# Why doesn't the no-cloning theorem make lasers impossible?

As I understand lasers, you start off with a few photons that are in an identical state, and other photons that are created later tend to have the same quantum numbers due to Einstein-Bose statistics. Isn't each photon that "joins" the group of preexisting ones a clone of the previous ones? Why doesn't this violate the no-cloning theorem?

• Without getting into the specific way that lasing works, this is possible because the no-cloning theorem only forbids one from copying an arbitrary (or unknown) state. So it wouldn't be possible to build a laser-like device where you inject any photon and it streams out a bunch of them. But a laser is always creating a particular (roughly coherent) state, so that's no problem. – Rococo Aug 18 '13 at 20:23
• @Rococo Injecting a photon of which I know the state could lead to replication, but injecting one whose state I don't know couldn't? Why would the laser work/not work depending on my knowledge of the photon's state? – yippy_yay Aug 18 '13 at 22:06
• Sorry, I should be more careful with my wording. It is not the content of your mind that matters, of course. Rather, what I mean is that you can engineer the system to replicate a particular set of orthogonal states, but not a general state. Wataya gives the proof of this. So, for example, it's possible to build a machine that spits out two vertically polarized photons when one goes in, and does the same thing for horizontally polarized photons, but if you feed it a circularly polarized photon it will not be able to copy that (I'd encourage you to try thinking about what will happen instead). – Rococo Aug 19 '13 at 23:57

As Rococo already pointed out, the no-cloning theorem doesn't forbid cloning of all specific states. It just states that you cannot make copies of arbitrary (general) states.

Let me (briefly) reiterate the core of the theorem: To clone a state you need a linear operator C that maps a state $|a\rangle|0\rangle$ to $|a\rangle|a\rangle$. This is not possible for general states: $$C |\lambda a+ \mu b\rangle|0\rangle$$ would have to map to $$|\lambda a+\mu b\rangle|\lambda a+ \mu b\rangle$$ per definition of the operator. But linearity (and homogenity) leads to $$C |\lambda a+\mu b\rangle|0\rangle = \lambda C|a\rangle|0\rangle + \mu C|b\rangle|0\rangle = \lambda|a\rangle|a\rangle + \mu |b\rangle|b\rangle$$ while $$|\lambda a+ \mu b\rangle|\lambda a+ \mu b\rangle = \lambda^2|a\rangle|a\rangle +\lambda \mu (|a\rangle|b\rangle+|b\rangle|a\rangle)+\mu^2|b\rangle|b\rangle$$

So you see that for e.g. $\lambda=1, \mu=0$ (i.e. a base state) there is no contradiction. But you can't clone a general superposition of your base states.

• 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? – yippy_yay Aug 20 '13 at 8:00
• 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. – wataya Aug 20 '13 at 9:52
• 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. – yippy_yay Aug 20 '13 at 19:42
• @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. – yippy_yay Aug 20 '13 at 22:40
• 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. – Piotr Migdal Jun 1 '14 at 14:14

A short history of the laser:

1917: Einstein describes theory of stimulated emission

Following decades: Lasers (and masers) are thought to be impossible due to quantum theory (Everyone knows Einstein was wrong about quantum theory :)

1951 Charles Townes builds a maser - some people are just awkward (John von Neumann: 'That can't be right', Niels Bohr: 'But that is not possible' - see How the Laser Happened: Adventures of a Scientist)

1982 Nick Herbert invents FLASH - A superluminal communicator based upon a new kind of quantum measurement (see How the hippies saved physics : science, counterculture, and the quantum revival )

Later in 1982: No-cloning theorem is devised, explaining why FLASH won't work.

So how was FLASH supposed to work? Well two entangled photons were produced with the "same" polarization (Actually, I think, they will be opposite polarizations, but that can be dealt with, and I'll ignore it from now on) One of the photons is measured at one place, thus collapsing the wave function. At another place the other photon is fed into a laser, which produces lots of photons with the "same" polarization. Measuring several of these allows you to work out how the wave function has collapsed, and thus what measurement was made at the other location, allowing superluminal information transfer.

The trouble is that, in reality the word "same" has two different meanings. In the entanglement case it means that the result will be the same whatever polarization measurement you do. In the case of the laser it doesn't - as wataya says, it might clone the linear polarization, but not the circular polarization.

I think there is a problem here, but I'm not sure what it is. It may be our ideas of quantum states - saying two states are both represented by |H> doesn't actually tell you how much the same they are. Or it could be our ideas of lasers - that the common idea of one photon stimulation the emission of another is inaccurate, and that the working of lasers has to be thought of as a collective way.