Ashish Panigrahi

# A primer to the no-cloning theorem

## Prerequisites

### 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).

## Introduction

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.

1. Quantum Computing lectures by Dr. Umesh Vazirani

2. Principles of Quantum Mechanics - Shankar

3. Quantum Computation and quantum information - Nielsen & Chuang