What is the no-cloning theorem in Quantum Computing?

No-Cloning states that it is not possible to copy an arbitrary state onto another state. For example, suppose we have a psi state in system A and a random state (0) in system B; then, we won’t be able to clone the psi state in both systems A and B.

$$|\psi\rangle_a|0\rangle_b \neq |\psi\rangle_a |\psi\rangle_b$$

To prove this, we will be taking a contradiction that occurs in quantum computing and clone transformation gate (CG), which copies the arbitrary state.

$CG(|\psi\rangle_a \otimes |0\rangle_b)=|\psi\rangle_a \otimes |\psi\rangle_b$ => equation 1

Proof

Let’s evaluate equation 1 to see if its L.H.Squbit (quantum bit) is equal to R.H.Sa unit vector on the plane where alpha and beta are complex numbers..

$$|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$$

Alpha and beta are amplitude probabilities. The sum of these probabilities is:

$$|\alpha|^2 + |\beta|^2 = 1$$

Let’s take the L.H.S of equation 1:

= $CG(|\psi\rangle \otimes |0\rangle)$.

Put the values of $|\psi\rangle$ into equation 1:

= $CG((\alpha|0\rangle + \beta|1\rangle) \otimes |0\rangle)$ = $CG(\alpha|0\rangle \otimes |0\rangle+ \beta|1\rangle \otimes |0\rangle)$

= $CG\alpha(|0\rangle \otimes |0\rangle)+ CG\beta(|1\rangle \otimes |0\rangle)$

= $\alpha(|00\rangle)+ \beta(|10\rangle)$

equation 2

Now, lets take the R.H.S of equation 1:

= $|\psi\rangle\otimes |\psi\rangle$

= $(\alpha|0\rangle + \beta|1\rangle) \otimes (\alpha|0\rangle + \beta|1\rangle)$

= $\alpha^2 |00\rangle +\alpha\beta|01\rangle + \alpha\beta|10\rangle + \beta^2 |11\rangle$

equation 3

By comparing equation 2 and equation 3:

$\alpha(|00\rangle)+ \beta(|10\rangle) \neq \alpha^2 |00\rangle +\alpha\beta|01\rangle + \alpha\beta|10\rangle + \beta^2 |11\rangle$

Hence, can conclude that L.H.S $\neq$ R.H.S, which is a contradiction. There is no cloning transformation gate in existence that perfectly copies the arbitrary state.

This is known as the No-Cloning theorem.