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Lower bound on survival probability

Let \(| \psi(t) \rangle\) be the time-dependent state of a quantum system evolving under the action of a time-independent Hamiltonian \(H\), i.e. \[i \hbar \frac{d | \psi(t) \rangle}{d t} = H | \psi(t) \rangle \,.\] Suppose that initially, at \(t=0\), the system is in some state \(| \psi(0) \rangle = | \psi_0 \rangle\). The autocorrelation function \[P(t) = \big| \langle \psi(t) | \psi_0 \rangle \big|^2\] quantifies the survival probability: \(P(t)\) is the probability that the system would be found in its original state after time \(t\). The survival probability equals unity at \(t=0\) and, generally, decays as \(t\) increases. How fast can \(P(t)\) decay? In particular, can the decay be exponential, i.e. \[P(t) \stackrel{?}{=} e^{-\gamma t}\] with some decay rate \(\gamma>0\) on a time interval \(0 \le t \le T\)? (The assumption of exponential decay is commonplace in back-of-the-envelope arguments. For instance, the number of atoms in a sample undergoing radioactive d

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 (usin