After not understanding the explanation of the no-cloning theorem proof in my lecture notes I turned to Wikipedia, this explanation made more sense to me however it had an extra phase factor that is neither mentioned in my lecture notes, or the no-cloning proof in Michael A. Nielsen's "Quantum Computation and Quantum Information". Why is it safe to ignore this extra factor? Is it only present in unitary operations or have all the operations I have been working with secretly had this phase factor that has been conveniently ignored?
2
-
1$\begingroup$ Is perhaps this question better suited for the physics board? $\endgroup$– redpanda2236Commented Dec 14, 2022 at 23:58
-
1$\begingroup$ does quantumcomputing.stackexchange.com/q/2036/55 or quantumcomputing.stackexchange.com/q/17760/55 answer your question? $\endgroup$– glS ♦Commented Dec 15, 2022 at 11:28
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Proofs of the no-cloning theorem vary in quality. It's often something you meet early on in your studies of quantum information, so people want to keep it simple, but those simplifications can miss out on some aspects of a more complete full proof. The one on Wikipedia will do, but it ignores one class of possibilities in which your unitary could be acting over a larger Hilbert space, and it's just a part of it that has to exhibit the cloning transformation. Other proofs that you see will take into account this possibility. For example, you might see that the required transformation as being something like $$ U|\phi\rangle_A|e\rangle_{BC}=|\phi\rangle_A|\phi\rangle_B|s(\phi)\rangle_C $$ In such a case, there's no phase factor $e^{i\alpha(\phi)}$ out the front because you can easily incorporate it into the definition of $|s(\phi)\rangle$.
If the proof doesn't have the extra system $C$, then it probably should have the phase in it, although it will make no difference to the proof.
answered Dec 15, 2022 at 7:50