A Primer to the No-Cloning Theorem

Before we dive into the no-cloning theorem, it is important to glance over certain methods and concepts apriori.

Unitary transform:

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 $U$ .

Tensor product:

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}

Disclaimer

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.

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.

References

Quantum Computing lectures by Dr. Umesh Vazirani
Principles of Quantum Mechanics - Shankar
Quantum Computation and quantum information - Nielsen & Chuang

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