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I have a little trouble with the simulated emission. I know of the no-cloning theorem which states that it is not possible to duplicate any state.

One the other hand, I know about the stimulated emission which out of a photon produce exactly the same (wavelength, polarisation, etc...). Maybe the fact is that the excited atom do a measure. (Let's say it can stimulate emission with only one direction of polarisation.)

But now if I have a statistical number of atoms with random "polarisation" direction, I should be able to copy any incident photon. I've made a cloning machine.

This cannot be true because of the no-cloning theorem. But I can't figure why.

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Sure you can clone a state. If you know how to produce it, you can just produce one more copy.

The answer to your question therefore lies in the specifics of the no-cloning theorem. It states that it is not possible to build a machine that clones an arbitrary (previously unknown!) state faithfully.

Stimulated emission does not fulfill this. Given an atom, only a certain range of frequencies, etc. can actually be used to produce stimulated emission, so you can't faithfully clone an arbitrary state. It's just an approximation to cloning, which is not prohibited.

See also: http://arxiv.org/abs/quant-ph/0205149 and references therein.

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In stimulated emission the field starts in a state containing $n$ photons, and ends up in a state containing $n+1$ photons. The system has made a transition from one state to another. It looks to me like nothing has been cloned. Same system, different states.

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I think @garyp has given the correct answer, but it is worth expanding on it a bit.

Stimulated emission is not cloning a state, but rather changes the state of a mode (characterized by wavelength, direction, polarization, etc.): $$ |n_{\mathbf{k},\lambda}\rangle \longrightarrow |n_{\mathbf{k},\lambda}+1\rangle . $$

The conceptual misunderstanding here is due to the possible lack of background in quantum field theory, and therefore thinking of photons as if they were particles. This leads to thinking that a new photon is another particle in a state identical to the states of the previous ones. Once it is understood that photon is not a particle (not in the first quantization sense at least), but a level of excitation of a single mode, the contradiction disappears.

The most direct "particle" parallel is an electron in a harmonic potential, changing its state from $n$ to $n+1$ - the two states are not identical, even though the potential frequency is the same for them.

Update
Perhaps an explanation is needed why no-cloning theorem does not appear in my answer. Word cloning in its most generic sense means producing two identical copies of something (cloning is just a buzz word for copying, duplicating). Thus, my answer could be summarized as stimulated emission is not cloning (in any sense of this word). This is a more general statement than saying that stimulated emission is not equivalent to a specific type of cloning in the domain of physics, e.g., such as implied by the no-cloning theorem.

Implementing no-cloning theorem would require that we have two identical systems which then could be excited into the same state. I.e., we could have two identical lasers (laser A and laser B) initially in different states and then synchronize them: $$ |n_{\mathbf{k},\lambda}\rangle_A \otimes |m_{\mathbf{q},\mu}\rangle_B \longrightarrow |n_{\mathbf{k},\lambda}\rangle_A \otimes |n_{\mathbf{k},\lambda}\rangle_B. $$ We could also define cloning af two modes (but not the whole systems) being in identical state: $$ |n_{\mathbf{k},\lambda}\rangle_A \otimes |m_{\mathbf{q},\mu}\rangle_B \longrightarrow |n_{\mathbf{k},\lambda}\rangle_A \otimes |m_{\mathbf{q},\mu},n_{\mathbf{k},\lambda}\rangle_B. $$ We could even concievably talking about cloning some of the photons to the corresponding mode of the other laser $$ |n_{\mathbf{k},\lambda}\rangle_A \otimes |m_{\mathbf{q},\mu}\rangle_B \longrightarrow |n_{\mathbf{k},\lambda}\rangle_A \otimes |m_{\mathbf{q},\mu},l_{\mathbf{k},\lambda}\rangle_B. $$

In other words, there are many things that one could conceivably call $cloning*, but the term is hardly applicable to the "identical" photons, since they are not really different entities that could be identical, but rather different states of the same system.

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    $\begingroup$ Very good points well expressed. Thanks for the "expansion". My vote is that the bounty goes to you! $\endgroup$ – garyp Jan 14 at 14:17
  • $\begingroup$ OK, so you've explained that excitation of an already excited mode is not cloning. But what is cloning then? $\endgroup$ – Ruslan Jan 14 at 23:25
  • $\begingroup$ This doesn't actually prove why this transformation cannot be used to clone a state. I think to answer this clearly, you have to explictly answer something like the following: If I have a state $a|1_H\rangle + b|1_V\rangle$ and it becomes $a |2_H\rangle + b|2_V\rangle$. Why CAN'T a process subsequently send one photon in one direction and the second in another direction and you have two copies of the superposition in polarization. Basically, you need to answer why having a second identical photon is not sufficient to be able to clone the information. $\endgroup$ – Steven Sagona Jan 15 at 4:35
  • $\begingroup$ Here you're basically saying "having two of something isn't the same as having one of something" therefore it's not a cloned state because the state went from one to two. $\endgroup$ – Steven Sagona Jan 15 at 4:39
  • $\begingroup$ @StevenSagona Precisely! To speak of cloning one needs to have "two of somethong", and this is not the case in stimulated emission. $\endgroup$ – Vadim Jan 15 at 8:48
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I think I just figured this out. For cloning you want to do the following:

$(A|p\rangle+B|q\rangle)|0\rangle$ goes to $(A|p\rangle+B|q\rangle)(A|p\rangle+B|q\rangle)$ where $|p\rangle$ and $|q\rangle$ are orthogonal.

