneous emission, as is also stated in the paper you mentioned. So does this mean that if I start a cascade by sending a single photon into an inverted laser medium, we cannot statistically infer its polarization by performing a tomography on all the produced photons? First of all, since stimulated emission creates a photon moving in the same direction as the first one and spontaneous emission is isotropic, so it should be easy to filter out most of the noise.

bhobba
Mentor
Well you can encode an arbitrary two-dimensional state into the polarization of a photon, so the frequency isn't an issue.

Photons don't have a two dimensional state.

E=hf. Change the frequency and you have a different energy and a different state.

I have zero idea what you are trying to say in the second bit. Spontaneous emission means that there is a possibility of getting more than one photon out if you send one in so obviously its not a clone.

Thanks
Bill

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So? As I said you can encode the full information of a qubit into only the polarization of a photon, regardless of its frequency. This has been done in QKD or quantum teleportation or Bell experiments and numerous other cases. You could easily choose a frequency that matches the laser medium. If you can copy the polarization of a single photon you can copy a qubit, which should not be possible according to no-cloning.

bhobba
Mentor
So? As I said you can encode the full information of a qubit into only the polarization of a photon,

That's irrelevant. Photons are not described by a qubit.

In fact its described by QFT where even the number of photons is not fixed - its a superposition of different numbers of photons. Send one in and because you can get more than one back it's not a clone of what went in.

Thanks
Bill

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I don't get it. Why should I care about all the other properties of a photon if I can use the polarization only to perform quantum computing.

And as I said, if i can create a second photon with identical polarization as the first one then superluminal communication is possible, regardless that this might not be a cloned state from the view of QFT.

bhobba
Mentor
I don't get it. Why should I care about all the other properties of a photon if I can use the polarization only to perform quantum computing.

Because that's not what the no cloning theorem says. It refers to an arbitrary state - not one parameter of that state ie its spin.

And no - you cant use stimulated emission to send information FTL via spin - the emission does not happen in a non-local way. Its described by QFT which explicitly incorporates SR, and right at its foundations is local. You cant violate the principles its based on.

Thanks
Bill

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Still: This would make superluminal communication possible.

Also, what is an arbitrary state in this context? We can hardly expect a device capable of clonig the state of a single atom to also be capable of copying the quantum state of the whole earth. So why should we expect a device capable of cloning the polarization of a photon at a specific frequency to do so on the whole frequency range?

bhobba
Mentor
Still: This would make superluminal communication possible.

What precisely do you not get that a theory cant violate the premises its based on? QFT is based on SR - you cant do it - end of story.

Thanks
Bill

Let's put QFT and the no-cloning theorem aside for the moment. Is it possible to shoot a polarized photon into an inverted laser medium and deduce its polarization from the produced photons or is it not. And if it's not, what exactly forbids it.

bhobba
Mentor
Let's put QFT and the no-cloning theorem aside for the moment.

You mean put aside the theory that describes photons to discuss photons. That makes no sense.

This is my last post on this - you have got your answer - and a link that gives more detail. But that is not what you now want to discuss - you want to discuss obvious fallacies such that it can be used to send information FTL. You cant.

Thanks
Bill

No, that's absolutely not what I want. Two facts:
1. The capability of copying the polarization of a photon would allow superluminal communication by using two entangled photons: Alice can choose to either measure her photon in the + or in the x basis and thereby make Bob's photon collapse into the same basis. If Bob is capable of cloning his photon, he can perform a tomography and deduce the basis Alice used. He can do so instantly after Alice's measurement, regardless the distance.
2. Stimulated emission creates a second photon of identical polarization and direction.
These two facts can be formulated without going any further into QFT or no-cloning (which is what I meant by putting them aside for a moment). All I want to know is what physical process forbids the second to be used for the protocol formulated in the first. Is it spontaneous emission? And if yes, why is it impossible to statistically infer the polarization of the first photon of a cascade even though there are more photons created in the same polarization than in any other?
Assume the polarization of the first photon to be an eigenstate of either the basis + or x. Let's measure half the produced photons in the basis + and the other one in the basis x. Since the spontaneously emitted photons produce only statistical noise, in one of the basis we should measure one eigenstate significantly more often than the other which would allow to deduce the state of the first photon.
All I want to know is the error in this train of thought.

I already asked this in #4 and #12, which you chose to ignore. I have zero idea why one should need to invoke QFT to answer those questions.

bhobba
Mentor
What precisely do you not get that a theory cant violate the premises its based on? QFT is based on SR - you cant do it - end of story.

