Quantum Computing 3.4 -- Copy a quantum state

We assume, that there is a quantum gate $U$ with the property $U |\phi\rangle \otimes |0\rangle = |\phi \rangle \otimes |\phi \rangle$ . If we apply $U$ on the basis states $|00\rangle$ and $10\rangle$ we get the results $\begin{aligned} U |00\rangle &= U |0\rangle \otimes |0\rangle = |0\rangle \otimes |0\rangle = |00\rangle \qquad (1)\\ U |10\rangle &= U |1\rangle \otimes |0\rangle = |1\rangle \otimes |1\rangle = |11\rangle \qquad (2) \end{aligned}$ Now we consinder the case, that the first qubit is given by a superpositon $|+\rangle = (|0\rangle + |1\rangle)/\sqrt{2}$ . We calculate $U|+\rangle \otimes |0\rangle$ with the help of $(1)$ and $(2)$ : $\begin{aligned} U |+\rangle \otimes |0\rangle &= U \frac{1}{\sqrt{2}} \left(|0\rangle + |1\rangle \right) \otimes |0\rangle \\ &= U \frac{1}{\sqrt{2}} \left(|00\rangle + |10\rangle \right) & & |U \text{ is linear}\\ &= \frac{1}{\sqrt{2}} \left(U |00\rangle + U|10\rangle \right) & &|(1)\text{ and } (2)\\ &= \frac{1}{\sqrt{2}} \left(|00\rangle + |11\rangle \right) \qquad (3) \end{aligned}$ We used the fact, that $U$ is a linear operatior (i.e. $U(\alpha |\phi\rangle + \beta |\psi \rangle) = \alpha U |\phi\rangle + \beta U |\psi\rangle$ ). This is true, because matrix multiplication is linear. Now we calculate the tensor product of the state $|+\rangle$ : $\begin{aligned} |+\rangle \otimes |+\rangle &= \frac{1}{\sqrt{2}} (|0\rangle + |1\rangle) \otimes \frac{1}{\sqrt{2}} (|0\rangle + |1\rangle) \\ &= \frac{1}{2} (|0\rangle \otimes (|0\rangle + |1\rangle) + |1\rangle \otimes (|0\rangle + |1\rangle)) \\ &= \frac{1}{2} (|0\rangle \otimes |0\rangle + |0\rangle \otimes |1\rangle + |1\rangle \otimes |0\rangle + |1\rangle \otimes |1\rangle)) \\ &= \frac{1}{2} (|0 0\rangle + |0 1\rangle + |1 0\rangle + |1 1\rangle)) \qquad (4) \end{aligned}$ Obviously, we get different results in $(3)$ and $(4)$ , so that $U |+\rangle \otimes |0\rangle \stackrel{(3)}{=} \frac{1}{\sqrt{2}} \left(|00\rangle + |11\rangle \right) \not= \frac{1}{2} (|0 0\rangle + |0 1\rangle + |1 0\rangle + |1 1\rangle)) \stackrel{(4)}{=} |+\rangle \otimes |+\rangle$ in contradiction to our assumption $U|\psi\rangle |0\rangle = |\psi\rangle |\psi\rangle$ . Therefore, an operator $U$ , that can copy any quantum state, does not exist. This fact is called No-Cloning-Theorem .

In conclusion, there is no copy operation for qubits in total contrast to classical bits, which can be copied.