Sunday, 21 June 2020

quantum mechanics - 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?



Answer



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.


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