Why is it that copying an unknown qubit is impossible but creating a copy of the standard computational basis is possible by entangling it to existing qubits? And why one is possible and the other isn't?

  • $\begingroup$ Doing a general unitary that copies any unknown state is not possible due to the no-cloning theorem. However, doing unitaries that do copy specific qubits are possible to construct. I do not understand the part of your question where you say that computational basis is possible to copy by "entangling it to existing qubits". A clarification on that would be helpful for answering the question. $\endgroup$ – Josu Etxezarreta Martinez Mar 7 at 8:47

The no cloning theorem can be stated in the form:

If you are given an unknown state $|\psi\rangle$, which is promised to be one of a set of distinct states $\{|\phi_i\rangle\}$, it is impossible to create a second copy of $|\psi\rangle$ if there is a pair of $|\phi_i\rangle$ which are not orthogonal.

Computational basis states are orthogonal, so they're not covered by the theorem. Indeed, you want them to not be covered by the theorem because we can easily copy classical data.

There's a very obvious way that you can copy computational basis states. Start with your unknown qubit $|\psi\rangle$, introduce an extra qubit in the $|0\rangle$ state, and perform a controlled-not controlled off the first qubit. What do you get? $$ |0\rangle|0\rangle\mapsto|0\rangle|0\rangle\qquad|1\rangle|0\rangle\mapsto|1\rangle|1\rangle $$ The state of the first qubit has been cloned. There's no entanglement in this process itself. Of course, controlled-not is an entangling gate, and this is the essence of the proof of no cloning (the actual proof requires taking care of a few more details): if we wish to clone a state in superposition as well as the possibility of these two basis states, then we know by linearity that if the above two maps hold, it must be that $$ (\alpha|0\rangle+\beta|1\rangle)|0\rangle\mapsto\alpha|00\rangle+\beta|11\rangle. $$ This is an entangled state, and not the separable state that you want, $(\alpha|0\rangle+\beta|1\rangle)(\alpha|0\rangle+\beta|1\rangle)$.


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