Quantum Zeno Paradox. Suppose a system starts out in an excited state \psi_{b} , which has a natural lifetime τ for transition to the ground state \psi_{a} . Ordinarily, for times substantially less than τ, the probability of a transition is proportional to t (Equation 11.49):
P_{b \rightarrow a}(t) \approx \frac{\pi|\wp|^{2}}{\epsilon_{0} \hbar^{2}} \rho\left(\omega_{0}\right) t (11.49).
P_{b \rightarrow a}=\frac{t}{\tau} . (11.145).
If we make a measurement after a time t, then, the probability that the system is still in the upper state is
P_{b}(t)=1-\frac{t}{\tau} (11.146).
Suppose we do find it to be in the upper state. In that case the wave function collapses back to \psi_{b} , and the process starts all over again. If we make a second measurement, at 2t, the probability that the system is still in the upper state is
\left(1-\frac{t}{\tau}\right)^{2} \approx 1-\frac{2 t}{\tau} , (11.147).
which is the same as it would have been had we never made the first measurement at t (as one would naively expect).
However, for extremely short times, the probability of a transition is not proportional to t, but rather to t² (Equation 11.46):
P_{b \rightarrow a}(t)=\frac{2}{\epsilon_{0} \hbar^{2}}|\wp|^{2} \int_{0}^{\infty} \rho(\omega)\left\{\frac{\sin ^{2}\left[\left(\omega_{0}-\omega\right) t / 2\right]}{\left(\omega_{0}-\omega\right)^{2}}\right\} d \omega (11.46)
P_{b \rightarrow a}=\alpha t^{2} (11.148).
(a) In this case what is the probability that the system is still in the upper state after the two measurements? What would it have been (after the same elapsed time) if we had never made the first measurement?
(b) Suppose we examine the system at n regular (extremely short) intervals, from t = 0 out to t = T (that is, we make measurements at T/n , 2T/n, 3T/n,……T) What is the probability that the system is still in the upper state at time T? What is its limit as n→∞? Moral of the story: Because of the collapse of the wave function at every measurement, a continuously observed system never decays at all!