# No-Cloning Theorem

## Introduction

In the classical world, it is quite common that we could make an exactly the same copy of something. However, in the quantum world, the laws of physics impose a severe restriction on copying: It is impossible to make a perfect copy of an unknown state.

In this blog post, I would like to discuss the No-Cloning Theorem in quantum theory.

## Prerequisites

### Kronecker Product Inverse-Transpose Property

Conjugate transposition are distributive over the Kronecker product:

$$(A \otimes B)^{\dagger} = A^{\dagger} \otimes B^{\dagger}$$

\begin{align} (A \otimes B)^{\dagger} &= \begin{bmatrix} A_{0,0}B & A_{0,1}B & \cdots & A_{0,n-1}B \\ A_{1,0}B & A_{1,1}B & \cdots & A_{1,n-1}B \\ \vdots & \vdots & \ddots & \vdots \\ A_{m-1,0}B & A_{m-1,1}B & \cdots & A_{m-1,n-1}B \\ \end{bmatrix}^{\dagger} \nonumber\\ &= \begin{bmatrix} (A_{0,0}B)^{\dagger} & (A_{1,0}B)^{\dagger} & \cdots & (A_{m-1,0}B)^{\dagger} \\ (A_{0,1}B)^{\dagger} & (A_{1,1}B)^{\dagger} & \cdots & (A_{m-1,1}B)^{\dagger} \\ \vdots & \vdots & \ddots & \vdots \\ (A_{0,n-1}B)^{\dagger} & (A_{1,n-1}B)^{\dagger} & \cdots & (A_{m-1,n-1}B)^{\dagger} \\ \end{bmatrix} \nonumber\\ &= \begin{bmatrix} \overline{A_{0,0}}B^{\dagger} & \overline{A_{1,0}}B^{\dagger} & \cdots & \overline{A_{m-1,0}}B^{\dagger} \\ \overline{A_{0,1}}B^{\dagger} & \overline{A_{1,1}}B^{\dagger} & \cdots & \overline{A_{m-1,1}}B^{\dagger} \\ \vdots & \vdots & \ddots & \vdots \\ \overline{A_{0,n-1}}B^{\dagger} & \overline{A_{1,n-1}}B^{\dagger} & \cdots & \overline{A_{m-1,n-1}}B^{\dagger} \\ \end{bmatrix} \nonumber\\ &= \begin{bmatrix} A_{0,0}^{\dagger}B^{\dagger} & A_{0,1}^{\dagger}B^{\dagger} & \cdots & A_{0, m-1}^{\dagger}B^{\dagger} \\ A_{1,0}^{\dagger}B^{\dagger} & A_{1,1}^{\dagger}B^{\dagger} & \cdots & A_{1, m-1}^{\dagger}B^{\dagger} \\ \vdots & \vdots & \ddots & \vdots \\ A_{n-1,0}^{\dagger}B^{\dagger} & A_{n-1,1}^{\dagger}B^{\dagger} & \cdots & A_{n-1,m-1}^{\dagger}B^{\dagger} \\ \end{bmatrix} \nonumber\\ &= A^{\dagger} \otimes B^{\dagger} \\ \end{align}

This concludes the proof.

### Kronecker Product Mixed-Product Property

Let $A \in \mathbb{C}^{m \times n}$, $B \in \mathbb{C}^{r \times s}$, $C \in \mathbb{C}^{n \times p}$, and $D \in \mathbb{C}^{s \times t}$, then

$$(A \otimes B)(C \otimes D) = (AC) \otimes (BD)$$

This has been proved in Kronecker Product In Circuits.

### Inner Product Expansion Property

Suppose $|x\rangle$ is any unit vector whose $|x|^2 = \langle x | x \rangle = 1$. We have $\langle \phi | \psi \rangle = (\langle \phi | \otimes \langle x |) (| x \rangle \otimes | \psi \rangle)$

\begin{align} (\langle \phi | \otimes \langle x |) (| \psi \rangle \otimes | x \rangle) &= (| \psi \rangle \otimes | x \rangle)^{\dagger} (|\phi\rangle \otimes |x\rangle) \\ &= (| \psi \rangle^{\dagger} \otimes | x \rangle^{\dagger}) (|\phi\rangle \otimes |x\rangle) \\ &= (| \psi \rangle^{\dagger}|\phi\rangle) \otimes (| x \rangle^{\dagger}|x\rangle) \\ &= \langle \phi | \psi \rangle \otimes \langle x | x \rangle\\ &= \langle \phi | \psi \rangle \otimes 1\\ &= \langle \phi | \psi \rangle\\ \end{align}

This concludes the proof.

