An operator \(U\) is unitary if

\[ U U^\dagger = U^\dagger U = I \]where \(U^\dagger\) is the complex conjugate (interchange rows & columns and replace \(i\) with \(-i\)) of \(U\).

Let there be 2 matrices \(A\) and \(B\) where

\[ A = \begin{pmatrix} a_{00} & a_{01} \\ a_{10} & a_{11} \\ \end{pmatrix} \] \[ B = \begin{pmatrix} b_{00} & b_{01} \\ b_{10} & b_{11} \\ \end{pmatrix} \]The tensor product of \(A\) and \(B\) is represented as \(A \otimes B\) and is evaluated as

\[ A \otimes B = \begin{pmatrix} a_{00} \cdot \begin{pmatrix} b_{00} & b_{01} \\ b_{10} & b_{11} \\ \end{pmatrix} & a_{01} \cdot \begin{pmatrix} b_{00} & b_{01} \\ b_{10} & b_{11} \\ \end{pmatrix} \\ a_{10} \cdot \begin{pmatrix} b_{00} & b_{01} \\ b_{10} & b_{11} \\ \end{pmatrix} & a_{11} \cdot \begin{pmatrix} b_{00} & b_{01} \\ b_{10} & b_{11} \\ \end{pmatrix} \\ \end{pmatrix} \]I have used the term unitary transform and unitary operation interchangeably (since they mean the same thing).

Given a quantum state \(\left| \psi \right> = \alpha \left| 0 \right> + \beta \left| 1 \right>\) with unknown complex coefficients \(\alpha\) and \(\beta\), is it possible to replicate the state onto another qubit? In other words, is it possible to go from \(\left| \psi \right>\) to \(\left| \psi \right> \otimes \left| \psi \right>\)? Let us form a hypothesis.

Given an unknown quantum state \(\left| \psi \right>\), there exists a unitary transform \(U\) such that it replicates the state \(\left| \psi \right>\) to \(\left| \psi \right> \otimes \left| \psi \right>\).

From the postulates of quantum mechanics, any transformation of a quantum system must be unitary. Hence, we are trying to look for a unitary operation \(U\) to make the following transform:

\[ U \left| \psi \right> \otimes \left| 0 \right> = \left| \psi \right> \otimes \left| \psi \right> \]Here the inputs are \(\left| \psi \right>\) and an ancilla qubit (in our case \(\left | 0 \right>\)).

The no cloning theorem tells us that such a unitary transform does not exist. Let us see how this is true, algebraically. We will do this by disproving the hypothesis. Let us look at the cases when \(\left| \psi \right> = \left| 0 \right>\) and when \(\left| \psi \right> = \left| 1 \right>\). On applying the operation \(U\), the 2-qubit state becomes,

\[ \left| 0 \right> \otimes \left| 0 \right> = \left| 00 \right> \xrightarrow[]{U} \left| 00 \right> \] \[ \left| 1 \right> \otimes \left| 0 \right> = \left| 10 \right> \xrightarrow[]{U} \left| 11 \right> \]Now let us apply \(U\) to the original state \(\left| \psi \right> (= \alpha \left| 0 \right> + \beta \left| 1 \right>)\) based on the above transformations.

\[ (\alpha \left| 0 \right> + \beta \left| 1 \right>) \otimes \left| 0 \right> = \alpha \left| 00 \right> + \beta \left| 10 \right> \xrightarrow[]{U} \alpha U \left| 00 \right> + \beta U \left| 10 \right> = \alpha \left| 00 \right> + \beta \left| 11 \right> \]Based on our original hypothesis, the output is

\[ (\alpha \left| 0 \right> + \beta \left| 1 \right>) \otimes (\alpha \left| 0 \right> + \beta \left| 1 \right>) = \alpha ^2 \left| 00 \right> + \alpha \beta \left| 01 \right> + \beta \alpha \left| 10 \right> + \beta ^2 \left| 11 \right> \]In matrix notation this looks like

\[ \begin{pmatrix} \alpha ^2 \\ \alpha \beta \\ \beta \alpha \\ \beta ^2 \\ \end{pmatrix} = U \begin{pmatrix} \alpha \\ 0 \\ \beta \\ 0 \\ \end{pmatrix} \]Now comparing the coefficients of the output states, such a case is only possible if \(\alpha = 1\) or \(\beta = 1\). This disproves the hypothesis, hence proving the no cloning theorem.

Principles of Quantum Mechanics - Shankar

Quantum Computation and quantum information - Nielsen & Chuang