The above is the principle that forbids it. The reason I introduce QFT is it immediately shows what you want to do cant be done. I suggest you think about why you want to ignore it when doing so means you do not want to examine your ideas using the theory that describes photons.

As to if you can shoot photons into a laser or something like that and from what it shoots out you can determine the spin of what goes in I have zero idea - you need to ask an experimental type. What I can say, and there is no room for doubt on this, if you could do it then the information could not be sent FTL. This is, as I have said a number of times, because the theory that describes photons, QED, the most accurately verified theory of all time, includes SR, where FTL signalling is explicitly forbidden.

How your Allice and Bob example uses stimulated emission has me beat. But any way you come up with will not allow FTL.

Thanks
Bill

I know that QFT forbids it. That's why there must be an error in my train of thought. And that's what I'm trying to find. Arguing with premises doesn't really help, I know them. I want to know what doesn't work with that specific implementation I suggested.

About the Alice and Bob example: If Bob can shoot his photon into a laser and determine the polarization, he gains information about which basis Alice used. If they previously agreed that the + basis stands for 1 and x for 0, they managed to transmit one bit of information without being bound to the speed of light. Since we know that this is not possible, some process needs to forbid that Bob can determine the polarization by using stimulated emission, and I'd like to know what.

bhobba
Mentor
If Bob can shoot his photon into a laser and determine the polarization, he gains information about which basis Alice used.

How? You are claiming shooting it in there means it shoots out a clone. So how does it tell us anything of Alice?

There is a way if you can clone a state you can send information FTL with your idea:
http://www.scientificamerican.com/article/mistakes-faster-than-light-telegraph-that-wasnt/

If that's the path you want to go down then give the details, and when I say details I mean exactly that, the same level of detail Nick Herbert gave, about doing that. The no cloning theorem forbids Herberts set-up - as it must yours as well. But the exact error depends on your exact set up.

Thanks
Bill

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How? You are claiming shooting it in there means it shoots out a clone. So how does it tell us anything of Alice?
Suppose they share the entangled state $$\left|\psi\right\rangle=\frac{1}{\sqrt{2}}\left(\left|\updownarrow\right\rangle\left|\updownarrow\right\rangle+\left|\leftrightarrow\right\rangle\left|\leftrightarrow\right\rangle\right)=\frac{1}{\sqrt{2}}\left(\left|\nearrow\right\rangle\left|\nearrow\right\rangle+\left|\searrow\right\rangle\left|\searrow\right\rangle\right)$$(sorry I don't know how to make two tips on the diagonal arrows). If Alice measures in the ##\left\{\left|\updownarrow\right\rangle,\left|\leftrightarrow\right\rangle\right\}## basis, her state will be projected onto ##\left|\updownarrow\right\rangle## or ##\left|\leftrightarrow\right\rangle##. The same happens to Bob's photon since they are entangled. If Alice measures in the ##\left\{\left|\nearrow\right\rangle,\left|\searrow\right\rangle\right\}## basis, her state will be projected onto ##\left|\nearrow\right\rangle## or ##\left|\searrow\right\rangle##, and the same for Bob's photon.
In a single, measurement, Bob cannot determine the basis. However if he is able to clone the photon, he can perform a tomography and determine if his state is in ##\left\{\left|\updownarrow\right\rangle,\left|\leftrightarrow\right\rangle\right\}## or ##\left\{\left|\nearrow\right\rangle,\left|\searrow\right\rangle\right\}## and thereby which basis Alice used.

Since this is impossible, there must be a mechanism that forbids Bob to shoot his photon into an inverted laser medium and recover its polarization the way I described in #14, and still don't know what that mechanism possibly could be.

bhobba
Mentor
However if he is able to clone the photon, he can perform a tomography and determine if his state is in ##\left\{\left|\updownarrow\right\rangle,\left|\leftrightarrow\right\rangle\right\}## or ##\left\{\left|\nearrow\right\rangle,\left|\searrow\right\rangle\right\}## and thereby which basis Alice used..

Please give the detail of how this tomography thing does something so marvellous and violating our known physical laws.

BTW its not cloning the photon - lets be clear - its amplifying a property it has - namely spin. And that is the whole problem - it simply amplifies a superposition - that was the undoing of Nick Herberts approach.

After thinking about it it looks virtually the same as Nick Herberts proposal with your 'Tomography' thing taking the place of the beam splitter determining plane or circular polarisation - that would correspond to your up and down polarisation (plane) and diagonal polarisation (circular). In fact I betting its the same thing.