### Unitary Matrix Preserves Inner Product

Given two complex vectors $x$ and $y$, multiplication by unitary matrix $U$ preserves their inner product.

\begin{align} \langle Ux, Uy \rangle &= \langle x, y \rangle \end{align}

Using the definition of inner product,

\begin{align} \langle Ux, Uy \rangle &= (Uy)^{\dagger} (Ux) \\ &= y^{\dagger}U^{\dagger} U x \\ &= y^{\dagger}(U^{\dagger} U) x \\ &= y^{\dagger} I x \\ &= y^{\dagger} x \\ &= \langle x, y \rangle \\ \end{align}

This concludes the proof.

## No-Cloning Theorem

In quantum mechanics, copy, as it is the same to other quantum operators except measurement operators, is an reversible and linear operator.

Given any unknown normalized quantum state $| \phi \rangle$, and the copy operator $U$, we are supposed to have the following mathematical expression if we are going to copy the quantum state $| \phi \rangle$ to another system whose quantum state is $|x \rangle$ before copy.

$$| \phi \rangle \otimes |x\rangle \xrightarrow[]{U} | \phi \rangle \otimes |\phi\rangle$$

We would like to write this transformation into equation.

$$U (| \phi \rangle \otimes |x\rangle) = | \phi \rangle \otimes |\phi\rangle$$

Without loss of generality, we could have another state $| \psi \rangle$ that is applied to the copy operator.

$$U (| \psi \rangle \otimes |x\rangle) = | \psi \rangle \otimes |\psi\rangle$$

Note that for $| \psi \rangle$ we also used $|x\rangle$ which $| \phi \rangle$ is using, meaning that before copy, the “vacant” system is always the same. This is something we could guarantee.

We examine the inner product of $| \phi \rangle$ and $| \psi \rangle$ using the copy property and all the properties in the prerequisite section.

\begin{align} \langle \phi | \psi \rangle &= (\langle \phi | \otimes \langle x |) (| \psi \rangle \otimes | x \rangle) \\ &= \big(U (\langle \phi | \otimes \langle x |)\big) \big(U (| \psi \rangle \otimes | x \rangle)\big) \\ &= (\langle \phi | \otimes \langle \phi |) (| \psi \rangle \otimes | \psi \rangle) \\ &= (| \psi \rangle \otimes | \psi \rangle)^{\dagger} (| \phi \rangle \otimes | \phi \rangle) \\ &= (| \psi \rangle^{\dagger} \otimes | \psi \rangle^{\dagger}) (| \phi \rangle \otimes | \phi \rangle) \\ &= (| \psi \rangle^{\dagger} | \phi \rangle ) \otimes (| \psi \rangle^{\dagger} | \phi \rangle ) \\ &= \langle \phi | \psi \rangle \otimes \langle \phi | \psi \rangle \\ \end{align}

Because $\langle \phi | \psi \rangle$ is scalar, we further have

\begin{align} \langle \phi | \psi \rangle &= \langle \phi | \psi \rangle \otimes \langle \phi | \psi \rangle \\ &= \langle \phi | \psi \rangle ^2 \\ \end{align}

Solving the above equation, we have $\langle \phi | \psi \rangle = 0$ or $1$.

Because $| \phi \rangle$ and $| \psi \rangle$ could be any arbitrary unknown states, $\langle \phi | \psi \rangle = 0$ or $1$ could never be satisfied.

Therefore, there is no general quantum copy operator that makes copy of any unknown state.

## More Restrict No-Cloning Theorem

In quantum theory, $| \phi \rangle$ and $c | \phi \rangle$, where $c$ is a non-zero complex number, represent the same physical state. If $|c \phi|^2 = 1$, then $| \phi \rangle$ and $c | \phi \rangle$ only have phase difference.

