No-cloning Theorem

In a classical computer there is no problem to copy an arbitrary piece of data. However, in quantum computer this is not the case. This is the consequence of a straightforward observation in linear algebra, called the "No-cloning Theorem". In the formalism of quantum computation, we use a unit vector in a Hilbert space to represent the state of a quantum bit. Operations on quantum bits are represented by unitary operators on the Hilbert space. To represent a system with more than one quantum bit, we use tensor products. Here is the No-cloning Theorem in this formulation.


No-cloning Theorem. Let $\mathcal{H}$ be a vector space (over $\mathbb{R}$ or $\mathbb{C}$) of dimension larger than 1. There do not exist a linear operator $U$ on $\mathcal{H} \otimes \mathcal{H}$ and a nonzero vector $v_0 \in \mathcal{H}$ such that $U(x \otimes v_0) = x \otimes x$ for all $x \in \mathcal{H}$.
Proof: Suppose such a $U$ and $v_0$ exist. Choose a nonzero vector $v \in \mathcal{H}$ such that $v, v_0$ are linearly independent. Then $U(v_0 \otimes v_0) = v_0 \otimes v_0, U(v \otimes v_0) = v \otimes v$ and $U((v + v_0) \otimes v_0) = (v + v_0) \otimes (v + v_0) = v \otimes v + v_0 \otimes v + v \otimes v_0 + v_0 \otimes v_0$. But linearity of $U$ demands $U((v + v_0) \otimes v_0) = U(v \otimes v_0) + U(v_0 \otimes v_0) = v \otimes v + v_0 \otimes v_0$. This is a contradiction.