### No-cloning theorem

Consider two quantum systems of the same nature. For concreteness, let us take them to be two hydrogen atoms. Suppose that the first H-atom is in an arbitrary unknown state $$| \alpha \rangle$$, while the second H-atom is in the ground state $$| 0 \rangle$$. Is it possible to construct a perfect cloning machine operating as follows: The machine changes the state of the second H-atom from $$| 0 \rangle$$ to $$| \alpha \rangle$$ without altering (or destroying) the state of the first H-atom? More specifically, the machine takes the initial state of the composite system, $| \alpha 0 \rangle = | \alpha \rangle \otimes | 0 \rangle \qquad \text{(initial state)}$ where $$\otimes$$ denotes the tensor product, and transforms it into $| \alpha \alpha \rangle = | \alpha \rangle \otimes | \alpha \rangle \qquad \text{(final state)}$ module perhaps some physically irrelevant global phase. The no-cloning theorem states that constructing such a machine is impossible.

### Proof 1 (using unitarity)

Let us assume that it is possible to construct the perfect cloning machine described above. Let $$U$$ be the operator representing the action of the cloning machine, i.e. $U | \alpha 0 \rangle = e^{i \varphi_{\alpha}} | \alpha \alpha \rangle \,,$ where $$\varphi_{\alpha}$$ is some (generally $$\alpha$$-dependent) phase.

We stress that the state $$| \alpha \rangle$$ is assumed to be completely arbitrary. Thus, if we performed the cloning process with another state $$| \beta \rangle$$ (arbitrarily different from $$| \alpha \rangle$$), the result would be $U | \beta 0 \rangle = e^{i \varphi_{\beta}} | \beta \beta \rangle \,.$ Let us investigate whether the assumption about the arbitrariness of $$| \alpha \rangle$$ and $$| \beta \rangle$$ is consistent with the last two equations. Taking the inner product between the corresponding sides of the equations, we obtain $\langle \alpha 0 | U^{\dagger} U | \beta 0 \rangle = e^{i \left( \varphi_{\beta} - \varphi_{\alpha} \right)} \langle \alpha \alpha | \beta \beta \rangle \,,$ or, equivalently, $\langle \alpha 0 | U^{\dagger} U | \beta 0 \rangle = e^{i \left( \varphi_{\beta} - \varphi_{\alpha} \right)} \langle \alpha | \beta \rangle^2 \,.$ We now take into account the fact that the operator $$U$$ must be unitary, i.e. $U^{\dagger} U = I \,,$ where $$I$$ is the identity operator. (This requirement follows from the fact that $$U$$ represents a transformation that conserves probabilitiy. Indeed, if for any input state $$| \Psi_{\text{in}} \rangle$$ the output state $$| \Psi_{\text{out}} \rangle = U | \Psi_{\text{in}} \rangle$$ is such that $$\langle \Psi_{\text{out}} | \Psi_{\text{out}} \rangle = \langle \Psi_{\text{in}} | \Psi_{\text{in}} \rangle$$, then $$U^{\dagger} U = I$$.) Then, the above inner product relation becomes $\langle \alpha | \beta \rangle = e^{i \left( \varphi_{\beta} - \varphi_{\alpha} \right)} \langle \alpha | \beta \rangle^2 \,.$ The only two solutions to the last equation are $\langle \alpha | \beta \rangle = 0 \,,$ meaning that the states $$| \alpha \rangle$$ and $$| \beta \rangle$$ are orthogonal to each other, and $e^{i \varphi_{\alpha}} | \alpha \rangle = e^{i \varphi_{\beta}} | \beta \rangle \,,$ meaning that $$| \alpha \rangle$$ and $$| \beta \rangle$$ represent (up to a global phase) the same physical state of the H-atom. Both solutions contradict the assumption of arbitrariness of $$| \alpha \rangle$$ and $$| \beta \rangle$$. This contradiction proves that a perfect cloning machine cannot be constructed.

### Proof 2 (using linearity)

Another way to prove the no-cloning theorem is to use the fact that $$U$$, as any evolution operator in quantum mechanics, must be liner. To this end, let us suppose that, initially, the first H-atom happened to be in a superposition of the ground and first excited states, say $| \alpha \rangle = \frac{1}{\sqrt{2}} \Big( | 0 \rangle + | 1 \rangle \Big) \,,$ and let us apply the cloning operator $$U$$ in the following two ways.

On the one hand, according to the definition of $$U$$, we have \begin{align} U | \alpha 0 \rangle &= e^{i \varphi_{\alpha}} | \alpha \alpha \rangle \\ &= e^{i \varphi_{\alpha}} \frac{1}{\sqrt{2}} \Big( | 0 \rangle + | 1 \rangle \Big) \otimes \frac{1}{\sqrt{2}} \Big( | 0 \rangle + | 1 \rangle \Big) \\ &= \frac{e^{i \varphi_{\alpha}}}{2} \Big( | 0 0 \rangle + | 0 1 \rangle + | 1 0 \rangle + | 1 1 \rangle \Big) \,. \end{align}

On the other hand, in view of the linearity of $$U$$, we have \begin{align} U | \alpha 0 \rangle &= U \frac{1}{\sqrt{2}} \Big( | 0 \rangle + | 1 \rangle \Big) \otimes | 0 \rangle \\ &= \frac{1}{\sqrt{2}} U \Big( | 0 0 \rangle + | 1 0 \rangle \Big) \\ &= \frac{1}{\sqrt{2}} \Big( U | 0 0 \rangle + U | 1 0 \rangle \Big) \\ &= \frac{1}{\sqrt{2}} \Big( e^{i \varphi_0} | 0 0 \rangle + e^{i \varphi_1} | 1 1 \rangle \Big) \,. \end{align}

The two results obtained for $$U | \alpha 0 \rangle$$ are incompatible: The first one has the cross terms $$| 0 1 \rangle$$ and $$| 1 0 \rangle$$, while the second one does not. Thus we have a contradiction and conclude that there can exist no operator $$U$$ with the desired cloning properties.