# A proof of no-cloning theorem in the case of pures states in a qubit system

I am working on this problem

Consider a qubit $$\scr H =\Bbb C^2$$ and pure states. Prove the no-clonning theorem ( hint: Use the linearity of the channel to arrive to a contradiction)

I wonder if the following proof of the non-cloning theorem for pure states I came up with is valid:

My try:

I have to prove that: There is no quantum channel that clones all pure states.

I am using the definitions written below.

A state $$\rho$$ is pure if there exist a unit vector $$|\psi\rangle$$ such that $$\rho=|\psi\rangle\langle \psi|$$. I can complete this vector to an orthonormal basis $$B=\{|\psi\rangle,|\phi\rangle|\}$$. So that

$$\rho= \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$$ with respect to this basis

and if by the sake of contradiction I assume that it can be cloned, there exists a quantum channel $$\Phi$$ such that $$\Phi(\rho)=\rho \otimes \rho =\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 &0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0\end{pmatrix}$$.

Now I am supposed to use the lineary of the channel to arrive at a contradiction (that was a hint of the problem)but I think I can more easily observe that the trace is 2 and not 1 as it should, so it is not trace-preserving. A contradiction.

Is this approach correct? I am specially doubtful about the Hadamard product being correct despite the matrix is not using the standard basis and if the Hadamard product should have been be normalized at some point.

Definition of quantum channel: A quantum chanel is a superoperator $$\Phi$$ (a map in $$L(L(\scr {H_A}),L(\scr {H_B})$$) that is trace preserving and completely positive( For all $$\scr H_R$$ and $$M_{AR}\ge 0$$, it holds that $$(\Phi\otimes I_R)[M_{AR}]\ge 0$$ )

Definition of cloning:

A quantum channel $$\Phi \in C(\scr H,\scr H\otimes \scr H)$$ (C is just the set of all quantum channels between the specified spaces)clones a state $$\rho \in D(\scr H)$$ if $$\Phi[\rho]=\rho\otimes \rho$$

• This must be a Kronecker, and not a Hadamard product. The Kronecker product $\rho\otimes\rho$ is surely $$\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 &0 & 0 & 0 \\ 0 & 0 & \color{red}{0} & 0 \\ 0 & 0 & 0 & 0\end{pmatrix}\,.$$ Commented Mar 8 at 12:20
• oops you are right I messed up the name and that entry. I guess this invalidates my proof, how would I fix it then? Commented Mar 8 at 12:29
• I don't know. No-cloning theorems are nothing new. Perhaps you look up a standard reference. Commented Mar 8 at 12:31
• @KurtG. Is the Kronecker product calculated the same if I use any orthogonal basis instead of the standard one? That is multiplying each entry of the first matrix by the whole second matrix and putting the results as blocks. Commented Mar 8 at 12:35
• That's how a Kronecker product is defined. Look it up in Wikipedia. It has nothing to do with a basis. Commented Mar 8 at 17:18

There are a number of ways to arrive at a contradiction using linearity, but a quite explicit one would be as follows:

Assume there existed a channel $$\Phi$$ which satisfies $$\tag{1} \Phi(|\psi\rangle\langle\psi|)=|\psi\rangle\langle\psi|\otimes|\psi\rangle\langle\psi|$$ for all unit vectors $$\psi$$. Starting from any two orthogonal unit vectors $$\psi_1,\psi_2$$ show that

1. $$\langle \psi_1\otimes \psi_1\,|\,\Phi(|\psi_1\rangle\langle \psi_1|)\,|\,\psi_1\otimes \psi_1\rangle=1$$ and $$\langle \psi_1\otimes \psi_1\,|\,\Phi(|\psi_2\rangle\langle \psi_2|)\,|\,\psi_1\otimes \psi_1\rangle=0$$ by applying (1).
2. both $$\langle \psi_1\otimes \psi_1\,|\, \Phi(|\tfrac{\psi_1+\psi_2}{\sqrt2}\rangle\langle \tfrac{\psi_1+\psi_2}{\sqrt2}|) \,|\,\psi_1\otimes\psi_1\rangle$$ and $$\langle \psi_1\otimes \psi_1\,|\, \Phi(|\tfrac{\psi_1-\psi_2}{\sqrt2}\rangle\langle \tfrac{\psi_1-\psi_2}{\sqrt2}|) \,|\,\psi_1\otimes\psi_1\rangle$$ are equal to $$\frac14$$, again, by applying$${}^1$$ (1).
3. because $$\Phi$$ is linear, it has to satisfy $$\Phi(|\psi_1\rangle\langle \psi_1|)+\Phi(|\psi_2\rangle\langle \psi_2|)=\Phi(|\tfrac{\psi_1+\psi_2}{\sqrt2}\rangle\langle \tfrac{\psi_1+\psi_2}{\sqrt2}|)+\Phi(|\tfrac{\psi_1-\psi_2}{\sqrt2}\rangle\langle \tfrac{\psi_1-\psi_2}{\sqrt2}|)$$

Doing so and combining these facts lets us arrive at the following contradiction: \begin{align*} 1=1+0&=\langle\psi_1\otimes\psi_1\,|\,\Phi(|\psi_1\rangle\langle \psi_1|)+\Phi(|\psi_2\rangle\langle \psi_2|)\,|\,\psi_1\otimes\psi_1\rangle\\ &=\langle\psi_1\otimes\psi_1\,|\,\Phi(|\tfrac{\psi_1+\psi_2}{\sqrt2}\rangle\langle \tfrac{\psi_1+\psi_2}{\sqrt2}|)+\Phi(|\tfrac{\psi_1-\psi_2}{\sqrt2}\rangle\langle \tfrac{\psi_1-\psi_2}{\sqrt2}|)\,|\,\psi_1\otimes\psi_1\rangle=\tfrac14+\tfrac14=\tfrac12 \end{align*}

$${}^1$$: Note that doing so is allowed because both $$\frac{\psi_1+\psi_2}{\sqrt2}$$, $$\frac{\psi_1-\psi_2}{\sqrt2}$$ are unit vectors as $$\langle\psi_1|\psi_2\rangle=0$$ by assumption