For stimulated emission one gets:

$(A|p\rangle+B|q\rangle)|0\rangle = A|p\rangle|0\rangle+B|q\rangle|0\rangle$ goes to $A|p\rangle|p\rangle+B|q\rangle|q\rangle$ which is different from the cloning case.

Hence, it does not violate the no-cloning theorem.

It also seems to me (judging from Mandel and Wolf) that $|p\rangle$ and $|q\rangle$ must be in the momentum basis for stimulated emission. One cannot use an arbitrary basis.


Update:

Not withstanding the recognition that @StevenSagona has given my old answer, for which I am grateful, I am not convinced that it is the best way to understand the issue. It has been several years since I provided this answer and in the meantime I have developed a different understanding. So, let me present that understanding.

The question is how stimulated emission works. One often finds the idea that the medium must have some intelligence that can determine the incoming state and then reproduces it. Assuming the input state is some Fock state where all photons have the angular spectrum $F$, then such a process would be represented by: $$ |n_F\rangle \rightarrow |(n+1)_F\rangle . $$ It assumes that the stimulated process can be modeled by an individual creation operator $\hat{a}_F$, which is not a unitary operator. Instead, the process should be represented by a unitary operator. Is it possible to find such a unitary operator $\hat{U}$ for this process? If such a unitary operator does exist, it would imply $$ \langle n_G|n_F\rangle = \langle G,F\rangle^n = \langle n_G|\hat{U}^{\dagger}\hat{U}|n_F\rangle = \langle (n+1)_G|(n+1)_F\rangle = \langle G,F\rangle^{(n+1)} . $$ It means that, either $|\langle G,F\rangle|=1$ or $|\langle G,F\rangle|=0$, which is inconsistent with the requirement that $F$ and $G$ be arbitrary angular spectra. Here, we have reproduced the no-cloning theorem, but for a slightly different scenario. So, we see that the first expression does not give a suitable representation of the stimulated emission process.

Then how does stimulated emission work? The thing is that tendency for the emission to have the same state as the incoming state is a result of a bosonic enhancement. It is not that the medium only reproduces the incoming states. It produces everything that the structure allows, but the part the corresponds to the incoming state receives a bosonic enhancement.

To model this process, let's assume that a spontaneous emission would produce a single photon state: $$ |\psi\rangle = |1_a\rangle C_a + |1_b\rangle C_b + |1_c\rangle C_c + ... , $$ where the subscripts represent the different modes$^{\star}$ that can be radiated and the $C$'s are arbitrary coefficients. To compare it with the incoming state $|n_F\rangle$, we can recast the expression as $$ |\psi\rangle = |1_F\rangle \alpha + |1_{\bar{F}}\rangle \beta , $$ where $\bar{F}$ is the orthogonal complement of $F$, and $\alpha$ and $\beta$ are the associated coefficients so that $|\alpha|^2+|\beta|^2=1$. In the stimulated process one would nominally get $$ |\psi\rangle|n_F\rangle = |1_F\rangle|n_F\rangle \alpha + |1_{\bar{F}}\rangle|n_F\rangle \beta . $$ However, this state is not properly normalized. If we look at the two terms, we see that $|1_{\bar{F}}\rangle|n_F\rangle$ is normalized, $$ \langle 1_{\bar{F}}|1_{\bar{F}}\rangle \langle n_F|n_F\rangle = 1 $$ but $|1_F\rangle|n_F\rangle$ is not a normalized state. Instead it becomes $$ |1_F\rangle|n_F\rangle = |(n+1)_F\rangle\sqrt{n+1} . $$ The factor $\sqrt{n+1}$ represents the bosonic enhancement. It causes the latter term to dominate over the term with the orthogonal complement. So the stimulated emission process is better represented as $$ |n_F\rangle \rightarrow |(n+1)_F\rangle\sqrt{n+1} + |1_{\bar{F}}\rangle|n_F\rangle , $$ where I assumed $\alpha=\beta=1$ and ignored the overall normalization. Obviously this is not a simple cloning of the incoming state.

$\star$ Here the term mode is used in its true sense as a solution of the equations of motion satisfying the boundary conditions.

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  • $\begingroup$ What is the interpretation of the states |p>, |q>, and |0>, and of the two systems in the tensor product, in the context of the question? $\endgroup$ – Norbert Schuch Jan 14 at 14:21
  • $\begingroup$ Thanks for the updated answer. In this new case you've described. I think there's still a high likelihood that a collective qubit encoded in the state $|n_F\rangle$ can cause the qubit mode to be selected in the spontaneous photon. So you can go from $(a | n\rangle_H+b | n\rangle_V)$ to (almost) $(a | n+1\rangle_H+b | n+1\rangle_V)$. If you could create a state that could turn this into $(a | n\rangle_H+b | n\rangle_V) \otimes (a | 1\rangle_H+b | 1\rangle_V)$ then you've copied your qubit information to another state. $\endgroup$ – Steven Sagona Jan 18 at 9:51
  • $\begingroup$ So the answer is either that this small error is what explains why it doesn't violate no cloning theorm, or that it is impossible to go from $(a | n+1\rangle_H+b | n+1\rangle_V)$ to $(a | n\rangle_H+b | n\rangle_V) \otimes (a | 1\rangle_H+b | 1\rangle_V)$ $\endgroup$ – Steven Sagona Jan 18 at 9:54
  • $\begingroup$ @StevenSagona. In my answer I assumed that the input state is a Fock state. In the case of a superposition of Fock states, as in your example, one needs to perform the process on each term individually. So the result would not be a simple tensor product as you indicated. $\endgroup$ – flippiefanus Jan 19 at 4:13

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