For example, $| \phi \rangle$ and $| \psi \rangle$ have exactly the same probability of collapsing to $| 0 \rangle$ and $| 1 \rangle$. However, $| \phi \rangle$ has phase $\varphi_1$ while $| \psi \rangle$ has phase $\varphi_2$.

\begin{align} | \phi \rangle &= \cos \frac{\theta}{2} | 0 \rangle + e^{i\varphi_1} \sin \frac{\theta}{2} | 1 \rangle \\ | \psi \rangle &= \cos \frac{\theta}{2} | 0 \rangle + e^{i\varphi_2} \sin \frac{\theta}{2} | 1 \rangle \\ \end{align}

More restrict No-Cloning Theorem states that “copying” any unknown state while abandoning the phase is not possible either.

Suppose we have such “copy” operator, the mathematical expression for “copy” will be as follows.

$$U (| \phi \rangle \otimes |x\rangle) = e^{\varphi} | \phi \rangle \otimes |\phi\rangle\\ U (| \psi \rangle \otimes |x\rangle) = e^{\varphi^{\prime}} | \psi \rangle \otimes |\psi\rangle$$

Similarly, we have

\begin{align} \langle \phi | \psi \rangle &= e^{\varphi} e^{\varphi^{\prime}}\langle \phi | \psi \rangle ^2 \\ &= e^{\varphi + \varphi^{\prime}} \langle \phi | \psi \rangle ^2 \\ \end{align}

The norm of the two sides should be equivalent.

\begin{align} |\langle \phi | \psi \rangle| &= |e^{\varphi + \varphi^{\prime}} \langle \phi | \psi \rangle ^2| \\ &= |e^{\varphi + \varphi^{\prime}}| |\langle \phi | \psi \rangle ^2| \\ &= 1 |\langle \phi | \psi \rangle ^2| \\ &= |\langle \phi | \psi \rangle ^2| \\ &= |\langle \phi | \psi \rangle |^2 \\ \end{align}

Solving the above equation, we have $|\langle \phi | \psi \rangle | = 0$ or $1$.

Because $| \phi \rangle$ and $| \psi \rangle$ could be any arbitrary unknown states, $|\langle \phi | \psi \rangle | = 0$ or $1$ could never be satisfied.

Therefore, there is no general quantum “copy” operator that makes “copy” of any unknown state that has lost the phase information.

## Transportation

Since it is not possible to do copy in quantum world, how about transportation?

Concretely, given any unknown normalized quantum state $| \phi \rangle$, and the transportation operator $U$, we are supposed to have the following mathematical expression if we are going to transport the quantum state $| \phi \rangle$ to another system whose quantum state is $|x \rangle$ before transportation.

$$U(| \phi \rangle \otimes |x\rangle) = | x \rangle \otimes |\phi\rangle$$

Note that although it is called transportation, it is more like a switch, where the two system states, one unknown and one known, got switched.

Without loss of generality, we could have another state $| \psi \rangle$ that is applied to the transportation operator.

$$U(| \psi \rangle \otimes |x\rangle) = | x \rangle \otimes |\psi\rangle$$

Similarly, we compute the inner product of $| \phi \rangle$ and $| \psi \rangle$.

\begin{align} \langle \phi | \psi \rangle &= (\langle \phi | \otimes \langle x |) (| \psi \rangle \otimes | x \rangle) \\ &= \big(U (\langle \phi | \otimes \langle x |)\big) \big(U (| \psi \rangle \otimes | x \rangle)\big) \\ &= (\langle x | \otimes \langle \phi |) (| x \rangle \otimes | \psi \rangle) \\ &= (| x \rangle \otimes | \psi \rangle)^{\dagger} (| x \rangle \otimes | \phi \rangle) \\ &= (| x \rangle^{\dagger} \otimes | \psi \rangle^{\dagger}) (| x \rangle \otimes | \phi \rangle) \\ &= (| x \rangle^{\dagger} | x \rangle ) \otimes (| \psi \rangle^{\dagger} | \phi \rangle ) \\ &= \langle x | x \rangle \otimes \langle \phi | \psi \rangle \\ &= 1 \otimes \langle \phi | \psi \rangle \\ &= \langle \phi | \psi \rangle \\ \end{align}

Unlike copy, we did not find anything usual that prevents transportation.

Lei Mao

06-19-2020

06-19-2020