# What does it mean that copying a state is impossible but creating a copy of by entangling is possible?

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?

• 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. – Josu Etxezarreta Martinez Mar 7 '19 at 8:47

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.
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, $$(\alpha|0\rangle+\beta|1\rangle)(\alpha|0\rangle+\beta|1\rangle)$$.