Suppose a beam of particles or a single particle, or a single particle. Is it possible to make a duplication of the beam or a single particle without collapse the states of the original particles?

  1. I think it might be possible as long as one does not measure the beam or particle.

  2. Suppose one want to perform the measurement, then the argument in question 1 definitely cause issues in definiteness. However, I recently thought that, if we perform the measurement in a state space, say discrete states space of 10, then we may void the violation of quantum mechanics if we only duplicate the original beam once. Because the measurement only determinate 2 of 10 states. Further, if we perform the degenerate procedure before the measurement, say make a two states system into a 4 sates system, then we may also "trick the QM" and extract an ambiguous amount of information without total destroy the entanglement.

Therefore, is it possible to duplicate the states of quantum and extract some information in secondary system without total screw up the original beams/particle?


There is a well-known result called the no-cloning theorem that forbids this. The proof is as follows:

Suppose you have an operator $\hat{A}$ which can clone an arbitrary state. It takes as its input a "blank" state which we'll call $|k\rangle$ and an arbitrary state $|\psi\rangle$. We'll also assume that these two states are part of a common Hilbert space (otherwise they couldn't be clones of each other in a meaningful sense). The system of these two states is represented as the tensor product $|k\rangle\otimes|\psi\rangle$. If $\hat{A}$ does what we say it does, then its action on that system is as follows:


In other words, the operator $\hat{A}$ copies an arbitrary state into a "blank" state without destroying the copied state (note that we have allowed $\hat{A}$ to alter the system's global phase, which has no impact on measurements). Since $\hat{A}$ represents the evolution of the system from one time to another, we require it to be unitary. Now, we choose two arbitrary states, $|a\rangle$ and $|b\rangle$ from the same Hilbert space, and we observe:

\begin{align} \langle a|b\rangle&=\langle a|b\rangle \langle k|k\rangle\\ &=(\langle k|\otimes \langle a|)(|k\rangle\otimes |b\rangle)\\ &=(\langle k|\otimes\langle a|)\hat{A}^\dagger \hat{A}(|k\rangle \otimes |b\rangle) \\ &=(\langle a|\otimes\langle a|)e^{-i\alpha(k,a)}e^{i\alpha(k,b)}(|b\rangle\otimes|b\rangle)\\ &=e^{i(\alpha(k,b)-\alpha(k,a))}\langle a|b\rangle ^2 \end{align}

From this we can see that $|\langle a|b\rangle|=|\langle a|b\rangle|^2$. But the only real numbers $x$ for which $x=x^2$ are $0$ and $1$. Therefore, either $|\langle a|b\rangle|=0$, which means $|a\rangle$ and $|b\rangle$ have to be orthogonal, or $|\langle a|b\rangle|=1$, which means $|a\rangle=e^{i\theta}|b\rangle$ for some $\theta$. But this makes no sense - $\hat{A}$ is supposed to be able to copy any state, not just a certain subset of states! (To be clear, it is always possible to choose two states in a nontrivial Hilbert space which are neither orthogonal nor indistinguishable by measurements, so there are indeed pairs of states which don't satisfy either of the two conditions above). This is a contradiction, and so the operator $\hat{A}$ cannot exist.

  • $\begingroup$ Thank you (For the single particle cases.) But for multiple particle cases:"the theorem does not hold if more than one copy of the initial state are provided: for example, broadcasting six copies starting from four copies of the original state is allowed, even if the states are drawn from a non-commuting set."(en.m.wikipedia.org/wiki/No-broadcast_theorem ) $\endgroup$ – user9976437 Jan 5 at 3:17
  • 1
    $\begingroup$ @user9976437 Such an action isn't really a duplication, as the system you started with (four copies of the original state) is not duplicated (because that would give you eight copies of the original state). If you want to duplicate, as you say, a "beam" of identical particles, then you will only be able to duplicate part of the original beam. You will lose a certain fraction each time. $\endgroup$ – probably_someone Jan 5 at 3:57

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