### 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 decay is often assumed to decrease exponentially with time. But is such exponential decay consistent with quantum theory?)

It turns out that it is impossible for the survival probability to decay exponentially fast. In fact, the short-time decay of $$P(t)$$ is bounded by the so-called Mandelstam-Tamm inequality, $P(t) \ge \cos^2 \left( \frac{\Delta E}{\hbar} t \right) \,,$ that holds on the time interval $$0 \le t \le \frac{\pi \hbar}{2 \Delta E}$$. Here $\Delta E = \sqrt{ \langle \psi_0 | H^2 | \psi_0 \rangle - \langle \psi_0 | H | \psi_0 \rangle^2 }$ is the energy uncertainty of the initial state.

### Proof

The Mandelstam-Tamm uncertainty relation states that, for any time-independent observable $$A$$ and at any instant $$\tau$$, $\Delta E \Delta t_A \ge \frac{\hbar}{2} \,,$ where $\Delta t_A = \left| \frac{d \langle A \rangle}{d\tau} \right|^{-1} \sqrt{\langle A^2 \rangle - \langle A \rangle^2} \qquad \text{and} \qquad \langle \cdot \rangle \equiv \langle \psi(\tau) | \cdot | \psi(\tau) \rangle \,.$ $$\Delta t_A$$ has the following physical meaning: is the time required for the expectation value of $$A$$ to change by a significant amount.

Let us take $A = | \psi_0 \rangle \langle \psi_0 | \,.$ Since $$A^2 = A$$ and $$\langle A \rangle = P(\tau)$$, we have $\Delta t_A = \left| \frac{d P(\tau)}{d\tau} \right|^{-1} \sqrt{P(\tau) - P^2(\tau)} \,.$ Substituting the last expression into the energy-time uncertainty relation, we find $\left| \frac{d P(\tau)}{d\tau} \right| \le \frac{2 \Delta E}{\hbar} \sqrt{P(\tau) - P^2(\tau)} \,.$ Assuming that $$P$$ decays monotonically, i.e. $$d P(\tau) / d \tau \le 0$$, we rewrite the last statement as $-\frac{d P(\tau)}{d\tau} \le \frac{2 \Delta E}{\hbar} \sqrt{P(\tau) - P^2(\tau)} \,,$ or $-\int_0^t \frac{d P(\tau)}{\sqrt{P(\tau) - P^2(\tau)}} \le \frac{2 \Delta E}{\hbar} t \,.$ Evaluating the integral, we get $2 \arccos \sqrt{P(\tau)} \, \bigg|_0^t \le \frac{2 \Delta E}{\hbar} t \,,$ and taking into account that $$P(0) = 1$$, we obtain $\arccos \sqrt{P(t)} \le \frac{\Delta E}{\hbar} t \,.$ This implies $\sqrt{P(t)} \ge \cos \left( \frac{\Delta E}{\hbar} t \right) \,,$ and, for $$0 \le \frac{\Delta E}{\hbar} t \le \frac{\pi}{2}$$, $P(t) \ge \cos^2 \left( \frac{\Delta E}{\hbar} t \right) \,.$