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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 statiscal 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 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.

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