'If Alice chose to measure circular polarization and happened to find L, then the entangled photon heading toward Bob would instantly go into the state R prior to entering the laser gain tube. Out of the laser would burst a stream of R photons heading toward Bob. He could then send half the beam toward a detector to measure plane polarization and half toward a detector to measure circular polarization. In this case, Herbert concluded, Bob would find half the photons in state R, none in state L, and a quarter each in states H and V. In an instant, Bob would know that Alice had chosen to measure circular polarization. Alice's choice—plane or circular polarization—would function like the dots and dashes of Morse code. She could signal Bob simply by alternating her choice of what type of polarization to measure. Bob could decode each bit of Alice’s code faster than light could have travelled between them.'

The error is as the article says - it simply amplifies the superposition.

Thanks
Bill

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Assume Alice decides to measure in the ##\left\{\left|\updownarrow\right\rangle,\left|\leftrightarrow\right\rangle\right\}## basis and gets ##\left|\updownarrow\right\rangle##. Bob's state gets projected onto the same state (but he doesn't know that yet). Now he makes many clones of that state (say, using stimulated emission). If he measures half of them in the basis ##\left\{\left|\updownarrow\right\rangle,\left|\leftrightarrow\right\rangle\right\}##, he will always find ##\left|\updownarrow\right\rangle##, whereas measuring the other half in the basis ##\left\{\left|\nearrow\right\rangle,\left|\searrow\right\rangle\right\}##, about half of the measurement will give ##\left|\nearrow\right\rangle## and the other half ##\left|\searrow\right\rangle##. So he can be pretty certain that Alice used the ##\left\{\left|\updownarrow\right\rangle,\left|\leftrightarrow\right\rangle\right\}## basis.

PeterDonis
Mentor
Stimulated emission creates a second photon of identical polarization and direction.

This is not correct. (I know it's what the Wikipedia article says, but Wikipedia is not a reliable source.) This claim says that we start with a single photon quantum state ##| 1 >_p## and we end up with a two-photon quantum state ##| 2 >_p##, where ##p## is shorthand for a particular set of photon quantum numbers (frequency, polarization, and direction) and the number indicates the number of photons (more precisely, it indicates that the state is an eigenstate of the photon number operator with the given number as its eigenvalue). But that's not what happens with stimulated emission; given an initial state ##| 1 >_p##, the final state is not just ##| 2 >_p## but a superposition of ##| 2 >_p## with other states that differ either in photon number or in at least one of the quantum numbers in ##p## (or possibly more than one of those). So the initial state has not been exactly cloned.

bhobba
bhobba
Mentor
Assume Alice decides to measure in the ##\left\{\left|\updownarrow\right\rangle,\left|\leftrightarrow\right\rangle\right\}## basis and gets ##\left|\updownarrow\right\rangle##. Bob's state gets projected onto the same state (but he doesn't know that yet). Now he makes many clones of that state (say, using stimulated emission). If he measures half of them in the basis ##\left\{\left|\updownarrow\right\rangle,\left|\leftrightarrow\right\rangle\right\}##, he will always find ##\left|\updownarrow\right\rangle##, whereas measuring the other half in the basis ##\left\{\left|\nearrow\right\rangle,\left|\searrow\right\rangle\right\}##, about half of the measurement will give ##\left|\nearrow\right\rangle## and the other half ##\left|\searrow\right\rangle##. So he can be pretty certain that Alice used the ##\left\{\left|\updownarrow\right\rangle,\left|\leftrightarrow\right\rangle\right\}## basis.

That's exactly the same as the link I gave. The error is as it details - ie it simply amplifies a superposition.

Thanks
Bill

Strilanc
The no cloning theorem does apply to single spins.

The no-cloning theorem does apply to single spins. A qubit is isomorphic to an electron's spin, and you can't clone qubits. The required operation is not unitary.

Do you have an example of a spin that can be cloned? (Not counting spins known ahead of time or restricted to orthogonal possibilities, of course.)

bhobba
Mentor
The no-cloning theorem does apply to single spins.

I clarified what I meant there - spin is simply one of the independent properties a photon has - it has spin and energy. To clone the state you clone both.

It has become apparent this is simply Nick Herberts proposal on FTL where a laser is used to amplify spin. It doesn't clone the photon for the reasons I and Peter Donas (with even more detail) gave - and you are correct - it doesn't strictly clone the spin either - but its close to it.

