Talk:No cloning theorem
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[edit] Clarification of implications?
This is generally a very clear article, but the principle raises some questions for newcomers. The second paragraph deals with quantum entanglement, but what about Bose–Einstein condensates and lasers? In the case of a condensate, I guess the same quantum state is reached because it is a known and definite state forced by cooling, rather then a superposition. But does a photon in a laser not stimulate emission of another identically-polarised photon, even if that polarisation is not known? --Cedderstk 08:25, 16 May 2006 (UTC)
- I think that would be an example of entanglement. E.g., if |0> and |1> are the two possible polarization states of the inital photon, then the final state can be |0>|0> or |1>|1>. So, if we start with a superposition of |0> and |1> (with coefficients a and b), we have:
- a|0> + b|1> --> a|0>|0> + b|1>|1>
- Whereas, for cloning we would need:
- a|0> + b|1> --> (a|0> + b|1>)x(a|0> + b|1>) = aa|0>|0> + ab|0>|1> + ba|1>|0> + bb|0>|0>
- In other words, cloning refers to the creation of two separable states, and that's what the theorem forbids. -- Tim314 00:37, 19 February 2007 (UTC)
