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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.