Thanks
Bill

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Strilanc
I clarified what I meant there - spin is simply one of the independent properties a photon has - it has spin an energy. To clone the state you clone both.

Thanks
Bill

Right, but cloning just the spin (or polarization), and not the other properties, is already a violation of the no-cloning theorem. Cloning the entire photon is a violation, but so is cloning certain subsets of it. That's what I was objecting to. (I probably should have included your "It refers to an arbitrary state - not one parameter of that state ie its spin." part of the quote in my reply.)

bhobba
Mentor
Right, but cloning just the spin (or polarization), and not the other properties, is already a violation of the no-cloning theorem. Cloning the entire photon is a violation, but so is cloning certain subsets of it. That's what I was objecting to. (I probably should have included your "It refers to an arbitrary state - not one parameter of that state ie its spin." part of the quote in my reply.)

Yes - you cant clone a subset either.

I also checked what I wrote just to refresh what I said and I couldn't locate your quote.

Thanks
Bill

Strilanc
Yes - you cant clone a subset either.

I also checked what I wrote just to refresh what I said and I couldn't locate your quote.

Thanks
Bill

Alright, sounds like we agree.

I pulled the quote from post #9 in this thread where you said:

Because that's not what the no cloning theorem says. It refers to an arbitrary state - not one parameter of that state ie its spin.

I think I may have simply misunderstood your "arbitrary" as meaning "entire".

bhobba
bhobba
Mentor
I have been investigating this a bit further.

I am not satisfied with the explanation in the article I linked to - I don't think it resolves it.

After searching I think I have finally got the correct answer:
http://www2.cruzio.com/~quanta/flash.html
'Shortly after publication of FLASH Wooters and Zurek published in Nature a paper entitled "A Single Quantum Cannot be Cloned" arguing that polarization-neutral laser amplifiers cannot be built. However Mandel a few weeks later published an explicit design for a polarization-neutral laser amplifier. Mandel in addition showed that such an amplifier would produce just enough noise that the polarization eigenstate of the input photon will be unmeasurable. For example when you put, say, an H photon into Mandel's device, two H photons come out--a single photon CAN be cloned--two times out of three. But one time out of three when an H photon goes in, an H and a V photon come out. This looks good--a signal-to-noise ratio of 2 to 1--but just this amount of noise exactly suffices to prevent you from discovering the difference between a random mixture of V, H photons (ZERO) and a random mixture of S, D photons (ONE). Thus Alice can send Boris a faster-than-light message but Boris cannot decode it.'

Interesting. I have only seen this FTL proposal with electrons and the no cloning theorem directly used to resolve it. Its not clear here with a laser how its resolved - it looked like a reasonably faithful amplifier. But really it isn't - and it not faithful hust enough to foul this up - interesting.

Thanks
Bill

Thank you. I guess that's the "mechanism" I was looking for preventing FTL communication.

Unfortunately it doesn't go into detail much. So one time out of three stimulated emission produces a V photon from an H photon? Does this mean the textbook treatment of stimulated emission is only true in two out of three times?

And what about spontaneous emission? In the paper in #3 they say "it is spontaneous emission that limits the achievable quality of the quantum cloning and ensures that the no-cloning theorem is not violated", but the other link doesn't mention it in that context.

bhobba
Mentor
So one time out of three stimulated emission produces a V photon from an H photon?

It mentions one time out of 3 a V and a H photon comes out. Thus one goes in - two comes out. This is QFT spontaneous emission which would seem the bottom line issue.

Thanks
Bill

Demystifier
Gold Member
So one time out of three stimulated emission produces a V photon from an H photon? Does this mean the textbook treatment of stimulated emission is only true in two out of three times?
Yes, that's probably the simplest way to understand why stimulated emission does not allow perfect cloning. In an attempt of cloning there is always some probability for an error, even if a very small one. But imperfect cloning which tollerates an error is, of course, possible.

Do we need QFT to prove this? I guess so, since photons are described by QFT.

Interestingly though, the proof of the general no-cloning theorem only uses Hilbert space properties and doesn't need special relativity.

bhobba
Mentor
Do we need QFT to prove this? I guess so, since photons are described by QFT.

I think its more that in QFT particle numbers are not fixed so its required to explain particle creation.

This is actually a deep requirement of combining relativity and QM. See:
https://www.amazon.com/dp/019969933X/?tag=pfamazon01-20

It has a chapter giving the gory mathematical detail - contour integration and all.

Thanks
Bill