4.6: Radioactive decay and imaginary potentials
- Page ID
- 28765
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If the number of particles in a given state is reduced in time, then the total intensity of that state is reduced. Consider a particle moving in a region of imaginary potential \(V (r) = −iV_0\). The TDSE is:
\[i\hbar \frac{ \partial}{\partial t}|\Phi , t \rangle = [H_0 − iV_0]|\Phi , t \rangle \nonumber\]
Assume that the time independent part of the state is an combination of eigenstates of the real part of the Hamiltonian:
\[|\Phi , t \rangle = \sum_n c_n(t) \text{ exp}(−iE_nt/\hbar )|\Phi_n \rangle; \quad \text{ where } \quad H_0|\Phi_n \rangle = E_n|\Phi_n \rangle \nonumber\]
Following the same analysis as for TDSE, premultiplying by \(\langle m|\), and for constant \(V_0\), \(V_{mn} = \delta_{mn}V_0\) we obtain:
\[i\hbar \dot{c}_m = −iV_0c_m \quad \Rightarrow \quad |c_m(t)|^2 = |c_m(0)|^2 e^{−2V_0t/\hbar} \nonumber\]
Thus the probability amplitude of the state decreases in time. An imaginary potential can be used to represent destruction of particles, either by absorption (in a scattering process, perhaps) or by radioactive decay. Obviously the ket is not a full description of the system, since that should include information about the decay products. The lifetime of the state is \(\tau = \hbar /2V_0\).
Notice that \(−iV_0\) is not a Hermitian operator, and so it is not possible to perform a single measurement of half life.