I was reading a proof of No-cloning theorem, there are a couple of steps that are not clear to me, but the book does not give explanation for them. So here it is: Theorem: It is impossible to create an identical copy of an arbitrary unknown quantum state. Proof (by contradiction): Suppose that there exists a unitary $C$ (that copies an arbitrary unknown q-state). Then: $C(|\psi\rangle\otimes|0\rangle)=|\psi\rangle\otimes\psi\rangle$

This is the first thing I have trouble with, why do we take only one state ($\left| 0 \right>$ state) to prove that it is impossible to copy any arbitrary state into $any$ qubit? The proof goes on stating: enter image description here

Here I don't understand how (116) is the same as (117) (don't mind explaining the identity matrix in the middle). I will appreciate it if you make these 2 steps clearer to me.


1 Answer 1


For step (116), the equivalence between both of them is proved by

\begin{equation} (\langle\psi_1|\otimes\langle0|)C^\dagger C(|\psi_2\rangle\otimes|0\rangle) = (\langle\psi_1|\otimes\langle0|)(|\psi_2\rangle\otimes|0\rangle) = \langle\psi_1|\psi_2\rangle\otimes\langle0|0\rangle=\langle\psi_1|\psi_2\rangle\langle0|0\rangle, \end{equation} where in the second step I used the property of tensor products that $(A\otimes B)(C\otimes D)=AC\otimes BD$, where obviously the dimensions of the multiplying matrices should match; and for the third step I used the fact that $\langle\psi_1|\psi_2\rangle$ and $\langle0|0\rangle$ are scalars, implying that their tensor product is just a multiplication between them.

For your first question, state $|0\rangle$ is used because it is the typical ancillary system used for quantum algorithms, that is, the zero state is the one used a auxilliary variable for doing the corresponding operations. However, note that the proof is not state dependent, meaning that you can take an arbitrary ancillary qubit, name it $|\rho\rangle$, and prove the theorem. Here I leave you such proof: \begin{equation} C(|\psi_1\rangle\otimes|\rho\rangle) = |\psi_1\rangle\otimes|\psi_1\rangle\\ C(|\psi_2\rangle\otimes|\rho\rangle) = |\psi_2\rangle\otimes|\psi_2\rangle \end{equation} And so $C$ would be the unitary we are looking for, so now the proof: \begin{equation} \langle\psi_1|\psi_2\rangle=\langle\psi_1|\psi_2\rangle\langle\rho|\rho\rangle =(\langle\psi_1|\otimes\langle\rho|)C^\dagger C(|\psi_2\rangle\otimes|\rho\rangle) = (\langle\psi_1|\otimes\langle\psi_1|)(|\psi_2\rangle\otimes|\psi_2\rangle)=(\langle\psi_1|\psi_2\rangle)^2. \end{equation} And so the same contradiction as before is obtained, and the theorem is proved.


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