Quantum reversibility and the no-cloning principle

Undoing a gate

An ideal unitary $U$ has inverse $U^\dagger$. To reverse a sequence, reverse its order and invert each gate. If a circuit applies H, then S, its state is $SH\lvert\psi\rangle$. Undo it by applying $S^\dagger$ first and H second.

X, Z, H and CX are their own inverses. S is not: $S^\dagger=\operatorname{diag}(1,-i)$.

Copying known basis information

CX maps $\lvert0\rangle\lvert0\rangle$ to $\lvert00\rangle$ and $\lvert1\rangle\lvert0\rangle$ to $\lvert11\rangle$. This copies a known computational-basis label into a fresh zero target.

For a superposition, however,

$$CX\left(\frac{\lvert0\rangle+\lvert1\rangle}{\sqrt2}\otimes\lvert0\rangle\right) =\frac{\lvert00\rangle+\lvert11\rangle}{\sqrt2}.$$

That is an entangled Bell state, not $\lvert+\rangle\otimes\lvert+\rangle$.

Why no universal perfect cloner exists

Suppose a unitary copied both unknown states $\lvert a\rangle$ and $\lvert b\rangle$ into a blank target. Preservation of inner products would require

$$\langle a|b\rangle=(\langle a|b\rangle)^2.$$

This only holds when the overlap is 0 or 1. It fails for general distinct, nonorthogonal states. A single universal operation therefore cannot perfectly clone arbitrary unknown pure states. This does not forbid preparing many copies of a state whose description you already know.

Practise

Prepare a Bell state in the gate playground. Undo CX and then H. Verify that all probability returns to 00. Explain why measurement halfway through would change this reversal experiment.

Reference: IBM Quantum — Basics of quantum information.