In Walter Thirring book Quantum mechanics of Atoms and Molecules he says that the probability that a initial state $$\Psi$$ be again measured at later time is $$|\langle \Psi|\exp(-iHt)\Psi\rangle|^2$$ and then he defines the lifetime, $$\tau$$ of this state by $$\tau(\Phi)=\frac{1}{2}\int|\langle \Psi|\exp(-iHt)\Psi\rangle|^2dt$$ My question is what is the intuition to the define the lifetime as he did?

• the factor half comes from the fact that the integral goes from $-\infty$ to $+\infty$. My problem here is that since $|\langle \Psi|\exp(-iHt)\Psi\rangle|^2$ that at time the system remain in the state $\Psi$ why life time is not defined as time expectation value that is $\tau(\Phi)=\frac{1}{2}\int t|\langle \Psi|\exp(-iHt)\Psi\rangle|^2dt$ – amilton moreira Dec 30 '19 at 17:18

The quantity $$P(t)=|\langle \Psi|\exp(-iHt)\Psi\rangle|^2$$ is the probability of survival of your state, and not the probability of survival per unit time! It is a dimensionless quantity. For a vast number of quantum systems, exponentially decaying it is something like $$e^{-t/\tau}$$, for a (mean) lifetime τ.
It is then apparent how the lifetime is gotten by (note the units of time!) $$\int_0^\infty dt ~ P(t) \to \int_0^\infty dt ~ e^{-t/\tau}=\tau ~.$$ (Mercifully you specify in your comment that your text integrates from minus infinity to plus infinity, which the reality of your probability allows to be halved by symmetry.)
But significantly, your never need consider an integral like $$\int dt ~t P(t)$$, in your comment, with dimensions of time-squared, as this is not a characteristic time. I believe many confusions in this are stem from contrasting survival (persistence) probability P, dimensionless, to the probability of decay per unit time interval, which has dimensions of inverse time.