Timeline for Why doesn't the no-cloning theorem make lasers impossible?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Jun 1, 2014 at 14:14 | comment | added | Piotr Migdal | While true, it is not related to how laser works. Laser typically works arbitrarily well for all polarizations. The reason is that at low intensities amplifications of single photons is not much stronger than amplification of quantum vacuum. | |
| Aug 20, 2013 at 22:40 | comment | added | yippy_yay | @SebastianHenckel I think I understand now: I could change the basis and go through the proof assuming that C clones every state, and would then get the result that some other state could not be cloned, proving that no such C can exist. Changing the basis doesn't mean that C can now act as a cloning operator on the state - because it's only an assumption that proves to be contradictory and therefore false. | |
| Aug 20, 2013 at 22:31 | vote | accept | yippy_yay | ||
| Aug 20, 2013 at 19:42 | comment | added | yippy_yay | But changing the basis doesn't change the machine. If I choose to represent the state vector in a different basis, that's a purely mathematical choice which shouldn't influence the workings of the copying machine. | |
| Aug 20, 2013 at 9:52 | comment | added | wataya | If you switch to a new basis c,d you can clone the new basis states but you'd lose the ability to clone the old states a,b. As I said: the no-cloning theorem is about arbitrary states. It says you can't build a machine (however complex it may be) that if you feed it with an arbitrary state always gives you two copies of that state as an output. You can build a machine that clones linearly polarized photons and you can build a machine that clones circularly polarized photons. But you can't build a machine that clones the state of any possible photon you feed it with. | |
| Aug 20, 2013 at 8:00 | comment | added | yippy_yay | Can't I just relabel $|\lambda a+\mu b\rangle$ as $|c\rangle$ (a vector in a new basis), and then avoid the no-go? The proof seems very mathematical and I can't see it physically, it leaves the feeling of being a trick-proof, like the one where you end up with all horses in the universe being white. I'm not saying this is the case, but I'm still doubtful. Also, in the same line, the no-cloning theorem should make a quantum computer impossible, as one will have to initialize the computer to a definite state. If I build 2, and initialize them to the same state, I've cloned the state? | |
| Aug 19, 2013 at 5:42 | history | edited | N. Virgo | CC BY-SA 3.0 |
corrected markup. (It's \rangle, not >.
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| Aug 19, 2013 at 5:34 | review | First posts | |||
| Aug 19, 2013 at 6:05 | |||||
| Aug 19, 2013 at 5:15 | history | answered | wataya | CC BY-SA 3.0 |