In many worlds interpretation of the quantum mechanics all possible outcomes of a measurement are realized, however, in different universes. Everytime a measurement occurs we register one outcome and for the others our universe is copied and possible outcomes appears in these copies.

In quantum mechanics also holds no cloning theorem which tells that it is not possible to copy a quantum state. For example, on a quatum computer, it is not possible to construct a gate mapping state $|\psi\rangle \otimes |0\rangle$ to $|\psi\rangle \otimes |\psi\rangle$. It is only possible to use CNOT gate as a fan-out but in this case, the state $|\psi\rangle$ is in the end entangled with its "copy" (they are not independent).

Imagine, we do a measurement of a qubit for example, so only two possibe outcome can occur - $|0\rangle$ and $|1\rangle$. Assume that we measured $|0\rangle$ and rest of our universe remains unchanged. In many world interpretation, our universe has been copied with only one exception - the result of the qubit measurement is $|1\rangle$. These two universes are independent. So, how is this copying of universes compatible with no cloning theorem?

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    $\begingroup$ Are you suggesting that the two universes have been cloned? $\endgroup$ Commented Apr 16, 2021 at 18:20
  • $\begingroup$ @MoziburUllah: Maybe, I did not understand something properly, however, in many world interpretation, when a measurement (a collapse of a wave function) occurs, the universe is splitted to two or more. Assuming only one qubit collapsed, other things have to be cloned. Right? $\endgroup$ Commented Apr 16, 2021 at 22:00
  • $\begingroup$ @Vesely: Ok, that is what I thought you were asking. Excellant question. $\endgroup$ Commented Apr 17, 2021 at 4:57

1 Answer 1


In the many-worlds interpretation, the measurement process amounts to the mapping \begin{align*} &\lvert\phi_\mathrm{system}\rangle\otimes\lvert\mbox{observer hasn't measured yet}\rangle \otimes \lvert \psi_{\mathrm{rest}}\rangle \\ &\qquad\mapsto\ \ \ a\lvert\mathrm{system}=0\rangle\otimes\lvert\mbox{observer sees }0\rangle \otimes \lvert \psi_{\mathrm{rest}}\rangle \\&\qquad\quad + b\lvert\mathrm{system}=1\rangle\otimes\lvert\mbox{observer sees }1\rangle \otimes \lvert \psi_{\mathrm{rest}}\rangle\ , \end{align*} with $\lvert\phi_\mathrm{system}\rangle=a|0\rangle+b|1\rangle$. This is perfectly unitary and entirely comparible with the no-cloning theorem.

  • $\begingroup$ Perfect answer, it is clear now. Thanks. $\endgroup$ Commented Apr 17, 2021 at 6:30

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